Heat Exchange Models

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ENERGY CONVERSION

ES 832a

Eric Savorywww.eng.uwo.ca/people/esavory/es832.htm

Lecture 8 Basics of heat exchangers

Department of Mechanical and Material Engineering

University of Western Ontario


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Heat Exchangers

The most common types of energy conversionsystems (e.g. internal combustion engines,

gas/steam turbines, boilers) consist of three parts:

1. a combustion process generating heat andkinetic energy (K.E.)

2. a device for converting K.E. to mechanical

(useful) energy

3. heat exchangers to recuperate the heat

either for heating purposes or to increase

efficiency.


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The different applications of heat exchangersrequire different designs (geometries):


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Heat Exchangers are classified according totheir function and geometry:

Function:

1. Recuperative: two fluids separated by a solidwall (this is the most common type)

2. Evaporative: enthalpy of evaporation of one

fluid is used to heat or cool the other fluid(condensers/evaporators and boilers)

3. Regenerative: use a third material whichstores/releases heat

Geometry: 1. Double Tube

2. Shell and Tube

3. Cross-flow Heat Exchangers

4. Compact Heat Exchangers


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Underlying calculation approach

The heat transfer rate for most heat exchangers

can be calculated using the LMTD-method (LogMean Temperature Difference), if the inlet (T1) and

outlet (T2) temperatures are known:

TAUQ (!

F

T/Tln

TTT

12

12

((((

!(

U = Overall heat transfer coefficient [ W/m2-oC ]

A = Effective heat transfer surface area [ m2 ]

F = Geometry correction factor

= Log mean temperature differenceT(


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Otherwise, the Effectiveness (\) Number of

TransferUnits (NTU) method may be used:

minmax CmAU

NTUQ

Q

!!\


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General Formulation forHeat Exchanger

Analysis (LMTD-method)

Most heat exchangers are characterized relative to adouble-pipe heat exchanger (H = Hot, C = Cold):

(T2

(T1


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CiCoCHoHiH TTCmTTCmTAUQ !!(!

We now want to derive the expression for LMTD

for a counter-flow double-pipe heat exchanger.

This will be done by considering the first law (forcounter flow):

First globally:

Then locally: Apply the first law between points 1and 2 (for counter-flow)

Heat lost by hot side = Heat gained by cold side


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For counter-flow:

By using the notation 1 and 2, as shown on thegraphs, this definition is valid for both Counter-

current and Co-flow (parallel) double-pipe heat

exchangers.

!

!

CoHi

CiHo

CoHiCiHo

CoHi

CiHo

CiCoHiHo

TT

TTln

TTTTAU

TT

TTln

TTTTAUQ

((

((!

1

2

12

T

Tln

TTAUQ


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-NTU (Effectiveness Number of Transfer

Units) Method

maxQ

Q

transferheat.maxlTheoretica

transferheatActual!!

\

If the inlet or outlet temperatures are not given,the LMTD-method becomes cumbersome to use.

It is thus advisable to use the Effectiveness-NTU

method. The method can be formulated from

the following definitions:

Effectiveness:

min

Cm

AUNTU

! Minimum thermal capacity

p max. temp. difference


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In general:

Actual heat transfer is given by

Theoretical maximum heat transfer by:

Hence, we obtain the effectiveness as:

CiCoCHoHiH TTCmTTCmQ !!

CiHimin TTCmQ !

CiHimin

CiCoC

CiHimin

HoHiH

TTCm

TTCm

TTCm

TTCm

!

!\


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For a counter-flow heat exchanger:

Let

and

Which, on using the definition for LMTD, leads to

an expression for the effectiveness as:

minH CmCm !

CH

max

min

Cm

Cm

Cm

CmR

!!

? A

? AR1NTU

R1NTU

eR1

e1

!\


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minC CmCm !

HC

max

min

Cm

Cm

Cm

CmR

!!

? A

? AR1NTU

R1NTU

eR1

e1

!\

If, instead

then

We end up with the same effectiveness:


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Counter flow Parallel flow

H

C

CmCm " H

C

CmCm


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Similar expressions are used for other types of

geometry.

For example, for a parallel double-pipe heat

exchanger, the effectiveness is:

? A

R1

e1R1NTU

!\

Next we shall look at some applications of these

concepts.


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Typical thermal design problems

Problem #1

Given the entrance temperature of the two

streams, given one exit temperature;

Find heat transfer area, A.

Problem #2

Given entrance temperature of the two

streams, given the heat transfer area, A;

Find the exit temperatures of the two

streams.


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Objective: Calculation procedure and advantages

/ disadvantages of:

Double pipeShell and tube

Cross flow heat exchangers

1. Double PipeH

eat Exchangers:


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Double Pipe Heat Exchangers


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Arrangements:


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Advantages:

- low pressure loss

- small applications (simple, cheap to build)

- counter flow: high effectiveness; parallel flow:

quick (short) fetches.

Disadvantage:

- requires large surface area (footprint on floor) if

large heat transfer rates are needed.


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2. Shell-and-Tube Heat Exchangers:


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Advantages:

- ideal for large scale applications

- commonly used in petrochemical industry where

dangerous substances are present (protective

shell)

- compact design or double tube heat exchanger.

Disadvantages:

- very bulky (heavy construction), baffles are used

to increase mixing

- subject to water hammer and corrosion (behind

baffles)

- high pressure loses (recirculation behind baffles)


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Heat transfer calculations:

Using counter flow, double pipe heat exchanger

definition for the temperatures

TAUQ (!

