HEAT EXCHANGERS
Heat exchangers transfer heat from one working fluid to another. For instance, steam generators, feedwater heaters,
reheaters and condensers are all examples of heat exchangers found in nuclear power systems. The heat transfer rate
across a heat exchanger is usually expressed in the form
Q& = UAΔTm
where:
Q&
U
A
ΔTm
=
=
=
=
(1)
heat transfer rate
overall heat transfer coefficient
heat exchanger area
average temperature difference between the fluids
The overall heat transfer coefficient is a function of the flow geometry, fluid properties and material composition of
the heat exchanger. The average temperature difference between the fluids is in general a function of the fluid
properties and flow geometry as well. Heat exchanger design requires consideration of each of these factors.
Overall Heat Transfer Coefficient
The overall heat transfer coefficient represents the total resistance to heat transfer from one fluid to another. The
functional form of U or the product UA, may be derived for any particular geometry by performing a standard
conduction analysis on the system of interest. To illustrate this, consider first a planar wall of thickness L, subject to
convection on both sides.
TH
TC
T1
z+ Δ z
T2
z
L
Figure 1: Planar Wall Heat Exchanger
The heat transfer rate from the hot fluid to the wall for a surface area defined by the length segment Δz is given by
Newton’s Law of Cooling as
q = hc1 As [TH − T1 ]
(2)
such that
TH − T1 =
q
hc1 As
(3)
114
The temperature drop across the wall is found by solution of the conduction equation
d ⎛ dT ⎞
⎜k
⎟=0
dx ⎝ dx ⎠
(4)
subject to the boundary condition
−k
dT
dx
=
0
q
As
(5)
Integrating Equation 4,
x′
∫0
k
d ⎛ dT ⎞
⎜k
⎟dx = 0
dx ⎝ dx ⎠
dT
dT
−k
dx′
dx
=0
(6)
(7)
0
and applying the boundary condition gives
k
dT
q
+
=0
dx′ As
(8)
Integrate Equation 8 over the wall thickness to obtain the temperature drop across the wall
L
L
∫ 0 dx′dx′ + ∫ 0 kAs dx′ = 0
dT
T1 − T2 =
q
qL
kAs
The temperature drop from the wall to the cold fluid is also found from Newton’s Law of Cooling, similar to
Equation 3 above
q
T2 − TC =
hc 2 As
(9)
(10)
(11)
Adding Equations 3, 10, and 11 gives the temperature drop from the hot to cold fluid as
TH − TC =
1 ⎤
q ⎡ 1
L
+ +
⎢
⎥
As ⎣ hc1 k hc 2 ⎦
(12)
or
q = UAs [TH − TC ]
(13)
where
115
⎡ 1
1 ⎤
L
U ≡⎢
+ +
⎥
⎣ hc1 k hc 2 ⎦
−1
(14)
is the Overall Heat Transfer Coefficient. In actual heat exchanger design, the planar wall is seldom used. A more
common design involves heat transfer across a tube wall as illustrated in Figure 2.
