heat exchanger design guide: a practical guide for planning, selecting and designing of shell

HEAT EXCHANGERDESIGN GUIDEA Practical Guide for Planning,Selecting and Designingof Shell and Tube Exchangers

M. NITSCHE AND R.O. GBADAMOSI

With numerous practical Examples

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Heat Exchanger

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FOREWORD

Dear Reader,This book is not an academic treatise but rather a book for solving daily practical

problems easily and for the illustration of essential influencing variables in the designof heat exchangers, condensers, or evaporators.

All calculations are explained with examples that I am using in my seminars sinceseveral years.

In this book, you will be shown how to proceed in the design of a heat exchanger inthe daily practice, how to determine the effective temperature difference for the heattransfer, and how to calculate the heat transfer coefficient using simple equations.

The most important influence parameters for the heat transfer coefficient are intro-duced. Different calculation models are compared. It is shown how to calculate therequired dew point and bubble point lines for mixtures.

From the wide range of published calculation methods, I have chosen the models thatcan simply be calculated using the hand calculator and deliver sure results. I refer to themodels which I have chosen from several existing literatures as Nitsche methods becauseI recommend these.

During the time from 1966 to 2007, I designed, planned, and built several chemicalplants: distillation plants with evaporators, condensers, and heat exchangers, for fattyalcohols, fatty acids, nitrochlorobenzenes, amine and hydrocarbons and tar oils; storagetanks and vessels with filling stations for tank truck, rail tank car, and barrels; stirredtank plants for reactions with decanters, centrifuges, and filters; plants for exhaust air pu-rification and gasoline recovery, methanol and ethyl acetate etc.; stripper for of sour waterpurification or for methyl isobutyl ketone recovery.

Since 1980, I report in seminars about piping, heat exchangers, and chemical plantsabout my practical experiences in the Design and Planning. At the latest, I realized duringplant startup what I wrongly calculated or what I wrongly planed. This book should helpyou to minimize the mistakes in the design of heat exchangers.

HamburgDr Manfred Nitsche

ix

CHAPTER 1

Heat Exchanger Design

Contents

1.1 Procedure in Heat Exchanger Design 21.2 Information about Heat Exchangers 12

1.2.1 Tube Pattern 121.2.2 Bypass and Leakage Streams 131.2.3 Baffles 141.2.4 Technical Remarks 151.2.5 Selection of a Shell and Tube Exchanger 16

1.2.5.1 Which Heat Exchanger Types Can Be Cleaned? 17Nomenclature 18References 19

In heat exchanger design the required heat exchanger area A (m2) is determined for acertain heat load Q (W) at a given temperature gradient Dt (�C).

A ¼ QU � Dt

�m2�

The overall heat transfer coefficient, U, is calculated as follows:

1U

¼ 1ai

þ 1ao

þ slþ fi þ fo

fi ¼ inner fouling factor (m2 K/W)fo ¼ outer fouling factor (m2 K/W)U ¼ overall heat transfer coefficient (W/m2 K)s ¼ tube wall thickness (m)l ¼ thermal conductivity of the tube material (W/m K)ai ¼ inner heat transfer coefficient in the tubes (W/m2 K)ao ¼ outer heat transfer coefficient on the shell side (W/m2 K)Figure 1.1 shows which overall heat resistances have to be overcome and how the

temperature profile in a heat exchanger looks like.The overall heat transfer coefficients and the temperature profile will be calculated in

Chapter 6.Reference values for heat transfer coefficients and overall heat transfer coefficients are

listed in Table 1.1.

Heat Exchanger Design Guidehttp://dx.doi.org/10.1016/B978-0-12-803764-5.00001-8

© 2016 Elsevier Inc.All rights reserved. 1

http://dx.doi.org/10.1016/B978-0-12-803764-5.00001-8

1.1 PROCEDURE IN HEAT EXCHANGER DESIGN

In order to calculate the convective heat transfer coefficients, the Reynolds number isneeded.

The heat transfer coefficients, a, are dependent on the Reynolds number, Re, hencethe flow velocity, w, on the tube and shell side, respectively.

Tube side: af w0.8 Shell side: a f w0.6

Therefore the cross-sectional areas must be known in order to determine the flowvelocities and the Reynolds numbers. For an existing heat exchanger, this is not a prob-lem if a drawing is available. In the case of the design of a new heat exchanger, the flowcross sections are not known. So, initially an estimation of the required area has to bedone and then an appropriate equipment has to be selected.

For the selection, the following criteria should be applied:• The flow velocity on both sides should be in the order of 0.5e1 m/s for liquids and in

the range of 15e20 m/s for gases.• The required heat exchanger area should be achieved with tube length of 3e6 m.

In Figure 1.3 the flow chart for the heat exchanger design is provided [1].In the following the procedure of heat exchanger design is explained in some more

detail:1. Determine flow rates, temperatures, and the fluid property data2. Determination of the heat loads on tube and shell side

Shellside: QS req ¼ MS � cS � ðT1 � T2ÞðW ÞTubeside: QT req ¼ MT � cT � ðt2 � t1ÞðW Þ

Figure 1.1 Heat transfer resistances and temperature profile.

2 Heat Exchanger Design Guide

Table 1.1 Reference values for heat transfer coefficients, a

Natural convection a (W/m2 K)

Gases at atmospheric pressure 4e6Oil (viscosity ¼ 100 mm2/s) 10e20Water 250e500Hydrocarbons, low viscosity 170e300

Condensation

Steam 5000e10000Organic solvents 1000e3000Light oils 1000e1500Heavy oils (vacuum) 100e300

Vaporization

Water 4000e10000Organic solvents 1000e2500Light oils 700e1400

Flowing media

Atmospheric gases 40e200Gases under pressure 150e300Organic solvents 300e1000Water 2500e4000

Guiding values for overall heat transfer coefficients, U

Condensation U (W/m2 K)

Water Water 1000e2000Organic solvent 600e1000Organic solvent þ inert gases 100e500Heavy hydrocarbons 50e200

Evaporation

Steam Water 2000e4000Organic solvent 500e1000Light oils 250e800Heavy oils 120e400

Flowing media

Steam Water 1500e4000Organic solvents 600e1000Gases 30e250

Water Water 1000e2000Organic solvents 250e800Gases 15e300

Organic solvents Organic solvents 100e300

Heat Exchanger Design 3

Table 1.2 Geometric data of heat exchangers according to DIN 28,184, part 1, for 25 � 2 tubes with 32 mm triangular pitchTypnr.

DNe

Ze

Da(mm)

B(mm)

ne

AE(mm2)

AR(mm2)

AS(m2/m)

fwe

VR(m3/h)

VM(m3/h)

1 150 2 168 30 14 1770 2425 1.1 0.251 8.73 6.372 200 2 219 40 26 2288 4503 2 0.366 16.21 8.243 250 2 273 50 44 5520 7620 3.5 0.259 27.43 19.874 300 2 324 60 66 5088 11,430 5.2 0.375 41.15 18.325 350 2 355 70 76 6230 13,162 6 0.397 47.38 22.436 350 4 355 70 68 6230 5888 5.3 0.381 21.20 22.437 400 2 406 80 106 11,072 18,357 8.3 0.319 66.09 39.868 400 4 406 80 88 9072 7620 6.9 0.382 27.43 32.669 500 2 508 100 180 14,600 31,172 14.1 0.360 112.22 52.5610 500 4 508 100 164 12,100 14,201 12.9 0.430 51.12 43.5611 600 2 600 120 258 19,560 44,681 20.3 0.389 160.85 70.4212 600 8 600 120 232 22,560 10,044 18.2 0.348 36.16 81.2213 700 2 700 140 364 22,260 63,038 28.6 0.456 226.94 80.1414 700 8 700 140 324 25,760 14,028 25.4 0.395 50.50 92.7415 800 2 800 160 484 29,440 83,819 38 0.454 301.75 105.9816 800 8 800 160 432 37,440 18,703 33.9 0.367 67.33 134.7817 900 2 900 180 622 41,400 107,718 48.9 0.407 387.79 149.0418 900 8 900 180 556 41,400 24,072 43.7 0.416 86.66 149.0419 1000 2 1000 200 776 46,000 134,388 61 0.452 483.80 165.6020 1000 8 1000 200 712 56,000 30,826 55.9 0.373 110.97 201.6021 1100 2 1100 220 934 55,220 161,750 73.4 0.460 582.30 198.7922 1100 8 1100 220 860 60,720 37,234 67.5 0.420 134.05 218.5923 1200 2 1200 240 1124 72,240 194,655 88.3 0.416 700.76 260.0624 1200 8 1200 240 1048 78,240 45,373 82.3 0.390 163.34 281.66

DN ¼ Nominal shell diameter; Z ¼ number of tube passes; Da ¼ shell diameter (mm); B ¼ baffle spacing (mm); n ¼ number of tubes; AE ¼ shell-side flow cross section(mm2); AR ¼ tube-side flow cross section (mm2); AS ¼ Heat exchanger area per m tube length (m2/m); VR ¼ Required flow rate for 1 m/s flow velocity on the tubeside (m3/h); VM ¼ Required flow rate for 1 m/s flow velocity at the shell side (m3/h).

4Heat

ExchangerDesign

Guide

For condensers and evaporators, the condensation and the vaporizationenthalpies need to be considered.

3. Calculation of the corrected effective temperature difference (CMTD) for the heatload

First, the logarithmic mean temperature difference (LMTD) is determined forideal countercurrent flow.

Most heat exchangers have multiple passes in order to increase flow velocity inthe tubes.

For example, the heat exchanger in Figure 1.2 has two passes. The medium onthe tube side flows forward and backward. In one tube pass the fluid flows cocurrentto the flow on the shell side. In the other pass the flow is in countercurrent to theshell-side flow.

Ideal countercurrent flow does not occur. The driving temperature gradient isworse. Therefore for heat exchangers with multiple passes, due to the nonidealcountercurrent flow, the temperature efficiency factor, F, must be calculated. Fshould be >0.75!

Using the temperature efficiency factor, F, the CMTD is determined.

CMTD ¼ F � LMTD:

The calculation of the effective temperature difference is shown in Chapter 2.

Figure 1.2 Heat exchanger for convective heat transfer.


With nonlinear condensation or evaporation curves, the average weightedtemperature difference must be determined.

Zones with approximate linear range of the condensation temperature areestablished for which the CMTDs are determined. Finally, the weighted averageof the effective temperature differences in the zones is determined.

4. Estimation of the required heat exchanger areaFor the calculated heat load and the available effective temperature difference,

the required heat exchanger area is estimated using the estimated overall heattransfer coefficient, U, from Table 1.1:

A ¼ Qreq

U � CMTD

�m2�

5. Selection of an appropriate equipment from Table 1.2 for the required heat exchangearea, A, from column, AS, with the area of the heat exchanger per m length (m2/m)

Figure 1.3 Procedure flow chart for the thermal design of a heat exchanger [1].


6. Determination of the flow velocity using the columns, VR and VM, in Table 1.2The volumetric flows on the tube and shell side are listed in columns, VR and

VM, which are required for a flow velocity of 1 m/s.

Example 1: Selection of an appropriate heat exchanger for a required areaA ¼ 55 m2

Tube-side flow rate Vtube ¼ 40 m3/h, shell-side flow rate Vshell ¼ 80 m3/hSelected: Type 12 with 18.2 m2/m tube length and 232 tubes in 8 passes, DN 600Tube length ¼ 4 m, Heat exchanger area ¼ 4 � 18.2 ¼ 72.8 m2 (32% excess)VR ¼ 36.16 m3/h for 1 m/s and VM ¼ 81.22 m3/h for 1 m/s from Table 1.2Determination of the flow velocity, wt, on the tube side:

wt ¼ VtubeVR

� 1 ¼ 4036:16

� 1 ¼ 1:1 m=s

Determination of the flow velocity, wsh, on the shell side:

wsh ¼ VshellVM

� 1 ¼ 8081:22

� 1 ¼ 0:985 m=s

Primary condition for a good convective heat transfer is an adequately high flow velocity.That is why both columns for VR and VM in Table 1.2 are important.Figures 1.4 and 1.5 show that the heat transfer coefficients increase with rising flow velocities.On the shell side, the baffle spacing, B, can be shortened, for instance, if the flow velocity shall be

increased in order to achieve a better heat transfer coefficient.Since the flows on which the design is based are known the flow velocities can be easily determined

with VR and VM (see Example 1).

7. Calculation of the convective heat transfer coefficients on the tube and shell sideThe Reynolds number can be calculated once the flow velocity is determined.With convective heat transfer, the heat transfer coefficient is dependent on the

Reynolds number, Re, and the Prandtl number, Pr (see Chapter 3).

a ¼ Const�Rem � Pr0:33 � l

dRe ¼ w � d

nPr ¼ n� c � r� 3600

l

n ¼ kinematic viscosity (mm2/s)d ¼ tube diameter (m)w ¼ flow velocity (m/s)l ¼ heat conductivity (W/m K)r ¼ density (kg/m3)c ¼ specific heat (Wh/kg K)


The Prandtl number, Pr, is only dependent on the physical properties, that is,density, viscosity, heat conductivity, and specific heat.

The calculated heat transfer coefficient, ai, in the tubes is converted withrespect to the outer area because the heat exchanger area refers to the tube outerdiameter.

aio ¼ ai � dido

�W�m2 K

�

8. Determination of the overall heat transfer coefficient, U, considering the resistance of thetube wall and potential coating and fouling layers

Figure 1.4 Tube-side heat transfer coefficient as a function of the flow velocity.


9. Calculation of the actual heat load of the selected heat exchanger and comparison withthe U-value for the required heat load

Q ¼ U � A� CMTD ðW Þ10. Checking the excess and the fouling reserve

�UUreq

� 100

�� 100 ¼ % reserve

Figure 1.5 Shell-side heat transfer coefficient as function of flow velocity.


Example 2: Design of a water cooler1. Required data

Shell side Tube sideFlow rate (kg/h) 50,000 36,800Inlet temperature (�C) 65 15Outlet temperature (�C) 50 35.4Density r (kg/m3) 983 994Specific heat capacity, c (Wh/kg K) 1.16 1.16Kinematic Viscosity, n (mm2/s) 0.47 0.92Heat conductivity, l (W/m K) 0.654 0.605Fouling factor, f (m2 K/W) 0.002 0.002

2. Determining the heat duty

Shell side Q ¼ m � c � Dt ¼ 50,000 � 1.16 � (65.0 � 50) ¼ 870,000 WTube side Q ¼ m � c � Dt ¼ 36,800 � 1.16 � (35.4 � 15) ¼ 870,000 W

3. Temperature difference, CMTDThe calculation follows the scheme described in Chapter 2. First the LMTD is determined for ideal

countercurrent flow.

T1 / T2 65:0 / 50

t2 z t1 35:4 z 15

29:6 35

LMTD ¼ 35� 29:6

ln 3529:6

¼ 32:2 �C

Since we intend to use a multipass heat exchanger, the temperature efficiency factor, F, for thecorrection of the logarithmic temperature gradients LMTD must be used.

F can be determined using Figure 1.6.

0.9

0.8

0.7

0.6

0.50.1

R = 4.0 3.0 2.0 1.5 1.0 0.8 0.6 0.4 0.2

0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.00

1.0

T2

T1

T1 – T

2

t2 – t

1

t2 – t

1

t2

t1

P = ────

T1 – T

2R = ────

Figure 1.6 Temperature efficiency factor, F, as a function of P and R [3].


The parameters P and R are calculated from the temperatures T1, T2, t1, and t2.

P ¼ 0:41 R ¼ 0:735

From the diagram in Figure 1.6, the value of F is read off at the intersection of P and R: F ¼ 0.948

CMTD ¼ F � LMTD ¼ 0:948� 32:2 ¼ 30:55 �C

4. Estimation of the required heat exchanger area, AHeat load, Q ¼ 870,000 WTemperature gradient, CMTD ¼ 30.55 �CEstimated overall heat transfer coefficient, U ¼ 716 W/m2 KRequired heat exchanger area, A, is given below:

A ¼ QU� CMTD

¼ 870; 000716� 30:55

¼ 39:8 m2

5. Selection of an appropriate heat exchanger from Table 1.2A heat exchanger type having the required area and sufficient flow velocities with the flows on the

tube side and the shell side must be selected.Chosen: type 10 DN 500 with 4 passes and 164 tubes of 25 � 2.Determination of the heat exchanger area with 3 m tube length: A ¼ 3 � 12.9 ¼ 38.7 m2

6. Determination of the flow velocity using the columns VR and VM in Table 1.2The volumetric flows on the tube and shell side are listed in columns, VR and VM, which are

required for a flow velocity of 1 m/s.Checking the tube side:

On the tube side there must be a flow of VR ¼ 51.12 m3/h for a flow velocity of 1 m/s.In our example, the flow on the tube side is only Vtube ¼ 37 m3/h.The resulting real flow velocity, wt, is as follows:

wt ¼ VtubeVR

� 1 ¼ 3751:12

� 1 ¼ 0:72 m=s

Checking the shell side:On the shell side VM ¼ 43.56 m/h for 1 m/s with a baffle spacing, B ¼ 100 mm.For B ¼ 200 mm the flow is 2 � 43.56 ¼ 87.1 m3/h for 1 m/s on the shell side.In the example, the flow on the shell side, Vsh ¼ 50 m/h.This gives a real flow velocity, wsh, as follows:

wsh ¼ VshellVM

� 1 ¼ 5087:1

� 1 ¼ 0:58 m=s

7. Calculation of the heat transfer coefficients on the tube side and the shell sideUsing the found flow velocities on tube and shell side, the Reynolds numbers can be determined.The calculation of the heat transfer coefficients will be described in Chapter 3.

Tube-side: wt ¼ 0:72 m=s f€ur 37 m3�hRe w � d

n¼ 0:72� 0:021

0:92� 10�6 ¼ 16; 435 Pr ¼ c � n� r� 3600l

¼ 6:32

Nu ¼ 0:023� 16; 4350:8 � 6:320:33 ¼ 99:67

ai ¼ 99:67� 0:6050:021

¼ 2872 W�m2 K aio ¼ 2872� 21

25¼ 2412 W

�m2 K


Shell-side: wshell ¼ 0:58 m=s for 50:85 m3�hRe ¼ 0:58� 0:025

0:47� 10�6 ¼ 30; 851 Pr ¼ 0:47� 10�6 � 1:16� 983� 36000:654

¼ 2:95

Nu ¼ 0:196� 30; 8510:6 � 2:950:33 ¼ 138:3

ao ¼ 138:3� 0:6540:025

¼ 3618 W�m2 K

8. Determination of the overall heat transfer coefficient, U, considering the resistances of the tube wall andthe fouling factor.

Tube wall thickness, s ¼ 0.002 m; wall heat conductivity, l ¼ 50 W/m K; tube-side fouling,fi ¼ 0.0002; shell-side fouling, fo ¼ 0.0002

1U

¼ 1aio

þ 1ao

þ slþ fi þ fo ¼ 1

2412þ 13618

þ 0:00250

þ 0:0004 ¼ 0:00113

U ¼ 884 W�m2 K

9. Calculation of the actual heat load of the selected heat exchanger and comparison with the U-valuefor the required heat load, Qreq

Q ¼ U� A� CMTD ¼ 884� 38:7� 30:55 ¼ 1; 045; 140 W

Ureq ¼ Qreq

A� CMTD¼ 870; 000

38:7� 30:55¼ 736 W

�m2 K

10. Checking the excess area and the fouling reserve

Excess area ¼�

UUreq

� 100

�� 100 ¼

�884736

� 100

�� 100 ¼ 20:1%

The heat exchanger has an excess area of 20.1%.

fres ¼ U� Ureq

U� Ureq¼ 884� 736

884� 736¼ 0:000227 m2 K=W

The excess for the fouling is 0.000227 m2 K/W.

1.2 INFORMATION ABOUT HEAT EXCHANGERS

1.2.1 Tube PatternAbetter heat transfer coefficient can be achieved through a triangular or rotated tube patternin the tube bundle because the flow is directly against the tube. In a quadratic tube pattern,the medium flows between the tube rows and the heat transfer coefficient are worse thanwith a rotated tube pattern. If a severe fouling is expected on the shell side, for instance, in akettle-type evaporator, the quadratic patternmust be chosen to facilitate the cleaning of the


shell side.A compromise is the rotated quadratic pitch.Theheat transfer is better because theflow is directly against the tubes and cleaning can take place between the tube rows.

Normally, the triangular pitch is chosen because the heat transfer is good and withtriangular pitch more tubes can fit in the shell. With triangular pitch the heat exchangerarea is greater in a pipe shell (Figure 1.7).

1.2.2 Bypass and Leakage StreamsOn the shell side the medium should flow through the tube bundle between thebaffles.

If, however, large bypass lanes are installed, for instance, large gaps between the shelland the tube bundle, the product flows through the free lanes and this bypass flow doesnot take part in the heat exchange process.

If the holes in the baffles are much larger than the tube diameters, leakage streamsthrough this gap results. The leakage stream flow does not take part in the heat exchangeprocess.

In order to improve the efficiency of a heat exchanger, the bypass gap between theouter tube rows and the shell should be minimized or blocked with blind tubes in orderto force the medium to flow across through the tube rows.

With floating head heat equipments which have a larger gap in the shell due to theconditional construction, sealing strips must block the outer gap (Figure 1.8).

Figure 1.7 Tube arrangements in shell and tube heat exchangers (1.9).

Sealing strips

Figure 1.8 Sealing strips to minimize bypass stream.


Also free tube lanes through pass partitions in flow direction or with U-tubebundles must be avoided. The tolerances between the tubes and the baffle holes andbetween baffles and the shell should be as small as possible in order to minimize thebypass flow.

Recommended Tolerances: TEMA or DIN 28008, DIN 28182 and DIN 28185

1.2.3 BafflesOn the shell side, the flow cross section is given through the baffle spacing.

Some design forms are shown in Figure 3.12. Normally, segmental baffles are used(Figure 1.9).

For convective heat transfer on the shell side, the following design is recommended:

Baffle spacing, B ¼ 0.2 � Di Segmental height, H ¼ 0.2 � Di

With convective heat transfer an upedown flow at the shell side with horizontalbaffle edges should be recommended because in this way stratified flow will be avoided.

For ventilation and draining, small slots can be arranged in the baffles which of courseincrease the bypass flows.

Segmental cut

Segmental height

Slot in the baffle

Baffle

Segmental height

Baffle spacing

Figure 1.9 Shell and tube exchanger with segmental baffles.


In horizontal condensers with the vapors on the shell side, one should try to achievea horizontal lefteright flow with vertical baffle edges so that the condensate at thebottom can safely drain out at the bottom and the noncondensables can flow out atthe top.

Segmental cut

Baffle

Condensate drain

The baffle spacings can be adjusted for the decreasing vapor stream, that is, smallerwith increasing flow path. This is important with vapors containing inert gases whichmust be convectively cooled before condensing.

Vent

Condensate

1.2.4 Technical RemarksWhich product should be to the tube or shell side?

Through the tubes: the medium with higher pressure, corrosive media and productswith high fouling tendency,On the shell side: viscous products.

How are the different expansions accommodated?Through an expansion joint in the shell of fixed tube sheet bundles,Through a U-tube construction with free tube expansion,With a floating head for the free expansion.

The different possibilities are shown in Figure 1.10.


1.2.5 Selection of a Shell and Tube ExchangerAn overview of the different heat exchanger types according to TEMA [3] for differentbonnets, shells, and baffle chambers is given in Figure 1.11.

Fixed tubesheetwith expansion joint

U-tube exchanger

U-tube evaporator

Stuffing box flange

Pull throughfloating head

Floating headevaporator

Split ring floatinghead

Figure 1.10 Expansion possibilities in shell and tube heat exchangers.


With the help of the logic flowchart in Figure 1.12 the required heat exchanger typecan be determined for the given problem definition. Thermal expansion stresses andshell-side fouling are of overriding importance.

1.2.5.1 Which Heat Exchanger Types Can Be Cleaned?Tube-side cleaning: TEMA-types, AEL and NEN, with tube diameters >20 mm

Shell-side cleaning:Removable bundle with square pitch: TEMA-types, CEU þ CES þ AES

Figure 1.11 TEMA systematics for different shells and bonnets constructions [3].


NOMENCLATUREU Calculated overall heat transfer coefficient (W/m2 K)Ureq Required overall heat transfer coefficient (W/m2 K)MS Shell-side flow rate (kg/h)cS Specific heat capacity shell side (Wh/kg K)

Severethermal stress

No

Fixed tubesheetor removable bundle

Severeshellside fouling ?

No

Perodic inspectionrequired?

No

Fixed tubesheet

Yes

U-Tubes, bayonettubes, floating heador fixed tubesheetwith expansion jointRemovable?

Yes

Chemical cleaningfeasible inplace ?

Yes

NoYes

Removable bundleFloating headsquare arrangement

Selection scheme for the choiceof the right exchanger-type

Figure 1.12 Selection scheme for heat exchangers [2].


MT Tube-side flow rate (kg/h)cT Specific heat capacity tube side (Wh/kg K)Qreq Required heat load (W)wsh Shell-side flow velocity (m/s)wt Tube-side flow velocity (m/s)Sf Fouling reserve (m2 K/W)aio Heat transfer coefficient ai in the tube relative to the outer area (W/m2 K)ao Heat transfer coefficient of the shell-side (W/m2 K)s Tube wall thickness (m)l Thermal conductivity of the tube material (W/m K)fi Tube-side fouling factor inside the tube (m2 K/W)fo Tube fouling factor at the outer side of the tube (m2 K/W)CMTD Corrected effective mean temperature difference ¼ F � LMTD (�C)LMTD Logarithmic mean temperature difference for countercurrent (�C)

REFERENCES[1] A.E. Jones, Thermal design of the shell and tube, Chem. Eng. 109 (2002) 60e65.[2] A. Devore, J. Vago, G.J. Picozzi, Specifying and selecting, Chem. Eng. 87 (1980) 133e148.[3] TEMA, Standards of the Tubular Exchanger Manufacturers Association, eighth ed., 1999.


CHAPTER 2

Calculations of the TemperatureDifferences LMTD and CMTD

Contents

2.1 Logarithmic Mean Temperature Difference for Ideal Countercurrent Flow 212.2 Corrected Temperature Difference for Multipass Heat Exchanger 222.3 Influence of Bypass Streams on LMTD 292.4 Mean Weighted Temperature Difference 302.5 Determination of the Heat Exchanger Outlet Temperatures 32

2.5.1 Calculation of the outlet temperatures in a multipass heat exchanger under considerationof the temperature efficiency factor F for nonideal countercurrent flow 332.5.1.1 Calculation of the outlet temperature Tc2 of the cold medium 33

2.5.2 Calculation of the outlet temperature th2 for ideal countercurrent without F 33References and Further Reading 35

2.1 LOGARITHMIC MEAN TEMPERATURE DIFFERENCE FOR IDEALCOUNTERCURRENT FLOW

The logarithmic mean temperature difference (LMTD) for ideal countercurrent flow isdetermined from the two temperature differences Dt1 and Dt2.

LMTD ¼ Dt1 � Dt2ln Dt1

Dt2

Dt1 ¼ T1 � t2 Dt2 ¼ T2 � t1

T1 ¼ shell-side inlet temperature (�C)T2 ¼ shell-side outlet temperature (�C)t1 ¼ tube-side inlet temperature (�C)t2 ¼ tube-side outlet temperature (�C)

Example 1: Calculation of the logarithmic mean temperature difference

Tube side: t1 ¼ 30 �C t2 ¼ 60 �CShell side: isothermal heating Nonisothermal heating

T1 ¼ T2 ¼ 100 �C T1 ¼ 100 �C T2 ¼ 80 �C100 / 100

60 z 30Dt1 ¼ 40 Dt2 ¼ 70

100 / 8060 z 30

Dt1 ¼ 40 Dt2 ¼ 50

LMTD ¼ 40�70ln 40

70¼ 53:6 �C LMTD ¼ 40�50

ln 4050

¼ 44:8 �C

Heat Exchanger Design Guidehttp://dx.doi.org/10.1016/B978-0-12-803764-5.00002-X


http://dx.doi.org/10.1016/B978-0-12-803764-5.00002-X

2.2 CORRECTED TEMPERATURE DIFFERENCE FOR MULTIPASS HEATEXCHANGER

In Chapter 1, it was already pointed out that with a multipass heat exchanger, there is noideal countercurrent but a mixture of co- and countercurrent flow.

This makes the effective temperature gradient worse.In Figures 2.2 and 2.3, temperature efficiency diagrams are shown, it is clear that the

driving temperature difference in the one-pass equipment with ideal countercurrent isbetter than in the two-pass heat exchanger.

Procedure for the determination of the corrected effective mean temperature difference(CMTD) for heat exchangers with nonideal countercurrent flow:

First, the LMTD is calculated from the inlet and outlet temperatures on the tube andshell side.With nonideal countercurrent, the LMTD must be corrected with a temperature ef-ficiency factor F.

CMTD ¼ F � LMTD

F ¼ temperature efficiency factorLMTD ¼ logarithmic mean temperature difference (�C) for countercurrent flowCMTD ¼ corrected effective mean temperature difference (�C) for multipass heatexchangers

Figure 2.1 Heat exchanger TEMAdtype E with two tube passes.


The temperature efficiency factor F can be calculated for a heat exchanger shell type Eaccording to TEMA using the following equation:

F ¼ ffiffiffiffiffiffiffiffiffiffiffiffiffiffi

R2 þ 1pR � 1

!ln½ð1� PzÞ=ð1� RPzÞ�

ln

"ð2=PzÞ � 1� R þ

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiR2 þ 1

pð2=PzÞ � 1� R �

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiR2 þ 1

p#

Pz ¼1�

�RP � 1P � 1

�1=N

R ��RP � 1P � 1

�1=N

Figure 2.2 Temperature profile as function of the heat load in a one-pass heat exchanger with idealcountercurrent flow.

Calculations of the Temperature Differences LMTD and CMTD 23

P ¼ t2 � t1T1 � t1

R ¼ T1 � T2

t2 � t1

N ¼ number of heat exchangers in seriesThe following table shows how large the temperature approach or a temperature

cross deteriorates the CMTD, and that one can improve the temperature efficiency factorwith multiple heat exchangers in series.

T1 T2 t1 t2 N LMTD F CMTD

58 42 15 35.1 1 24.9 0.906 22.658 42 20 40.1 1 19.9 0.844 16.858 42 25 45.1 1 14.9 0.670 10

Figure 2.3 Temperature profile as function of the heat load in a two-pass heat exchanger with co- andcountercurrent flow.


T1 T2 t1 t2 N LMTD F CMTD

58 42 25 45.1 2 14.9 0.936 13.958 42 30 50.1 2 9.8 0.840 8.258 42 35 55.1 4 4.7 0.818 3.858 42 37 57.1 6 2.4 0.631 1.558 42 37 57.1 7 2.4 0.764 1.8

Corollary:1. With large approach of temperatures at the tube and shell side, the temperature effi-

ciency factor F and hence the CMTD deteriorates.2. Through the arrangement of multiple heat exchangers one after the other, one can

improve the temperature efficiency factor F because with this arrangement one ap-proaches the countercurrent flow.

3. A large temperature cross is only possible with multiple equipments in series.The temperature efficiency factor F can also be determined graphically using the diagrams

in Figure 2.4 with the calculation variables P and R.Alternatively, the CMTD in multipass heat exchangers can be determined with the

calculation variables O and M.

CMTD ¼ M

lnO þMO �M

ð�CÞ

O ¼ ðT1 � t2Þ þ ðT2 � t1Þ

M ¼ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiðT1 � T2Þ2 þ ðt2 � t1Þ2

q

Example 2: Calculation of LMTD and CMTD for the heat exchangers in Figures 2.2and 2.3

Shell inlet temperature T1 ¼ 80 �CShell outlet temperature T2 ¼ 50 �CTube inlet temperature t1 ¼ 20 �CTube outlet temperature t2 ¼ 40 �CDt1 ¼ T1 � t2 ¼ 80 � 40 ¼ 40 �CDt2 ¼ T2 � t1 ¼ 50 � 20 ¼ 30 �CCalculation of the LMTD for ideal countercurrent in Figure 2.1:

LMTD ¼ 40� 30

ln 4030

¼ 34:8 �C

Determination of the CMTD for the two-pass heat exchanger in Figure 2.2.


Figure 2.4 Diagrams with the temperature correction factors as function of P and R. (From Fraas,Ozisik, Heat Exchanger Design, Wiley, 1965 [9].)


Determination of the calculation variables P and R:

P ¼ t2 � t1T1 � t1

¼ 40� 2080� 20

¼ 0:333 R ¼ T1 � T2t2 � t1

¼ 80� 5040� 20

¼ 1:5

For the variables P and R, one takes from the above diagram in Figure 2.4, the efficiency temperaturefactor F ¼ 0.91.

Calculation of the CMTD for nonideal countercurrent:

CMTD ¼ F � LMTD ¼ 0:91� 34:8 ¼ 31:6 �C

In the two-pass heat exchanger with nonideal countercurrent flow, the effective mean temperaturedifference of LMTD ¼ 34.8 �C deteriorates to CMTD ¼ 31.6 �C.

Alternative calculation of CMTD with M and O:

O ¼ ð80� 40Þ þ ð50� 20Þ ¼ 70

M ¼ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffið80� 50Þ2 þ ð40� 20Þ2

q¼ 36

CMTD ¼ 36

ln70þ 3670� 36

¼ 31:6 �C

Example 3: Correction factor F for the arrangement of heat exchangers in series

T1 ¼ 58 �C T2 ¼ 42 �C t1 ¼ 25 �C t2 ¼ 45:1 �C

P ¼ t2 � t1T1 � t1

¼ 45:1� 2558� 25

¼ 0:6 R ¼ T1 � T2t2 � t1

¼ 58� 4245:1� 25

¼ 0:8

The correction factor F follows from the intersection of P ¼ 0.6 on the x-axis with the curve for R ¼ 0.8in Figure 2.4.

For a single heat exchanger F ¼ 0.67Two exchangers in series F ¼ 0.94Three exchangers in series F ¼ 0.97Four exchangers in series F ¼ 1

Through the arrangement of heat exchangers in series, the countercurrent effect is improved andtherefore achieves a better usage of the temperature difference.

Example 4: Calculation of the temperature factor F and the CMTD for four heatexchangers in series

T1 ¼ 58 �C T2 ¼ 42 �C t1 ¼ 35 �C t2 ¼ 55:1 �C


Dt1 ¼ 58� 55:1 ¼ 2:9 �C Dt2 ¼ 42� 35 ¼ 7 �C

LMTD ¼ 7� 2:9

ln 72:9

¼ 4:65 �C R ¼ 58� 4255:1� 35

¼ 0:796 P ¼ 55:1� 3558� 35

¼ 0:8739

F ¼ 0.8186 for four heat exchangers in series from Figure 2.4.With the temperature efficiency factor F, the CMTD is determined as follows:

CMTD ¼ 0:8186� 4:65 ¼ 3:8 �C

Example 5: Calculation of the CMTD for isothermal and nonisothermal heating

Tube side: t1 ¼ 30 �C t2 ¼ 60 �CShell side: Isothermal heating (F ¼ 1) Nonisothermal heating (F < 1)T1 ¼ T2 ¼ 100 �C T1 ¼ 100 �C T2 ¼ 80 �CDt1 ¼ 100 � 60 ¼ 40 �C Dt1 ¼ 100 � 60 ¼ 40 �CDt2 ¼ 100 � 30 ¼ 70 �C Dt2 ¼ 80 � 30 ¼ 50 �C

LMTD ¼ 70� 40

ln 7040

¼ 53:6 �CLMTD ¼ 50� 40

ln 5040

¼ 44:8 �C

F ¼ 1 with isothermal heating F ¼ 0.948CMTD ¼ 1 � 53.6 ¼ 53.6 �C CMTD ¼ 0.948 � 44.8 ¼ 42.5 �C

Alternative calculation for CMTD with isothermal heating:

O ¼ ð100� 60Þ þ ð100� 30Þ ¼ 110

M ¼ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffið100� 100Þ2 þ ð60� 30Þ2

q¼ 30

CMTD ¼ 30

ln110þ 30110� 30

¼ 53:6 �C

Alternative calculation for CMTD with nonisothermal heating:

O ¼ ð100� 60Þ þ ð80� 30Þ ¼ 90

M ¼ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffið100� 80Þ2 þ ð60� 30Þ2

q¼ 36

CMTD ¼ 36

ln90þ 3690� 36

¼ 42:5 �C

In Figure 2.5, it is shown how to connect multiple heat exchangers in series.


2.3 INFLUENCE OF BYPASS STREAMS ON LMTD

An important issue that has not been considered so far is the temperature profiledistortion.

On the shell side not the total product stream flows across through the tubebundle and becomes heated or cooled. A part of the product stream flows in bypassor through leakage away from the heat exchanger tubes and does not take part inheat exchange.

Therefore, only a part of the shell-side product stream is heated or cooled and this partis heated or cooled more because the flow is smaller.

Hence the driving temperature gradient is reduced.The LMTD becomes smaller with increasing leakage.This is shown in Figure 2.6 for the following conditions:Heating medium: Steam at 100 �C, isothermProduct heating from 70 �C to 90 �C

1-2 exchanger

2-4 exchanger

3-6 exchanger

Figure 2.5 Arrangement in series of heat exchangers.


Inlet temperature difference Dt1 ¼ T1 � t2 ¼ 100 � 70 ¼ 30 �COutlet temperature difference Dt2 ¼ T2 � t1 ¼ 100 � 90 ¼ 10 �C

LMTD ¼ 30� 10

ln 3010

¼ 18:2 �C at 0% bypass

Figure 2.6 shows that the increasing number n of change in reversing the flow direc-tion the temperature efficiency becomes better. This is because with each shell-side pass,both streams mix.

Recommendation: Baffle spacing ¼ 0.2 � shell diameter or maximum 2 �C tempera-ture change per shell pass.

Figure 2.7 shows the deficiency in the mean temperature difference with increasingleakage.

Figure 2.8 shows how the logarithmic differential temperature difference (LDT) be-tween the hot and the cold stream in the individual zones along the length of the heatexchanger decreases. The entering cold product has a high driving temperature differenceLDT which decreases with progressing product heating over the length of the heatexchanger. The ideal LDT line without leakage and the LDT line for 50% bypassflow are shown. The driving temperature difference is considerably reduced by thebypass leakage.

2.4 MEAN WEIGHTED TEMPERATURE DIFFERENCE

For nonstraight line temperature curves in the temperatureeheat load diagram,for instance in multicomponents condensation, large heat loads are required at high-temperature gradients and smaller heat loads with low-temperature differences.

Figure 2.6 Logarithmic mean temperature difference (LMTD) as a function of the bypass leakage.


A normal calculation of the driving temperature difference CMTD is wrong.The CMTDs in the different load zones are determined and lastly, the WMTD is

calculated.An example is shown in Figure 2.9.The difference between CMTD ¼ 67.32 K and WMTD ¼ 103.78 is

considerable.The condenser would have been oversized with CMTD ¼ 67.38 �C by 54%.

Figure 2.8 LDT curve over the length of the heat exchanger.

Figure 2.7 Temperature efficiency as function of the bypass stream.


2.5 DETERMINATION OF THE HEAT EXCHANGER OUTLETTEMPERATURES

In a normal case, the required heat exchanger area for a specified heating or cooling,that is, for a specified temperature difference, is determined.

In this chapter, it will be shown how to calculate the outlet temperatures of a specifiedheat exchanger for different product stream rates.

Required data:Wh ¼ hot stream flow rate (kg/h)th1 ¼ inlet temperature of the hot stream (�C)th2 ¼ outlet temperature of the hot stream (�C)ch ¼ specific heat capacity of the hot medium (Wh/kg K)Wc ¼ cold stream flow rate (kg/h)Tc1 ¼ inlet temperature of the cold medium (�C)Tc2 ¼ outlet temperature of the cold medium (�C)Cc ¼ specific heat capacity of the cold medium (Wh/kg K)A ¼ heat exchanger area (m2)U ¼ overall heat transfer coefficient (W/m2 K)F ¼ temperature efficiency factor for nonideal countercurrent

Figure 2.9 Example for the determination of the mean weighted temperature difference (WMTD).


2.5.1 Calculation of the outlet temperatures in a multipass heatexchanger under consideration of the temperature efficiencyfactor F for nonideal countercurrent flow

th2 ¼ ð1� RÞ � th1 þ ð1� B2Þ � R � Tc1

1� ðR � B2Þ ð�CÞ

R ¼ Wc � ccW � ch

B2 ¼ exp

�F � U � AWc � cc

� ðR � 1Þ�

2.5.1.1 Calculation of the outlet temperature Tc2 of the cold medium

Tc2 ¼ th1 � th2R

þ Tc1 ð�CÞ

2.5.2 Calculation of the outlet temperature th2 for ideal countercurrentwithout F

th2 ¼ ð1� RÞ � th1 þ ð1� B1Þ � R � Tc1

1� ðR � B1Þ ð�CÞ

B1 ¼ exp

�U � AWc � cc

� ðR � 1Þ�

Example 6: Calculation of the heat exchanger outlet temperatures th2 and Tc2

Hot side: Wh ¼ 10 t/h th1 ¼ 58 �C th2 ¼ 42 �C ch ¼ 0.5 Wh/kg KCooling heat load Q ¼ 10000 � 0.5 � (58 � 42) ¼ 80,000 WCold side: cooling water Tc1 ¼ 25 �C Tc2 ¼ 45.1 �C cc ¼ 1.16 Wh/kg K

Overall heat transfer coefficient U ¼ 400 W/m2 KTemperature efficiency factor F ¼ 0.67

Determination of the required cooling water rate for the cooling heat load Q ¼ 80 kW

Wc ¼ Qcc � ðTc2 � Tc1Þ ¼ 80000

1:16� ð45:1� 25Þ ¼ 3431 kg=h

Calculation of LMTD and CMTD

58 / 4245:1 ) 2512:9 17


LMTD ¼ 17� 12:9

ln 1712:9

¼ 14:9 �C

P ¼ 45:1� 2558� 25

¼ 0:609 R ¼ 58� 4245:1� 25

¼ 0:796 F ¼ 0:67

Calculation of the heat exchanger area A:

A ¼ QU� F � LMTD

¼ 80000400� 0:67� 14:9

¼ 20 m2

Calculation of the auxiliary parameters R and B2:

R ¼ Wc � ccWh � ch

¼ 3431� 1:1610000� 0:5

¼ 0:796

B2 ¼ exp

�F � U� AWc � cc

� ðR� 1Þ�¼ exp

�0:67� 400� 203431� 1:16

� ð0:796� 1Þ�

¼ 0:7598

Calculation of the outlet temperature th2 of the hot medium:

th2 ¼ ð1� RÞ � th1 þ ð1� B2Þ � R� Tc11� ðR� B2Þ

th2 ¼ ð1� 0:796Þ � 58þ ð1� 0:7598Þ � 0:796� 251� ð0:796� 0:7598Þ ¼ 42 �C

Calculation of the outlet temperature Tc2 of the cold stream

Tc2 ¼ th1 � th2R

þ Tc1 ¼ 58� 420:796

þ 25 ¼ 45:1 �C

Example 7: Calculation of the cooling water outlet temperature Tc2 for a smallercooling water flow rate

Wc ¼ 3135 kg/h instead of 3431 kg/h Data: see example 6

The cooling water outlet temperature rises from 42 �C to 47 �C by reducing the cooling water flow ratefrom 3431 kg/h to 3135 kg/h.

Thereby reducing the CMTD from 9.95 �C to CMTD ¼ 6.42.

LMTD ¼ 13.78 �C R ¼ 0.727 P ¼ 0.666 F ¼ 0.465 CMTD ¼ 6.42 �C

The lower effective temperature difference reduces the heat duty within the heat exchanger from80 kW for CMTD ¼ 9.95e51.36 kW for CMTD ¼ 6.42 �C:

Q ¼ k � A� CMTD ¼ 400� 20� 6:42 ¼ 51360 W


In addition, it has to be considered that through the smaller product rate, the flow velocity and hencealso the heat transfer coefficient will be reduced and so that the overall heat transfer coefficient U becomessmaller.

REFERENCES AND FURTHER READING[1] R.A. Bowman, Ind. Engr. Chem. 28 (1936), 541/544.[2] R.A. Bowman, Trans. ASME 62 (1940), 283/294.[3] J. Fernandez, Trans. ASME (1957), 287/297.[4] F.K. Fisher, Ind. Engr. Chem. 30 (1938), 377/383.[5] J. Bowman, R. Turton, Chem. Eng ( Juli 1990).[6] Ch.A. Plants, Chem. Eng. ( Juni 1992 þ Juli 1992).[7] R. Mukherjes, Chem. Eng. Prog. (April 1996 þ Februar 1998).[8] J. Fisher, R.O. Parker, Hydrocarbon Process. ( Juli 1969).[9] A.P. Fraas, M.N. Ozisik, Heat Exchanger Design, John Wiley, N.Y., 1965.


CHAPTER 3

Calculations of the Heat TransferCoefficients and Pressure Lossesin Convective Heat Transfer

Contents

3.1 Tube-Side Heat Transfer Coefficient 413.2 Shell-Side Heat Transfer Coefficient 423.3 Comparison of Different Calculation Models 493.4 Pressure Loss in Convective Heat Exchangers 51

3.4.1 Tube-side pressure drop DPT 513.4.2 Shell-side pressure loss 53

3.5 Heat Exchanger Design with Heat Exchanger Tables 56Nomenclature 63References and Further Reading 63

How are the heat transfer coefficients calculated? [1,3e5,11,14].The required calculation equations for the determination of the heat transfer coeffi-

cients on the tube side and on the shell side are listed in Figure 3.2. The application of theequations is shown in Examples 1 and 2. The calculation of the heat transfer coefficients issimple and the results are on the safe side.

The relationships are valid for the following geometrical conditions:Baffle spacing/diameter B/D ¼ 0.2e0.3Segmental height/diameter H/D ¼ 0.2e0.3These geometrical conditions have proved to be effective in practice.The calculation equations in Figure 3.2 indicate that the Nusselt number Nu,

that is, the heat transfer coefficient a is dependent on the Prandtl and the Reynoldsnumber.

The Prandtl number Pr contains only the physical properties that cannot be changed.For the design in general, the properties in the heat exchanger at average temperature arechosen.

If, however, the properties in the heat exchanger greatly change with temperature asegmental calculation of the individual zones is recommended.



http://dx.doi.org/10.1016/B978-0-12-803764-5.00003-1

The improvement of the heat transfer coefficients can only be achieved by increasingthe Reynolds number Re, that is, by increasing the flow velocity w.

With increasing Reynolds number the laminar thickness of the boundary layer onthe tube wall decreases and the quotient az l/s becomes higher. The following depen-dencies of the heat transfer coefficient a on the flow velocity w follow from the calcu-lation equations in Figure 3.2:

Tube side: a f w0.8 Shell side: a f w0.6

A sufficiently high flow velocity is the required condition for a good heat transfercoefficient.

The recommended flow velocities are 0.5e2 m/s for liquids and 10e25 m/s forgases. With viscous media the shell-side heat transfer coefficients are higher thanthe a-values on the tube side. This is why viscous products are preferably located atthe shell side.

Shellside inlet

t1

Tubeside inletFigure 3.1 Heat exchangers with two tube passes.


Example 1: Calculation of the tube-side heat transfer coefficientBasic data:

Heat exchanger DN 500 with 164 tubes25 � 2

di ¼ 21 mm nP ¼ four tube passes

41 tubes per passTube-side throughput VT ¼ 51.12 m3/hDensity r ¼ 995 kg/m3 Specific heat capacity c ¼ 1.16 Wh/kg KKinematic viscosity n ¼ 0.92 mm2/s Heat conductivity l ¼ 0.605 W/m K

1. Calculation of the flow cross section aT of 41 tubes

aT ¼ nnP

� d2i �p

4¼ 164

4� 0:0212 � 0:785 ¼ 0:0142 m2

Nu

Nu

Nu Nu

w × di

0.023 × Re0.8 × Pr0.33

0.196 × Re0.6 × Pr0.33 0.156 × Re0.6 × Pr0.33

Nu

Nu

1.86 × 0.33Re × Pr × di

× Pr0.42 0.037 × Re0.75 – 6.66

3600 × v × c ×

0.6 < Pr < 500

α λ

Nuα λ

v

w × do

v

λρ

owcross

V (m3/h)B × (Di – nacr × do)

Figure 3.2 Calculation equations for the convective heat transfer.

Calculations of the Heat Transfer Coefficients and Pressure Losses in Convective Heat Transfer 39

2. Determination of the flow velocity wT in the tubes

wT ¼ V�m3

�h�

aT � 3600¼ 51

0:0142� 3600¼ 1 m=s

3. Determination of the Reynolds number

Re ¼ wT � din

¼ 1� 0:0210:92� 10�6 ¼ 22826

4. Calculation of the Pr number

Pr ¼ n� c � r� 3600l

¼ 0:92� 10�6 � 1:16� 995� 36000:605

¼ 6:32

5. Determination of the Nusselt number

Nu ¼ 0:023� Re0:8 � Pr0:33 ¼ 0:023� 228260:8 � 6:320:33 ¼ 129:62

6. Calculation of the heat transfer coefficient

a ¼ Nu� l

di¼ 129:62� 0:605

0:021¼ 3734 W

�m2 K

Example 2: Calculation of the shell-side heat transfer coefficientBasic data:

Shell diameter Di ¼ 1.1 m 270 tubes 38 � 2 mm do ¼ 38 mmTriangular pitch T ¼ 50 mm Baffle spacing B ¼ 200 mmFlow throughput VShell ¼ 200 m3/hDensity r ¼ 803 kg/m3 Specific heat capacity c ¼ 2.59 kJ/kg K ¼ 0.721 Wh/kg KKinematic viscosity n ¼ 051 mm2/s Heat conductivity l ¼ 0.108 W/m K

1. First, the flow cross section across for the cross stream with the pitch T is determined

across ¼ Di � B��1� do

T

�¼ 1:1� 0:2�

�1� 38

50

�¼ 0:0528 m2

Alternative calculation with the number of tubes nacr in the cross stream

nacr ¼ Di

T¼ 1100

50¼ 22

across ¼ B� ðDi � nacr � doÞ ¼ 0:2� ð1:1� 22� 0:038Þ ¼ 0:0528 m2

2. Thereafter, the flow velocity wcross for the cross stream is determined

wcross ¼ VM�m3

�h�

across � 3600¼ 200

0:0528� 3600¼ 1:052 m=s

3. Calculation of the Reynolds number

Re ¼ wcross � don

¼ 1:052� 0:0380:51� 10�6 ¼ 78; 398


4. Calculation of the Pr number

Pr ¼ n� c � r� 3600l

¼ 0:51� 10�6 � 0:721� 803� 36000:108

¼ 9:84

5. Determination of the Nusselt number for triangular pitch

Nu ¼ 0:196� Re0:6 � Pr0:33 ¼ 0:196� 783980:6 � 9:840:33 ¼ 360:18

6. Calculation of the heat transfer coefficient

a ¼ Nu� l

do¼ 360:18� 0:108

0:038¼ 1023:7 W

�m2 K

3.1 TUBE-SIDE HEAT TRANSFER COEFFICIENT [3,6,12,14,16]

The flow velocity and the heat transfer coefficient on the tube side can be increasedthrough a multipass arrangement. The medium flows thereby in multiple passes throughthe tubes and the tube-side flow cross section is reduced. In Figure 3.1 a tube bundle withtwo tube-side passes is depicted. With two passes the tube-side flow cross section ishalved and therefore the flow velocity is doubled. The higher flow velocity in thetube improves the heat transfer coefficient. In Figure 3.3 the dependency of the heattransfer coefficient on the flow velocity and on the tube diameter is shown.

