Heat Exchanger Design -IX-mod - 中興大學-機械工程學系

Principles of Heat Exchanger Design (ix)
----- Design Examples Using
the LMTD Method
沈

: Heat Transfer with Applications by Kirk D. Hagen
理
• (1). 理
--- (i) (ii) Fourier’s law (iii) 流 Newton’s law of cooling (iv) 輻 Stefan-
Boltzmann law (v) 例 說
• (2)
• --- (i) 串 聯 念 - (ii) 串 聯 念 - (iii) 例 說 (iv)
• (3) 理
--- (i) (ii) 理 論 (iii) 狀 理 論 (iv)
(v) 狀
• (4)
• (5) 流 流
-- (i) 流 流 (ii) 流 流 雷 諾 數 (iii) 數 (C f,x
) —friction coefficient
(iv) 念 流 數 (v) 紐 塞 數 朗 數 Reynolds Analogy and Chilton-Colburn
Analogy (vii) (viii) 量 (ix) (Tube Bank)
• (6) 流 流
-- (i) 流 流 雷 諾 數 (ii) 流 流 流 (iii) Darcy 數 -
--friction factor (iv) 流 數 紐 塞 數 (v)
• (7) 冷 流
• (8) --- LMTD 理 論
• (9) LMTD 例 說
• (10) LMTD 例 說
• (11) --- Ntu 理 論
• (12) Ntu 例 說

(9) LMTD
例 說

例 1: 冷 冷
(Shell and Tube Heat Exchanger)
例 2: 更 例 1 冷 --- 冷 度 35 0 C 45 0 C
例 3: 良 例 2 冷 --- 數
例 4: 冷 (shell-and-tube type)
------ Oil on Tube Side
例 5: 冷 (shell-and-tube type)
------- Oil on Shell Side (3/8” steel pipe)
例 6: 冷 (shell-and-tube type)
------- 例 5 更 (3/4” steel pipe)
例 7: 冷

例 1: 冷 冷
(Shell and Tube Heat Exchanger)
冷 冷 (water-cooled condenser) 冷
路 (two-tube-passes shell-and-tube condenser ()
冷 ( 量 ) 78.3 kW, 冷 度 35 0 C 冷 R-22
(i) 冷 流 (m/s)
(ii) 冷 ( 見 ) 度 (m)
冷
35 0 C (sat. vapor)
冷
25 0 C
Water cooled condenser
Two-passes shell-and-tube type
(L)
冷
35 0 C (sat. liquid)
冷
30 0 C

: refrigerant (R-22, 35 0 C): k = 0.083 (W/m-K), µ = 0.000188(pa-s)
water(28 0 C): ρ w
= 996 (kg/m 3 ), µ w
= 0.00078 (pa-s), k w
= 0.61 (W/m-K),
c p,w =4190 (J/kg-K),
F i = fouling factor (water to copper)=0.000176 (m 2 -K/W)
copper: k c = 390 W/m-K

LMTD = ∆T
∆T − ∆T (35 − 25) − (35 − 30)
=
ln( )
∆T
冷
1 2
0
lm
= =
7. 21
∆T1
ln[(35 − 25) /(35 − 30)]
2
冷 量 流 率 = 冷 量 / (c p,w
∆T)
= 78.3 kW / [4.19 (kJ/kg-K) x (30-25)(K) = 3.737 (kg/s)
冷 流 率 = 量 流 率 / 度 = 0.003752 (m 3 /s) = 225.12 (L/min)
冷 流 (V) = 流 率 / [22 () x ] = 1.1 (m/s)
C
• Dittus-Bolter Equation: Nu d =0.023 Re 0.8 Pr 0.4 (Turbulent Flow)
Re = ρ w
Vd i
/ µ w
= 996 (kg/m 3 ) x 1.1 (m/s) x 0.014 (m) / 0.00078 (pa-s) = 19664
Pr = c p,w
µ w
/ k w
= 4190 (J/kg-K) x 0.00078 (pa-s) / 0.61 (W/k-K) = 5.35
h i = (k w / d i )(0.023) Re 0.8 Pr 0.4 = 5338 (W/m 2 -K)

