me421lec2

7/28/2019 ME421Lec2

http://slidepdf.com/reader/full/me421lec2 1/49

ME421 Heat ExchangerDesign

Lecture 2Basic Design Methods

(for indirect contact, two-fluid recuperators)

7/28/2019 ME421Lec2


Task• Heat Exchanger (HEX) Rating

Checking the existing design for compatibility with theuser requirements (outlet temperature, heat load etc.)given: flow rates, inlet temperatures, allowable pressuredrop; thus HT area and passage dimensionsfind: heat transfer rate, fluid outlet temperatures, actualpressure drop

• HEX Sizing Thermal and pressure drop considerations, maintenancescheduling with fouling consideration.given: inlet and outlet temperatures, flow rates, pressuredropfind: dimensions - type and size of HEX

7/28/2019 ME421Lec2


HEX Classification According to Flow Arrangement

7/28/2019 ME421Lec2


Mixed Flow Arrangements

Multipass Crossflow Arrangement

7/28/2019 ME421Lec2


Parallel Flow (PF)Counter Flow (CF)

Cold fluid evaporating at constant T Hot fluid condensing at constant T

Fluid Temperature Variation

7/28/2019 ME421Lec2


7/28/2019 ME421Lec2


Assumptions for Basic Design

Equations for Sizing• steady-state, steady flow• no heat generation in the HEX• negligible ∆PE, ∆KE• adiabatic processes• no phase change (later)• constant specific heats

7/28/2019 ME421Lec2


Basic Design Equations for Sizing

mQ UA T= ∆1 2 2 1( ) ( ) ( ) ( )p h h h p c c cQ mc T T mc T T= − = −& &

• Q : HT Rate• ∆ T

m: mean temperature difference

• A: total hot-side or cold-side HT area• U: average overall HT coefficient based on

A

7/28/2019 ME421Lec2


Overall Heat Transfer Coefficient

Single Smooth and Clean Plane Wall The overall heat transfer coefficient for a single

smooth and clean plane wall can be written as

1 11 1

t

i i o o

UA tR

hA kA h A

= = + +

7/28/2019 ME421Lec2


Tubular geometryFor the unfinned and clean tubular HEX, the

overall heat transfer coefficient is given by

1 1ln( / )1 1

2

o o i it o i

i i o o

U A U A R r r

hA kL h Aπ

= = =⎡ ⎤+ +⎢ ⎥

⎣ ⎦

7/28/2019 ME421Lec2


FoulingFor the HEX whose walls are fouled by deposit

formation on both the inside and outsidesurfaces, the total thermal resistance can beexpressed as

Rw : wall resistanceR

f : fouling factor / unit fouling resistance

1 1 1 1 1fi fot w

o o i i i i i o o o

R RR R

UA U A U A h A A A h A= = = = + + + +

7/28/2019 ME421Lec2


Wall resistance The wall resistance is obtained from thefollowing equations:

o i

(for a plane wall)

ln( r r )(for a tube wall)

2

w

t

kAR

kLπ

⎧⎪

⎪=⎨⎪

⎪⎩

7/28/2019 ME421Lec2


U based on outside surface area

U is usually based on the outer area. U based onthe outside surface area of the wall for anunfinned, tubular HEX is given by

If fins are present on the wall(s), fin efficiency andfin area should be considered in calculating U.See Eqns. (2.13 - 2.19)

Simple examples on p.42 of book

1ln( )1 1

oo o o o i

fi foi i i o

Ur r r r r

R Rr h r k h

=⎡ ⎤

+ + + +⎢ ⎥⎣ ⎦

7/28/2019 ME421Lec2


Orders of Magnitude for h [w/m 2K]