A disadvantage of the multipass arrangement is the reduction of the effectivetemperature difference for the heat transfer because there is no ideal countercurrent flow.

In the two passes on the tube side the medium flows in the first pass in counter-current flow and in the second pass in cocurrent flow to the product stream on theshell side.

Figure 3.3 Tube-side heat transfer coefficient of heptane as function of flow velocity and tubediameter.


Therefore, the logarithmic mean temperature difference (LMTD) for countercurrentflow must be corrected with the temperature effective factor F.

CMTD ¼ F � LMTD

CMTD is the corrected effective mean temperature difference for the heat transfer.The calculation of CMTD is shown in Chapter 2.An additional disadvantage of the multipass arrangement is the increase of the pressure

loss through the multiple flow redirection and because of the higher flow velocity whichgoes into the pressure loss calculation in a square function.

InFigures 3.4 and 3.5 the heat transfer coefficients for some liquids and gases are shown.

3.2 SHELL-SIDE HEAT TRANSFER COEFFICIENT[1,2,4,9,10e12,14e16]

With the convective heat exchanger shown in Figure 3.1, the medium flows on the shellside across the tube bundle and lengthways through the baffle cuts. The flow cross-

Figure 3.4 Tube-side heat transfer coefficients as function of the flow velocity for different liquids.


sectional areas should be preferably equal for the cross stream through the bundle and thelengthways stream in the baffle cuts. In this case, there are equal flow velocities withoutacceleration turbulences with associated pressure losses.

With increasing flow velocity, the heat transfer coefficient a becomes better.The shell-side stream velocity for the cross flow can be calculated with the following

equations:

wcross ¼V

�m3

�h�

3600� B� ðDi � nacr � doÞ ðm=sÞ

wcross ¼V

�m3

�h�

3600�Di � B��1� do

T

� ðm=sÞ

By decrease of the baffle spacing B the flow velocity is increased and the heat transfercoefficient becomes better.

Figure 3.5 Tube-side heat transfer coefficients as function of the flow velocity for different gases.


In Figures 3.6 and 3.7 it is shown how the shell-side heat transfer coefficient of liquidsand gases rises with increasing flow velocity.

Through the arrangement of more baffles to improve the flow velocity, the pressureloss in the cross stream across the tube bundle is increased so that the product flow in-creases and hence the product flow rate flowing through the leak and bypass cross sectionaside the tube bundle and does not take part in heat change.

This deteriorates the heat transfer.Figure 3.8 shows that according to VDI-W€armeatlas [11] the correction factor fW

for the shell-side heat transfer coefficient becomes smaller with decreasing B/D ratio,that is, increasing number of baffles because each baffle causes additional bypass and leak-ing streams.

The correction factor fW is defined as follows:fW ¼ fG � fL � fBfG is the geometrical factor which essentially considers how many heat exchangertubes lie in the baffle window.

Figure 3.6 Shell-side heat transfer coefficients as a function of the flow velocity for different liquids.


Figure 3.7 Shell-side heat transfer coefficients as a function of the flow velocity for different gases.

Figure 3.8 Correction factor according to VDI-W€armeatlas [11] for the shell-side heat transfer as afunction of B/D ratio. B ¼ baffle spacing, D ¼ shell diameter.


fL is the leakage factor for the leaking stream through the gap between the tube andthe baffle holes and also between the baffles and the shells.The fL factor is dependent on the specified tolerance and the number of the baffles.fB is the bypass factor for the flow through the tube lanes and the annular gap betweenthe tube bundle and the shell. With sealing strips and dummy tubes, the bypass streamcan be minimized such that fB ¼ 1.In Figure 3.9 the different streams on the shell side of a heat exchanger according to

Tinker [9] are shown.Stream A: leakage stream through the gap between the tubes and the baffle holesStream B: cross stream through the bundleStream C: bypass stream through the annular gap between the bundle and the shellStream E: bypass stream between the baffle and the shellStream F: bypass stream due to the nature of the construction caused tube lanesIn Figure 3.10 it is shown for a DN 500 heat exchanger that stream B, that is, the flow

part flowing across the tube bundle increases with increasing baffle spacing. Less bafflesresult in the occurring of less leakage and bypass streams but less baffles with wider bafflespacings decrease the flow velocity and deteriorate the heat transfer.

These curves are valid for the following heat exchanger data:Shell diameter DN 500 with 256 tubes 20 � 2, 6 m long, triangular pitch 26 mm,four passes.Shell-side product: 15 m3/h kerosine, which is cooled with water from 114 to 40 �C.From Figure 3.10 the influence of the pass partition of a multipass equipment on the

bypass stream can be seen. A ribbon pass arrangement across the flow gives more crossflow than a pass arrangement in quadrants because the tube lanes in flow direction ina quadrant partition permit a higher stream F through the tube lanes.

In Figure 3.11 it can be seen that the heat transfer coefficients rise with reducing bafflespacing in spite of the increase in bypass and leakage streams. The values are valid for aconstant flow throughput of 15 m3/h.

The improvement due to the higher flow velocity is greater than the deterioration bythe increasing bypass and leakage streams.

Figure 3.9 Shell-side product streams according to Tinker [3,9].


The leakage and bypass streams also worsen the temperature efficiency for the heat ex-change because a part of the product stream does not take part in the heat exchange processand therefore reduces the effective driving temperature difference for the heat exchange.

The reducedproduct ratewithout thebypass and leakage streampart is furtherheatedandtherefore reduces the temperature difference between the product and the heatingmedium.

In each pass through the baffle chamber a mixed temperature of the cross stream andthe bypass/leakage stream results.

Figure 3.10 The flow part of the cross stream through the tube bundle of a DN 500 heat exchanger asfunction of the baffle spacing with ribbon and quadrant arrangement.

Figure 3.11 Shell-side heat transfer coefficients as function of baffle spacing for ribbon and quadrantarrangement.


In Figure 3.12 different baffle types are shown.In practice the segmental baffles with horizontal chord shown below are preferably

used for the convective heat transfer because the medium is better mixed by the upand down streams and possible temperature differences are compensated.

With condensers vertical baffle, cutswith vertical chord are advantageous inorder to avoidcondensate built-up and allow the noncondensable gases to escape and flow out upward.

If the pressure loss on the shell side is to be reduced disc and doughnutdor doublesegmental baffles are used.

The full circle baffles have the advantage that there is no bypass streams through thediametrical gap between the tube bundle and the shell; however, the mainly horizontalstreams deteriorate the heat transfer coefficient.

With the double segmental baffles, the flow cross-sectional area is doubled and hencethe flow velocity halved so that the pressure loss is considerably reduced at the cost of apoorer heat transfer coefficient.

Disc and doughnut

DoubleSegmental

Segmental

Figure 3.12 Different baffle types.


Fundamentally the means of reducing the pressure loss on the shell side of heatexchangers are as follows:• altering the baffle spacing and baffle arrangement• larger pitch or quadratic pitch• shorter tube lengths• larger nozzle diameter• multiple heat exchangers in parallel arrangement• installation of a pure cross stream heat exchanger, for instance, according to TEMA J

or TEMA X.What is important for a working heat exchanger?High flow velocity at the allowable pressure drops.Preferably low bypass and leakage streams.For that it is necessary:Multiple passes on the tube side.Small clearance tube/baffle plate holes þ baffle/shell.Small gap shell/bundle outer tube limit with small diametrical gap.Pass arrangement in ribbons without uniformed tube lanes for the bypass.Constructive measures:Longitudinal plates or sealing strips in the shell in the case of large diametrical gaps orspaces between the shell and the bundle tube limit diameter.This is especially the case for floating head equipments with large free flow crosssections between the bundle and the shell (Figure 3.13).Dummy tubes for reducing the bypass cross sections in the tube lanes and in thediametrical gap between bundle and shell.

3.3 COMPARISON OF DIFFERENT CALCULATION MODELS

In Figures 3.14 and 3.15 the heat transfer coefficients calculated according to differentmodels for cooling water on the tube side and ethanol on the shell side are shown.

The ethanol is cooled from 58 to 42 �C with cooling water which is heated from 15to 35.1 �C.

The calculations were made for the heat exchanger type 10 in Table 1 in Chapter 3.6:DN 500 with 164 tubes in four passes, tube length 3 m, baffle spacing 100 mm.Model Nitsche method (see Figure 3.2)Model TPLUS program: software for the design and rating of heat exchangersModel Kern [16]Model VDI-W€armeatlas [11]Corollary: From the overall heat transfer coefficients, shown in Figure 3.16, it can be

seen that the deviations are not large and that one is on the safe side with the simpleNitsche method.


Figure 3.13 Floating head heat exchanger DN 500 with sealing strips to reduce the bypass streamand to improve the heat transfer coefficient.

Figure 3.14 Tube-sideheat transfer coefficients in a type10heat exchanger for coolingwater in the tubes.


3.4 PRESSURE LOSS IN CONVECTIVE HEAT EXCHANGERS

3.4.1 Tube-side pressure drop DPTThe pressure loss in the tubes is calculated as follows:

DPT ¼ w2T � r

2� nP �

�f � Ldi

þ 4

�ðPaÞ Friction factor f ¼ 0:216

Re0:2

Nozzle pressure loss:

Inlet nozzle: DPNi ¼ w2Ni � r

2ðPaÞ Outlet nozzle: DPNo ¼ 0:5� w2

No � r

2ðPaÞ

Figure 3.15 Shell-side heat transfer coefficients in a type 10 heat exchanger for ethanol at 50 �C.

Figure 3.16 Overall heat transfer coefficients for an ethanolewater heat exchanger.


DPT ¼ pipe pressure lossDPtot ¼ total pressure loss ¼ DPT þ DPni þ DPnowni ¼ flow velocity in inlet nozzle (m/s)wno ¼ flow velocity in outlet nozzle (m/s)wT ¼ flow velocity in the tube (m/s)L ¼ tube length (m)di ¼ tube diameter (m)nP ¼ number of tube passesr ¼ density (kg/m3)

Example 3: Tube-side pressure loss

V ¼ 14.35 m3/h wT ¼ 0.28 m/s Re ¼ 6406 L ¼ 3 mdi ¼ 21 mm nP ¼ four passes 41 tubes per pass r ¼ 995 kg/m3

f ¼ 0:216

64060:2¼ 0:03742 DPT ¼ 0:282 � 995

2� 4�

�0:03742� 3

0:021þ 4

�¼ 1458 Pa

Inlet nozzle: DN100 wNi ¼ 0:5077 m=s DPNi ¼ 0:50772 � 9952

¼ 128 Pa

Outlet nozzle: DN80 wNo ¼ 0:793 m=s DPNo ¼ 0:5� 0:7932 � 9952

¼ 156 Pa

DPtot ¼ 1458þ 128þ 156 ¼ 1742 Pa

In Figure 3.17 the calculated pressure losses using the different models for water in aheat exchanger type 10 in Table 1.1 (Chapter 1) are shown.

Figure 3.17 Tube-side pressure loss as function of the tube-side throughput in a heat exchanger type10 according to Table 1.1 in Chapter 1.


3.4.2 Shell-side pressure lossThe total pressure loss DPtot on the shell side is made of• the pressure loss DPW in the flow through the baffle windows• the pressure loss DPcr in the cross stream through the tubes• the nozzle pressure loss DPn in inlet and outlet nozzle

DPtot ¼ DPw þ DPcr þ DPnDPw ¼ BF� nB � 9� w2

w

DPcr ¼ BF� ðnB þ 1Þ � f � ncross � w2cr � 9=2

f ¼ friction factor for cross stream ¼ f(Re)ncross ¼ tube rows in the cross stream between both baffle cutsnB ¼ number of baffles wcr ¼ cross stream velocity (m/s)r ¼ density (kg/m3) wW ¼ window velocity (m/s)BF ¼ bypass factor considering bypass and leakage streams ¼ 0.36

Example 4: Shell-side pressure loss calculation

V ¼ 50 m3/h 9 ¼ 763.4 kg/m3 n ¼ 0.91 mm2/sDi ¼ 496 mm 164 tubes 25 � 2 Triangular pitch 32 mmBaffle spacing B ¼ 100 mm Segment height H ¼ 100 mmNumber of baffles nB ¼ 28 Flow through tube rows ncross ¼ 9Flow cross section in cross stream across ¼ 0.0121 m2 ➔ wcr ¼ 1.148 m/sFlow cross section in the windows aW ¼ 0.0165 m2 ➔ wW ¼ 0.842 m/s

1. Pressure loss DPW in the 28 baffle windows

DPW ¼ BF� nB � r� w2w ¼ 0:36� 28� 763:4� 0:8422 ¼ 5455 Pa

The bypass factor BF ¼ 0.36 comes from the assumption that because of the bypass streams only 60%of the geometrically calculated flow velocity will be achieved.

2. Pressure loss DPcr with cross stream through nine tube rows

DPcr ¼ BF��ðnB þ 1Þ � f � ncross � w2

cr � r

2

�

DPcr ¼ 0:36� ð28þ 1Þ � f � 9� 1:1482 � 763:42

¼ 47; 266� f

The friction factor f is calculated according to different models with the help of the Reynoldsnumber.

Re ¼ wcr � don

¼ 1:148� 0:0250:91� 10�6 ¼ 31; 538


2.1 According to Bell with fB [10]

fB ¼ 2:68� Re�0:182 ¼ 2:68� 31538�0:182 ¼ 0:407

DPcrB ¼ 47266� 0:407 ¼ 19237 Pa

DPw þ DPcr ¼ 5455þ 19; 237 ¼ 24; 692 Pa

2.2 According to ClarkeDavidson with fCD [17]

fCD ¼ Re�0:2 � 3:12ffiffiffiffiffiffiffiffiffiffiT=do

p ¼ 31; 538�0:2 � 3:12ffiffiffiffiffiffiffiffiffiffiffiffi32=25

p ¼ 0:347

DPcrCD ¼ 0:347� 47; 266 ¼ 16; 418 Pa

DPw þ DPcr ¼ 5455þ 16; 418 ¼ 21; 873 Pa

2.3 According to Jakob with fJ [13]

fJ ¼ Re�0:16 ��1þ 0:47

ðT=do � 1Þ1:08�

¼ 2:85� 31; 538�0:16 ¼ 0:545

DPcrJ ¼ 0:545� 47; 266 ¼ 25; 755 Pa

DPw þ DPcr ¼ 5455þ 25; 755 ¼ 31; 210 Pa

2.4 According to Donohue with fD [7]

fD ¼ Re�0:2 � 8�T � dodo

�0:2 ¼ 3:86� 31; 538�0:2 ¼ 0:487

DPcrD ¼ 0:487� 47; 266 ¼ 23; 039 Pa

DPw þ DPcr ¼ 5455þ 23; 039 ¼ 28; 494 Pa

2.5 According to Chopey with the bypass factor BF as a function of Di and B [15].Number of rows for cross stream ncross:

ncross ¼ b� Di

T¼ 0:7� 0:496

0:032¼ 10:85

Reynolds number and friction factor:

ReCh ¼ wcr � ðT � doÞn

¼ 1:148� ð0:032� 0:025Þ0:91� 10�6 ¼ 8830

fCh ¼ 4� Re�0:25Ch ¼ 0:4126

Pressure loss calculation:

DPcrCh ¼ BF� ðnB þ 1Þ � ncross � fCh �wcr � r

2ðPaÞ

DPcrCh ¼ BF� 29� 10:85� 0:4126� 1:1482 � 763:42

¼ 65312� BF


Calculation of the bypass factor BF [15] and the pressure loss:

BF ¼ R1 � R2

R1 ¼ 0:75�ffiffiffiffiffiBDi

r¼ 0:75�

ffiffiffiffiffiffiffi100496

r¼ 0:3368

R2 ¼ 0:85� D0:08i ¼ 0:85� 0:4960:08 ¼ 0:8036

BF ¼ 0:3368� 0:8036 ¼ 0:2706

DPcrCh ¼ 0:2706� 65; 312 ¼ 17; 675 Pa

DPw þ DPcr ¼ 5455þ 17; 675 ¼ 23; 130 Pa

2.6 Pressure loss according to Kern(new) with the bypass correction in fK [16]

DPMK ¼ ðnB þ 1Þ � fK � Di

de� w2

K � r

2ðPaÞ

First the special “Kern data” must be determined:

wK ¼ 1:28 m=s de ¼ 0:0198 m ReK ¼ 27; 649

fK ¼ 0:436� lg Re�1:4485 ¼ 0:05

DPMK ¼ 29� 0:05� 49619:8

� 1:28� 763:42

¼ 22; 849 Pa

DPMK ¼ DPw þ DPcr ¼ 22; 849 Pa

3. Nozzle pressure loss

Flow velocity in DN 150 wn ¼ 0:786 m=s

DPn ¼ 1:5� w2n � r

2¼ 1:5� 0:7862 � 763:4

2¼ 354 Pa

4. Comparison of result for DPtot ¼ DPw þ DPcr þ DPn

Bell DPtot ¼ 25,046 PaClarkeDavidson DPtot ¼ 22,227 PaJakob DPtot ¼ 31,564 PaDonohue DPtot ¼ 28,848 PaChopey DPtot ¼ 23,484 PaKern DPtot ¼ 23,203 Pa

In Figure 3.18 the calculated pressure losses using the different models for ethanol in aheat exchanger type 10 with the baffle spacing of 100 mm are shown.


3.5 HEAT EXCHANGER DESIGN WITH HEAT EXCHANGER TABLES

In Tables 3.1e3.4 are listed the most important data for heat exchangers according toDIN 28184 and DIN 28191.

In order to simplify the selection of the correct heat exchanger type for a specificproblem definition with respect to a good flow velocity the volumetric flow ratesare listed in both columns of the right side which are necessary for the flow velocityof 1 m/s:

VR ¼ necessary flow throughput for 1 m/s on the tube side (m3/h)VM ¼ necessary flow throughput for 1 m/s on the shell side (m3/h)The VM value on the shell side is valid for the given baffle spacing B.The heat exchanger area per meter tube length is listed under A (m2/m).Procedure:First during the design an estimated overall heat transfer coefficient for the required

heat exchanger area is determined. The heat exchanger is then selected from thetables.

Criteria for the selection:• The heat exchanger type must have an area greater than the estimated.• The tube-side and shell-side flow rates should lie in the order of the flows VR and

VM that are listed in the tables for a flow velocity of 1 m/s.The determination of the required flow velocities for the design of the heat exchanger

is very simple using the listed values of VR and VM in the table.The application of the tables is shown in the following examples.

Figure 3.18 Shell-side pressure loss as function of the flow velocity in a heat exchanger type 10according to Table 1.1 in Chapter 1.


Table 3.1 Geometrical data of heat exchangers according to DIN 28184 part 1 (triangular pitch 60�), tubes 25 � 2, pitch 32 mm triangular, tubetolerance Dd ¼ 0.08 mm, tube hole in the baffle dhole ¼ 26 mmTypeNr. DN nP

Da

mmDi

mmsmm

SMUmm

DH

mmBmm n nW nacr

Am2/m fW

VRm3/h

VMm3/h

1 150 2 168 159 4.5 1.25 143 50 14 6 4 1.09 0.3624579 8.73 8.892 200 2 219 207.2 5.9 1.25 191 50 26 9 6 2.04 0.4663294 16.21 9.233 250 2 273 260.4 6.3 1.25 248 50 44 12 6 3.45 0.3133393 27.43 18.804 300 2 324 309.8 7.1 1.25 298 60 66 25 9 5.18 0.4285523 41.15 17.515 350 2 355 339 8 1.5 316 70 76 22 10 5.96 0.4681409 47.38 21.596 350 4 355 339 8 1.5 325 70 68 22 10 5.34 0.4751083 21.20 21.597 400 2 406 388.4 8.8 1.5 373 80 106 35 10 8.32 0.3562161 66.09 38.908 400 4 406 388.4 8.8 1.5 369 80 88 26 11 6.91 0.4420558 27.43 31.799 500 2 508 496 6 1.75 478 100 180 62 14 14.13 0.4098792 112.22 51.7210 500 4 508 496 6 1.75 482 100 164 45 15 12.88 0.5205206 51.12 42.7811 600 2 600 588 6 1.75 576 120 258 82 17 20.26 0.4489858 160.85 69.6012 600 8 600 588 6 1.75 573 120 232 64 16 18.22 0.4110614 36.16 80.3513 700 2 700 684 8 2.25 672 140 364 113 21 28.58 0.5294837 226.94 79.3714 700 8 700 684 8 2.25 666 140 324 88 20 25.44 0.4712546 50.50 91.9315 800 2 800 784 8 2.25 771 160 484 155 24 38.01 0.5295184 301.75 105.2316 800 8 800 784 8 2.25 772 160 432 116 22 33.92 0.4339129 67.33 133.9617 900 2 900 880 10 2.25 868 180 622 174 26 48.85 0.4824717 387.79 148.2618 900 8 900 880 10 2.25 868 180 556 156 26 43.66 0.4962607 86.66 148.2619 1000 2 1000 980 10 2.75 966 200 776 216 30 60.94 0.5353497 483.80 164.8520 1000 8 1000 980 10 2.75 966 200 712 164 28 55.92 0.4528635 110.97 200.7921 1100 2 1100 1076 12 2.75 1058 220 934 264 33 73.35 0.5440129 582.30 198.0422 1100 8 1100 1076 12 2.75 1055 220 860 242 32 67.54 0.5009315 134.04 217.8223 1200 2 1200 1176 12 2.75 1159 240 1124 328 35 88.27 0.4919161 700.76 259.3024 1200 8 1200 1176 12 2.75 1160 240 1048 302 34 82.3 0.4633054 163.34 280.87

Calculations

ofthe

Heat

TransferCoefficients

andPressure

Lossesin

Convective

Heat

Transfer57

Table 3.2 Geometrical data of heat exchangers according to DIN 28184 part 4, triangular pitch 60�, tube 20 � 2, pitch: 25 mm triangular,dhole ¼ 21 mmTypeNr. DN nP

Da

mmDi

mmsmm

SMUmm

DH

mmBmm n nW nacr

Am2/m fW

VRm3/h

VMm3/h

1 250 2 273 260.4 6.3 1.25 248 50 76 30 10 4.78 0.3983235 27.51 10.272 250 4 273 260.4 6.3 1.25 250 50 64 24 8 4.02 0.2880986 11.58 17.303 300 4 324 309.8 7.1 1.25 297 60 92 36 12 5.78 0.4417207 16.65 14.484 350 8 355 339 8 1.5 328 70 100 28 12 6.28 0.3765783 9.05 24.255 400 8 406 388.4 8.8 1.5 373 80 144 32 14 9.05 0.4011764 13.03 30.556 500 8 508 496 6 1.75 486 100 256 74 18 16.09 0.3862411 23.16 48.327 600 8 600 588 6 1.75 574 120 376 114 22 23.62 0.4213845 34.02 63.318 700 8 700 684 8 2.25 671 140 532 176 26 33.43 0.4240278 48.13 82.049 800 8 800 784 8 2.25 770 160 724 208 30 45.49 0.4430131 65.51 105.3810 900 8 900 880 10 2.25 868 180 936 286 34 58.81 0.4594987 84.69 129.0011 1000 8 1000 980 10 2.75 970 200 1196 390 38 75.15 0.4505112 108.21 157.8112 1100 8 1100 1076 12 2.75 1060 220 1436 412 42 90.23 0.4751299 129.93 186.3213 1200 8 1200 1176 12 2.75 1162 240 1736 528 46 109.08 0.4792821 157.07 220.60

58Heat

ExchangerDesign

Guide

Table 3.3 Geometrical data of heat exchangers according to VDI-W€armeatlas 1984 chap. GDIN 28184 (triangular pitch 60�), tubes 16 � 2, pitch20 mm triangular, dhole ¼ 17 mmTypeNr. DN nP

Da

mmDi

mmsmm

SMUmm

DH

mmBmm n nW nacr

Am2/m fW

VRm3/h

VMm3/h

1 150 2 168 159 4.5 1.25 147.4 50 36 10 7 1.81 0.3852249 7.33 7.562 200 4 219 207.2 5.9 1.25 198.1 50 60 22 8 3.02 0.2937709 6.11 13.483 250 4 273 260.4 6.3 1.25 253.9 50 96 32 10 4.83 0.2797606 9.77 17.474 300 8 324 309.8 7.1 1.25 297.7 60 128 42 12 6.43 0.2947632 6.51 24.855 350 8 355 339 8 1.5 330.8 70 164 56 14 8.24 0.3090594 8.35 28.396 400 8 406 388.4 8.8 1.5 377.8 80 228 68 16 11.46 0.3157251 11.60 7.337 500 8 508 496 6 1.75 485 100 404 134 22 20.31 0.3392087 20.56 51.32

Calculations

ofthe

Heat

TransferCoefficients

andPressure

Lossesin

Convective

Heat

Transfer59

Table 3.4 Geometrical data of heat exchanger with flanged floating head according to DIN 28191, tubes 25 � 2, pitch 32 mm triangular,dhole ¼ 26 mm (tube hole and the baffle)TypeNr. DN nP

Da

mmDi

mmsmm

SMUmm

DH

mmBmm n nW nacr

Am2/m fW

VRm3/h

VMm3/h

1 150 2 168.3 159 4.5 1.25 125.6 50 10 2 3 0.8 0.823 6.2 11.12 200 2 219.1 207 6.3 1.25 167.7 50 18 3 4 1.4 0.774 11.2 14.53 250 2 273 260 6.3 1.25 226.8 50 32 10 5 2.5 0.625 20 16.14 300 2 323.9 310 7.1 1.25 268.2 60 46 10 6 3.6 0.656 28.7 24.15 350 4 355.6 340 8 1.5 292.8 70 42 5 6 3.3 0.69 13.1 38.86 400 4 406.4 389 8.8 1.5 349.3 80 70 20 8 5.5 0.677 21.8 40.97 500 4 508 490 8.8 1.75 450 100 122 40 10 9.6 0.709 38 62.88 600 8 600 588 6 1.75 539.4 120 172 58 12 13.5 0.723 26.8 96.69 700 8 700 684 8 2.25 639 140 248 66 14 19.5 0.71 38.7 127.110 800 8 800 784 8 2.25 738.9 160 344 90 16 27 0.709 53.6 169.911 900 8 900 880 10 2.25 827.4 180 432 130 18 33.9 0.723 67.3 228.612 1000 8 1000 980 10 2.75 931.8 200 560 142 20 44 0.711 87.3 241.313 1100 8 1100 1076 12 2.75 1025.2 220 688 174 22 54 0.715 107.2 289.614 1200 8 1200 1176 12 2.75 1125.4 240 840 238 24 66 0.704 130.9 351

60Heat

ExchangerDesign

Guide

Example 5: Design of a convective heat exchanger

Tube side Shell sideFlow (kg/h) 42,828 28,100Density (kg/m3) 995 763.4Specific heat capacity (Wh/kg K) 1.16 0.74Kinematic viscosity (mm2/s) 0.92 0.9Conductivity (W/m K) 0.605 0.162Inlet temperature (�C) 25 58Outlet temperature (�C) 31.7 42Fouling factor (m2 K/W) 0.0002 0.0002Heat duty (kW) 332.8 332.8Prandtl number Pr 6.32 11.3Volumetric flow (m3/h) 43 36.8

Chosen: Type 10 in Table 1, DN 500 with 164 tubes 25 � 2, triangular pitch 32 mm

Flow rate at 1 m/s VR ¼ 51.1 m3/h VM ¼ 42.8 m3/h

Flow velocity 4351:1 ¼ 0:84 m=s 36:8

42:8 ¼ 0:86 m=s

Reynolds number Re 19,218 23,472Nusselt number Nu 112.96 182.85Heat transfer coefficient a 3254 W/m2 K 1185 W/m2 KOverall heat transfer coefficient U 628 W/m2 KCorrected temperature difference CMTD ¼ 20.43 �C

Required heat exchange area Areq:

Areq ¼ QU� CMTD

¼ 332; 800628� 20:43

¼ 25:9 m2

Chosen: tube length L ¼ 2.5 mChosen heat exchange area A ¼ 2.5 m � 12.88 m2/m ¼ 33.2 m2

Reserve: 24%

Example 6: Design of a heat exchanger

Tube side Shell sideFlow (kg/h) 14,276 28,100Density (kg/m3) 995 763.4Specific heat capacity (Wh/kg K) 1.16 0.74Kinematic viscosity (mm2/s) 0.92 0.9Conductivity (W/m K) 0.605 0.162Inlet temperature (�C) 15 58Outlet temperature (�C) 35.1 42Fouling factor (m2 K/W) 0.0002 0.0002


Heat load (kW) 332.8 332.8Prandtl number Pr 6.32 11.3Volumetric flow (m3/h) 14.35 36.8LMTD ¼ 24.89 �CCMTD ¼ 22.56 �CLMTD, logarithmic mean temperature difference; CMTD, corrected effective mean temperature difference.

Chosen: Type 10 in Table 1: DN 500 with 164 tubes 25 � 2, triangular pitch 32 mm

Flow rate at 1 m/s VR ¼ 51.1 m3/h VM ¼ 42.8 m3/h

Flow velocity 14:3551 ¼ 0:28 m=s 36:8

42:8 ¼ 0:86 m=s

The flow velocity on the tube side is too small.

New choice: Type 4 in Table 2: DN 350 with 100 tubes 20 � 2, triangular pitch 25 mm

Flow rate at 1 m/s VR ¼ 9 m3/h VM ¼ 24.2 m3/h

Flow velocity 14:359 ¼ 1:56 m=s 36:8

24:2 ¼ 1:53 m=s

Reynolds number Re 27,130 34,074Nusselt number Nu 148.8 228.7Heat transfer coefficient a (W/m2 K) 5600 1852Overall heat transfer coefficient U ¼ 863 W/m2 K

Required heat exchange area Areq:

Areq ¼ 332; 800863� 22:56

¼ 17 m2

Chosen tube length L ¼ 3.5 mChosen heat exchange area A ¼ 3.5 � 6.28 ¼ 22 m2

29 % reserve.

Notes for the heat exchanger tables:

A ¼ heat exchanger surface area per m tube length (m2/m)B ¼ baffle spacing (mm)Da ¼ outer diameter (mm)Di ¼ inner diameter (mm)DH ¼ bundle outer limit (mm)dhole ¼ hole diameter in baffle plate (mm)fW ¼ correction factor according to VDI-W€armeatlasn ¼ number of tubesnacr ¼ number of tubes in cross streamnP ¼ number of tube passesnw ¼ number of tubes in baffle windows ¼ shell wall thickness (mm)SMU ¼ width of gap between baffle and shell inner diameter (mm)VR ¼ required volumetric flow (m3/h) for 1 m/s flow velocity on the tube sideVM ¼ required volumetric flow (m3/h) for 1 m/s flow velocity on the shell side


NOMENCLATUREacross flow cross section shell side (m2)aT flow cross section tube side (m2)B baffle spacing (m)Di shell inner diameter (m)di tube inner diameter (m)do tube outer diameter (m)l tube length (m)nacr number of tubes in cross streamnP number of tube passesw flow velocity (m/s)wcross cross flow velocity shell side (m/s)wT tube-side velocity (m/s)V volumetric throughput (m3/h)Nu Nusselt numberRe Reynolds numberPr Prandtl numberT pitch (m)a heat transfer coefficient (W/m2 K)c specific heat capacity (Wh/kg K)l heat conductivity (W/m K)n kinematic viscosity (m2/s)r density (kg/m3)

REFERENCES AND FURTHER READING[1] R. Mukherjes, Chem. Eng. Prog. 92 (1996) 72e79 þ 94 (1998) 21e37.[2] J. Fisher, R.O. Parker, Hydrocarbon Process. 48 (1969) 147e154.[3] A.P. Fraas, M.N. Ozisik, Heat Exchanger Design, John Wiley, N.Y, 1965.[4] A.E. Jones, Thermal design of the shell-and-tube, Chem. Eng. 109 (2002) 60e65.[5] A. Devore, G.J. Vago, G.J. Picozzi, Specifying and selecting, Chem. Eng. (Octomber 1980).[6] E.N. Sieder, G.E. Tate, Ind. Eng. Chem. 28 (1936) 1429.[7] D.A. Donohue, Heat transfer and pressure drop in heat exchangers, Ind. Eng. Chem. 41 (1949) 2499.[8] Petr.Ref. 34 (1955), No. 10 und No. 11.[9] T. Tinker, Shell-side characteristics of shell and tube heat exchangers, Trans. ASME 80 (1958) 36 (f f ).[10] K.J. Bell, Thermal design of heat transfer equipment, in: Perry’s Handbook, 1973.[11] VDI-W€armeatlas, Aufkage, 5, VDI-Verlag, D€usseldorf, 1988.[12] Lord, Minton, Slusser, Design of heat exchangers, Chem. Eng, 26 ( January 1970).[13] M. Jakob, Trans. ASME 60 (1938), 381/392.[14] Coates, B.S. Pressburg, Heat transfer to moving fluids, Chem. Eng. Dez. 66 (1959) 67e72.[15] N.P. Chopey, Heat transfer, in: Handbook of Chemical Engineering Calculations, McGraw-Hill,

New York, 1993.[16] D.Q. Kern, Process Heat Transfer, McGraw-Hill, N.Y, 1950.[17] L. Clarke, R.L. Davidson, Manual for Process Engineering Calculations, McGraw-Hill, N.Y, 1962.


CHAPTER 4

Geometrical Heat ExchangerCalculations

Contents

4.1 Calculation Formula 654.1.1 Heat exchanger area A 654.1.2 Tube-side flow cross section aT 664.1.3 Flow velocity wT in the tubes 664.1.4 Shell-side cross section across 664.1.5 Shell-side cross stream velocity wC 664.1.6 Shell-side longitudinal stream cross area aw in the baffle window 664.1.7 Number of tube rows in cross stream ncross between the windows 68

4.2 Tube-Side Calculations 684.2.1 Flow velocity wT in the tubes for a volumetric flow VT (m

3/h) 684.2.2 Number of tubes per pass tP for a given flow velocity wT 694.2.3 Required number of tubes per pass tP for a given Reynolds number Re 694.2.4 Required flow velocity wT for a given Reynolds number Re 69

4.3 Shell-Side Calculations 704.3.1 Flow cross section across for the flow across the tubes 704.3.2 Flow cross section aw for the longitudinal flow in the baffle cuts 704.3.3 Average flow cross section aav on the shell side 724.3.4 Shell-side flow velocities wav 724.3.5 Required baffle spacing B for a given cross stream velocity wcross 724.3.6 Number of baffles nB with given baffle spacing B 724.3.7 Determination of the total vertical tube rows ntv 724.3.8 Determination of the tube rows ncross in the cross stream between the baffle

windows with the segment height H 73

4.1 CALCULATION FORMULA

4.1.1 Heat exchanger area A

A ¼ n� do � p� L�m2�

n ¼ total number of tubesdo ¼ tube outer diameter (m)L ¼ tube length (m)



http://dx.doi.org/10.1016/B978-0-12-803764-5.00004-3

4.1.2 Tube-side flow cross section aT

aT ¼ nnP

� d2i �p

4

�m2�

tP ¼ tubes per passdi ¼ tube inner diameter (m)

4.1.3 Flow velocity wT in the tubes

wT ¼ VT

aT � 3600ðm=sÞ

VT ¼ tube-side volumetrical flow (m3/h)

4.1.4 Shell-side cross section across

across ¼ B� ðDi � nacr � doÞ�m2� nacr ¼ Di � sR

sq¼ DH

sq

B ¼ baffle spacing, ca. 0.2 � Di (m)Di ¼ shell inner diameter (m)nacr ¼ number of tubes in the middle across the flow directionsR ¼ edge strip (m)sq ¼ tube middle gap across the flow direction (m) ¼ pitch TDH ¼ outer limit diameter (m) ¼ envelope of tube pattern

4.1.5 Shell-side cross stream velocity wC

wcross ¼ VShell

across � 3600ðm=sÞ

VShell ¼ shell-side volumetric flow (m3/h)

4.1.6 Shell-side longitudinal stream cross area aw in the baffle window

aw ¼ D2i � C � nw � d2o �

p

4

�m2� C ¼ 16� x2 � 13� x3

12�ffiffiffiffiffiffiffiffiffiffiffiffiffix� x2

p with x ¼ HDi

C ¼ geometrical calculation factorH ¼ segment height (m)nw ¼ number of tubes in the baffle windowThe number of tubes nw lying in the baff le cut can be determined by calculation from

the segment height H and the distance of the tube rows from the middle.


A tube layout is more appropriate.According to the equation above, the tubes lying in the segment have to be subtracted

from the segment area.As a first approximation, the window area which is not covered with the tubes can be

estimated with the free room factor NF.

NF ¼ 1� n� d2oD2

iaw ¼ D2

i � C �NF�m2�

n ¼ total number of tubesThe calculation with NF gives a little lower flow cross sections in the window because

the edge strip was not considered.In order to reduce the effect of the edge strip, especially with small shell diameters, the

outer limit diameter DH should be used instead of the shell inner diameter Di.

NF ¼ 1� n� d2oD2

Haw ¼ D2

i � C �NF�m2�

For the shell side, the following designs should be strived for:

across ¼ aw x ¼ H/Di ¼ 0.2 bis 0.25 B/Di ¼ 0.2

Calculation factors C for different x-values:

x[ H/Di C

0.20 0.11170.21 0.11970.22 0.12790.23 0.13630.24 0.14480.25 0.15340.26 0.16210.27 0.17090.28 0.17980.29 0.18890.30 0.1980

Longitudinal stream flow velocity ww in the window ww ¼ VShell

aw � 3600ðm=sÞ

Average shell-side flow cross section aav aav ¼ ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiacross � aw

p ðm2ÞAverage shell-side flow velocity wav wav ¼ VShell

aav � 3600ðm=sÞ

Number of baffles nB nB ¼ L � BBþ a

Geometrical Heat Exchanger Calculations 67

a ¼ baffle thickness (m)B ¼ baffle spacing (m)L ¼ tube length (m)

Check calculation for the baffle spacing B B ¼ L � nB � anB þ 1

ðmÞ

4.1.7 Number of tube rows in cross stream ncross between the windows

ncross ¼ Di � 2H � SG � ðnP � 1Þsl

þ 1

nP ¼ number of tube passesSG ¼ strips of the pass partition plates for the pass partitionsl ¼ tube middle clearance longitudinal to flow directionCross and longitudinal clearances (T ¼ tube pitch)

Tube middle spacing Triangular Square Rotated square 45�

Cross flow sq sq ¼ T sq ¼ T sq ¼ 1.414 TLongitudinal flow sl sl ¼ 0.866 T sl ¼ T sl ¼ 0.707 T

Flo

w

Triangular Square Rotated

square

Flo

w

Flo

w

4.2 TUBE-SIDE CALCULATIONS (FIGURE 4.1)

4.2.1 Flow velocity wT in the tubes for a volumetric flow VT (m3/h)

wT ¼ VT

2827� d2i � tPðm=sÞ VT ¼ MT

rT

�m3�h� tP ¼ n

nP


Figure 4.1 Tube-side flow in heat exchanger.

4.2.2 Number of tubes per pass tP for a given flow velocity wT

tP ¼ VT

2827� d2i � wT

4.2.3 Required number of tubes per pass tP for a given Reynoldsnumber Re

tP ¼ VT

2827� n�Re� di

4.2.4 Required flow velocity wT for a given Reynolds number Re

wT ¼ Re� n

diðm=sÞ

di ¼ tube inner diameter (m)r ¼ density (kg/m3)n ¼ kinematic viscosity (m2/s)VT ¼ volumetric throughput (m3/h)MT ¼ mass throughput (kg/h)Re ¼ Reynolds number


nP ¼ number of tube passestP ¼ Tubes per pass

Example 1: Tube-side calculations

MT ¼ 5000 kg/h r ¼ 850 kg/m3 VT ¼ 5000/850 ¼ 5.88 m3/hn ¼ 2 mm2/s di ¼ 16 mm

Calculation of the required flow velocity for Re [ 5000

wT ¼ Re� n

di¼ 5000� 2� 10�6

0:016¼ 0:625 m=s

Required number of tubes tP per pass for wT [ 0.625 m/s

tP ¼ 5:882827� 0:0162 � 0:625

¼ 13 Rohre mit di ¼ 16 mm

Required number of tubes tP per pass for Re [ 5000

tP ¼ 5:882827� 2� 10�6 � 5000� 0:016

¼ 13 Rohre

4.3 SHELL-SIDE CALCULATIONS (FIGURE 4.2)

4.3.1 Flow cross section across for the flow across the tubes (Figure 4.2)

across ¼ B� ðDi � nacr � doÞ�m2�

B ¼ baffle spacing (m)do ¼ outer diameter of the tubes (m)Di ¼ shell inner diameter (m)DH ¼ outer limit diameter (m)nacr ¼ number of tubes in cross flow on the center lineT ¼ pitch (m)s ¼ Di � DH ¼ edge strips between shell and tube bundle

4.3.2 Flow cross section aw for the longitudinal flow in the baffle cuts

aw ¼ AS � nW � d2o � p=4

AS ¼ area of the baffle cut (m2)

For H=Di ¼ 0:2/AS ¼ 0:112�D2i

For H=Di ¼ 0:25/AS ¼ 0:154�D2i


Number of verticaltube rows

Number of tubes inthe baffle window

Segment height H = 80 mm

Segment height H = 80 mm Number of tube rowsbetween windows

Segment height

Baffle spacing B

Number of tubes in crossstream on the center line

Figure 4.2 Shell-side flow in heat exchanger.


nw ¼ number of tubes in the baffle windowH ¼ segmental height ¼ height of the baffle cutIf the number of tubes in the segment is not known, the longitudinal stream cross

section aw can be determined with the NF factor.

NF ¼ 1� n� d2oD2

i

n ¼ total number of tubes

aW ¼ NF� AS�m2�

Notice: Equal flow cross sections will be preferably desired for the cross and longitu-dinal flow.

4.3.3 Average flow cross section aav on the shell side

aav ¼ ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiacross � aW

p �m2�

4.3.4 Shell-side flow velocities wav

Average velocity wav ¼ VShell

3600� aav¼ VShell

3600� ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiaacross � aw

p ðm=sÞ

wcross ¼ VShell

3600� acrossðm=sÞ ¼ crossflow velocity through the bundle

ww ¼ VShell

3600� awðm=sÞ ¼ longitudinal flow velocity through the window

4.3.5 Required baffle spacing B for a given cross stream velocity wcross

B ¼ VShell

3600� wcross � ðDi � nacr � doÞ ðmÞ

4.3.6 Number of baffles nB with given baffle spacing B

nB ¼ L � BBþ a

B ¼ L � nB � anB þ 1

ðmÞ

4.3.7 Determination of the total vertical tube rows ntv

ntv ¼ Di � do � SG � �np � 1

�sl

SG ¼ stripes for pass partitionsnP ¼ number of tube passes


4.3.8 Determination of the tube rows ncross in the cross streambetween the baffle windows with the segment height H

ncross ¼ DH � do � 2H � ðnP � 1Þ � SGsl

þ 1

Example 2: Shell-side Calculations

Heat exchanger DN 400 with 106 tubes 25 � 2 in triangular pitch T ¼ 32 mmDi ¼ 400 mm L ¼ 4 m a ¼ 5 mm nP ¼ 4 SG ¼ 25 mmDH ¼ 373 mm for nP ¼ 2 n ¼ 106 A ¼ 8.32 m2/mDH ¼ 369 mm for nP ¼ 4 n ¼ 88 A ¼ 6.91 m2/mBaffle spacing B ¼ 0.2 � Di ¼ 0.2 � 400 ¼ 80 mmSegment height H ¼ 0.2 � Di ¼ 0.2 � 400 ¼ 80 mmH/Di ¼ 0.2 ➡ AS ¼ 0.112 � D2

i ¼ 0.112 � 0.42 ¼ 0.0179 m2

Calculation of the flow sections:

nacr ¼ Di

T¼ 400

32¼ 12 Tubes NF ¼ 1� n� d2o

D2i

¼ 1� 106� 0:0252

0:42¼ 0:586

across ¼ B� ðDi � nacr � doÞ ¼ 0:08� ð0:4� 12� 0:025Þ ¼ 0:008 m2

aw ¼ NF� AS ¼ 0:586� 0:0179 ¼ 0:0105

aav ¼ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi0:008� 0:0105

p¼ 0:0092 m2

Calculation of the shell-side flow velocities for VShell [ 20 m3/h

wcross ¼ 203600� 0:008

¼ 0:694 m=s ww ¼ 203600� 0:0105

¼ 0:53 m=s

wav ¼ 203600� 0:0092

¼ 0:604 m=s

Determination of the number of baffles nB for a [ 5 mm

nB ¼ L� BBþ a

¼ 4000� 8080þ 5

¼ 46 baffles

Check calculation of the baffle spacing B for a [ 5 mm

B ¼ L� nB � anB þ 1

¼ 4000� 46� 546þ 1

¼ 80 mm

Estimation of the total vertical tube rows ntv for nP [ 2

nTV ¼ DH � do � SG � ðnP � 1Þ0:866� T

¼ 373� 25� 10� ð2� 1Þ0:866� 32

¼ 12:2

Only even numbers are possible: ntv ¼ vertical 12 tube rows.


Determination of the vertical tube rows ncross between the baffle cuts for nP [ 2

ncross ¼ DH � do � ðnP � 1Þ � SG � 2� H0:866� T

þ 1

¼ 373� 25� ð2� 1Þ � 10� 2� 800:866� 32

þ 1 ¼ 7:42

Only even numbers are possible: ncross ¼ vertical seven tube rows between the windows.

Example 3: Calculation of ncross for the data in Figure 4.3

ncross ¼ 442� 25� ð4� 1Þ � 15� 2� 109:622:6

þ 1 ¼ 7:76

Only even numbers are possible: ncross ¼ vertical seven tube rows between the windows.

Figure 4.3 Floating Head Heat Exchanger DN 500 with sealing strips to reduce the bypass stream.


CHAPTER 5

Dimensionless CharacterizationNumber for the Heat Transfer

Contents

5.1 Reynolds Number Re for the Characterization of the Flow Condition 755.2 Prandtl Number Pr, Peclet Number Pe, and Temperature Conductivity a 755.3 Nusselt Number Nu for the Calculation of the Heat Transfer Coefficient 765.4 Stanton Number St for the Calculation of the Heat Transfer Coefficient 765.5 Colburn Factor JC for the Calculation of the Heat Transfer Coefficient 775.6 Kern Factor JK for the Calculation of the Heat Transfer Coefficient 775.7 Graßhof Number Gr for the Calculation of the Heat Transfer Coefficient in Natural Convection 78References 78

5.1 REYNOLDS NUMBER RE FOR THE CHARACTERIZATION OF THEFLOW CONDITION

Re ¼ w � dv

w ¼ flow velocity (m/s)d ¼ tube diameter (m)n ¼ kinematic viscosity

Example 1: Calculation of the Reynolds number Re

Data: w ¼ 1.2 m/s di ¼ 21 mm n ¼ 0.95 � 10�6 m2/s

Re ¼ 1:2� 0:0210:95� 10�6 ¼ 26; 526

5.2 PRANDTL NUMBER PR, PECLET NUMBER PE, AND TEMPERATURECONDUCTIVITY A [1e8]

Pr ¼ n

a¼ Pe

Re¼ 3600� h� cp

l¼ 3600� n� cp � r

l

Pe ¼ w � da

¼ Pr �Re a ¼ l

cp � r� 3600

�m2�s�



http://dx.doi.org/10.1016/B978-0-12-803764-5.00005-5

a ¼ temperature conductivity (m2/s)Pe ¼ Peclet number (�)r ¼ density (kg/m3)cp ¼ specific heat capacity (Wh/kg K)l ¼ thermal heat conductivity (W/m K)h ¼ dynamic viscosity (Pas)n ¼ kinematic viscosity (m2/s)The factor 3600 is needed to convert hours to seconds in order to make the dimen-

sions coherent.