TH
TC
T1
z+ Δ z
T2
z
ri
ro
r=0
Figure 2. Circular Tube Heat Exchanger
In this case, the heat transfer rate from the hot fluid to the wall within the length segment Δz is
q = hc1 2πri Δz[TH − T1 ]
(15)
such that
TH − T1 =
q
hc1 2πri Δz
(16)
For the tube wall, the conduction equation is
1 d ⎛ dT ⎞
⎜ rk
⎟=0
r dr ⎝ dr ⎠
(17)
subject to the boundary condition
− 2πrkΔz
dT
dx
=q
(18)
ri
Integrating Equation 17,
∫
r′
ri
d ⎛ dT ⎞
⎜ rk
⎟dr = 0
dr ⎝ dr ⎠
(19)
116
r ′k
dT
dT
− kr
′
dr
dr
=0
(20)
ri
and applying the boundary condition gives
r ′k
q
dT
+
=0
dr ′ 2πΔz
(21)
Integrate Equation 21 over the wall thickness to obtain the temperature drop across the wall
∫
ro
ri
dT
dr ′ +
dr ′
∫
T1 − T2 =
ro
ri
q dr ′
=0
2πkΔz r ′
⎛r
q
ln⎜ o
2πkΔz ⎜⎝ ri
(22)
⎞
⎟
⎟
⎠
(23)
The temperature drop from the wall to the cold fluid is
T2 − TC =
q
hc 2 2πro Δz
(24)
Adding Equations 16, 23, and 24 gives the temperature drop from the hot to cold fluid as
TH − TC =
q
Δz
⎡ 1
⎛r
1
+
ln⎜⎜ o
⎢
2
π
r
h
2
π
k
⎝ ri
⎣⎢ i c1
⎞
1 ⎤
⎟⎟ +
⎥
⎠ 2πro hc 2 ⎦⎥
(25)
or
q = UA[TH − TC ]
(26)
where
⎡
⎛r
1
1
+
UA ≡ ⎢
ln⎜⎜ o
⎢⎣ 2πri Δzhc1 2πkΔz ⎝ ri
⎤
⎞
1
⎟⎟ +
⎥
⎠ 2πro Δzhc 2 ⎥⎦
−1
(27)
Since the heat transfer area on the interior of the tubes is different from that on the exterior in cylindrical geometry,
the product UA is normally used to describe heat exchanger performance. Equations 13 and 26 represent the heat
transfer across a small length segment Δz where the hot and cold fluid temperatures can be considered constant. In
reality, the hot and cold fluid temperatures change continuously along the length of the heat exchanger as illustrated
in Figure 3 below for the Double Pipe Heat Exchanger given in Figure 4.
117
Th
1
Th
q
1
q
Th2
Tc
Th2
1
Tc2
dA
dA
Tc
Tc2
1
A
1
2
1
A
2
Counter Flow
Parallel Flow
Figure 3: Fluid Temperature Profiles in a Double Pipe Heat Exchanger
If we let Δz be a differential length, then the heat transfer across the differential length segment is
q = dQ& = UdA(TH − TC )
(28)
where dA is the differential area associated with length dz. The total heat transferred across the heat exchanger is
then obtained by integrating Equation 28 over the length of the heat exchanger
∫
∫
2
2
dQ& = UdA(TH − TC )
1
(29)
1
Q& = UAΔTm
(30)
∫ (T
(31)
where
ΔTm ≡
1
A
2
H
− TC )dA
1
is the average temperature difference between the hot and cold fluids over the length of the heat exchanger. Since
the area in Equation 30 is now the total area of the heat exchanger instead of that associated with a small length
segment of a single tube, the expression given for UA in Equation 27 is modified to reflect the actual area of the heat
exchanger
UA =
n
⎛r
1
1
ln⎜⎜ o
+
hi Ai 2πkL ⎝ ri
⎞
1
⎟⎟ +
h
o Ao
⎠
(32)
where:
n = number of tubes
118
hi
Ai
ho
Ao
ri
ro
k
L
=
=
=
=
=
=
=
=
convective heat transfer coefficient on the tube interior
interior surface area of one tube
convective heat transfer coefficient on the tube exterior
exterior surface area of one tube
tube inner radius
tube outer radius
tube thermal conductivity
tube length
The value of ΔTm depends on the heat exchanger design, with analytic expressions only available in special cases.
One such case is that of the simple double pipe heat exchanger illustrated below.
