Chapter 03
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3. HEAT EXCHANGER DESIGN
3.1
Can you have a cross-flow exchanger in which both flow are mixed? Discuss.
Answer: Yes, as long as there is a mixing portion in the line of flow like the illustration below.
Find the appropriate mean radius, r , that will make Q = kA(r )∆T (r o − r i ) , valid for the
3.2
one dimensional heat conduction through a thick spherical shell, where A(r ) = 4π r 2 (cf . Example 3.1) Solution: 1 ∂ (rT ) 2
2
r ∂r 2 ∂ (rT ) 2
∂r
∂ (rT ) ∂r
=0
=0
= C 1
rT = C 1r + C 2 T = C 1 +
C 2
r At r = = r o, T = = T o At r = = r i, T = T i
Then, T o = C 1 + T i = C 1 +
C 2 r o C 2 r i
1
3. HEAT EXCHANGER DESIGN
T i > T o
1
T i − T o = ∆T = C 2
−
r i
C 2 =
r o
(T i − T o )(r i r o ) (r o − r i )
C 1 = T o − C 1 =
1
C 2 r o
= T o −
(T i − T o )r i T o r o − T o r i − T i r i + T o r i = (r o − r i ) r o − r i
T o r o − T i r i r o − r i
(T i − T o )(r i r o ) (r o − r i )r r o − r i (T − T o )(r i r o ) dT =− i (r o − r i )r 2 dr (T i − T o )(r i r o ) dT 2 = − k (4π r )− Q = −kA 2 dr (r o − r i )r k 4π r i r o (T i − T o )
T =
T o r o − T i r i
Q=
Q=
+
r o − r i k 4π r i r o ∆T r o − r i
Equating: Q=
kA(r )∆T
=
k 4π r i r o ∆T
r o − r i 2
k 4π r ∆T
=
r o − r i k 4π r i r o ∆T
r o − r i
r o − r i
r = r i r o ← Answer.
3.3
Rework Problem 2.14, using the methods of Chapter 3.
Solution: & a = 0.5 kg/s m ρ air At
20 C = 1.205 kg/m
3
c pa At 20 C = 1006 J/kg.K
& a c pa = (0.5 kg/s)(1006 J/kg.K) = 503 W/K C min = m 2
U = 300 W/m .K
Equation 3.22: 2
3. HEAT EXCHANGER DESIGN
T i > T o
1
T i − T o = ∆T = C 2
−
r i
C 2 =
r o
(T i − T o )(r i r o ) (r o − r i )
C 1 = T o − C 1 =
1
C 2 r o
= T o −
(T i − T o )r i T o r o − T o r i − T i r i + T o r i = (r o − r i ) r o − r i
T o r o − T i r i r o − r i
(T i − T o )(r i r o ) (r o − r i )r r o − r i (T − T o )(r i r o ) dT =− i (r o − r i )r 2 dr (T i − T o )(r i r o ) dT 2 = − k (4π r )− Q = −kA 2 dr (r o − r i )r k 4π r i r o (T i − T o )
T =
T o r o − T i r i
Q=
Q=
+
r o − r i k 4π r i r o ∆T r o − r i
Equating: Q=
kA(r )∆T
=
k 4π r i r o ∆T
r o − r i 2
k 4π r ∆T
=
r o − r i k 4π r i r o ∆T
r o − r i
r o − r i
r = r i r o ← Answer.
3.3
Rework Problem 2.14, using the methods of Chapter 3.
Solution: & a = 0.5 kg/s m ρ air At
20 C = 1.205 kg/m
3
c pa At 20 C = 1006 J/kg.K
& a c pa = (0.5 kg/s)(1006 J/kg.K) = 503 W/K C min = m 2
U = 300 W/m .K
Equation 3.22: 2
3. HEAT EXCHANGER DESIGN
lim lim ε = 1 − e − NTU
C max → ∞
NTU =
UA C min Q
T cout = T cin +
C c
C c = C min = 503 W/K T cin = 20 C Q = ε C min T hin − T cin T hin = 120 C 2
(a) q = U T hin − T cin = (300 W/m .K)(120 C – 20 C) = 30,000 W/m 2
(b) at x :
A = (1 m)( x) = x m V = (1 m)(1 m)( x) = x m3
At the entrance: dT ma c pa = UA(T hin − T cin ) dt dT ρ Vc pa = qA dt dT = qx ρ xc pa dt 30000 dT q =
dt
2
=
ρ c pa
(1.205)(1006)
= 24.75 C/s
x = 2 m
(c)
Q = ε C min T hin − T cin )
NTU = ε = 1 − e
UA C min
=
(300)(2) 503
=1.19284
−1.19284
= 0.69664 Q = (0.69664 )(503)(120 − 20) = 35,041 W 35041 Q = 20 + = 89.7 C T cout = T cin + C c 503 3.4
2.4 kg/s of a fluid have a specific heat of 0.81 kJ/kg.K enters a counterflow heat exchanger at 0 C and are heated to 400 C by 2 kg/s of a fluid having a specific heat of 0.96 kJ/kg.K entering the unit at 700 C. Show that to heat the cooled fluid to 500C, all other conditions remaining unchanged, would require the surface area for a heat transfer to be increased by 87.5 %. 3
3. HEAT EXCHANGER DESIGN
Solution: T cin = 0 C T cout = 400 C
C c = (2.4 kg/s)(0.81 kJ/kg.K) = 1.944 kW/K T hin = 700 C
C h = (2 kg/s)(0.96 kJ/kg.K) = 1.92 kW/K Q = C c (T cout − T cin ) = (1.944 kW/K)(400 – 0 K) = 777.6 kW
T hout = T hin −
Q C h
= 700 −
777.6 1.92
= 295 C
For counterflow heat exchanger
∆T a = T hin − T c out = 700 C – 400 C = 300 C ∆T b = T hout − T cin = 295 C – 0 C = 295 C
LMTD =
∆T a − ∆T b
∆T a T ∆ b
=
ln A =
300 − 295 = 297.5 C 300 ln 295
Q
U ( LMTD ) Assume U is constant or unchanged at any new conditions.
AU =
Q
=
LMTD
777.6 297.5
= 2.6138 kW/K
if T cout = 500 C Q = C c (T cout − T cin ) = (1.944 kW/K)(500 – 0 K) = 972 kW T hout = T hin −
Q C h
= 700 −
992 1.92
= 193.75 C
∆T a = T hin − T c out = 700 C – 500 C = 200 C ∆T b = T hout − T cin = 193.75 C – 0 C = 193.75 C
4
3. HEAT EXCHANGER DESIGN
LMTD =
∆T a − ∆T b
∆T a T ∆ b
=
ln AU =
Q
972
=
LMTD
196.86
Percentage Increase = 3.5
200 − 193.75 = 196.86 C 200 ln 193.75 = 4.9375 kW/K
4.9375 − 2.6138 2.6138
(100%) = 88.9 %
A cross-flow heat exchanger with both fluids unmixed is used to heat water ( c p = 4.18 kJ/kg.K) from 40 C to 80 C, flowing at the rate of 1.0 kg/s. What is the overall heat transfer coefficient if the hot engine oil ( c p = 1.9 kJ/kg.K), flowing at the rate of 2.6 kg/s, 2 enters at 100 C? The heat transfer area is 20 m . (Note that you can use either an effectiveness or an LMTD method. It would be wise to use both as check.)
Solution: 2
Cross-flow heat exchanger: A = 20 m . & w cw = (1.0 kg/s)(4.18 kJ/kg.K) = 4.18 kW/K Water: C c = m T cin = 40 C T cout = 80 C
& o co = (2.6 kg/s)(1.9 kJ/kg.K) = 4.94 kW/K Hot Engine Oil: C h = m T hin = 100 C
Heat Balance: Q = C c (T cout − T cin ) = C h (T hin − T hout )
(4.18)(80 − 40) = (4.94)(100 − T hout )
T hout = 66.15 C
Use Effectiveness Method: Eq. (3.16a) C c T cout − T cin ε =
(4.18)(80 − 40) = 0.667 C min (T hin − T cin ) (4.18)(100 − 40) =
For cross-flow heat exchanger, both unmixed flow. Fig. 3.17, ε = 0.667 C min C 4.18 = 0.85 = c = C max C h 4.94 NTU =
UA C min
= 2.3 5
3. HEAT EXCHANGER DESIGN U (20 )
= 2.3 4.18 2 U = 0.48 W/m .K
Use LMTD Method: LMTD ≡
∆T a − ∆T b
∆T a ∆ T b ∆T a = 66.15 C – 40 C = 26.15 C ln
∆T b = 100 C – 80 C = 20 C
LMTD ≡
26.15 − 20 = 22.94 C 26 . 15 ln 20
Q = UAF ( LMTD )
Fig. 3.14 x, cross-flow heat exchanger, both passes unmixed. P= R =
T t out −T t in T sin − T t in T sin −T sout T t out − T t in
use, T sin =100 C T sout = 66.15 C T t out = 80 C T t in = 40 C R = P=
100 − 66.15 80 − 40 80 − 40
= 0.85
= 0.667
100 − 40 Then F = 0.81 Q = C c T cout − T cin ) = UAF ( LMTD )
(4.18 )(80 − 40 ) = U (20 )(0.81)(22.94 ) 2
U = 0.45 W/m .K
3.6
Saturated non-oil-bearing steam at 1 atm enters the shell pass of a two-tube-pass shell condenser with thirty 20 ft tubes in each tube pass. They are made of schedule 160, ¾ in. 3 steel pipe (nominal diameter). A volume flow rate of 0.01 ft /s of water entering at 60 F 2 enters each tube. The condensing heat transfer coefficient is 2000 Btu/h.ft .F, and we
6
3. HEAT EXCHANGER DESIGN 2
calculate h = 1380 Btu/h.ft .F for the water in the tubes. Estimate the exit temperature of the water and the mass rate of condensate. Solution: Properties of water at 60 F (15.56 C). 3 3 ρ w = 999 kg/m = 62.37 lbm/ft . c pw = 4.1887 kJ/kg.K = 1.000 Btu/lbm.F
For 30 tubes, & w = (0.01 ft3 /s)(62.37 lbm/ft3)(30) = 18.711 lbm/s = 67,360 lbm/hr m T cin = 60 F
Properties of steam at 1 atm (14.696 psia), from steam table of other references. h fg = 1150.4 Btu/lbm – 180.2 Btu/lbm = 970.2 Btu/lbm T hin = T hout = 212 F Properties of Sch. 160 Steel Pipe, ¾ in Nom. Diameter, from other references.
