Heat Transfer/Heat Exchanger• How is the heat transfer? • Mechanism of Convection• Applications . • Mean fluid Velocity and Boundary and their effect on the rate of heat

transfer.• Fundamental equation of heat transfer• Logarithmic-mean temperature difference.• Heat transfer Coefficients.• Heat flux and Nusselt correlation • Simulation program for Heat Exchanger

How is the heat transfer?

• Heat can transfer between the surface of a solid conductor and the surrounding medium whenever temperature gradient exists.ConductionConvection

Natural convection Forced Convection

Natural and forced Convection

Natural convection occurs whenever heat flows between a solid and fluid, or between fluid layers.

As a result of heat exchange

Change in density of effective fluid layers taken place, which causes upward flow of heated fluid.

If this motion is associated with heat transfer mechanism only, then it is called Natural Convection

Forced Convection

If this motion is associated by mechanical means such as pumps, gravity or fans, the movement of the fluid is enforced.

And in this case, we then speak of Forced convection.

Heat Exchangers• A device whose primary purpose is the transfer of energy

between two fluids is named a Heat Exchanger.

Applications of Heat Exchangers

Heat Exchangers prevent car engine

overheating and increase efficiency

Heat exchangers are used in Industry for

heat transfer

Heat exchangers are used in AC and

furnaces

• The closed-type exchanger is the most popular one.

• One example of this type is the Double pipe exchanger.

• In this type, the hot and cold fluid streams do not come into direct contact with each other. They are separated by a tube wall or flat plate.

Principle of Heat Exchanger• First Law of Thermodynamic: “Energy is conserved.”

generatedsin out

outin ewqhmhmdt

dE

ˆ.ˆ.

outin

hmhm ˆ.ˆ. h

hphh TCmQ ..

ccpcc TCmQ ..

0 0 0 0

•Control Volume

Qh

Cross Section Area

HOT

COLD

Thermal Boundary Layer

Q hot Q cold

Th Ti,wall

To,wall

Tc

Region I : Hot Liquid-Solid Convection

NEWTON’S LAW OF CCOLING

dqx hh . Th Tiw .dA Region II : Conduction Across Copper Wall

FOURIER’S LAW

dqx k.dT

dr

Region III: Solid – Cold Liquid Convection

NEWTON’S LAW OF CCOLING

dqx hc . Tow Tc .dA

THERMAL

BOUNDARY LAYER

Energy moves from hot fluid to a surface by convection, through the wall by conduction, and then by convection from the surface to the cold fluid.

• Velocity distribution and boundary layer

When fluid flow through a circular tube of uniform cross-suction and fully developed,

The velocity distribution depend on the type of the flow.

In laminar flow the volumetric flowrate is a function of the radius.

V u2rdrr0

rD / 2

V = volumetric flowrate

In turbulent flow, there is no such distribution.

• The molecule of the flowing fluid which adjacent to the surface have zero velocity because of mass-attractive forces. Other fluid particles in the vicinity of this layer, when attempting to slid over it, are slow down by viscous forces.

r

Boundary layer

• Accordingly the temperature gradient is larger at the wall and through the viscous sub-layer, and small in the turbulent core.

• The reason for this is 1) Heat must transfer through the boundary layer by conduction.2) Most of the fluid have a low thermal conductivity (k)3) While in the turbulent core there are a rapid moving eddies, which they are equalizing the temperature.

heating

cooling

Tube wall

Twh

Twc

Tc

Metalwall

Warm fluid

cold fluid

qx hAT

qx hA(Tw T)

qx k

A(Tw T)

h

Region I : Hot Liquid – Solid Convection

Th Tiw qx

hh .Ai

qx hhot . Th Tiw .A

Region II : Conduction Across Copper Wall

qx kcopper .2L

lnro

ri

To,wall Ti,wall qx .ln

ro

ri

kcopper .2L

Region III : Solid – Cold Liquid Convection

To,wall Tc qx

hc .Ao

qx hc To,wall Tc Ao

+

Th Tc qx

1

hh .Ai

ln

ro

ri

kcopper .2L

1

hc .Ao

qx U.A. Th Tc 1

1

.

ln.