冷
!
44 tubes
∆ ≈ = = = =
13 rows
2 3
gρ hfgk
1/4
hcond
= 0.725 [ ]
µ(∆T)Nd
2
assume T 5K ; g 9.81 (m/s ) ; N 3.38 tubes/row ; do
0.016 m
o
9.81 (m/s ) × (1153) (kg/m ) × 172513 (J/kg) × (0.083) (W/m −K)
= 0.725[
3
0.000188 (J − s/m ) × 5 (K) × (3.38) × 0.016 (m)
= −
2
1626 (W/m K)
2 2 3 2 3 3
: 1 (J) = 1 (N-m) ; 1 (N) =1 (kg-m/s 2 ) ; 1 (W) = 1 (J/s)
]
1/4
1
U
o
=
1
h
o
流
+
F
o
+
do
(
2k
c
d
)ln(
d
o
i
)
+
d
(
d
o
i
)F
i
+
d
(
d
o
i
)(
1
h
流
i
)

1
U
o
=
Hence,
1
=
1626
= 0.000615
=
1
h
o
+
m
0.0010328 (
U
o
F
+
o
0
+
+
+
do
(
2k
0
= 968 (
m
d
)ln(
d
0.016 0.016
[ ]ln( )
(2)(390) 0.014
0.0000027
− K
)
W
W
)
2
- K
2
c
+
o
i
)
+
d
(
d
+
o
i
)F
i
+
+
d
(
d
o
0.000201
i
)(
1
h
i
)
0.016
( )(0.000176)
0.014
+
0.000214
+
0.016
( )(
0.014
1
5338
流
)
Q = U o A o (LMTD)
Hence A o = Q / [U o (LMTD)] = 78300 (W) / [968 (W/m 2 -K) x 7.21(K)]
= = 11.2 (m 2 )
度 = A o / [ 數 x ]
= 5.07 (m) !

∆T 5 0 C
Q = 78.3 kW T B
T water
T A
=35 0 C 1/h o
F o
(d o
/2k)ln(d o
/d i
) (d o
/d i
)(F i
) (d o
/d i
)(1/h i
)
略
Q = h o A ∆T = h o A (T A –T B )
度
∆T = Q / (h o A)
= 78300 (W) / [11.2 (m 2 ) x 1626 (W/m-K)]
= 4.3 0 C
: ∆T
不

Pressure Drop (∆p, 流 力 降 )
2
∆p
G v1
2 v2
A vm
2 v
= ( )( )[( K
c
+ 1−σ
) + 2( −1)
+ f ' − (1 −σ
− K
e
)
p1 2 p
v A v
v
1
(i)
v 2 ≈ v 1
σ = free-flow area / frontal area = 率 = 0.3
K c = 0.35 ; K e =0.48 ; σ 2 = 0.09
1
(ii) Flow
acceleration
Fanning friction factor
c
1
(iii) Core friction
~1.0
(iv)
~1.0
2
1
]
∆p
p
1
G
= (
2
2
v
)(
p
( ρV
)
= [
2
( ρV
)
= [
2
2
2
1
1
)[( K
v
](
p
1
v
](
p
1
1
1
)
c
2
+ 1−σ
) +
[0.83
)[0.83
+
+
f
f
'
f
A
A
c
L'
]
D
'
]
A
A
c
=
2
− (1 −σ
−
( ρV
)
[
2
2
K
v
](
p
1
1
e
)]
)[0.83
+
f
'
4L'
]
D

L’ = 2 L = 10.14 (m) ; D = = d i = 14 (mm) = 0.014 (m)
V = 1.1 (m/s) ; ρ = 996 kg/m 3
Re = 19664 ; ε = 0.0015 mm f = 0.0265
∆p
2
( ρV
)
L'
= [ ]( v1
)[0.83 + f ]
2
D
2
ρV
L'
不 !
= [ ][0.83 + f ]
2
D
3
2
996(kg/m ) × [1.1( m / s)]
=
[0.83+
(0.0265)(724.3)]
2
2
= 12066 (kg/m -s ) = 12066 (pa) = 12.07 (kpa)
(Pump)
!