Figure 4: Double Pipe Heat Exchanger
Log Mean Temperature Difference
We wish to calculate the total heat transfer in the double pipe heat exchanger through an expression of the type
Q& = UAΔTm
where ΔTm is some appropriate average temperature difference between the hot and cold fluids. If we assume
parallel flow, the steady state heat transferred through a differential area dA is
(
)
(
)
dQ& = − m& C p h dTh = m& C p c dTc
(33)
where the subscripts “h” and “c” denote the hot and cold fluids respectively. We have already seen that we can
write the heat transfer across this differential area in terms of the Overall Heat Transfer Coefficient as
dQ& = UdA(Th − Tc )
From Equation 33, dTh = −
dQ&
m& C p
(
)h
and dTc =
dQ&
m& C p
(
)c
(34)
such that
⎧⎪ 1
dTh − dTc = d (Th − Tc ) = −dQ& ⎨
⎪⎩ m& C p
(
+
1
m& C p
)h (
⎫⎪
⎬
c⎪
⎭
)
(35)
119
Substitute for dQ& from Equation 34
d (Th − Tc )
⎪⎧ 1
= −UdA⎨
(Th − Tc )
⎪⎩ m& C p
(
+
1 ⎪⎫
⎬
m& C p c ⎪
⎭
)h (
)
(36)
Assuming all terms on the right hand side of Equation 36 are constant, we can integrate from point (1) to point (2)
along the length of the heat exchanger
∫
2
1
d (Th − Tc )
=−
(Th − Tc )
∫
2
⎧⎪ 1
UdA⎨
⎪⎩ m& C p
(
1
⎧⎪ 1
2
ln (Th − Tc )1 = −UA⎨
⎪⎩ m& C p
(
1
&
mC p
+
)h (
+
1
m& C p
)h (
⎧
⎛ (T − Tc )2 ⎞
⎟ = −UA⎪⎨ 1
ln⎜⎜ h
⎟
⎪⎩ m& C p
⎝ (Th − Tc )1 ⎠
+
⎫⎪
⎬
c⎪
⎭
)
(37)
⎫⎪
⎬
c⎪
⎭
(38)
)
1 ⎪⎫
⎬
m& C p c ⎪
⎭
(
)h (
)
− Th2 = m& C p
) (
)c (Tc
Th1 − Th2
1
and
m& C p
Q&
=
(39)
We further take advantage of the fact
(
Q& = m& C p
)h (Th
1
2
− Tc1
)
(40)
such that
Tc2 − Tc1
Q&
(41)
⎛ (T − Tc )2 ⎞
⎟ = − UA Th − Th + Tc − Tc
ln⎜⎜ h
1
2
2
1
⎟
Q&
⎝ (Th − Tc )1 ⎠
(42)
1
m& C p
(
)h
=
(
)c
Substituting into Equation 39
[(
) (
) (
)(
)
)]
)]
and rearranging gives
(
Th2 − Tc 2 − Th1 − Tc1
Q& = UA
= UAΔTm
ln Th2 − Tc 2 / Th1 − Tc1
[(
(43)
where
ΔTm ≡
(Th − Tc ) − (Th − Tc )
ln[(Th − Tc ) / (Th − Tc )]
2
2
2
1
2
1
1
(44)
1
is called the Log Mean Temperature Difference (LMTD). The LMTD represents the effective, average temperature
difference between the two heat transfer fluids over the length of the heat exchanger and though derived here for
parallel flow, Equation 44 is also valid for counter flow heat exchangers. The LMTD given in Equation (44) is only
strictly valid for single phase fluids in double pipe heat exchangers. In a heat exchanger other than a double-pipe
120
type, the heat transfer is calculated by using a correction factor applied to the LMTD for a counterflow double-pipe
heat exchanger with the same hot and cold fluid temperatures. Equation (1) is then modified such that
Q& = UAFΔTm
(45)
Correction factors for a single shell pass, and multiple tube passes are provided in Figure 5 below (from Holman,
Heat Transfer 3rd Edition).
Figure 5: Correction Factors for a single shell pass, and even number of tube passes
(Holman, Heat Transfer 3rd Ed.)
Steam Generators and Condensers
In steam generators and condensers, the shell side temperature can be assumed constant at the saturation
temperature. This simplifies the derivation of the LMTD as follows. Assume the constant temperature is that of the
cold fluid.