.

coldicopper

i

oo

ihot

o

hrk

rr

r

rh

rU

U = The Overall Heat Transfer Coefficient [W/m.K]

Th Tc qx

R1 R2 R3

U 1

A.R

r

o

r

i

Calculating U using Log Mean Temperature

coldhot dqdqdq

ch TTT ch dTdTTd )(

hhphh dTCmdq ..

ccpcc dTCmdq ..

Hot Stream :

Cold Stream:

cpc

chph

h

Cm

dq

Cm

dqTd

..)(

dATUdq ..

cpc

hph CmCm

dATUTd.

1

.

1...)(

2

1

2

1

..

1

.

1.

)( A

Acpc

hph

T

TdA

CmCmU

T

Td

outc

inc

outh

inhch TTTT

q

AUTT

q

AU

T

T

...

ln1

2

2

1

2

1

..)( A

Ac

c

h

hT

TdA

q

T

q

TU

T

Td

1

2

12

ln

.

TT

TTAUq

Log Mean Temperature

16

Heat Exchanger (HX) Design MethodsHX designers usually use two well-known methods for calculating the heat transfer rate between fluid streams—the UA-LMTD and the effectiveness-NTU (number of heat transfer units) methods.

Both methods can be equally employed for designing HXs. However, the -NTU method is preferred for rating problems where at least one exit temperature is unknown. If all inlet and outlet temperatures are known, the UA-LMTD method does not require an iterative procedure and is the preferred method.

Energy Balance

Overall Energy Balance

• Assume negligible heat transfer between the exchanger and its surroundings and negligible potential and kinetic energy changes for each fluid.

, ,h i h ohq m i i

, ,c c o c iq m i i

fluid enthalpyi

• Assuming no l/v phase change and constant specific heats, , , ,p h h i h ohq m c T T

, ,h h i h oC T T

, , ,c p c c o c iq m c T T

, ,c c o c iC T T

, Heat capacity r s ateh cC C

• Application to the hot (h) and cold (c) fluids:

18

LMTD (Log Mean Temperature Difference)The most commonly used type of heat exchanger is the recuperative heat exchanger. In this type the two fluids can flow in counter-flow, in parallel-flow, or in a combination of these, and cross-flow.

The true mean temperature difference is the Logarithmic mean Temperature difference (LMTD), is defined as

2

1

21

tt

ln

ttLMTD

CON CURRENT FLOW

1

2

12

lnT

T

TTTLn

731 TTTTT inc

inh

1062 TTTTT outc

outh

COUNTER CURRENT FLOW

1062 TTTTT inc

outh

731 TTTTT outc

inh

U Ý m h . Ý C p

h . T3 T6 A.TLn

Ý m c . Ý C p

c . T7 T10 A.TLn

T1T2

T4 T5

T3

T7 T8 T9

T10

T6

Counter - Current Flow

T1 T2T4 T5

T6T3

T7T8 T9

T10

Parallel Flow

Log Mean Temperature evaluation

T1

A

1 2

T2

T3

T6

T4 T6

T7 T8

T9

T10

Wall∆ T1

∆ T2

∆ A

A

1 2

T1

A

1 2

T2

T3

T6

T4 T6

T7 T8

T9

T10

Wall

q hh Ai Tlm

Tlm (T3 T1) (T6 T2)

ln(T3 T1)

(T6 T2)

q hc Ao Tlm

Tlm (T1 T7) (T2 T10)

ln(T1 T7)

(T2 T10)

Exchanger Fouling

Electron microscope image showing fibers, dust, and other deposited material on aresidential air conditioner coil and a fouled water line in a water heater.

Exchanger Fouling

Overall Coefficient

Overall Heat Transfer Coefficient

• An essential requirement for heat exchanger design or performance calculations.

• Contributing factors include convection and conduction associated with the two fluids and the intermediate solid, as well as the potential use of fins on both sides and the effects of time-dependent surface fouling.

