Chapter 5

Heat Exchanger Design Methods

Learning Objectives

At the end of this chapter students will:

• Be able to determine the mean temperature difference for a heat exchanger

• Be familiar with the assumptions made when defining mean temperature

difference, and hence know the limitations of the concept.

• Understand the concept of heat exchanger effectiveness

• Be familiar with the assumptions made when formulating effectiveness-NTU

relationships, and hence know the limitations of these relationships.

• Be able to size a heat exchanger or determine its performance using the mean

temperature and effectiveness-NTU approaches

5.1 Mean Temperature Difference

We have seen (Section 3.2) that for heat transfer between two fluids separated by a

wall the rate of heat transfer is given by the expression:

(&Q UA T Th c= − − ) (5.1)

This can be applied over the whole area of the wall if Th and Tc are constant, but in

most heat exchangers the temperature of at least one fluid stream varies as it flows

through the exchanger. It is therefore useful, particularly for hand calculations to

define a mean temperature such that:

&Q UA Tm=− ∆ (5.2)

An appropriate mean temperature may easily be calculated for parallel and

countercurrent flows, subject to a number of simplifying assumptions. Crossflow and

more complex flow arrangements are not amenable to simple analysis but correction

factors for a wide range of flow arrangements are available in the literature and these

may be applied to the counterflow case.

5.1

Logarithmic mean temperature difference

A mean temperature difference applicable to parallel and countercurrent flow heat

exchangers may be calculated if the following conditions hold, or may be assumed to

hold:

1. There is no external heat transfer to or from the heat exchanger

2. Axial conduction (along the heat exchanger walls or the fluid streams) is negligible

3. Changes in potential and kinetic energy are negligible

4. The overall heat transfer coefficient (U) is constant throughout the heat

exchanger. This implies, in practice, that each individual heat transfer coefficient must

be constant.

5. The specific heat capacity of each fluid is constant throughout the heat exchanger

(we shall see later that the analysis is valid for one fluid being at constant

temperature, e.g. a phase change at constant pressure)

6. The temperature of each fluid is constant over any cross section of its path

through the heat exchanger.

7. The flow rate of each fluid is constant throughout the heat exchanger and there is

no bypassing of sections

8. In shell-and-tube heat exchangers the temperature change of the shell-side fluid

within one baffle space is small compared to its overall temperature change within

the unit. This implies a large number of baffles.

Let us now consider parallel and counter flow heat exchangers, the temperature

profiles of the two streams are represented in Fig. 5.1. (a) and (b), respectively.

5.2

(b) Counterflow

Th,i

Th,o

Tc,i

Tc,o

1 2

∆T

∆T1

∆T2

dTh

dTc

dQ&

Distance along heat exchanger

Th,i

Th,o

Tc,i

Tc,o

1 2(a) Parallel Flow

∆T

∆T1

∆T2

dTh

dTcdQ&

Distance along heat exchanger

Figure 5.1 Temperature profiles in heat exchangers

5.3

We can examine a small slice of the heat exchanger and apply an energy balance:

− = =dQ dQ dQh c& & & (5.3)

dQ m c dT m c dTh p h h c p c c& & &, ,= = (5.4)

dQ U TdA& =− ∆ (5.5)

where ∆T T Th c= −

( )d T dT dTh∆ = − c

,

(5.6)

Now let us consider the appropriate signs: For parallel flow, taking the positive

direction of flow as from left to right and noting that Tc increases with distance along

the heat exchanger and Th decreases, we can write equation 5.4 as:

dQ m c dT m c dTh p h h c p c c& & &,= − = (5.7a)

and

dTdQ

m cdT

dQm ch

h p hc

c p c= − =

&

&

&

&, ,, (5.8a)

Combining equations 5.6 and 5.8a gives:

d TdQ

m cdQ

m cdQ

m c m ch p h c p c h p h c p c( )

&

&

&

&&& &, , , ,

∆ =−

− = − +⎛

⎝⎜

⎞

⎠⎟1 1

(5.9a)

Similar logic may be applied for the counterflow case, however both Tc and Th

decrease with distance along the heat exchanger:

dQ m c dT m c dTh p h h c p c c& & &,= − = − , (5.7b)

5.4

dTdQ

m cdT

dQm ch

h p hc

c p c= − = −

&

&

&

&, ,, (5.8b)