1 Btu / h. ft .F = 24.85 Btu/h.ft.F 1 . 7307 W / m . K
k s = 43 W/m.K
r i = (1/2)(0.612 in) = 0.306 in = 0.0255 ft r o = (1/2)(1.050 in) = 0.525 in = 0.04375 ft
Based on outside area: 1
=
U o
r o r i + 1
r o ln
r o
+
r i hi
k s
ho
hi = 1380 Btu/h.ft.F, ho = 2000 Btu/h.ft.F
1 U o
=
0.04375
(0.0255 )(1380 )
0.04375 0.0255 + 1
0.04375 ln +
24.85
2000
2
U o = 371 Btu/h.ft .F NTU =
U o Ao C min
Ao = 2π r o L L = 30 (20 ft)(2) = 1200 ft 2 Ao = 2π (0.04375 )(1200 ) = 330 ft .
& w c pw = (67,360 lbm/h)(1.000 Btu/lbm.F) = 67,360 Btu/h.F C min = m 7
3. HEAT EXCHANGER DESIGN
NTU = ε =
U o Ao
=
(371)(330)
C min
67,360
= 1.818
lim ε = 1 − e − NTU = 1 − e −1.818 = 0.838
C max →∞
Q = ε C min T hin − T cin = (0.838)(67,360)(212 – 60) = 8,580,047 Btu/h
Exit Temperature of Water & wc pw T cout − T cin Q = C c T cout − T cin = m 8,580,047 = (67,360) T cout − 60 T cout = 187.4 F
Mass flow rate of Condensate: & c h fg Q=m
& c (970.2 ) 8,580,047 = m & c = 8,844 lbm/h m 3.7
Consider a counterflow heat exchanger that must cool 3000 kg/h of mercury from 150 F 2 to 128 F. The coolant is 100 kg/h of water, supplied at 70 F. If U is 300 W/m .K, complete the design by determining reasonable value for the area and the exit-water temperature.
Solution: Mercury T hin = 150 F = 65.56 C T hout = 128 F = 53.33 C
& m = 3000 kg/h = 0.833333 kg/s m c pm @ 60 C, Table A.3, pg.707 = 138.244 J/kg.K
Water T cin = 70 F = 21.11 C
& w = 100 kg/h = 0.027778 kg/s m c pw @ 21.11 = 4187.3 J/kg.K 2
U = 300 W/m .K C c = mwc pw = (0.027778)(4183.7) = 116.2 W/K C h = mm c pm = (0.833333)(138.244) = 115.2 W/K
Effectiveness Method 8
3. HEAT EXCHANGER DESIGN
For counterflow heat exchanger:
1 − exp− 1 −
ε =
C min NTU C max
C C 1 − min exp − 1 − min NTU C max C max C min = 115.2 W/K C max = 116.2 W/K
Solve for ε . Q = ε C min T hin − T cin = C h T hin − T hout ε =
ε =
C h T hin − T hout
C min (T hin − T cin )
(115.2)(65.56 − 53.33) = 0.275 (115.2)(65.56 − 21.11)
Then:
115.2 NTU 116.2 ε = 0.275 = 115.2 115.2 1− exp − 1 − NTU 116.2 116.2 1 − exp− 1 −
0.275(1 − 0.9914e −0.008606 NTU ) = 1 − e −0.008606 NTU e −0.008606 NTU = 0.996749 NTU = 0.37843 NTU =
UA
=
(300) A 115.2
C min
= 0.37843
2
A = 0.1453 m . T cout = T cin +
Q C c
T cout = 21.11 +
3.8
= T cin +
C h T hin − T hout C c
115.2(65.56 − 53.33) 116.2
= 33.24 C = 91.83 F
An automobile air-conditioner gives up 18 kW at 65 km/h if the outside temperature is 35 C. The refrigerant temperature is constant at 65 C under these conditions, and the air rises 6 C in temperature as it flows across the heat exchanger tubes. The heat exchanger is of 2 0.7 the finned-tube type shown in Fig. 3.6b, with U ≅ 200 W/m .K. If U ~ (air velocity) and the mass flow rate increases directly with the velocity, plot the percentage reduction of heat transfer in the condenser as a function of air velocity between 15 and 65 km/h.
9
3. HEAT EXCHANGER DESIGN
Solution: Let C 1 and C 2 are constant,
V = air velocity, km/h.
& a = C 1V m U = C 2V 0.7 Solving for C 2. U = 200 = C 2 (65) C 2 = 10.76442
0.7
Solving for LMTD. ∆T a − ∆T b LMTD = ∆T a ln ∆ T b ∆T a = T r − T cin = 65 C – 35 C = 30 C ∆T b = T r − T cout
T cout − T cin = 6 C T cout = 35 C + 6 C = 41 C ∆T b = T r − T cout = 65 C – 41 C = 24 C
LMTD =
∆T a − ∆T b
∆T a T ∆ b
=
ln
30 − 24 = 26.89 C 30 ln 24
Solving for A. Q = UA( LMTD ) 18000 = (200)A(26.89 ) 2 A = 3.347 m .
&a Solving for m c pa = 1007 J/kg.K
& a c pa (T cout − T cin ) Q=m & a (1007 )(6 ) Q=m & a = 2.979 kg/s m Solving foe C 1. & a = C 1V m 2.979 = C 1 (65) C 1 = 0.04583 Then at any air velocity: & a = 0.04583V m 10
3. HEAT EXCHANGER DESIGN
U = 10.76442V 0.7
Solving for NTU . UA NTU = C min
& a c pa = (0.04583V )(1007 ) = 46.15V C min = m
(10.76442V )(3.347 ) 0.7
NTU =
46.15V
=
0.7807 V
0.3
Solving for ε : − NTU ε = lim ε = 1 − e C max→ ∞
0.7807 0. 3 V
ε = 1 − exp −
Q′ = ε C min T hin − T cin
0.7807 (46.15V )(65 − 35) 0.3 V 0.7807 Q′ = 1384.5V 1 − exp − 0.3 V Q′ = 1 − exp −
Percentage reduction , x, %. x =
Q − Q′ Q
(100% ) = 1 −
1384.5V 0.7807 1 exp − − (100 ) 0.3 18000 V
0.7807 x = 1001 − 0.07692V 1 − exp − 0.3 V Tabulation:
,%
V , km/h 15 20 25 30 35 40 45 50 55 60 65
x
66.2 58.1 50.6 43.4 36.6 30.0 23.7 17.5 11.5 5.7 0.0
11
3. HEAT EXCHANGER DESIGN
Plot:
3.9
Derive eqn. (3.21).
Solution: Eq. (3.21), Counter-flow:
1 − exp− 1 −
ε =
C min NTU C max
C C 1 − min exp − 1 − min NTU C max C max
Derivation: Equation (3.17) Q = ε C min (T hin − T cin ) = C c (T cout − T cin ) Equation (3.18) UA NTU = C min Equation (3.9) ln
∆T a
C − 1 + c (T cin − T cout ) + ∆T a C h
C − 1 + c (T cin − T cout ) + ∆T a C h ln ∆T a
1 C c
= −UA
1 C c
= UA
−
−
1
C h
1
C h
∆T a = T hin − T c out from Eq. 3.8 , counterflow.
12
3. HEAT EXCHANGER DESIGN
Then ∆T a = T hin − T cin − T cout − T cin ∆T a =
Q
−
ε C min
Q C c
Let C c = C min , C h = C max ∆T a =
Q(1 − ε ) ε C min
C min Q − − 1 + C C UA C min max min 1 − + 1 = ln C min C max Q (1 − ε ) ε C min C ε C min NTU ln 1 − min + 1 = 1 − − 1 ε C C max max C min 1 ε 1 ε − + − C max C = 1 − min NTU ln C max 1 − ε C min ε + 1 − C max C = 1 − min NTU ln 1 − ε C max C 1 − ε = − 1 − min NTU ln C min C max − C ε + 1 max C 1 − ε = exp− 1 − min NTU C C min max ε + 1 − C max
1 − ε = −ε
C min C max
exp− 1 −
1 − exp− 1 −
ε =
C C min NTU + exp− 1 − min NTU C max C max
C min NTU C max
C C 1 − min exp − 1 − min NTU C max C max 13
3. HEAT EXCHANGER DESIGN
Derive the infinite NTU limit of the effectiveness of parallel and counterflow heat exchangers at several values of C min / Cm ax. Use common sense and the First Law of Thermodynamics, and refer to eqn. (3.20) and eqn. (3.21) only to check your results.
3.10
Solution: Eq. (3.20) – Parallel
C min NTU C max
1 − exp − 1 +
ε =
1+
C min C max
Eq. (3.21) – Counterflow
1 − exp− 1 −
ε =
C min NTU C max
C C 1 − min exp − 1 − min NTU C max C max
By energy balance, First Law of Thermodynamics. Q T hout = T hin − C h T cout = T cin +
Q C c
For Parallel Flow: T hout = T cout Then T cout = T hin − T cout = T cin +
Q C h Q C c
1
Q C min = 1 + C C C C h min max c C T hin − T cin = ε (T hin − T cin )1 + min C max T hin − T cin = Q
ε =
1+
+
1
1 C min C max
14
3. HEAT EXCHANGER DESIGN
Check with Eq. (3.20) with 1 − exp − 1 +
1 − exp− 1 +
ε =
1+ ε =
1+
C min NTU C max
C min
=
C max
NTU = 0 C max C min
1+ 0 C min
1+
C max
1 C min C max
For Counterflow T hin = T cout Q = ε C min T hin − T cin = C c T cout − T cin Q = ε C min T hin − T cin = C min T hin − T cin ε
=1
Check with Eq. (3.21) with 1 − exp − 1 +
1 − exp − 1 −
ε =
C min NTU C max
C C 1 − min exp− 1 − min NTU C max C max ε
3.11
=
NTU = 0 C max C min
1+ 0 1+ 0
=1 Derive the equation ε = ( NTU , C min C max ) for the heat exchanger depicted in Fig. 3.9.