dQ& = UdA(Th − Tsat )
and dTh = −
dQ&
m& C p
(
)h
(46)
such that
⎧⎪ 1
dTh = d (Th − Tsat ) = − dQ& ⎨
⎪⎩ m& C p
(
⎫⎪
⎬
h⎪
⎭
)
(47)
Substitute for dQ& from Equation 46
121
⎧⎪ 1
d (Th − Tsat )
= −UdA⎨
(Th − Tsat )
⎪⎩ m& C p
(
⎫⎪
⎬
h⎪
⎭
)
(48)
Assuming all terms on the right hand side of Equation 48 are constant, we can integrate from point (1) to point (2)
along the length of the heat exchanger
∫
2
1
d (Th − Tsat )
=−
(Th − Tsat )
∫
2
⎧⎪ 1
UdA⎨
⎪⎩ m& C p
(
1
⎧
⎛ (T − Tsat )2 ⎞
⎟ = −UA⎪⎨ 1
ln⎜⎜ h
⎟
⎪⎩ m& C p
⎝ (Th − Tsat )1 ⎠
(
⎫⎪
⎬
h⎪
⎭
)
(49)
⎫⎪
⎬
h⎪
⎭
(50)
)
Recall
(
Q& = m& C p
)h (Th
1
− Th2
)
(51)
such that
⎛ (T − Tsat )2 ⎞
⎟ = − UA Th − Th
ln⎜⎜ h
1
2
⎟
T
T
(
)
−
Q&
sat 1 ⎠
⎝ h
[(
)]
(52)
which may be rearranged to give
Q& = UA
(Th
[(
2
− Th1
)(
)
ln Th2 − Tsat / Th1 − Tsat
)] = UAΔTm
(53)
where
ΔTm ≡
(Th
(Th − Tsat ) − (Th − Tsat )
ln[(Th − Tsat ) / (Th − Tsat )] ln[(Th − Tsat ) / (Th − Tsat )]
2
2
− Th1
)
=
1
2
1
2
(54)
1
Note, that this is identical to Equation 44 with
Tc1 = Tc 2 = Tsat .
A similar expression can be developed for condensers with the hot fluid temperature replaced by the saturation
temperature. In Once Through Steam Generator designs, the secondary water usually enters subcooled, boils, and in
some cases leaves superheated. For more precise heat transfer area calculations, an overall heat transfer coefficient
and LMTD should be determined for each region. U-tube steam generator designs actually contain one shell side
pass and two tube-side passes. This would normally require the use of Equation 45 with an appropriate value of the
correction factor F. Since the shell side temperature is constant however, the F factor becomes 1.0 and the LMTD
can be used without correction. This also holds for condensers. U-tube designs incorporating integral preheaters are
more complicated and require multiple LMTDs. The overall heat transfer coefficient can be calculated from
Equation 32 with the tube side heat transfer coefficient calculated from the Dittus-Boelter or some other appropriate
single-phase correlation. The shell side boiling heat transfer coefficient can be computed from the Jens-Lottes
correlation. It should be noted, that in this derivation it was assumed that the UA value was constant. We have seen
in our discussion of boiling heat transfer that the boiling heat transfer coefficient is a strong function of the wall
122
temperature. The same is true for condensation. To accurately predict the performance of heat exchangers subject
to either boiling or condensation generally requires Equation 49 to be integrated numerically.
Fouling Factors
After a period of time, the heat transfer surfaces of a heat exchanger may become coated with deposits from the
heat transfer fluids or corrosion. In either case, the additional resistance to heat transfer due to these materials
decreases the performance of the heat exchanger and must be accounted for. This is done through the use of
experimentally determined fouling factors. These fouling factors are then used to modify the overall heat transfer
coefficient according to
R f = 1/U ′ −1/U
where:
R f = fouling factor
U ′ = overall heat transfer coefficient for the "dirty" heat exchanger
U = overall heat transfer coefficient for the clean heat exchanger
Recommended values for fouling factors are given in Table 1 for some common fluids.
Type of Fluid
Sea water below 125 F
Sea water above 125 F
Treated boiler feedwater above 125 F
Fuel oil
Quenching oil
Alcohol vapors
Steam, non-oil bearing
Industrial air
Refrigerating liquid
Fouling Factor
(hr-ft2-F/Btu)
0.0005
0.001
0.001
0.005
0.004
0.0005
0.0005
0.002
0.001
Table 1: Fouling Factor for Common Fluids (Holman, Heat Transfer 3rd Ed.)
123
Example:
Estimate the tube bundle length required for a U-Tube steam generator having the following characteristics.