Combining equations 5.6 and 5.8b gives:

d TdQ

m cdQ

m cdQ

m c m ch p h c p c h p h c p c( )

&

&

&

&&& &, , , ,

∆ =−

− = − −⎛

⎝⎜⎜

⎞

⎠⎟⎟

1 1 (5.9b)

Integrating equation 5.9 from the end 1 to end 2 of the heat exchanger thus yields

∆ ∆T T Qm c m ch p h c p c

2 1

1 1− = − ±

⎛

⎝⎜⎜

⎞

⎠⎟⎟&

& &, , (5.10)

∆ ∆T T Qm c m ch p h c p c

1 2

1 1− = ±

⎛

⎝⎜⎜

⎞

⎠⎟⎟&

& &, , (5.11)

where the positive sign in equation 5.11 refers to the parallel flow case and the

negative sign to the counterflow case.

Rearranging equation 5.11 gives:

( )dQ

d Tm c m ch p h c p c

&& , ,

=−

±∆

1 1 (5.12)

and substituting for from equation 5.5 yields: dQ&

( )−±

=d T

m c m cU TdA

h p h c p c

∆∆

1 1& , , (5.13)

or

( )−= ±

⎛

⎝⎜⎜

⎞

⎠⎟⎟

d TT

Um c m c

dAh p h c p c

∆∆

1 1& , ,

(5.14)

which may then be integrated over the entire length of the heat exchanger to give

5.5

−⎛

⎝⎜

⎞

⎠⎟ = ±

⎛

⎝⎜⎜

⎞

⎠⎟⎟ln

& &, ,

∆∆

TT

UAm c m ch p h c p c

2

1

1 1 (5.16)

Equation 5.11 gives:

1 1 1 2

& & &, ,m c m c

T TQh p h c p c

±⎛

⎝⎜⎜

⎞

⎠⎟⎟ =

−∆ ∆ (5.17)

which may be substituted into equation 5.16 to give

−⎛⎝⎜

⎞⎠⎟ =

−ln &

∆∆

∆ ∆TT

UAT T

Q2

1

1 2 (5.18)

A final rearrangement yields:

&

lnQ UA

T TTT

= −−

⎛⎝⎜

⎞⎠⎟

∆ ∆∆∆

1 2

1

2

(5.19)

and comparison with equation 5.2 allows us to define a mean temperature

difference:

&Q UA Tlmtd=− ∆ (5.20)

∆∆ ∆

∆∆

TT T

TT

lmtd =−

⎛⎝⎜

⎞⎠⎟

1 2

1

2ln

(5.21)

It should be noted that the expression for Logarithmic Mean Temperature

Difference (LMTD) derived above is identical for parallel and counter flow

configurations, when expressed in terms of the fluid temperature differences at the

two ends of the heat exchanger. However, the value of LMTD obtained is always

higher for the counterflow arrangement (except in special cases where the values

are equal). Indeed, the LMTD for the counterflow arrangement is the highest

attainable for given process conditions.

5.6

For other flow configurations analytic solutions for the mean temperature difference

are more complex. It is convenient to consider the more general case such that:

&Q UA Tm=− ∆ (5.2)

and:

∆ ∆T F Tm lmtd counterflow= , (5.22)

Where F is a factor depending on the flow configuration in the heat exchanger which

can be expressed as a function of the temperatures at inlet and outlet.

F=f(P,R, flow configuration) (5.23)

Where P and R are related to the inlet and outlet temperatures of the heat

exchanger stream as indicated in Fig. 5.2.

5.7

Figure 5.2 Correction factors, F, for use in determining ∆Tm (Adapted from Fraas A.P. Heat Exchangers Design, Wiley, 2nd Edition, 1988)

5.8

Figure 5.2 (continued) Correction factors, F, for use in determining ∆Tm (Adapted from Fraas A.P. Heat Exchangers Design, Wiley, 2nd Edition, 1988)

5.9

5.2 Heat Exchanger Effectiveness and NTU

An alternative approach, based on the same assumptions as those defined when

calculating the LMTD, is to employ a parameter known as the effectiveness of the

heat exchanger and relate this to the fluid conditions and the heat exchanger area

and geometry.