Solution: ε = ( NTU , C min C max ) Eq. (3.22), C max → ∞ ε =
lim ε = 1 − e − NTU
C max→ ∞
Derivation for Parallel Flow
1 − exp − 1 + ε =
1+
C min NTU C max
C min C max
15
3. HEAT EXCHANGER DESIGN
ε =
1 − exp[− (1 + 0)NTU ]
1+ 0 − NTU ε = lim ε = 1 − e C max→ ∞
Derivation for Counterflow
1 − exp− 1 −
ε =
C min NTU C max
C C 1 − min exp − 1 − min NTU C max C max 1 − exp[− (1 − 0 ) NTU ]
ε =
1 − (0) exp[− (1 − 0 ) NTU ] − NTU ε = lim ε = 1 − e C max→ ∞
3.12
A single-pass heat exchanger condenses steam at 1 atm on the shell side and heats water from 10 C to 30 C on the tube side with U = 2500 W/m2.K The tubing is thin-walled, 5 cm in diameter, and 2 m in length. (a) Your boss asks whether the exchanger should be counterflow or parallel flow. How do you advise her? Evaluate: (b) the LMTD; (c) the & H 2O ; (d) ε . m
Solution: (a) I will advise here that counterflow or parallel flow configuration is irrelevant. The temperature on the hot side is constant. (b) For steam, T h = 100 C For water, T cin = 10 C T cout = 30 C c p = 4187 J/kg.K
For tubing, D = 5 cm = 0.05m L = 2 m A = π DL ∆T a − ∆T b
LMTD =
∆T a ∆ T b
ln
∆T a =100 C – 10 C = 90 C ∆T b = 100 C – 30 C = 70 C
LMTD =
90 − 70 = 79.58 C 90 ln 70
16
3. HEAT EXCHANGER DESIGN
& H 2O c p T cout − T cin (c) Q = UA( LMTD ) = m
(2500 )(π )(0.05)2 (2)(79.58) = m& H 2O (4187 )(30 − 10) & H 2O = 0.7464 kg/s m (d) ε = lim ε = 1 − e − NTU C max → ∞
NTU = ε = 1 − e
3.13
UA
=
C min −0. 2513
UA
& H 2O c p m
(2500 )(π )(0.05)2 (2) = = 0.2153 (0.7464 )(4187)
= 0.2222
Air at 2 kg/s and 27 C and a stream of water at a1.5 kg/s and 60 C each enter a heat 2 exchanger. Evaluate the exit temperatures if A = 12 m2, U = 185 W/m .K, and: a. The exchanger is parallel flow; b. The exchanger is counterflow; c. The exchanger is cross-flow, one stream mixed; d. The exchanger is cross-flow. Neither stream mixed.
Solution: & a = 2 kg/s m
& w = 1.5 kg/s m T cin = 27 C T hin = 60 C 2
A = 12 m 2 U = 185 W/m .K
Specific heat of air at 27 C = c pa = 1007 J/kg.K Specific heat of water at 60 C = c pw = 4186.2 J/kg.K
& a c pa = (2 kg/s)(1007 J/kg.K) = 2014 W/K = C min C c = m & w c pw = (1.5 kg/s)(4186.2 J/kg.K) = 6279.3 W/K = C max C h = m (a)
Eq. 3.20 for parallel flow
1 − exp − 1 + ε =
1+ NTU =
UA C min
=
C min NTU C max
C min C max
(185)(12) = 1.1023 (2014 ) 17
3. HEAT EXCHANGER DESIGN
1 − exp− 1 + ε =
1+
2014 (1.1023) 6279.3 = 0.5806 2014 6279.3
Q = ε C min (T hin − T cin )
Q = (0.5806 )(2014 )(60 − 27 ) = 38,587.8 W T hout = T hin − T cout = T cin +
(b)
Q C h Q C c
= 60 − = 27 +
38,587.8 6279.3 38,587.8 2014
= 53.86 C = 46.16 C
Eq. 3.21 for counterflow
1 − exp− 1 −
C min NTU C max
C C 1 − min exp − 1 − min NTU C max C max 2014 1 − exp − 1 − (1.1023) 6279.3 ε = = 0.6213 2014 2014 ( 1− exp− 1 − 1.1023) 6279 . 3 6279 . 3 ε =
Q = ε C min (T hin − T cin )
Q = (0.6213)(2014 )(60 − 27 ) = 41,292.8 W T hout = T hin − T cout = T cin +
(c)
Q C h Q C c
= 60 − = 27 +
41,292.8 6279.3 41,292.8 2014
= 53.42 C = 47.50 C
For cross-flow, one stream mixed, Figure 3.17b NTU = 1.103 C mixed C 2014 = min = = 0.32 C unmixed C max 6279.3 Then, ε = 0.60 Q = ε C min T hin − T cin Q = (0.60 )(2014 )(60 − 27 ) = 39,877.2 W
18
3. HEAT EXCHANGER DESIGN
T hout = T hin − T cout = T cin +
(d)
Q C h Q C c
= 60 − = 27 +
39,877.2 6279.3 39,877.2 2014
= 53.65 C = 46.80 C
For cross-flow, neither stream mixed, Figure 3.17a NTU = 1.103 C mixed C 2014 = min = = 0.32 C unmixed C max 6279.3 Then, ε = 0.60 Q = ε C min T hin − T cin Q = (0.60 )(2014 )(60 − 27 ) = 39,877.2 W
T hout = T hin − T cout = T cin +
3.14
Q C h Q C c
= 60 − = 27 +
39,877.2 6279.3 39,877.2 2014
= 53.65 C = 46.80 C
Air at 0.25 kg/s and 0 C enters a cross-flow heat exchanger. It is to be warmed to 20 C by 0.14 kg/s of air at 50 C. The streams are unmixed. As a first step in the design process, plot U against A and identify the approximate range of area for the heat exchanger.
Solution:
& a1 = 0.25 kg/s m T cin = 0 C T cout = 20 C
& a 2 = 0.14 kg/s m T hin = 50 C
Specific heat of air at 10 C (average of 0 C and 20 C) = c pa1 = 1006 J/kg.K Specific heat of air at 50 C = c pa 2 = 1008 J/kg.K
& a1c pa1 = (0.25 kg/s)(1006 J/kg.K) = 251.5 W/K = C max C c = m & a 2c pa 2 = (0.14 kg/s)(1008 J/kg.K) = 141.12 W/K = C min C h = m Q = ε C min T hin − T cin = C c T cout − T cin = C h T hin − T hout C c T cout − T cin = C h T hin − T hout
19
3. HEAT EXCHANGER DESIGN
(251.5)(20 − 0) = (141.12) 50 − T hout T hout = 14.36 C ε C min
T hin − T cin = C c T cout − T cin
ε (141.12 )(50 − 0 ) =
251.5(20 − 0)
= 0.713
ε
Fig. 3.17a, cross-flow unmixed, C min
=
141.12
C max ε
251.5
= 0.56
= 0.713
Then, NTU =
UA C min
=2
UA = 2(141.12) = 282.24 W/K 2
Table 2.2, Typical range of magnitudes of U for Air-to-Air is U = 60 – 550 W/m .K Plot:
2
At 60 W/m .K, A = 2
282.24
At 550 W/m .K, A =
2
= 4.704 m
60 282.24
2
= 0.513 m
550 2 2 Range of area = 0.513 m to 4.704 m .
20
3. HEAT EXCHANGER DESIGN
3.15
A particular two shell-pass, four tube-pass heat exchanger uses 20 kg/s of river water at 10 C on the shell side to cool 8 kg/s of processed water from 80 C to 25 C on the tube 2 side. At what temperature will the coolant be returned to the river? If U is 800 W/m .K, how large must the exchanger be?
Solution: Shell side: & r = river water = 20 kg/s m T sin = T cin = 10 C
Tube side: & p = processed water = 8 kg/s m T t in = T hin = 80 C T t out = T hout = 25 C
Specific heat of water at 52.5 C (average of 80 C and 25 C), c pp = 4183.2 J/kg.K Specific heat of water at 10 C , c pr = 4196.2 J/kg.K
& r c pr = (20 kg/s)(4196.2 J/kg.K) = 83,924 W/K C c = m & p c pp = (8 kg/s)(4183.2 J/kg.K) = 33,465.6 W/K C h = m Using LMTD Method: T h − T cout − T hout − T cin LMTD = in T h − T cout ln in T h − T c in out Solving for T cout ; Q = C h T hin − T hout = C c T cout − T cin
(33,465.6 )(80 − 25) = (83,924) T cout − 10 T cout = 31.93 C = T sout LMTD =
T hin − T cout − T hout − T cin
T hin − T cout T h − T c in out
ln LMTD =
(80 − 31.93) − (25 − 10) 80 − 31.93 ln 25 − 10
= 28.4 C
2
U = 800 W/m .K T t − T t in 25 − 80 P = out = 0.786 = T sin − T t in 10 − 80 R =
T sin − T sout T t out − T t in
=
10 − 31.93 25 − 80
= 0.40 21
3. HEAT EXCHANGER DESIGN
For two shell-pass, four tube-pass heat exchanger, Fig. 3.14b With P = 0.786, R = 0.4, then F = 0.936 Q = UAF ( LMTD ) = C h T hin − T hout
(800) A(0.936)(28.4) = (33,465.6)(80 − 25) 2
A = 86.552 m .
3.16
A particular cross-flow process heat exchanger operates with the fluid mixed on one side 2 only. When it is new, U = 2000 W/m .K, T cin = 25 C, T cout = 80 C, T hin = 160 C and T hout = 70 C. After 6 months of operation, the plant manager reports that the hot fluid is only being cooled to 90 C and that he is suffering a 30% reduction in total heat transfer. What is the fouling resistance after 6 months of use? (Assume no reduction of cold-side flow rate by fouling.)