61 x 106
595
540
830
8519
0.75
0.048
10
Primary Coolant Flow Rate
Tube Side Inlet Temperature
Tube Side Outlet Temperature
Secondary (shell) Side Pressure
Number of Tubes
Tube Outer Diameter
Tube Wall Thickness
Tube Thermal Conductivity
lbm/hr
F
F
psia
inches
inches
Btu/hr-ft-F
SOLUTION
The total heat transfer rate through the steam generator is given by
Q& = m& C p ΔT = UAΔTm
such that
UA =
Q&
ΔTm
In addition, we have
UA =
UA =
n
⎛r
1
1
+
ln⎜⎜ o
hi Ai 2πKL ⎝ ri
⎞
1
⎟⎟ +
h
o Ao
⎠
n
⎛r
1
1
+
ln⎜ o
hi 2πri L 2πKL ⎜⎝ ri
UA =
2πnL
1
1 ⎛r
+ ln⎜ o
hi ri K ⎜⎝ ri
⎞
1
⎟⎟ +
h
2
o πro L
⎠
⎞
1
⎟⎟ +
h
o ro
⎠
from which L can be determined.
Fluid Properties
The primary side fluid properties are taken at the average fluid temperature and a pressure of 2000 psia.
Tave =
595 + 540
≅ 570 F
2
C p = 1.31 Btu/lbm-F
k
μ
= 0.319 Btu/hr-ft-F
= 0.2265 lbm/hr-ft
Log Mean Temperature Difference
At a secondary side pressure of 830 psia, the saturation temperature is 522.4 F.
124
ΔTm =
(Th − Tsat )− (Th − Tsat )
ln[(Th − Tsat )/ (Th − Tsat )]
2
1
2
ΔTm =
ΔTm =
[(
1
(Th
2
− Th1
)(
)
ln Th2 − Tsat / Th1 − Tsat
)]
(540 − 595)
= 38.3F
ln[(540 − 522.4) / (595 − 522.4)]
Convective Heat Transfer Coefficient
The tube side convective heat transfer coefficient can be obtained from the Dittus-Boelter correlation
⎛G D
hi Di
= 0.023⎜⎜ i i
k
⎝ μ
Gi =
⎞
⎟⎟
⎠
0.8
⎛ Cpμ ⎞
⎜
⎟
⎜ k ⎟
⎝
⎠
0.3
61 × 106
m&
m&
=
=
= 3.069 × 10 6 lbm hr - ft 2
Ax nπ Di2 4 (8519)(π )(0.654 12) 2 / 4
(0.319) ⎛⎜ (3.069 × 106 )(0.654 12) ⎞⎟
hi = 0.023
⎟
(0.654 12) ⎜⎝
0.2265
⎠
0.8
⎛ (1.31)(0.2265) ⎞
⎜
⎟
0.319
⎝
⎠
0.3
= 6506Btu hr - ft 2 - F
Boiling Heat Transfer Coefficient
The boiling heat transfer coefficient is computed using the Jens-Lottes correlation
q′′ = 10 6
exp(4 P / 900)
(Tw − Tsat )4
60 4
ho = 10 6
exp(4 P / 900)
(Tw − Tsat )3 .
4
60
such that
The calculation of the boiling heat transfer coefficient is complicated by the nonlinearity in tube wall temperature
which is unknown. We can estimate the boiling heat transfer coefficient by making some reasonable estimate of the
wall temperature however, and for this example assume
Tw = Tsat + (0.5)ΔTm ≅ 542F
giving for the boiling heat transfer coefficient
ho ≅ 10 6
exp(4 × 830 / 900)
(542 − 522)3 = 24,691Btu hr - ft 2 - F
60 4
Tube Bundle Length
The UA value is found from the total heat transfer rate and the LMTD
125
UA =
UA =
m& C p ΔT
Q&
=
ΔTm
ΔTm
(61 × 10 6 )(1.31)(595 − 540)
= 113.14 × 10 6 Btu hr - F .
38.8
The bundle length is then obtained from the UA value through the relationship.