F

T/Tln

TTT

12

12

((

((!(

CoHi1CiHo2 TTTTTT !(!(


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Heat exchanger correction factor plot for one

shell pass and an even number of tube passes

= +


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Heat exchanger correction factor plot for two shell

passes and twice an even number of tube passes


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For n-shell passes with an even number of tubes:

Again, for boiling or evaporation R p 0

so that \ = 1 e-NTU


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Cross flow and compact heat exchangers

Overview: Cross-flow and compact heatexchangers are used where space is limited. These

aim to maximize the heat transfer surface area.

Cross-flow Heat Exchangers:Commonly used in gas (air) heating applications.

The heat transfer is influenced by whether the

fluids are unmixed (i.e. confined in a channel) or

mixed (i.e. not confined, hence free to contactseveral different heat transfer surfaces).

e.g.: both fluids unmixed: air-conditioning devices

e.g.: both fluids mixed: boilers


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In a cross-flow heat exchangerthe direction of fluids are

perpendicular to each other. The required surface area,

Across for this heat exchanger is usually calculated by

using tables. It is between the required surface area for

counter-flow (Acounter) and parallel-flow (Aparallel) i.e.

Acounter< Across


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Both fluids unmixed

Both fluids unmixed

One fluid unmixed


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TAUQ (!

F

T/TlnTTT

12

12

(( ((!(

Cross-flow heat exchangers have the same

analysis equations as before:

with F as the correction factor (see graphs). The\-NTU method may also be used


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Heat exchanger correction factor plot for single

pass, cross-flow with one fluid mixed


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Heat exchanger correction factor plot for single

pass, cross-flow with both fluids unmixed


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Compact heat exchangers: These are cross-flow

heat exchangers characterized by very large heat

transfer area per unit volume. In fact, the contact

area is so large that much of the flow behaves as

duct or channel flow.

For this reason, the heat-transfer is dominated by

wall effects and the characteristics cannot be

evaluated as for the other types.

For these heat exchangers, the heat transfer rate is

directly related to pressure loss.


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Advantages:

- very small

- ideal for transferring heat to/from fluids with

very low conductivity or where the

heat transfer must be done in very small

spaces (e.g. electronic component cooling,

cryogenic cooling, domestic furnaces).

Disadvantages:

- high manufacturing costs

- very heavy

- extremely high pressure losses.


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Examples of compact heat exchangers


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To solve problems involving design and selection

(sizing) of compact heat exchangers it is first

required to find the effective pressure (static)loss. This loss can be shown, based on

fundamental heat transfer principles, to be

directly related to the heat transfer rate based on

Colburns analogy:

f friction factor, St Stanton number,

Pr Prandtl number and jH = Colburn factor

32

HPrSt

8

fj !!


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These calculations can be quite involved and so

most design or sizing applications use data in

tables and graphs.

All material properties are calculated at the bulk

average temperature, i.e. at (T1+T2)/2, if T1 = inlet,

T2 = exit

CpPrnumberandtlPr

V

Q!

ER

!

2

max

H

U

Ddx

dP

ffactorFrictionV

!


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Q! GDRe H

m

Reynolds number at the smallest diameter:

DH = hydraulic diameter at smallest cross-section= 4 Ac/ P

Ac = smallest cross-sectional area

P = perimeter (circumference) of tube

= dynamic viscosityE = thermal diffusivityG = maximum mass flow rate flux

= mass flow rate

cAmG

!


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A

Ac!WCpG

hSt !

W = ratio of open area to total frontal area (A)h = heat transfer coefficient

Cp = specific heat capacity

-

W!( 1m

c1

222

1

V

V

A

A

f1V

V

12

GV

p

(p = pressure loss through heat exchangerVm = (V2 + V1) / 2


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Overall heat transfer coefficient UA is computed

from:

hc

Ah1

Ah1

AU1 !

(h A)h = hot fluid(h A)c = cold fluid

A = effective heat transfer area

Then the heat transferQ is:

TAUQ (!


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Heat transfer and friction factor for a finned flat tube

heat exchanger


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Heat transfer and friction factor for a finned circulator-tube

heat exchanger (details on next slide)


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Summary


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Summary of effectiveness equations

Heat exchanger Effectiveness:

type:


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type:

= +


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type:


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Example questions


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Example 1 Finned flat tube heat exchanger

Air at 1 atm and 300 K enters a finned flat tubeheat exchanger (as in graph in an earlier slide)

with a velocity of 15 m/s. Calculate the heat

transfer coefficient (h).

Note at this temperature the air properties

(found from tables) are:

V = 1.1774 kg/m3

Q = 1.983 x 10-5 kg/ms

Cp = 1.0057 kJ/KgoC

Pr = 0.708


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Example 2 Shell and tube heat exchanger

Hot oil at 100oC is used to heat air in a shell and

tube heat exchanger. The oil makes 6 tube passes

and the air makes 1 shell pass. 2.0 kg/s of air

(specific heat of 1009 J/kgoC) is to be heated from20 to 80oC. The specific heat of the oil is 2100

J/kgoC and its flow rate is 3.0 kg/s. Calculate the

area required for the heat exchanger forU = 200

W/m2oC.


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Example 3 Finned-tube (both fluids unmixed)

cross-flow heat exchanger

A finned-tube exchanger is used to heat 2.36 m3

/sof air (specific heat of 1006 J/kgoC) at 1 atm from

15.55 to 29.44oC. Hot water enters the tubes at

82.22oC and the air flows across the tubes,

producing an average overall heat transfercoefficient of 227 W/m2oC. The total surface area of

the exchanger is 9.29m2. Calculate the heat transfer

rate (kW) and the exit water temperature.

Note: We dont know whether the air or the water is

the minimum thermal capacity fluid. So try with the

air as the minimum fluid first and see if the \-NTU

equations give a possible solution. If not then we

have to use water as the minimum and iterate to a

solution

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Heat Exchange Models