Solution: 2
U = 2000 W/m .K T cin = 25 C T cout = 80 C T hin = 160 C T hout = 70 C
Use LMTD Method: T h − T cout − T hout − T cin (160 − 80) − (70 − 25) LMTD1 = in = 60.83 C = T hin − T cout 160 − 80 ln ln 70 − 25 T h − T c in out Q = UAF ( LMTD ) Q1 = U 1 AF 1 (LMTD1 ) P= R =
T t out − T t in T sin − T t in T sin − T sout T t out − T t in
=
=
70 − 160 25 − 160 25 − 80 70 − 160
= 0.67 = 0.611
Fig. 3.17d, cross-flow, fluid mixed on one side only, F 1 = 0.76 Q1 = U 1 A(0.76 )(60.83) = 46.23U 1 A with Q2 = (1 − 0.30 )Q1 = 0.70Q1 T hout = 90 C
22
3. HEAT EXCHANGER DESIGN
LMTD2 =
T hin − T cout − T hout − T cin
T h − T cout ln in T h − T c in out
=
(160 − 80) − (90 − 25) 160 − 80 ln 90 − 25
= 72.24 C
Q2 = U 2 A(0.90 )(72.24 ) = 65.02U 2 A Q2 = 0.70Q1
65.02U 2 A = 0.70(46.23)U 1 A U 2 = 0.4977U 1 R f =
1
−
U 2
R f =
3.17
1 U 1
1.00924 2000
1 1 1.00924 − 1 = U U 1 0 . 4977 1
=
2
= 0.0005 m .K/W
Water at 15 C is supplied to a one shell-pass, two tube-passes heat exchanger to cool 10 kg/s of liquid ammonia from 120 C to 40 C. You anticipate a U on the order of 1500 2 2 W/m .K when the water flows in the tubes. If A is to be 90 m , choose the correct flow rate of water.
Solution:
& a = 10 kg/s m T sin = T hin = 120 C T sout = T hout = 40 C T t in = T cin = 15 C 2
U = 1500 W/m .K 2 A = 90 m .
Specific heat of water at 15 C, c pw = 4189.4 J/kg.K Specific heat of liquid ammonia at 80 C (average of 120 C and 40 C). c pa = 5838.1 J/kg.K
& a c pa = (10)(5838.1) = 58,381 W/K C c = m Try C min = C c = 58,381 W/K Use Fig. 3.17c (one shell-pass, two tube-passes) UA (1500 )(90) NTU = = = 2.312 C min 58,381 Q = ε C min (T hin − T cin ) = C h (T hin − T hout ) C max = C h
23
3. HEAT EXCHANGER DESIGN ε C min (120 − 15) = C max (120 − 40)
C min C = 0.762 max
ε
From Fig. 3.17c, NTU = 2.312 Tabulation:
C min C max
ε
0.25 0.50 0.75 1.00
0.80 0.72 0.64 0.58
C min C max
ε
0.20 0.36 0.48 0.58
C Since ε min < 0.762, C min ≠ C c C max Use C min = C h Q = ε C min T hin − T cin = C min T hin − T hout ε (120 − 15) = (120 − 40) ε =
0.762 C max = C c = 58,381 W/K
NTU =
UA
=
(1500 )(90) 135,000 =
C min
C min
C min
Fig. 3.17 Tabulation: ε = 0.762
C min C max
NTU
0.25 0.375 0.5
1.9 2.5 3.6
By extrapolation: x − 21893 =
C min C max
C min from
14595 21893 29191
x − 54000
29191 − 21893 38571 − 54000 x − 54000 = −2.11414( x − 21893) x = 32,203 = C min
or C min
(32203 − 21893) (0.5 − 0.375) = 0.552 (29191 − 21893) C max C min = 0.552(58,381) = 32,226 W/K = 0.375 +
24
C min from NTU
71053 54000 38571
3. HEAT EXCHANGER DESIGN
Or NTU = 2.5 +
29191 − 21893 135,000
NTU = C min =
(32203 − 21893)
(3.6 − 2.5) = 4.05
C min
135,000
=
135,000 4.05
NTU
= 33,333
Then: C min =
1 3
(32203 + 32226 + 33333) = 32,587 W/K
and & w c pw C min = m
& w (4189.4 ) 32,587 = m & w = 7.8 kg/s m 3.18
Suppose that the heat exchanger in Example 3.5 had been a two shell-pass, four tube-pass exchanger with the hot fluid moving in the tubes. (a) What would be the exit temperature in this case? (b) What would be the area if we wanted the hot fluid to leave at the same temperature that it does in the example?
Solution: From Example 3.5 T cin = 40 C C c = 20,000 W/K T hin = 150 C C h = 10,000 W/K 2
A = 30 m 2 U = 500 W/m .K
Answer from Ex. 3.5, T hout = 84.44 C (a) Fig. 3.17d, two shell-pass, four tube-pass heat exchanger C min = C h = 10,000 W/K C max = C c = 20,000 W/K NTU = C min C max
=
UA
=
C min
10,000 20,000
(500)(30 ) 10,000
= 1.5
= 0.5
ε = 0.673
25
3. HEAT EXCHANGER DESIGN
Q = ε C min T hin − T cin = (0.673)(10,000)(150 – 40) = 740,300 W Q
T hout = T hin −
C h
= 150 −
740,300 10,000
= 75.97 C
(b) T hout = 84.44 C Q = ε C min T hin − T cin = C min T hin − T hout ε =
T hin − T hout T hin − T cin
C min C max
=
150 − 84.44 150 − 40
= 0.596
= 0.5
Fig. 3.17d, two shell-pass, four tube-pass heat exchanger, NTU =
UA C min
= 1.14
(500)( A) = (1.14)(10,000) 2
A = 22.8 m .
3.19
Plot the maximum tolerable fouling resistance as a function of U new for a counterflow exchanger, with given inlet temperature, if a 30 % reduction in U is the maximum that can be tolerated.
Solution: 1 U old
= R f +
1 U new
if U old = (1 − 0.30)U new = 0.70U new Then: 1
1
U old
0.70U new
=
= R f +
1 U new
1 1 − 1 = R f U 0 . 70 new R f =
3 7U new
26
3. HEAT EXCHANGER DESIGN
3.20
Water at 0.8 kg/s enters the tubes of a two-shell-pass, four-tube-pass heat exchanger at 17 C and leaves at 37 C. It cools 0.5 kg/s of air entering the shell at 250 C with U = 432 2 W/m .K. Determine: (a) the exit air temperature; (b) the area of heat exchanger; and (c) the exit temperature if, after some time, the tubes become fouled with R f = 0.0005 2 m K/W.
Solution:
& w = 0.8 kg/s m T cin = 17 C T cout = 37 C
& a = 0.5 kg/s m T hin = 250 C 2
U = 432 W/m .K c pw at 27 C = 4181 J/kg.K c pa at 250 C = 1035 J/kg.K C c = mwc pw = (0.8)(4181) = 3344.8 W/K C h = ma c pa = (0.5)(1035) = 517.5 W/K
(a) Solve for exit air temperature. C c (T cout − T cin ) = C h (T hin − T hout )
(3344.8)(37 − 17) = (517.5) 250 − T hout T hout = 120.74 C
(b) Solving for the area of the heat exchanger. Q = UAF ( LMTD )
27
3. HEAT EXCHANGER DESIGN
LMTD =
T hin − T cout − T hout − T cin
=
T h − T cout ln in T h − T c in out
(250 − 37) − (120.74 − 17) 250 − 37 ln 120.74 − 17
= 151.88 C
For two-shell-pass, four-tube-pass heat exchanger, use Fig. 3.14b T t out = 37 C T t in = 17 C T sin = 250 C T sout = 120.74 C P= R =
T t out − T t in T sin − T t in T sin − T sout T t out − T t in
=
=
37 − 17 250 − 17
= 0.086
250 − 120.74 37 − 17
= 6.463 > 1
Since R > 1, use reciprocal rule. P = PR = (0.086)(6.463) = 0.56 R = 1/ R = 1/(6.463) = 0.155 Fig. 3.14b, F ~ 1.0 Then, Q = C c (T cout − T cin ) = (3344.8)(37 − 17 ) = 66,896 W 2
U = 432 W/m .K Q = UAF ( LMTD ) 66,496 = (432)( A)(1.0)(151.88) 2 A = 1.02 m . 2
(c) Exit temperature if R f = 0.0005 m K/W 1 1 = R f + U ′ U 1 1 = 0.0005 + U ′ 432 2 U ′ = 355.3 W/m .K C min = C h = 517.5 W/K C max = C c = 3344.8 W/K NTU =
′ U A
C min
=
(355.3)(1.02) 517.5
= 0.7003
Fig. 3.17b, two-shell-pass, four-tube-pass heat exchanger. C min 517.5 = = 0.155 C max 3344.8 NTU = 0.7003 Then, ε = 0.47 28
3. HEAT EXCHANGER DESIGN
Q = ε C min T hin − T cin = (0.47 )(517.5)(250 − 17 ) = 56,671 W T cout = T c in + T hout = T h in −
3.21
Q C c
= 17 +
Q C h
56,671 3,344.8
= 250 −
= 34 C
56,671 517.5
= 140.5 C
You must cool 78 kg/min of a 60%-by-pass mixture of glycerin in water from 108 C to 50 C using cooling water available at 7 C. Design a one-shell-pass, two-tube-pass heat 2 exchanger if U = 637 W/m .K. Explain any design decision you make and report the area, T H 2Oout , and any other relevant features.