Example 2: Calculation of the Prandtl number Pr

Data: r ¼ 903 kg/m3 cp ¼ 0.594 Wh/kg K l ¼ 0.118 W/m Kh ¼ 858 mPas n ¼ 0.95 � 10�6 m2/s

a ¼ l

cp � r� 3600¼ 0:118

0:594� 903� 3600¼ 6:1� 10�8 m2�s

Pr ¼ n

a¼ 0:95� 10�6

6:1� 10�8 ¼ 15:5

Pr ¼ 3600� n� cp � r

l¼ 3600� 0:95� 10�6 � 0:594� 903

0:118¼ 15:5

Pr ¼ 3600� h� cpl

¼ 3600� 858� 10�6 � 0:5940:118

¼ 15:5

5.3 NUSSELT NUMBER NU FOR THE CALCULATION OF THE HEATTRANSFER COEFFICIENT [6]

Nu ¼ a� dl

¼ St�Re� Pr a ¼ Nu� l

d

�W=m2 K

�

a ¼ heat transfer coefficient (W/m2 K)

5.4 STANTON NUMBER ST FOR THE CALCULATION OF THE HEATTRANSFER COEFFICIENT

G ¼ w � r (kg/m2 s) G ¼ mass stream density (kg/m2 s)


St ¼ NuRe� Pr

¼ a

G � c¼ a

3600� c � w

a ¼ St� 3600� cp � w � r�m2�s�

5.5 COLBURN FACTOR JC FOR THE CALCULATION OF THE HEATTRANSFER COEFFICIENT

JC ¼ St� Pr2=3 a ¼ JCPr2=3

� 3600� w � r�W�m2 K

�

5.6 KERN FACTOR JK FOR THE CALCULATION OF THE HEATTRANSFER COEFFICIENT [1]

JK ¼ Nu

Pr1=3a ¼ JK � Pr1=3 � l

d

�W�m2 K

�

Example 3: Calculation of the heat transfer coefficient with Nu, St, JC, and JK

Data: r ¼ 903 kg/m3 cp ¼ 0.594 Wh/kg K l ¼ 0.118 W/m Kn ¼ 0.95 � 10�6 m2/s di ¼ 21 mm w ¼ 1.2 m/sRe ¼ 26526 Pr ¼ 15.5 Nu ¼ 196.55

a ¼ Nu� l

d¼ 196:55� 0:118

0:021¼ 1104 W

�m2 K

St ¼ NuRe� Pr

¼ 196:5526526� 15:5

¼ 0:00048

a ¼ St� 3600� w � r ¼ 0:00048� 3600� 0:594� 1:2 ¼ 1108 W�m2 K

JC ¼ St� Pr2=3 ¼ 0:00048� 15:52=3 ¼ 0:003

a ¼ JCPr2=3

� 3600� cp � w � r ¼ 0:003

15:52=3� 3600� 0:594� 1:2� 903

¼ 1112 W�m2 K

JK ¼ Nu

Pr1=3¼ 196:55

15:51=3¼ 79:55

a ¼ JK � Pr1=3 � l

d¼ 79:55� 15:51=3 � 0:118

0:021¼ 1104 W

�m2 K

Dimensionless Characterization Number for the Heat Transfer 77

5.7 GRAßHOF NUMBER GR FOR THE CALCULATION OF THE HEATTRANSFER COEFFICIENT IN NATURAL CONVECTION [8]

Gr ¼ 9:81� b� Dt � l3

nb ¼ 1

v� v1 � v2

t1 � t2¼ r�

1r1� 1

r2

t1 � t2ð1=KÞ

b ¼ volumetric heat expansion coefficient (1/K)Dt ¼ temperature difference productdwall (K)l ¼ characteristic length (m)v1 ¼ specific volume (m3/kg) at t1v2 ¼ specific volume (m3/kg) at t2Product Gr$Pr

Gr � Pr ¼ 35316� r� cp � b� l3 � Dtl� n

Example 4: Calculation of the Graßhof number

Data: r ¼ 800 kg/m3 cp ¼ 0.6 Wh/kg K l ¼ 0.13 W/m K

n ¼ 50 � 10�6 m2/s b ¼ 0.001 K�1 Dt ¼ 50 K d ¼ 50 mm

Pr ¼ 3600� 50� 10�6 � 0:6� 8000:13

¼ 664:6

Gr ¼ 9:81� 0:001� 50� 0:053

50� 10�6 ¼ 24:53� 103

Gr� Pr ¼ 24:53� 103 � 664:6 ¼ 16:3� 106

Gr� Pr ¼ 35316� 800� 0:6� 0:002� 0:053 � 50

50� 10�6 � 0:13¼ 16:3� 106

REFERENCES[1] D.Q. Kern, Process Heat Transfer, McGraw-Hill, N.Y., 1950.[2] A.P. Fraas, M.N. Ozisik, Heat Exchanger Design, John Wiley, N.Y., 1965.[3] W.H. McAdams, Heat Transmission, McGraw-Hill, N.Y., 1954.[4] Perry’s Chemical Engineer’s Handbook, McGraw-Hill, 1984.[5] G.F. Hewitt, G.L. Shires, T.R. Bott, Process Heat Transfer, CRC Press, Boca Raton, 1994.[6] V.D.I. W€armeatlas, 5. Auflage, VDI-Verlag, D€usseldorf, 1988.[7] W.M. Rohsenow, J.P. Hartnett, Handbook of Heat Transfer, McGraw-Hill, N.Y., 1973.[8] E.R.G. Eckert, R.M. Drake, Heat and Mass Transfer, McGraw Hill, N.Y., 1959.


CHAPTER 6

Overall Heat Transfer Coefficient andTemperature Profile

Contents

6.1 Calculation of the Overall Heat Transfer Coefficient 796.2 Calculation of the Temperature Gradient in a Heat Exchanger 846.3 Viscosity Correction 86

Calculation procedure 866.4 Calculation of the Heat Transfer Coefficient from the Overall Heat Transfer Coefficient 88Nomenclature 89References 89

6.1 CALCULATION OF THE OVERALL HEAT TRANSFERCOEFFICIENT [1e10]

The equation for the heat transfer is as follows:

Q ¼ U � A� LMTD A ¼ QU � LMTD

Normally the required heat exchanger surface area A is calculated for a given problemdefinition. The amount of heat Q to be transferred will result from the problemdefinition.

Likewise, the logarithmic mean temperature difference, LMTD, or for nonidealcountercurrent flow the corrected mean effective temperature difference (CMTD).

The main problem lies in establishing the heat transfer coefficients on the tube andshell side of the heat exchanger.

If the heat transfer coefficients are known, the overall heat transfer coefficient can beestablished.

Without considering the different areas on the inside and outside of the heatexchanger tubes and the fouling and the conduction through the tube wall the followingUa-value follows from the two heat transfer coefficients:

Ua ¼ 11aiþ 1

ao

¼ ai � ao

ai þ ao



http://dx.doi.org/10.1016/B978-0-12-803764-5.00006-7

Example 1: Calculation of Ua from the heat transfer coefficients

Data: ai ¼ 620 W/m2 K ao ¼ 699 W/m2 K

Ua ¼ 620� 699620þ 699

¼ 328 W�m2 K

Alternatively the overall heat transfer coefficient Ua can be determined from the diagram in Figure 6.1.

Figure 6.1 Diagram for the determination of Ua from the a-values.

Example 2:

Data: ai ¼ 4000 W/m2 K ao ¼ 1500 W/m2 K

Ua ¼ 4000� 15004000þ 1500

¼ 1091 W�m2 K

The good heat transfer coefficient in the tube does not help much. The overall heat transfer coefficientis always determined by the worse heat transfer coefficient.

The inner heat transfer coefficient ai must be converted to aio because with plain tubes the heatexchanger surface area generally refers to the outer area of the tube.

aio ¼ ai � dido

¼ ai � AiAo

�W�m2 K

�

Example 3: Conversion of ai to the outside area

Data: ai ¼ 620 W/m2 K di ¼ 16 mm da ¼ 20 mm

aio ¼ 620� 1620

¼ 496 W�m2K

The heat transfer coefficient aio is the tube side heat transfer coefficient based on the outside area ofthe tubes.


Next the heat conduction through the tube wall is to be considered especially with tube materials hav-ing bad heat conductivity l, for instance, stainless steel.

Uclean ¼ 11aio

þ 1aoþ s

l

�W�m2 K

�

Example 4: Calculation of Uclean

Data: aio ¼ 496 W/m2 K ao ¼ 699 W/m2 K s ¼ 2 mm l ¼ 56 W/m K

Uclean ¼ 11496þ 1

699 þ 0:00256

¼ 287 W�m2 K

If chromiumenickelesteel is used with a heat conductivity of 16 W/m K, this deteriorates the overallheat transfer coefficient.

Uclean ¼ 11496þ 1

699 þ 0:00216

¼ 280 W�m2 K

Finally, the fouling fi on the tube inside and fa on the tube outside have to be considered.

Udirty ¼ 11aio

þ 1aoþ s

lþ fi þ fa

¼ 11

Ucleanþ fi þ fa

�W�m2 K

�

The fouling is dependent on the products, the operation period, the flow velocity, and the performedcleanings. The values for the fouling fi on the tube inside and fa on the shell side lie in the range of 0.0001e0.001 m2 K/W depending on the problem definition.

Material

<1 m/s >1 m/s

<50 �C >50 �C <50 �C >50 �C

Drinking water 0.0002 0.0004 0.0002 0.0004Distilled water 0.0001 0.0001 0.0001 0.0001River water 0.0006 0.0008 0.0004 0.0006Cooling water 0.0002 0.0004 0.0002 0.0004Brackish water 0.0004 0.0006 0.0002 0.0004Sea water 0.0001 0.0002 0.0001 0.0002Vessel-desalted water 0.0004 0.0004 0.0004 0.0004

Material f (m2 K/W)

Fuel oil 0.001Lubricating oil 0.0002Thermal oil 0.0002Vegetable oil 0.0006Organic liquids 0.0002Organic vapors 0.0001Diesel-exhaust gas 0.002Oil-free vapor 0.0001Oil-rich vapor 0.0002Air 0.0004

Overall Heat Transfer Coefficient and Temperature Profile 81

The values are valid for an operation period of 1 year.If the tubes will be coated for corrosion reasons, the heat conductivity resistance of the coating must

be additionally considered.

Example 5: Calculation of Udirty

Data: aio ¼ 496 W/m2 K ao ¼ 699 W/m2 K s ¼ 2 mm l ¼ 56 W/m Kfi ¼ 0.0002 m2 K/W fa ¼ 0.0002 m2 K/W Uclean ¼ 287 W/m2 K

Udirty ¼ 11496 þ 1

699þ 0:00256 þ 0:0002þ 0:0002

¼ 257 W�m2 K

Udirty ¼ 11287 þ 0:0002þ 0:0002

¼ 257 W�m2 K

It might be interesting to establish the fouling reserve of a new heat exchanger.Thereby, the determined Uclean-value from the two heat transfer coefficients and the heat conduction

through the tubes is compared with that Ureq-value required for the problem definition.

Xf ¼ Uclean � Ueq

Uclean � Ureq

Example 6: Calculation of the existing fouling reserve Sf

Data: aio ¼ 496 W/m2 K ao ¼ 699 W/m2 K Uclean ¼ 287 W/m2 Ks ¼ 2 mm l ¼ 56 W/m K Ureq ¼ 230 W/m2 KX

f ¼ Uclean � Ueq

Uclean � Ureq¼ 287� 230

287� 230¼ 0:0009

m2 KW

In this case, for both fouling values fi þ fa ¼ 0.0004 m2 K/W is used so that an additional fouling reserveof 0.0009 � 0.0004 ¼ 0.0005 m2 K /W is available.

Inf luences of foulingIn Figure 6.2 it is shown that a good overall heat transfer coefficient Uclean is more highly reduced throughfouling than a bad overall heat transfer coefficient Uclean.

Through the fouling the U-values are leveled.The calculations so far are based on the assumption of plain walls.If the tube form in shell and tube heat exchangers is considered, then the following exact equation for

the overall heat transfer coefficient Uex in tubes results:

Uex ¼ 1�fi � 1

ai

�� do

diþ do

2�l� ln do

diþ 1

aoþ fa


Example 7: Calculation of Uex for the data of Example 5

Uex ¼ 1�0:0002� 1

620

�� 20

16þ 0:022�56� ln 20

16 þ 1699þ 0:0002

¼ 254 W�m2 K

The overall heat transfer coefficient Uex for tubes is a little bit lower, but the difference between Uex andUdirty is small, so that a simple equation for plain walls can be used for the calculation.

Recommended calculation procedure1. Calculation of the heat transfer coefficients ai in the tube and ao on the shell side2. Correction of ai to aio: aio ¼ ai � di

doðW=m2 KÞ

3. Calculation of Uclean: Uclean ¼ 11aio

þ 1aoþ s

l

ðW=m2 KÞ

4. Calculation of Udirty Udirty ¼ 11

Ucleanþ fi þ fa

ðW=m2 KÞ

5. Calculation of the fouling reserve of the new heat exchanger:P

f ¼ Uclean � Ueq

Uclean � Ureq

Example 8: Calculation of the overall heat transfer coefficient and the foulingreserve

Data: ai ¼ 620 W/m2 K ao ¼ 699 W/m2 K s ¼ 2 mm l ¼ 56 W/m Kdi ¼ 16 mm do ¼ 20 mm fi ¼ fa ¼ 0.0002 m2 K /W Ureq ¼ 230 W/m2 K

aio ¼ 620� 1620

¼ 496 W�m2 K

Uclean ¼ 11496 þ 1

699 þ 0:00256

¼ 287 W�m2 K

Figure 6.2 The fall of the overall heat transfer coefficient with increasing fouling.


Udirty ¼ 11287 þ 0:0002þ 0:0002

¼ 257 W�m2 K

Xf ¼ Uclean � Ueq

Uclean � Ureq¼ 287� 230

287� 230¼ 0:0009

m2 KW

A clear fouling reserve of 0.0005 m2 K /W is available.Often it can be worthwhile for the selection or the design of a heat exchanger to calculate the overall

heat transfer coefficient per meter of the heat exchanger and consequently determine the required totallength. The corresponding equation for the calculation of the overall heat transfer coefficient UL per metertube is as follows:

UL ¼ p1

ai�diþ 1

2�l� ln do

diþ 1

ao�doþ fi

diþ fa

do

ðW=m KÞ

From UL the required total length Lreq is calculated for a certain heat load Q as follows:

Q ¼ UL� Lreq � CMTD ðWÞ Lreq ¼ QUL � CMTD

ðmÞ

Example 9: Calculation of the overall heat transfer coefficient per meter heatexchanger length

Data: ai ¼ 620 W/m2 K ao ¼ 699 W/m2 K s ¼ 2 mm l ¼ 56 W/m Kdi ¼ 16 mm do ¼ 20 mm fI ¼ fa ¼ 0.0002 m2 K /WQ ¼ 180,000 W LMTD ¼ 51 K

UL ¼ p1

620� 0:016þ 12� 56

� ln2016

þ 1699� 0:02

þ 0:00020:016

þ 0:00020:02

¼ 15:96 W=m K

Lreq ¼ 180; 00015:96� 51

¼ 221 m

6.2 CALCULATION OF THE TEMPERATURE GRADIENT IN A HEATEXCHANGER

The temperature fall in the individual overall heat transfer resistances can be calculatedusing the heat flux density q (W/m2):

q ¼ Udirty � Dttot ¼ ao � Dto ¼ aio � Dti ¼ l

s� Dtwall ¼ 1

fi� Dtfi

¼ 1fa� Dtfa

�W�m2�

Dto ¼ qao

Dti ¼ qaio

Dtwall ¼ q� sl

Dtfi ¼ q� fi Dfa ¼ q� fa

1Udirty

¼ 1ao


Example 10: Calculation of the temperature gradient in the overall heat transferresistances

Data: ao ¼ 699 W/m2 K ai ¼ 620 W/m2 K l ¼ 56 W/m Kfa ¼ 0.0002 fi ¼ 0.0002 s ¼ 2 mmDttot ¼ 20 �C ¼ LMTD di ¼ 20 mm do ¼ 25 mm

aio ¼ 620� 2025

¼ 496 W�m2 K

1Udirty

¼ 1699

þ 1496

þ 0:00256

þ 0:0002þ 0:0002 ¼ 0:0039 Udirty ¼ 257:6 W�m2 K

q ¼ U� Dttot ¼ 257:6� 20 ¼ 5151 W�m2

Dto ¼ qao

¼ 5151699

¼ 7:4 �C Dti ¼ qaio

¼ 5151496

¼ 10:4 �C

Dtwall ¼ q� sl¼ 5151� 0:002

56¼ 0:2 �C

Dtfi ¼ q� fi ¼ 5151� 0:0002 ¼ 1 �C Dtfa ¼ q� fa ¼ 5151� 0:0002 ¼ 1 �C

Check : Dttot ¼ 7:4þ 10:4þ 1þ 1þ 0:2 ¼ 20 �C

The calculated temperature fall is shown in Figure 6.3.

Out

side

foul

ing

tubewall

Insi

de fo

ulin

g

Δ tfa = 1,0 °C

Δ tfi = 1,0 °C

Δ ti = 10,4 °C

Δ tWall = 0,2 °C

Δ to = 7,4 °C

Figure 6.3 Calculation of the temperature gradient in the heat exchanger.


6.3 VISCOSITY CORRECTION

The viscosity has a predominant influence on the thickness of the laminar border film onthe wall and this border layer determines the heat transfer coefficient. The higher the vis-cosity on the wall, thicker is the border film and worse the heat transfer coefficient.

The correction of the calculated heat transfer coefficient am for the average producttemperature tm follows from the viscosity ratio of the product at average fluid tempera-ture and at wall temperature tW.

akorr ¼ F� am

Cooling :hm

hW< 1 F ¼

�hm

hW

�0:14

Heating :hm

hW> 1 F ¼

�hW

hm

�0:14

akorr ¼ corrected heat transfer coefficient (W/m2 K)am ¼ calculated a-value at tm (W/m2 K)F ¼ correction factorhm ¼ viscosity at tm (mPas)hW ¼ viscosity at wall temperature tW (mPas)The wall temperature is smaller for cooling than the average product temperature and

the viscosity increases. This deteriorates the heat transfer coefficient.It is the opposite for heating. The lower viscosity on the wall improves the heat trans-

fer coefficient.

Calculation procedure1. Calculation of the heat transfer coefficient at average product temperature tm

tm ¼ tin þ tout2

tin ¼ inlet temperature (�C)tout ¼ outlet temperature (�C)

2. Determination of the wall temperature tw

Cooling : tw ¼ tm � Dt ð�CÞHeating : tw ¼ tm þ Dt ð�CÞ

Calculation of the temperature difference Dt between tm and tW (see Chapter 6.2)

Dt ¼ qam

ð�CÞ


q ¼ heat f lux density (W/m2)3. Derivation of the viscosity at the calculated wall temperature4. Calculation of the viscosity correction F

Example 11: Calculation of the viscosity correction at thin media

Data: aio ¼ 336 W/m2 K ao ¼ 700 W/m2 K U ¼ 202.11 W/m2 KLMTD ¼ 16.83 �C q ¼ 3401.92 W/m2

Shell side: Water 25 �C ➜ 30 �C tm ¼ 27.5 �C

Dto ¼ 3401:92700

¼ 4:9 �C

tw ¼ 27:5þ 4:9 ¼ 32:4 �C

The wall temperature is higher at the shell side than the average product temperature and the viscosityis smaller.

Tube side: Isopropanol 70 �C ➜ 30 �C tm ¼ 50 �C

Dti ¼ 3401:92336

¼ 10:1 �C

tw ¼ 50� 10:1 ¼ 39:9 �C

The wall temperature is smaller than the average product temperature and the viscosity on the wall ishigher.

From Figure 6.4 the viscosities of water on the shell side and isopropanol in the tubes at different tem-peratures are taken.

Water: At 27.5 �C ➔ hm ¼ 0.89 mPas At 32.4 �C➔ hW ¼ 0.8 mPas�hm

hW

�0:14

¼�0:890:8

�0:14

¼ 1:01

Figure 6.4 Dynamic viscosity of some liquids as function of the temperature.


akorr ¼ 1:01� 336 ¼ 339 W�m2 K

The lower viscosity at the wall improves the heat transfer coefficient.

Isopropanol: At 50 �C ➔ hm ¼ 1.08 mPas At 39.9 �C ➔ hW ¼ 1.37 mPas�hm

hW

�0:14

¼�1:081:37

�0:14

¼ 0:97

akorr ¼ 0:97� 700 ¼ 679 W�m2 K

The higher viscosity on the wall worsens the heat transfer coefficient.

Example 12: Calculation of the viscosity correction for a high viscous oil

aio ¼ 50 W/m2 K ao ¼ 1500 W/m2 K U ¼ 47.15 W/m2 K tm ¼ 70 �CLMTD ¼ 15 �C q ¼ 707.229 W/m2

Dt ¼ 707:22950

¼ 14:1 �C tw ¼ 70� 14:1 ¼ 55:9 �C

hm ¼ 110 mPas at 70 �C hW ¼ 192 mPas at 55:9 �C�hm

hW

�0:14

¼�110192

�0:14

¼ 0:925 akorr ¼ 0:925� 50 ¼ 46:2 W�m2K

U ¼ 43:8 W�m2 K

Due to the high viscosity of the oil on the wall, the heat transfer coefficients and the overall heat trans-fer coefficients become worse.

6.4 CALCULATION OF THE HEAT TRANSFER COEFFICIENT FROM THEOVERALL HEAT TRANSFER COEFFICIENT

The following equation is valid for the overall heat transfer coefficient.

1Udirty

¼ 1ai

þ 1ao

þ slþ fi þ fa

After a simple conversion, the heat transfer coefficients can be determined.

1ai

¼ 1Udirty

� 1ao

� sl� ð fi þ faÞ 1

ao¼ 1

Udirty� 1ai

� sl� ð fi þ faÞ


Example 13: Determination of the a-values from the overall heat transfercoefficient U

Data: U ¼ 300 W/m2 K aa ¼ 1000 W/m2 K fi þ fa ¼ 0.0004 s/l ¼ 0.000141ai

¼ 1300

� 11000

� 0:00014� 0:0004 ai ¼ 557:6 W�m2 K

1ao

¼ 1300

� 1557:6

� 0:00014� 0:0004 ao ¼ 1000 W�m2 K

Kontrolle

1Udirty

¼ 1ai

þ 1ao

þ slþ fi þ fa ¼ 1

557:6þ 11000

þ 0:00014þ 0:0004 U ¼ 300 W�m2 K

NOMENCLATUREA area of the exchanger (m2)ai tube-side heat transfer coefficient (W/m2 K)aio ai based on the outer area (W/m2 K)ao shell-side heat transfer coefficient (W/m2 K)di tube inner diameter (m)do tube outer diameter (m)fi inner fouling resistance (m2 K/W)fa outer fouling resistance (m2 K/W)U general overall heat transfer coefficient (W/m2 K)Ua overall heat transfer coefficient calculated from both a-values (W/m2 K)Uclean overall heat transfer coefficient without fouling (W/m2 K)Udirty overall heat transfer coefficient with fouling (W/m2 K)Ureq required overall heat transfer coefficient (W/m2 K)Uex overall heat transfer coefficient considering fouling þ tube form (W/m2 K)UL overall heat transfer coefficient per meter tube (W/m2 K)L heat conductivity of the tube material (W/m K)L length of the heat exchanger tubes (m)Lreq required total length of the tubes (m)CMTD corrected effective mean temperature difference (K)Q heat load (W)s tube wall thickness (m)

REFERENCES[1] D.Q. Kern, Process Heat Transfer, McGraw-Hill, NY, 1950.[2] W.H. McAdams, Heat Transmission, McGraw-Hill, NY, 1954.[3] Perry’s Chemical Engineer’s Handbook, McGraw-Hill, 1984.[4] G.F. Hewitt, G.L. Shires, T.R. Bott, Process Heat Transfer, CRC Press, Boca Raton, 1994.[5] Lord, Minton, Slusser, Design of heat exchangers, Chem. Eng. 26 ( Jan. 1970).


[6] N.P. Chopey, Heat transfer, in: Handbook of Chemical Engineering Calculations, McGraw-Hill,New York, 1993.

[7] V.D.I. W€armeatlas, 5. Aufkage, VDI-Verlag, D€usseldorf, 1988.[8] F. Hell, Grundlagen der W€arme€ubertragung, VDI-Verlag, D€usseldorf, 1982.[9] W.M. Rohsenow, J.P. Hartnett, Handbook of Heat Transfer, McGraw-Hill, NY, 1973.[10] E.R.G. Eckert, R.M. Drake, Heat and Mass Transfer, McGraw Hill, NY, 1959.


CHAPTER 7

Chemical Engineering Calculations

Contents

7.1 Vapor Pressure Calculations 917.2 Equilibrium between the Liquid and the Vapor Phase 937.3 Bubble Point Calculation 947.4 Dew Point Calculation 957.5 Calculation of Dew and Bubble Lines of Ideal Binary Mixtures 957.6 Flash Calculations 977.7 Condensation or Flash Curve of Binary Mixtures 997.8 Calculation of Nonideal Binary Mixtures 1017.9 Flash Calculations for Multicomponent Mixtures 103References 104

In the design of condensers and evaporators for multicomponent mixtures, the dew pointand dew point line and the bubble point and bubble point line must be calculated.

In condensers the first droplet is condensed at the dew point of the mixture, whereasthe mixture is totally liquid at the bubble point.

The vapor mixture must be cooled from the dew point to the bubble point.In evaporators of liquid mixtures, the first droplet evaporates at the bubble point and

at the dew point the total mixture is in vapor state.The mixture must be heated from the bubble point to the dew point.For the design of condensers and evaporators, the condensation, that is, flash curves

and the heat load curve with the heat load requirement as a function of the temperatureis needed (Figure 7.1).

7.1 VAPOR PRESSURE CALCULATIONS [5]

The vapor pressure is calculated using the Antoine equation and the Antoine constants A,B, and C which are available in the literature. Some vapor pressure curves are shown inFigure 7.2.

Antoine equation:

lg p0 ¼ A� BCþ tð�CÞ ðmbarÞ



http://dx.doi.org/10.1016/B978-0-12-803764-5.00007-9

Example 1: Vapor pressure calculation for benzene and toluene

Benzene Toluene

A ¼ 7.00481 A ¼ 7.07581B ¼ 1196.76 B ¼ 1342.31C ¼ 219.161 C ¼ 219.187

Temperature Vapor pressures:50 �C: p0B ¼ 360.4 mbar p0T ¼ 122.5 mbar98 �C: p0B ¼ 1710 mbar p0T ¼ 696 mbar103 �C: p0B ¼ 1960 mbar p0T ¼ 810 mbar

Figure 7.1 Condensing curve.

Figure 7.2 Vapor pressure of different components as function of temperature.


7.2 EQUILIBRIUM BETWEEN THE LIQUID AND THE VAPOR PHASE [5]

Dalton (Figure 7.3):

p1 ¼ y1 � Ptot Ptot ¼ p1 þ p2 þ p3 þ.pi

Raoult (Figure 7.3):

p1 ¼ x1 � p01 p2 ¼ x2 � p02

Ptot ¼ x1 � p01 þ x2 � p02 þ x3 � p03 þ.xi � p0i

Equilibrium equation: yi � Ptot ¼ pi ¼ xi � p0iCalculation of the equilibrium factor K:

K ¼ yixi

¼ p0iPtot

yi ¼ composition of the component i in the vapor phase (mole fraction)xi ¼ composition of the component i in the liquid phase (mole fraction)Ptot ¼ total pressure (mbar)p0i ¼ vapor pressure of the component i (mbar)pi ¼ partial pressure of the component i (mbar)

Example 2: Calculation of the partial pressures according to Raoult and the vaporcomposition according to Dalton and the equilibrium factors

x1 ¼ 0.6 p01 ¼ 800 mbarx2 ¼ 0.4 p02 ¼ 1300 mbar

Calculation of the partial pressures according to Raoult:

p1 ¼ x1 � p01 ¼ 0.6 � 800 ¼ 480 mbarp2 ¼ x2 � p02 ¼ 0.4 � 1300 ¼ 520 mbar

Ptot ¼ 1000 mbar

Figure 7.3 Graphical representation of the laws of Dalton and Raoult.

Chemical Engineering Calculations 93

Calculation of the vapor composition according to Dalton:

y1 ¼ p1Ptot

¼ 4801000

¼ 0:48 ¼ 48 Mol%

y2 ¼ p2Ptot

¼ 5201000

¼ 0:52 ¼ 52 Mol%

Calculation of the equilibrium factors K:

K1 ¼ y1x1

¼ 0:480:6

¼ p01Ptot

¼ 8001000

¼ 0:8

K2 ¼ y2x2

¼ 0:520:4

¼ p02Ptot

¼ 13001000

¼ 1:3

7.3 BUBBLE POINT CALCULATION [5]

The bubble point of a mixture is defined as follows:Xyi ¼

XKi � xi ¼ 1

The bubble point pressure Pbub is calculated directly.

Pbub ¼ x1 � p01 þ x2 � p02

The bubble point temperature for a particular total pressure is determined iterativelyby calculating the bubble point pressure at different temperatures.

The following example shows the procedure.

Example 3: Iterative bubble point calculation for a benzeneetoluene mixture30 Mol% benzene in the liquid phase (x ¼ 0.3)70 Mol% toluene in the liquid phase (x ¼ 0.7)Ptot ¼ 1000 mbar

1. Choice: t ¼ 95 �C p0B ¼ 1573 mbar p0T ¼ 634.4 mbarPbub ¼ 0.3 � 1573 þ 0.7 � 634.4 ¼ 916 mbar Pbub < Ptot

2. Choice: t ¼ 100 �C p0B ¼ 1807 mbar p0T ¼ 740 mbarPbub ¼ 0.3 � 1807 þ 0.7 � 740 ¼ 1060 mbar Pbub > Ptot

3. Choice: t ¼ 98 �C p0B ¼ 1710 mbar p0T ¼ 696 mbarPbub ¼ 0.3 � 1710 þ 0.7 � 696 ¼ 1000 mbar Pbub ¼ Ptot ¼ 1000 mbar

Check calculation at 98 �C:

K1 ¼ p0BPtot

¼ 17001000

¼ 1:71

K2 ¼ p0TPtot

¼ 6961000

¼ 0:696

XKi � xi ¼ 1:71� 0:3þ 0:696� 0:7 ¼ 1

Thus, the bubble point specification is fulfilled.


7.4 DEW POINT CALCULATION [5]

The dew point of a vapor mixture is defined as follows:Xxi ¼

Xyi�Ki ¼ 1

The dew point pressure Pdew is calculated directly.

1Pdew

¼ y1p01

þ y2p02

The dew point temperature for a particular total pressure is calculated iteratively.The following example shows the procedure.

Example 4: Iterative dew point calculation for a benzeneetoluene mixture30 Mol% benzene in the vapor phase (y ¼ 0.3)70 Mol% toluene in the vapor phase (y ¼ 0.7)Ptot ¼ 1000 mbar

1. Choice: t ¼ 100 �C p0B ¼ 1807 mbar p0T ¼ 740 mbar1

Pdew¼ 0:3

1807þ 0:7740

¼ 0:0011 Pdew ¼ 899 mbar < Ptot ¼ 1000 mbar

2. Choice: t ¼ 104 �C p0B ¼ 2013 mbar p0T ¼ 834.4 mbar1

Pdew¼ 0:3

2013þ 0:7834:4

¼ 0:00099 Pdew ¼ 1012 mbar > Ptot ¼ 1000 mbar

3. Choice: t ¼ 103.8 �C p0T ¼ 2002 mbar p0T ¼ 829 mbar1

Pdew¼ 0:3

2002þ 0:7829

¼ 0:001 Pdew ¼ 1006 mbarz Ptot ¼ 1000 mbar

Check calculation at 103.8 �C:

K1 ¼ 20021006

¼ 1:99 K2 ¼ 8291006

¼ 0:824

X yiKi

¼ 0:31:99

þ 0:70:824

¼ 1

The condition for the dew point is reached.

7.5 CALCULATION OF DEW AND BUBBLE LINESOF IDEAL BINARY MIXTURES [5]

The bubble point of a liquid mixture is defined such that the sum of the partial pressuresof the mixture reaches the system pressure and the first droplet is evaporated.


Due to the preferential evaporation of light components, the heavier components inthe liquid mixture increase and therefore the bubble point increases. The representationof the bubble point temperature as a function of the composition of a lighter componentin the mixture is called the bubble line.

Equation of the bubble line:

x1 ¼ f ðtÞ ¼ Ptot � p02p01 � p02

ðMolfraction of the light component in the liquidÞ

The bubble point temperature increases with decreasing concentration of the lightboiling components.

The dew point of a vapor mixture is the temperature in which the first droplet con-denses. Due to the preferential condensation of the heavier component, the contents ofthe lighter components in the vapor mixture increase and the dew point temperature falls.

The representation of the dew point temperature as a function of the vapor compo-sitions is known as the dew line.

Equation of the dew line:

y1 ¼ f ðtÞ ¼ p01Ptot

� Ptot � p02p01 � p02

ðMolfraction of the light component in the vaporÞ

A diagram with bubble and dew line is known as the phase diagram (Figure 7.4).

Example 5: Construction of the phase diagram for benzene (1)etoluene (2)at 1.013 barProcedure:1. Calculation of both bubble points with the Antoine equation. The two bubble points are the boundary

points in the phase diagram.

Boiling point of benzene: 80.1 �C Boiling point of toluene: 110.6 �C

2. Calculation of the vapor pressures at different temperatures with the Antoine equation.3. Determination of the liquid composition x and the vapor composition y at different temperatures.

Temperature (�C) p01 (mbar) p02 (mbar) x1 y1

80.1 1013 390 1.000 1.00083 1107 430 0.861 0.94186 1211 476 0.731 0.87489 1322 525 0.612 0.79992 1442 578 0.504 0.71795 1569 636 0.404 0.62698 1705 698 0.313 0.527101 1850 765 0.229 0.418104 2004 836 0.151 0.300107 2168 913 0.080 0.170110.6 2378 1013 0.000 0.000

Bubble line: x1 ¼ f (temperature) Dew line: y1 ¼ f (temperature)

In Figure 7.4 the phase diagram of the benzeneetoluene mixture is presented.


The dew point of a vapor mixture with 30 mol% benzene and 70 mol% toluene is around 104 �C.The bubble point of the mixture is around 98 �C.The first liquid droplet condenses at 104 �C at the dew point.At the bubble point at 98 �C, the total mixture is liquid.In order to condense the total mixture, the mixture must be cooled down from 104 �C to 98 �C.On the contrary, in vaporizing the mixture must be heated from the bubble point to the dew point.

7.6 FLASH CALCULATIONS [1,2,5]

The vapor and liquid rates change in the region between the bubble and the dew point.When condensing the vapor rate decreases as the temperature becomes lower.When evaporating, more vapor is formed with increased heating.With help of flash calculation the amount of vapor at temperatures between the bub-

ble and the dew point is determined (Figure 7.5).The compositions in the vapor and liquid phase when cooling or heating are also

determined using the equilibrium factors (Figure 7.5).

Figure 7.4 Phase diagram for the benzeneetoluene mixture.

Figure 7.5 Flash separation.


At the bubble point, the total mixture is in the liquid phase.At the dew point, the total mixture is in the vapor phase.How much of the mixture is in the vapor phase at temperatures between the bubble point and the

dew point? Figure 7.5.The amount of vapor V in the feed F is calculated using the equilibrium constants K1

and K2 as follows:

VF

¼ z1 � K1�K21�K2

� 1

K1 � 1

What are the compositions of the liquid and the vapor phase of the mixture?

x1 ¼ 1� K2

K1 � K2ðMolfractionÞ y1 ¼ K1 � x1ðMolfractionÞ

V ¼ vapor rate (kmol/h)F ¼ feed rate (kmol/h)K1 ¼ equilibrium constant of the lighter component 1K2 ¼ equilibrium constant of the heavier component 2z1 ¼ composition of the lighter component 1 in feed (molfraction)x1 ¼ composition of the lighter component in the liquid phase (molfraction)y1 ¼ composition of the lighter component in vapor (molfraction)

Example 6: Flash calculations for a benzeneetoluene mixtureAt the bubble point (98 �C), all is liquidAt the dew point (104 �C), all is vaporFeed composition: 30 mol% benzene in the vapor (z1 ¼ 0.3 mol fraction)

The amount in the vapor phase at 100 �C is to be determined

t ¼ 100 �C Ptot ¼ 1000 mbarVapor pressure benzene

p0B ¼ 1807 mbarVapor pressure toluene

p0T ¼ 740 mbar

Ptot ¼ 1000 mbar

Benzene : K1 ¼ p0BPtot

¼ 18071000

¼ 1:807

Toluene : K2 ¼ p0TPtot

¼ 7401000

¼ 0:74

VF¼

0:3� 1:807� 0:741� 0:74

1:807� 1¼ 0:2864

28.64 mol% of the feed is vapor!Calculation of the compositions in liquid and vapor at t ¼ 100 �C

x1 ¼ 1� K2K1 � K2

¼ 1� 0:741:807� 0:74

¼ 0:2437 ➔24.37 mol% benzene in the liquid

y1 ¼ K1 � x1 ¼ 1.807 � 0.2437 ¼ 0.44 ➔44 mol% benzene in the vapor


7.7 CONDENSATION OR FLASH CURVE OF BINARY MIXTURES

For the design of condensers and evaporators for mixtures, the condensation or flash curveis required.

In these curves, the vapor fraction V/F of the mixture based on the feed rate F isplotted over the temperature of the mixture.

The calculations follow the equations in Section 7.6.

Example 7: Calculation of the condensation curve for the mixture benzene-o-xyleneInlet composition: z1 ¼ 0.576 ¼ 57.6 mol% benzene in the vapor mixture

T (�C) K1 K2 V/F x1 y196 1.57 0.22 0 0.576 0.90698.9 1.74 0.25 0.2 0.502 0.874104 2 0.3 0.4 0.41 0.82110.2 2.34 0.37 0.6 0.318 0.748116.2 2.72 0.448 0.8 0.242 0.66121 3.07 0.52 1 0.188 0.576

T ¼ 96 �C: VF¼

z1 � K1 � K21� K2

� 1

K1 � 1¼

0:576� 1:57� 0:221� 0:22

1:57� 1¼ 0

x1 ¼ 1� K2K1 � K2

¼ 1� 0:221:57� 0:22

¼ 0:576 y1 ¼ K1 � x1 ¼ 1:57� 0:576 ¼ 0:906

T ¼ 98.9 �C VF¼

0:576� 1:74� 0:251� 0:25

� 1

1:74� 1¼ 0:2

x1 ¼ 1� 0:251:74� 0:25

¼ 0:502 y1 ¼ 1:74� 0:502 ¼ 0:874

T ¼ 116.2 �C VF¼

0:576� 2:72� 0:4481� 0:448

� 1

2:72� 1¼ 0:8

x1 ¼ 1� 0:4482:72� 0:448

¼ 0:242 y1 ¼ 2:72� 0:242 ¼ 0:66

T ¼ 121 �C VF¼

0:576� 3:07� 0:521� 0:52

� 1

3:07� 1¼ 1

x1 ¼ 1� 0:523:07� 0:52

¼ 0:188 y1 ¼ 3:07� 0:188 ¼ 0:576


The condensation curve for the mixture benzene-o-xylene is shown in Figure 7.6.The molar vapor fraction V/F drops from V/F ¼ 1 at the dew point at 121 �C to V/F ¼ 0 at the bubble

point 96 �C.After conversion from mole to weight units, the vapor and condensate rate in kilogram per hour at

different temperatures shown in Figure 7.7 results for a feed rate of 2000 kg/h with 50 weight% benzeneand 50 weight% o-xylene.

The required heat loads for the condensation of the benzene-o-xylene mixture result from the en-thalpies for cooling the vapor mixture, the condensation and the cooling of the condensate.

Figure 7.6 Condensation curve of the mixture benzene-o-xylene of Example 7.

Figure 7.7 Vapor and liquid rates of the benzene-o-xylene mixture at different temperatures.


In Figure 7.8 the determined heat loads for the condensing of a mixture of 2000 kg/h benzene-o-xylene are shown as a function of the temperature.

In condensing and cooling from dew point to bubble point of the mixture, the required heat load risesfrom 0 to 230 kW.

7.8 CALCULATION OF NONIDEAL BINARY MIXTURES [3,4,5]

In the calculation of phase equilibriums of nonideal mixtures the activity coefficient gmust be considered.

Ptot ¼ g1 � x1 � p01 þ g2 � x2 � p02 ðmbarÞy1 � Ptot ¼ g1 � x1 � p01

y1 ¼ x1 � p01 � g1

PtotðMolfractionÞ

K1 ¼ p01 � g1

PtotK2 ¼ p02 � g2

Ptot

The activity coefficient g is strongly dependent on the liquid composition and can be calculatedwith different models: Wilson, NRTL, Uniquac, Unifac [3,5]

Example 8: Bubble point calculation with activity coefficient for the mixturemethanol (1)ewater (2)

x1 ¼ 0.1 x2 ¼ 0.9 t ¼ 87.8 �Cg1 ¼ 1.705 g2 ¼ 1p01 ¼ 2444 mbar p02 ¼ 648 mbar

Pbub ¼ x1 � g1 � p01 þ x2 � g2 � p02 ¼ 0:1� 1:705� 1833þ 0:9� 1� 486 ¼ 1000 mbar

Figure 7.8 Heat load curve for the condensation of 2 t/h of the benzene-o-xylene mixture as afunction of the temperature.


K1 ¼ g1 � p01Ptot

¼ 1:705� 24441000

¼ 4:167

K2 ¼ 1� 6481000

¼ 0:64

Bubble point check: S Ki � xi ¼ 0.1 � 4.167 þ 0.9 � 0.64 ¼ 1The bubble point condition is reached at 87.8 �C!Without considering the activity coefficient g the bubble point temperature of 93.1 �C instead of 87.8 �C

results.

Example 9: Dew point calculation with activity coefficient g for the mixturemethanol (1)ewater (2)

y1 ¼ 0.1 y2 ¼ 0.9 t ¼ 97.1 �Cg1 ¼ 2.2 g2 ¼ 1.0p01 ¼ 3343 mbar p02 ¼ 910 mbar

1Pdew

¼ y1g1 � p01

þ y2g2 � p02

¼ 0:12:2� 3343

þ 0:9910

¼ 0:001

Pdew ¼ 1 bar

K1 ¼ 2:2� 33431000

¼ 7:35 K2 ¼ 1� 9101000

¼ 0:91

X yiKi¼ 0:1

7:35þ 0:90:91

¼ 1

The dew point condition is reached at 97.1 �C!

Example 10: Flash calculation for the mixture methanol (1)ewater (2)

P ¼ 1 bar ¼ 1000 mbar t ¼ 92.1 �C Feed rate ¼ 100 kMol/hInlet vapor compositions: z1 ¼ 0.1 z2 ¼ 0.9

g1 ¼ 1.848 g2 ¼ 1.004p01 ¼ 2817 mbar p02 ¼ 756 mbar

K1 ¼ 1:848� 28171000

¼ 5:2 K2 ¼ 1:004� 7561000

¼ 0:76

VF¼

z1 � K1 � K21� K2

� 1

K1 � 1¼

0:1� 5:2� 0:761� 0:76

� 1

5:2� 1¼ 0:20

At 92.1 �C 20% of the feed are vapor.Vapor rate V ¼ 0.20 � 100 ¼ 20 kmol/hLiquid rate L ¼ 100 � 20 ¼ 80 kmol/h


Calculation of the composition in the liquid and in the vapor:

x1 ¼ 1� K2K1 � K2

¼ 1� 0:765:2� 0:76

¼ 0:054 Molfraction

x2 ¼ 1 � 0.054 ¼ 0.946 (molfraction) ¼ 94.6 Mol%y1 ¼ K1 � x1 ¼ 5.2 � 0.054 ¼ 0.28 (molfraction) ¼ 28 mol%y2 ¼ K2 � x2 ¼ 0.76 � 0.946 ¼ 0.72 (molfraction) ¼ 72 mol%Without considering the activity coefficient g, the calculation gives the following wrong results at

92.1 �C:

K1 ¼ 2.81 K2 ¼ 0.756V ¼ 0 L ¼ 100 kmol/h

Nothing is vaporized because the bubble point lies with g ¼ 1 at 93.1 �C, that is, over 92.1 �C.

7.9 FLASH CALCULATIONS FOR MULTICOMPONENT MIXTURES [1,2]

The calculation must be performed iteratively for mixtures with more than twocomponents.

The individual value for V/F of the different components is calculated with an esti-mated value for V/F. The sum of the V/F values of the components must equal the V/Festimated value.

VF

¼X zi

1þ LV � Ki

LV

¼ FV

� 1 ¼ 1V=L

yi ¼ FV

�

0BB@ zi

1þ LV � Ki

1CCA xi ¼ F

V�

0BB@ zi

Ki þ LV

1CCA

Figure 7.9 Vapor composition of a hydrocarbon mixture as function of the temperature.


A description of the procedure with example is found in references (1) and (2).From Figure 7.9, it can be seen that the dew point temperature greatly falls with

increasing concentration of the low-boiling components.

REFERENCES[1] E.J. Henley, J.D. Seader, Equilibrium-stage Separation Operations in Chemical Engineering,

John Wiley, New York, 1981.[2] B.D. Smith, Design of Equilibrium Stage Processes, McGraw-Hill, New York, 1963.[3] J. Gmehling, B. Kolbe, M. Kleiber, J. Rarey, Chemical Thermodynamics for Process Simulation,

Wiley-VCH-Verlag, Weinheim, 2012.[4] G. Mehos, Estimate binary equilibrium coefficients, Chem. Eng. 103 (1996) 131.[5] M. Nitsche, Kolonnen-fibel, Springer Verlag, Berlin, 2014.


CHAPTER 8

Design of Condensers

Contents

Construction Types of Condensers 1068.1 Condenser Construction Types 106

8.1.1 Condensation in the shell chamber on the horizontal tube bundle 1078.1.2 Condensation on horizontal water-cooled tubes in a top condenser 1098.1.3 Condensation shell side on a vertical bundle 1098.1.4 Condensation in vertical tubes 1108.1.5 Vertical countercurrent condensation in reflux coolers on stirred tanks 1108.1.6 Condensation in horizontal tubes 110

8.2 Heat Transfer Coefficients in Isothermal Condensation 1118.2.1 Calculation of the heat transfer coefficients 111

8.2.1.1 Condensation at a horizontal tube bundle 1168.2.1.2 Condensation at a vertical tube bundle 1178.2.1.3 Condensation in horizontal tubes 1188.2.1.4 Condensation in vertical tubes 119

8.2.2 Parameters influencing the heat transfer coefficients 1208.3 Comparison of Different Calculation Models 1228.4 Condensation of Vapors with Inert Gas 1258.5 Condensation of Multicomponent Mixtures 129

Differential and integral condensation of multicomponent mixtures 1348.6 Miscellaneous 135

8.6.1 Influence of pressure loss in the condenser 1358.6.2 Condensate drain pipe 1358.6.3 Maximal vapor flow capacity 1388.6.4 Top condensers on columns 1398.6.5 Calculation of the pressure loss DPT in condensing in the tubes 1418.6.6 Maximum condensate load in horizontal tubes 1428.6.7 Flood loading in reflux condensers 143

Nomenclature 145References 145

Media in vapor state are condensed on cold surfaces if the cooling surface is colder thanthe dew point of the vapors. In film condensation a condensate film forms on the coolingarea.

In technical plants the vapors are liquefied by film condensation.In isothermal condensation of single components, condensation occurs at the dewpoint.In multicomponent mixtures, in condensing, the mixture must be cooled and lique-

fied according to the condensation line from the dew point to the bubble point.



http://dx.doi.org/10.1016/B978-0-12-803764-5.00008-0

CONSTRUCTION TYPES OF CONDENSERS

The following indirect condensation types are distinguished:Condensation on horizontal tube bundle (Figure 8.1(a))Condensation in horizontal tubes (Figure 8.1(b))Condensation in vertical tubes (Figure 8.1(c))Condensation on vertical tube bundle (Figure 8.1(d))The four principal different condenser types are shown in Figure 8.1.In Figure 8.2 it can be seen how the heat transfer coefficients in the different

condenser types differ in the condensation of ethanol and cyclohexane and how theheat transfer coefficients change with increasing condensate rate.

The heat transfer coefficients are best in the case of horizontal condensation.With increasing condensate rate or growing condensate film thickness the heat trans-

fer coefficients fall. When condensation takes place in the tubes the a-value becomes bet-ter for higher vapor loadings because the condensate film becomes turbulent and thevapor shear stress improves the heat transfer.

8.1 CONDENSER CONSTRUCTION TYPES

In the following the operation of the different condenser construction types shown inFigure 8.4 is described.

E = Vent D = VaporK = Condensate W = Cooling water

(a)

(b)

(c) (d)

Figure 8.1 Different condenser types.


8.1.1 Condensation in the shell chamber on the horizontal tube bundle(Figure 8.4(a))

The vapors enter from the top and flow through a horizontal tube bundle.Cooling water flows through the tubes. The vapors condense on the cold tubes and

the condensate runs over the tubes down to the bottom.Due to the small condensate film thickness, good heat transfer coefficients result

which reduce with increasing condensate rate.Advantages: Low subcooling and more adequate for the condensation in vacuumbecause of the low pressure drop and the lack of an adiabatic mass flux capacity limitby a sudden cross-sectional constriction.

Figure 8.2 Heat transfer coefficients in different condenser construction types as function of thecondensate rate.

Design of Condensers 107

Figure 8.3 Low pressure drop cross condenser for the vacuum condensation.

Figure 8.4 Different condenser constructions.


Tube pitch and baffle spacings can be adjusted to the vapor stream.Low pressure loss and easy tube-side cleaning.Disadvantages: Noncondensable gases must be removed, otherwise one part of thecooling area is covered. With a wide boiling mixture a differential condensationresults. The high-boiling components are initially condensed, run down, and areno longer available for an equilibrium condensation. Hence the condensation ofthe concentrated low-boiling components becomes more difficult, for instance, inthe condensation of a watereammonia mixture.It comes to a condensate congestion if no free flow area is provided at the bottom inthe baffles. The baffles should be best arranged sideways.In Figure 8.3, a cross stream condenser with low pressure drop for the condensation invacuum is shown.

8.1.2 Condensation on horizontal water-cooled tubes in a topcondenser (Figure 8.4(b))

The vapors flow through the shell chamber from the bottom to the top and arecondensed on the horizontal tubes through which cooling water is flowing. To avoidshort circuit flows a separating baffle plate must be installed. The condensate runningfrom the tubes is collected in the bottom and is returned to the column as reflux or takenout as distillate product.

Advantages: Low holdup in distillate cycle and therefore absolutely necessary in batchdistillations. High-boiling components are washed back by the later condensing light-boiling components.Simple construction with the inert gas outlet on the top outmost and coldest point.Dephlegmator effect by partial condensation of multicomponent mixtures is possible.Disadvantages: The vapors running upward are flowing against the condensed liquidand can therefore cause evaporation of light boiling components in the bottomregion.The light-boiling components accumulate toward the top and are then more difficultto condense because instead of an equilibrium a differential condensation occurs. Itmust be cooled deeper. If the vapor velocities are high, liquid entrainment may occur.

8.1.3 Condensation shell side on a vertical bundle (Figure 8.4(c))The vapors are liquefied on the shell side outside the vertical tubes.

Advantages: This arrangement is required for a vertical thermosiphon evaporator withthe two-phase flow in the tubes and the steam heating on the shell side.Disadvantages: Noncondensable gases cannot be easily removed and therefore cover apart of the heat exchanger surface area. This causes a CO2 corrosion in steam-heatedevaporators.


Segmental baffles cause condensate buildup and constitute trap bags for inert gas.Disc baffles are more adequate.

8.1.4 Condensation in vertical tubes (Figure 8.4(d))The vapors and the condensate flow in cocurrent direction from the top to the bottomthrough the tubes which are cooled in the countercurrent direction on the shell side.

Advantages: Countercurrent arrangement with maximum effective temperaturedifference.Particularly adequate for condensate subcooling and optimal for avoiding differentialcondensation of wide boiling mixtures and streams containing inert gases.The vapors enter the tubes with high velocity so that the heat transfer is improved bythe shear stress, however this occurs at the cost of the pressure loss.Due to the turbulence in the condensate film the heat transfer coefficient with Rey-nolds numbers over 1800 will be clearly better than with the laminar film condensa-tion according to Nusselt.Disadvantages: Not suitable as vacuum condenser because of the limited adiabatic massflow capacity (Section 8.6.3) and not suitable as top condenser because of the requiredconstruction height for gravity reflux.The inevitable subcooling of the reflux is undesirable, because then the first distilla-tion stages in the column must operate as direct heat exchanger.Sufficient baffles must be installed on the shell side in order to enable a cooling watervelocity of 0.8e1.2 m/s.

8.1.5 Vertical countercurrent condensation in reflux coolers on stirredtanks (Figure 8.4(e))

The vapors flow from the bottom to the top through the tubes and the condensate runsback from the top to the bottom. The cooling water flows through the shell chamber.

Advantage: Greatly suitable as reflux cooler on stirred tanks.Disadvantage: Danger of differential condensation because the high-boiling compo-nents are condensed out at the bottom.A check calculation must be performed in order to avoid flooding when the outflow-ing condensate is blocked by the upward flowing vapors (see Section 8.6.7).

