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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