Solution: & g = 78 kg/min m T hin = 108 C T hout = 50 C T cin = 7 C c pg at 79 C of 60 % glycerin in water = 3474 J/kg.K c pw at 7 C = 4201 J/kg.K
One-shell-pass, two-tube-pass heat exchanger. Fig. 3.17c & g c pg = (78)(3474)/60 = 4516.2 W/K C h = m Q = C h T hin − T hout = (4516.2)(108 − 50) = 261,940 W Q = ε C min T hin − T cin
say C min = C h = 4516.2 W/K Q = ε (4516.2)(108 − 7 ) = 261,940 ε = 0.5743
assume C min = C max C min C max
= 1.0
then, NTU = 2.32 NTU =
2.32 =
UA C min
(637 )A
4516.2 2 A = 16.45 m . Check for F-factor, LMTD = 50 C – 7 C = 43 C 29
3. HEAT EXCHANGER DESIGN
Q = UAF ( LMTD ) 261,940 = (637)(16.45)( F )(43) F = 0.58, very small
Say
C min C max
= 0.5, ε = 0.5743
then, NTU = 1.091 NTU =
1.091 =
UA C min
(637)A
4516.2 2 A = 7.735 m . Check for F-factor C max = 2C min = 2(4516.2) = 9032.4 W/K T cout = T c in + LMTD =
Q C c
261,940
= 7+
9,032.4
= 36 C
T hin − T cout − T hout − T cin
T h − T cout ln in T h − T c in out
=
(108 − 36) − (50 − 7 ) 108 − 36 ln 50 − 7
Q = UAF ( LMTD ) 261,940 = (637)(7.735)( F )(56.26) F = 0.95 From. Fig. 3.14a T t − T t in P = out T sin − T t in R =
T sin − T sout T t out − T t in
T t out = 50 C T t in = 108 C T sin = 7 C T sout = 36 C P= R =
50 − 108 7 − 108 7 − 36
= 0.574
= 0.5 50 − 108 Then F = 0.90 < 0.95
30
= 56.26 C
3. HEAT EXCHANGER DESIGN
Say
C min C max
= 0.25
= 0.5743 then, NTU = 0.9545 ε
NTU =
UA C min
0.9545 =
(637)A
4516.2 A = 6.7672 C max = 4C min = 4(4516.2) = 18,064.8 W/K T cout = T c in + LMTD =
Q C c
= 7+
261,940 18,064.8
= 21.5 C
T hin − T cout − T hout − T cin
T h − T cout ln in T h − T c in out
=
(108 − 21.5) − (50 − 7 ) 108 − 21.5 ln 50 − 7
Q = UAF ( LMTD ) 261,940 = (637)(6.7672)( F )(62.24) F = 0.976 From. Fig. 3.14a T t − T t in P = out T sin − T t in R =
T sin − T sout T t out − T t in
T t out = 50 C T t in = 108 C T sin = 7 C T sout = 21.5 C P= R =
50 − 108 7 − 108 7 − 21.5
= 0.574
= 0.25 50 − 108 Then F = 0.96 < 0.976 By extrapolation: x − 0.95 x − 0.90 0.976 − 0.95
=
0.96 − 0.90 31
= 62.24 C
3. HEAT EXCHANGER DESIGN
2.3077( x − 0.95) = x − 0.90 x = 0.99 From Fig. 3.14a, for F = x = 0.99 at P = 0.574 R ~ 0.099 C But R = min = 0.099 C max C min = 4516.2 W/K C max = 4516.2 / 0.099 = 45,618 W/K Q = 261,940 W
Fig. 3.17c, ε = 0.5743, NTU =
0.895 =
UA C min
C min C max
= 0.099
= 0.895
(637)A
4516.2 2 A = 6.35 m . Check for F. Q 261,940 T cout = T c in + = 12.74 C = 7+ C c 45,618 LMTD =
T hin − T cout − T hout − T cin
T h − T cout ln in T h − T c in out
=
(108 − 12.74) − (50 − 7) 108 − 12.74 ln 50 − 7
= 65.70 C
Q = UAF ( LMTD ) 261,940 = (637)(6.35)( F )(65.70) F = 0.99 = 0.99 o.k.
Therefore: 2 A = 6.35 m . T H 2Oout = 12.74 C C min C max
= 0.099
ε = 0.5743
3.22
A mixture of 40%-by-weight glycerin, 60%-water, enters a smooth 0.113 m I.D. tube at & mixture = 8 kg/s. The heat transfer coefficient inside the 30 C. The tube is kept at 50 C, and m 2
pipe is 1600 W/m .K. Plot the liquid temperature as a function of position in the pipe. Solution: T cin = 30 C 32
3. HEAT EXCHANGER DESIGN
& g = 8 kg/s m 2
hi = 1600 W/m .K T hin = T hout = 50 C c pg at 30 C, 40%-glycerin = 3480 J/kg.K
& g c pg = (8)(3480) = 27,840 W/K C min = C c = m C max → ∞ C min C max
=0
lim ε = 1 − e − NTU
C max → ∞
Q = ε C min T hin − T cin = C c T cout − T cin ε C min
T hin − T cin = C min T cout − T cin
T cout = ε T hin − T cin + T cin NTU =
UA
=
C min
hi A C min
A = π Dx
where x – position in pipe from T cin . T cout = (1 − e − NTU ) T hin − T cin + T cin NTU = NTU = T cout
hi A C min
=
hi (π Dx ) C min
(1600 )(π )(0.113) x
= 0.0204 x 27840 −0.0204 x )(50 − 30) + 30 = (1 − e
T cout = 50 − 20e −0.0204 x
Tabulation: x
,m
T cout = 50 − 20e −0.0204 x , C
0 1 2 3 4 5 6 7 8 9
30 30.4 30.8 31.2 31.6 31.9 32.3 32.7 33.0 33.4 33
3. HEAT EXCHANGER DESIGN
10 20 30 40 50
33.7 36.7 39.2 41.2 42.8 50.0
∞
Plot:
3.23
Explain in physical terms why all effectiveness curves Fig. 3.16 and Fig. 3.17 have the same slope as NTU → 0. Obtain this slope from eqns. (3.20) and (3.21).
Solution: For parallel flow, Eq. (3.20)
1 − exp − 1 + ε =
1+
C min NTU C max
C min C max
For counterflow, Eq. (3.21)
1 − exp− 1 −
ε =
C min NTU C max
C C 1 − min exp − 1 − min NTU C max C max
Slope for parallel flow:
34
3. HEAT EXCHANGER DESIGN
1 − exp − 1 +
ε =
1+
d ε
=
C min NTU C max
C min C max
C C min 1 + exp − 1 + min NTU C max C max
dNTU d ε dNTU
1+
= exp − 1 +
C min C max
NTU C max C min
if NTU = 0 d ε dNTU
= 1 , independent of
C min C max
, therefore, the same for all.
Slope for counterflow:
1 − exp− 1 −
ε =
C min NTU C max
C C 1 − min exp − 1 − min NTU C max C max
d ε
=
dNTU
C C C C C C C C Substitut min exp − 1 − min NTU 1 − min exp − 1 − min NTU − 1 − exp− 1 − min NTU min 1 − min exp− 1 − min NTU 1 − C max C max C max C max C max C max C max C max 2 C min C min NTU exp − 1 − 1 − C max C max
e, NTU = 0.
C min C min C C 1 − 1 − − (1 − 1) min 1 − min d ε C max C max C max C max = 2 dNTU C min 1 − C max C min C min C C 1 − 1 − − (1 − 1) min 1 − min d ε C max C max C max C max = 2 dNTU C min 1 − C max d ε dNTU
= 1 , also independent of
C min C max
, therefore, the same for all.
35
3. HEAT EXCHANGER DESIGN
3.24
You want to cool air from 150 C to 60 C but you cannot afford a custom-built heat exchanger. You find a used cross-flow exchanger (both fluids unmixed) in storage. It was previously used to cool 136 kg/min of NH 3 vapor from 200 C to 100 C using 320 kg/min 2 of water at 7 C; U was previously 480 W/m . K . How much air can you cool with this exchanger, using the same water supply, if U is approximately unchanged? (Actually, you would have to modify U using the methods of Chapters 6 and 7 once you had the new air flow rate, but that is beyond our present scope.)
Solution: & N = 136 kg/min m T hin = 200 C T hout = 100 C
& w = 320 kg/min m T cin = 7 C 2
U = 480 W/m .K. Use c pN at 107 C = 7818 J/kg.K, since values at 127 C and above from table. c pw at 7 C = 4201 J/kg.K
& wc pw = (320)(4201)/(60) = 22,405 W/K C max = m & N c pN = (136)(7818)/(60) = 17,721 W/K C min = m C min
=
C max
17,721 22, 405
= 0.79
Q = ε C min (T hin − T cin ) = C h (T hin − T hout )
C h = C min = 17,721 J/kg.K Q = ε (17,721)(200 − 7 ) = (17,721)(200 − 100) ε = 0.52
From Fig. 3.17a, Cross-flow C min ε = 0.52, = 0.79 C max then, NTU = 1.05 UA NTU = C min 1.05 =
(480) A
17,721 2 A = 38.765 m . For the new conditions: 36
3. HEAT EXCHANGER DESIGN
c pa at 105 C (average of 150 C and 60 C) = 1012 J/kg.K
Using the same water temperature is used as 7 C T hin = 150 C T hout = 60 C T cin = 7 C
& a c pa = 1012 m& a W/K C min = C h = m NTU =
UA
=
(480)(38.765) 18.3866 =
&a 1012m
C min
&a m
Q = ε C min T hin − T cin = C min T hin − T hout = C h T hin − T hout ε
T hin − T cin = T hin − T hout
ε (150 − 7 ) = (150 − 60)
= 0.63
ε
& wc pw = 22,405 W/K C max = m &a C min = 1012m C min
=
&a 1012m
C max
22,405
&a = 0.04517 m
Fig. 3.17a, cross-flow C min &a = 0.04517 m C max NTU = ε
18.3866
&a m
= 0.63
Tabulation of trial and error method: &a C min m
NTU
NTU (graph)
0.91933 1.22577 1.1492
1.4 1.143 1.2
C max
20 15 16
0.9034 0.67755 0.72272
By interpolation: x − 1.22577
=
x − 1.143
1.1492 − 1.2257 1.2 − 1.143 − 1.342( x − 1.22577 ) = x − 1.143 x = 1.19 Then: 37
3. HEAT EXCHANGER DESIGN
1.19 − 1.143 (16 − 15) + 15 = 15.8 kg/s − 1 . 2 1 . 143
&a = m 3.25
≈ 950 kg/min
A one tube-pass, one shell-pass, parallel-flow, process heat exchanger cools 5 kg/s of gaseous ammonia entering the shell side at 250 C and boils 4.8 kg/s of water in the tubes. 2 The water enters subcooled at 27 C and boils when it reaches 100 C . U = 480 W/m .K 2 2 before boiling begins and 964 W/m .K there-after. The area of the exchanger is 45 m , 6 and hfg for water is 2.257 x 10 J/kg. Determine the quality of the water at the exit.