8.1.6 Condensation in horizontal tubes (Figure 8.4(f))The vapors are condensed while flowing through the horizontal tubes.

The cooling water flows in countercurrent direction on the shell side.


Advantages: Very good heat transfer coefficients because of the thin condensate filmthickness and the improvement by the shear stress. Also adequate as top condenseror air cooler.Integral or equilibrium condensation because of the equilibrium between vapor andcondensate.Disadvantages: Not adequate for vacuum condensation because of the limited adiabaticmass flow capacity (Section 8.6.3).Subcooling of the condensate cannot be avoided. The collected condensate reducesthe cooling surface. Flooding of some tubes can occur in the case of sudden load var-iations (see Section 8.6.6).Noncondensable gases cover a part of the tube surface.Liquid drops can be entrained by the inert gases in the outlet accumulator if the flowcross-section for the gas is set too low.

8.2 HEAT TRANSFER COEFFICIENTS IN ISOTHERMAL CONDENSATION[1e6,8,9]

8.2.1 Calculation of the heat transfer coefficientsUsing the equations listed in Tables 8.1 and 8.2 the heat transfer coefficients for theisothermal condensation can be easily calculated.

Owing to the fact that the condensate film becomes thicker with increasing conden-sation and the heat transfer coefficient decreases with increasing condensate film thick-ness, a condenser should be calculated zone wise. In common practice the division is in10 zones.

Lumped calculations give an average condensate film thickness if the condensate loadMliq is set to 50% of the total vapor rate W.

The application of the calculation equations is shown in the Example 3.In the condensation in horizontal tubes the gravity-driven heat transfer coefficient de-

creases with increasing vapor rate because the condensate film gets thicker.On the other hand the shear stress-driven heat transfer coefficient increases with

increasing vapor rate. This is shown in Figure 8.5.The shear stress-driven heat transfer coefficient decreases with decreasing vapor

rate. The gravity-driven heat transfer coefficient decreases with increasing condensaterate.

The condenser becomes flooded if the condensate loads in horizontal tubes are toohigh.

In order to avoid this, the condensate height h in the tube at the outlet should notexceed a fourth of the tube diameter di.


Table 8.1 Equations for the calculation of the heat transfer coefficients of isothermal condensationaccording to the Nitsche method

Gravity-driven heat transfer coefficient with laminar condensate film:

Equation 1.1: ag ¼ 1:47� lL �

�1

Reliq

�1=3 ¼ 1:47� l��

gn2� 1

Reliq

�1=3ðW=m2 KÞFor lumped calculations the Reynolds number Re is calculated with Mliq ¼ W

2 ðkg=hÞ

Horizontal in tubes Horizontal at tube bundle

Reliq ¼ Mliq�41800�n�ltube�hliq

Reliq ¼ Mliq�43600�n0:75�ltube�hliq

Vertical in tubes Vertical at tube bundle

Reliq ¼ Mliq�43600�n�p�di�hliq

Reliq ¼ Mliq�43600�n�p�do�hliq

Heat transfer coefficient for vertical condensation with wavy or turbulent condensate film:Wave formation in range: 40 < Re > 1600

Equation 1.2: aw ¼ 1:007� lL �

�1

Reliq

�2=9ðW=m2 KÞTurbulent condensate film from Re ¼ 1600

Equation 1.3: at ¼ 0:01725� lL �

ReliqðRe0:75liq �235Þ�Pr�0:5þ159 ðW=m2 KÞ

Heat transfer coefficient for the shear stress-controlled condensation in the tubesHorizontal in the tubesEquation 1.4

as ¼ 0:024� l

di� Gtot � di

hliq

!0:8

� Pr0:43 �

0BBB@1þ

�rliq

rV

�0:5

2

1CCCAðW=m2 KÞ

Gtot ¼ 4�W

3600� n� p� d2iðkg=m2 sÞ Chosen : a ¼ as if

as

ag> 1:5

If the quotient as/ag is higher than 1.5, as is valid for the calculationVertical in the tubesEquation 1.5

as ¼ 0:0099� l

di� Gm � dihliq

!0:9

� Pr0:5 ��rliq

rV

�0:5

� hV

hliq

!0:1

Gm ¼ffiffiffiffiffiffiffiffiW 2

3

s� 4

3600� n� p� d2iðkg=m2 sÞ

Chosen: as > ag 0 a ¼ asIf as is greater than ag, as is valid for the calculation.


For the condition h/di ¼ 0.25 the allowable condensate loading Wallow according toSection 8.6.6 is determined as follows [7]:

Wallow ¼ 1570� n� r� d2:56i ðkg=hÞ

Table 8.2 Kern equations for the calculation of the heat transfer coefficients [1]

Heat transfer coefficient ag for gravitation-driven isothermal condensation:

Equation 2.1: ag ¼ 0:925� l� r2�gh�G ðW=m2 KÞ

G ¼ condensate loading per unit periphery (kg/m s) with Mliq ¼W/2

Vertical in the tubes Vertical outside tubes

GVi ¼ Mliq

n�p�di�3600 ðkg=m sÞ GVo ¼ Mliq

n�p�do�3600 ðkg=m sÞHorizontal in the tubes Horizontal outside tubes

GHi ¼ Mliq

0:5�n�ltube�3600 ðkg=m sÞ GHi ¼ Mliq

n0:66�ltube�3600 ðkg=m sÞTurbulence correction factor fC for wavy vertical condensation at Re > 40:

Equation 2.2: fC ¼ 0:8��Re4

�0:11

¼ 0:8��GV

h

�0:11

aw ¼ fC � agðW=m2 KÞTurbulent vertical condensation at Re > 2100 with equation 2.3:

Re ¼ 4�Gh

> 2100 at ¼ 0:0076��l3 � r2 � g

h2

�1=3

��4�GV

h

�0:4

ðW=m2 KÞ

Characteristic length L ¼�n2

g

�1=3

Figure 8.5 Shear stress- and gravity-driven heat transfer coefficients in horizontal tubes as functionof vapor rate.


Example 1: Calculation of the allowable condensate load in horizontal tubes

Density r ¼ 614.3 kg/m3

Condenser with 186 tubes 25 � 2, di ¼ 21 mm, L ¼ 4 m, A ¼ 58 m2

Wallow ¼ 1570� 186� 614:3� 0:0212:56 ¼ 9092 kg=hCondenser with 80 tubes 57 � 3.5, di ¼ 50 mm, L ¼ 4 m, A ¼ 57.3 m2

Wallow ¼ 1570� 80� 614:3� 0:052:56 ¼ 36; 036 kg=h

The condenser capacity can be increased with larger tube diameters.In vertical tubes the heat transfer becomes better if the condensate stream gets wavy or turbulent.At high vapor velocities the heat transfer is highly improved by the shear stress.This is shown in Figure 8.6.The heat transfer coefficient reduces with increasing vapor rate because of the thicker condensate film.

For wavy streams the heat transfer coefficient falls less greater and the heat transfer coefficient increaseswith increasing condensate rate for turbulent condensate stream.

Very good heat transfer coefficients are achieved with high vapor velocities due to the shearstress.

As a criterion for the improvement of the heat transfer by the shear force of the vapor flow the Wallisfactor J is used [10].

J ¼ mffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffig� d � rV �

�rliq � rV

�q

An improvement of the heat transfer by the shear stress is achieved at J > 1.5.m ¼ mass stream density (kg/m2 s) ¼ rV � wV

rV ¼ vapor density (kg/m3)rliq ¼ liquid density (kg/m3)d ¼ tube diameter (m)wV ¼ vapor stream flow velocity (m/s)

Figure 8.6 Heat transfer coefficients in vertical tubes as function of the vapor rate.


The mass stream density m is calculated as follows:In the tubes:

m ¼ W

3600� n� ðp=4Þ � d2i¼ wV � rV

�kgm2 s

�

On the shell side:

m ¼ W

B� Di ��1� do

T

� �kgm2 s�

B ¼ baffle spacing (m)Di ¼ shell inner diameter (m)di ¼ inner diameter (m)do ¼ outer diameter (m)n ¼ number of tubesW ¼ vapor rate (kg/h)The condensation on the shell side of horizontal and vertical tube bundles is mostly gravity driven. The

improvement of the heat transfer coefficient by the vapor shear stress in the first tube rows occur only atmass stream densities >30 kg/m2 s.

Example 2: Calculation of the mass stream density m and the Wallis factor J

Data: Shell diameter Di ¼ 0.55 m 186 tubes 25 � 2 L ¼ 4 mPitch T ¼ 32 mm triangular Baffle spacing B ¼ 0.94 mrD ¼ 3.19 kg/m3 rF ¼ 614 kg/m3 Vapor rate W ¼ 10,000 kg/h

Shell side calculation:

m ¼ W

3600� B� Di ��1� da

T

� ¼ 10; 000

3600� 0:94� 0:55��1� 25

32

� ¼ 24:56 kgm2 s

Shell side cross-section fc ¼ B� Di ��1� da

T

�¼ 0:94� 0:55�

�1� 25

32

�¼ 0:113 m2

wV ¼ W3600� rD � fq

¼ 10; 0003600� 3:19� 0:113

¼ 7:7 m=s

m ¼ wV � rV ¼ 7:7� 3:19 ¼ 24:56 kgm2 s

J ¼ mffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffig� d � rV �

�rliq � rV

�q ¼ 24:56ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi9:81� 0:025� 3:19� ð614� 3:19Þp ¼ 1:12


Tube-side calculation:

m ¼ W

3600� n� 0:785� d2i¼ 10; 000

3600� 186� 0:785� 0:0212¼ 43:1kg

m2 s

Tube-side cross-section fT ¼ n� d2i � ðp=4Þ ¼ 186� 0:0212 � 0:785 ¼ 0:06439 m2

wV ¼ 10; 0003600� 3:19� 0:06439

¼ 13:52 m=s

m ¼ wV � rV ¼ 13:52� 3:19 ¼ 43:1 kgm2s

J ¼ 43:1ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi9:81� 0:021� 3:19� ð614� 3:19Þp ¼ 2:15

Example 3: Calculations of the heat transfer coefficients according to theequations listed in Tables 8.1 and 8.2Heat exchanger data: A ¼ 11.7 m2 D ¼ 0.28 m 50 tubes 25 � 2, ltube ¼ 3 mPhysical data: Vapor CondensateDensity (kg/m3) 2.897 720Dynamic viscosity (Pas) 8.28 � 10�6 0.000415 (0.415 mPas)Kinematic viscosity (m2/s) 2.85 � 10�6 0.576 � 10�6 (0.576 mm2/s)Vaporization enthalpy (kJ/kg) 358.1 358.1Heat conductivity (W/m � K) e 0.106Specific heat (kJ/kg K) e 2.1567Prandtl number e 8.44Characteristic length (m) e 32.35 � 10�6

Total vapor rate W ¼ 2000 kg/h cyclohexaneSet for lumped calculation with the average condensate rateMliq ¼ 0.5 � 2000 ¼ 1000 kg/h condensateNotice: The examples are calculated according to the lumped method, i.e., to the average of the

condensate rate.For comparison the results of the zone calculation are also listed in the result comparison.

8.2.1.1 Condensation at a horizontal tube bundleEquation 1.1:

Mfl ¼ 0:5� 2000 ¼ 100 kg=h

Reliq ¼ 1000� 4

3600� 500:75 � 3� 0:415� 10�3 ¼ 47:46

ag ¼ 1:47� 0:106

32:35� 10�6 ��

147:46

�1=3

¼ 1330Wm2 K


Kern equation 2.1:

W ¼ 0:5� 2000 ¼ 1000 kg=h ¼ Mf 1

Gho ¼ 1000500:66 � 3� 3600

¼ 0:007 kg=m s

ag ¼ 0:925� 0:106��

7202 � 9:810:415� 10�3 � 0:007

�1=3

¼ 1182 Wm2 K

8.2.1.2 Condensation at a vertical tube bundleEquation 1.1:

Mliq ¼ 1000 kg=h

Reliq ¼ 1000� 4

3600� 50� p� 0:025� 0:415� 10�3 ¼ 681:8

ag ¼ 1:47� 0:106

32:35� 10�6 � 681:8�1=3 ¼ 547:5 Wm2 K

Since the Reynolds number Refl ¼ 681 is higher than 40 the Equation 1.2 for wavycondensate flow must be used.

Equation 1.2:

aw ¼ 1:007� 0:10632:35� 10�6 � 681:8�2=9 ¼ 774 W

m2 K

The heat transfer coefficient for wavy flow according to Equation 1.2 is higher.Kern equation 2.1:

W ¼ 1000 kg=h ¼ Mfl

GVo ¼ 100050� p� 0:025� 3600

¼ 0:0707 kg=m s

ag ¼ 0:925� 0:106��

7202 � 9:810:415� 10�3 � 0:0707

�1=3

¼ 547 Wm2 K

Controlling the Reynolds number:

Re ¼ 4� 0:07070:415� 10�3 ¼ 641 > Re ¼ 40

For Re > 40 the turbulence correction is recommended according to Equation 2.2:

fc ¼ 0:8��Re4

�0:11

¼ 0:8��6814

�0:11

¼ 1:408

aw ¼ 1:408� 547 ¼ 770 Wm2 K


8.2.1.3 Condensation in horizontal tubesEquation 1.1:

Mliq ¼ 1000 kg=h

Reliq ¼ 1000� 4

1800� 50� 3� 0:415� 10�3 ¼ 35:7

ag ¼ 1:47� 0:106

31:35� 10�6 ��

135:7

�1=3

Wm2 K

ag ¼ 1:47� 0:106�

9:81

ð0:576� 10�6Þ2 � 35:7

!1=3

¼ 1463 Wm2 K

Calculation according to Equation 1.4 with vapor shear stress

wV ¼ 4� 1000

3600� 2:897� p� 50� 0:0212¼ 5:53 m=s

Gtot ¼ 4� 2000

3600� 50� p� 0:0212¼ 32:08 kg

m2 s

as ¼ 0:024� 0:1060:021

��32:08� 0:0210:415� 10�3

�0:8

� 8:440:43

�1þ

�7202:897

�0:5

2¼ 940 W

m2 K

The higher value is valid ag ¼ 1463 W/m2 K.The Wallis factor J is clearly lower than 1.5.

m ¼ wV � rV ¼ 5:53� 2:897 ¼ 16 kgm2 s

J ¼ 169:81� 0:021� 2:897� ð720� 2:897Þ ¼ 0:77 < 1:5

Kern equation 2.1:

W ¼ 1000 kg=h ¼ Mliq

GHi ¼ 10000:5� 50� 3� 3600

¼ 0:0037 kg=m s

ag ¼ 0:925� 0:106��

7202 � 9:810:415� 10�3 � 3:7� 10�3

�1=3

¼ 1461Wm2 K


8.2.1.4 Condensation in vertical tubesEquation 1.1:

Mliq ¼ 1000 kg=h

Reliq ¼ 1000� 4

3600� 50� p� 0:021� 0:415� 10�3 ¼ 811:6

ag ¼ 1:47� 0:106

31:35� 10�6 ��

1811:6

�1=3

¼ 516 Wm2 K

Calculation with equation 1.2 for Re > 40 for wavy condensate film:

aw ¼ 1:007� 0:10632:35� 10�6 � 811:6�2=9 ¼ 744 W

m2 K

Calculation according to equation 1.5 for shear stress

Gav ¼ffiffiffiffiffiffiffiffiffiffiffiffi20002

3

s� 4

3600� 50� p� 0:0212¼ 18:52 kg

m2 s

as ¼ 0:0099� 0:1060:021

��18:52� 0:0210:415� 10�3

�0:9

� 8:440:5 ��

7202:897

�1=2

��8:28� 10�6

0:415� 10�3

�0:1

¼ 730 Wm2 K

It is valid: aw ¼ 744 W/m2 KKern equation 2.1 for vertical condensation in tubes: Mliq ¼ 1000 kg/hCalculation for laminar condensate film:

GVi ¼ 100050� p� 0:021� 3600

¼ 0:0842 kg=m s

ag ¼ 0:925� 0:106��

7202 � 9:810:415� 10�3 � 0:0842

�1=3

¼ 516 Wm2 K

With turbulence correction factor fc according to Equation 2.2 for turbulence

Re ¼ 4� 0:0842

0:415� 10�3 ¼ 811:6Re4

¼ 811:64

¼ 202:9

fc ¼ 0:8� 202:90:11 ¼ 1:43

at ¼ 1:43� 516 ¼ 740 Wm2 K


Comparison of result for 2 t/h vapors according to the Nitsche and the Kern method

Lumped Zones

Horizontal at the tube bundle Nitsche a ¼ 1330 W/m2 K a ¼ 1422 W/m2 KKern a ¼ 1182 W/m2 K a ¼ 1205 W/m2 K

Vertical at the tube bundle Nitsche a ¼ 774 W/m2 K a ¼ 815 W/m2 KKern a ¼ 770 W/m2 K a ¼ 822 W/m2 K

Horizontal in the tubes Nitsche a ¼ 1463 W/m2 K a ¼ 1564 W/m2 KKern a ¼ 1461 W/m2 K a ¼ 1490 W/m2 K

Vertical in the tubes Nitsche a ¼ 744 W/m2 K a ¼ 794 W/m2 KKern a ¼ 740 W/m2 K a ¼ 788 W/m2 K

8.2.2 Parameters influencing the heat transfer coefficientsThe heat transfer in film condensation is dependent on the condensate film thickness, theturbulence in the condensate film and the vapor shear stress.

In laminar region the heat transfer is only dependent on the heat conductionthrough the film thickness d and the gravitation-driven heat transfer coefficient ag isas follows:

ag ¼ l

d

�Wm2 K

�

In condensation in horizontal tubes and on a horizontal tube bundle there are thincondensate films and therefore the heat transfer coefficients are very high.

The film becomes thicker with increasing condensate load and the a-values fall.In Figure 8.7 it is shown that the heat transfer coefficient in the individual zones of a

condenser falls because the condensate film becomes thicker with increasing zonenumber.

In condensation in vertical tubes or on vertical surfaces the condensate film getsthicker downward.

This makes the heat transfer in laminar region worse but with increasing film thick-ness from Re > 40 waves built up on the condensate film and improves the heat transfer.

The heat transfer coefficient decreases much less with increasing condensate load.From a Reynolds number Re ¼ 1600 the turbulences in the condensate film are so

large that the heat transfer coefficient increases.With larger condenser vapor load or higher Reynolds numbers of the condensate the

heat transfer in horizontal and vertical tubes is considerably improved by the vapor shearstress or the turbulent flow.

In Figure 8.8 the heat transfer coefficients of ethanol and cyclohexane as a function ofthe vapor rate is given.

A distinction is made between the shear stress-controlled and the gravitation-drivenheat transfer.


In gravitation region, three different flow types are formed depending on the conden-sate Reynolds number.

Zone I up to Re ¼ 40 with laminar condensate flow without wavesZone II between Re ¼ 40 and Re ¼ 1600 with laminar wavy flowZone III from Re ¼ 1600 with turbulent condensate flowIn zone I and II the heat transfer coefficient falls with increasing vapor rate because the

condensate film becomes thicker. In section II the fall of the heat transfer coefficients isless than in section I because the heat transfer is improved by the waviness of the conden-sate film.

In zone III the heat transfer coefficient increases because the heat transfer resistancegets reduced by the turbulent condensate stream.

The curves 2 and 4 illustrate the improvement of the heat transfer by the vapor shearstress at larger vapor rates or higher flow velocities.

It also comes to an improvement of the heat transfer in condensation in horizontaltubes at high vapor loads by the vapor shear stress.

The condensation on the shell side of horizontal and vertical tube bundles is mostlygravitational driven. It comes to an improvement of the heat transfer by the vapor shearstress in the first rows only at a mass stream density of m > 30 kg/m2 s.

Figure 8.7 Heat transfer coefficients of ethanol and cyclohexane in the individual zones of acondenser.


8.3 COMPARISON OF DIFFERENT CALCULATION MODELS

In a technical study, different calculation models for the determination of heat transfercoefficients at isothermal condensation for a certain water-cooled condenser were used.

The results are listed in Table 8.3.Condenser dimensions:Shell diameter 270 mm with 50 tubes 25 � 2, 6 m long, A ¼ 23.5 m2

Condensed material: Cyclohexane Cooling: cooling water 25/35 �C

Heat resistance by fouling and heat conductivity through the wall: 0.00035.In the table there is a distinction between the lumped and the zone wise calculation.In the lumped calculation 50% of the total flow rate for the condensate load is input as

average for the condensate load because the condensate rate of zero at the inlet rises up to100% at the outlet.

An average condensate film thickness results from the average value between inlet andoutlet, that is, at 50% of the total vapor rate.

Figure 8.8 Heat transfer coefficients of ethanol and cyclohexane in the condensation in vertical tubes.


In the zone wise approach the condenser is subdivided into 10 sections and for eachzone the heat transfer coefficient for the condensate and vapor loading in this section iscalculated and the required heat exchanger area for the zone is determined.

The heat transfer coefficients determined according to the lumped method are poorerin the gravitation-driven condensation because the average condensate film is thickerthan that in the zone wise calculation. If however the shear stress determines the heattransfer, a too high value can result from the lumped method because of the greater vaporrate.

A zone wise calculation is recommended for the design.The results of the calculated heat transfer coefficients with different models are shown

in the following figures:Figure 8.9: In condensation on horizontal tube bundle the results according to the

four models are almost identical.Figure 8.10: In condensation in horizontal tubes the results according to

VDI-W€armeatlas, Kern, and Nitsche only deviate marginally from another.Figure 8.11: In condensation in vertical tubes the methods according to VDI and

Nitsche are recommended since both consider the improvement by the shear stress.The models according to Kern and Huhn give too low heat transfer coefficients.The Colburn method should be avoided.

Table 8.3 Heat transfer coefficients in isothermal film condensation


5000

4000

3000

2000

1000

0

Condensate Rate (kg/h)

Hea

t Tra

nsfe

r Coe

ffici

ent (

W/m

2 K)

1000 2000 3000 4000 5000

1 VDI-Wӓrmeatlas2 Kern3 Nitsche4 TPLUS-Program

14

32

Figure 8.9 Heat transfer coefficients for the condensation on horizontal tube bundle. (Literaturesources: Colburn, Applied Mechanics Reviews, Vol. 5, 1952, Rev. 247; Huhn, VerfahrenstechnischeBerechnungsmethoden Teil1, VCA Verlag, 1987; Kern, Process Heat Transfer, McGraw Hill, N.Y., 1950; VDI,VDI-W€armeatlas, Ja1 e Ja19, VDI-Verlag, 1994.)

Figure 8.10 Heat transfer coefficients for the condensation in horizontal tubes. (Literature sources:Colburn, Applied Mechanics Reviews, Vol. 5, 1952, Rev. 247; Huhn, Verfahrenstechnische Berechnungs-methoden Teil1, VCA Verlag, 1987; Kern, Process Heat Transfer, McGraw Hill, N.Y., 1950; VDI,VDI-W€armeatlas, Ja1 e Ja19, VDI-Verlag, 1994.)


Figure 8.12: The calculation methods according to VDI, Huhn, and Nitsche givesimilar results.

Figure 8.13: In the calculation of the overall heat transfer coefficient U the differentvalues of the heat transfer coefficients calculated according to different models aresmoothed to a great extent by the heat transfer coefficient on the cooling water sideand the fouling factors.

8.4 CONDENSATION OF VAPORS WITH INERT GAS [2]

Small amounts of noncondensable inert gases in the vapors to be condensed drasticallyreduce the heat transfer coefficient for the condensation of the vapors because the vaporsto be condensed must diffuse to the cooled condensation surface by the noncondensableinert gas (see Figure 8.14).

The inert gas accumulates to the cooling wall because the vapors condense out so thatan inert film builds up on the surface.

The vapors must diffuse through this gas film to the cold condensation surface.This prevention in transportation deteriorates the heat transfer coefficient, for

instance, for vapor in the presence of 1e10 Vol% air to 10e30% of the heat transfercoefficient of saturated vapor without inert gas.

Figure 8.11 Heat transfer coefficients for the condensation in vertical tubes. (Literature sources:Colburn, Applied Mechanics Reviews, Vol. 5, 1952, Rev. 247; Huhn, Verfahrenstechnische Berechnungs-methoden Teil1, VCA Verlag, 1987; Kern, Process Heat Transfer, McGraw Hill, N.Y., 1950; VDI,VDI-W€armeatlas, Ja1 e Ja19, VDI-Verlag, 1994.)


Figure 8.12 Heat transfer coefficients for the condensation in vertical tube bundle. (Literature sources:Colburn, Applied Mechanics Reviews, Vol. 5, 1952, Rev. 247; Huhn, Verfahrenstechnische Berechnungs-methoden Teil1, VCA Verlag, 1987; Kern, Process Heat Transfer, McGraw Hill, N.Y., 1950; VDI,VDI-W€armeatlas, Ja1 e Ja19, VDI-Verlag, 1994.)

Figure 8.13 Overall heat transfer coefficients for the different calculation models. (Literature sources:Colburn, Applied Mechanics Reviews, Vol. 5, 1952, Rev. 247; Huhn, Verfahrenstechnische Berechnungs-methoden Teil1, VCA Verlag, 1987; Kern, Process Heat Transfer, McGraw Hill, N.Y., 1950; VDI,VDI-W€armeatlas, Ja1 e Ja19, VDI-Verlag, 1994.)


In condensing out of the vapors the inert gas partial pressure toward the condensationsurface increases and the vapor partial pressure decreases.

The reducing vaporsdpartial pressure results in the lowering of the dew or conden-sation temperature (see Figure 8.14).

Therefore, the effective temperature difference for the condensation is decreased andhence the condensation efficiency.

The inert gas functions as carrier for the vapors and deteriorates the efficiency in theliquidizing.

It must be considerably cooled deeper in order to liquefy the vapors. This is shown inFigure 8.15.

Ptot = Total Pressure

Partial Vapor Pressure

Partial Inert gas Pressure

Δ tidealΔ trealTK

TW

PK

PGK

TW

TDew

TVaporGas film

Condensatefilm

Tubewall

Water

Figure 8.14 Condensation with inert gas.

Figure 8.15 Condensed xylene rate as a function of the temperature.


Therefore, there are three effects that make the condensation efficiency worse in thepresence of inert gases:1. The transportation prevention by the gas film reduces the a-value for the

condensation.2. The reduction of the dew point reduces the effective Dt for the heat transfer.

Dtreal < Dtideal

Example for the reduction of dew point by inert gasTotal pressure: 950 mbarAir content H2Oepartial pressure Condensation temperatureVol% mbar �C0 950 98.25.2 900 96.7110.5 850 95.1515.8 800 93.5121 750 91.78

3. It must be cooled considerably deeper in order to condense out the vapors(Figure 8.15).In Figure 8.14 it is shown how the vapor partial pressure and the condensation

temperature reduce toward the cooling wall and how the gas film builds up by theincrease of the inert gas partial pressure.

The calculation method according to Colburn and Drew is very complex and is nor-mally performed with a computer program.

An example for a hand calculation is given in Hewitt [3].For hand calculations the following very simple calculation method for the determi-

nation of the heat transfer coefficient in the presence of inert gas can be adopted.

acorr ¼ 1Qgas

Qtot�agasþ 1

acond

Wm2 K

Qgas ¼ convective heat duty for gaseor vapor coolingQtot ¼ total heat duty for the cooling and condensingagas ¼ heat transfer coefficient for the convective coolingacond ¼ heat transfer coefficient for the condensationIn addition to the condensation it must convectively cooled along the dew line and

the heat transfer coefficient for the condensing is drastically reduced because of the lowheat transfer coefficients for convective cooling of gases and vapors.

Example 4: Condensation of 2000 kg/h xylene vapors with 40 kg/h nitrogenThe condensation of pure xylene vapors occurs at the dew point of xylene at 143.9 �C.

The heat transfer coefficient is acond ¼ 1680 W/m2 K and the overall heat transfer coefficient

U ¼ 632 Wm2 K


In cooling with cooling water from 25 to 35 �C a driving temperature difference of 113.8 �C results forthe isothermal condensation of xylene vapor.

Hence the required heat exchanger surface area A for Q ¼ 195 kW is determined:

A ¼ QU� Dt

¼ 195; 000632� 113:8

¼ 2:7 m2

For a mixture of 2000 kg/h xylene and 40 kg/h nitrogen the following calculation results:The mixture must be cooled down to 40 �C in order to liquefy the 2000 kg/h xylene.This reduces the effective temperature difference from 113.8 to 71.7 �C.Calculation of the corrected heat transfer coefficient: acorr

Qgas/Qtot ¼ 0.316 agas ¼ 195 W/m2 K acond ¼ 1680 W/m2 K

acorr ¼ 10:316195 þ 1

1680

¼ 451 Wm2 K

Calculation of the overall heat transfer coefficient U with aWater ¼ 2956 W/m2 K

1U

¼ 12956

þ 1451

þ 0:00034 U ¼ 345 Wm2 K

Required heat exchanger surface area A for Q ¼ 312 kW:

A ¼ 312; 00071:7� 345

¼ 12:6 m2

8.5 CONDENSATION OF MULTICOMPONENT MIXTURES (SEECHAPTER 7)

For the design of condensers for multicomponent mixtures the condensation curvebetween the dew and bubble point is needed. The condensation function is not isotherm.

The calculation of such condensation curves of multicomponent mixtures is not sim-ple, especially for nonideal mixtures.

The condensation curve for the binary mixture benzeneetoluene is shown inFigure 8.16.

The vapors enter the condenser with y1 ¼ 30 mol% benzene. The dew point is at104 �C.

The out-condensed liquid is enriched with the high-boiling component toluene andthe composition of the lighter component benzene drops from y1 in the vapor to x1 inthe condensate. Due to the preferred out-condensation of the high-boiling toluene thebenzene composition of the vapors rises to y2 with the equilibrium composition x2 in thecondensate.

After cooling to the boiling point at 98 �C the vapor composition y1 and the exitingcondensate have the same composition as the vapors at the inlet: y1 ¼ x3.

The temperature drop from dew point to bubble point has an influence on the effec-tive temperature difference for the heat transfer.


In nonlinear shape of the temperature in the temperature heat load diagramthe average weighted temperature difference, WMTD, must be determined (seeExample 8).

In comparison with the isothermal condensation of individual components the heattransfer coefficients in condensing of mixtures are clearly worse because the light-boilingcomponents on the cooling surface prevent the condensing of the higher boiling com-ponents or because the mixture must be convectively cooled from the dew point to thebubble point.

The deterioration of the heat transfer coefficients is shown in the followingoverview.

Components Benzene Benzene þ toluene Benzene þ toluene þ xyleneHeat transfer coefficient(W/m2 K)

1983 1662 1410

Components Butane Butane þ toluene Butane þ toluene þ xyleneHeat transfer coefficient(W/m2 K)

2211 1405 1218

Figure 8.16 Bubble and dew line for the mixture of benzeneetoluene.


2500

2000

1500

1000

500

Butane

Heat T

ransfe

r C

oe

ffic

ien

t (W

/m2K

)

1852

2211

1405

1218

Butane +

Benzene

Butane +

Toluene

Butane + Toluene

+ Xylene

0

The heat transfer coefficient for the convective cooling of gases or vapors is muchworse than the a-value for the condensing.

Therefore the corrected a-value considering the convective cooling and condensa-tion is less than the heat transfer coefficient for pure condensing.

The deterioration of the heat transfer coefficient by the convective cooling is depen-dent on the heat load Qgas for the gas or vapor cooling.

An improvement of the convective heat transfer coefficient for the vapor cooling isachieved by the higher flow velocity of the vapors.

In the following it is shown how to estimate the correction a-value for the conden-sation with a simple method if the ratio Qgas/Qtot is known.

In Figure 8.17 the reduction of the heat transfer coefficient with increasing gas cool-ing is depicted.

acorr ¼ acond

1þ Qgas�acond

Qtot�agas

¼ 1Qgas

Qtot�agasþ 1

acond

�Wm2 K

�

Qgas ¼ convective heat duty for the gas and vapor coolingQtot ¼ total heat duty for the cooling and condensationagas ¼ heat transfer coefficient for the convective coolingacond ¼ heat transfer coefficient for the condensation

Example 5: Condensation of a multicomponent mixture considering the gascooling

agas ¼ 50 W/m2 K acond ¼ 2500 W/m2 K Qgas/Qtot ¼ 0.05

acorr ¼ 10:0550

þ 12500

¼ 714 W=m2 K acorr ¼ 2500

1þ 0:05� 250050

¼ 714 W=m2 K

By the vapor cooling the high a-value of 2500 W/m2 K for the condensation is reduced to 714 W/m2 K.


Example 6: Condensation of a mixture of 1000 kg/h benzene and 1000 kg/hxylene

Dew point: 121.8 �C Bubble point: 96.1 �CQgas ¼ 25.5 kW Qtot ¼ 238 kW Qgas/Qtot ¼ 0.109acond ¼ 2000 W/m2 K agas ¼ 238 W/m2 K

acorr ¼ 10:109238 þ 1

2000

¼ 1041 Wm2 K

The convective cooling from the dew point 121.8 �C to the bubble point of 96.1 �C reduces the heattransfer coefficient from 2000 to 1041 W/m2 K.

Figure 8.17 Heat transfer coefficients for the condensation as function of the partial gas cooling.


Example 7: Condensation of a hydrocarbon mixture containing inert gas

Nitrogen 50 kg/h Inlet: 132 �CPentane 1000 kg/h Outlet: 30 �COctane 2000 kg/h Condensed: 5968 kg/h liquidNonane 3000 kg/h Gas outlet: 82 kg/hTotal: 6050 kg/h

Qgas ¼ 350.2 kW Qtot ¼ 881.2 kW Qgas/Qtot ¼ 0.397 A ¼ 66.6 m2

agas ¼ 260 W/m2 K acond ¼ 1500 W/m2 K WMTD ¼ 45.26 �C

acorr ¼ 10:397260 þ 1

1500

¼ 455:5 Wm2 K

Cooling water: a ¼ 2690 W/m2 K

1U

¼ 12690

þ 1455:5

þ 0:00034 U ¼ 344 Wm2 K

A ¼ Qtot

U�WMTD¼ 881; 200

344� 42:26¼ 60:6 m2z10% reserve

Example 8: Condensation of a mixture of 500 kg/h pentane and 4000 kg/hdecane

Dew point: 167.6 �C Bubble point: 96.1 �C Heat duty Qgas ¼ 588 kW

Cooling water: a ¼ 4900 W/m2 K.

acond ¼ 1700 W/m2 K agas ¼ 370 W/m2 K Qgas/Qtot ¼ 0.366 A ¼ 14 m2

Calculation of the corrected heat transfer coefficient:

acorr ¼ 10:366370 þ 1

1700

¼ 634 Wm2 K

The heat transfer coefficient deteriorates from 1700 to 634 W/m2 K by the convective cooling of thevapors from dew point to bubble point.

Calculation of the overall heat transfer coefficient:

1U

¼ 14900

þ 1634

þ 0:00034 U ¼ 471 Wm2 K

Calculation of the average weighted temperature difference, WMTD:Due to the nonlinear shape of the temperature over the heat duty the weighted

average temperature must be determined (see Chapter 2).Hence the temperatureeheat duty curve shown in Figure 8.18 is then subdivided

into four zones with an approximate linear shape.


For each zone the heat duty and the corrected temperature difference, CMTD, isdetermined and then the average weighted temperature difference is calculated.

Calculation of the weighted temperature difference, WMTD:

Q1 ¼ 81.3 kW CMTD1 ¼ 136.5 �C Q1/CMTD1 ¼ 0.595Q2 ¼ 120 kW CMTD2 ¼ 132.2 �C Q2/CMTD2 ¼ 0.908Q3 ¼ 106 kW CMTD3 ¼ 123.1 �C Q3/CMTD3 ¼ 0.86Q4 ¼ 274 kW CMTD4 ¼ 89.35 �C Q4/CMTD4 ¼ 3.066H Q ¼ 588 kW H Q/CMTD ¼ 5.43

WMTD ¼P

QPQ=CMTD

¼ 5885:43

¼ 108:3 �C

Under the assumption of a linear temperature curve a CMTD ¼ 97.5 �C results.This value is wrong!Calculation of the required area:

A ¼ 588; 000471� 108:3

¼ 11:5 m2

Differential and integral condensation of multicomponent mixturesThe condensation on horizontal tube bundles is not an equilibrium or integral conden-sation in which vapor and liquid are in equilibrium. The high-boiling components thatare condensed out at the inlet of the condenser exit from the tubes at the bottom of theheat exchanger and no longer participate in equilibrium.

Figure 8.18 Temperatureeheat load curve for the pentaneedecane mixture.


This differential or nonequilibrium condensation is subjected to worse condensationconditions.

The light-boiling components build up in the condensation process and reduce thedew point.

The effective temperature difference becomes smaller and it must be cooled deeper.This is highly predominant in condensation of wide boiling components, for instance,

ammoniaewater.In Figure 8.19 an equilibrium condenser with condensation in vertical tubes and a

differential condenser with stage condensation on horizontal tube bundle is shown.In Figure 8.20 the condensation process for a mixture of the light hydrocarbons pro-

pane, butane, and pentane in air in integral and stage condensation is depicted.In the stage condensation it must be cooled much deeper in order to liquefy the gas-

oline vapor.

8.6 MISCELLANEOUS

In the following, some problems in the selection and design of condensers are treated.

8.6.1 Influence of pressure loss in the condenserDew and bubble point of multicomponent mixtures are determined by the compositionof the vapors and the pressure.

Due to the pressure loss in the condenser the pressure in the system drops.Therefore the dew and bubble point are lowered.This reduces the effective temperature difference for the condensation.

Example 9: Condensation of 2 t/h benzene and 2 t/h toluene with cooling water25/35 �C at different pressures

Pressure ¼ 1 bar: Dew point ¼ 97.48 �C Bubble point ¼ 90.94 �C CMTD ¼ 64.02 �CPressure ¼ 0.95 bar: Dew point ¼ 95.68 �C Bubble point ¼ 89.12 �C CMTD ¼ 62.2 �CPressure ¼ 0.5 bar: Dew point ¼ 74.56 �C Bubble point ¼ 67.76 �C CMTD ¼ 40.9 �C

8.6.2 Condensate drain pipeThe exit stream from the condenser to the condensate accumulator should be self-aerated.

This is achieved with a Froude number Fr < 0.31.

Fr ¼ wffiffiffiffiffiffiffiffiffiffig � d

p ¼ 2:4��Hd

�1:5

< 0:31


Figure 8.19 Integral condenser with equilibrium condensation and differential condenser with stage-wise nonequilibrium condensation.


The required nozzle diameter d for a condensate exit rate V (m3/h) is determined asfollows:

d ¼ 0:0422� V 0:4ðmÞThe ratio liquid height H over the nozzle to the nozzle diameter is thereby

H/d ¼ 0.25.In order to avoid after evaporation by the static negative pressure in the exit pipe a

liquid height Hreq over the exit nozzle is necessary.

Hreq ¼ 2� w2

g

w ¼ liquid velocity (m/s)

Example 10: Required condensate exit pipe diameter for V ¼ 4 m3/h condensate

d ¼ 0.0422 � 40.4 ¼ 0.0735 m ¼ 73.5 mm H ¼ 0.25 � 0.0735 ¼ 0.018 m ¼ 18.4 mm

w ¼ 4=3600

0:07352 � ðp=4Þ ¼ 0:263m=s

Fr ¼ 0:263ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi9:81� 0:0735

p ¼ 0:31 ¼ 2:4��18:473:5

�1:5

¼ 0:3

Hreq ¼ 2� 0:2632

9:81¼ 0:014 m ¼ 14 mm

Figure 8.20 Condensation process for the integral and differential condensation of a gasoline vapormixture in an airehydrocarbon mixture.


8.6.3 Maximal vapor flow capacityIn condensation in vertical or horizontal tubes the maximal adiabatic mass flux G mustbe examined at the inlet from a large vapor pipe into the several small pipes of thecondenser. The adiabatic flow capacity is calculated with the adiabatic exit flow equationfor nozzles.

G ¼ ε� F � j� ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi2� P1 � r1

p ðkg=sÞ

j ¼ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffik

k� 1�"�

P2P1

�2=k

��P2P1

�kþ1k

#vuut

F ¼ flow cross-section (m2)G ¼ adiabatic flow capacity (kg/s)P1 ¼ pressure before the inlet into the pipes (Pa)P2 ¼ pressure at the outlet from the tubes (Pa)ε ¼ flow contraction (0.6e0.8)k ¼ adiabatic exponentj ¼ adiabatic flow factorThe condensation efficiency is limited by the adiabatic flow capacity especially in

condensation in vacuum.

Example 11: Adiabatic flow capacity of a condenser DN 700 with 364 tubes25 � 2

Vapor mole weight M ¼ 200 Gas density r1 ¼ 0.129 kg/m3 k ¼ 1.4P1 ¼ 2000 Pa P2 ¼ 1500 Pa P2/P1 ¼ 0.75j ¼ 0.4279 3¼ 0.8

Calculation of the adiabatic flow capacity:

G ¼ 0:8� 0:000346� 0:4279�ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi2� 2000� 0:129

p¼ 0:00269 kg=s pro Rohr ¼ 9:68 kg=h

Maximum total throughput in 364 tubes: 3525 kg/hCalculation of the condensation performance M (kg/h):A ¼ 114.3 m2 area in condenser (L ¼ 4 m)

U ¼ 600 W/m2 K LMTD ¼ 25 �C Latent heat ¼ 150 Wh/kg


Q ¼ U� A� LMTD ¼ 600� 114:3� 25 ¼ 1714; 500 W

M ¼ Qr¼ 1714; 500

150¼ 11; 430 kg=h

Thermally 11,430 kg/h can be condensed, but through the adiabatic flow capacity the throughput islimited to 3525 kg/h.

8.6.4 Top condensers on columns [11]The arrangement of condensers at the top of distillation columns has some advantages(Figure 8.21):

The pumping of the reflux to the column top is not necessary.No pumping is required and no pipe lines and no steel supports for the condenser andthe accumulator.No energy is required for the back pumping of the reflux.Due to the elimination of the long vapor pipes the pressure loss is reduced.This is a great advantage especially in vacuum distillations.The low holdup in the condenser simplifies the separation in discontinuous batchdistillation.Most important for the proper function of a top condenser is the correct hydraulic

design of the reflux line. It is a process of gravity-driven boiling liquid from the condenserto the column.

The driving height H must be greater than the pressure loss between the pointsA and B.

H is the difference in height between the condenser outlet nozzle of the condenserand the inlet nozzle of the column. The vapor density rV is neglected.

H � rliq � g ¼ DPcond þ DPRL þ DPM þ DPCV

A

B

p2p1

H

Z1

Z1

d

DZ2

Z2

a. Z1 > Z2 b. Z1 < Z2, D > d

Figure 8.21 Arrangement of horizontal and vertical top condensers [11].


DPcond ¼ pressure loss in the condenser (Pa)DPRL ¼ pressure loss in the reflux line (Pa)DPM ¼ pressure loss of a flow measuring device (Pa)DPCV ¼ pressure loss of a control valve (Pa)In order to avoid large additional heights for the increase of the difference in heightH

the pressure losses in the reflux line is minimized. Flow measuring devices with low pres-sure loss are used and only in special cases a control valve is installed. The reflux line isdimensioned for a self-venting flow.

Example 12: Design of a reflux line for V ¼ 28.26 m3/hCalculation of the diameter for a self-venting flow:

d ¼ 0:0422� 28:260:4 ¼ 0:160 m ¼ 160 mm

Calculation of the pressure loss for the reflux:Flow velocity w ¼ 0.39 m/s.

Pipe length L ¼ 10 m Liquid density r ¼ 800 kg/m3

DPR ¼�0:04� 10

0:16

�� 0:392 � 800

2¼ 152 Pa

Pressure loss in the condenser and in the flow measuring device:

DPcond ¼ 200 Pa DPM ¼ 6000 PaTotal pressure loss H DP ¼ 200 þ 152 þ 6000 ¼ 6352 Pa

Required height difference Hreq:

Hreq ¼P

DPg� rliq

¼ 63529:81� 800

¼ 0:81 m Chosen: H ¼ 1.5 m

Due to load variations in the column disturbances in the hydraulic equilibrium and the reflux variationscan occur. This reduces the efficiency of the fractionation.

The reflux variations can be prevented by installing a control valve in the reflux line, however sufficientheight H for the additional pressure loss in the control valve must be provided.

Example 13: Design of a reflux line with control valveData of Example 12

Additional pressure loss in the control valve DPCV ¼ 0.2 bar ¼ 20,000 PaH DP ¼ DPcond þ DPRL þ DPM þ DPCV ¼ 200 þ 152 þ 6000 þ 20,000 ¼ 26,352 Pa

Hreq ¼ 26; 3529:81� 800

¼ 3:36 m


In order to reduce the required height H a flow measuring device and a control valve of low pressureloss is used.

DPM ¼ 1000 Pa DPRV ¼ 5000 Pa∍DP ¼ DPKon þ DPR þ DPM þ DPRV ¼ 200þ 152þ 2000þ 6000 ¼ 8352 Pa

Hreq ¼ 83529:81� 800

¼ 1:06 m Chosen: H ¼ 1.5 m

The chosen driving height H should always be substantially higher than the calculated requiredheight Hreq.

A locking siphon is installed at the reflux inflow in the column in order to prevent vapor short circuit.The height Z1 must be greater than Z2 or the diameter D of Z2 must be greater than the diameter of thesiphon in order to prevent the canceling out of the siphon.

This lock can cause problems if after condensing two liquid phases with severely different densitiesform.

Siphon in flow disturbances occurs if the two siphon arms are filled with liquids of different weights.This is why draining facilities must be provided.

8.6.5 Calculation of the pressure lossDPT in condensing in the tubes [4]The friction pressure loss DPT is calculated as follows:

DPT ¼ w2m � r

2��f � L

dþ 1:5

�ðPaÞ

wm ¼ win þ wout

2ðm=sÞ

d ¼ tube diameter (m) f ¼ friction factor L ¼ tube length (m)wm ¼ average flow velocity (m/s) win ¼ inlet velocity (m/s)wout ¼ outlet velocity (m/s) rV ¼ gas density (kg/m3)

Due to the condensation of the vapors a pressure regain is achieved by the negativeacceleration.

The vapor velocity is reduced to 0 of the inlet stream velocity.The pressure regain by the negative acceleration is calculated as follows:

DPacc ¼ m� ðwout � winÞ

m ¼ M

3600� d2 � p=4� n¼ win � r

�kgm2 s

�

M ¼ vapor inlet rate (kg/h)m ¼ mass stream density (kg/m2 s)n ¼ number of tubes


Example 14: Pressure loss calculation in a condenserM ¼ vapor inlet rate to condenser ¼ 4 t/h

Horizontal condenser with 72 tubes25 � 2. 4 m long

aTube ¼ 0.0249 m2

rliq ¼ 720 kg/m3 rV ¼ 1.623 kg/m3 Friction factor f ¼ 0.04

GV ¼ vapor rate at the inlet ¼ 4000/1.623 ¼ 2464.5 m3/h

win ¼ GV3600� aTube

¼ 2464:53600� 0:0249

¼ 27:5 m=s wout ¼ 0 m=s

wm ¼ 27:5þ 02

¼ 13:75 m=s

Calculation of the friction pressure loss DPT:

DPR ¼ 13:752 � 1:6232

��0:04� 4

0:021þ 1:5

�¼ 1399 Pa ¼ 14 mbar

Calculation of the pressure regain DPacc:

m ¼ 27:5� 1:623 ¼ 44:6 kgm2 s DPacc ¼ 44:6� ð0� 27:5Þ ¼ 1226 Pa ¼ 12:3 mbar

Effective pressure loss DPeff ¼ DPT � DPacc ¼ 14 � 12.3 ¼ 1.7 mbar.

8.6.6 Maximum condensate load in horizontal tubes [7]If the condensate load is too high in horizontal tubes the condenser becomes flooded.

In order to avoid this, the condensate height h in the tube at the outlet should notexceed a fourth of the tube diameter di.

The allowable condensate load Wall for the condition di ¼ 0.25 is determined asfollows:

Wall ¼ 1570� n� r� d2:56i ðkg=hÞn ¼ number of tubesr ¼ condensate density (kg/m3)

Example 15: Allowable condensate load in horizontal tubesCondenser with 186 tubes 25 � 2

di ¼ 21 mm L ¼ 4 m A ¼ 58 m2 r ¼ 614.3 kg/m3

Wall ¼ 1570� 186� 614:3� 0:0212:56 ¼ 9092 kg=h m ¼ 39:2 kg=m2 sCondenser with 80 tubes 57 � 3.5di ¼ 50 mm L ¼ 4 m, A ¼ 57.3 m2 r ¼ 614.3 kg/m3

Wall ¼ 1570� 80� 614:3� 0:052:56 ¼ 36; 036 kg=h m ¼ 408 kg=m2 s

By the selection of larger pipe diameters the capacity in condensing in horizontal tubes can beincreased.


8.6.7 Flood loading in reflux condensers [4,12e14]In the vertical reflux condensers installed on stirred tanks for the boiling in the reflux theallowable flood loading must be checked alongside with the thermal design. In theseapparatus the vapors flow upward and the condensate runs downward in the oppositedirection to the vapors.

Congestion of the condensate occurs if the vapor velocity is too high.

The allowable vapor load G is determined according to English and others asfollows:

G ¼ 246� d0:3i � r0:46liq � s0:09 � r0:5V

h0:14liq � cos d0:32 � ðC=V Þ0:07�kgm2 h

�

Gcond ¼ G � Atubeðkg=hÞThe following is valid for the allowable vapor load of the reflux nozzle on the stirred

tank:

Gnozz ¼ G � d0:3noz

d0:3i

�kgm2 h

�


di ¼ inner tube diameter (mm)dnozz ¼ reflux nozzle diameter (mm)V ¼ vapor rate (kg/h)G ¼ allowable vapor load for the inner tubes (kg/m2 h)Gcond ¼ allowable vapor load of the condenser (kg/h)Gnozz ¼ allowable nozzle load (kg/m2 h)C ¼ condensate rate (kg/h)rV ¼ vapor density (kg/m3)rliq ¼ liquid density (kg/m3)hliq ¼ liquid viscosity (mPas)s ¼ surface tension (N/m)d ¼ tube chamfered angle (�)Notice: The allowable flood load can be increased if the inner tubes of the

condenser are pushed through the tube sheet and chamfered. The same holds for thereflux nozzle.