Contraction Loss Coefficient
(K c ) and Expansion Loss
Coefficient (K e )

論 :
冷 度 度
列 :
(i) 數
(ii) 例

: 冷 :
h fg ≈ 2400 (kJ/kg)
(100 0 C) --- k w ≈ 0.68 (kg/m-K) ; µ w = 0.00028 (pa-s)
h cond ( h o ) !
2
assume T 40 K ; g 9.81 (m/s ) ; N 30 tubes/row ; do
0.0158 m
T
cond
∆ ≈ = = =
= 580 K
gρ h k
hcond
= 0.725 [ ]
µ(∆T)Nd
2 3
fg 1/4
o
9.81 (m/s ) × (699) (kg/m ) × 1353000 (J/kg) × (0.594) (W/m −K)
= 0.725[
3
0.000088 (J − s/m ) × 40 (K) × (30) × 0.0158 (m)
= −
2
3873 (W/m K)
2 2 3 2 3 3
]
1/4

例 2: 更 例 1 冷
--- 冷 度 35 0 C45 0 C
冷
45 0 C (sat. vapor)
冷
25 0 C
Water cooled condenser
Two-passes shell-and-tube type
(L)
冷
45 0 C (sat. liquid)
冷
30 0 C

力 , g
: refrigerant (R-22, 45 0 C): k = 0.078 (W/m-K), µ = 0.000177(pa-s)
h fg = 160912 (J/kg), ρ =1109 (kg/m 3 )
water(28 0 C): ρ w
= 996 (kg/m 3 ), µ w
= 0.00078 (pa-s), k w
= 0.61 (W/m-K),
c p,w =4190 (J/kg-K),
F i = fouling factor (water to copper)=0.000176 (m 2 -K/W)
copper: k c = 390 W/m-K

LMTD = ∆T
∆T − ∆T (45 − 25) − (45 − 30)
=
ln( )
∆T
冷
1 2
0
lm
= =
17. 38
∆T1
ln[(45 − 25) /(45 − 30)]
2
冷 量 流 率 = 冷 量 / (c p,w
∆T)
= 78.3 kW / [4.19 (kJ/kg-K) x (30-25)(K) = 3.737 (kg/s)
冷 流 率 = 量 流 率 / 度 = 0.003752 (m 3 /s) = 225.12 (L/min)
冷 流 (V) = 流 率 / [22 () x ] = 1.1 (m/s)
C
• Dittus-Bolter Equation: Nu d =0.023 Re 0.8 Pr 0.4 (Turbulent Flow)
Re = ρ w
Vd i
/ µ w
= 996 (kg/m 3 ) x 1.1 (m/s) x 0.014 (m) / 0.00078 (pa-s) = 19664
Pr = c p,w
µ w
/ k w
= 4190 (J/kg-K) x 0.00078 (pa-s) / 0.61 (W/k-K) = 5.35
h i = (k w / d i )(0.023) Re 0.8 Pr 0.4 = 5338 (W/m 2 -K)

冷
!
2
assume ∆T ≈ 12K ; g = 9.81 (m/s ) ; N = = 3.38 tubes/row ; do
= 0.016 m
gρ h k
hcond
= 0.725 [ ]
µ(∆T)Nd
2 3
fg 1/4
o
44 tubes
13 rows
9.81 (m/s ) × (1109) (kg/m ) × 160912 (J/kg) × (0.078) (W/m −K)
= 0.725[
3
0.000177 (J − s/m ) × 12 (K) × (3.38) × 0.016 (m)
= −
2
1220 (W/m K)
2 2 3 2 3 3
]
1/4
1
U
o
=
1
h
o
流
+
F
o
+
do
(
2k
c
d
)ln(
d
o
i
)
+
d
(
d
o
i
)F
i
+
d
(
d
o
i
)(
1
h
流
i
)

1
U
o
=
1
h
o
+
1
=
1220
F
+
= 0.0008196
2
m − K
= 0.001237 ( )
W
W
Hence, Uo
= 808 ( )
2
m - K
o
0
+
+
do
(
2k
+
0.016 0.016
[ ]ln( )
(2)(390) 0.014
0
c
d
)ln(
d
+
o
i
)
+
d
(
d
0.0000027
o
i
+
)F
i
+
+
d
(
d
o
)(
0.016
( )(0.000176)
0.014
0.000201
i
1
h
+
i
)
0.000214
+
0.016
( )(
0.014
1
5338
流
)
Q = U o A o (LMTD)
Hence A o = Q / [U o (LMTD)] = 78300 (W) / [808 (W/m 2 -K) x 17.38 (K)]
= = 5.57 (m 2 )
度 = A o / [ 數 x ]
= 2.52 (m)