Firstly we shall define two terms, namely the Capacity Rate and the Capacity Rate

Ratio: The capacity rate for a stream in the heat exchanger is defined by:

CQ

T Tmc

in out

p

=−

≡

≡ ∞

&

& (for single - phase) (for constant temperature phase change, condensation or boiling)

(5.24)

Remembering that

& & &, ,Q m c T m c Th p h h c p c c= =∆ ∆ (5.25)

The capacity rate ratio may then be defined:

CRCC

TT

CRc

c= = ≤ ≤min

max

max

min,

∆∆

0 1

c

(5.26)

where Cmin and Cmax are the lower and higher capacity rates, respectively and

are the corresponding fluid temperature changes. ∆ ∆T Tc min max and

5.10

(b) Cold stream has Cmin

`

Th,i

Th,o= Tc,i

Tc,o

1 2

Distance along heat exchanger

Tc,i

Th,i= Tc,o

Th,o

1 2

Distance along heat exchanger

(a) Hot stream has Cmin

Figure 5.3 Temperature profiles in ideal counterflow heat exchangers

5.11

The Second Law of Thermodynamics tells us that all heat transfer must be from the

higher to the lower temperature, therefore the limiting performance of a

counterflow heat exchanger must be as shown in figs 5.3(a) or 5.3(b). That is, either

the hot fluid is cooled to the inlet temperature of the cold fluid or the cool fluid is

heated to the inlet temperature of the hot fluid. More generally we can say that in an

ideal heat exchanger the fluid with the lower value of capacity rate (and hence

subject to the larger temperature change) will exit the heat exchanger at the same

temperature as the other fluid enters. Thus we can determine the maximum possible

rate of heat transfer for given process conditions:

(&min , ,Q C T Tideal h in c in= − ) (5.27)

and the effectiveness, ε, of a real heat exchanger may be defined:

( )( )

( )( )

( )

ε = =−

−=

−

−

=−

&

&, ,

min , ,

, ,

min , ,

min

, ,

Q

Q

C T T

C T T

C T T

C T T

T

T T

actual

ideal

h h in h out

h in c in

c c out c in

h in c in

c

h in c in

∆ (5.28)

where ∆Tc min is the temperature change for the stream having the smaller capacity

rate, remember, this implies that it is the larger temperature change.

The effectiveness of a heat exchanger may be expressed:

( )ε = f U A C C, , ,min max , flow configuration (5.29)

or

(ε = f NTU CR, , flow configuration) (5.30)

where the Number of Transfer Units, NTU, is defined:

NTUUAC

=min

(5.31)

5.12

Rearrangement of equation 5.21, together equations 5.25, 5.27 and 5.31 yields

expressions for the effectiveness of counterflow and parallel flow heat exchangers:

For counterflow:

( )( )(( ))ε =

− − −− − −1 1

1 1exp

expNTU CR

CR NTU CR (5.32)

For Parallel flow:

( )( )ε =

− − ++

1 11

exp NTU CRCR (5.33)

Expressions or approximate solutions may be derived for other configurations. Some

are presented graphically in Fig. 5.4.

For all flow conditions if CR=0, i.e. one side of the heat exchanger is isothermal

(normally when one side of the heat exchanger is condensing or boiling, or it is

electrically heated) then

e N= − −1 exp( )TU (5.34)

It is worth noting that for most heat exchanger configurations the effectiveness

increases with increasing NTU, asymptotically approaching a maximum value. For

counter flow and for CR=0 this maximum value is unity. For all other flow

configurations, if CR>0, the asymptotic value is less than one. In some configurations,

however, for example in split flow heat exchangers with mixing of one fluid, as

shown in Fig. 5.4, the effectiveness reaches a maximum value and then declines with

increasing NTU.