Example 16: Calculation of the allowable vapor load for pushed through andchamfered tubesReflux condenser with 110 tubes 20 � 2, di ¼ 16 mm, chamfered angle d ¼ 75�

rliq ¼ 958 kg/m3 rV ¼ 0.58 kg/m3 s ¼ 5.8 N/m hliq ¼ 0.28 mPas C/V ¼ 1

G ¼ 246� 160:3 � 9580:46 � 5:80:09 � 0:580:5

0:280:14 � 0:2590:32 � 10:07¼ 21835 kg

m2h

Rohrquerschnitt F ¼ 110� 0:0162 � 0:785 ¼ 0:022 m2

GK ¼ G� F ¼ 21; 835� 0:022 ¼ 480 kg=h

Allowable nozzle load for a chamfered nozzle with dnozz ¼ 150 mm

Gnozz ¼ 21; 835� 1500:3

160:3¼ 42; 732 kg

m2h

Anozz ¼ 0:152 � 0:785 ¼ 0:0177 m2

Gd¼150 ¼ 42; 732� 0:0177 ¼ 756 kg=h

Without chamfered tubes and nozzles the following allowable loadings result:

Chamfered angle d ¼ 0 cos d ¼ 1G ¼ 14,171 kg/m2 h Gcond ¼ 14,171 � 0.022 ¼ 312 kg/hGnozz ¼ 27,734 kg/m2 h for d ¼ 150 mm Gd¼150 ¼ 28,177 � 0.0177 ¼ 491 kg/h


NOMENCLATUREcP specific heat capacity (Wh/kg K)do tube outer diameter (m)di inner tube diameter (m)G condensate rate per unit periphery (kg/m s)g gravitational acceleration (m/s2)Gm average mass stream density (kg/m2 s)L characteristic lengthltube tube length (m)Mliq condensate mass stream (kg/h)Mtot total mass stream (kg/h)n number of tubesPr Prandtl numberReliq condensate Reynolds numberW vapor rate (kg/h)wV vapor velocity (m/s)ag gravity-controlled heat transfer coefficient (W/m2 K)at heat transfer coefficient for turbulent condensate flow (W/m2 K)aw heat transfer coefficient for wavy flow (W/m2 K)as shear stress-controlled heat transfer coefficient (W/m2 K)hliq dynamic viscosity of the liquid phase (Pas)hV dynamic viscosity of the vapor phase (Pas)l condensate heat conductivity (W/m K)rliq density of the liquid phase (kg/m3)rV density of the vapor phase (kg/m3)nliq kinematic viscosity of the liquid phase (m2/s)

REFERENCES[1] D.Q. Kern, Process Heat Transfer, McGraw-Hill, New York, 1950.[2] W.M. Rohsenow, J.P. Hartnett, Handbook of Heat Transfer, McGraw-Hill, New York, 1973.[3] G.F. Hewitt, G.L. Shires, T.R. Bott, Process Heat Transfer, CRC Press Boca Raton, 1994.[4] N.P. Chopey, Heat Transfer, in: Handbook of Chemical Engineering Calculations, McGraw-Hill,

New York, 1993.[5] W. Nusselt, Ver. Deutsch. Ing. 60 (1916) 541.[6] G.H. Gilmore, Chem. Eng. (April 1953).[7] R.C. Lord, P.E. Minton, R.P. Slusser, Design parameters for condensers and reboilers, Chem. Eng. 23

(3) (1970) 127.[8] A. Devore, How to design multitube condensers, Pet. Ref. 38 (June 1959) 205.[9] M. Hadley, Kondensation bin€arer Dampfgemische unter dem Einfluß der vollturbulenten €omung,

in: Fortschr.Ber.VDI Reihe, 3, VDI-Verlag D€usseldorf, 1997. Nr. 468.[10] A.A. Durand, C.A.A. Guerrero, E.A. Ronces, Gravity effects in horizontal condensers, Chem. Eng.

109 (April 2002) 91e94.[11] R. Kern, How to design overhead condensing systems, Chem. Eng. (September 1975).[12] K.G. English, u. andere, Chem. Eng. Progr. 59 (1963) 7.[13] K. Feind, Verfahrenstechnik 5 (1971) 10.[14] F.A. Zenz, R.E. Lavin, Hydrocarbon Process. 44 (3) (1965).


CHAPTER 9

Design of Evaporators

Contents

9.1 Evaporation Process 1489.1.1 Pool boiling 1489.1.2 Flow boiling 1489.1.3 Flash evaporation 149

9.2 Evaporator Construction Types 149Reboiler types 149

9.2.1 Thermosiphon evaporator 1509.2.1.1 Vertical thermosiphon circulation evaporator 1509.2.1.2 Horizontal thermosiphon circulation evaporator 1519.2.1.3 Once-through thermosiphon reboiler 152

9.2.2 Forced circulation reboiler 153Advantages 153Disadvantages 153

9.2.3 Flash evaporator 153Advantages 154Disadvantages 154

9.2.4 Kettle reboiler 155Recommendations 155Advantages 155Disadvantages 155

9.2.5 Internal evaporator 155Advantages 156Disadvantages 156

9.2.6 Falling film evaporator 156Advantages 157Disadvantages 157

9.3 Design of Evaporators for Nucleate Boiling 158Heat transfer coefficient aNB in the tube bundle 161Maximum allowable heat flux density in the tube bundle 161

9.3.1 Practical design with examples 1629.3.2 Dimensioning of kettle reboilers 166

9.3.2.1 Hydraulic design 1669.3.2.2 Thermal design 1689.3.2.3 Allowable bubble rising velocity 1699.3.2.4 Allowable heat flux density in the kettle reboiler 1709.3.2.5 Entrainment 172

9.4 Design of Falling Film Evaporators 172References 179



http://dx.doi.org/10.1016/B978-0-12-803764-5.00009-2

9.1 EVAPORATION PROCESS

In evaporation, depending on the difference between the hot wall and the boiling tem-perature, the physical properties and the stream conditions, three different forms ofboiling can occur: pool boiling, flow boiling, and flash evaporation.

9.1.1 Pool boilingThe evaporation of a stationary liquid in a vessel that is heated from the bottom is calledpool boiling.

If the temperature differences between the heating wall TW and the boiling mediumTB are low, the evaporation takes place from the surface area. For this “surface flash evap-oration,” the heat transfer coefficients are determined with the relationships for free con-vection (Figure 9.8).

The heat transfer coefficients are small in comparison to nucleate boiling.At higher temperature differences, TW�TB, bubbles build up at the heating surface,

which break up at a certain size and flow upward. By further increasing of the temper-ature difference or the heat flux density, larger turbulence occurs in the liquid and theheat transfer increases. This situation is called nucleate boiling.

In technical apparatus, work is done in pool boiling at temperature differences TW�TB

of 10e30 �C. The heat transfer coefficients lie in the range of 1000e3000 W/m2 K fororganic media and at 6000e12,000 W/m2 K for water at a pressure of 1 bar (Figure 9.9).

In a too high temperature difference between the wall and the boiling medium, acoherent layer of vapor is build up on the heating surface. Due to the poor thermalheat conductivities of the vapors compared with liquids, the heat transfer coefficientsbecome severely deteriorate.

This mechanism is called film evaporation (Figure 9.8).In unstable film evaporation, the vapor film is broken by the liquid in certain points.The stable film evaporation is achieved if the unbroken vapor film is widespread.This occurrence is known as the Leidenfrost phenomenon.In order to avoid this region in technical plants, the critical heat flux density must not

be exceeded.

9.1.2 Flow boilingIn flow boiling, the evaporation occurs while streaming through tubes or the channels.

This application is common in technical plants, for example, in thermosiphonreboilers.

Due to the partial evaporation, a two-phase stream with lower density occurs and theflow velocity in the channels becomes strongly increased.

Consequently, the heat transfer coefficients get better.


Due to the heat input, bubbles evolve on the heating surface which get freed andreach the main streaming region. The streaming form of the two-phase stream changesits shape with increasing vapor composition from the bubble flow (5% vapor) to plugflow and to annular flow.

In technical reboilers, the plug flow and annular flow with high flow velocity anda liquid film on the heated wall should be strived for.

With increasing vaporization, the liquid film on the wall disappears and droplet flowoccurs.

In this region, the heat transfer coefficient decreases dramatically because the heattransportation must take place through the poorly conducting vapor layer on the wall.

The region of the droplet flow can be avoided if the critical vapor rate is determinedin which the transition from annular flow in droplet flow occurs.

No droplet flow occurs for vapor rates <30% in the evaporator outlet.A special type of flow boiling is the falling film evaporation in which the evaporation

occurs from a thin liquid film on the wall. This kind of evaporation is especially suitablefor vacuum evaporations because the bubble point is not increased by the static height ofa liquid or of a two-phase mixture.

9.1.3 Flash evaporationIn the flash evaporation, the liquid in the evaporator is not vaporized but it is heated un-der pressure to a temperature higher than the bubble point and the evaporation processtakes place thereafter at the pressure relief in the pressure-reducing valve.

The evaporation of the superheated liquid takes place at the expansion valve outlet.The latent heat for evaporation is drawn from the superheated liquid.

This kind of vaporization is particularly suitable in the vaporization of dirty productswith high flow velocities in order to reduce the fouling of the heating surface.

9.2 EVAPORATOR CONSTRUCTION TYPES

The industrially mostly used separation operation for fluid mixtures is the distillation orrectification. In these separation operations, the required heat is provided by an evapo-rator at the bottoms of the fractionation columns.

These evaporators in the bottoms of distillation columns are known as reboilers.As opposed to the reboilers, solutions are concentrated in evaporators by vaporizing.

The concentrate of the solution, for instance, sugar, can thereby be the required productor the evaporated exhaust vapor, for instance, water from sea water.

Reboiler typesIn the following, the different reboiler construction types are treated:• Vertical thermosiphon evaporators

Design of Evaporators 149

• Horizontal thermosiphon evaporators• Forced circulation evaporators• Flash evaporators• Kettle reboilers• Internal evaporators• Falling film evaporators

9.2.1 Thermosiphon evaporatorThe process fluid flows through the evaporator from the bottom to the top.

As a result of the partial evaporation, a two-phase mixture with lower density occursin the evaporator and due to the difference in static heights, natural circulation occurs.The driving pressure gradient must be greater than the pressure loss in the thermosiphoncircle. It then comes to equilibrium between the driving height and the thermosiphoncirculating rate.

A thermosiphon circle does not work,• if the driving height for the thermosiphon circulation is too low• if the driving temperature gradient is too low for a sufficient evaporation• if the composition of the light boiling components in the mixture is too low• if a large portion of the evaporator surface area is needed for the heating of the prod-

uct up to the bubble point.Normal heat flux density: q z 20,000e30,000 W/m2

Normal vaporization rate: 10e20% of the circulation at atmospheric pressureNormal vaporization rate: 30e50% of the circulation at vacuum evaporationNormal heat transfer coefficient at atmospheric pressure: az 1500e2500 W/m2 KNormal heat transfer coefficient at vacuum evaporation: az 1000e1500 W/m2 KThe design of thermosiphon evaporators will be shown in Chapter 10.

9.2.1.1 Vertical thermosiphon circulation evaporator (Figure 9.1)The evaporation occurs in the flow through vertical tubes that are heated on the shellsidefrom the bottom to the top.

Tube lengths: 2 m (vacuum) to 4 m, normal 2.5e3 mTube diameter: 25e50 mm (vacuum)Downcomer cross-section: 50% of the flow cross-section in the evaporatorRiser cross-section: 100% of the flow cross-section in the evaporatorLiquid height at atmospheric pressure: upper tube sheetLiquid height at vacuum evaporation: 1 m under upper tube sheet.

Advantages• Simple apparatus and easy to clean.• Short residence time at the heating surface.


Figure 9.1 Thermosiphon circulation reboilers.

Disadvantages• Bubble point increase because of the static height over the evaporator surface.• The heat exchanger surface is constructively limited.• Less suitable for vacuum evaporation.• Not suitable for products with higher viscosity and dirty media and mixtures with

wide boiling ranges which contain small amount of light boiling components.• Pulsing stream at flow velocities in riser under 5 m/s.

9.2.1.2 Horizontal thermosiphon circulation evaporator (Figure 9.1)The product is evaporated on the shellside and heated on the tubeside.

Recommendations• Good distribution of the circulating liquid over several inlet nozzles.• Sufficient flow velocity >1 m/s on the shellside.• Control valve in the inlet to avoid over flooding.• TEMA-construction types: AJL, AHL, AEL.

Advantages• Lower bubble point increase than for vertical apparatus because the liquid height over

the evaporator surface is less than in the vertical construction.• No constructive dimension limitation of the heat exchanger surface.• Less sensitive with regard to circulation for process fluctuations.• Suitable for more viscous media up to 20 mPas.

Disadvantages• Pull-through bundle with square pitch is necessary for cleaning.• Difficult cleaning.


9.2.1.3 Once-through thermosiphon reboiler (Figure 9.2)In once-through evaporator, the outflowing bottom liquid flows from the last bottom traydirectly into the reboiler and the light componentsmust be vaporized according to the spec-ification in an evaporation cycle. This is more difficult for the production than with a circu-lating evaporator where the bottom rates are recirculated several times and hence theproduct specification of the bottom can be keptmore easily through a larger buffer volume.

Advantages• An additional theoretical stage for the separation.• Low thermal damage because the product flows only once through the evaporator

and stays very shortly in the hot zone.• No bubble point increase by the high boiling components in the bottom.

Figure 9.2 Thermosiphon once-through reboiler.


Figure 9.3 Forced circulation reboiler.

Disadvantages• Difficult product setting without circulation.

9.2.2 Forced circulation reboiler (Figure 9.3)The product is pumped through the evaporator with a pump. This forced circulation isnecessary if a thermosiphon circulation is not possible because the product is viscous orthe product contains little volatile components in the vapor.

Heat flux density: q z 30,000e50,000 W/m2

Evaporation rate: 1e5% of the circulationDetermination of the a values according to the equations for the convective heattransfer for the corresponding flow velocity.

Advantages• Suitable for high heat flow densities and dirty materials• The circulation rate and the vaporization rate can be set individually• Suitable for small evaporation rates.

Disadvantages• High circulation rate necessary (energy costs)• Circulating pump is required with sufficient net positive suction head (NPSH) value

for boiling products• Complex piping and danger of leakage in the pump.

9.2.3 Flash evaporator (Figure 9.4)The liquid is heated under pressure in the evaporator.

The evaporation occurs after the flashing at a lower pressure in the reducing valve.Large quantities circulated are necessary.


Example 1: Calculation of the vapor rate G, which is released in flashing theheated circulated liquid F

F ¼ circulating rate ¼ 10; 000 kg=h c ¼ spec: heat capacity ¼ 2 kJ=kg K

r ¼ latent heat ¼ 250 kJ=kg Dt ¼ heating temperature difference ¼ 20 K

G ¼ F � c � Dtr

¼ 10; 000� 2� 20250

¼ 1600 kg=h

Recommended flow velocity in the tubes: 1.5e2.5 m/s.The a value is determined with the equations for the convective heat transfer.

AdvantagesEspecially suitable for polymerizing or dirty products because the fouling danger can bereduced by high flow velocities in the tubes.

DisadvantagesPump is necessary with sufficient NPSH value.

Figure 9.4 Flash reboiler.


9.2.4 Kettle reboilerThis evaporator consists of a horizontal vessel with built-in tube bundle and an overflowweir in order that the tubes are constantly covered with the liquid (Figure 9.5).

Pure vapors flow back into the column, no two-phase mixture.The liquid flows out over the weir and is drawn at the bottom. The inflow height to

the evaporator must be sufficiently large.

RecommendationsHeat flux density: q z 20,000e30,000 W/m2

Design: see “Dimensioning of kettle reboilers” in Section 9.3.2.Vessel diameter: z1.5e2 times heating bundle diameter.Vapor velocity over tube bundle: <0.5 m/s.

Advantages• No sensitive thermosiphon circle• Simple design with one additional theoretical stage.

Disadvantages• Heavy fouling and high boiling components accumulation• Long residence time with the danger of thermal decomposition• Difficult cleaning of the shellside• Purge nozzle for dirt removing is necessary.

9.2.5 Internal evaporator (Figure 9.6)In this construction, the tubes or plates heating bundle is built directly in the bottom ofthe column.

Figure 9.5 Kettle reboiler.


Advantages• An internal evaporator is cost-effective because there is no need for a vessel and piping• Lower fouling tendency than in kettle-type evaporator.

Disadvantages• The column diameter determines the heating bundle length and constrains the

possible evaporator surface• For cleaning or repair, the column must be shut down, emptied, and gas freed.

9.2.6 Falling film evaporatorThe liquid product is directed as irrigating liquid falling film on the inside of the tubes(Figure 9.7).

Figure 9.6 Internal U-tube evaporator in a column.


A good distribution is important. In order to avoid dry spaces, the irrigating rate of1e2 m3/h and m tube circumference must be kept. The liquid does not vaporize onthe heating surface but on the liquid surface. No liquid height over the evaporatorarea exists and therefore no bubble point increase by a static height occurs.

This is why the falling film evaporator is especially suitable for evaporation invacuum.

Recommended tube diameter: 40e80 mm, in order to minimize the pressure loss.

Advantages• No bubble point increase and short residence times.

Disadvantages• Complex construction with distributors and circulation pump is necessary• Foulings by crack residues deteriorate the liquid distribution.

Cocurrent falling film reboiler

Countercurrent falling film reboiler

Steam

Condensate

Feed

Vapor

Feed

Residue

Steam

Condensate

Figure 9.7 Falling film reboiler.


9.3 DESIGN OF EVAPORATORS FOR NUCLEATE BOILING [1e5]

In Figure 9.8, the heat flux density q of heated water as a function of the temperature dif-ference, TW�TB, between the heated wall and the boiling liquid is shown. Thefollowing zones are noted:

At low temperature differences, no vaporization occurs but only a heating of theliquid by natural convection. This is the region up to point A. The heat transfer coeffi-cients are small and lie in the range of 250e350 W/m2 K.

The normal range of nucleate boiling lies between the points A and B.The heat flux density increases strongly with increasing temperature difference,

TW�TB.The heat transfer coefficients for organic media lie in the range 2000e3000 W/m2 K.Some heat transfer coefficients are given in Figure 9.9.The unproductive film boiling occurs for the large temperature differences in the

range C to E.The heat transfer coefficients are poor and due to the high wall temperatures, organic

materials are decomposed.The heat transfer coefficients for the vaporization are dependent on the temperature dif-

ference, TW�TB, and the physical properties of the medium being vaporized. InFigure 9.9, the heat transfer coefficients of some materials for the nucleate boiling as afunction of the temperature difference between the heating wall and medium beingvaporized are given.

The curves are valid for the boiling of single components at normal pressure.

Figure 9.8 Heat flux density of water as a function of temperature difference between heating walland the boiling medium.


In evaporation in vacuum, the heat transfer coefficients are clearly less because thevapor bubbles are larger and cover more heating surface. The heat transfer coefficients ofmixtures are smaller because the high boiling components increase the bubble point onthe heating surface. The mixture must be heated in the evaporation from the bubblepoint up to the dew point.

In the equation of Mostinski [1], the different influences on the heat transfer coefficient innucleate boiling at a single tube aNS are covered:

aNS ¼ 35:5� 10�6 � P0:69c ðPaÞ � q0:7 � A� B

�W�m2 K

�aNS ¼ 0:00417� P0:69

c ðkPaÞ � q0:7 � A� B�W�m2 K

�Pc ¼ critical pressure (Pa or kPa)A ¼ correction factor for the boiling pressureB ¼ correction factor for mixtures (B ¼ 1 for single components)Calculation of the correction factor A for the pressure and B for mixtures with a boiling range, BR:

A ¼ 1:8� P0:17R þ 4� P1:2

R B ¼ 1

1þ 0:027� q0:1 � BR0:7

BR ¼ boiling range (K) ¼ temperature difference between dew and boiling pointPc ¼ critical pressure (bar)PR ¼ reduced pressure ¼ P/PcP ¼ working pressure (bar)q ¼ heat flux density (W/m2)

Figure 9.9 Heat transfer coefficient for the nucleate boiling as a function of the temperature differ-ence between the heating wall and the boiling medium.


Example 2: Calculation of the heat transfer coefficient of boiling benzene on asingle tube at different pressures

Pc ¼ 48:9 bar q ¼ 30; 000 W�m2 B ¼ 1 ðsingle componentÞ

aNS ¼ 0:00417� 48900:69 � 30; 0000:7 � A ¼ 1994� A

P ¼ 1 bar

PR ¼ 148:9

¼ 0:0204 A ¼ 1:8� 0:02040:17 þ 4� 0:02041:2 ¼ 0:967

aNS ¼ 0:987� 1994 ¼ 1927 W�m2 K

P ¼ 0:1 bar

PR ¼ 0:148:9

¼ 0:00204 A ¼ 1:8� 0:002040:17 þ 4� 0:002041:2 ¼ 0:63

aNS ¼ 0:63� 1994 ¼ 1257 W�m2 K

In vacuum at 100 mbar, the heat transfer coefficient is clearly less than at 1 bar.

Example 3: Calculation of the heat transfer coefficient of boiling ethanol on asingle tube at 1 bar and at different heat flux densities

Pc ¼ 63:8 bar ¼ 6380 kPa PR ¼ 1=63:8 ¼ 0:0157 A ¼ 0:915 B ¼ 1

q ¼ 10; 000 W�m2

aNS ¼ 0:00417� 63800:69 � 10; 0000:7 � 0:915 ¼ 1016 W�m2 K

q ¼ 20; 000 W=m2

aNS ¼ 0:00417� 63800:69 � 20; 0000:7 � 0:915 ¼ 1650 W=m2 K

q ¼ 30; 000 W=m2

aNS ¼ 1:6102� 30; 0000:7 ¼ 2192 W=m2 K

q ¼ 60; 000 W=m2

aNS ¼ 1:6102� 60; 0000:7 ¼ 3561 W=m2 K

The heat transfer coefficient becomes greater with increasing heat flux density.

Example 4: Calculation of the heat transfer coefficient for a benzeneetoluenemixture with a boiling range, BR ¼ 10 �C

aNS ¼ 1927 W/m2 K from Example 1 for benzene.

B ¼ 11þ 0:027� 30; 000� 100:7

¼ 0:725

aNS ¼ 0:725� 1927 ¼ 1397 W�m2 K


The heat transfer coefficient for the mixture is smaller than the value for the singlecomponent.

Heat transfer coefficient aNB in the tube bundleThe heat transfer coefficient aNB in the tube bundle is clearly better than the heat transfercoefficient on a single tube aBS due to the larger turbulence. With the correction factorFB ¼ 1.5 for the tube bundle, the following equation results:

aNBz1:5� aNS ¼ 0:00626� P0:69c � q0:7 � A� B

�W�m2 K

�

Example 5: Calculation of the heat transfer coefficient of the boiling benzeneetoluene mixture from Example 4 in a tube bundle

aNB ¼ 1:5� 1397 ¼ 2095 W�m2 K

Maximum allowable heat flux density in the tube bundle [6,7]The maximum allowable heat flux density for a tube bundle evaporator results from theallowable vapor velocity over the projected bundle surface area AP.

Too high bubble velocities give rise to a vapor bubble congestion in the bundle.A vapor bubble film then disturbs the heat transfer. Film boiling occurs with poorheat transfer coefficients.

For the allowable vapor rising velocity, wallow, over a tube bundle or a kettle-type evap-orator in order to avoid bubble congestion between the tubes, the following holds:

wallow ¼ 0:18�0@�rliq � rV

�� s� 9:81

r2V

1A0:25

ðm=sÞ

wallow ¼ flow velocity based on the projection of the bundle (m/s)rV ¼ vapor density (kg/m3)rliq ¼ liquid density (kg/m3)s ¼ surface tension (N/m)From the allowable vapor stream velocity, wallow, the practically allowable heat flux

density without vapor bubble congestion between the tubes can be calculated.

qallow ¼ wallow � AP � 3600� rV � rA

�W�m2�

A ¼ heat exchanger-heat transfer surface area (m2)AP ¼ dB � LB ¼ projected bundle surface area (m2)dB ¼ bundle width (m)LB ¼ bundle length (m)r ¼ heat of vaporization (Wh/kg)


Example 6: Determination of the allowable vapor velocity over the bundleand the allowable heat flux density for avoiding bubble congestion

Bundle diameter dB ¼ 0.325 m Bundle length LB ¼ 4 mAP ¼ 0.325 � 4 ¼ 1.3 m2 A ¼ 21.2 m2

s ¼ 0.0155 N/m r ¼ 234 Wh/kg rV ¼ 1.57 kg/m3 rliq ¼ 719 kg/m3

wallow ¼ 0:18��ð719� 1:57Þ � 0:0155� 9:81

1:572

�0:25¼ 0:46 m=s

qallow ¼ 0:46� 1:3� 3600� 1:57� 23421:2

¼ 37; 652 W�m2

9.3.1 Practical design with examplesThe heat transfer coefficients can be very easily determined with the diagram according toKern [5] given in Figure 9.10. The heat transfer coefficients for the nucleate boiling and theheating by natural convection are shown as function of the temperature difference, TW�TB.

In practice, the heat flux density q is limited in order to minimize the product damage inthe evaporator and to avoid vapor bubble congestion.

q ¼ QA

�W�m2�

Q ¼ heat load (W)A ¼ heating area (m2)Recommendation for water and aqueous solutions:qmax z 90,000 bis 120,000 W/m2

aNB z 5500 bis 11,000 W/m2KRecommendation for organic media:Forced circulation: qmax z 60,000e95,000 W/m2

Natural circulation: qmax z 30,000e50,000 W/m2

aNB z 1700e2500 W/m2KOften the product must be first heated to the boiling temperature before the real

evaporation. This is the case especially for product mixtures with a boiling range. Insuch cases, an average weighted heat transfer coefficient, aa, is used for the supply of sen-sible heat by natural convection and the evaporation. This average heat transfer coeffi-cient is determined as follows:

aa ¼ Qtot

QH

aheatingþ QE

aboiling

�W�m2 K

�


QH ¼ heat load for convective heating (W)QE ¼ heat load for the evaporation (W)Qtot ¼ QH þ QE (W)aboiling ¼ heat transfer coefficient for the evaporation (W/m2 K)aheating ¼ heat transfer coefficient for the heating by natural convection (W/m2 K)

Example 7: Overall heat transfer coefficient for heating up and evaporation ofgasoline

Feed to the evaporator: 17,500 kg/h Evaporated rate: 13,000 kg/hHeating in the evaporator from 157 to 168 �CEvaporation: aboiling ¼ 2500 W/m2 KHeating: aheating ¼ 335 W/m2 KEvaporation load: QE ¼ 1060 kW Heating load: QH ¼ 150 kW

aa ¼ 150þ 1060150335

þ 10602500

¼ 1388 W�m2 K

1

104

103

102

98765

4

max for organic liquids

3

2

98765

4

3

2

2 3 4 5

Temperature difference TW-TB (°C)

Natural Convection

Nuc

leat

e Bo

iling

Hea

t Tra

nsfe

r Coe

ffici

ent (

W/m

2 K)

6 7 8 9 2 3 4 5 6 7 8 9102101

α

max for waterα

Figure 9.10 Heat transfer coefficients for nucleate boiling and convective heating.


This example makes clear how strong the heat transfer coefficient for the nucleateboiling is reduced by the heating process.

Guide line figures for the overall heat transfer coefficients of steam-heated reboilers:Light hydrocarbons: Uz 900e1200 W/m2 KHeavy hydrocarbons: Uz 550e900 W/m2 KWater and aqueous solutions: Uz 1250e2000 W/m2 KGuide line figures for thermal oil-heated reboilers: Uz 400e600 W/m2 KThe overall heat transfer coefficient U of evaporators is predominantly determined by the

fouling of the heating surfaces.

Example 8: Steam-heated kettle-type evaporatoraio ¼ 5000 W/m2 K for the steam condensation in the tubes.Fouling factor for both sides ftot ¼ fi þ fa ¼ 0.0004.aNB ¼ 2000 W/m2 K for the evaporation outside the tubes with ftot ¼ 0.0004

1U

¼ 15000

þ 12000

þ 0:0004 U ¼ 909 W�m2 K

aNB ¼ 2000 W/m2 K for the evaporation outside the tubes with ftot ¼ 0.001

1U

¼ 15000

þ 12000

þ 0:001 U ¼ 588 W�m2 K

Example 9: Design of a tube bundle for a distillation stillDetermination of the necessary heating area for a tube bundle of 25 � 2 tubes.Required heat load Q ¼ 850,000 W for the distillation.

Bubble point temperature TS ¼ 107 �C Heating steam temperature Tsteam ¼ 140 �CDt ¼ 140�107 ¼ 33 �C

Calculation of the wall temperature TW and of the overall heat transfer coefficient U:ai ¼ 6000 W/m2K for the heating with condensing steam in the 25 � 2 tubes

aio ¼ ai � dido

¼ 6000� 2125

¼ 5040 W�m2 K based on the tube outside

aiof ¼ 11

5040þ 0:0001

¼ 3351 W�m2 K with fouling

Estimated value for the heat transfer by nucleate boiling: aNB ¼ 1700 W/m2 K on the tube outside.For the determination of aNB from Figure 9.10, the difference between the wall and the boiling tem-

perature, TW�TB, is needed:

TW ¼ TB þ aiof

aiof þ aNB� ðTsteam � TBÞ ¼ 107þ 3351

3351þ 1700� ð140� 107Þ ¼ 128:9 �C

TW � TB ¼ 128:9� 107 ¼ 21:9 �C


From Figure 9.10 for TW�TB ¼ 21.9 �C, a clearly higher value results than the recommended maximumvalue of aNB z 2500 W/m2 K.

Chosen: aNB ¼ 2000 W/m2 KThe overall heat transfer coefficient U is derived as follows:

1U

¼ 1aiof

þ 1aNB

þ slþ fboiling

aiof ¼ heat transfer coefficient in the tube with fouling ¼ 3351 (W/m2 K)aNB ¼ heat transfer coefficient for the boiling ¼ 2000 (W/m2 K)s ¼ tube wall thickness ¼ 0.002 (m)l ¼ thermal heat conductivity of the tube material ¼ 14 (W/m K)fboiling ¼ fouling factor on the vaporization side ¼ 0.0002 (m2 K/W)

1U

¼ 13351

þ 12000

þ 0:00214

þ 0:0002 ¼ 1876

U ¼ 876 W�m2K

Hence the following required heating area Areq results:

Areq ¼ QU� Dt

¼ 850; 000876� 33

¼ 29:4 m2

The heat flux density is: q ¼ U � Dt ¼ 876 � 33 ¼ 28,908 W/m2

The recommended heat flux densities for the different evaporator types should preferably be main-tained in order not to thermally damage the product.

The temperature difference, TW�TB, between the heating wall and the boiling liquid should be in theorder of ca. 20 �C.

Example 10: Butane evaporation at 20 bar in an evaporator with 33 m2 area151 tubes 15.6/19 mm diameter, 3.66 m long.Evaporated butane rate W ¼ 18,523 kg/h

TB ¼ 117 �C Tsteam ¼ 157 �C P ¼ 20 bar PC ¼ 37.6 barrliq ¼ 433 kg/m3 rV ¼ 36.4 kg/m3 c ¼ 0.73 Wh/kg Ks ¼ 0.003 N/m l ¼ 0.09 W/m K h ¼ 0.1 mPas r ¼ 62 Wh/kg

Heat load Q and heat flux density q:

Q ¼ 18; 523� 62 ¼ 1; 148; 426 W

q ¼ 114; 42633

¼ 34; 800 W�m2

Calculation of the heat transfer coefficient, aNB, according to Mostinski for a heating bundle with pressurecorrection factor A:

aNB ¼ 0:00626� 37600:69 � 34; 8000:7 � A ¼ 2767:5� A

A ¼ 1:8��

2037:6

�0:17þ 4�

�2037:6

�1:2¼ 3:49

aNB ¼ 3:49� 2767:5 ¼ 9658:6 W�m2 K


Check of the maximum allowable heat flux density qallow:

wallow ¼ 0:18� �

rliq � rV�� s� 9:81

r2V

!0:25¼ 0:18�

�ð433� 36:4Þ � 0:003� 9:8136:42

�0:25

¼ 0:055 m=s

Projected bundle surface area AP: AP ¼ dB � LB ¼ 0.4 � 3.66 ¼ 1.46 m2

Maximum allowable heat flux density qallow:


¼ 0:055� 1:46� 3600� 36:4� 6233

¼ 19; 876 W�m2

Allowable butane evaporation Wallow:

Wallow ¼ qallow � Ar¼ 19; 876� 33

62¼ 10; 579 kg=h

18,523 kg/h cannot be evaporated but only 10,579 kg/h.

9.3.2 Dimensioning of kettle reboilers [5e7]9.3.2.1 Hydraulic designFor the operation of a kettle-type reboiler, the required height DH between the liquidheight in the column and the overflow weir in the reboiler must be given (Figure 9.11).If the height DH is not sufficiently dimensioned, the liquid height in the bottom of thecolumn rises up to the maximum of the height of the vapor returning nozzle in whichcase the function of the kettle-type reboiler is blocked.

The necessary height DH results from the pressure losses.Calculation of the required height DH with the security factor 2:

DH ¼2�

�DPliq þ DPV þ DPA

�g � rliq

ðm static headÞ

DPliq ¼�f � L

dþ Kliq

�� w2

liq � rliq

2ðPaÞ

DPV ¼�f � L

dþ KV

�� w2

V � rV

2ðPaÞ

DPA ¼ m2V

rV� m2

liq

rliqzm2V

rVðPaÞ

mV ¼ wV � rV�kg�m2 s

�mliq ¼ wReb � rliq

�kg�m2 s

�


DPliq ¼ pressure loss in the liquid line from the column to the reboiler (Pa)DPV ¼ pressure loss in the vapor line from the reboiler to the column (Pa)DPA ¼ acceleration pressure loss in the reboiler (Pa)GV ¼ vapor rate to the column (kg/h)Gliq ¼ total liquid rate to the reboiler (kg/h)GS ¼ liquid draw-off rate from the reboiler (kg/h)mV ¼ mass flux density of the vapors in the vapor line (kg/m2 s)mliq ¼ mass flux density of the liquid in the reboiler (kg/m2 s)f ¼ friction factor of the pipingL ¼ piping length (m)Kliq ¼ resistance factors of the fittings in the liquid lineKV ¼ resistance factors of the fittings in the vapor linewV ¼ flow velocity in the vapor line (m/s)wliq ¼ flow velocity in piping to the reboiler (m/s)wReb ¼ liquid velocity in the reboiler (m/s)rV ¼ vapor density (kg/m3)rFl ¼ liquid density (kg/m3)

Example 11: Calculation of the required height difference for a kettle reboiler

Gliq ¼ 20 t/h GV ¼ 15 t/h GS ¼ 5 t/h Kliq ¼ 3 KV ¼ 2rliq ¼ 750 kg/m3 Vliq ¼ 26.7 m3/h d ¼ 0.1 m wliq ¼ 0.94 m/s L ¼ 10 mrV ¼ 5 kg/m3 VV ¼ 3000 m3/h d ¼ 0.25 m wV ¼ 17 m/s L ¼ 10 mwReb ¼ 0.01 m/s

Figure 9.11 Required height dif-ference DH for a kettle reboiler


DPliq ¼�0:04� 10

0:1þ 3

�� 0:942 � 750

2¼ 2319 Pa

DPV ¼�0:04� 10

0:25þ 2

�� 172 � 5

2¼ 2601 Pa

mV ¼ 17� 5 ¼ 85 kg�m2 s mliq ¼ 0:01� 750 ¼ 7:5 kg

�m2 s

DPA ¼ 852

5¼ 1445 Pa

DH ¼ 2� ð2319þ 2601þ 1445Þ9:81� 750

¼ 1:73 m

What is to be considered?1. The vapor height above the liquid should be at least 0.5 m in order that preferably no entrainment

occurs. The entrained droplets increase the vapor density and hence the pressure drop in the vaporline DPV.

2. The weir overflow height must be considered in establishing the vapor height because the free heightabove the liquid is thereby reduced.

3. A good distribution of the liquid rate entering the reboiler at the bottom without resistances at the inletin the reboiler, for instance, in restrictions by the support for the tube bundle which increase the liquidpressure loss DPliq.

4. The liquid should be introduced over two or three inlet nozzles in order to preferably achieve uniformevaporation.

9.3.2.2 Thermal designFirst the heat transfer coefficient, aNB, is determined for the nucleate boiling.

Calculation of the heat transfer coefficient, aNB, according to Mostinski for nucleate boilingor pool boiling in the tube bundle with the correction factors A for the pressure and B forthe boiling range in product mixtures:

aNB ¼ 0:00626� P0:69c � q0:7 � A� B

�W�m2 K

�Calculation of the pressure correction factor A:

A ¼ pressure factor ¼ 1:8� P0:17R þ 4� P1:2

R

Calculation of the correction factor B for mixtures with the boiling range, BR (�C):B ¼ exp (�0.027 � BR)Pc ¼ critical pressure (kPa)q ¼ heat flux density (W/m2)


Example 12: Calculation of a steam-heated kettle-type evaporator

q ¼ 33,000 W/m2 Pc ¼ 3900 kPa P ¼ 1 bar BR ¼ 10 �CB ¼ expð� 0:027� 10Þ ¼ 0:76

PR ¼ 1003900

¼ 0:0256 A ¼ 1:8� 0:02560:17 þ 4� 0:02561:2 ¼ 1:015

aNB ¼ 0:00626� 39000:69 � 33; 0000:7 � 1:015� 0:76 ¼ 2111 W�m2 K

Calculation of the overall heat transfer coefficient:aio ¼ 8000 W/m2K for condensing steam.

s ¼ 2 mm lWall ¼ 50 W/m K Fouling factor f ¼ 0.00021U

¼ 12111

þ 18000

þ 0:00250

þ 0:0002 U ¼ 1192 W�m2 K

Q ¼ 0:7 MW LMTD ¼ 22 �C

A ¼ QU� LMTD

¼ 700; 0001192� 22

¼ 26:7 m2

9.3.2.3 Allowable bubble rising velocity [6,7]If the heat flux density is too high, vapor bubble congestion between the tubes of thekettle-type evaporator occurs. This results in an undesired film evaporation becausethe vapor bubbles block the heating area. The danger of vapor bubble congestion canbe reduced by using a square tube arrangement with large pitch.

In the design of a kettle-type evaporator, the maximum allowable rising velocity,wallow, of the evaporated vapors over the projected area of the evaporator bundle AP

should not be exceeded.

AP ¼ dB � LB ¼ bundle width� bundle length

The maximum vapor rising velocity, wallow, above the projected area AP is deter-mined as follows:

wallow ¼ 0:18�0@�rliq � r *

V

�� s� 9:81

r2V

1A0:25

ðm=sÞ

rliq ¼ liquid density (kg/m3)rV ¼ vapor density (kg/m3)s ¼ surface tension (N/m)


9.3.2.4 Allowable heat flux density in the kettle reboilerThe allowable bubble rising velocity limits the allowable heat flux density of the evap-orator surface area A as follows:


�W�m2�

r ¼ heat of vaporization (Wh/kg)

Example 13: Calculation of the allowable bubble rising velocity and the heat fluxdensity for a kettle-type evaporator

rliq ¼ liquid density ¼ 775 kg/m3 rV ¼ vapor density ¼ 3 kg/m3

s ¼ surface tension ¼ 0.018 N/m r ¼ 86.4 Wh/kg A ¼ 31.6 m2

dB ¼ 0.45 m L ¼ 6 m

wallow ¼ 0:18��ð775� 3Þ � 0:018� 9:81

32

�0:25

¼ 0:355 m=s

Pojected area AP ¼ dB � L ¼ 0:45� 6 ¼ 2:7 m2

Vapor rate V ¼ 2700 m3�hBubble velocity wbubble ¼ 2700

3600� 2:7¼ 0:277 m=s

The real rising velocity of the vapor bubbles lies with 0.277 m/s under the maximum allowable bubblerising velocity of 0.355 m/s.

Allowable heat flux density due to the bubble rising velocity:

qallow ¼ 0:355� 2:7� 3600� 3� 86:431:6

¼ 28; 303 W�m2

Calculation of the real heat flux density q for the heat load Q ¼ 0.7 MW:

q ¼ 700; 00031:6

¼ 22; 152 W�m2

The available heat flux density of q ¼ 22,152 W/m2 is less than the allowable heat flux densityqzul ¼ 28,303 W/m2.

Example 14: Ethanol evaporation with the kettle-type evaporator

Data: A ¼ 31.6 m2 FB ¼ 2.7 m2 rliq ¼ 757 kg/m3 rV ¼ 1.435 kg/m3

r ¼ 267 Wh/kg s ¼ 17.7 mN/m Pc ¼ 6390 kPa q ¼ 30,000 W/m2

Calculation of the heat transfer coefficient, aNB, for nucleate boiling:

aNB ¼ 0:00626� 63900:69 � 30; 0000:7 ¼ 3600 W�m2 K



ai ¼ 8000 W/m2 K s ¼ 2 mm lWall ¼ 50 W/m K ftot ¼ 0.0003 A ¼ 31.6 m2

LMTD ¼ 22 �C1U

¼ 13600

þ 18000

þ 0:00250

þ 0:0003 U ¼ 1346 W�m2 K

Heat flux density q ¼ U� LMTD ¼ 1346� 22 ¼ 29; 620 W�m2

Heat duty Q ¼ q� A ¼ 29; 620� 31:6 ¼ 935; 988 W

Vapor rate V ¼ Qr � rV

¼ 935; 988267� 1:435

¼ 2442:9 m3�hBubble velocity wbubble ¼ 2442:9

2:7� 3600¼ 0:25 m=s

Calculation of the allowable vapor rising velocity, wallow

wallow ¼ 0:18��ð757� 1:435Þ � 0:0177� 9:81

1:4352

�0:25¼ 0:51 m=s > wbubble ¼ 0:25 m=s

Calculation of the allowable heat flux density, qallow

qallow ¼ 0:51� 2:7� 3600� 1:435� 26731:6

¼ 59; 933 W�m2 > q ¼ 29; 620 W

�m2

Example 15: Water evaporation with the kettle-type evaporator

Water data: rliq ¼ 958 kg/m3 rV ¼ 0.597 kg/m3 r ¼ 626.9 Wh/kgs ¼ 58.9 mN/m Pc ¼ 22,129 kPa Q ¼ 40,000 W/m2

A ¼ 31.6 m2 LMTD ¼ 22 �C ftot ¼ 0.0003s ¼ 2 mm l ¼ 50 W/m K asteam ¼ 8000 W/m2 KAP ¼ 2.7 m2

aNB ¼ 0:00626� 22; 1290:69 � 40; 0000:7 ¼ 10; 376 W�m2 K

1U

¼ 110; 376

þ 18000

þ 0:00250

þ 0:0003 U ¼ 1781 W�m2 K

q ¼ k � LMTD ¼ 1781 � 22 ¼ 39,289 W/m2 Q ¼ q � A ¼ 39,189 � 31.6 ¼ 1.238 MW

V ¼ 1:238� 106

626:9� 0:597¼ 3308:9 m3�h wbubble ¼ 3308:9

2:7� 3600¼ 0:34 m=s

Calculation of the allowable bubble rising velocity and the allowable heat flux density:

wallow ¼ 0:18��ð958� 0:597Þ � 0:0589� 9:81

0:5972

�0:25¼ 1:13 m=s > wbubble ¼ 0:34 m=s

qallow ¼ 1:13� 2:7� 3600� 0:597� 626:931:6

¼ 130; 086 W�m2 > q ¼ 39; 189 W

�m2


9.3.2.5 EntrainmentIn order to avoid too much entrainment of droplets in the rising vapor, the shell diameterDM of the kettle-type evaporator is enlarged by the factor 1.5 to 2 and is greater than theheating bundle diameter dB.

DMz1:5� dB

The required flow cross-section Areq above the boiling liquid to avoid the dropletentrainment is determined from the allowable vapor velocity, wdroplet:

wdroplet ¼ 0:04�ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffirliq

rV� 1

rðm=sÞ

Areq ¼ VVapor

wdroplet

�m2� Lreq ¼ Areq

DMðmÞ

V ¼ vapor rate (m3/s) Lreq ¼ required shell vessel length (m)rliq ¼ liquid density (kg/m3) rV ¼ vapor density (kg/m3)

Example 16: Calculation of the required flow cross-section against dropletentrainment

V ¼ 3000 m3/h rliq ¼ 580 kg/m3 rV ¼ 6 kg/m3 dB ¼ 0.4 m DM ¼ 0.8 m

wdroplet ¼ 0:04�ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi5806

� 1

r¼ 0:39 m=s

Areq ¼ 30003600� 0:39

¼ 2:13 m2

Lreq ¼ 2:130:8

¼ 2:66 m

9.4 DESIGN OF FALLING FILM EVAPORATORS [8e10]

In evaporation in vacuum, falling film evaporators are used because in these equipmentsnot the bubble point is raised by a static liquid height. For a given heat duty Q, initially,the required evaporator area A is determined with an estimated overall heat transfer co-efficient U for the effective temperature difference Dt.

A ¼ QU � Dt

�m2�

Then a number n of the required tubes for a chosen tube dimension is established andthe maximum vapor velocity wV in the tube outlet is calculated.


n ¼ Ado � p� L

wV ¼ Qr � at � rV � 3600

ðm=sÞ

at ¼ n� d2i �p

4

�m2�

do ¼ tube outer diameter (m) di ¼ tube inner diameter (m)L ¼ tube length (m) at ¼ tubes flow cross-section (m2)rV ¼ vapor density (kg/m3) A ¼ reboiler area (m2)

The pressure drop in the tubes should be preferably low in vacuum vaporization inorder to avoid a boiling point increase due to the higher pressure in the evaporator.

From Figure 9.12, it follows that the pressure loss for the toluene evaporation invacuum increases with increasing vapor rate and that a higher liquid feed rate results inhigher pressure loss.

Figure 9.13 shows how the pressure loss increases with the vapor rate in a hexadecaneevaporator.

From Figure 9.13, it can be derived:• that the pressure loss greatly increases with increasing vapor rate,• that at a lower pressure, the pressure losses are greater,• that a higher liquid feed increases the pressure loss.

In calculating the pressure losses for the two-phase stream in the evaporator, theincreasing vapor fraction of the two-phase mixture due to the increasing tube lengthmust be taken into account.

Figure 9.12 Pressure loss in a toluene evaporator as function of the vapor rate at different liquidfeed rates.


The total pressure drop in the evaporator is composed of three parts:The friction pressure drop DPFr, the one that is increasing with the tube length.The acceleration pressure drop DPA, which increases with the evaporator length.The negative hydrostatic pressureDPH, which is increasing toward the end of the tube.

Dptot ¼ DPFr þ DPA � DPH

In Figure 9.14, the function of the pressure drop composites in a falling film evapo-rator for hexadecane at a pressure of 33 mbar is shown.

Figure 9.13 Pressure loss in a falling film evaporator for hexadecane at different pressures and liquidfeeds as function of the vapor rate.

Figure 9.14 Pressure drop parts in the different zones of a falling film evaporator.


In order to avoid higher pressure losses, the mass stream density m in the evaporatortubes should be in the range of 15e18 kg/m2 s maximum (see Figure 9.15).

m ¼ wV � rV (kg/m2 s)wV ¼ flow velocity in the tubes rV ¼ vapor density (kg/m3)

Also, the vapor flow velocity wV at the tube end should be at maximum 50% of thesonic velocity, wsonic.

wsonic ¼ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffik� R

M� T

rðm=sÞ

R ¼ gas constant ¼ 8314 (N/m2)(m3)/(kMol)(K) T ¼ temperature (K)k ¼ adiabatic exponent M ¼ mole weight (kg/kmol)

The heat transfer coefficient in turbulent falling film can be calculated according tothe following equations:

Numrich ð9:8Þ a ¼ 0:0055� l

l�Re0:44 � Pr0:4

�W�m2 K

�

Scholl ð9:9Þ a ¼ 0:0054� l

l�Re0:38 � Pr0:57

�W�m2 K

�

Hadley ð9:10Þ a ¼ 0:00454� l

l�Re0:44 � Pr0:48

�W�m2 K

�

l ¼�v2

g

�1=3Re ¼ VU

3600� vVU ¼ Vliq

LU

�m3�h m

�LU ¼ n� p� diðmÞ

Figure 9.15 Pressure loss in a hexadecane evaporator as function of the mass flux density.


Re ¼ Reynolds number Pr ¼ Prandtl numberLU ¼ circumferential length of the tubes (m) Vliq ¼ liquid feed (m3/h)VU ¼ liquid jet loading (m3/h m) l ¼ heat conductivity of the product (W/m K)n ¼ kinematic viscosity of the product (m2/s)

In Figure 9.16, the heat transfer coefficients calculated according to different modelsare shown as a function of the Reynolds number. With increasing Reynolds number orincreasing liquid jet loading VU, the heat transfer coefficient becomes greater.

Design recommendation: VU ¼ 2e4 m3/h m tube circumference.From Figure 9.17, it follows that the heat transfer coefficient becomes better with

increasing vapor velocity and higher liquid loading.With regard to thermal damage of organic products, the residence time tR in the

evaporator is important. The residence time tR in the falling film with the film thicknessd is determined as follows:

tR ¼ n� di � p� L � d

Vliq

.3600

ðsÞ d ¼ 0:3035�Re0:583 � l ðmÞ

For a uniform liquid distribution on the individual tubes of the evaporator, a liquiddistributor is installed on the top tube sheet.

The distributor has on the outside an edge of ca. 100 mm and at least so manydischarge pipes as evaporator tubes.

An additional predistributor is installed for large diameters.The dimensioning of the liquid distributor is according to the equation for the vessel

outlet rate Vdistr:

V distr ¼ahole � nT � d2T � p=4� ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi2� g �H

p �m3�s�

Figure 9.16 The heat transfer coefficients as a function of the Reynolds number in the falling filmcalculated with different models.