Solution: & a = 5 kg/s m
Parallel Flow:
T hin = 250 C
& w = 4.8 kg/s m T cin = 27 C
From Appendix A, extrapolated. Specific heat of gaseous ammonia at 250 C, c pa = 2524 J/kg.K Specific heat of water at 63.5 C, c pw = 4188 J/kg.K 2
Boiling water at 100 C, U 1 = 480 W/m .K T cout = 100 C
& wc pw T cout − T cin Q1 = m Q1 = (4.8)(4188)(100 – 27) Q1 = 1,467,475 W Q1 = U 1 A 1( LMTD1 ) LMTD1 =
T hin − T cin − T hout − T cout
T hin − T cin T h − T c out out
ln
& a c pa (T hin − hhout ) Q1 = m 1,467,475 = (5)(2524 )(250 − T hout ) T hout = 133.72 C LMTD1 =
(250 − 27 ) − (133.72 − 100) 250 − 27 ln 133.72 − 100
= 100.2 C
Q1 = U 1 A 1( LMTD1 )
1,467,475 = (480 ) A 1(100.2 ) 38
3. HEAT EXCHANGER DESIGN 2
A 1 = 30.5 m . 2
Total Area, A = 45 m . Remaining area for evaporation at 100 C. A 2 = A − A 1 2
A 2 = 45 − 30.5 = 14.5 m . 2
At evaporation, U 2 = 964 W/m .K. T hin =133.72 C
Specific heat of gaseous ammonia at 133.72 C c pa = 2297 J/kg.K, from Appendix A
& w h fg ( x ) Q2 = ε C min (T hin − T cin ) = m & a c pa =(5)(2297) = 11,485 W/K C min = m T cin = 100 C 6
h fg = 2.257 x 10 J/kg ε = lim lim ε = 1 − e − NTU C max → ∞
NTU =
U 2 A2
ε = 1 − e
=
C min
(964)(14.5) 11485
= 1.2171
−1. 2171
= 0.704 & w h fg ( x ) Q2 = ε C min (T hin − T cin ) = m
Q2 = (0.704)(11485 )(133.72 − 100 ) = (4.8)(2.257 × 106 )( x ) x = 0.025 or 2.5 % vapor
3.26
0.72 kg/s of superheated steam enters a cross-flow heat exchanger at 240 C and leaves at 120 C. It heats 0.6 kg/s of water entering at 17 C. U = = 612 W/m2.K. By what percentage will the area differ if a both-fluids-unmixed exchanger is used instead of a one-fluidunmixed exchanger?
Solution: & s = 0.72 kg/s m T hin = 240 C T hout = 120 C
& w = 0.60 kg/s m T cin = 17 C
Specific heat of superheated steam at 180 C, c ps = 1976 J/kg.K, Table A.6 39
3. HEAT EXCHANGER DESIGN
Specific heat of water at 17 C, c pw = 4187 J/kg.K. Solving for T cout .
& s c ps T hin − T hout = m& wc pw T cout − T cin Q=m
(0.72)( )(1976 )(240 − 120) = (0.60)( )(4187 ) T cout − 17 T cout = 85 C
LMTD =
T hin − T cout − T hout − T cin
T h − T cout ln in T h − T c in out
P= R =
=
(240 − 85) − (120 − 17) 240 − 85 ln 120 − 17
= 127.23 C
T t out − T t in T sin − T t in T sin − T sout T t out − T t in
T t out = 120 C T t in = 250 C T sin = 17 C T sout = 85 C P= R =
120 − 85 120 − 240 17 − 85
= 0.54
= 0.57 120 − 240 & s c ps T hin − T hout = (0.72 )(1976 )(240 − 120) = 170,727 W Q=m For cross-flow, both fluid unmixed, Fig. 3.14c. F 1 = 0.95 Q = UA 1 F 1 ( LMTD ) 170,727 = (612 ) A 1(0.95)(127.23) 2
A 1 = 2.308 m . For cross-flow, one fluid unmixed, Fig. 3.14d F 2 = 0.933 Q = UA 2 F 2 ( LMTD )
170,727 = (612) A 2 (0.933)(127.23) 2
A 2 = 2.350 m .
% area =
3.27
A 2 − A 1 A 2
(100%) =
2.308 − 2.350 2.350
(100%) = - 1.8 %
Compare values of F from from Fig. 3.14c and Fig. 3.14d for the same conditions of inlet and outlet temperatures. Is the one with higher F automatically automatically the more desirable exchanger? Discuss. 40
3. HEAT EXCHANGER DESIGN
Discussion: At given the same conditions of inlet and outlet temperatures, value of F from Fig. 3.14c is higher than values from Fig. 3.14d. Higher values of F automatically automatically will be the more desirable heat exchanger because it will give lesser heat transfer area, A. 3.28
Compare values of ε for the same NTU and
C min C max
in parallel and counterflow heat
exchangers. Is the one with higher ε automatically the more desirable exchanger? Discuss. Discussion: Reference to Fig. 3.16, ε -value for counterflow gives higher values than parallel-flow for the C min same NTU and . Higher values of ε will automatically be the more desirable heat C max exchanger because it will give higher heating capacity, Q = ε C min min (T hin − T cin ) . 3.29
The irreversibility rate of a process is equal to the rate of entropy production times the lowest absolute sink temperature accessible to the process. Calculate the irreversibility (or lost work) for the heat exchanger in Example 3.4. What kind of configuration would reduce the irreversibility given the same end temperatures?
Solution: From Ex. 3.4, two-shell pass, four-tube passes oil cooler. & o = 5.795 kg/s Oil: m T hin = 181 C T h out = 38 C
Water: T cin = 32 C T c out = 49 C c poil = 2282 J/kg.K 2
U = 416 W/m .K Using P = 0.959, R = 0.119, F = = 0.92 2 A = 121.2 m .
Solving for irreversibility: & = m& c ln T hout + m& c ln T cout S un o po w pw T hin T cin Specific heat of water at 40 C, c pw = 4180 J/kg.K
& o c po T hin − T hout = m& wc pw T cout − T cin Q=m 41
3. HEAT EXCHANGER DESIGN
(5.795)(2282 )(181 − 38) = m& w (4180)(49 − 32 ) & w = 26.612 kg/s m
38 + 273 49 + 273 + (26.612)(4180 ) ln = 1031 W/K + + 181 273 32 273
& = (5.795)(2282 )ln S un
& = (32 + 273)(1031) = 314,455 W Irreversibility = T sink S un Any configuration will give the same irreversibility for the given the same end temperatures. Plot T oil and T H 2O as a function of position in a very long counterflow heat exchanger
3.30
where water enters at 0 C, with C H 2O = 460 W/K, and oil enters at 90 C, with C oil = 920 2
2.
W/K, U = 742 W/m .K, and A = 10 m Criticize the design.
Solution: For counterflow case:
1 − exp− 1 −
ε =
C min NTU C max
C C 1 − min exp − 1 − min NTU C max C max UA
NTU =
C min
C min = C H 2O = 460 W/K C max = C oil = 920 W/K T hin = 90 C T cin = 0 C
NTU = C min C max ε =
=
UA
=
(742)(10) 460
C min
460 920
= 16.13
= 0.5
1 − exp[− (1 − 0.5)(16.13)]
= 0.9997 1 − (0.5) exp[− (1 − 0.5)(16.13)] Q = ε C min (T hin − T cin ) = (0.9997 )(460)(90 − 0) = 41,287.6 W Q =C H 2O (T cout − T cin )
41,387.6 = (460)(T H 2O − 0)
T H 2O = 89.973 C
Q =C oil (T hin − T hout )
42
3. HEAT EXCHANGER DESIGN
41,387.6 =(920)(90 − T oil ) T oil = 45.014 C ≈ 45 C
Plot:
•
3.31
Since NTU =
UA C min
2
> 5 , the area of 10 m is very large.
Liquid ammonia at 2 kg/s is cooled from 100 C to 30 C in the shell side of a two shellpass, four tube-pass heat exchanger by 3 kg/s of water at 10 C. When the exchanger is 2 new, U = 750 W/m .K. Plot the exit ammonia temperature as a function of the increasing tube fouling factor.
Solution: U new = 750 W/m2.K
& a = 2 kg/s m T hin = 100 C T hout = 30 C
& w = 3 kg/s m T cin = 10 C
Specific heat of liquid ammonia at 65 C, c pa = 5348 J/kg.K Specific heat of water at 10 C, c pw = 4196 J/kg.K
& a c pa = (2)(5348) = 10,696 W/K C min = C h = m & wc pw = (3)(4196) = 12,588 W/K C max = C c = m R f =
1 U old
−
1 U new
For new unit: two-shell-pass, four-tube-pass Q = C h T hin − T hout = C c T cout − T cin 43
3. HEAT EXCHANGER DESIGN
(10,696)(100 − 30) = (12,588) T cout − 10 T cout = 69.48 C
T hin − T cout − T hout − T cin
LMTD =
T h − T cout ln in T h − T c in out
=
(100 − 69.48) − (30 − 10) 100 − 69.48 ln 30 − 10
T sin = 100 C T sout = 30 C T t in = 10 C T t out = 69.48 C P= R = P= R =
T t out − T t in T sin − T t in T sin − T sout T t out − T t in
69.48 − 10 100 − 10 100 − 30 69.48 − 10
= 0.66 = 1.18 > 1
Use Reciprocal Rule: P = PR = (0.66)(1.18) = 0.78 R = 1/ R = 1/1.18 = 0.85 From Fig. 3.14b, F = 0.60 Q = C h T hin − T hout = UAF ( LMTD )
(10,696)(100 − 30) = (750)A(0.60)(24.89) 2
A = 66.85 m .
For old unit: 1 1 − R f = U old U new 2
U new = 750 W/m .K
1 U old
= R f +
U old =
1 U new
= R f +
1 750
750 750 R f + 1 44
= 24.89 C
3. HEAT EXCHANGER DESIGN
750 (66.85) + 750 1 R U old A f = NTU = 10,696
C min
NTU = C min C max
=
4.6875 750 R f + 1
10,696 12,588
= 0.85
Fig. 3,17d, Tabulation for
C min C max
= 0.85
NTU 5 4 3 2
ε
0.778 0.760 0.720 0.66
By curve fitting: ε = 0.4761 + 0.1129( NTU ) − 0.0105(NTU )
2
, for 5 > NTU > 2
Then, Q = ε C min T hin − T cin =C h T hin − T hout C min = C h T hout = T hin − ε (T hin − T cin ) T hout = 100 − ε (100 − 10) = 100 − 90ε
[
T hout = 100 − 90 0.4761 + 0.1129( NTU ) − 0.0105( NTU )
2
]
T hout = 57.151 − 10.161( NTU ) + 0.945( NTU )
2
T hout
4.6875 + 0.945 4.6875 = 57.151 − 10.161 750 R + 1 750 R + 1 f f
T hout = 57.151 −
47.63 750 R f + 1
+
2
20.764
(750 R
f
)
+1
2
Tabulation of Values: 2 R f , m .K/W
T hout , C
0 0.0005 0.0010 0.0015
30 33.5 36.7 39.3
Plot: 45
3. HEAT EXCHANGER DESIGN
3.32
A one shell-pass, two tube-pass heat exchanger cools 0.403 kg/s of methanol from 47 C to 7 C on the shell side. The coolant is 2.2 kg/s of Freon 12, entering the tubes at –33 C 2 with U = 538 W/m .K. A colleague suggests that this arrangement wastes Freon. She thinks you could do almost as well if you cut the Freon flow rate all the way down to 0.8 kg/s. Calculate the new methanol outlet temperature that would result from this flow rate, and evaluate her suggestion.