• With subscripts c and h used to designate the hot and cold fluids, respectively, the most general expression for the overall coefficient is:

, ,

1 1 1

1 1

c h

f c f hw

o o o oc c h h

UA UA UA

R RR

hA A A hA

Overall Coefficient

o,

Overall surface efficiency of fin array (Section 3.6.5)

1 1

o

fc or h f

c or h

A

A

total surface area (fins and exposed base) surface area of fins only

t

f

A AA

Assuming an adiabatic tip, the fin efficiency is

,

tanhf c or h

c or h

mL

mL

2 /c or h p w c or hm U k t

, partial overall coe1

fficientp c or hf c or h

hUhR

2 for a unit surfFouling fact ace area (m W)or K/fR

Table 11.1

conduction resistan Wall (K/Wce )wR

25

Heat Exchanger UA-LMTD Design Method

Where U is the overall heat transfer coefficient (and is assumed to be constant over the whole surface area of the heat exchanger).

Lhr

RLk

r/rlnR

LhrR

UA

R

LMTDLMTDUAQ

ooo,f

p

ioi,f

ii

n

i

i

n

i

i

2

1

22

11

1

1

Heat Transfer Duty Overall Heat TransferCoefficient (W/m2.K)

Water to condensingR-12

440-830

Steam to water 960-1650

Water to water 825-1510

Steam to gases 25-2750

26

Calculation of heat transfer——heat transfer coefficient

Table - data of heat resistance from scales

fluidRs

m2· /W℃fluid

Rs

m2· /W℃

water (u<1m/s, t<50 )℃   vapors  

sea 0.0001 organic 0.0002

river 0.0006 steam (no oil) 0.0001

well 0.00058 waste (with oil) 0.0002

distilled 0.0001 freezing (with oil) 0.0004

boiler supply 0.00026 gases  

for cooling tower

(untreated)0.00058 air 0.0003

for cooling tower

(treated)0.00026 compressed air 0.0004

with solid particles 0.0006 nature gas 0.002

brine 0.0004 coke oven gas 0.002

For reducing this resistance, （ 1 ） cleaning the heat exchanger; （ 2 ） treat the fluid before use.

27

2. Using empirical data or equation to calculate U

Calculation of heat transfer——heat transfer coefficient

Fluid type U W/(m2·K)water—gas 12~60

water—water 800~1800

water—kerosene around 350

water—organic solvent 280~850

gas—gas 12~35

saturated steam—water 1400~4700

saturated steam—gas 30~300

saturated steam—oil 60~350

saturated steam—boiling oil 290~870

Table - empirical data of U for shell - pipe heat exchangers

Empirical equations are given by manufacturers.

28

Effect of h to U

Calculation of heat transfer——heat transfer coefficient

e. g.: using saturated steam with （ h2 ） to heat air (h1) : the pipe size is 25×2.5, air is in the pipe. The conductive heat resistance can be ignored. We have h1

(W/m2.k)

h2

(W/m2.k)

UO

(W/m2.k)comparison

40 5000 31.8 UO 2close to h1;

40 10000 31.9 h2 doubles ， UO almost no change

80 5000 63.2 h1 doubles ， UO doubles

—— the relatively importance of h to U

the results show that the increase of relative smaller h is key to increase U—— the key resistance is important.

If the value of the two h are not quite different ， the increase of any one h is effective to raising the K .

211

2 11

hdh

d

UO

LMTD Method

A Methodology for Heat ExchangerDesign Calculations

- The Log Mean Temperature Difference (LMTD) Method -• A form of Newton’s Law of Cooling may be applied to heat exchangers by using a log-mean value of the temperature difference between the two fluids:

1mq U A T

1 2

11 21n /m

T TT

T T

Evaluation of depends on the heat exchanger type.1 2 and T T

• Counter-Flow Heat Exchanger:

1 ,1 ,1

, ,

h c

h i c o

T T TT T

2 ,2 ,2

, ,

h c

h o c i

T T TT T

LMTD Method (cont.)

• Parallel-Flow Heat Exchanger:

1 ,1 ,1

, ,

h c

h i c i

T T TT T

2 ,2 ,2

, ,

h c

h o c o

T T TT T

Note that Tc,o can not exceed Th,o for a PF HX, but can do so for a CF HX.