Figure 9.17 Heat transfer coefficients in a toluene evaporator as a function of vapor velocity atdifferent liquid jet loadings.

ahole ¼ out-streaming factor ¼ 0.6e0.8 dT ¼ diameter of the outflow pipes (m)H ¼ liquid height in the distributor (m) nT ¼ number of the discharge pipes

For a diameter dT ¼ 15 mm for the outflow tubes and an outflow velocity ofwT ¼ 0.6 m/s in the discharge pipes, the number of the outflow tubes nT of the distrib-utor and the liquid height H in the distributor can be determined as follows:

nT ¼ Vliq

wT� d2T � p=4� 3600¼ Vliq

0:6� 0:000175� 3600¼ Vliq

0:378

H ¼ w2T

2� g � a2hole¼ w2

T

2� 9:81� 0:62¼ w2

T

7:06ðmÞ

Example 17: Design of a falling film evaporator for hexadecane

Data: P ¼ 33 mbar t ¼ 170.35 �C Q ¼ 100 kW M ¼ 226r ¼ 75 Wh/kg Dt ¼ 20 �C rV ¼ 0.203 kg/m3 rliq ¼ 662 kg/m3

Pr ¼ 12.73 l ¼ 0.094 W/m K n ¼ 0.672 mm2/s Tubes 50 � 2Tube length L ¼ 5 m

Estimation of the required area and the required number of tubes for an estimated U value.U ¼ 600 W/m2 K:

A ¼ QU� Dt

¼ 100; 000600� 20

¼ 8:3 m2

n ¼ Ado � p� L

¼ 8:30:05� p� 5

¼ 11 Rohre 50� 2


Chosen:16 tubes 50 � 2 with the tube length L ¼ 5 mHeat exchanger surface A ¼ n � L � p ¼ 16 � 5 � p ¼ 12.6 m2

Flow cross-section in the tubes aT ¼ n � di2 � p/4 ¼ 16 � 0.0462 � 0.785 ¼ 0.0266 m2

Calculation of the vapor flow velocity wV:

Vapor density rV ¼ M� P þ 27322:4� 1013� ð273þ tÞ ¼ 226� 33� 273

22:4� 1013� ð237þ 170Þ ¼ 0:203 kg�m3

Vapor velocity wV ¼ Qr � aT � rV � 3600

¼ 100; 00075� 0:0266� 0:203� 3600

¼ 68:6 m=s

Alternative calculation of the vapor flow velocity:

Vapor rate GV ¼ Qr¼ 100; 000

75¼ 1333 kg=h Vapor volume V ¼ GV

rV¼ 1333

0:203¼ 65:66 m3�h

Vapor velocity wV ¼ V3600� aT

¼ 65:663600� 0:0266

¼ 68:6 m=s

Calculation of the sonic velocity for hexadecane vapors:

wsonic ¼ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffik� R

M� T

r¼

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi1:0164� 8314

226� 443

r¼ 128:7 m=s

Recommendation: wVmax z 0.5 � wSchall ¼ 0.5 � 128.7 ¼ 64.4 m/s.Recommended maximum velocity of the vapors with the mass stream density.mmax ¼ 15 kg/m2 s:

wVmax ¼ mmax

rV¼ 15

0:203¼ 73:9 m=s

The vapor velocity wD ¼ 68.6 m/s lies in the range of the allowable maximum flow velocity.Calculation of the required liquid feed Vfl for a liquid jet loading of VU ¼ 3 m3/h m volume:

Vliq ¼ VU � LU ¼ VU � ðn� di � pÞ ¼ 3� 16� 0:046� p ¼ 6:93 m3�h ¼ 4:6 t=h

Chosen: Gliq ¼ 8 t/h Vliq ¼ 12.08 m3/h VU ¼ 5.23 m3/h m

Calculation of the heat transfer coefficient for VU ¼ 5.23 m3/h m:

Re ¼ VU3600� n

¼ 5:23

3600� 0:672� 10�6 ¼ 2161

l ¼�n2

g

�1=3¼ �

0:672� 10�6�2

9:81

!1=3¼ 36� 10�6

a ¼ 0:0055� 0:094

36� 10�6 � 21610:44 � 12:730:4 ¼ 1165 W�m2 K

Calculation of the overall heat transfer coefficient and the heat load:Heat transfer coefficient for the heating on the shellside ao ¼ 6000 W/m2 K

Fouling rate ftot ¼ 0.0003 Heat conductivity of the tube wall lWall ¼ 14 W/m K


1U

¼ 1

1165� 4650

þ 16000

þ 0:00214

þ 0:0003 U ¼ 648 W�m2 K

Q ¼ U� A� Dt ¼ 648� 12:6� 20 ¼ 163; 296 W

Calculation of the film thickness d, the residence time tR, and the falling film velocity wfilm:

d ¼ 0:3035� Re0:583 � l ¼ 0:3035� 21610:583 � 36� 10�6 ¼ 0:00096 m ¼ 0:96 mm

tR ¼ LU � d� L

Vliq�3600

¼ 2:31� 0:00096� 512:08=3600

¼ 3:3 s wfilm ¼ LtR

¼ 53:3

¼ 1:5 s

Design of the liquid distributor for 12 m3/h liquid feed:Number of the required outflow tubes nT of the distributor:

nT ¼ Vliq0:378

¼ 120:378

¼ 32 discharge pipes

Liquid level H in the distributor:

H ¼ w2T

7:06¼ 0:62

7:06¼ 0:05 m ¼ 50 mm

Check calculation:

Vdistr ¼ ahole � nT � d2T � 0:785�ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi2� g� H

p¼ 0:6� 32� 0:0152 � 0:785�

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi2� 9:81� 0:05

p

¼ 0:00336 m3�sVliq ¼ Vdistr ¼ 3600� 0:00336 ¼ 12:08 m3�hwT ¼ ahole �

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi2� g� H

p¼ 0:6�

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi2� 9:82� 0:05

p¼ 0:6 m=s

REFERENCES[1] I.L. Mostinski, Calculations of heat transfer in boiling liquids, Teploenergetika 10 (1953) 4.[2] M.J. McNelly, J. Imp. Coll. Chem. Eng. Soc. 7 (1953) 18.[3] J. Starczewski, Heat transfer to nucleate boiling liquids, Brit. Chem. Eng. 10 (8) (1965).[4] M.G. Cooper, Saturation nucleate pool boiling, J. Chem. Eng. Symp. Ser. 86 (2) (1984) 785e793.[5] D.Q. Kern, Process Heat Transfer, McGraw-Hill, N.Y., 1950.[6] J.W. Palen, W.M. Small, Kettle and internal reboilers, Hydrocarb. Proc. 43 (11) (1964) 199e208.[7] Lord M. Slusser, Design parameters for condenser and reboilers, Chem. Eng. 23 (1970) 3.[8] R.Numrich: CIT 68 (11/96), 1466 ff.[9] S. Scholl, Verdampfung und Kondensation, in: Fluid-Verfahrenstechnik Bd. 2, Wiley-VCH-Verlag,

Weinheim, 2006.[10] M. Hadley, VDI-Fortschrittsberichte Reihe 3, Nr. 468, VDI-Verlag, 1997.


CHAPTER 10

Design of Thermosiphon Reboilers

Contents

10.1 Thermal Calculations 18210.1.1 Required circulating rate W in the thermosiphon cycle 18210.1.2 Estimation of the required reboiler area A 183

10.2 Calculation of the Heat Transfer Coefficient 18410.2.1 Calculation of the heat transfer coefficient aconv for the convective boiling at the

increased flow velocity of the two-phase mixture 18410.2.2 Calculation of the heat transfer coefficient aN for the nucleate boiling with the

correction factor S for confined nucleate boiling 18610.2.3 Calculations for a vertical thermosiphon evaporator 18710.2.4 Calculations for a horizontal thermosiphon evaporator with cross flow between the

tubes from the bottom to the top 18810.2.5 Calculations for a horizontal thermosiphon evaporator with longitudinal flow around

the tubes 18910.3 Calculation of the Two-Phase Density and the Average Density in the Reboiler 191

10.3.1 Calculation of the two-phase density r2Ph of a vaporeliquid mixture 19110.3.2 Calculation of the average density rReb in the reboiler from the density of the liquid rliq

and the density of the two-phase mixture r2Ph 19110.4 Flow Velocity wReb in the Reboiler 19310.5 Determination of the Required Height H1 for the Thermosiphon Circulation or the Maximum

Allowable Pressure Loss DPmax in Thermosiphon Circulation 19410.5.1 Calculation of the required elevation height H1 for overcoming the pressure loss DP in

the thermosiphon circulation with the safety factor S 19410.5.2 Calculation of the maximum allowable pressure loss DPallow in the thermosiphon

circulation at given heights H1, H2, and H4 19410.5.3 Calculation of the thermosiphon reboiler feed height H1 in practice 197

10.5.3.1 Vertical circulating reboiler 19710.5.3.2 Horizontal circulation evaporator 19810.5.3.3 Vertical once-through evaporator 19910.5.3.4 Horizontal once-through evaporator 201

10.5.4 Effects of the elevation height H1 on the circulating rate, the pressure loss, thetemperature difference, and the heat load of thermosiphon evaporators 202

10.6 Design of Riser and Downcomer Diameter 20310.7 Calculation of the Pressure Losses in the Thermosiphon Circulation 206

10.7.1 Liquid pressure loss in the downcomer 20610.7.2 Friction pressure loss for two-phase flow in the evaporator 20710.7.3 Accelerated pressure loss in the evaporator 20810.7.4 Friction pressure loss in the riser for the two-phase flow 209



http://dx.doi.org/10.1016/B978-0-12-803764-5.00010-9

10.7.5 Total pressure loss in the thermosiphon circulation 21010.7.6 Reboiler characteristic curves 211

10.8 Calculation of the Required Reboiler Length or Area for the Heating up to the BoilingTemperature and for the Evaporation in Vertical Thermosiphon Evaporators 21210.8.1 Required reboiler length and area for heating up to the boiling temperature 21310.8.2 Required tube length Levap and area Aevap for evaporation 215

10.9 Required Heating Length for Vertical Thermosiphon Reboilers According to Fair 21510.10 Calculation of the Pressure and Boiling Point Increase by Means of the Liquid Height H1 21810.11 Average Overall Heat Transfer Coefficient for Heating þ Vaporizing 22110.12 Calculation of the Vapor Fraction x of the Two-Phase Mixture in a Vertical Reboiler 22110.13 Thermosiphon Reboiler Design Example 222References 226

10.1 THERMAL CALCULATIONS [1e7]

10.1.1 Required circulating rate W in the thermosiphon cycleInitially the heat loadQ for the heating and evaporation for the given problem definitionis determined (Figure 10.1).

Figure 10.1 Thermosiphon circulation reboiler.


Then the required circulation rateW of the thermosiphon cycle for the chosen vapormass fraction x is determined.

Calculation of the heat load Q for heating and evaporating:

Q ¼ Qheat þQevap ¼ W � c � Dtheat þW � x� rðWÞRequired thermosiphon circulating rate W for the chosen evaporation rate x:

W ¼ Qc � Dtheat þ x� r

ðkg=hÞ Vapour rate GV ¼ x�W ðkg=hÞ

Q ¼ heat load (W)W ¼ liquid inlet rate into the circulating evaporator (kg/h)c ¼ specific heat capacity (Wh/kg K)Dtheat ¼ required heating up to the bubble point (K)r ¼ latent heat (Wh/kg)x ¼ vapor mass fraction in two-phase mixture (kg/kg)

Example 1: Calculation of the circulating and vapor rate in the thermosiphonevaporator

Q ¼ 600 kW r ¼ 150 Wh/kg x ¼ 0.2 c ¼ 0.6 Wh/kg K Dtheat ¼ 4 �C

W ¼ 600; 0000:6� 4þ 0:2� 150

¼ 18; 518 kg=h

Vapor rate V ¼ x �W ¼ 0:2� 18; 518 ¼ 3704 kg=h vapor

Cross-checking with heat balance:

Q ¼ W � c � Dt þW � x � r ¼ 18; 518� 0:6� 4þ 18; 518� 0:2� 150 ¼ 600; 000 W

10.1.2 Estimation of the required reboiler area AThe required area A for the required heat load Q is determined with an estimated U-value for the overall heat transfer.

A ¼ QU � Dt

�m2�

U ¼ estimated overall heat transfer coefficient (W/m2 K)

Example 2: Determination of the reboiler area and tube number

Q ¼ 600 kW U ¼ 1200 W/m2 K Dt ¼ 20 �C

A ¼ QU� Dt

¼ 600; 0001200� 20

¼ 25 m2

Design of Thermosiphon Reboilers 183

Calculation of the required number of tubes n for the reboiler area A ¼ 25 m2

Chosen tube length L ¼ 3 m Chosen tube dimension 25 � 2 mm Pitch ¼ 32 mm D

n ¼ Ap� di � L

¼ 25p� 0:025� 3

¼ 106 tubes 25� 2

Chosen: n ¼ 127 tubes 25 � 2 with triangular pitch 32 mmThe shell inner diameter Di ¼ 384 mm then results.

10.2 CALCULATION OF THE HEAT TRANSFER COEFFICIENT

The total heat transfer coefficient aReb is composed of the convective heat transfer co-efficient aconv for the convective boiling at the increased stream velocity of the two-phase mixture and an for the corrected nucleate boiling.

aReb ¼ aconv þ an�W=m2 K

�

10.2.1 Calculation of the heat transfer coefficient aconv forthe convective boiling at the increased flow velocityof the two-phase mixture

Initially the aliq-value for the convective heat transfer of the streaming liquid isdetermined.

Then the heat transfer coefficient for the proportional liquid flow aliq is multiplied bythe multiplicator M for the two-phase stream velocity.

aconv ¼ M � aliq�W=m2 K

�Calculation of the heat transfer coefficient aliq for the streaming liquid in the tube:

Reynolds number 2300 > Re < 8000 Nu ¼�0:037�Re0:75liq � 6:66

�� Pr0:42

Reynolds number > 8000 Nu ¼ 0:023�Re0:8liq � Pr0:33

a ¼ Nu� l

diReliq ¼ w liq � di

nliq

Reliq ¼ Reynolds numberPr ¼ Prandtl numberNu ¼ Nusselt numberl ¼ heat conductivity (W/m K)di ¼ tube inner diameternliq ¼ kinematic viscosity (m2/s)wliq ¼ flow velocity of the liquid (m/s)


The multiplicator M for the conversion of aliq to aconv at the two-phase streaming iscalculated as follows:

M ¼ 2:5��

1Xtt

�0:7

Xtt ¼�1� xx

�0:9

� rV

rliq

!0:5

��hliq

hV

�0:1

¼�Gliq

GV

�0:9

� rV

rliq

!0:5

��hliq

hV

�0:1

Xtt ¼ LockharteMartinelli parameter for turbulent streamingx ¼ vapor weight composition of the two-phase streamingrliq ¼ liquid density (kg/m3)rV ¼ vapor density (kg/m3)hliq ¼ dynamic liquid viscosity (mPa s)hV ¼ dynamic vapor viscosity (mPa s)Gliq ¼ liquid rate (kg/h)GV ¼ vapor rate (kg/h)

Example 3: Calculation of the heat transfer coefficient for corrected convectiveboiling in a vertical thermosiphon reboiler for toluene

Circulating rate W ¼ 40 t/h x ¼ 0.4 ¼ 40% evaporation at the outlet ¼ 16 t/h vapor

The vapor mass fraction increases from x ¼ 0 at the reboiler inlet to x ¼ 0.4 at the reboiler outlet.The calculation is made with the average vapor mass fraction xa and the average vapor rate GVa and the

average liquid rate GliqaAverage vapor mass fraction over the reboiler tube length xa ¼ 0.2

Average liquid rate Gliqa ¼ 32 t/h ¼ 41 m3/h Average vapor rate GVa ¼ 8 t/hrV ¼ 3 kg/m3 rliq ¼ 781 kg/m3 hV ¼ 0.009 mPa s hliq ¼ 0.244 mPa slliq ¼ 0.116 W/m K Pr ¼ 4.12 di ¼ 21 mm nliq ¼ 0.3073 mm2/s

Calculation of the flow cross-section at of the tubes and the flow velocity wliq:

at ¼ n� d2i �p

4¼ 120� 0:0212 � 0:785 ¼ 0:04154 m2 wliq ¼ 41

3600� 0:04154¼ 0:274 m=s

Calculation of aliq:

Reliq ¼ wliq � din

¼ 0:274� 0:021

0:3073� 10�6 ¼ 18; 723

Nu ¼ 0:023� Re0:8 � Pr0:33 ¼ 0:023� 18; 7230:8 � 4:120:33 ¼ 96:06

aliq ¼ Nu� l

di¼ 96:06� 0:116

0:021¼ 530 W=m2 K


Calculation of M and acovn

Xtt ¼�GliqGV

�0:9

� rV

rliq

!0:5

��hliq

hV

�0:1

¼�328

�0:9

��

3781

�0:5

��0:2440:009

�0:1

¼ 0:3

M ¼ 2:5��1Xtt

�0:7

¼ 2:5��

10:3

�0:7

¼ 5:8

aconv ¼ M� aliq ¼ 5:8� 530 ¼ 3074 W=m2 K

10.2.2 Calculation of the heat transfer coefficient aN for the nucleateboiling with the correction factor S for confined nucleate boiling

Initially the heat transfer coefficient aNB for the nucleate boiling is determined.This value will be corrected with the correction factor S according to Chen for the

confined nucleate boiling at the two-phase streaming.

aNB ¼ 0:00626� P0:69c � q0:7

�W=m2 K

�an ¼ S � aNB

�W=m2 K

�

S ¼ 1

1þ �2:53� 10�6 �Re1:172Ph

� Re2Ph ¼ Reliq �M1:25

q ¼ heat flux density (W/m2)Pc ¼ critical pressure (kPa)S ¼ correction factor for the confined nucleate boiling according to ChenRe2Ph ¼ Reynolds number at two-phase streaming according to Chen

Example 4: Heat transfer coefficient for corrected nucleate boiling

Pc ¼ 4100 kPa q ¼ 25,000 W/m2 Re ¼ 18,723 M ¼ 5.8

aNB ¼ 0:00626� 41000:69 � 25; 0000:7 ¼ 2333 W=m2 K

Re2Ph ¼ 18; 723� 5:81:25 ¼ 168; 523 S ¼ 1

1þ ð2:53� 10�6 � 168; 5231:17� ¼ 0:23

an ¼ 2333� 0:23 ¼ 536 W=m2 K

The total heat transfer coefficient for the evaporation in the vertical thermosiphon evaporator is thesum of the heat transfer coefficient for the convective heat transfer of the two-phase streaming(Example 3) and the corrected heat transfer coefficient for the prevented nucleate boiling by thestreaming (Example 4):

aReb ¼ aconv þ aN ¼ 3074þ 536 ¼ 3610 W=m2 K


10.2.3 Calculations for a vertical thermosiphon evaporatorThe calculation is performed in the following steps:• determination of the flow cross-section in the tubes at• determination of the flow velocity wliq for the liquid rate• calculation of the Reynolds number Reliq• determination of the heat transfer coefficient aliq for the liquid flowing in the

tubes

Example 5: Thermal design of a steam-heated vertical thermosiphon evaporatortype AEM with evaporation in the tubesData: average liquid rate V ¼ 85 m3/h

Reboiler DN 700 with 394 tubes 25 � 2 L ¼ 2.5 m, flow Cross-section at ¼ 0.136 m2

Pr ¼ 4.33 l ¼ 0.117 W/m K N ¼ 0.318 mm2/s

Calculation of the heat transfer coefficient aconv for the convective boiling:

wliq ¼ 853600� 0:136

¼ 0:173 m=s Reliq ¼ 0:173� 0:021

0:318� 10�6 ¼ 11; 431

aliq ¼ 0:023� Re0:8 � Pr0:33 � l

di¼ 0:023� 11; 4310:8 � 4:330:33 � 0:117

0:021¼ 366 W=m2 K

Two-phase multiplicator M ¼ 4

aconv ¼ 4� 366 ¼ 1466 W=m2 K

Calculation of the heat transfer coefficient for the nucleate boiling:

Pc ¼ 4100 kPa q ¼ 20,000 W/m2

aNB ¼ 0:00626� P0:69c � q0:7 ¼ 0:00626� 41000:69 � 20; 0000:7 ¼ 1996 W=m2 K

Re2Ph ¼ 11; 431� 41:25 ¼ 64; 663 S ¼ 0:48

an ¼ 0:48� 1996 ¼ 958 W=m2 K

Total heat transfer coefficient aReb ¼ aconv þ aN ¼ 1466 þ 958 ¼ 2424 W/m2 KCalculation of the overall heat transfer coefficient U:

ftot ¼ 0.0003 s ¼ 2 mm lWall ¼ 14 W/m K

Tubeside aio ¼ 2424� 2125

¼ 2036 W=m2 K Shell side ao ¼ 8000 W=m2 K

1U

¼ 12036

þ 18000

þ 0:00214

þ 0:0003 U ¼ 944 W=m2 K

Calculation of the heat load Q with logarithmic mean temperature difference (LMTD) ¼ 20 �C:

A ¼ 394� 0:025� p� 2:5 ¼ 77 m2

Q ¼ U� A� LMTD ¼ 944� 77� 20 ¼ 1:454 MW

q ¼ U� LMTD ¼ 944� 20 ¼ 18; 880 W=m2


From Figure 10.2 it follows that the heat transfer coefficient aReb increases with increasing vaporizationbecause the heat transfer coefficient akon for the convective boiling of the two-phase mixture stronglyincreases.

Dependences on the vapor composition x• The two-phase density decreases with increasing vaporization and the flow velocity of the two-phase

mixture increases. The higher flow velocity improves the heat transfer coefficient for the convectiveboiling.

• The overall heat transfer coefficient increases at higher evaporation rates because the a-value for theconvective boiling increases with increasing vapor rate.

• The effective temperature difference LMTD increases at larger evaporation rates because due to thelower density in the reboiler at higher vapor compositions the boiling pressure and hence the boilingtemperature are reduced.

10.2.4 Calculations for a horizontal thermosiphon evaporator withcross flow between the tubes from the bottom to the top [8,9]

Calculation steps:• determination of the flow cross-section for the cross streams around the tubes across• determination of the flow velocity wcross for the liquid rate• calculation of the Reynolds number Reliq• determination of the heat transfer coefficient for the streaming liquid aliq across the

tubes in the square arrangement.

aliq ¼ 0:158�Re0:6 � Pr0:33 � l

do

�W=m2 K

�aconv ¼ M � aliq

�W=m2 K

�

aReb ¼ aconv þ an�W=m2 K

�

Figure 10.2 Heat transfer coefficients in a vertical thermosiphon evaporator as a function of theevaporation rate.


Example 6: Thermal design of a horizontal thermosiphon evaporator type AJL orAHL with shell-side cross flow evaporationTube bundle DN 450 with 132 tubes 25 � 2, 6 m long, 32 mm quadratic pitch

Physical data of Example 5Calculation of the heat transfer coefficient aconv for the convective boiling:

Number of tubes in cross flow ncross ¼ Di

T¼ 450

32¼ 14

across ¼ Di � ncross � do ¼ 0:45� 14� 0:025 ¼ 0:1 m2 cross flow area

wcross ¼ 853600� 0:1

¼ 0:236 m=s Re ¼ 0:236� 0:025

0:318� 10�6 ¼ 18; 562

aliq ¼ 0:158� Re0:6 � Pr0:33 � l

do¼ 0:158� 18; 5620:6 � 4:330:33 � 0:117

0:025¼ 436:6 W=m2 K

Two-phase multiplicator ¼ 4 aconv ¼ 4 � 436.6 ¼ 1746 W/m2 K

Calculation of the heat transfer coefficient aN for the nucleate boiling:

Pc ¼ 4100 kPa q ¼ 20,000 W/m2

aNB ¼ 0:00626� 41000:69 � 25; 0000:7 ¼ 1996 W=m2 K

Re2Ph ¼ 18; 562� 41:25 ¼ 105; 002 S ¼ 0:345 an ¼ 0:345� 1996 ¼ 688 W=m2 K


Tubeside aio ¼ 8000� 2125

¼ 6720 W=m2 K Shell side aReb ¼ 2434 W=m2 K

1U

¼ 12434

þ 16720

þ 0:00214

þ 0:0003 U ¼ 998 W=m2 K

Calculation of the heat load Q for LMTD ¼ 20 �C:

Reboiler area A ¼ 132� 0:025� p� 6 ¼ 62 m2

Q ¼ U� A� LMTD ¼ 998� 62� 20 ¼ 1:237 MW

Heat flux density q ¼ U� LMTD ¼ 998� 20 ¼ 19; 960 W=m2

10.2.5 Calculations for a horizontal thermosiphon evaporator withlongitudinal flow around the tubes [8,9]

Calculation steps:• determination of the flow cross-section for the longitudinal streaming around the

tubes ashell• determination of the flow velocity wliq for the liquid rate• determination of the hydraulic diameter dH• calculation of the Reynolds number• determination of the heat transfer coefficient for the liquid aliq flowing shell side


ashell ¼ p

4� �D2

i � n� d2o��m2� dH ¼ D2

i � n� d2on� do

ðmÞ

aliq ¼ 0:023�Re0:8 � Pr0:33 � l

do

�W=m2 K

�for Re > 8000

aconv ¼ M � aliq�W=m2 K

�aReb ¼ aconv þ an

�W=m2 K

�

Example 7: Thermal design of a horizontal thermosiphon evaporator type AELwith longitudinal flow through the shell without bafflesDN 450 with 132 tubes 25 � 2, 6 m long, 32 mm quadratic pitch

Calculation of the heat transfer coefficient for the convective boiling on the shell side:

ashell ¼p

4� �D2

i � n� d2o� ¼ 0:785� �0:452 � 132� 0:0252

� ¼ 0:094 m2

wliq ¼ 853600� 0:094

¼ 0:25 m=s

dH ¼ D2i � n� d2on� do

¼ 0:452 � 132� 0:0252

132� 0:025¼ 0:036 m

Re ¼ wliq � dHv

¼ 0:25� 0:036

0:318� 10�6 ¼ 28; 301

aliq ¼ 0:023� 28; 3010:8 � 4:330:33 � 0:1170:036

¼ 441:7 W=m2 K

Multiplicator for two-phase flow M ¼ 4

aconv ¼ 4� 441:7 ¼ 1767 W=m2 K

Calculation of the heat transfer coefficient for the nucleate boiling:

Pc ¼ 4100 kPa q ¼ 20,000 W/m2

aNB ¼ 0:00626� 41000:69 � 20; 0000:7 ¼ 1996 W=m2 K

Re2Ph ¼ 28; 301� 41:25 ¼ 160; 095 S ¼ 0:243 an ¼ 0:243� 1996 ¼ 486 W=m2 K


Tubeside aio ¼ 6720 W=m2 K Shell side aReb ¼ 2253 W=m2 K

1U

¼ 12253

þ 16720

þ 0:00214

þ 0:0003 U ¼ 966 W=m2 K

Calculation of the heat load Q for LMTD ¼ 20 �C:

Reboiler area A ¼ 132� 0:025� p� 6 ¼ 62 m2

Q ¼ U� A� LMTD ¼ 966� 62� 20 ¼ 1:2 MW

Heat flux density q ¼ U � LMTD ¼ 960 � 22 ¼ 19,320 W/m2


10.3 CALCULATION OF THE TWO-PHASE DENSITYAND THE AVERAGE DENSITY IN THE REBOILER

10.3.1 Calculation of the two-phase density r2Ph of a vaporeliquidmixture

r2Ph ¼ 1xrVþ 1�x

rliq

�kg=m3�

x ¼ vapor fraction in the mixturerV ¼ vapor density (kg/m3)rliq ¼ liquid density (kg/m3)

10.3.2 Calculation of the average density rReb in the reboiler fromthe density of the liquid rliq and the density of the two-phasemixture r2Ph

rReb ¼ 11

r2Ph� 1

rliq

� ln

0BBB@

�1

r2Ph� 1

rliq

�1rliq

1CCCA

Calculation of the average density rReb in the reboiler from the vapor density rV and the liquiddensity rliq and the vapor fraction x:

rReb ¼ 1

x��

1rV� 1

rliq

�� ln

0BBB@x�

�1rV� 1

rliq

�1rliq

1CCCA

Example 8: Calculation of the two-phase density r2Ph and the average densityrReb in the reboiler (Figure 10.3)

x ¼ 0.136 Vapor density rV ¼ 0.68 kg/m3 Liquid density rliq ¼ 750 kg/m3

Calculation of the two-phase density of the mixture:

r2Ph ¼ 1xrVþ 1�x

rliq

¼ 10:1360:68 þ 0:864

750

¼ 4:97 kg=m3


Calculation of the average density in the reboiler:

rReb ¼ 11

4:97� 1750

� ln

1

4:97� 1750

1750

!¼ 25:07 kg=m3

rReb ¼ 1

0:136��

10:68 � 1

750

�� ln

0B@0:136�

�1

0:68 � 1750

�1750

1CA ¼ 25:07 kg=m3

Figure 10.3 Density of the two-phase mixture at the evaporator outlet and average reboiler density inthe evaporator as a function of the vapor rate in the two-phase mixture for two different vapordensities.


10.4 FLOW VELOCITY WREB IN THE REBOILER

wReb ¼ Gr� a� 3600

ðm=sÞ

a ¼ flow cross-section (m2)G ¼ mass flow rate through the evaporator (kg/h)r ¼ density of the two-phase mixture (kg/m3) or the liquid or the average density inthe reboiler

Example 9: Calculation of the two-phase flow velocity in the vertical reboilerReboiler inlet: 20 t/hDensity of the liquid: 750 kg/m3

Flow cross-section ¼ 0.06474 m2 for 187 tubes 25 � 2Flow velocity for the incoming liquid: wein ¼ 0.114 m/sOutlet from the reboiler ¼ 20 t/h with 20 weight% vaporVapor density rG ¼ 5 kg/m3

Density of the two-phase mixture r2Ph ¼ 24.35 kg/m3 at the evaporator outletFlow velocity for the outflowing two-phase mixture: wout ¼ 3.52 m/sLogarithmic averaged velocity wav in the 3-m-long evaporator tubes:

wav ¼ wout � win

ln woutwin

¼ 3:52� 0:114

ln 3:520:114

¼ 1 m=s

Average density rReb in the reboiler:

rReb ¼ 11

24:35 � 1750

� ln

1

24:35 � 1750

1750

!¼ 85:43 kg=m3

Hence the average flow velocity in the reboiler tubes

wav ¼ 20; 00085:43� 0:06474� 3600

¼ 1 m=s

Alternative calculation for the vapor density rV ¼ 0.68 kg/m3

Two-phase density at the evaporator outlet for x ¼ 0.2: r2Ph ¼ 3.388 kg/m3

Flow velocity at the evaporator outlet: 25.32 m/sAverage flow velocity in the reboiler: wa ¼ 4.67 m/sAverage density in the reboiler rReb ¼ 18.36 kg/m3

wav ¼ 20; 00018:36� 3600� 0:06474

¼ 4:67 m=s


10.5 DETERMINATION OF THE REQUIRED HEIGHT H1 FOR THETHERMOSIPHON CIRCULATION OR THE MAXIMUM ALLOWABLEPRESSURE LOSS DPMAX IN THERMOSIPHON CIRCULATION [10]

10.5.1 Calculation of the required elevation height H1 for overcomingthe pressure loss DP in the thermosiphon circulation with thesafety factor S

H1 ¼ S � DP þ g � �H2� r2ph þH4� rReb�

g � rliqðm liquid heightÞ

Calculation of the required driving pressure DPreq for thermosiphon circulation:

DPreq ¼ rliq � g �H1 ¼ S � DP þ g � ðr2Ph �H2þ rReb �H4ÞðPaÞ

10.5.2 Calculation of the maximum allowable pressure loss DPallow inthe thermosiphon circulation at given heights H1, H2, and H4

DPallow ¼ g ��H1� rliq �H2� r2Ph �H4� rReb

�Pa

DP ¼ pressure loss in the thermosiphon circulation (Pa)H1 ¼ driving liquid height (m) (see Figure 10.4)rliq ¼ liquid density (kg/m3)H2 ¼ height of the two-phase mixture in the riser (m)r2Ph ¼ two-phase density (kg/m3)H4 ¼ height of the reboiler (m)rReb ¼ average density in the reboiler (kg/m3)S ¼ safety factor ¼ 2

Example 10: Calculation of the required height H1 for thermosiphon circulation

DP ¼ 69.2 mbar ¼ 6920 Pa rliq ¼ 750 kg/m3

H2 ¼ 2 m r2Ph ¼ 4.97 kg/m3 H4 ¼ 3 m rReb ¼ 25.07 kg/m3

Safety factor S ¼ 2Calculation of the required elevation height H1 for a pressure loss of DP ¼ 69.2 mbar:

H1 ¼ 2� 6920þ 9:81� ð2� 4:97þ 3� 25:07Þ9:81� 750

¼ 1:99 m elevation height with S ¼ 2

Required driving pressure for circulation:

DPreq ¼ rliq � g� H1 ¼ 750� 9:81� 1:99 ¼ 14; 641 Pa

Maximum allowable pressure loss at H1 ¼ 1.99 m:

DPallow ¼ 9:81� ð1:99� 750� 2� 4:97� 3� 25:07Þ ¼ 13; 840 Pa ¼ 2� DP


Figure 10.4 Thermosiphon evaporator with the heights H1 þ H2 þ H4 for the determination of thedriving pressure for the thermosiphon circulation.


Notice:If a part of the tube length is required for heating up the liquid to the boiling temperature this length mustbe considered when determining H1netto.

H1netto ¼ H1� Hheat

Hheat ¼ required heating length ðmÞ

Example 11: Calculation of the available driving pressure Dpavail for thethermosiphon circulation

The calculation is performed for the heights H1 ¼ 2.5 m and H1 ¼ 3.5 m.

H2 ¼ 1.83 m r2 ¼ 24.3 kg/m3

H4 ¼ 3 m r3 ¼ 85.3 kg/m3

r1 ¼ 750 kg/m3

DPavail ¼ g� �H1� rliq � H2� r2Ph � H4� rReb�

H1 ¼ 2.5 m

DPavail ¼ 9:81� ð2:5� 750� 1:83� 24:3� 3� 85:3Þ ¼ 15; 447 Pa ¼ 154 mbar

H1 ¼ 3.5 m

DPavail ¼ 9:81� ð3:5� 750� 1:83� 24:3� 3� 85:3Þ ¼ 22; 805 Pa ¼ 228 mbar


10.5.3 Calculation of the thermosiphon reboiler feed height H1in practice

10.5.3.1 Vertical circulating reboiler (see Figure 10.5)Reference points for H1: weld seam torospherical head/column frame

Calculation of the minimum feed height H1min with the safety factor 2:

H1min ¼ 2� DP þH2� r2Ph þH4� rReb

rliq

¼ 2� DP þ r2Ph � ðH3�H4Þ þH4� rReb

rliq � r2PhðmÞ

DP ¼ pressure loss in the thermosiphon circuit (mm liquid height)

H2 ¼ H1þH3�H4ðmÞ

Figure 10.5 Vertical thermosiphon circulation reboiler.


Example 12: Calculation of the minimum feed height H1 for a vertical circulationreboiler

DP ¼ 300 mm liquid height ¼ 0.3 � 9.81 � 750 ¼ 2207 PaH3 ¼ 2.5 m H4 ¼ 3 mrliq ¼ 750 kg/m3 r2Ph ¼ 24.3 kg/m3 rReb ¼ 85.3 kg/m3

H1 ¼ 2� 300þ 24:3� ð2:5� 3Þ þ 85:3� 3750� 24:3

¼ 1:16 m

H2 ¼ H1þ H3� H4 ¼ 1:16þ 2:5� 3 ¼ 0:66 m

H1 ¼ 2� 300þ 24:3� 0:66þ 85:3� 3750

¼ 1:16 m

Cross-check:

DHallow ¼ H1� rliq � H2� r2Ph � H4� rReb ¼ 1:16� 750� 0:66� 24:3� 3� 85:3

¼ 600 mm liquid height

DPallow ¼ DHallow

1000� g� rliq ¼ 0:6� 9:81� 750 ¼ 4414 Pa ¼ 2� DP

10.5.3.2 Horizontal circulation evaporator (see Figure 10.6)Reference points for H1: weld seam torospherical head/column frame

Calculation of H1 with the safety factor 2:

H1 ¼ 2� DP þH3� r2Ph

rliq � r2Ph¼ 2� DP þH2� r2Ph

rliqðmÞ H2 ¼ H1þH3ðmÞ

Example 13: Calculation of the minimum feed height H1 for a horizontalcirculation reboiler

DP ¼ 500 mm liquid height ¼ 3679 Pa H3 ¼ 2.5 mrliq ¼ 750 kg/m3 r2Ph ¼ 24.3 kg/m3

H1 ¼ 2� 500þ 24:3� 2:5750� 24:3

¼ 1:46 m H2 ¼ H1þ H3 ¼ 1:46þ 2:5 ¼ 3:96 m

H1 ¼ 2� 500þ 24:3� 3:96750

¼ 1:46 m

Cross-check:

DH ¼ rliq � H1� r2Ph � H2 ¼ 750� 1:46� 24:3� 3:96 ¼ 1000 mm liquid height

DPallow ¼ DH1000

� g� rliq ¼ 1� 9:81� 750 ¼ 7358 Pa ¼ 2� DP


10.5.3.3 Vertical once-through evaporator (see Figure 10.7)Reference point for H1: height of the liquid draw support nozzle at the column

With the once-through evaporator, more driving height H1 mostly exists thannecessary.

Calculation of H1 with safety factor S ¼ 2:

H1 ¼ S � DP � r2Ph � ðH4þ DHÞ þ rReb �H4rliq � r2Ph

¼ S � DP þ r2Ph �H2þ rReb �H4rliq

ðmÞ

DP ¼ pressure loss in the thermosiphon circuit (mm liquid height)The distance between the draw nozzle and the vapor return nozzle is normally

DH ¼ 1 m.

H2 ¼ H1�H4� 1 ðmÞ H1 ¼ H2þH4þ 1

Figure 10.6 Horizontal thermosiphon circulation reboiler.


Example 14: Calculation of the minimum feed height H1 for a verticalonce-through reboiler

DP ¼ 2000 mm liquid height ¼ 14,715 Pa H4 ¼ 3 m DH ¼ 1 mrliq ¼ 750 kg/m3 r2Ph ¼ 24.3 kg/m3 rReb ¼ 85.3 kg/m3

H1 ¼ 2� 2000� 24:3� ð3þ 1Þ þ 85:3� 3750� 24:3

¼ 5:73 m H2 ¼ 5:73� 3� 1 ¼ 1:73 m

H1 ¼ 2� 2000þ 24:3� 1:73þ 85:3� 3750

¼ 5:73 m

Figure 10.7 Vertical thermosiphon once-through reboiler.


Cross-check:

DHallow ¼ rliq � H1� r2Ph � H2� rReb � H4 ¼ 750� 5:73� 1:73� 24:3� 3� 86:3

¼ 4000 mm liquid height

DPallow ¼ 4� 9:81� 750 ¼ 29; 430 Pa ¼ 2� DP

10.5.3.4 Horizontal once-through evaporator (see Figure 10.8)Reference point for H1: height of the liquid draw nozzle at the column

Constructive requirement: H2 ¼ H1 � 1 mIn most cases, with the once-through evaporator, the given driving height H1 is

much more than required.Calculation of the elevation height H1 with the safety factor 2:

H1 ¼ 2� DP þ r2Ph �H2rliq

¼ 2� DP � r2Ph � 1rliq � r2Ph

ðmÞ

Figure 10.8 Horizontal thermosiphon once-through reboiler.


Example 15: Calculation of the minimum feed height H1 for a horizontalonce-through reboiler

DP ¼ 1000 mm liquid height ¼ 7358 Pa H1 � H2 ¼ 1 mrliq ¼ 750 kg/m3 r2Ph ¼ 24.3 kg/m3 rReb ¼ 85.3 kg/m

H1 ¼ 2� 1000� 24:3� 1750� 24:3

¼ 2:72 m H2 ¼ H1� 1 ¼ 2:72� 1 ¼ 1:72

H1 ¼ 2� 1000� 24:3� 1:72750

¼ 2:72 m

Cross-check:

DHallow ¼ rliq � H1� r2Ph � H2 ¼ 750� 2:72� 24:3� 1:72 ¼ 2000 mm liquid height

DPallow ¼ DHallow

1000� g� rliq ¼ 2000

1000� 9:81� 750 ¼ 14; 715 Pa ¼ 2� DP

10.5.4 Effects of the elevation height H1 on the circulating rate, thepressure loss, the temperature difference, and the heat load ofthermosiphon evaporators

In the following Figures 10.9e10.13 it is shown how the driving height H1 influences:• the circulating rate W through the thermosiphon evaporator,• the pressure loss by thermosiphon circulation,• the effective temperature difference LMTD and the heat load.

Figure 10.9 With increasing driving height H1 the circulating rate W increases through a verticalthermosiphon evaporator.


10.6 DESIGN OF RISER AND DOWNCOMER DIAMETER

Recommended riser diameter DR for a vertical thermosiphon evaporator:

DR ¼ di �ffiffiffin

p ðmÞ

Figure 10.11 The effective temperature difference LMTD of a steam-heated vertical thermosiphonevaporator decreases with increasing driving height H1, because the higher hydrostatic pressureincreases the boiling pressure.

Figure 10.10 The pressure loss in the evaporator circulation increases with increasing circulating rate.


Figure 10.12 The heat load of a vertical thermosiphon evaporator decreases with increasing heightH1 because the effective temperature difference LMTD decreases.

Figure 10.13 Circulating rates in horizontal thermosiphon evaporators as function of the drivingheight H1 for the evaporator types AJL, AHL, and AEL. Due to the high pressure loss in the evaporatortype AEL a higher driving height H1 is needed.


Recommended minimum flow velocity of the two-phase flow in the riser to avoid pulsingpiston flow with pressure variations in the column:

wmin ¼ffiffiffiffiffiffiffiffi100rV

sðm=sÞ

Recommended maximum flow velocity of the two-phase flow in the riser:

wmax ¼ffiffiffiffiffiffiffiffiffiffi6000r2Ph

sðm=sÞ

Recommended diameter DD for the downcomer:

DD ¼ 12�DR ðmÞ

Recommended flow velocities: in the downcomer: 0.5e1 m/s:di ¼ evaporator tube inner diameter (m)n ¼ number of evaporator tubesrV ¼ vapor density (kg/m3)r2Ph ¼ density of the two-phase mixture (kg/m3)

Example 16: Determination of downcomer and riser diameter

n ¼ 211 tubes di ¼ 21 mm Circulating rate W ¼ 40 t/hrliq ¼ 750 kg/m3 r2Ph ¼ 24.3 kg/m3 rV ¼ 5 kg/m3 x ¼ 0.2Vliq ¼ 53.3 m3/h V2Ph ¼ 1664 m3/h

DR ¼ di �ffiffiffin

p ¼ 0:021�ffiffiffiffiffiffiffi211

p¼ 0:305 m w2Ph ¼ 6:3 m=s

DD ¼ 12� DR ¼ 0:5� 305 ¼ 0:15 m wFl ¼ 0:84 m=s

wmin ¼ffiffiffiffiffiffiffi100rV

s¼

ffiffiffiffiffiffiffi1005

r¼ 4:5 m=s

wmax ¼ffiffiffiffiffiffiffiffiffiffi6000r2Ph

s¼

ffiffiffiffiffiffiffiffiffiffi600024:3

r¼ 15:7 m=s

Recommended nozzle spacing in the columnHeight between the liquid head in the column and the two-phase inlet nozzle of the reboiler: ca. 0.5 m

Height between the last bottom tray in the column and the two-phase inlet nozzle of the reboiler:0.5e0.8 m

Liquid height in the column bottom: 1e1.5 m


10.7 CALCULATION OF THE PRESSURE LOSSES IN THETHERMOSIPHON CIRCULATION

The total pressure loss in a thermosiphon circulation consists of the following(Figure 10.14):• pressure loss in the downcomer for the liquid inflow• friction pressure loss for the two-phase flow in the reboiler tubes• acceleration pressure loss in the reboiler tubes• friction pressure loss in the riser for the two-phase flow

10.7.1 Liquid pressure loss in the downcomer

DPDownc ¼ f � Ld� w2

Downc � rliq

2ðPaÞ

f ¼ friction factor ¼ 0:216Re0:2

L ¼ downcomer length (m)d ¼ diameter (m)wDownc ¼ flow velocity in the downcomer (m/s)rliq ¼ liquid density (kg/m3)

Figure 10.14 Pressure losses in a thermosiphon reboiler.


Example 17: Calculation of the pressure loss in the downcomer (Figure 10.15)

Downcomer length L ¼ 25 m Downcomer diameter ¼ 0.125 mFlow rate ¼ 30 m3/h Flow velocity wDownc ¼ 0.68 m/sKinematic viscosity n ¼ 0.4 mm2/s Reynolds number Re ¼ 212,314Friction factor f ¼ 0.0185 Liquid density rF ¼ 750 kg/m3

DPDownc ¼ 0:0185� 250:125

� 0:682 � 7502

¼ 641 Pa ¼ 6:4 mbar

10.7.2 Friction pressure loss for two-phase flow in the evaporatorAccording to LockharteMartinelli the fractional single-phase flow pressure loss of gas orliquid is converted to the pressure loss of the two-phase flow with the correction factors(Figure 10.16)

F2V or F2

liq

DP2ph ¼ F2V � DPV ¼ F2

liq � DPliq

DPV ¼ pressure loss for fractional vapor flow (Pa)DPliq ¼ pressure loss for fractional liquid flow (Pa)

X ¼ffiffiffiffiffiffiffiffiffiffiDPliqDPV

s

Figure 10.15 Pressure loss in the downcomer for two nominal widths as function of the circulating rate.


For the correction factors F2G and F2

F it follows:

F2liq ¼ 1þ C

Xþ 1

X2 F2V ¼ 1þ C � X þ X2

Turbulent liquid and gas flow: C ¼ 20Laminar liquid and turbulent gas flow: C ¼ 12Turbulent liquid and laminar gas flow: C ¼ 10Laminar liquid and gas flow: C ¼ 5

10.7.3 Accelerated pressure loss in the evaporatorDue to the evaporation the flow volume increases and the flow in the evaporator tubesgets accelerated. The hereby caused pressure loss is calculated as follows:

DPacc ¼ m2 �

1r2Ph

� 1rliq

!¼ m� ðwout � winÞðPaÞ

m ¼ mass flow density (kg/m2 s)r2Ph ¼ density of the two-phase mixture

rliq ¼ liquid density (kg/m3)win ¼ inlet flow velocity of the liquid (m/s)wout ¼ outlet flow velocity of the two-phase mixture (m/s)

Figure 10.16 Two-phase pressure loss in the reboiler for different vapor densities as function of thevapor fraction in the two-phase mixture.


Example 18: Calculation of the accelerated pressure lossIncoming liquid rate: 22 t/h in 187 tubes 25 � 2 (Figure 10.17)

rliq ¼ 750 kg/m3 r2Ph ¼ 4.97 kg/m3 win ¼ 0.125 m/s wout ¼ 18.99 m/s

m ¼ 22; 000

3600� 0:0212 � 0:785� 187¼ 94:39 kg=m2 s

DPacc ¼ 94:392 ��

14:97

� 1750

�¼ 1781 Pa ¼ 17:8 mbar

DPacc ¼ 94:39� ð18:99� 0:125Þ ¼ 1780 Pa ¼ 17:8 mbar

10.7.4 Friction pressure loss in the riser for the two-phase flowAccording to LockharteMartinelli the pressure loss of the fractional single-phase flow ofgas or liquid is converted into the pressure loss of the two-phase flow with the correctionfactors F2

V or F2liq.

DP2Ph ¼ F2V � DPV ¼ F2

liq � DPliq

DPV ¼ pressure loss for fractional vapor flowDPliq ¼ pressure loss for fractional liquid flow

X ¼ffiffiffiffiffiffiffiffiffiffiDPliqDPV

s

Figure 10.17 Accelerated pressure losses in the thermosiphon evaporator as function of the vapor-ization rate for different vapor densities.


For the correction factors F2V and F2

liq it follows:

F2liq ¼ 1þ C

Xþ 1

X2 F2liq ¼ 1þ C � X þ X2

Turbulent liquid and gas flow: C ¼ 20Laminar liquid and turbulent gas flow: C ¼ 12Turbulent liquid and laminar gas flow: C ¼ 10Laminar liquid and gas flow: C ¼ 5

Example 19: Calculation of the two-phase pressure loss in the riser

Riser diameter: 250 mm Equivalent length with elbow and reducers: L ¼ 30 m

The two-phase mixture consists of 5 t/h vapor and 15 t/h liquid.

5 t/h vapor rV ¼ 2 kg/m3 wV ¼ 14.15 m/s nV ¼ 0.085 mm2/s15 t/h liquid rliq ¼ 750 kg/m3 wliq ¼ 0.113 m/s nliq ¼ 0.85 mm2/s

Initially, the pressure losses for the liquid and the vapor flow are determined:

Reliq ¼ 33; 235 Friction factor f ¼ 0:216

33; 2350:2¼ 0:027

DPliq ¼ 0:027� 300:25

� 0:1132 � 7502

¼ 15:4 Pa

ReV ¼ 41; 617 Friction factor f ¼ 0:216

41; 6170:2¼ 0:0257

DPV ¼ 0:0257� 300:25

� 14:152 � 22

¼ 617:5 Pa

X ¼ffiffiffiffiffiffiffiffiffiffiffi15:4617:5

r¼ 0:158

For turbulent gas and liquid flow C ¼ 20.

F2V ¼ 1þ 20� 0:158þ 0:1582 ¼ 4:18

Fliq ¼ 1þ 200:158

þ 1

0:1582¼ 167:6

DP2Ph ¼ 4:18� 617:5 ¼ 167:6� 15:4 ¼ 2581 Pa

10.7.5 Total pressure loss in the thermosiphon circulationThe total pressure loss in the thermosiphon circuit is shown in Figure 10.18.

With increasing vapor rate the pressure loss in the thermosiphon circulation increasesbecause due to the growing vapor fraction the flow velocity in the reboiler tubes increases.


Figure 10.19 shows that the pressure loss at lower vapor densities is clearly larger thanat higher vapor densities because with small vapor densities the volumetric fraction of thevapor is larger and hence also the flow velocity of the two-phase flow.

Design data:Circulation rate: 20 t/hReboiler with 187 tubes 25 � 2, 3 m long

10.7.6 Reboiler characteristic curvesFor the design of a thermosiphon circle it is recommended to create the so-called“reboiler characteristic curves.” Therein the calculated pressure loss in the thermosiphoncirculation is plotted against the liquid circulation. In the same diagram the available pres-sure potential for the liquid circulation at different driving heightsH1 is plotted. From thereboiler characteristic curves it can be read off which elevation height H1 for a certainliquid circulation is required (Figure 10.20).

Figure 10.18 Pressure losses in the thermosiphon circulation at a vapor density of 5 kg/m3 as a func-tion of the vapor rate at 20 t/h liquid circulation.


10.8 CALCULATION OF THE REQUIRED REBOILER LENGTH OR AREAFOR THE HEATING UP TO THE BOILING TEMPERATURE AND FORTHE EVAPORATION IN VERTICAL THERMOSIPHONEVAPORATORS

In Figure 10.21 it is shown that in the thermosiphon evaporator, two processes occur:heating of the liquid up to the boiling temperature at the pressure in the evaporatorand subsequently the actual evaporation. From point B to point C the liquid is heated

Figure 10.19 Pressure losses in the thermosiphon circulation for different vapor densities and 20 t/hcirculation rate.

Figure 10.20 Reboiler characteristic curves for the evaporation of 2 t/h vapor.


up to the boiling temperature. The temperature rises and the pressure decreases becausethe static pressure is lower due to the superimposed liquid height at C. The evaporationstarts at point C. From point C to point D the static pressure falls and the boiling tem-perature decreases.

10.8.1 Required reboiler length and area for heating up to the boilingtemperature

Lheat ¼Gliq � cliq � Dtheat

n� d � p� Uheat � LMTDheatðmÞ Aheat ¼ Qheat

Uheat � LMTDheat

�m2�

Gliq ¼ liquid inlet into the evaporator (kg/h)cliq ¼ specific heat capacity of the liquid (Wh/kg K)Dtheat ¼ required heating temperature up to the boiling point (�C)n ¼ number of tubesd ¼ tube diameter

Figure 10.21 Heating and evaporation zones in the vertical thermosiphon reboiler with pressure andtemperature curve.


