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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Heat exchanger Effectiveness:
type:
= +
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Heat exchanger Effectiveness:
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