Gases (natural convection) 3-25

Engine oil (natural convection) 30-60

Flowing liquids (nonmetal) 100-10,000

Flowing liquid metals 5000-250,000

Film boiling 300-400

Dropwise condensation 60,000-120,000

7/28/2019 ME421Lec2


Log Mean Temperature Difference (LMTD)Method for HEX Analysis

Parallel and Counterflow HEX

Differential energy balance

where C h and C c are the hot and cold fluid heatcapacity rates+ and - signs correspond to PF and CF,

respectively

( ) ( )p h h p c cQ mc dT mc dTδ = − = ±& &

h h c cQ C dT C dTδ = − = ±

7/28/2019 ME421Lec2


• Also in terms of overall heat transfer coefficientand temperature differences

• From both energy balances (previous slide)

• Using δQ

( )h cQ U T T dAδ = −

1 1( )h c h c

c h

d T T dT dT QC C

δ ⎛ ⎞

− = − = −⎜ ⎟

⎝ ⎠

( ) 1 1( )

h c

h c c h

d T T U dA T T C C

⎛ ⎞− = −⎜ ⎟− ⎝ ⎠

7/28/2019 ME421Lec2


• Integrating the previous equation with constantU, C h, and C c over HEX length

or, for CF

• For PF,

• Note that temperature distribution along the HEXis an exponential function of A.

2 1

1 2

1 1ln h c

h c c h

T T UA T T C C

⎛ ⎞− = −⎜ ⎟− ⎝ ⎠

2 1 1 21 1

( )exph c h cc h

T T T T UAC C

⎡ ⎤⎛ ⎞− = − −⎢ ⎥⎜ ⎟

⎝ ⎠⎣ ⎦

2 1 1 2

1 1( )exph c h c

c h T T T T UA C C

⎡ ⎤⎛ ⎞

− = − −⎢ ⎥⎜ ⎟⎝ ⎠⎣ ⎦2 1 - +

7/28/2019 ME421Lec2


After obtaining C h and C c and solving for Q

or

where ∆ T 1 is the temperature difference betweenthe two fluids at one end of the HEX and ∆ T 2 is the

one at the other end of the HEX.

1 2 2 1

1 2

2 1

( ) ( )

ln

h c h c

h c

h c

T T T TQ UA T T

T T

− − −=⎛ ⎞−⎜ ⎟

−⎝ ⎠

1 2

1 2ln( ) T TQ UA

T T∆ − ∆= ∆ ∆

7/28/2019 ME421Lec2


Temperature Variation for a Counter Flow HEX

7/28/2019 ME421Lec2


• Defining LMTD as

• The total heat transfer rate for all single-passflow arrangements (PF, CF, evaporator,condenser)

• In case of CF HEX with Ch=C c, ∆ Tlm isindeterminate

• Therefore from L’Hospital: ∆ Tlm= ∆ T1= ∆ T2

1 2

1 2ln( )lm T T

T T T∆ − ∆

∆ = ∆ ∆

lmQ AU T= ∆

1 2 2 1 1 2( ) ( ) andh h c c T T T T T T− = − ∆ = ∆

7/28/2019 ME421Lec2


• In case of PF HEX, the same definition of LMTD(previous slide) is valid. But obviously

• LMTD represents the maximum temperaturepotential for heat transfer that can only beobtained in a counter flow heat exchanger.

Therefore, the surface area required to obtain aprescribed heat transfer rate Q is smaller for acounter flow arrangement than that for a parallelflow arrangement, assuming that U is the samefor both cases.

• Also note that T c2 can exceed T h2 for counter

flow but not for parallel flow.• Example 2.4 in book

)( T and )( 222111 chch T T T T T −=∆−=∆

7/28/2019 ME421Lec2


Multipass and Crossflow HX

• LMTD defined for single pass (PF or CF) HEX is not

valid for multipass and crossflow HEX. This time bydefining a mean temperature difference ∆ Tm

• We can determine ∆ Tm in terms of LMTD of CF HEX,

and two quantities P and R.

mQ AU T= ∆

7/28/2019 ME421Lec2


[ ]2 1 1 2,

2 1 1 2

( ) ( )ln ( ) ( )

h c h clmcf h c h c

T T T T T T T T T

− − −∆ = − −

2 1

1 1 max

c c c

h c

T T TP

T T T− ∆= =− ∆

1 2

2 1

c h h

h c cC T TR C T T−= = −

7/28/2019 ME421Lec2


Meanings of ∆ T lm,cf P and R

• ∆ Tlm,cf is the log-mean temperature differencefor a counter flow arrangement with the samefluid inlet and outlet temperatures.

