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)
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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)
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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.
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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
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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)
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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
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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
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⎧⎪ 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
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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.)
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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.
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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
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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) ⎠
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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 )
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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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