Solution: First condition, & m = 0.403 kg/s m T hin = 47 C = T sin T hout = 7 C = T sout
46
3. HEAT EXCHANGER DESIGN
& f = 2.2 kg/s m T cin = -33 C = T t in 2
U = 538 W/m .K
Specific heat of Methanol at 27 C, c pm = 2534 J/kg.K Specific heat of Freon at –33 C, c pf = 1100 J/kg.K Q = mm c pm (T hin − T hout ) = m f c pf (T cout − T cin )
(0.403)(2534 )(47 − 7 ) = (2.2)(1100 ) T cout − (− 33) T cout = -16.12 C = T t out
Q = (0.403)(2534 )(47 − 7 ) = 40,848 W Q = UAF ( LMTD ) LMTD =
T hin − T cout − T hout − T cin
T hin − T cout T h − T c in out
=
ln P= R =
T t out − T t in T sin − T t in T sin − T sout T t out − T t in
=
=
[47 − (− 16.12)] − [7 − (− 33)] = 50.684 C 47 − (− 16.12) ln 7 − (− 33)
− 16.12 − (− 33)
47 − (− 33) 47 − 7 − 16.12 − (− 33)
= 0.211 = 2.37 > 1
Use Reciprocal Rule: P = PR = (0.211)(2.37) = 0.50 R = 1/ R = 1/(2.37) = 0.42 From Fig. 3.14b, F = 0.95 Q = UAF ( LMTD ) 40,848 = (538)A(0.95)(50.684) 2 A = 1.577 m . Second Condition: U = 538 W/m2.K, A = 1.577 m2. & m = 0.403 kg/s m c pm = 2534 J/kg.K
& f = 0.8 kg/s m c pf = 1100 J/kg.K T hin = 47 C T cin = -33 C
& m c pm = (0.403)(2534) = 1021.2 W/K C h = m & f c pf = (0.80)(1100) = 880 W/K C c = m 47
3. HEAT EXCHANGER DESIGN
C min = C c C max = C h C min C max
880
=
1021.2
NTU =
UA
= 0.86
=
(538)(1.577) 880
C min
= 0.964
Fig. 3.17c, one shell-pass, two-tube pass heat exchanger, ε = 0.48 Q = ε C min T hin − T cin = (0.48)(880)[47 − (− 33)] = 33,792 W T hout = T hin −
Q C h
= 47 −
33,792 1021.2
= 13.9 C
Check for T cout : T cout = T cin +
Q C c
= −33 +
33,792 880
= 5.4 C
New Methanol outlet temperature, T hout = 13.9 C > 7 C. Therefore her suggestion will not attain 7 C methanol temperature. 3.33
The factors dictating the heat transfer coefficients in a certain two shell-pass, four tube-
& shell ) . The exchanger cools 2 kg/s of pass heat exchanger are such that U increase as (m 0.6
2
air from 200 C to 40 C using 4.4 kg/s of water at 7 C, and U = 312 W/m .K under these circumstances. If we double the airflow, what will its temperature be leaving the exchanger? Solution: & a = m& shell = 2 kg/s m T hin = 200 C = T sin T hout = 40 C = T sout
& w = 4.4 kg/s m T cin = 7 C = T t in 2
U = 312 W/m .K
& shell ) U ∝ (m
0.6
& shell ) U = k (m
0.6
312 = k (2.0) k = 205.84
0.6
& shell ) U = 205.84(m
0.6
&a) = 205.84(m
0.6
48
3. HEAT EXCHANGER DESIGN
Specific heat of air at 120 C, c pa = 1013 J/kg.K Specific heat of water at 7 C, c pw = 4201 J/kg.K First condition: & a c pa (T hin − T hout ) = m& w c pw (T cout − T cin ) Q=m
(2 )(1013)(200 − 40) = (4.4)(4201) T cout − 7 T cout = 24.55 C = T t out
Q = (2 )(1013)(200 − 40 ) = 324,160 W Q = UAF ( LMTD ) LMTD =
T hin − T cout − T hout − T cin
=
T h − T cout ln in T h − T c in out T t out − T t in
P=
T sin − T t in T sin − T sout
R =
T t out − T t in
=
24.55 − 7
=
(200 − 24.55) − (40 − 7 ) 200 − 24.55 ln 40 − 7
= 85.256 C
= 0.090933
200 − 7 200 − 40
=9.11681 > 1
24.55 − 7
Use Reciprocal Rule: P = PR = (0.090933)(9.11681) = 0.83 R = 1/ R = 1/(9.11681) = 0.11 From Fig. 3.14, two shell-pass, four tube-pass heat exchanger F = 0.975 Q = UAF ( LMTD ) 324,160 = (312)(A)(0.975)(85.256) 2 A = 12.5 m . Second Condition: & a = m& shell = 2(2 kg/s) = 4 kg/s m T hin = 200 C
& w = 4.4 kg/s m T cin = 7 C
& shell ) U = 205.84(m
0.6
&a) = 205.84(m
0.6
U = 205.84(4.0) = 472.9 W/m .K 0. 6
2
2
A = 12.5 m . & a c pa = (4)(1013) = 4052 W/K C h = m
& wc pw = (4.4)(4201) = 18,485 W/K C c = m C min = C h = 4052 W/K C max = C c = 18,485 W/K
49
3. HEAT EXCHANGER DESIGN
C min C max
4052
=
18485
NTU =
UA
= 0.22
=
(472.9)(12.5)
C min
4052
= 1.46
Fig. 3.17d, two shell-pass, four tube-pass heat exchanger ε = 0.72 Q = ε C min T hin − T cin
Q = (0.72 )(4052 )(200 − 7 ) = 563,066 W T hout = T hin −
3.34
Q C h
= 200 −
563,066 4052
= 61.0 C
A flow rate of 1.4 kg/s of water enters the tubes of a two-shell-pass, four-tube-pass heat exchanger at 7 C. A flow rate of 0.6 kg/s of liquid ammonia at 100 C is to be cooled to 30 2 C on the shell side; U = 573 W/m .K. (a) How large must the heat exchanger be? (b) How large must it be if, after some months a fouling factor of 0.0015 will build up in the tubes, and we still want to deliver ammonia at 30 C? (c) If we make it large enough to accommodate fouling, to what temperature will it cool the ammonia when it is new? (d) At what temperature does water leave the new, enlarged exchanger?
Solution: (a) Two-shell-pass, four-tube-pass
& w = 1.4 kg/s m T cin = T t in = 7 C
& a = 0.6 kg/s m T hin = T sin = 100 C T hout = T sout = 30 C 2
U = 573 W/m .K
Specific heat of liquid ammonia at 65 C, c pa = 5348 J/kg.K Specific heat of water at 7 C, c pw = 4201 J/kg.K
& a c pa = (0.6)(5348) = 3208.8 W/K C h = C min = m & wc pw = (1.4)(4201) = 5881.4 W/K C c = C max = m Q = C h T hin − T hout = (3208.8)(100 – 30) = 224,616 W
Q = UAF ( LMTD )
50
3. HEAT EXCHANGER DESIGN
LMTD =
T hin − T cout − T hout − T cin
T hin − T cout T h − T c in out
ln
Q = C c (T cout − T cin )
224.616 = (5881.4 )(T cout − 7 )
T cout = 45.2 C = T t out LMTD =
(100 − 45.2) − (30 − 7 )
P= R =
100 − 45.2 ln 30 − 7 T t out − T t in T sin − T t in T sin − T sout T t out − T t in
=
=
= 36.63 C
45.2 − 7 100 − 7
= 0.4108
100 − 30 45.2 − 7
= 1.8325 > 1
Use Reciprocal Rule: P = PR = (0.4108)(1.8325) = 0.75 R = 1/ R = 1/(1.8325) = 0.55 From Fig. 3.14, two shell-pass, four tube-pass heat exchanger F = 0.915 Q = UAF ( LMTD ) 224,616 = (573)(A)(0.915)(36.63) 2 A = 11.7 m . (b) if R f = 0.0015 1 U old
1
= R f +
1 U new
= 0.0015 +
1
U 573 2 U = 308.15 W/m .K Q = UAF ( LMTD ) 224,616 = (308.15)( A)(0.915)(36.63) 2 A = 21.75 m . 2
2
(c) if A = 21.75 m , U = 573 w/m .K C min = 3208.8 W/K C max = 5881.4 W/K T hin = 100 C
51
3. HEAT EXCHANGER DESIGN
T cin = 7 C
C min
=
3208.8 5881.4
C max
UA
NTU =
=
= 0.55
(573)(21.75) 3208.8
C min
= 3.9
Fig. 3.17d, two shell-pass, four tube-pass heat exchanger ε = 0.85 Q = ε C min T hin − T cin
Q = (0.85)(3208.8)(100 − 7 ) = 253,656 W T hout = T hin −
Q
= 100 −
C h
Q
(d) T cout = T cin + 3.35
C c
253,656 3208.8
=7−
= 21 C
253,656 5881.4
= 50 C
Both C ’s in a parallel-flow heat exchanger are equal to 156 W/K, U = 327 W/m2.K and A 2 = 2 m . The hot fluid enters at 140 C and leaves at 90 C. The cold fluid enters at 40 C. If both C’s are halved, what will be the exit temperature of the hot fluid?