Uheat ¼ overall heat transfer coefficient for the convective heating (W/m2 K)LMTDheat ¼ temperature difference for the heating (�C)Qheat ¼ heat duty for the heating (W)The required temperature difference for the heating up to the boiling temperature is

initially estimated, for example, 3e10 �C.The overall heat transfer coefficient Uheat for the heating is determined from the heat

transfer coefficients aliq for the heating of the liquid product and asteam for condensingsteam (asteam z 5000 W/m2 K):

The heat transfer coefficient aliq for the heating of the product in the tube is deter-mined with the formula for the convective heat transfer in the tube.

Due to the low flow velocity at the inflow of the liquid into the evaporator(w z 0.1e0.25 m/s) the aliq-values are very low:

w ¼ 0.1 m/s Re ¼ 5250 Pr ¼ 5.4 Nu ¼ 32.81 aliq ¼ 187.5 W/m2 Kw ¼ 0.15 m/s Re ¼ 7875 Pr ¼ 5.4 Nu ¼ 49.28 aliq ¼ 281 W/m2 Kw ¼ 0.2 m/s Re ¼ 10,500 Pr ¼ 5.4 Nu ¼ 66.13 aliq ¼ 377.9 W/m2 Kw ¼ 0.25 m/s Re ¼ 13,125 Pr ¼ 5.4 Nu ¼ 79.05 aliq ¼ 451.7 W/m2 K

Example 20: Required tube length for convective heating to the boiling point

Gliq ¼ 20,000 kg/h cliq ¼ 0.6 Wh/kg K n ¼ 187 tubes d ¼ 0.025 m

Heating steam temperature tsteam ¼ 165 �C Boiling temperature ¼ 155 �CProduct inlet temperature ¼ 150 �C Estimated value for aliq ¼ 350 W/m2 K

Required heating Dtheat ¼ 155 � 150 �C ¼ 5 �C asteam ¼ 5000 W/m2 K

Product temperature 150 / 155

Steam temperature 165 z 165

Differences 15 10

LMTDheat ¼ 15� 10

ln1510

¼ 12:3 �C1

Uheat¼ 1

350þ 15000

þ 0:00025 Uheat ¼ 302 W=m2 K

Chosen : Uheat ¼ 300 W=m2 K

Lheat ¼ 20; 000� 0:6� ð155� 150Þ187� 0:025� p� 300� 12:3

¼ 1:11 m

Aheat ¼ 20; 000� 0:6� 5300� 12:3

¼ 16:3 m2

Check: Aheat ¼ 1.11 � 187 � 0.025 � p ¼ 16.3 m2


10.8.2 Required tube length Levap and area Aevap for evaporation

L evap ¼ GV � rn� d � p� Uevap � LMTDevap

ðmÞ

Aevap ¼ GV � rUevap � LMTDevap

�m2�

GV ¼ evaporated rate (kg/h) r ¼ latent heat (Wh/kg)Uevap ¼ overall heat transfer coefficient for the evaporation (W/m2 K)LMTDevap ¼ temperature difference for the evaporation (K)

Example 21: Calculation of the required tube length for the evaporation at 155 �C

GV ¼ 3000 kg/h r ¼ 80 Wh/kg Uevap ¼ 1000 W/m2 KHeating steam temperature: 165 z 165 �CBoiling temperature: 155 ➔ 155 �C LMTD ¼ 10 �C

Levap ¼ GV � rn� d � p� Uevap � LMTDevap

¼ 3000� 80187� 0:025� p� 1000� 10

¼ 1:63 m

Aevap ¼ GV � rUevap � LMTDevap

¼ 3000� 801000� 10

¼ 24 m2

Cross-check: Aevap ¼ 1.63 � 187 � 0.025 � p ¼ 24 m2

Alternative calculation for Uheat ¼ 180 W/m2 K and Uevap ¼ 800 W/m2 K

Lheat ¼ 1.84 m Aheat ¼ 27.1 m2

Levap ¼ 2.04 m Aevap ¼ 30 m2

The fractional area for the heating requires almost 50%!

10.9 REQUIRED HEATING LENGTH FOR VERTICAL THERMOSIPHONREBOILERS ACCORDING TO FAIR [2]

The required tube length fraction Xheat for the heating is calculated as follows (seeFigure 10.22):

PB � PPB � PA

¼ X ¼ Dt=DPDt=DLDP=DL

þ DtDP

DPDL

¼ rliq � g

Lðmbar=mÞ

DtDL

¼ n� d � p� LMTDheat � Uheat

Gliq � cliqð�C=mÞ

Dt/DL ¼ heating gradient of the product per meter tube length L (�C/m)DP/DL ¼ pressure increase due to the liquid height in the tube (mbar/m)Dt/DP ¼ temperature gradient of the vapor pressure curve (�C/mbar)


The value Dt/DP must be determined from the vapor pressure values.In Table 10.1 the temperature gradients for some products are listed.With increasing temperature or at higher pressures the temperature gradient

decreases.In vacuum the temperature gradient is larger and so a greater heating surface area for

the heating up to the boiling temperature is required.

Example 22: Calculation of the fractional tube length Xheat for the heating

n ¼ 187 tubes d ¼ 0.025 m LMTD ¼ 12.3 �C Uheat ¼ 300 W/m2 KGliq ¼ 20,000 kg/h cliq ¼ 0.6 Wh/kg K rliq ¼ 750 kg/m3 Ltot ¼ 3 m

DtDL

¼ 187� 0:025� p� 12:3� 30020; 000� 0:6

¼ 4:51 �C=m

DPDL

¼ 750� 9:811

¼ 7357 Pa=m ¼ 73:57 mbar=m

Dt/DP ¼ 0.03 �C/mbar from the vapor pressure curve

Xheat ¼ 0:034:5173:57þ 0:03

¼ 0:328

Required tube length for the heating Lheat ¼ Xheat � Ltot ¼ 0.328 � 3 ¼ 0.98 mThe fractional length for the heating becomes shorter if the heating rate for Dt/DL (�C/m tube) is

enlarged, for instance, by more tubes or a better overall heat transfer coefficient for the heating ormore heating area by structured surfaces (fins) or by the installation of turbulators.

Example:Dt/DL ¼ 4.5 �C/m➔ Lheat ¼ 0.98 mDt/DL ¼ 12 �C/m➔ Lheat ¼ 0.47 m

Table 10.1 Temperature gradients of the vapor pressure curve

ProductTemperaturerange (�C)

Temperaturegradient (�C/mbar)

Benzene 50/70 0.05480/100 0.02594/104 0.021

Hexane 60/68 0.03570/80 0.02782/92 0.021

Toluene 90/100 0.05110/120 0.031126/136 0.022

Heptane 68/80 0.06990/100 0.036103/123 0.024

Octane 70/100 0.097120/130 0.036132/152 0.025


The temperature gradient Dt/DP is especially large in the low-pressure region. Therefore, in vacuumevaporators the required length for the heating is particularly large.

In Figure 10.22 the Xheat-value as the function of the heating gradient for different temperature gradi-ents of the vapor pressure curve is shown. The Xheat-value increases with rising temperature gradient of thevapor pressure curve because more has to be heated.

The required heating area is reduced by a better heating.

Example 23: Determination of the required heating length for different heatinggradientsReboiler height H4 ¼ 3 m

Temperature gradient of the vapor pressure curve Dt/DP ¼ 2.2 �C/m FS

Dt/DL ¼ 2 �C/m Xheat ¼ 0.52 Hheat ¼ 1.57 mDt/DL ¼ 4 �C/m Xheat ¼ 0.35 Hheat ¼ 1.06 mDt/DL ¼ 6 �C/m Xheat ¼ 0.27 Hheat ¼ 0.8 m

Figure 10.22 The fractional heating tube length as a function of the temperature gradient of thevapor pressure.


10.10 CALCULATION OF THE PRESSURE AND BOILING POINTINCREASE BY MEANS OF THE LIQUID HEIGHT H1

The pressure in the evaporator is increased due to the required driving height H1 for thethermosiphon circulation and the required height for heating up to the boilingtemperature.

In Figure 10.23 it is shown how the pressure in the reboiler rises with increasingdriving height H1.

Due to the higher pressure above the evaporating area the boiling point becomeshigher and the effective temperature difference for the heat transfer is reduced.

For the actual function of the reboiler, the evaporation, the heating wall temperaturemust lie above the boiling point. In multicomponent mixtures additionally the temper-ature rise from boiling point to dew point must be considered. Therefore the liquidcirculation through the thermosiphon evaporator must be correspondingly heated takingthe boiling point increase into account (Figure 10.24).

The calculations are performed as follows:Initially the pressure is calculated at point C and the pressure increase between point A

and C is determined. The required boiling point increase for this pressure rise is thendetermined.

PA ¼ pressure above the liquid in the column bottom (Pa) or (mbar)PB ¼ pressure at the bottom in the evaporatorPC ¼ pressure above the liquid level in the reboilerIn Table 10.2 it is shown that a higher inflow height H1 increases the pressure in the

evaporator and hence also the boiling point (Figure 10.23).

Figure 10.23 Pressure rise and boiling point increase in a thermosiphon evaporator as function of thedriving height.


The required heating length is derived from the quotient boiling point increasedivided by the temperature increase by heating per meter tube.

PB ¼ PA þH1� rliq � 9:81ðPaÞ PC ¼ PB �Hheat � r1 � 9:81ðPaÞPC ¼ PA þ DP þ 9:81� ½H2� r2Ph þ ðH4�HheatÞ � r3�ðPaÞDPAC ¼ PC � PAðPaÞ DP ¼ Pressure loss in the reboiler ðPaÞ

Boiling point increase DtAC ¼ DPAC � Dt/DP (�C)Required heating length Hheat ¼ DtAC/Dt/DL (m)In the normal case at atmospheric evaporation H1 ¼ H4.The liquid level in the column is at the level of the top tube sheet of the evaporator.H1netto ¼ H1 � Hheat ¼ H1 � (1 � Xheat)H4 ¼ Hheat þ Hevap

Example 24: Calculation of the pressure and boiling point increase

PA ¼ 1 bar H1 ¼ 3 m H2 ¼ 2 m H4 ¼ 4 m Hheat ¼ 1 mrliq ¼ 750 kg/m r2Ph ¼ 47.1 kg/m rReb ¼ 135.8 kg/m DP ¼ 9800 Pa

PB ¼ 100,000 þ 3 � 750 � 9.81 ¼ 122,072 PaPC ¼ 122,072 � 9.81 � (1 � 750) ¼ 114,715 PaPC ¼ 100,000 þ 9800 þ 9.81 � (2 � 47.1 þ 3 � 135.8) ¼ 114,720 PaPressure increase between A and C ¼ DPAC ¼ 114,720 � 100,000 ¼ 14,720 Pa ¼ 147.2 mbarTemperature increase at pressure rise Dt/DP ¼ 0.03 �C/mbarBoiling temperature increase between A and C ¼ DtAC ¼ 0.03 � 147.2 ¼ 4.41 �C

Table 10.2 Boiling point increase at different H1-valuesH1 (m) Hheat (m) PB (mbar) DPBC (mbar) PC (mbar) DPAC (mbar) DtAC (�C)

2 0.65 1147 47.8 1099 99 2.973 0.98 1220 72 1148 148 4.444 1.31 1294 96.3 1198 198 5.94

Figure 10.24 Pressures PA, PB, and PC in the reboiler circuit.


Required heating length for Dt/DL ¼ 4.5 �C/m tubeHheat ¼ 4.41/4.5 ¼ 0.98 m tube length

Basic data: XHeiz ¼ 0.328 Dt/DP ¼ 0.03 �C/mbar Dt/DL ¼ 4.5 �C/m tube

The pressures at the points A, B, and C as a function of the driving height H1 is shown in Figure 10.25.In evaporations in vacuum the driving height H1 is reduced to 50% of the reboiler height H4 in order to

reduce the pressure increase in the evaporator and the required heating height. The liquid level in the col-umn is at half of the reboiler height. The following Example 25 makes it clear.

Example 25: Boiling point increase and required heat length at vacuumevaporationHeating gradient Dt/DL ¼ 0.054 �C/mbar ¼ 4 �C/m tube length

Temperature gradient of the vapor pressure curve Dt/DP ¼ 0.06796 �C/mbar ¼ 5 �C pro m FS

Xheat ¼ 54þ 5

¼ 0:555

Case 1: H1 ¼ 3 m H4 ¼ 3 m Hheat ¼ Xheat � H4 ¼0.555 � 3 ¼ 1.67 m

rliq ¼ 750 kg/m3 r2Ph ¼ 25.07 kg/m3 PA ¼ 100 mbar ¼ 10,000 Pa

PB ¼ 10,000 þ 9.81 � 3 � 750 ¼ 32,072 PaPC ¼ 32,072 � 9.81 � 1.67 � 750 ¼ 19,785 PaDPAC ¼ 19,785 � 10,000 ¼ 9785 Pa ¼ 9785

9:81�750 ¼ 1:33 m FS ¼ 97.85 mbarTemperature increase due to the pressure rise of 97.85 mbar:

DtAC ¼ 0.06796 � 97.85 ¼ 6.65 �C DtAC ¼ 5 � 1.33 ¼ 6.65 �C

Hheat ¼ DtACDt=DL

¼ 6:65ð�CÞ4ð�C=m tube lengthÞ ¼ 1:66 m

Figure 10.25 Pressure curve at the point A, B, and C as a function of the driving height H1.


Case 2: H1 ¼ 1.5 m Hheat¼ Xhat � H1¼ 0.555� 1.5¼ 0.83 m

PB ¼ 10,000 þ 9.81 � 1.5 � 750 ¼ 21,036 PaPC ¼ 21,036 � 9.81 � 0.83 � 750 ¼ 14,930 PaDPAC ¼ 14,930 � 10,000 ¼ 4930 Pa ¼ 49.3 mbar ¼ 0.67 m FSTemperature increase by the pressure rise of 49.3 mbar:

DtAC ¼ 0.06796 � 49.3 ¼ 3.35 �C DtAC ¼ 5 � 0.67 ¼ 3.35 �C

Hheat ¼ DtACDt=DL

¼ 3:35ð�CÞ4ð�C=m tube lengthÞ ¼ 0:83 m

10.11 AVERAGE OVERALL HEAT TRANSFER COEFFICIENT FORHEATING D VAPORIZING

For the calculation of an average overall heat transfer coefficient Ua the following is validwith Xheat:

Ua ¼ Xheat � Uheat þ ð1� XheatÞ � Uevap�W=m2 K

�Since the driving temperature differences for the heating and the vaporizing are

different, the calculation of an average overall heat transfer coefficient is not quite correct.

Example 26: Calculation of the average overall heat transfer coefficient Ua

Xheat ¼ 0.4 Uheat ¼ 300 W/m2 K Uevap ¼ 2000 W/m2 K

Ua ¼ Xheat � Uheat þ (1 � Xheat) � Uevap ¼ 0.4 � 300 þ 0.6 � 2000 ¼ 1320 W/m2 KThe good U-value for the vaporizing is substantially deteriorated by the heating.

10.12 CALCULATION OF THE VAPOR FRACTION X OF THE TWO-PHASE MIXTURE IN A VERTICAL REBOILER

The evaporated product rate changes above the height of the evaporator. At the inlet ofthe liquid in the evaporator the vapor fraction x ¼ 0. The calculation of the vapor frac-tion in different heights L is carried out as follows:

x ¼ q� L � n� d � p

r �W

GV ¼ x�W ðkg=hÞ

GV ¼ q� Ar

¼ q� ðn� d � p� LÞr

ðkg=hÞ


W ¼ circulating rate through the thermosiphon evaporator (kg/h)GV ¼ vapor rate (kg/h)q ¼ heat flux density (W/m2) ¼ U � Dtr ¼ heat of vaporization (Wh/kg)x ¼ vapor content in the two-phase mixture (fraction)n ¼ number of tubesd ¼ tube diameter (m)L ¼ tube length for the evaporation (m)The x-values at different heights are required for the determination of the heat trans-

fer coefficients and the pressure losses at the different zones of the evaporator. The calcu-lation should be subdivided into several zones because the conditions change stronglyabove the height due to the different vapor content x in the two-phase mixture.

Example 27: Calculation of the vapor content at different tube lengths

q ¼ 10,000 W/m2 n ¼ 187 tubes d ¼ 0.025 m r ¼ 80 Wh/kgL ¼ 1.63 m W ¼ 20,000 kg/h

x ¼ q� L� n� d � p

r �W¼ 10; 000� 1:63� 187� 0:025� p

80� 20; 000¼ 0:15

GV ¼ x �W ¼ 0:15� 20; 000 ¼ 3000 ðkg=hÞ

GV ¼ q� Ar

¼ q� ðn� d � p� LÞr

ðkg=hÞ ¼ 10; 000� 0:025� 187� 1:63� p

80¼ 3000 ðkg=hÞ

With these equations the vapor content x in the different heights of the evaporator above the heatinglength can be determined:

L ¼ 0.5 m x ¼ 0.0458 916 kg/h vaporL ¼ 1.0 m x ¼ 0.0918 1836 kg/h vaporL ¼ 1.4 m x ¼ 0.128 2560 kg/h vaporL ¼ 1.63 m x ¼ 0.150 3000 kg/h vapor

In Figure 10.26 it is shown how the two-phase density and the flow velocity of the two-phase mixturechange with increasing vaporization.

10.13 THERMOSIPHON REBOILER DESIGN EXAMPLE

Heat load Q ¼ 600 kW r ¼ 150 Wh/kg c ¼ 0.6 Wh/kg �CRequired heating Dt ¼ 4 �C x ¼ 0.2 kg/kg vapor fractionVapor density rD ¼ 5 kg/m3 Liquid density rF ¼ 750 kg/m3

Calculation of the circulating rate for x ¼ 0.2:

W ¼ 600; 0000:6� 4þ 0:2� 150

¼ 18; 518 kg=h

GV ¼ 0:2� 18; 518 ¼ 3704 kg=h


Thermosiphon circulation W ¼ 18,518 kg/hEvaporation rate D ¼ 3704 kg/h vapors

Circulating rates at other evaporation rates x:

x ¼ 0.1 W ¼ 34,483 kg/h D ¼ 3448 kg/hx ¼ 0.15 W ¼ 24,096 kg/h D ¼ 3614 kg/hx ¼ 0.2 W ¼ 18,518 kg/h D ¼ 3704 kg/hx ¼ 0.25 W ¼ 15,038 kg/h D ¼ 3759 kg/hx ¼ 0.3 W ¼ 12,658 kg/h D ¼ 3797 kg/h

Estimation of the reboiler area A:

Uest ¼ 700 W/m2 K Dt ¼ 15 �C L ¼ 3.5 m Tubes 25 � 2

A ¼ 600; 000700� 15

¼ 57:1 m2

n ¼ 57:1p� 0:025� 3:5

¼ 207 tubes 25� 2

Chosen: 211 tubes with triangular pitch 32 mmShell inner diameter Di ¼ 483.2 mm for 211 tubes with pitch 32 mmReboiler area A ¼ n � da � p � L ¼ 211 � 0.025 � p � 3.5 ¼ 58 m2

Calculation of the two-phase density and the reboiler density and the flow velocities in thereboiler:

r2Ph ¼ 1xrG

þ 1�xrF

¼ 10:15 þ 0:9

750

¼ 47:1 kg=m3

Figure 10.26 Two-phase density and flow velocity as a function of the evaporation rate.


r3 ¼ 11

47:1 � 1750

� ln

"1

47:1� 1750

1750

#¼ 135:8 kg=m3

Two-phase and reboiler densities for other x-values:

x (kg/kg) r2Ph (kg/m3) r3 (kg/m

3)0.1 47.1 135.80.15 32.1 104.20.2 24.3 85.30.25 19.6 72.80.3 16.4 63.7

Flow velocity in the 211 tubes for W ¼ 34,483 kg/h and x ¼ 0.1:Flow cross-section at ¼ 211 � 0.0212 � p/4 ¼ 0.07308 m2

Liquid velocity at the inlet:

wliq ¼ 34; 483750� 0:07308� 3600

¼ 0:175 m=s

Two-phase flow velocity at the outlet:

w2Ph ¼ 34; 48347:1� 0:07308� 3600

¼ 2:78 m=s

Average flow velocity in the reboiler:

wm ¼ 34; 483135:8� 0:07308� 3600

¼ 0:96 m=s

Calculation of the heat transfer and overall heat transfer heat coefficientHeat transfer coefficient for convective heat transfer

wliq ¼ 0.175 m/s Re ¼ 9187 Pr ¼ 5.4

Nu ¼ 0:023�Re0:8 � Pr0:33 ¼ 0:023� 91870:8 � 5:40:33 ¼ 59:42

aliq ¼ Nu� l

di¼ 59:42� 0:12

0:021¼ 339:57

Xtt ¼�1� 0:10:1

�0:9

��

5750

�0:5

��

0:30:012

�0:1

¼ 0:8139

M ¼ 2:5��

10:8139

�0:7

¼ 2:88

aconv ¼ 2:88� 339:6 ¼ 980:6�W=m2 K

�


Heat transfer coefficient for the boiling of mixtures according to MostinskiaNB ¼ 822 W/m2 K

Re2Ph ¼ 9187� 2:881:25 ¼ 34; 468

S ¼ 11þ ð2:52� 10�6 � 34; 4681:17Þ ¼ 0:66

an ¼ 0:66� 822 ¼ 542:7 W=m2 K

Total heat transfer coefficient for the evaporation:aReb ¼ 980.6 þ 542.7 ¼ 1523 W/m2 KConversion of ai to the outer tube surface area:

aio ¼ ai � dida

¼ 1523� 2125

¼ 1279 W=m2 K

Calculation of the overall heat transfer coefficient U for steam heating withasteam ¼ 7000 W/m2 K:

aio ¼ aReb ¼ 1279 W/m2 K asteam ¼ 7000 W/m2 K Sf ¼ 0.0002

1U

¼ 1aReb

þ 1asteam

þ Sf ¼ 11279

þ 17000

þ 0:0002 U ¼ 889 W=m2 K

The determined overall heat transfer coefficient U ¼ 889 W/m2 K for the evaporation islarger then the estimated U-value Uest ¼ 700 W/m2 K but it must also be heated up.

Cross-checking of the heat duty Q:

Q ¼ U �A � Dt ¼ 889 � 58 � 15 ¼ 773,452 W Reserve ¼ 29%Heating duty Qheat ¼ 34,483 � 0.6 � 4 ¼ 82,759 W LMTD ¼ 19 �C

Required heating area AHeiz for Uheat ¼ 340 W/m2 K AHeiz ¼ 82;759304�19 ¼ 14:3 m2

Vaporization duty Qverd ¼ 3448 � 150 ¼ 517,200 W LMTD ¼ 15 �CRequired vaporization area Aevap for UReb ¼ 889 W/m2 K AVerd ¼ 517;200

889�15 ¼ 38:8 m2

Total required area Areq ¼ 14.3 þ 38.8 ¼ 53.1 m2 < 58 m2 9% reserve!

Recommendation: Enlargement of the area by longer tubes in order to improve thereserve.

L ¼ 4 m length A ¼ 66.3 m2 25% reserve!

Calculation of the required driving height H1 for an assumed pressure lossDP ¼ 40 mbarin the thermosiphon circulation by downcomer, evaporator, and riser:

Riser height H2 ¼ 2 m Reboiler height H4 ¼ 4 mrliq ¼ 750 kg/m3 r2Ph ¼ 47.1 kg/m3 rReb ¼ 135.8 kg/m3

H1 ¼ 2� 4000þ 9:81� ð2� 47:1þ 4� 135:8Þ9:81� 750

¼ 1:94 m


Chosen: H1 ¼ 2 mAllowable pressure loss DPzulDPallow ¼ 9.81 � (2 � 750 � 2 � 47.1 � 4 � 135.8) ¼ 8462 Pa ¼ 84.6 mbarUnder consideration of the heating height HHeiz in the reboiler:If a part of the tube length in the evaporator is required for the heating, the heating

length Hheiz must be considered when determining the h eight H1.

H4 ¼ 4 m H2 ¼ 2 m Hheat ¼ 0.5 m

H1 ¼ S � DP þ g � ðH2� r2Ph þ rReb � ðH4�HheatÞ þHheat � r1

g � r1

H1 ¼ 2� 4000þ 9:81� ð47:1� 2þ 135:8� 3:5þ 0:5� 750Þ9:81� 750

¼ 2:35 m

The required height H1 becomes higher.Cross-checking the allowable pressure loss:DPallow ¼ g � (H1 � rliq � H2 � r2ph � (H4 � Hheat) � rReb � Hheat � rliq)DPzul ¼ 9.81 � (2.35 � 750 � 2 � 47.1 � 3.5 � 135.8 � 0.5 � 750) ¼8025 Pa ¼ 80.2 mbarCalculation of the diameters for downcomer and riser for W ¼ 34,483 kg/h:

DRiser ¼ 0:21�ffiffiffiffiffiffiffiffi211

p¼ 0:3 m Chosen : D ¼ 0:25 m

DDownc ¼ 0:5� 0:3 ¼ 0:15 m

wRiser ¼ 732

3600� 0:252 � 0:785¼ 4:14 m=s

wDownc ¼ 46

3600� 0:152 � 0:785¼ 0:72 m=s

wmin Riser ¼ffiffiffiffiffiffiffiffi1005

r¼ 4:47 m=s wmax Riser ¼

ffiffiffiffiffiffiffiffiffiffi600047:1

r¼ 11:3 m=s

REFERENCES[1] N.P. Liebermann, E.T. Liebermann, Working Guide to Process Equipment, McGraw-Hill, N.Y.,

2003.[2] J.R. Fair, What you need to design thermosiphon reboilers, Petr. Ref. 39 (2) (1960).[3] G.A. Hughmark, Designing Thermosiphon Reboilers, CEP 57, 1961, 7, CEP 60 (1964), 7 und CEP

65 (1969), 7.[4] J.C. Chen, Correlation for boiling heat transfer to saturated liquids in convective flow, Ind. Eng. Proc.

Des. Dev. 58 (3) (1966).[5] D.L. Love, No hassle reboiler selection, Hydroc. Process. (October 1992).[6] E. Chen, Optimize reboiler design, Hydroc. Process. (July 2001).


[7] G.R. Martin, A.W. Sloley, Effectively design and simulate thermosyphon reboiler systems, Hydroc.Process. (JuneeJuly 1995).

[8] J.R. Fair, A. Klip, Thermal design of horizontal reboilers, Chem. Eng. Progr. 79 (3) (1983).[9] G.K. Collins, Horizontal thermosiphon reboiler design, Chem. Eng. (July 1976).[10] R. Kern, How to design piping for reboiler systems, Chem. Eng. (June 1975).


CHAPTER 11

Double Pipe, Helical Coil, and CrossFlow Heat Exchanger

Contents

11.1 Double Pipe and Multipipe Heat Exchangers 22911.1.1 Calculation of double pipe and multipipe heat exchangers 22911.1.2 Calculation of the overall heat transfer coefficient U and driving temperature gradient

LMTD and the required heat exchanger area A 23211.2 Helical Coil Heat Exchanger 23711.3 Cross Flow Bundle 241Nomenclature 244References 245

11.1 DOUBLE PIPE AND MULTIPIPE HEAT EXCHANGERS

The calculations of the heat transfer coefficient follow the equations for the heat transferin tubes.

On the shell side, the different hydraulic diameters for the determination of the heattransfer coefficient dh and the pressure loss dh

0must be considered.

Double pipe heat exchangers are applied for low heat loads, small volume flows, andhigh pressures (Figure 11.1).

Multitube heat exchangers with a pipe shell and several inner tubes without baffleshave a more exchange area than the double pipe equipments.

11.1.1 Calculation of double pipe and multipipe heat exchangers [1,2]The calculations are according to the equations for the heat transfer in the tube.

Laminar : Re < 2300

Nu ¼ 1:86�Re0:33 � Pr0:33 ��diL

�0:33

Intermediate region: 2300 < Re < 8000

Nu ¼ �0:037�Re0:75 � 6:66

�� Pr0:42



http://dx.doi.org/10.1016/B978-0-12-803764-5.00011-0

Turbulent: Re > 8000

Nu ¼ 0:023�Re0:8 � Pr0:33 Pr ¼ n� c � r� 3600l

ai ¼ Nu� l

di

�W

�m2 K

�

Figure 11.1 Double pipe heat exchanger.


Flow velocity and Reynolds number in the tube:

wt ¼ Vt

n� d2i �p

4

ðm=sÞ Re ¼ wt � din

Flow cross-section ashell and flow velocity wshell in the annulus or shell:

ashell ¼ p

4� �

D2i � n� d2o

��m2� wshell ¼ Vshell

ashell � 3600ðm=sÞ

For the calculation of the heat transfer coefficient and the pressure loss, the hydraulicdiameter dh for heat exchange and dh

0for pressure loss are needed.

Reynolds number for the heat transfer with dh: Re ¼ wshell � dhnHydraulic diameter for double tube heat exchanger:

dh ¼ D2i � n� d2on� do

ðmÞ

Hydraulic diameter for multitube heat exchanger:

dh ¼ D2i � d2odo

ðmÞ

Reynolds number Re0 for the pressure loss with dh0: Re0 ¼ wshell � d0h

nDouble tube heat exchanger: d0h ¼ Di � do

Multitube heat exchanger: d0h ¼ D2i � n� d2o

Di þ n� doPressure loss calculation:

Nozzle pressure loss DPnozz ¼ 1:5� wnozz � r

2

Tube-side pressure loss DPt ¼�f � L

diþ z

�� w2

t � r

2ðPaÞ

Shell-side pressure loss DPshell ¼ f � Ld0h

� w2shell � r

2ðPaÞ

Friction factor f ¼ 0:275Re0:2

Double Pipe, Helical Coil, and Cross Flow Heat Exchanger 231

11.1.2 Calculation of the overall heat transfer coefficient U and drivingtemperature gradient LMTD and the required heat exchangerarea A

1U

¼ 1aio

þ 1ao

þ slwall

þ fi þ fo

LMTD ¼ Dt1 � Dt2

lnDt1Dt2

ð�CÞ A ¼ QU � LMTD

�m2�

Example 1: Double pipe heat exchanger design

Heat load Q ¼ 3.53 kW LMTD ¼ 32.46 �CInner tube 21.3 � 2 (di ¼ 17.3 mm) Shell pipe 33.7 � 2 (Di ¼ 29.7 mm)VT ¼ 0.05 m3/h Vshell ¼ 0.3 m3/hr ¼ 1000 kg/m3 r ¼ 1000 kg/m3

n ¼ 0.41 mm2/s n ¼ 0.8 mm2/sc ¼ 1.16 Wh/kg K c ¼ 1.16 Wh/kg Kl ¼ 0.66 W/m K l ¼ 0.61Pr ¼ 2.594 Pr ¼ 5.476

Tube side:

w T ¼ 0:053600� 0:01732 � p=4

¼ 0:059 m=s

Re ¼ 0:059� 0:0173

0:41� 10�6 ¼ 2490

Nu ¼ �0:037� 24900:75 � 6:66

�� 2:5940:42 ¼ 9:52

ai ¼ 9:52� 0:660:0173

¼ 363 W�m2 K aio ¼ 363� 17:3

21:3¼ 295 W

�m2 K

Shell side:

ashell ¼p

4� �

0:02972 � 1� 0:02132� ¼ 336:3� 10�6 m2

dh ¼ 0:02972 � 0:02132

0:0213¼ 0:0201 Re ¼ 0:248� 0:0201

0:8� 10�6 ¼ 6230

Nu ¼ �0:037� Re0:75 � 6:66

�� 5:4760:42 ¼ 39:39


ao ¼ 39:39� 0:610:0201

¼ 1195 W�m2 K

Overall heat transfer coefficient U:

1U

¼ 11195

þ 1295

þ 0:00214

þ 0:0002 ¼ 0:00457 U ¼ 219 W�m2 K

Areq ¼ 353032:46� 219

¼ 0:5 m2 Lreq ¼ 0:5p� 0:0213

¼ 7:5 m pipe

Pressure loss calculation for 2 � 3.75 m long tubes:Tube side:

DPnozz ¼ 1:5� 0:0592 � 10002

¼ 2:6 Pa f ¼ 0:275

24900:2¼ 0:058

DPt ¼�0:058� 7:5

0:0173þ 1

�� 0:0592 � 1000

2¼ 46 Pa

DPtot ¼ 2:6þ 46 ¼ 48:6 Pa

Shell side:

2 inlet and 2 outlet nozzles DN 20 wnozz ¼ 0.265 m/s for 0.3 m3/h

DPnozz ¼ 3� 0:2652 � 10002

¼ 105 Pa

d0h ¼ Di � da ¼ 0:0297� 0:0213 ¼ 0:0084 m

Re0 ¼ 0:248� 0:00840:8� 10�6 ¼ 2604

Friction factor f ¼ 0:275

Re0:2¼ 0:275

26040:2¼ 0:057

DPshell ¼�0:057� 7:5

0:0084þ 1

�� 0:2482 � 1000

2¼ 1596 Pa

DPtot ¼ 105þ 1596 ¼ 1701 Pa


Example 2: Shell-side heat transfer coefficient in a multitube heat exchanger

Vshell ¼ 2.3 m3/h Pr ¼ 7.1 l ¼ 0.58 W/m K n ¼ 1 mm2/s

Shell diameter Di ¼ 84 mm with seven inner pipes 20 � 2, 3 m long

ashell ¼ p

4� �

0:0842 � 7� 0:022� ¼ 0:0033 m2

dh ¼ 0:0842 � 7� 0:022

7� 0:02¼ 0:0304 m d0h ¼ 0:0842 � 7� 0:022

0:084þ 7� 0:02¼ 0:019 m

wshell ¼2:3

0:0033� 3600¼ 0:194 m=s Re ¼ 0:194� 0:0304

1� 10�6 ¼ 5897

Calculation of the heat transfer coefficient on the shell side with Re ¼ 5897:

Nu ¼ �0:037� 58970:75 � 6:66

�� 7:10:42 ¼ 41:5

ao ¼ Nu� l

dh¼ 41:5� 0:58

0:0304¼ 792 W

�m2 K

Example 3: Comparison of double pipe or multitube heat exchangersCalculation of the double pipe heat exchanger:

Tube side:

5.2 m3/h 20/30 �C Q ¼ 51,480 Wr ¼ 1100 kg/m3 l ¼ 0.4 W/m K c ¼ 0.9 Wh/kg Kn ¼ 2.4 mm2/s wT ¼ 1.248 m/s Pr ¼ 21.38Inner pipe 42.4 � 2 mm di ¼ 38.4 mm

Re ¼ 1:248� 0:0384

2:4� 10�6 ¼ 19; 968 Nu ¼ 0:023� 19; 9680:8 � 21:380:33 ¼ 174:14

ai ¼ 174:14� 0:40:0384

¼ 1814 W�m2K aio ¼ 1814� 38:4

42:4¼ 1643 W

�m2 K

Shell side:

Vshell ¼ 5.2 m3/h 80/71.25 �C Pr ¼ 2.35l ¼ 0.666 W/m K n ¼ 0.385 mm2/s c ¼ 1.16 Wh/kg K r ¼ 974.65 kg/m3

Outer tube 70 � 2 di ¼ 66 mm

ashell ¼p

4� �

0:0662 � 0:04242� ¼ 0:002 m2 wshell ¼ 0:719 m=s

dh ¼ 0:0662 � 0:04242

0:0424¼ 0:06033 m Re ¼ 0:719� 0:06033

0:385� 10�6 ¼ 112; 668

Nu ¼ 335:4 a ¼ 335:4� 0:6660:06033

¼ 3703 W�m2 K


Calculation of the overall heat transfer coefficient U:

1U

¼ 11643

þ 13703

þ 0:00250

þ 0:0002 ¼ 0:001119 U ¼ 894 W�m2 K

LMTD ¼ 50:6 �C

Areq ¼ 51; 480894� 50:6

¼ 1:138 m2 Lreq ¼ 1:138p� 0:0424

¼ 8:54 m tube

Chosen: 3 � 3 m ¼ 9 m pipe.Pressure loss calculation for double pipe heat exchanger, 3 � 3 m ¼ 9 m tube length.Tube side:

Nozzle pressure loss DPnozz ¼ 1:5� 1:2482 � 11002

¼ 1285 Pa

Piping pressure loss DPT with friction factor f ¼ 0.034 for Re ¼ 19,968

DPt ¼�0:034� 9

0:0384þ 2

�� 1:2482 � 1100

2¼ 8540 Pa

DPtot ¼ 1285þ 8540 ¼ 9825 Pa

Shell side with three inlet and outlet nozzles DN 50Nozzle flow velocity wSt ¼ 0.736 m/s

DPnozz ¼ 3� 1:5� 0:7362 � 974:62

¼ 1188 Pa

d0h ¼ 0:066� 0:0424 ¼ 0:0236 m

Re0 ¼ 0:719� 0:0236

0:385� 10�6 ¼ 44; 073

Friction factor f ¼ 0.0278 for Re0 ¼ 44,073

DPshell ¼ 0:0278� 90:0236

� 0:7192 � 974:62

¼ 2674 Pa

DPtot ¼ 2674þ 1188 ¼ 3862 Pa

Calculation of the multipipe heat exchanger:Shell pipe 76.1 � 2, di ¼ 72.1 mm with seven inner tubes 16 � 1Tube side: Cross-section area aT ¼ 0.0011077 m2

wt ¼ 5:20:001077� 3600

¼ 1:34 m=s Re ¼ 1:34� 0:014

2:4� 10�6 ¼ 7816

Pr ¼ 21:38 Nu ¼ 87:2

ai ¼ 87:2� 0:40:014

¼ 2491 W�m2 K aio ¼ 2491� 14

16¼ 2179 W

�m2 K


Shell side:

ashell ¼p

4� �

0:07212 � 7� 0:0162� ¼ 0:00267 m2

wshell ¼ 0:54 m=s Pr ¼ 2:35

dh ¼ 0:07212 � 7� 0:0162

7� 0:016¼ 0:0304 m Re ¼ 0:54� 0:0304

0:385� 10�6 ¼ 42; 639

Nu ¼ 0:023� 42; 6390:8 � 2:350:33 ¼ 154:2

ao ¼ 154:2� 0:660:0304

¼ 3378 W�m2 K

1U

¼ 12179

þ 13378

þ 0:00250

þ 0:0002 ¼ 0:000995 U ¼ 1005 W�m2 K

Areq ¼ 51; 4801005� 50:6

¼ 1:01 m2 Lreq ¼ 1:01p� 7� 0:016

¼ 2:9 m

A heat exchanger with seven tubes of 16 � 1, 3 m long is required.

Pressure loss calculation for multipipe heat exchanger:Tube side:

Nozzles DN 32 wnozz ¼ 1:8 m=s

DPnozz ¼ 1:5� 1:82 � 11002

¼ 2673 Pa

Friction factor f ¼ 0.043 for Re0 ¼ 7816

DPT ¼ 0:043� 2:90:014

� 1:342 � 11002

¼ 8800 Pa

DPtot ¼ 2673þ 8800 ¼ 11; 473 Pa

Shell side:

Nozzles DN 50 Nozzle flow velocity wSt ¼ 0:736 m=s

DPnozz ¼ 1:5� 0:7362 � 974:62

¼ 396 Pa

d0h ¼ 0:0185 m wshell ¼ 0:54 m=s

Re0 ¼ 0:54� 0:0185

0:385� 10�6 ¼ 25; 948

f ¼ 0:032 for Re0 ¼ 25; 948

DPshell ¼ 0:032� 2:90:0185

� 0:542 � 974:62

¼ 708 Pa

DPtot ¼ 708þ 396 ¼ 1104 Pa


11.2 HELICAL COIL HEAT EXCHANGER [3,4]

These equipments are used for low and average heat duties or low flow rates in order toachieve a higher flow velocity for the heat transfer. The heat transfer areas are greater thanthat of the double pipe heat exchangers, and the heat transfer is a bit better due to thecentrifugal effect.

Geometrical calculations:

Spiral total tube length L L ¼ N �ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiðDH � pÞ2 þ T 2

qðmÞ

Spiral total surface area A A ¼ QU � Dt

ðm2Þ

Number of spiral windings N N ¼ Ap� do � L

NSpiral height H H ¼ N � T þ do (m)

Tube-side heat transfer coefficient calculation:

Re > 8000

Nu ¼ 0:023�Re0:8 � Pr0:33 ai ¼ Nu� l

diaSP ¼ ai �

�1þ 3:5� di

DH

�

aSPa ¼ aSP � dido



1U

¼ 1aSPo

þ 1ashell

þ sWlW

þ fi þ fo

Calculation of the shell-side flow cross-section ashell and the flow velocity wshell:

ashell ¼ p

4� ��

D2i �D2

K

�� D2Ho �D2

Hi

��m2�

wshell ¼ Vshell

ashell � 3600ðm=sÞ

Calculation of ashell with the hydraulic diameter De:

De ¼ 4� Vf

p� do � L¼ 4� ðVa � VcÞ

p� do � LðmÞ

Va ¼ p

4� �

D2i �D2

K

�� T �N�m3� Vc ¼ p

4� d2o � L

�m3�

Re ¼ wshell �De

n

Nu ¼ 0:6�Re0:5 � Pr0:31

ashell ¼ Nu� l

De

�W�m2 K

�

Alternative calculation of ashell with the outer tube diameter do:

Re ¼ wshell � don

Nu ¼ 0:196�Re0:6 � Pr0:33

ashell ¼ Nu� l

do

�W�m2 K

�

Example 4: Design of a helical coil heat exchanger

Di ¼ 0:46 m DK ¼ 0:34 m DH ¼ 0:4 m

do ¼ 30 mm di ¼ 25 mm

T ¼ 45 mm ¼ 1:5� da

Tube side : 1:55 m3�h 127 �C/100 �C

Shell side : 2:3 m3�h 30 �C/47 �C

Q ¼ 42; 400 W LMTD ¼ 74:9 �C F ¼ 0:99 CMTD ¼ 0:99� 74:9 ¼ 74:1 �C


Tube-side calculation:

di ¼ 25 mm V ¼ 1.55 m3/h r ¼ 870 kg/m3

n ¼ 0.719 � 10�6 m2/s c ¼ 1.16 Wh/kg K l ¼ 0.486 W/m Kw ¼ 0.877 m/s

Re ¼ 0:877� 0:025

0:719� 10�6 ¼ 30; 513 Pr ¼ 0:719� 10�6 � 1:16� 870� 36000:486

¼ 5:37

Nu ¼ 0:023� Re0:8 � Pr0:33 ¼ 0:023� 30; 5130:8 � 5:370:33 ¼ 155:7

ai ¼ 155:7� 0:4860:025

¼ 3027 W�m2 K

aSP ¼ 3027��1þ 3:5� 0:025

0:4

�¼ 3690 W

�m2 K

aSPo ¼ 3690� 2530

¼ 3075 W�m2 K

Shell-side calculation with the hydraulic diameter De:

L ¼ N �ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiðDH � pÞ2 þ T2

q¼ N �

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffið0:4� pÞ2 þ 0:0452

q¼ 1:257� N

Va ¼ p

4� �

0:462 � 0:342�� 0:045� N ¼ 0:0034� N

Vc ¼ p

4� 0:032 � 1:257� N ¼ 0:0009� N

Vf ¼ Va � Vc ¼ ð0:0034� 0:0009Þ � N ¼ 2:504� 10�3 � N

De ¼ 4� 2:504� 10�3 � Np� 0:03� 1:257� N

¼ 84:52� 10�3 m


Calculation of the flow cross-section ashell:DHa ¼ Di � da ¼ 460 � 30 ¼ 430 mmDHi ¼ DK þ da ¼ 340 þ 30 ¼ 370 mm

ashell ¼p

4� ��

0:462 � 0:342��

0:432 � 0:372� ¼ 0:0377 m2

Calculation of the flow velocity wshell for Vshell ¼ 2.3 m3/h:

wshell ¼2:3

0:0377� 3600¼ 0:017 m=s

Calculation of the heat transfer coefficient ashell with De:

v ¼ 1:71� 10�6 m2�s c ¼ 1:16 Wh=kg K r ¼ 935 kg�m3 l ¼ 0:473 W=m K

Pr ¼ 1:71� 10�6 � 1:16� 935� 36000:473

¼ 14:11

Re ¼ w � De

n¼ 0:017� 0:0845

1:71� 10�6 ¼ 840

Nu ¼ 0:6� Re0:5 � Pr0:31 ¼ 0:6� 8400:5 � 14:110:31 ¼ 39:5

ashell ¼39:5� 0:473

0:0845¼ 221 W

�m2 K


aSPo ¼ 3075 W/m2 K ashell ¼ 221 W/m2 K

1U

¼ 13075

þ 1221

þ 0:002550

þ 0:0002 ¼ 0:0051 U ¼ 196 W�m2 K

Areq ¼ 42; 40074:1� 196

¼ 2:92 m2 Nreq ¼ 2:92p� 0:03� 1:257

¼ 24:6

Chosen: N ¼ 25

L ¼ 25 � 1.257 ¼ 31.4 m H ¼ 25 � 0.045 þ 0.03 ¼ 1.155 m

Alternative calculation of ashell with do ¼ 30 mm:

Re ¼ 0:017� 0:03

1:71� 10�6 ¼ 298:2

Nu ¼ 0:196� Re0:6 � Pr0:33 ¼ 0:196� 298:20:8 � 14:110:33 ¼ 14:46

ashell ¼14:46� 0:473

0:03¼ 228 W

�m2K

U ¼ 201 W/m2 K


11.3 CROSS FLOW BUNDLE [5,6,7]

Cross stream bundles are rectangular tube bundles which are predominantly used for gascooling or heating. It will be shown how the heat transfer coefficients and pressure lossesin flows through the cross stream bundles are determined.

The convective heat transfer coefficient in the tubes is treated in Chapter 3.

Application cases:• air cooler for cooling or condensing of products in the tubes• air or water preheater with hot flue gases

Due to the poor heat transfer coefficients on the gas side, finned tubes are often usedto increase the efficiency.

Calculation of the heat transfer coefficients according to Grimison [5]:

Nu ¼ K �ReM � fR

a ¼ Nu� l

d

�W�m2 K

�

The constants K andM as a function of the cross and longitudinal ratio for inline andstaggered arrangement are listed in Table 11.1.

As from 10 tube rows, the correction factor is fR ¼ 1.

Cross pitch ratio sC ¼ PCdo

Longitudinal pitch ratio sL ¼ PLdo

PC ¼ Cross pitch PL ¼ Longitudinal pitch


For the pressure loss, the following holds:

DP ¼ 2� n� w2 � r

2ðPaÞ

2 ¼ pressure loss constant n ¼ number of tube rowsw ¼ velocity (m/s) r ¼ density (kg/m3)

According to Jakob [7], the following holds for aligning arrangement:

2 ¼ Re�0:15 �"0:176þ 0:32� sL

ðsC � 1Þð0:43þ1:13=sLÞ

#

According to Jakob, the following holds for the staggered arrangement:

2 ¼ Re�0:16 �"1þ 0:47

ðsC � 1Þ1:08#

Table 11.1 K- and M-values for air at different cross- and longitudinal pitch ratios accordingto Grimison [5]SC 1.25 1.5 2 3

K M K M K M K M

SL Inline tube banks

1.25 0.348 0.592 0.275 0.6080.100

0.704 0.0633 0.752

1.5 0.367 0.586 0.250 0.6200.101

0.702 0.0678 0.744

2 0.418 0.570 0.299 0.6020.229

0.632 0.198 0.648

3 0.290 0.601 0.357 0.5840.374

0.581 0.286 0.608

SL Staggered tube banks

0.6 e e e e e e 0.213 0.6360.9 e e e e 0.446 0.571 0.401 0.5811.0 e e 0.497 0.558 e e e e1.125 e e e e 0.478 0.565 0.5l8 0.5601.25 0.518 0.556 0.505 0.554 0.519 0.556 0.522 0.5621.5 0.451 0.568 0.460 0.562 0.452 0.568 0.488 0.5682 0.404 0.572 0.416 0.568 0.482 0.556 0.449 0.5703 0.310 0.592 0.356 0.580 0.440 0.562 0.421 0.574


Example 5: Heat transfer coefficients and the pressure drop in a cross flowbundles

air at 200 �C and 1 bar Pr ¼ 0.685 9 ¼ 0.7457 kg/m3

l ¼ 0.0368 W/mK n ¼ 34.63 mm2/s d ¼ 38 mmn ¼ 20 sC ¼ 1.5 sL ¼ 1.5

The heat transfer coefficients and the pressure losses at different flow velocities w are calculated.

w (m/s) 5 10 15 20 25Re 5486 10,973 16,460 21,946 27,433

Aligned arrangement: K ¼ 0.25 M ¼ 0.62Nu 52 79.97 102.8 122.9 141.1a (W/m2 K) 52.8 81.2 104.5 124.8 143.4z 0.348 0.313 0.295 0.282 0.273DP (Pa) 65 233.9 495 843 1274

Staggered arrangement: K ¼ 0.46 M ¼ 0.562Nu 58.1 85.8 107.7 126.6 143.6a (W/m2 K) 59 87.1 109.4 128.6 145.8z 0.5027 0.449 0.421 0.4027 0.3886DP (Pa) 93.7 335.5 707.5 1201.3 1811

Example 6: Calculation of the heat transfer coefficient for 18,000 m3/h air withthe physical properties of Example 5 for a tube bundle with Atot ¼ 1 m2 face areawith different cross and longitudinal pitches

a) sC ¼ 3 sL ¼ 1.5 K ¼ 0.0678 M ¼ 0.744 Ages ¼ 1 m2

Afree ¼ Atot � sC � 1sC

¼ 1� 3� 13

¼ 0:666 m2

w ¼ 18; 0000:666� 3600

¼ 7:5 m=s Re ¼ w � dn

¼ 7:5� 0:038

34:63� 10�6 ¼ 8230

Nu ¼ 55.5 a ¼ 56.4 (W/m2 K)

b) sC ¼ 3Nu ¼ 68.7

sL ¼ 3a ¼ 69.8 W/m2 K

K ¼ 0.286 M ¼ 0.608

c) sC ¼ 2Afrei ¼ 0.5 m2

sL ¼ 1.5w ¼ 10 m/s

K ¼ 0.101Re ¼ 10,973

M ¼ 0.702Nu ¼ 69.28

a ¼ 70.4 W/m2 K

d) sC ¼ 2Nu ¼ 81.9

sL ¼ 2a ¼ 83.2 W/m2 K

K ¼ 0.229 M ¼ 0.632

e) sC ¼ 1.5Afrei ¼ 0.333 m2

sL ¼ 1.5w ¼ 15 m/s

K ¼ 0.25Re ¼ 16,460

M ¼ 0.62Nu ¼ 102.8

a ¼ 104.5 W/m2 K

f) sC ¼ 1.5Nu ¼ 103.5

sL ¼ 3a ¼ 105 W/m2 K

K ¼ 0.357 M ¼ 0.584

g) sC ¼ 1.25Afrei ¼ 0.2 m2

sL ¼ 1.5w ¼ 25 m/s

K ¼ 0.367Re ¼ 27,433

M ¼ 0.586Nu ¼ 146.4

a ¼ 148.7 W/m2 K

h) sC ¼ 1.25Nu ¼ 134.8

sL ¼ 3a ¼ 137 m2 K

K ¼ 0.29 M ¼ 0.601

NOMENCLATUREAshell flow cross-section in the shell (m2)aio tube-side heat transfer coefficient based on the tube outer surface area (W/m2 K)ao shell-side heat transfer coefficient (W/m2 K)ai tube-side heat transfer coefficient (W/m2 K)h dynamic viscosity (mPa s)l heat conductivity of the flowing product (W/m K)lwall heat conductivity of the tube wall (W/m K)n kinematic viscosity (m2/s)r density (kg/m3)Areq required heat exchange area (m2)c specific heat capacity (Wh/kg K)do tube outer diameter (m)dh hydraulic diameter for heat transfer coefficient calculation (m)dh0hydraulic diameter for pressure drop calculation (m)

Di shell inner diameter (m)di tube inner diameter (m)f friction factor (-)fi, fo fouling factors tube and shell side (m2K/W)U overall heat transfer coefficient (W/m2 K)L tube length (m)LMTD logarithmic mean temperature difference (K)n number of inner tubes (-)Nu Nubelt number (-)Pr Prandtl number (-)Dpshell pressure loss in the shell (Pa)DpT pressure loss in the tube (Pa)Dpnozz pressure loss in the nozzle (Pa)Q heat load (W)Re Reynolds number (-)Vshell volumetric flow in the shell (m3/s)VT volumetric flow in the tube (m3/s)wshell flow velocity in the shell (m/s)wT flow velocity in the tube (m/s)wnozz flow velocity in the nozzle (m/s)z number of the 180�-turns (-)Di shell inner diameterDK core tube outer diameterDH average spiral diameterdo tube outer diameterdi tube inner diameterT spiral pitch ¼ 1.5 � doN number of spiral windingsDHo outer spiral diameter ¼ Di � doDHi inner spiral diameter ¼ DK þ doa inner gap w15 mm or do/2b outer gap w15 mm or do/2L total tube length


REFERENCES[1] G.P. Purohit, Thermal and hydraulic design of hairpin and finned-bundle exchangers, Chem. Eng.