UA =
L=
2πnL
1
1 ⎛r
+ ln⎜⎜ o
hi ri K ⎝ ri
113.14 × 10 6
2π (8519)
⎞
1
⎟⎟ +
⎠ ho ro
⇒L=
UA ⎛⎜ 1
1 ⎛r ⎞
1 ⎞⎟
+ ln⎜⎜ o ⎟⎟ +
⎜
2πn ⎝ hi ri K ⎝ ri ⎠ ho ro ⎟⎠
⎞
⎛
1
1 ⎛ 0.75 ⎞
1
⎟⎟ ≅ 44 ft
⎜⎜
+ ln⎜
⎟+
⎝ (6506)(0.654 / 24) 10 ⎝ 0.654 ⎠ (24,691)(0.75 / 24) ⎠
126
Example: (From Holman, Heat Transfer 3rd Edition)
Water with a mass flow rate of 30,000 lbm/hr is heated from 100 to 130 F in a tube and shell heat exchanger. A
single pass of water at 15,000 lbm/hr and 200 F is used on the shell side as the heating fluid. The overall heat
transfer coefficient is 250 Btu/hr-ft2-F and the average water velocity in the 0.75 inch I.D. tubes is 1.2 ft/sec.
Because of space limitations, the heat exchanger must be no longer than 8 feet. Consistent with this restriction,
calculate the number of tube passes, the number of tubes per pass, and the length of the tubes.
SOLUTION
Consider first a single tube pass. The exit temperature on the shell side is calculated from
(
)
(
)
Q& = m& C p c ΔTc = m& C p h ΔTh
ΔTh =
(m& C p )c ΔTc (30,000 × 1)(30)
(m& C p )h = (15,000 × 1) = 60F
Thexit = 200 − 60 = 140F
The LMTD is given by
ΔTm =
(Th − Tc )− (Th − Tc )
ln[(Th − Tc )/ (Th − Tc )]
2
2
2
1
2
1
1
1
which for this example (assuming a counterflow heat exchanger) gives
ΔTm =
(200 − 130) − (140 − 100) = 53.6F
ln[(200 − 130) / (140 − 100)]
The total heat transfer area is obtained from the equality
(
)
Q& = m& C p c ΔTc = UAΔTm
A=
(m& C p )c ΔTc (30,000 × 1)(30)
=
= 67.3 ft 2
UΔTm
(250)(53.6)
The number of tubes is obtained from the total tube side mass flow rate and the average velocity in the tubes through
the relationship
m& c = ρvAx = ρvnπDi2 / 4
n=
4m& c
4 × 30,000
=
= 36.5 ⇒ 37tubes
2
ρvπDi
(62)(1.2 × 3600)π (.75 / 12) 2
The total heat transfer area is A = nπDi L ⇒ L =
A
giving for the tube length ( and therefore the heat exchanger
nπDi
length) for a single pass heat exchanger
L=
67.3
= 9. 6 ft
(37 )π(. 75 / 12 )
127
This length is greater than the allowable 8 feet for the heat exchanger length, implying more than one tube pass is
required. Note, while this reduces the length of the heat exchanger it will not reduce the tube length. Examine next
two tube passes. The total heat transfer rate is now given by
Q& = UAFΔTm
The correction factor F is given in terms of the ratios
R=
T1 − T2
t 2 − t1
P=
t 2 − t1
T1 − t1
For this problem
T1
T2
t1
t2
= 200
= 140
= 100
= 130
R=
P=
200 − 140
=2
130 − 100
130 − 100
= 0.3
200 − 100
From the graph of correction factors, F = 0.88 which increases the total heat transfer area to
A=
(m& C p )c ΔTc (30,000 × 1)(30)
=
UFΔTm
(250)(.88)(53.6)
= 76.3 ft 2
The total number of tubes remains unchanged at 37. The total heat transfer area is now given by
A = nπDi 2 L ⇒ L =
A
2 nπDi
where the 2 is due to the two tube passes through the heat exchanger. The required length for the two pass heat
exchanger is then
L=
76.3
= 5. 25 ft
( 2 )(37 )π(. 75 / 12 )
which satisfies the design requirement.
Note: This problem is somewhat idealistic, as the overall heat transfer coefficient is assumed to be constant. In
reality, as you increase the number of passes, you reduce the mass flux on the shell side and correspondingly the
shell side convective heat transfer coefficient.
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