Solution: C min = C max For parallel flow, Eq. 3.20
1 − exp − 1 +
ε =
1+ C min C max
C min NTU C max
C min C max
=1
if C min = C max = (1/2)(156 W/K) = 78 W/K NTU = ε =
UA
=
C min
(327)(2) 78
= 8.385
1 − exp[− (1 + 1)(8.385)]
= 0.5 1+1 Q = ε C min T hin − T cin = (0.5)(78)(140 – 40) = 3900 W
T hout =T hin −
Q
= 140 −
3900
= 90 C C 78 This is still the same since NTU >5, ε remain the same. 52
3. HEAT EXCHANGER DESIGN
3.36
A 1.68 ft2 cross-flow heat exchanger with one fluid mixed condenses steam at 2 atmospheric pressure ( h = 2000 Btu/h.ft .F) and boils methanol ( T sat = 170 F and h = 2
1500 Btu/h.ft .F) on the other side. Evaluate U (neglecting resistance of the metal), LMTD, F , NTU , ε , and Q. Solution: Steam at atmosphere, T h = 212 F Methanol, T c = 170 F Solving for U , 1 1 1 =
+
U 2000 1500 2 U = 857 Btu/h.ft .F Solving for LMTD. F = 1.0 for isothermal fluid. Solving for NTU , UA NTU = C min
but C min →∞ NTU = 0 Solving for ε − NTU ε = lim ε = 1 − e C max→ ∞
but NTU = 0 ε = 1.0 3.37
Eqn. (3.21) is troublesome when
C min C max
= 1.0 . Develop a working equation for ε in this
case. Compare it with Fig. 3.16. Solution: Eq. (3.21)
1 − exp− 1 −
ε =
if
C C 1 − min exp− 1 − min NTU C max C max
C min C max
ε =
C min NTU C max
= 1.0
0
0 L’Hospital rule.
53
3. HEAT EXCHANGER DESIGN
1 − exp− 1 −
C min NTU C max
C min 0 0 C C C max 1 − min exp− 1 − min NTU C max C max C min ∂ 1 − exp− 1 − NTU C C min max C min 1 ∂ C max C max C min C ∂ exp− 1 − min NTU 1 − C C max C max C min 1 ∂ min C max C max C − (− NTU )(− 1) exp − 1 − min NTU C max C min
Lim ε = Lim
C min → C max
→
=
ε =
=
C max
ε =
=1
C min C min C min 1 exp 1 exp 1 − − − − − − ( ) NTU NTU C max C max C max C min
− (− NTU )
C max
− NTU exp− 1 −
ε =
C min NTU C max
C min =1 C max
C C min − 1 exp − 1 − min NTU − NTU C max C max C min C max
ε =
NTU
C min + 1 NTU C max C min C max
ε =
=1
=1
NTU NTU + 1
Comparison Table: NTU 5 4 3 2
ε =
NTU
Fig. 3.16, ε
NTU + 1 0.83 0.8 0.75 0.67
0.83 0.8 0.75 0.67
54
=1
3. HEAT EXCHANGER DESIGN
3.38
The effectiveness of a cross-flow exchanger with neither fluid mixed can be calculated NTU 0.22 0.78 from the following approximate formula: ε = 1 − exp[exp(− NTU r ) − 1] r where r ≡
C min C max
. How does this compare with correct values?
Solution: Fig. 3.17a Comparison C r ≡ min = 1.0 C max NTU 5 4 3 2 1 r ≡
C min C max
ε (approximate)
ε (Fig.
3.17a) 0.75 0.72 0.68 0.62 0.47
0.749 0.723 0.684 0.615 0.469
= 0.75 NTU 5 4 3 2 1
ε (approximate)
ε (Fig.
3.17a) 0.83 0.800 0.75 0.67 0.51
0.828 0.800 0.755 0.675 0.505
Therefore it gives very near values to two digits. If r ≡
C min C max
=0
The term
[exp(− NTU
0.78
0.22
NTU
r ) − 1]
=
0 0
r
Use L’Hospital rule: ∂
exp(− NTU r ) − 1 NTU 0.78
Lim r → 0
r
0.22
=
{[exp(− NTU
0.78
∂r
∂
0.22
(r )
∂r
55
r ) − 1] NTU
r = 0
}
r = 0
3. HEAT EXCHANGER DESIGN 0.78
=
− NTU
exp(− NTU r )r =0 0.78
(1) r 0
= − NTU 0.78
=
Then 0.78 ε = 1 − exp( − NTU ) Comparison r ≡
C min C max
=0, ε = 1 − exp( − NTU 0.78 )
NTU 5 4 3 2 1
ε (approximate)
0.970 0.948 0.905 0.820 0.632
At lower values of r ≡
3.39
C min C max
or r ≡
C min C max
ε (Fig.
3.17a) 0.99 0.98 0.95 0.87 0.64
→ 0 , this will not give correct values accurately.
Calculate the area required in a two-tube pass, one-shell pass condenser that is to 6 2 condense 10 kg/h of steam at 40 C using water at 17 C. Assume that U = 4700 W/m .K, the maximum allowable temperature rise of the water is 10 C and h fg = 2406 kJ/kg.
Solution: − NTU ε = 1 − e UA NTU = C min Specific heat of water at 17 C, c pw = 4187 J/kg.K T h = 40 C T cin = 17 C T cout = 27 C 6
& s = 10 kg/h m 2
U = 4700 W/m .K h fg = 2406 kJ/kg
& s h fg Q = C min T cout − T cin = m
1 (2406) 3600
C min (27 − 17 ) = (10 ) 6
C min = 66,833 W/K
& s h fg Q = ε C min T hin − T cin = m 56
3. HEAT EXCHANGER DESIGN
ε (66,833)(40 − 17 ) =
(10 ) 6
1 (2406 ) 3600
= 0.4348 − NTU ε = 1 − e
ε
0.4348 = 1 − e − NTU NTU = 0.5706 UA NTU = C min 0.5706 =
(4700 )A 66,833
2
A = 8.114 m .
3.40
An engineer wants to divert 1 gal/min of water at 180 F from his car radiator through a small cross-flow heat exchanger with neither flow mixed, to heat 40 F water to 140 F for shaving when he goes camping. If he produces a pint per minute of hot water, what will be the area of the exchanger and the temperature of the returning radiator coolant if U = 2 720 W/m .K?
Solution: T hin = 180 F T cin = 40 F T cout = 140 F
Specific heat of water at 180 F, c ph = 1.003 Btu/lb-F Specific heat of water at 90 F, c pc = 0.998 Btu/lb-F 3
Density of water at 180 F, ρ h = 60.56 lb/ft . 3 Density of water at 140 F, ρ c = 61.37 lb/ft .
2
U = 720 W/m .K =
720
5.6786 2 U = 126.8 Btu/hr-ft -F
Btu/hr-ft2-F
&h Solving for m & h = ρ hV h m 3
3
3
V h = (1 gal/min)(231 in /gal)(1 ft / 1728 in )( 60 min/hr) 3
V h = 8.0208 ft /hr
& h = (60.56)(8.0208) lb/sec = 485.74 lb/sec m 57
3. HEAT EXCHANGER DESIGN
&c Solving for m & c = ρ cV c m 3
3
3
V c = (1 pint/min)(0.125 gal/min)(231 in /gal)(1 ft / 1728 in )( 60 min/hr) 3
V c = 1.0026 ft /hr
& c = (61.37)(1.0026) lb/sec = 61.53 lb/sec m & c c pc T cout − T cin Q=m Q = (61.53)(0.998)(140 – 40) = 6141 Btu/hr
& h c ph T hin − T hout Q=m 6141 = (485.74)(1.003)(180 - T hout ) T hout = 167.4 F LMTD =
T hin − T cout − T hout − T cin
T hin − T cout T h − T c in out
ln LMTD =
(180 − 140 ) − (167.4 − 40 )
P= R =
180 − 140 ln 167.4 − 40 T t out − T t in T sin − T t in
T sin − T sout T t out − T t in
=
=
140 − 40 180 − 40
180 − 167.4 140 − 40
= 75.45 F
= 0.71
= 0.126
Fig. 3.14c, Cross-flow exchanger, neither mixed, F = 0.965. Q = UAF ( LMTD ) 6141 = (126.8)( A)(0.965)(75.45) 2 A = 0.665 ft And T hout = 167.4 F
3.41
In a process for forming lead shot, molten droplets of lead are showered into the top of a tall tower. The droplets fall through air and solidify before they reach the bottom of the tower. The solid shot is collected at the bottom. To maintain a steady state, cool air is introduced at the bottom of the tower and warm air is withdrawn at the top. For a particular tower, the droplets are 1 mm in diameter and at their melting temperature of 600 K when they are released. The latent heat of solidification is 850 kJ/kg. They fall with a mass flow rate of 200 kg/hr. There are 2430 droplets per cubic meter of air inside 58
3. HEAT EXCHANGER DESIGN
the tower. Air enters the bottom at 20 C with a mass flow rate of 1100 kg/hr. The tower has an internal diameter of 1m with adiabatic walls. a. Sketch, qualitatively, the temperature distribution of the shot and the air along the height of the tower. b. If it is desired to remove the shot at a temperature of 60 C, what will be the temperature of the air leaving the top of the tower? c. Determine the air temperature at the point where the lead had just finishing solidifying. d. Determine the height that the tower must have in order to function as desired. The 2 heat transfer coefficient between the air and the droplets is h = 318 W/m .K Solution: (a)
(b) Specific heat of lead shot = c pl = 130 J/kg.K Specific heat of air = c pa = 1008 J/kg.K
& l = 200 kg/hr m ∆h = 850 kJ/kg = 850,000 J/kg
& l ∆h + c pl T hin − T hout Q=m T hin = 600 K = 327 C T hout = 60 C
1 [850,000 + 130(327 − 60)]= 49,151 W 3600
Q = (200)
& a = 1100 kg/hr m T cin = 20 C
& a c pa T cout − T cin Q=m 59
3. HEAT EXCHANGER DESIGN
1 (1008 )(T cout − 20) 3600
49,151 = (1100 ) T cout = 179.6 C
(c) Cooling required after solidification
& l c pl T hin − T hout Q2 = m
1 (130)(327 − 60) = 1928.33 W 3600
Q2 = (200 )
& a c pa T cout 2 − T cin Q2 = m
1 (1008 )(T cout 2 3600
1928.33 = (1100 )
− 20
)
T cout 2 = 26.3 C
2 (d) h = 318 W/m .K
Q = UA( LMTD ) 2
U = h = 318 W/m .K Solidification: Q1 = 49,151 – 1928.33 = 47,233 W LMTD1 =
T hin − T cout 2 − T hin − T cout 1
T hin − T cout 2 T h − T c out 1 in
ln LMTD1 =
(327 − 26.3) − (327 − 179.6 ) 327 − 26.3 ln 327 − 179.6
= 215 C
Q1 = UA 1(LMTD1 )
47,223 = (318)( A 1 )(215) 2
A 1 = 0.691 m .
Cooling, Q2 = 1928.33 W LMTD2 =
T hin − T cout 2 − T hout − T cin
T hin − T cout 2 T h − T c in out
ln LMTD2 =
(327 − 26.3) − (60 − 20 ) 327 − 26.3 ln 60 − 20
= 129.24 C
60
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