5.13

Figure 5.4 Heat transfer effectiveness for various geometries (Adapted from Kays W.M. and Landon A.L., Compact Heat Exchangers, McGraw-Hill, 2nd Edition, 1964)

5.14

Figure 5.4 (cont) Heat transfer effectiveness for various geometries (Adapted from Kays W.M. and Landon A.L., Compact Heat Exchangers, McGraw-Hill, 2nd Edition, 1964)

5.15

Figure 5.4 (cont) Heat transfer effectiveness for various geometries (Adapted from Kays W.M. and Landon A.L., Compact Heat Exchangers, McGraw-Hill, 2nd Edition,1964)

5.3 Design and Performance Calculations

The mean temperature difference and ε-NTU approaches to heat exchanger design

and performance estimation are both based on the same set of assumptions. Both

approaches would be expected to yield the same results. It may be argued that the ε-

NTU approach introduces a factor (the effectiveness) which has some

thermodynamic significance. This is not apparent in the mean temperature

approach. More importantly, from a practical point of view, the

ε-NTU approach often results in the simpler algebra and may eliminate some of the

need for iteration which is frequently encountered in heat exchanger design and

analysis. It must, however, be remembered that both analyses are underpinned by

the assumption that a constant U value may be determined for the heat exchanger

(or for the section of heat exchanger concerned) and that the corresponding area

5.16

can be defined. Since the product UA is used in both approaches it does not matter

which area is chosen as the reference, providing the appropriate U value is used.

Two problems are typically encountered in heat exchanger calculations:

a) Given U, Cc,Ch and the terminal temperatures determine the area required.

b) Given U, Cc,Ch,,A and the inlet temperatures for both fluid streams determine the

outlet temperatures

Let us consider the two approaches to each of these problems:

Mean Temperature Difference approach to problem (a)

1. Calculate the rate of heat transfer from the temperature change of one

stream and the appropriate capacity rate (alternatively, if and the inlet conditions

are known calculate the outlet temperatures)

&Q

&Q

2. Calculate P and R from the known inlet and outlet temperatures

3. Determine the correction factor, F, from the appropriate curve or expression

4. Calculate the log mean temperature difference assuming counterflow using the

known inlet and outlet temperatures.

5. Calculate the area required:

AQ

UF Tlmtd=

&

∆

ε-NTU approach to problem (a)

1. Calculate the effectiveness, ε, and the capacity rate ratio, CR, from the known (or

calculated) inlet and outlet conditions and capacity rates.

2. Use the appropriate ε-NTU relationship or curve to determine the value of NTU.

3. Calculate A using:

A NTUCU

= min

As can be seen from the above steps, there is little to choose between the two

approaches for a problem of type (a).

5.17

Mean Temperature Difference approach to problem (b)

1. Calculate R from the capacity rates

2. Assume an outlet temperature for one stream and determine P.

3. Using the appropriate chart or relationship evaluate the correction factor F.

4. Calculate and the other outlet temperature using the assumed value of outlet

temperature

&Q

5. Evaluate the log mean temperature difference assuming counterflow using the

inlet and assumed outlet temperatures.

6. Calculate using: &Q

&Q UAF Tlmtd= ∆

7. Compare the values of calculated in steps 4 and 6. If they are acceptably close

then finish. Otherwise revise the assumption of outlet temperature in step 2 and

repeat steps 3-7.

&Q

ε-NTU approach to problem (b)

1. Calculate NTU and CR using the available data

2. Use the ε-NTU curve for the appropriate flow configuration and CR to determine

the effectiveness, ε.

3. Calculate from: &Q

( )&min , ,Q C T Th in c in= −ε

and hence determine the actual outlet temperatures.

In case (b) the ε-NTU approach is quicker and easier than the mean temperature

difference approach. It may also be argued that the effectiveness of a heat exchanger

has a more fundamental meaning that the mean temperature difference.

5.18

5.4 Design of a Heat Exchanger- Example

Design a heat exchanger to meet the following specification:

Fluid Hydrocarbon

gas

Hydrocarbon

liquid

Flow rate kg/s 0.6 3.0

Inlet temp oC 250 20

Outlet temp oC 25

Specific heat capacity kJ/kgK 2.219 2.378

Viscosity Ns/m2 1 x 10-5 5 x 10-4

Thermal conductivity W/mK 0.028 0.110

Prandtl No 0.793 10.81

Density kg/m3 4.0 748

Max. pressure drop* kPa 50 145

Fouling factor 0.0003 0.0002

Inlet Pressure kPa 106 2 x 106

*excluding nozzle losses

Calculate heat transfer coefficients using:

Nu = 0 023 0 8 0 4. Re Pr. .

and friction factors using:

25..0Re079.0 −=fc

and the frictional pressure drop is given by:

∆pV c LDm f

=2 2ρ

Losses in bends = 0.5 velocity heads

5.19

Design of a Heat Exchanger Example - solution

Step 1

Choose a type of heat exchanger, in practice iterative approach is required It may be

necessary to perform outline design calculations with more than one exchanger type

before selecting the most appropriate. Unless guided by experience of similar

applications (or very lucky!) it is unlikely that you will choose the best type without

performing some preliminary analysis, it is even less likely that you will choose the

optimum configuration (number and diameter of tubes, plate type etc.) at the first

attempt. The design presented here is not optimised but illustrates the steps

required.