• P is a measure of the ratio of the heat actuallytransferred to the cold fluid to the heat which

would be transferred if the same fluid were to beraised to the hot fluid inlet temperature;therefore is the temperature effectiveness of the

HEX on the cold fluid side.• R is the ratio of the mc p value of the cold fluid to

that of the hot fluid and it is called the heatcapacity rate ratio .

7/28/2019 ME421Lec2


Correction factor F

• With the definition of the correction factor F heattransfer rate for a multipass or crossflow HEXcan be given as

• F is a nondimensional term which depends onthe temperature effectiveness P, the heat

capacity rate ratio R, and the flow arrangement.

,lmcf Q UAF T= ∆

Ch f i f

7/28/2019 ME421Lec2


Charts for correction factors

• Depending on the flow arrangement there areseries of charts to obtain F using calculatedvalues of P and R.

• Such figures are:In our textbook: Figures 2.7 through 2.14.

In Incropera and DeWitt: Figures 11.10through 11.13.

• It is immaterial whether the cold fluid flowsthrough the shell or inside the tubes

• If temperature change of one fluid is negligible,

F = 1• Some examples of charts are...

7/28/2019 ME421Lec2


• LMTD correction factor F for a shell-and-tubeheat exchanger – one shell pass and two ormultiple of two tube passes

• Example 2.5 in book now, also study examples2.6-2.8

7/28/2019 ME421Lec2


7/28/2019 ME421Lec2


Eff ti N b f T f U it

7/28/2019 ME421Lec2


Effectiveness-Number of Transfer Units

(ε-NTU) for HEX Analysis• When inlet and exit temperatures are unknown, a

trial and error procedure may be needed. Instead,the method of number of transfer units (NTU)based on HEX effectiveness may be used.

• The ε - NTU method is based on the fact that theinlet or exit temperature differences of a heatexchanger are a function of UA/C c and C c/C h.

• The HEX heat transfer equations may be writtenin dimensionless form resulting in somedimensionless groups.

7/28/2019 ME421Lec2


Dimensionless groups

1. Heat capacity rate ratio: , C * ≤ 1

2. HEX heat transfer effectiveness:

ε is the ratio of the actual heat transfer rate in a HEXto the thermodynamically limited maximum possibleheat transfer rate if an infinite heat transfer areawere available in a counter flow HEX.

min

max

CCC

∗=

max

Q

Qε =

7/28/2019 ME421Lec2


• The actual heat transfer is obtained either by theenergy given off by the hot fluid or the energyreceived by the cold fluid

If Ch > C c, then (Th1 -Th2 ) < (T c2 -Tc1 )If Ch < C c, then (Th1 -Th2 ) > (T c2 -Tc1 )

• The above equations are valid for CF and PF. The fluid that might undergo the maximumtemperature difference is the fluid having theminimum heat capacity rate Cmin .

1 2 2 1( ) ( ) ( ) ( )p h h h p c c cQ mc T T mc T T= − = −& &

M i h f

7/28/2019 ME421Lec2


• Maximum heat transfer:

or

• Therefore, HEX effectiveness can be written as

• The above equation is valid for all heat

exchanger flow arrangements . The value of ε

ranges between 0 and 1.• For a given ε and Q max , the actual HT rate is

Q = ε (mc p)min (Th1 -Tc1)

max 1 1( ) ( ) if p c h c c hQ mc T T C C= − <&

max 1 1( ) ( ) if p h h c h cQ mc T T C C= − <&

1 2 2 1

min 1 1 min 1 1

( ) ( )