For equivalent values of UA and inlet temperatures, 1 , 1 ,m CF m PFT T

• Shell-and-Tube and Cross-Flow Heat Exchangers:1 1 ,m m CFT F T

Figures 11.10 - 11.13F

Special Conditions

Special Operating Conditions

Case (a): Ch>>Cc or h is a condensing vapor .hC

– Negligible or no change in , , .h h o h iT T T

Case (b): Cc>>Ch or c is an evaporating liquid .cC

– Negligible or no change in , , .c c o c iT T T

Case (c): Ch=Cc.1 2 1mT T T –

32

comparison of flow directions

Calculation of heat transfer——temperature difference

21

21

/ln tt

tttm

Given ： the inlet and outlet temperatures of two fluids Find ： logarithmic average T differences in different situations Solution ： cocurrent counter currenthot fluid 240 → 160 ℃ ℃

cold fluid - ） 100 → 140 ℃ ℃

t1=140 20= t2

240 → 160 ℃ ℃

- ） 140 ℃ ← 100 ℃

t1=100 60= t2

140 20

61.7ln 140 / 20mt C

,co

100 60

78.3 Cln 100 / 60mt

,counter

<It isclear thatt m,co < tm,counter

33

Comparison 1——heat transfer area needed

At same flow rate and same T1, T2, t1 and t2 , for same heat

transfer task

Q = w1Cp1(T1 - T2) = w2Cp2(t2 - t1) = UA t△ m

m co counter

co counter

* , for same Q and K from the comparison of Δt A >A ;

*for a given heat exchanger,A and K are constant then Q <Qm

QA

K t

， ，

，

task = capacity

Calculation of heat transfer——temperature difference

34

For a given heating task ， Q = W2Cp2(t2 - t1)

consumption of heating medium ： W1 = W2Cp2(t2 - t1) / Cp1(T1-T2 )

t

A

t1

t2

T1

T2

cocurrent

counter current ： T2min is limited by t1

T2t1

t

A

t2

T1

counter current

cocurrent ： T2min is limited by t2

It is obvious that t2 >t1 , then T2co > T2counter , W1co > W1counter

Therefore, for a given task, a counter current heat exchanger consumes little heating medium than a cocurent one.

1 2W T

Calculation of heat transfer——temperature difference

Comparison 2——consumption of heating medium

For a given T1

Problem: Overall Heat Transfer Coefficient

Determination of heat transfer per unit length for heat recoverydevice involving hot flue gases and water.

KNOWN: Geometry of finned, annular heat exchanger. Gas-side temperature and convection coefficient. Water-side flowrate and temperature.

FIND: Heat rate per unit length.

SCHEMATIC:

Do = 60 mm Di,1 = 24 mm Di,2 = 30 mm t = 3 mm = 0.003m L = (60-30)/2 mm = 0.015m

Problem: Overall Heat Transfer Coefficient (cont.)

ASSUMPTIONS: (1) Steady-state conditions, (2) Constant properties, (3) One-dimensional conduction in strut, (4) Adiabatic outer surface conditions, (5) Negligible gas-side radiation, (6) Fully-developed internal flow, (7) Negligible fouling.

PROPERTIES: Table A-6, Water (300 K): k = 0.613 W/mK, Pr = 5.83, = 855 10-6 Ns/m2.

ANALYSIS: The heat rate is

where

m,h m,ccq UA T T

w oc c h1/ UA 1/ hA R 1/ hA

i,2 i,1 4w

ln D / D ln 30 / 24R 7.10 10 K / W.

2 kL 2 50 W / m K lm

Problem: Overall Heat Transfer Coefficient (cont.)

the internal flow is turbulent and the Dittus-Boelter correlation gives

4 / 5 0.44 / 5 0.4 2c i,1 D

0.613 W / m Kh k / D 0.023Re Pr 0.023 9990 5.83 1883 W / m K

0.024m

11 2 3chA 1883 W / m K 0.024m 1m 7.043 10 K / W.

The overall fin efficiency is o f f1 A / A 1

2fA 8 2 L w 8 2 0.015m 1m 0.24m

2 2f i,2A A D 8t w 0.24m 0.03m 8 0.003m 0.31m .