The approach temperature ( )in,oilout,gas TT − is very close - this suggests that

counterflow is required.

The pressure levels and temperature are high for plate and plate-fin exchangers

If a compact heat exchanger is required PCHE, or SPFHE may be suitable.

Otherwise try double-pipe

Use double pipe

Choose standard unit (see Table at end of solution)

Determine necessary length

Try standard double-pipe section

4” Nominal diameter with 19.02mm tube OD

7 finned tube in shell 16 longtitudinal fins/tube

Dimensions

Shell internal diameter, Ds, 102.26mm

Number of tubes, Nt, 7

Tubes

Number of fins, Nf, 16

Thickness of fins, b, 0.0009m

Height of fins, lf, 0.0053m

Tube internal diameter, di, 0.0148m

Tube external diameter, di, 0.01902m

Extruded fins

5.20

Step 2

Since the liquid side heat transfer coefficient is likely to be higher than the gas side

coefficient liquid should be on the tube side. (i.e. the fluid with the lower heat

transfer coefficient goes on the side with the larger area)

Step 3

Internal heat transfer coefficient

Flow area inside tubes, Ad

Nti

t= =π π2

4 x

x 0.01484

x 7 = 1.204 x 10 m2

-3 2

Mass flux GmA

oil

t= = =&

.3

12042491

x 10kg / m s-3

2

7373410 x 50.0148 x 2491Re 4- ===

µi

iGd

hence flow is turbulent

Use Dittus-Boelter

Nu = 0.023 x 73734 x 10.81 = 467467 x 0.110

0.0148W / m K

0.8 0.4

2a =Nukdi

i= = 3472

Step 4

Wall resistance referred to inside area, kw=52W/mK

rdk

ddw

i

w

o

i= = =

20 0148 19 02

14 8ln

.ln

..2 x 52

36 x 10 m W / K-6 2

Step 5

5.21

Shell side heat transfer coefficient

Flow area As = Shell CSA-(tube+fin CSAs)

Shell CSA = π πDs

2

48 22= =

x 0.102264

x 10 m2

-3 2.

Tube CSA = π πd

Not

2

4199 x

x 0.019024

x 7 x 10 m2

-3 2= = .

Fin CSA = b l N Nf f t = 0.0009 x 0.00533 x 16 x 7 = 0.54 x 10 m -3 2

Flow area As=(8.22-1.99-0.54) x 10-3=5.7 x 10-3m2

Wetted Perimeter P= π Ds + ( π do + 2 x lf x Nf) x Nt

P= 0.1023π + ( 0.01902π + 2 x 0.00533 x 16) x 7= 1.934m

Hydraulic diameter de = = =4 x Flow area

Wetted perimeter4 x 5.7 x 10

1.935m

-3

0 0118.

GmA

gas

s= = =& .

..

0 65 7

105 3 x 10

kg / sm-32

12425410 x 10.0118 x 105.3Re 5- ===

µi

iGd

hence flow is turbulent

Use Dittus-Boelter to find clean coefficient

Nu = 0 023 0 8 0 3. Re Pr. .

Nu = 0.023 x 124254 x 0.793 = 249.4249.4 x 0.028

0.0118W / m K

0.8 0.3

2a =Nukdo

i= = 592

5.22

With fouling

1 1 1592

0 0003

1α α

α

o f of o

o f

r,

,

,

.= + = +

=

= 1.989 x 10 m K / W

1.989 x 10 = 502W / m K

-3 2

-32

Step 6

Fin Efficiency

( )η fin

f

f

ml

ml=

tanh

mkb

ml f

= = =

=

252

146 4

146 4

α 2 x 502 x 0.0009

x 0.00533 = 0.781

.