( ) ( )

h h h c c c

h c h c

C T T C T T

C T T C T Tε

− −= =− −

7/28/2019 ME421Lec2


3. Number of Transfer Units:

The third dimensionless number NTU shows thenondimensional heat transfer size of the HEX

min min

1NTU

A

AUUdA

C C= =

∫

Example: Single pass heatexchanger C >C

7/28/2019 ME421Lec2


Example: Single pass heat exchanger C c>C h

•C h=C min and C c=C max

•We had obtained

•By using definition of NTU

where the + is for counter flow and the – is forparallel flow

2 1 1 21 1

( )exph c h cc h

T T T T UA C C

⎡ ⎤⎛ ⎞

− = − −⎢ ⎥⎜ ⎟⎝ ⎠⎣ ⎦

min2 1 1 2

max

( )exp NTU 1-h c h cC

T T T TC

⎡ ⎤⎛ ⎞− = − − ±⎢ ⎥⎜ ⎟⎝ ⎠⎣ ⎦

• Using

7/28/2019 ME421Lec2


• Using

and

Th2 and Tc2 can be eliminated to obtain for CF

• If C c < C h, the result will be the same

1 2 2 1( ) ( ) ( ) ( )p h h h p c c cQ mc T T mc T T= − = −& &

1 2 2 1

min 1 1 min 1 1

( ) ( )( ) ( )

h h h c c c

h c h c

C T T C T T

C T T C T Tε

− −

= =− −

[ ]

[ ]min max

min max min max

1 exp NTU(1-

1 ( )exp NTU(1-

C C

C C C C )ε

− −=

− −

7/28/2019 ME421Lec2


• For C min /C max =1, this result is indeterminate, but

by applying L’Hospital’s rule, the following result isobtained for counter flow

• and for parallel flow

NTU1 NTU

ε = +

2NTU1(1 )2 eε

−

= −

F C /C 0 b h CF d PF HX

7/28/2019 ME421Lec2


• For C min /C max =0, both CF and PF HXs

• Similar expressions for all flow arrangementsincluding multipass and crossflow can beobtained following the same approach. Some of those results are presented as tables like thefollowing. Also there are charts like Fig 2.15 of the text book

NTU1 eε −= −

ε-NTU Expressions (Table 2.2 of the book, more detail in book)

7/28/2019 ME421Lec2


p ( )

Type of HEX ε(NTU,C*) NTU( ε,C*)

Counterflow

Parallel Flow

Cross flow, C min

mixed and C maxunmixed

Cross flow, C maxmixed and C minunmixed

1 to 2 shell-and-tube HEX

( )[ ]( )[ ]NTU1exp1

NTU1exp1∗−−∗−

∗−−−=

CC

Cε ⎟⎟ ⎠

⎞⎜⎜⎝ ⎛

−

∗−∗−

=ε

ε

1

1ln

1

1NTU

C

C

( )[ ][ ]NTU1exp11

1 ∗+−−∗+

= CC

ε ( )[ ]∗++∗+

−= CC

11ln1

1NTU ε

( )⎥⎦

⎤

⎢⎣

⎡

∗

∗−−

−= C

C NTUexp1

exp1ε

( )[ ]ε −∗

+∗−= 1ln1ln1

NTU CC

( )[ ]{ }[ ]NTUexp1exp11

−−∗−−∗

= C

C

ε ( )⎥⎦

⎤⎢⎣

⎡ ∗−∗

+= CC

ε 1ln1

1-lnNTU

( )⎭⎬

⎫

⎩⎨

⎧

⎭⎬⎫

⎩⎨⎧

⎟⎟ ⎠ ⎞

⎜⎜⎝ ⎛ ∗

++∗

+−

⎟⎟ ⎠ ⎞

⎜⎜⎝ ⎛ ∗+−∗+−

∗+=

2/12

112

2/12112

ln2/121

1NTU

CC

CC

Cε

ε

( )