90 (1983) 62e70.[2] D.A. Donohue, Heat transfer and pressure drop in heat exchangers”, Ind. Eng. Chem. 41 (1949), 2499

and Petrol.Ref. 34(1955), No. 10 and No. 11.[3] R.K. Patil, B.W. Shende, P.K. Ghosh, Designing a helical-coil heat exchanger, Chem. Eng. 13 (12)

(1982).[4] S.S. Haraburda, Consider helical-coil heat exchangers, Chem. Eng. 102 (1995) 149e152.[5] E.D. Grimison, Trans. ASME 59 (1937) 583e594.[6] O.L. Pierson, Trans. ASME 59 (1937) 563.[7] M. Jakob, Trans. ASME 60 (1938) 381e392.


CHAPTER 12

Finned Tube Heat Exchangers

Contents

12.1 Why Finned Tube Heat Exchangers? 24712.2 What Parameters Influence the Effectiveness of Finned Tubes? 24912.3 Finned Tube Calculations 251

12.3.1 Calculation of the fin efficiency hF 25112.3.2 Calculation of the weighted fin efficiency hW 25112.3.3 Calculation of the overall heat transfer coefficient Ui for the inner tube surface area

Ai without fouling 25312.3.4 Calculation of the overall heat transfer coefficient Uo based on the outer tube area

Ao without fouling 25412.3.5 Calculation of the overall heat transfer coefficient Uo based on the finned outer surface

area Ao considering the fouling factors ro and ri 25512.3.6 Calculation of the overall heat transfer coefficient Ui for the inner tube surface area

Ai considering the fouling factors ro and ri 25612.3.7 Fouling and Temperature Gradient 25612.3.8 Comparison of the specific heat duties Ui � Ai (W/m K) of different tubes 258

12.4 Application Examples 259References 264

12.1 WHY FINNED TUBE HEAT EXCHANGERS?

Finned tubes are used if the heat transfer coefficient on the outside of the tubes is verymuch lower than the heat transfer coefficient inside the tubes. The key point for theheat transfer coefficient is the heat conduction through the boundary layer d on the tube.

a ¼ l

d

�W=m2 K

�

In media having poor heat conductivity l, for instance, gases or for high viscousmaterials with larger boundary film thickness d, the heat load Q can be improved byan outer area enlargement.

Q ¼ ao � Ao � Dto ¼ ai � Ai � Dti

The required area ratio considering the fouling factors is calculated as follows:�Ao

Ai

�req

�1aoþ ro

1aiþ ri



http://dx.doi.org/10.1016/B978-0-12-803764-5.00012-2

Ao ¼ outer surface area of the tube (m2/m)Ai ¼ inner surface area of the tube (m2/m)ao ¼ outer heat transfer coefficient (W/m2 K)ai ¼ inner heat transfer coefficient (W/m2 K)ro ¼ outer fouling factorri ¼ inner fouling factor

Example 1: Calculation of the required area ratio

a0 ¼ 100 W�m2 K ai ¼ 1000 W

�m2 K r0 ¼ 0:0001 ri ¼ 0:0002

�AoAi

�req

¼1100 þ 0:00011

1000 þ 0:0002¼ 8:5

Due to the poor outer heat transfer coefficient, the outer surface area should be larger than the innersurface tube area by 8.5!

Advantages of finned tubes• higher heat load per m tube or per construction volume• smaller equipment dimensions/less tubes• smaller flow cross-section in the tubes/better heat transfer• less pressure loss

A distinction is made between the following types of finned tubes:• high-finned cross-finned tubes with finned heights of 10e16 mm• high-finned longitudinal-finned tubes with finned heights of 12.7e25 mm• low-finned cross-finned tubes with finned heights of 1.5e3 mm

The high-finned cross-finned tubes are used in air coolers and gas heat exchangers.The longitudinal-finned tubes are used as vessel heater or in double pipe heat exchangers for high

viscous media, for instance oil or bitumen, becausedas opposed to the cross-finned tubesdthe distancesbetween the fins are much greater than the laminar boundary layer on the tube so that a flow between thefins is possible.

For a face flow length of 10 mm, the following laminar boundary layer thickness results depending onthe Reynolds number Re at plates:

Re ¼ 100 j d ¼ 4:64 mm

Re ¼ 200 j d ¼ 3:28 mm

Re ¼ 500 j d ¼ 2:07 mm

The fin spacing must be larger than the boundary layer thickness! Low-finned tubes are used to increasethe efficiency in existing heat exchangers or to reduce the size of the equipment, for instance, for therefrigerant evaporation and condensation or for the decrease of the construction heights of heatingbundles.

At convective heat transfer on the shell side, the Reynolds number should be >500.


12.2 WHAT PARAMETERS INFLUENCE THE EFFECTIVENESSOF FINNED TUBES?

In Figure 12.1, it can be seen how strong the outer surface area of a tube with 20 mminner diameter increases depending on the fin height.

With increasing fin height, however, the fin efficiency hF, which includes the tem-perature drop from the core tube up to the fin tip falls.

This is shown in Figure 12.2. It can also be seen how strong the heat conductivity ofthe fin material has an influence on the fin efficiency.

The different curves for various ao-values in Figure 12.3 show the influence of theouter heat transfer coefficient ao on the fin efficiency hF.

Figure 12.1 Outer tube surface area as a function of the fin height for a core tube diameter of 20 mm.

Figure 12.2 Fin efficiency hF as a function of the fin height and the fin material.

Finned Tube Heat Exchangers 249

With increasing fin height and a rising ao-value, the fin efficiency falls.Larger fin heights are only of interest with small heat transfer coefficients on the outer

side of the tubes, for instance, with gases.• Large fin heights are economic for small outer heat transfer coefficients up to

ao ¼ 50 W/m2 K.• Low-finned tubes have an advantage at higher ao-values up to 1000 W/m2 K.

The deterioration by the fin efficiency hF of the effective Dt for the heat transfer isonly valid for the fin surface area and not for the core tube outer surface area.

This is considered in the weighted fin efficiency hW, which is lightly betterthan the fin efficiency hF. From economic view, the weighted fin efficiency shouldbe >80%.

The following listing shows the use of different steel-finned tubes with anexample:

Heat duty Q ¼ 100 kW Dt ¼ 50 K ai ¼ 3000 W/m2 K

Figure 12.3 Fin efficiency hF as a function of the fin height for different outer heat transfer coefficientsof ao ¼ 40e800 W/m2 K.


The required tube length L for the heat duty Q ¼ 100 kW is determined fordifferent ao-values and finned tube types.

ao ¼ 50 W/m2 K ao ¼ 500 W/m2 KType 1: hF ¼ 16 mm dC ¼ 38 mm L ¼ 340 m L ¼ 145 mType 2: hF ¼ 10 mm dC ¼ 20 mm L ¼ 905 m L ¼ 313 mType 3: hF ¼ 1.5 mm dC ¼ 22.2 mm L ¼ 1695 m L ¼ 317 mPlain tube25 � 2

L ¼ 4348 m L ¼ 623 m

hF ¼ fin heightdC ¼ core tube diameter

12.3 FINNED TUBE CALCULATIONS [1e3]

12.3.1 Calculation of the fin efficiency hF

hF ¼ tanh XX

tanh X ¼ eX � e�X

eX þ e�X

X ¼ hF �ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi2� ao

bF � lF

r

ao ¼ outer heat transfer coefficient (W/m2 K)lF ¼ heat conductivity of the fin material (W/m K)bF ¼ fin width (m)hF ¼ fin height (m)Correction for disk-finned tubes in which XDF is inserted for the calculation of the fin

efficiency hF instead of X:

XDF ¼ X ��1þ 0:35� ln

dFdC

�

dF ¼ fin diameter (mm)dC ¼ core tube diameter (mm)

12.3.2 Calculation of the weighted fin efficiency hW

hF ¼ hF � AF þ AC

Ao


Ao ¼ total outer surface area ¼ AF þ AC (m2)AC ¼ core tube surface area (m2)AF ¼ fin surface area (m2)

Example 2: Calculation of the fin efficiency for ao ¼ 40 W/m2 K

dC ¼ 20 mm dF ¼ 40 mm Ao ¼ 0.55 m2/mbF ¼ 0.3 mm lF ¼ 50 W/m K di ¼ 16 mmhF ¼ 10 mm AC ¼ 0.07 m2/m AF ¼ 0.48 m2/m

X ¼ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

2� 400:0003� 50

r� 0:01 ¼ 0:7303

XDF ¼ 0:7303��1þ 0:35� ln

4020

�¼ 0:9075 tanh XDF ¼ 0:7199

hF ¼ tanh XX

¼ 0:71990:9075

¼ 0:7933

hW ¼ 0:7933� 0:48þ 0:070:55

¼ 0:82

Example 3: The same data as in Example 2, but ao ¼ 300 W/m2 K

X ¼ 0:01�ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi2� 300

0:0003� 50

r¼ 2

XDF ¼ 2:48

hF ¼ 0:3968

hW ¼ 0:3968� 0:48þ 0:070:55

¼ 0:4736

Example 4: The same data as in Example 2, but ao ¼ 1000 W/m2 K

X ¼ 0:01�ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi2� 1000

0:0003� 50

r¼ 3:65

XDF ¼ 4:5373

hF ¼ 0:22 hW ¼ 0:32


12.3.3 Calculation of the overall heat transfer coefficient Ui for theinner tube surface area Ai without fouling

1Ui

¼ 1aoi

þ 1ai

þ slF

aoi ¼ ao � hF �Ao

Ai¼ ao

Ai� ðhF � AF þ ACÞ

Qi ¼ Ui � Ai � Dt ðWÞao ¼ outer heat transfer coefficient (W/m2 K)aoi ¼ outer heat transfer coefficient based on the core tube surface areaai ¼ inner heat transfer coefficient (W/m2 K)lt ¼ heat conductivity of the core tube (W/m K)s ¼ tube wall thickness of the core tube (m)Dt ¼ driving temperature gradient (K)Qi ¼ heat duty in relation to the inner tube surface area (W)

Example 5: Calculation of the overall heat transfer coefficient Ui based on thetube inner surface area Ai per m tube without fouling consideration

ai ¼ 3000 W/m2 K s ¼ 2 mm lt ¼ 50 W/m K Dt ¼ 30 KAi ¼ 0.05 m2/m Ao ¼ 0.55 m2/m Ao/Ai ¼ 11

1. ao ¼ 40 W/m2 K hF ¼ 0.7933 hW ¼ 0.82

aoi ¼ ao � hW � AoAi

¼ 40� 0:82� 11 ¼ 360 W�m2 K

aoi ¼ ao

Ai� ðhF � AF þ ACÞ ¼ 40

0:05� ð0:7933� 0:48þ 0:07Þ ¼ 360 W

�m2 K

1Ui

¼ 1360

þ 13000

þ 0:00250

¼ 0:0032

Ui ¼ 317 W�m2 K

Qi ¼ Ui � Ai � Dt ¼ 317� 0:05� 30 ¼ 475:5 W=m tube

2. ao ¼ 300 W/m2 K hW ¼ 0.4736

aoi ¼ 300� 0:4736� 11 ¼ 1563 W�m2 K

1Ui

¼ 11563

þ 13000

þ 0:00250

¼ 0:001 Ui ¼ 987 W�m2 K

Qi ¼ 987� 30� 0:05 ¼ 1480:5 W=m tube


3. ao ¼ 1000 W/m2 K hW ¼ 0.32

aoi ¼ 1000� 0:32� 11 ¼ 3520 W�m2 K

1Ui

¼ 13520

þ 13000

þ 0:00250

¼ 0:0007 Ui ¼ 1521 W�m2 K

Qi ¼ 1521� 30� 0:05 ¼ 2281:5 W=m tube

12.3.4 Calculation of the overall heat transfer coefficient Uo basedon the outer tube area Ao without fouling

1Uo

¼ 1aoW

þ Ao

Ai��slþ 1ai

�

aoW ¼ ao ��1� ð1� hFÞ �

AF

Ao

�¼ ao

Ao� ðhF � AF þ ACÞ ¼ ao � hW

Qo ¼ Uo � Ao � Dt ðWÞao ¼ outer heat transfer coefficient (W/m2 K)aoW ¼ weighted outer heat transfer coefficient under consideration of the fin effi-ciency for the fin surface areaQa ¼ heat duty based on the outer tube surface area (W)

Uo ¼ Ui � Ai

AoUi ¼ Uo � Ao

Ai

Example 6: Calculation of the overall heat transfer coefficient Ua based on theouter tube surface area per m tube without fouling considerationData as in Example 5

1. ao ¼ 40 W/m2 K hF ¼ 0.7933

aoW ¼ 40��1�

�1� 0:7933� 0:48

0:55

��¼ 32:8 W

�m2 K

1Uo

¼ 132:8

þ 0:550:05

��0:00250

þ 13000

�¼ 0:0346

Ua ¼ 28:82 W�m2 K Fo ¼ 0:55 m2�m tube

Qo ¼ 28:82� 30� 0:55 ¼ 475:5 W=m tube

Conversion:

Ui ¼ Uo � Ao=Ai ¼ 28:82� 11 ¼ 317 W�m2 K

Qi ¼ 317� 30� 0:05 ¼ 475:5 W=m tube


2. aA ¼ 300 W/m2 K hR ¼ 0.3963

aoW ¼ 300��1� ð1� 0:3963Þ � 0:48

0:55

�¼ 141:9 W

�m2 K

1Uo

¼ 1141:9

þ 0:550:05

��0:00250

þ 13000

�¼ 0:0112

Uo ¼ 89.72 W/m2 KQa ¼ 89.72 � 30 � 0.55 ¼ 1480.4 W/m tube

Conversion : Ui ¼ 89:72� 11 ¼ 986:9 W�m2 K

Qi ¼ 986:9� 30� 0:05 ¼ 1480:35 W=m tube

12.3.5 Calculation of the overall heat transfer coefficient Uo based onthe finned outer surface area Ao considering the fouling factorsro and ri

1Uo

¼1ao

þ ro

hFþ Ao

Ai��1ai

þ ri � sWlW

�

1Uo

¼ 1aoW

þ rohW

þ Ao

Ai��1ai

þ sWlW

þ ri

�

Qa ¼ Ua � AA � Dt

Example 7: Calculation of the overall heat transfer coefficient Ua consideringfouling

ao ¼ 800 W/m2 K ai ¼ 6000 W/m2 KAo ¼ 0.207 m2/m Ao/Ai ¼ 3.27 hW ¼ 0.9368ro ¼ ri ¼ 0.00015 sW ¼ 1 mm lW ¼ 50 W/m KDt ¼ 25 K

aoW ¼ ao � hW ¼ 800� 0:9368 ¼ 749:4 W�m2 K

1Uo

¼ 1749:4

þ 0:000150:9368

þ 3:27��

16000

þ 0:00150

þ 0:00015

�¼ 0:0026

Uo ¼ 385.3 W/m2 K

Qo ¼ Uo � Ao � Dt ¼ 385:3� 0:207� 25 ¼ 1994 W=m tube


12.3.6 Calculation of the overall heat transfer coefficient Ui for theinner tube surface area Ai considering the fouling factors roand ri

1Ui

¼ 1aoi

þ�

rohW

� Ai

Ao

�þ 1ai

þ ri þ sWlW

Qi ¼ Ui � Ai � Dt

Example 8: Calculation of Ui considering fouling data as in Example 7

aoi ¼ ao � hW � AoAi

¼ 800� 0:9368� 3:27 ¼ 2450:6 W�m2 K

1Ui

¼ 12450:6

þ�0:000150:9368

� 13:27

�þ 16000

þ 0:00015þ 0:00150

¼ 0:0008

Ui ¼ 1259.9 W/m2 K

Qi ¼ Ui � Ai � Dt ¼ 1259:9� 0:0633� 25 ¼ 1994 W=m tube

Conversion:

Uo ¼ Ui � AiAo

¼ 1259:9� 13:27

¼ 385:3 W�m2 K

Qo ¼ Uo � Ao � Dt ¼ 385:3� 0:207� 25 ¼ 1994 W=m tube

12.3.7 Fouling and Temperature GradientA further advantage of fin tubes is the smaller temperature drop by foulings on theenlarged outer tube surface area FA. The fouling does not have such a strong effect aswith plain tubes.

Dtro ¼ ro � qAo

ðKÞ Dtri ¼ ri � qAi

ðKÞ

Dtro ¼ temperature drop by fouling on Ao

Dtri ¼ temperature drop by fouling on Ai

q ¼ heat flux density (W/m tube)Since the surface area of finned tubes is much larger than the surface area of the inner

tube the fouling effect on the outer side is considerably less.

Example 9: Calculation of the temperature drop by fouling

Ao ¼ 0.207 m2/m Ai ¼ 0.0638 m2/mro ¼ ri ¼ 0.0002 Uo ¼ 531 W/m2 K Dt ¼ 25 K


Q ¼ Uo � Ao � Dt ¼ 531� 0:207� 25 ¼ 2748 W=m tube

Dtro ¼ 0:0002� 27480:207

¼ 2:6 K

Dtri ¼ 0:0002� 27480:0638

¼ 8:6 K

The temperature drop on the large outer surface area is much smaller!The temperature gradient at the outer side is also reduced due to the larger outer surface area.The temperature difference between the wall and the medium is smaller so that a more careful

heating with less Dt is possible, or at high temperatures on the tube outer side the material is treatedcarefully.

Example 10: Calculation of the temperature gradients for plain tube and finnedtube

Q ¼ 500 kW ao ¼ 800 W/m2 K ai ¼ 6000 W/m2 Kro ¼ ri ¼ 0.00015 sW ¼ 1 mm lW ¼ 50 W/m K Dt ¼ 25 K

1. Plain tube: 25 � 1 Fo ¼ 0.07854 m2/m Ao/Ai ¼ 1.087

1Uo

¼ 1800

þ 0:00015þ 0:00150

þ 1:087��

16000

þ 0:00015

�

Uo ¼ 566.8 W/m2 Kq ¼ Uo � Dt ¼ 566.8 � 25 ¼ 14,170 W/m2

Dtao ¼ 14170800

¼ 17:71 �K

Dtro ¼ 0:00015� 14170 ¼ 2:13 �K

Dtai ¼ 141706000

� 1:087 ¼ 2:57 �K

Dtri ¼ 0:00015� 14170� 1:087 ¼ 2:31 �K

DtW ¼ 0:00150

� 14170 ¼ 0:28 �K

2. Finned tube Trufin S/T

Ao ¼ 0.207 m2/m Ao/Ai ¼ 3.27 hW ¼ 0.9368

1Uo

¼1800

þ 0:00015

0:9368þ 3:27�

�1

6000þ 0:00015þ 0:001

50

�

U ¼ 385:3 W�m2 K

q ¼ Uo � Dt ¼ 385:3� 25 ¼ 9632:5 W�m2


Dtao ¼ 9632:5800� 0:9368

¼ 12:85 �K

Dtro ¼ 0:00015� 0632:50:9368

¼ 1:55 �K

Dtai ¼ 9632:56000

� 3:27 ¼ 5:25 �K

Dtri ¼ 0:00015� 9632:5� 3:27 ¼ 4:72 �K

DtW ¼ 0:00150

� 9632:5� 3:27 ¼ 0:63 �K

Dtao ¼ temperature gradient for aoDtai ¼ temperature gradient for aiDtro ¼ temperature gradient for the outer fouling roDtri ¼ temperature gradient for the inner fouling riDtW ¼ temperature gradient for the heat conductivity through the wall

12.3.8 Comparison of the specific heat duties Ui � Ai (W/m K)of different tubes

In Figure 12.4, the specific heat duties per m tube of different tubes as a function of theouter heat transfer coefficient are shown.

The curves are valid for an inner heat transfer coefficient of ai ¼ 3000 W/m2 K in thetube.• Plain steel tubes 25 � 2 made, Ao/Ai ¼ 1.19• Low-finned Trufin tubes 22.2 � 1.65 with 1.5 mm steel fin height, Ao/Ai ¼ 3.6• High-finned Applifin tubes 20 � 2 with 10 mm fin height, Ao/Ai ¼ 11

It can be clearly seen that by the finning, the heat duty can be increased.

Figure 12.4 Heat load (W/m K) as a function of the outer heat transfer coefficients.


The advantage is seen with the high-finned tubes in the area of lower outer heattransfer coefficients and also the possibilities for an increase of the efficiency by the useof low-finned tubes in the area of higher outer heat transfer coefficients.

12.4 APPLICATION EXAMPLES

From the discussions so far, it follows that finned tubes are preferably applied if a low heattransfer coefficient should be compensated by a large surface area. From the followingcalculated examples, the preferred application of different fin tube types for specific tasksresults:• From Example 11, the advantages of low-finned tubes for heating bundles for the

evaporation if low construction heights and low bundle widths are desired result.• Example 12 emphasizes the advantages of longitudinal fin tubes for heating coils for

heating storage tanks, because the heat transfer by natural convection is very poor.• From Example 13, it is clear that the high-finned cross-finned tubes are suitable for

the gas cooling or heating.• In Example 14, it is shown that double-tube heat exchangers with finned tubes and

multitube heat exchangers are very suitable for high viscous and gaseous media,because in small room large duties are possible [4].

Example 11: Evaporator heating bundle for a distillation still withoutconsideration of fouling and the heat conduction resistance of the tube wall

Q ¼ 500 kW ai ¼ 6000 W/m2 K ao ¼ 800 W/m2 K Dt ¼ 25 K

1. With plain tubes 25 � 1 L ¼ 4 m Ao ¼ 0.0785 m2/m1Uo

¼ 16000

� 2523

þ 1800

Uo ¼ 699 W�m2 K

Areq ¼ QU� Dt

¼ 500;000699� 25

¼ 28:6 m2 ¼ 364 m tube DN 25

Arrangement: 91 Rohre, 4 m long, pitch 32 mmBundle width B ¼ 90 � 32 þ 25 ¼ 2905 mm

2. With low-finned Trufin tubes S/T 60-197042

do ¼ 25.4 mm hF ¼ 1.5 mm bF ¼ 0.3 mm lF ¼ 50 W/m KAo ¼ 0.207 m2/m Ao/Ai ¼ 3.27 dF/dC ¼ 1.135X ¼ 0.4898 XDF ¼ 0.5116 hF ¼ 0.921 hW ¼ 0.9368

aoW ¼ hW � ao ¼ 0:9368� 800 ¼ 749:4 W�m2K

1Uo

¼ 1749:4

þ 16000

� 3:27 Uo ¼ 532 W�m2K

Areq ¼ 500;000532� 25

¼ 37:6 m2 ¼ 182 m tube DN 25

Arrangement: 46 tubes, 4 m long, pitch 32 mmBundle width B ¼ 45 � 32 þ 25.4 ¼ 1465.4 mm


3. Longitudinal-finned tubes, dC ¼ 25.4 mm, with 20 longitudinal fins 12.7 mm high,

bF ¼ 0.81 mm Ao ¼ 0.5869 m2/m Ao/Ai ¼ 8.838X ¼ 2.52 hF ¼ 0.391 hW ¼ 0.4559 aoW ¼ 364.7 W/m2 K

1Uo

¼ 1364:7

þ 16000

� 8:838 Uo ¼ 237 W�m2 K

Areq ¼ 500;000237� 25

¼ 84:4 m2 ¼ 144 m tube

Arrangement: 36 tubes, 4 m long, pitch 66 mmBundle width B ¼ 35 � 66 þ 50.8 ¼ 2360.8 mmCorollary: The smallest bundle results by applying the low-finned tubes!

Example 12: Heating bundle for the heating of a product in a storage tank with10 m diameter without considering the fouling and the heat conductionresistance of the tube wall

Q ¼ 200 kW ai ¼ 6000 W/m2 K ao ¼ 50 W/m2 K Dt ¼ 50 K

Tube data as in Example 11.

1. With plain tubes 25 � 1 Ao ¼ 0.0785 m2

1Uo

¼ 16000

� 2523

þ 150

Uo ¼ 49:5 W�m2 K

Areq ¼ 200;00049:5� 50

¼ 80:8 m2 ¼ 1029 m pipe DN 25

Arrangement: 128 tubes, 8 m long, single pass

2. With Trufin-finned tubes Ao ¼ 0.207 m2/m Ao/Ai ¼ 3.27

X ¼ 0.122 XDF ¼ 0.1279 hF ¼ 0.9945 hW ¼ 0.9957aAW ¼ 0.9957 � 50 ¼ 49.8 W/m2 K

1Uo

¼ 149:8

þ 16000

� 3:27 Uo ¼ 48:5 W�m2 K

Areq ¼ 200;00048:5� 50

¼ 82:5 m2 ¼ 398 m pipe

Arrangement: 50 tubes, 8 m long, single pass

3. With longitudinal-finned tubes

dC ¼ 25.4 mm hF ¼ 12.7 mm Ao ¼ 0.5869 m2/m Ao/Ai ¼ 8.838hF ¼ 0.885 hW ¼ 0.897 aoW ¼ 0.897 � 50 ¼ 44.85 W/m2 K

1Uo

¼ 144:85

þ 8:838� 16000

Uo ¼ 42 W�m2 K

Areq ¼ 200;00042� 50

¼ 95:2 m2 ¼ 162 m pipe

Arrangement: 20 tubes, 8 m long, single passRecommendation: Heating bundle consisting of longitudinal-finned tubes


Example 13: Gas cooling in a cross flow bundle with cooling water in the tubes

ai ¼ 5000 W/m2 K in the tubesGas flow rate VShell ¼ 904,000 m3/h Q ¼ 8 Mio W Dt ¼ 40 KAllowable pressure loss ¼ 3.8 mbarAllowable bundle width ¼ 6 mProperties of the gases: 9 ¼ 0.885 kg/m3 l ¼ 0.0332 W/m K

n ¼ 25 mm2/s Pr ¼ 0.68

1. With plain tubes 38 � 3.6 without fouling

Estimation of the required area A with U ¼ 90 W/m2 K:

A ¼ QU� Dt

¼ 8� 106

90� 40¼ 2222 m2

Lreq ¼ 19;000 m ¼ 3166 tubes with L ¼ 6 m

Due to the low allowable pressure losses, an aligning arrangement with PC ¼ PL ¼ 2 � da ¼ 76 mm ischosen.

PC ¼ cross pitch of the tubesPL ¼ longitudinal pitch of the tubesArrangement: 100 tubes one over the other, 6 m long, 36 tube rows one behind the other, aligned

arrangement

Total tube length ¼ 100 � 36 � 6 ¼ 21,600 m Bundle length L ¼ 6 mTotal surface area ¼ 2578 m2 Bundle height H ¼ 7.6 mBundle cross-sectional area Abundle ¼ 6 � 7.6 ¼ 45.6 m2 (width � height)

Free flow cross-sectional area Afree for the gas through the tube bundle:

Afree ¼ PC � doPC

� H � L ¼ 76� 3876

� 7:6� 6 ¼ 22:8 m2

wgas ¼ 904;00022:8� 3600

¼ 11 m=s

Re ¼ wgas � don

¼ 11� 0:038

25� 10�6 ¼ 16; 720

Pressure loss calculation according to Jakob (Section 11.3): z ¼ 0.189

DP ¼ w2 � r

2� nR � z ¼ 11� 0:88

2� 36� 0:189 ¼ 362 Pa

Heat transfer calculation according to Grimison (Section 11.3):

Nu ¼ 0:229� 16;7200:632 � 1 ¼ 106:8

a ¼ 106:8� 0:03320:038

¼ 93:3 W�m2 K

Calculation of the overall heat transfer coefficient for steam heating with ai ¼ 5000 W/m2 K:

aio ¼ 5000� 30:838

¼ 4428 W�m2 K

1U

¼ 14428

þ 193:3

þ 0:003650

¼ 0:0011


U ¼ 90.7 W/m2 K

Heat duty Q: Q ¼ k � A � Dt ¼ 90.7 � 2578 � 40 ¼ 9.35 W

Required heat duty Qreq ¼ 8 Mio W Fouling reserve: 16.9%Bundle dimensions: Height ¼ 7.6 m Width ¼ 6 m Depth ¼ 2.7 m

2. With finned tubes Applifin 95,725 without fouling

Outer surface area Ao ¼ 1.5 m2/m dC ¼ 25.4 mm dF ¼ 57.2 mmCross pitch PC ¼ 57.2 mm Longitudinal pitch PL ¼ 49.5 mm, staggered

Arrangement: 153 tubes one over the other, 6 m long, five rows one behind the other

Total tube length Ltot ¼ 153 � 6 � 5 ¼ 4590 m tube Total surface area A ¼ 6885 m2

Abundle ¼ 7.6 � 6 ¼ 45.6 m2 fproj ¼ 0.03 m2/m tube Afree ¼ 45.6 � 6 � 0.03 ¼ 18 m2

w ¼ 904;0003600� 18

¼ 13:95 m=s Re ¼ 13:95� 0:025425� 10�6 ¼ 14; 174

Pressure loss calculation with z ¼ 0.8 from manufacturer data

DP ¼ 13:952 � 0:882

� 5� 0:8 ¼ 343 Pa

Calculation of the heat transfer coefficients according to manufacturer data:

Nu

Pr1=3¼ 0:37� Re0:553 � fR ¼ 0:37� 14;1740:553 � 1 ¼ 73:1

Nu ¼ 73:1� 0:680:33 ¼ 64:4

ao ¼ 64:4� 0:03320:0254

¼ 84:2 W�m2 K

Calculation of the fin efficiency:

X ¼ 0:0159�ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi2� 84:2

50� 0:0004

r¼ 1:459 XDF ¼ 1:8735 hF ¼ 0:5092

hW ¼ 0:5092� 1:46þ 0:041:5

¼ 0:5222

aoW ¼ 0:5222� 84:2 ¼ 44 W�m2 K


ai ¼ 4000 W/m2 K aoW ¼ 44 W/m2 K ¼ effective a-value shell side1U

¼ 144

þ 23:5��0:002550

þ 14000

�¼ 0:0298 U ¼ 33:6 W

�m2K

Heat duty Q ¼ 33:6� 6885� 40 ¼ 9:25 Mio W

Required heat load: Qreq ¼ 8 Mio W Reserve: 15.7%Bundle dimensions: Height: 7.6 m Width: 6 m Depth: 255.2 mm


Comparison between plain tube and finned tube:

Plain tube Finned tubeHeight (m) 7.6 7.6Width (m) 6,- 6Depth (m) 2.7 0.255Tube length (m) 21,600 4590Surface area (m2) 2578 6885

Corollary: By the use of finned tubes, the tube length can be reduced from 21,600 to 4590 m and thebundle depth from 2.7 to 0.26 m.

Example 14: Double pipe oil cooler with an inner longitudinal-finned tubePipe shell with Di ¼ 77.9 mm Ao/Ai ¼ 5.5

Inner tube with longitudinal fins: do ¼ 48.2 mm, fin height 12.7 mm, surface area 0.76 m2/mTube-side product: cooling water

Required heat load Qreq ¼ 7 kW Effective Dt ¼ 23 �CsW ¼ 2 mm lW ¼ 59 W/m K ra ¼ ri ¼ 0.0001

Shell side product rate VShell ¼ 3.5 m3/h Flow cross-section Afree ¼ 0.0026 m2

Flow velocity wShell ¼ 0.374 m/s.

Oil data: r ¼ 846 kg/m3 c ¼ 0.58 Wh/kg K l ¼ 0.131 W/m K n ¼ 19 mm2/s

Pr ¼ 3600� 19� 10�6 � 0:58� 8460:131

¼ 256

Re ¼ 0:374� 0:0104

19� 10�6 ¼ 204:7/Laminar Flow

Calculation of the heat transfer coefficient:

Nu ¼ 1:86��204:7� 256� 0:0104

6

�1=3

¼ 8:36

ao ¼ 8:36� 0:1310:0104

¼ 105 W�m2 K

X ¼ 0:0127�ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi2� 105

0:001� 50

r¼ 0:823 hF ¼ 0:822

hW ¼ 0:822� 0:61þ 0:150:76

¼ 0:857

aoW ¼ 0:857� 105 ¼ 90 W�m2 K

Tube side: 3 m3/h cooling water.

di ¼ 44.6 mm wT ¼ 0.54 m/s Pr ¼ 6.94 l ¼ 0.604 W/m K

Re ¼ 0:54� 0:0442

1� 10�6 ¼ 23;868

Nu ¼ 0:023� 23;8680:8 � 6:941=3 ¼ 139:45

aT ¼ 139:45� 0:6040:0442

¼ 1905 W�m2 K


Calculation of the overall heat transfer coefficient Uo:

1Uo

¼ 1aoW

þ rohW

þ AoAi

��1ai

þ sWlW

þ ri

�

1Uo

¼ 190

þ 0:00010:857

þ 5:5��0:00259

þ 11905

þ 0:0001

�Uo ¼ 68:1 W

�m2 K

Required area Areq ¼ 700023� 68:1

¼ 4:47 m2

Required tube length Lreq ¼ 4:470:76

¼ 5:88 m pipe

Alternative calculation for a multipipe heat exchanger with plain tubes:Shell diameter Di ¼ 84 mm with seven inner tubes 20 � 2Shell side: 3.5 m3/h oil

Flow cross-sectionfShell ¼ 0.0033 m2

De ¼ 0.0304 m Surface area ¼ 0.4398 m2/m

wShell ¼ 0.295 m/s Re ¼ 471

Nu ¼ 1:86��471� 256� 0:0304

6

�¼ 15:78

ao ¼ 15:78� 0:1310:0304

¼ 68 W=m2K

Tube side: 3 m3/h cooling water wT ¼ 0.59 m/s Re ¼ 18,009Nu ¼ 111.3 ai ¼ 4202 W/m2 K aio ¼ 3362 W/m2 K


1U

¼ 13362

þ 168

þ 0:00250

þ 0:0002

U ¼ 65:6 W�m2 K

Required area Areq ¼ 700065:6� 23

¼ 4:64 m2

Required length Lreq ¼ 4:640:4398

¼ 10:6 m

REFERENCES[1] D.Q. Kern, A.D. Kraus, Extended Surface Heat Transfer, McGraw-Hill, N.Y., 1972.[2] Th.E. Schmidt, K€altetechnik 18 (4) (1966) 135e138.[3] W.M. Kays, A.L. London, Compact Heat Exchangers, McGraw-Hill, N.Y., 1964.[4] G.P. Purohit, Thermal and hydraulic design of hairpin and finned-bundle exchangers, Chem. Eng., 90

(1983) 62e70.


INDEX

‘Note: Page numbers followed by “f” indicate figures, “t” indicate tables, and “b” indicate boxes.’

AAccelerated pressure loss, 208e209, 211fAntoine equation, 91e92, 96be97b

BBaffle spacing, 14e15bypass streams, 30ribbon and quadrant arrangement, 46, 47f

Baffle types, 48, 48fBenzeneetoluene mixture, 97fBubble point calculation, 94e97Bypass streams, 13e14, 13fbypass leakage, 30, 30fLDT curve, length, 30, 31fshell-side pressure loss, 53be55btemperature efficiency, 30, 31f

CColburn factor JC, 77Condensation curve, 91, 92f, 99e101Condensersvs. calculation models, 126fhorizontal tube bundle, 123, 124fhorizontal tubes, 123, 124fisothermal film condensation, 122, 123tvertical tube bundle, 125, 126fvertical tubes, 123, 125f

condensate drain pipe, 135e137construction types, 106f, 108fheat transfer coefficients, 106, 107fhorizontal tube bundle, shell chamber,107e109, 108f

horizontal tubes, 108f, 110e111horizontal water-cooled tubes, top condenser,108f, 109

shell side, vertical bundle, 108f, 109e110vertical countercurrent condensation,108f, 110

vertical tubes, 108f, 110flood loading, reflux condensers, 143e144heat transfer coefficients, 116bcondensate loading, 113e116, 114be115bhorizontal tube bundle, 116e117

horizontal tubes, 118Kern equations, 111, 113tmass stream density m, 115be116bNitsche method, 111, 112tparameters, 120e121, 121fe122fshear stress and gravity, 111, 113fvertical tube bundle, 117vertical tubes, vapor rate, 114f,119e120

Wallis factor J, 115be116binert gas, 127f

dew point reduction, 128transportation prevention, 128

maximal vapor flow capacity, 138e139maximum condensate load, 142multicomponent mixtures

bubble and dew line, 129, 130fdifferential and integral condensation,134e135, 136fe137f

heat transfer coefficients, 130e131hydrocarbon mixture, inert gas, 133bpartial gas cooling, 131e133, 131b, 132ftemperatureeheat load curve, 134fWMTD, 130

pressure loss, 135, 141e142top condensers, 139e141, 139f

Convective heat exchangers, 42e43design, 61btube-side pressure drop DPT, 52f, 52b

Corrected effective mean temperature difference(CMTD)

vs. LMTD, 42multipass heat exchanger

arrangement in series, 28, 29fcorrection factor F, 27bisothermal and nonisothermal heating, 28bone-pass heat exchanger, ideal countercurrentflow, 22, 23f

temperature correction factors, 25, 26ftemperature efficiency factor, 23e24temperature factor F, 27be28btwo-pass heat exchanger, co- andcountercurrent flow, 22, 24f

265

Cross flow bundle, 243baligning arrangement, 242application cases, 241heat transfer coefficients and pressure drop, 243bK- and M-values, 241, 242tstaggered arrangement, 242e243

DDalton law, 93, 93fDephlegmator effect, 109Dew point calculation, 95e97Double pipe heat exchangers, 230f, 237fdesign, 232be233bhydraulic diameter, 231vs. multitube heat exchangers, 234be236bpressure loss calculation, 231

Droplet flow, 149

EEvaporation processflash evaporation, 149flow boiling, 148e149pool boiling, 148, 158fe159f

Evaporatorsfalling film evaporator, 156e157, 157fheat transfer coefficients, 176, 176fe177fpressure drop parts, 174, 174f

flash evaporator, 153e154, 154fforced circulation reboiler, 153, 153finternal evaporator, 155e156, 156fkettle reboiler, 155, 155fnucleate boiling, 161e162and convective heating, 162, 163fheat flux density, 158, 158fheat transfer coefficients, 158, 159fsteam-heated kettle-type evaporator, 164btube bundle, design, 164be165bvacuum evaporation, 159

thermosiphon evaporator, 150horizontal thermosiphon circulationevaporator, 151, 151f

once-through thermosiphon reboiler,152e153, 152f

vertical thermosiphon circulation evaporator,150e151, 151f

FFalling film evaporator, 156e157, 157fheat transfer coefficients, 176, 176fe177f

pressure drop parts, 174, 174fFinned tube heat exchangersapplicationdouble pipe oil cooler, inner longitudinal-finned tube, 263be264b

fouling and heat conduction resistance,259be260b

gas cooling, cross flow bundle,261be263b

calculationsfin efficiency hF, 251fouling, 256e258overall heat transfer coefficient, 253e256specific heat duties Ui � Ai (W/m K),258e259, 258f

temperature gradient, 256e258weighted fin efficiency hw, 251e252

fin efficiency hF, 249, 249ffouling factors, 247e248outer heat transfer coefficient ao, 249, 250fouter tube surface area, 249, 249frequired area ratio, 248bsteel-finned tubes, 250

Flash calculations, 97e98, 97fmulticomponent mixtures, 103e104, 103f

Flash curve, 99e101Flash evaporation/evaporator, 149, 153e154, 154fFlow boiling, 148e149Flow velocityshell-side heat transfer coefficient, 8ftube-side heat transfer coefficient, 8f

Forced circulation reboiler, 153, 153fFriction pressure loss, 207e210, 209f

GGeometrical heat exchanger calculationscalculation formulaflow velocity wT, 66heat exchanger area A, 65shell-side cross section across, 66shell-side cross stream velocity wC, 66shell-side longitudinal stream cross area aw,66e68

tube rows in cross stream ncross, 68tube-side flow cross section aT, 66

shell-side calculations, 70e74tube-side calculations, 68e70

Graßhof number Gr, 78

266 Index

HHeat exchangersbaffle spacing, 14e15bypass and leakage streams, 13e14, 13fdesign, procedure, 61be62bactual heat load, 9convective heat transfer, 5e6, 5fheat exchanger area, 4t, 6, 7b, 8fe9fheat transfer coefficient, 8Reynolds number, 2, 7e8thermal design, flow chart, 2, 6fwater cooler, 10be12b

geometrical data, 1, 4t, 56, 57te60tshell and tube heat exchangers, 16e17, 16fe18fthermosiphon reboilers, 184aconv, two-phase mixture, 184e186horizontal thermosiphon evaporator, 188e190aN, correction factor S, 186vertical thermosiphon evaporator, 187e188

tube pattern, 12e13, 13fwith two tube passes, 38f

Heat flux density, 158, 158fHeat load curve, 99be100b, 101f, 101bHeat resistances, 1, 2fHeat transfer coefficient, 1calculation equations, 37, 39fColburn factor JC, 77condensers, 116bcondensate loading, 113e116, 114be115bhorizontal tube bundle, 116e117horizontal tubes, 118Kern equations, 111, 113tmass stream density m, 115be116bNitsche method, 111, 112tparameters, 120e121, 121fe122fshear stress and gravity, 111, 113fvertical tube bundle, 117vertical tubes, vapor rate, 114f, 119e120Wallis factor J, 115be116b

Graßhof number Gr, 78Kern factor JK, 77Nusselt number Nu, 76overall heat transfer coefficient, 88e89reference values, 1, 3tStanton number St, 76e77

Helical coil heat exchangerdesign, 238be240bgeometrical calculations, 237

shell-side flow cross section and overall heattransfer coefficient, 238

Horizontal thermosiphon circulation evaporator,151, 151f

Horizontal tube bundle, shell chamber, 107e109,108f

Horizontal water-cooled tubes, top condenser,108f, 109

IIdeal binary mixtures, 95e97Ideal countercurrent flow, 21Inert gas, 127fdew point reduction, 128transportation prevention, 128

Internal U-tube evaporator, 155e156, 156fIsothermal heating, 28b

KKern equations, 111, 113tKettle reboilers, 155, 155fallowable bubble rising velocity, 169allowable heat flux density, 170e171entrainment, 172hydraulic design, 166e168thermal design, 168e169

LLeakage streams, 13e14Liquid pressure loss, 206e207, 208fLiquid/vapor phase equilibrium, 93e94, 93fLogarithmic mean temperature difference

(LMTD)bypass streams

bypass leakage, 30, 30fLDT curve, length, 30, 31ftemperature efficiency, 30, 31f

heat exchanger calculation, 25be27bideal countercurrent flow, 21

MMaximal vapor flow capacity, 138e139Mean weighted temperature difference (WMTD),

30e31, 32fMulticomponent mixturescondensers

bubble and dew line, 129, 130fdifferential and integral condensation,134e135, 136fe137f

Index 267

Multicomponent mixtures (Continued)heat transfer coefficients, 130e131hydrocarbon mixture, inert gas, 133bpartial gas cooling, 131e133, 131b, 132ftemperatureeheat load curve, 134fWMTD, 130

flash calculations, 103e104Multipass heat exchanger, CMTDarrangement in series, 28, 29fcorrection factor F, 27bisothermal and nonisothermal heating, 28bone-pass heat exchanger, ideal countercurrent

flow, 22, 23ftemperature correction factors, 25, 26ftemperature efficiency factor, 23e24temperature factor F, 27be28btwo-pass heat exchanger, co- and countercurrent

flow, 22, 24fMultitube heat exchangers, 230fvs. double pipe, 234be236bhydraulic diameter, 231pressure loss calculation, 231shell-side heat transfer coefficient, 234b

NNonideal binary mixtures, 101e103Nonisothermal heating, 28bNucleate boiling, 148, 161e162and convective heating, 162, 163fheat flux density, 158, 158fheat transfer coefficients, 158, 159fkettle reboilersallowable bubble rising velocity, 169allowable heat flux density, 170e171entrainment, 172hydraulic design, 166e168thermal design, 168e169

steam-heated kettle-type evaporator, 164btube bundle, design, 164be165bvacuum evaporation, 159

Nusselt number Nu, 76

OOutlet temperatures, 32e35Overall heat transfer coefficientconversion of ai, outside area, 80be81bfouling reserve Sf, 82be84bper meter heat exchanger length, 84bUa, heat transfer coefficients, 80b

Uclean calculation, 81be82bUdirty calculation, 82bUex calculation, 83b

PPeclet number Pe, 75e76Pool boiling, 148, 158fe159fPrandtl number Pr, 75e76

QQuadratic tube pattern, 12e13

RRaoult law, 93, 93fReboiler characteristic curves, 211, 213fReboiler types, 149e150Reflux condensers, 143e144Reynolds number Re, 75Rotated tube pattern, 12

SSegmental baffles, 14, 14fShell-side heat transfer coefficient, 8f,

40be41b, 234bbaffle types, 48, 48fgases, flow velocity, 44, 45fliquids, flow velocity, 44, 44fribbon and quadrant arrangement, 46, 47fribbon pass arrangement, 46shell-side stream velocity, 43Tinker, 46, 46fVDI-W€armeatlas, 44, 45f

Shell-side pressure loss, 53e55, 53t, 53be55b, 56fStanton number St, 76e77Steam-heated kettle-type evaporator, 164b

TTEMA systematics, 16, 17fTemperature conductivity A, 75e76Temperature efficiency factor F, 23e24Temperature gradient, 84e85Temperature profile, 1, 2fThermosiphon circulationpressure lossaccelerated pressure loss, 208e209, 211ffriction pressure loss, 207e210, 209fliquid pressure loss, 206e207, 208freboiler characteristic curves, 211, 213ftotal pressure loss, 210e211, 212f

268 Index

required height H1, DPmax, 206fAJL, AHL, and AEL, 204fcirculating rate W, vertical thermosiphonevaporator, 202f

DPalow, H1, H2, and H4 heights, 194e196,194be196b

horizontal circulation evaporator, 198, 199fhorizontal once-through evaporator,201e202, 201f

increasing circulating rate, 203fsafety factor S, pressure loss DP, 194steam-heated vertical thermosiphonevaporator, LMTD, 203f

vertical once-through evaporator, 199e201,200f

vertical thermosiphon evaporator, LMTD,197e198, 197f, 204f

Thermosiphon once-through reboiler,152e153, 152f

Thermosiphon reboilers, 150average overall heat transfer coefficient, 221design example, 222e226flow velocity WReb, 193heat transfer coefficient, 184aconv, two-phase mixture, 184e186horizontal thermosiphon evaporator, 188e190aN, correction factor S, 186vertical thermosiphon evaporator, 187e188

pressure and boiling point calculation, 223fdifferent H1-values, 218, 219tdriving height function, 218, 219fpressures PA, PB, and PC, 218, 220frequired heat length, vacuum evaporation,220be221b

required reboiler length and area, 213e214required tube length Levap and Aevap, 215riser/downcomer diameter, 203e205, 205bthermal calculationsrequired circulating rate W, 182e183, 182frequired reboiler area A, 183e184

thermosiphon circulation, 206, 206fe207f.See also Thermosiphon circulation

accelerated pressure loss, 208e209, 211fAJL, AHL, and AEL, 204fcirculating rate W, vertical thermosiphonevaporator, 202f

DPalow, H1, H2, and H4 heights, 194e196,194be196b

friction pressure loss, 207e210, 209f

horizontal circulation evaporator, 198, 199fhorizontal once-through evaporator,201e202, 201f

increasing circulating rate, 203fliquid pressure loss, 206e207, 208freboiler characteristic curves, 211, 213fsafety factor S, pressure loss DP, 194steam-heated vertical thermosiphonevaporator, LMTD, 203f

total pressure loss, 210e211, 212fvertical once-through evaporator,199e201, 200f

vertical thermosiphon evaporator, LMTD,197e198, 197f, 204f

two-phase/average densityliquid rliq, two-phase mixture r2Ph, 191e192,191be192b

r2Ph, vaporeliquid mixture, 191types

horizontal thermosiphon circulationevaporator, 151, 151f

once-through thermosiphon reboiler,152e153, 152f

vertical thermosiphon circulation evaporator,150e151, 151f

vertical thermosiphon reboilerheating and evaporation zones, 212e213, 217frequired heating length, 215e217vapor fraction X, two phase mixture,221e222, 222b

Total pressure loss, 210e211, 212fTube-side heat transfer coefficient, 8f, 39be40bdifferent gases, flow velocity, 42, 43fheptane, flow velocity and tube diameter,

41, 41fliquids, flow velocity, 42, 42fmultipass arrangement, 41type10 heat exchanger, 49, 50fe51f

Two-phase/average densityliquid rliq, two-phase mixture r2Ph, 191e192,

191be192br2Ph, vaporeliquid mixture, 191

VVapor pressure calculationsAntoine equation, 91function of temperature, 92f

Vertical thermosiphon circulation evaporator,150e151, 151f

Index 269

Vertical thermosiphon reboilerheating and evaporation zones, 212e213, 217frequired heating length, 215e217vapor fraction X, two phase mixture, 221e222,

222b

Viscosity correction, 86calculation procedure, 86e88

WWater cooler, design, 10be12b

270 Index

heat exchanger design guide: a practical guide for planning, selecting and designing of shell

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