.,

( ) ( )η fin

f

f

ml

ml= = =

tanh tanh ..

.0 781

0 7810836

Step 7

External heat transfer coefficient related to inside diameter

( ) ( ) (( ))α

α η ππo f i

o f fin fin tube

i

A A

A, ,, .

=+

=+502 0 836

2029

x 16 x0.00533 x 2 x 0.01902 -16 x 0.0009 x 0.0148

W / m K2

5.23

Step 8

External heat transfer coefficient with reference to inside area

1 1 1 13472

0 0002 0 0000361

2029

102983

Ur r

U

i if i w

o f i

i

= + + + = + + +

=

=

α α,, ,

. .

. x 10 m K / WW / m K

-3 2

2

Step 9

Log Mean Temperature difference and heat load

( ) ( )( ) ( )

( )

& .

& .

, ,

, ,

, ,

Q mc T T

mc T T

T TQ

mc

gas gas in gas out

oil oil out oil in

oil out oil inoil

= − =

= −

= + = + =

0 x 2.219 x (250 - 25) = 299.6 kW

3 x 2.378Co

6

20299 6

62

20oC62oC

25oC250oCGas

Oil

( ) ( )∆ ∆

∆ ∆∆∆

T TT T

TT

m lmtd= =−

⎛⎝⎜

⎞⎠⎟=

− − −−−

⎛⎝⎜

⎞⎠⎟

=1 2

1

2

250 62 25 20250 6225 20

50 5ln ln

. o C

5.24

Step 10

Area required

AQ

U Tm= = =

& ..

∆299 6

06 035

.983 x 50.5m2 (This is internal area of tubes)

Step 11

Length of heat exchanger

Internal area/metre length = π πd Ni t = x 0.0148 x 7 = 0.326m / m2

Length = Area requiredArea / metre

5 m= =6 0350 362

18..

.

Use 2 x 10m lengths of standard section

Gas in

Gas out

Liquid in

Liquid out

5.25

Step12

Pressure Drops

For Tube Side

Re. Re . .. .

= =

= = =− −

73734 24910 079 0 079 4 790 25 0 25

kg / m s x 73734 x 10

2

-3

Gf

Frictional pressure drop

∆pfLGdf

i= = =

42

1076182

ρ2 x 4.79 x 10 x 20 x 2491

2 x 0.0148 x748Pa

-3 2

Assume 0.5 velocity head in 180o bend

∆p VG

bend = = = =12

12

0 25 207322

x 12

x 12

x 2491748

Pa2

ρρ

.

Total 110kPa this is acceptable ∆p = 109691Pa ≡

For Shell side

Re .. Re . .. .

= =

= = =− −

124254 105 30 079 0 079 4 20 25 0 25

kg / m s x 124254 x 10

2

-3

Gf

Frictional pressure drop

∆pfLGdf

e= = =

42

394662

ρ2 x 4.2 x 10 x 20 x 105.3

2 x 0.0118 x 4Pa

-3 2

Assume 1 velocity head in link between shells

∆p VG

link = = = = 12

12

x 105

4Pa

2

ρρ

22

0 53

1386..

5.26

Total 41kPa this is acceptable ∆p = 40852Pa ≡

5.27

Note: Assume extruded fins 0.9mm thick

Design of a Heat Exchanger Example – Standard units

5.28

Design of a Heat Exchanger Example – useful equations

5.29

Summary Points

• Providing certain assumptions are valid, a mean temperature difference may

be determined for a heat exchanger based on the fluid inlet and outlet temperatures.

• The Logarithmic Mean Temperature Difference (LMTD) may be calculated

for counter-flow and parallel flow configurations.

• A correction factor is available in graphical form for other configurations and

is applied to the LMTD for the streams flowing in counter flow.

• The effectiveness of a heat exchanger is defined as the ratio of the rate of

heat transferred in the heat exchanger to the maximum thermodynamically possible

rate of heat transfer for the same fluid inlet conditions. The thermodynamic

maximum rate of heat transfer would be achieved in an infinitely long counter flow

heat exchanger.

• Providing certain assumptions are valid, the effectiveness of a heat exchanger

may be calculated or determined from published tables or graphs.

• Use of the Mean Temperature difference and Effectiveness-NTU approaches

to heat exchanger calculations will yield identical results.

5.30