⎭⎬⎫

⎩⎨⎧

⎭⎬⎫

⎩⎨⎧

⎟⎟ ⎠ ⎞

⎜⎜⎝ ⎛ ∗+−−

⎟⎟ ⎠ ⎞

⎜⎜⎝ ⎛ ∗+−+

∗++∗+

=

2/121NTUexp1

2/121NTUexp1

2/1211

2

C

C

CC

ε

7/28/2019 ME421Lec2


7/28/2019 ME421Lec2


LMTD and ε-NTU relations

7/28/2019 ME421Lec2


LMTD ε-NTU

F = φ (P, R, flow arrangement) ε = φ (NTU, C*, flow arrangement)

)11(min c Th TCQ −= ε cf lm TUAFQ ,∆=

( ) ( )11min

12

11min

21

c Th TCc Tc TcC

c Th TCh Th ThC

−

−

=−

−

=ε

( )21ln21

, T T

T T

cf lm TLMTD ∆∆

∆−∆=∆=

122,211 c Th T Tc Th T T −=∆−=∆

12

21,11

12

c Tc Th Th T

Rc Th Tc Tc T

P−

−=

−

−=

( )max

min

max

min

pcm

pcm

C

CC

&

&==∗

∫== AUdA

CC

UA

min

1

min

NTU

Example 29

7/28/2019 ME421Lec2


Example 2.9

A two-pass tube, baffled single-pass shell, shell-and-tube HEX is used as an oil cooler. Coolingwater flows through the tubes at 20 oC at a flow rate

of 4.082 kg/s. Engine oil enters the shell side at aflow rate of 10 kg/s. The inlet and outlettemperatures of oil are 90 oC and 60 oC,respectively. Determine the surface are of the HEXusing both the LMTD and ε-NTU methods, if theoverall heattransfer coefficient based on the outsidetube area is 262 W/m 2K. The specific heats of waterand oil are 4179 J /kgK and 2118 J /kgK,

respectively.

h

7/28/2019 ME421Lec2


Heat Exchanger Sizing

If inlet temperatures, one of the outlettemperatures and mass flow rates are known,we can use LMTD method for sizing problem:

1. Calculate Q and the unknown temperature

2. Calculate LMTD and obtain F if necessary3. Calculate U

4. Determine A from A=Q/(UF ∆ T lm,cf )

H E h R i

7/28/2019 ME421Lec2


Heat Exchanger RatingFor an available heat exchanger (size, massflow rates, inlet temperatures and materials areknown) using ε-NTU method we can rate theheat exchanger:

1. Calculate C*=C min /C max and NTU=UA/C min

2. Determine ε from appropriate charts or ε-NTUequations3. Calculate Q= ε Cmin (Th1 -Tc1 )

4. Calculate outlet temperatures

ε

7/28/2019 ME421Lec2


Sizing Using ε-NTU method

1. Calculate ε using C min , C max and temperatures

2. Calculate C*=C min /C max3. Calculate U4. Determine NTU from charts or equations5. When NTU is known calculate heat transfer

area from A=(C min NTU)/U

R ti i LMTD

7/28/2019 ME421Lec2


Rating using LMTD

• LMTD method can also be used for rating but forthe unknown temperature trial and errorapproach is required.

• In general LMTD is more appropriate for sizingand ε-NTU is more appropriate for ratingproblems.

• Variable U may complicate the calculations, nextfigure shows some cases where U is variable.Variation dependent on flow Reynolds number,HT surface geometry, fluid physical properties

7/28/2019 ME421Lec2


N i l l i f i bl U

7/28/2019 ME421Lec2


Numerical analysis of variable U case• Divide heat exchanger length to regions

conveniently and consider as HEX in series

(index i is for regions). Select a convenient ∆Ai• Use numerical/finite-difference methods• Calculate inner and outer HT coefficients and U

for the initial ∆A increment• Calculate ∆Q i for this increment• Calculate the values of T h and T c for the next

increment.• Continue until the entire area handled.

• More detail in Section 2.8 in book

me421lec2

Documents