From Eq. 11.4,

ftanh mL

mL

With

D 6 2i,1

4m 4 0.161 kg / sRe 9990

D 0.024m 855 10 N s / m

Problem: Overall Heat Transfer Coefficient (cont.)

Hence f 0.800 /1.10 0.907 o f f1 A / A 1 1 0.24 / 0.31 1 0.907 0.928

11 2 2o hhA 0.928 100 W / m K 0.31m 0.0347 K / W.

It follows that

1 3 4cUA 7.043 10 7.1 10 0.0347 K / W

cUA 23.6 W / K

and q 23.6 W / K 800 300 K 11,800 W < for a 1m long section.

where

1/ 2 1/ 22 1m 2h / kt 2 100 W / m K / 50 W / m K 0.003m 36.5m

1/ 2 1mL 2h / kt L 36.5m 0.015m 0.55

1/ 2tanh 2h / kt L 0.499.

Problem: Overall Heat Transfer Coefficient (cont.)

COMMENTS: (1) The gas-side resistance is substantially decreased by using the fins

,2f iA D and q is increased.

(2) Heat transfer enhancement by the fins could be increased further by using a material of larger k, but material selection would be limited by the large value of Tm,h.

III. Rate equation and TTM

The rate equation for a shell-and-tube heat exchanger is the same as for a concentric pipe exchanger:

However, Ui and TTM are evaluated somewhat differently for shell-and-tube exchangers. We will first discuss how to evaluate TTM and then a little later in the notes we will discuss how to evaluate Ui for shell-and-tube exchangers.In a shell-and-tube exchanger, the flow can be single or multipass. As a result, the temperature profiles for the two fluids in a shell-and-tube heat exchanger are more complex, as shown below.

TMTUAq ii

TMTUAq ii

Computation of TTM:

For the concentric pipe heat exchanger, we showed the following (parallel and countercurrent flow):

TTM =

When a fluid flows perpendicular to a heated or cooled tube bank, and if both of the fluid temperatures are varying, then the temperature conditions do not correspond to either parallel or countercurrent. Instead, this is called crossflow.

Tlm

F o r c ro s s f lo w a n d m u l t ip a s s h e a t e x c h a n g e d e s ig n s , w e m u s t in t ro d u c e a c o r re c t io n f o r th e lo g m e a n te m p e ra tu r e d i f f e re n c e ( L M T D ) :

T T M =

Z T ha T hb

T cb T ca

H Tcb Tca

T ha T ca

FG * TLM

The factor Z is the ratio of the fall in temperature of the hot fluid to the rise in temperature of the cold fluid.

The factor H is the heating effectiveness, or the ratio of the actual temperature rise of the cold fluid to the maximum possible temperature rise obtainable (if the warm-end approach were zero, based on countercurrent flow).

From the given values of H and Z, the factor FG can be read from the text book figures:

Therefore, as with the concentric pipe heat exchanger, the true mean temperature difference for the 1-1 exchanger is equal to the log mean temperature difference (TLM).

For multiple pass shell-and-tube designs, the flow is complex and the TLM is less than that for a pure countercurrent design.

We must account for the smaller temperature driving force using a correction factor, FG, which is less than 1 and typically greater than 0.8.

The rate of heat transfer in multiple pass heat exchangers is written as:

LMTUAFq Gwhere TLM is the log mean temperature difference for pure countercurrent flow

Textbook Figures 15.6 a, b

1-2 exchangers

2-4 exchangers

Ten Minute Problem -- FG for multiple pass HE

For a 2-4 heat exchanger with the cold fluid inside the tubes and the following temperatures:

Tca = 85°F Tha = 200°FTcb = 125°F Thb = 100°F

(a) What is the true mean temperature difference?(answer TTM = 31.7°F)

(b) What exchanger area is required to cool 50,000 lbm/hr of product (shell-side fluid) if the overall heat transfer coefficient is 100 Btu/hr-ft2-°F and Cp for the product is 0.45 Btu/lbm-°F?(answer A = 710 ft2)