∆T 12 0 C
Q = 78.3 kW T B Q = 78.3 kW
T water
T A
=45 0 C 1/h o
F o
(d o
/2k)ln(d o
/d i
) (d o
/d i
)(F i
) (d o
/d i
)(1/h i
)
略
Q = h o A ∆T = h o A (T A –T B )
度
∆T = Q / (h o A)
= 78300 (W) / [5.57 (m 2 ) x 1220 (W/m-K)]
= 11.5 0 C
: ∆T
不

論 :
• 冷 度 35 0 C45 0 C 冷
冷 度 度
冷 量

例 3: 良 例 2 冷 --- 數
圗 冷 ( 量 ) 78.3 kW, 冷 度 45 0 C
冷 R-22
(i) 冷 流 (m/s)
(ii) 冷 度 (m)
冷
45 0 C (sat. vapor)
冷
25 0 C
Water cooled condenser
Two-passes shell-and-tube type
(L)
冷
45 0 C (sat. liquid)
冷
30 0 C

d i = 14 mm
d o = 16 mm
74
冷
力 , g
: refrigerant (R-22, 45 0 C): k = 0.078 (W/m-K), µ = 0.000177(pa-s)
h fg = 160912 (J/kg)
water(28 0 C): ρ w
= 996 (kg/m 3 ), µ w
= 0.00078 (pa-s), k w
= 0.61 (W/m-K),
c p,w = 4190 (J/kg-K),
F i = fouling factor (water to copper) = 0.000176 (m 2 -K/W)
copper: k c = 390 W/m-K

冷
LMTD = ∆T
∆T − ∆T (45 − 25) − (45 − 30)
=
ln( )
∆T
1 2
0
lm
= =
17. 38
∆T1
ln[(45 − 25) /(45 − 30)]
2
冷 量 流 率 = 冷 量 / (c p,w
∆T)
= 78.3 kW / [4.19 (kJ/kg-K) x (30-25)(K) = 3.737 (kg/s)
冷 流 率 = 量 流 率 / 度 = 0.003752 (m 3 /s) = 225.12 (L/min)
流 (V) = 流 率 / [37 () x ] = 0.654 (m/s)
C
• Dittus-Bolter Equation: Nu d =0.023 Re 0.8 Pr 0.4 (Turbulent Flow)
Re = ρ w
Vd i
/ µ w
= 996 (kg/m 3 ) x 0.654 (m/s) x 0.014 (m) / 0.00078 (pa-s) = 11692
Pr = c p,w
µ w
/ k w
= 4190 (J/kg-K) x 0.00078 (pa-s) / 0.61 (W/k-K) = 5.35
h i = (k w / d i )(0.023) Re 0.8 Pr 0.4 = 3520 (W/m 2 -K)

冷
!
2
assume ∆T ≈ 11K ; g = 9.81 (m/s ) ; N = = 4.35 tubes/row ; do
= 0.016 m
gρ h k
hcond
= 0.725 [ ]
µ(∆T)Nd
2 3
fg 1/4
o
74 tubes
17 rows
9.81 (m/s ) × (1109) (kg/m ) × 160912 (J/kg) × (0.078) (W/m −K)
= 0.725[
3
0.000177 (J − s/m ) × 11 (K) × (4.35) × 0.016 (m)
= −
2
1171 (W/m K)
2 2 3 2 3 3
]
1/4
1
U
o
=
1
h
o
流
+
F
o
+
do
(
2k
c
d
)ln(
d
o
i
)
+
d
(
d
o
i
)F
i
+
d
(
d
o
i
)(
1
h
流
i
)

1 1 do do do do
1
= + F
o
+ ( )ln( ) + ( )F
i
+ ( )( )
Uo ho 2kc di di di hi
1 0.016 0.016 0.016 0.016 1
= + 0 + [ ]ln( ) + ( )(0.000176) + ( )( )
1171 (2)(390) 0.014 0.014 0.014 3520
= 0.0008542 + 0 + 0.0000027 + 0.000201 + 0.0003247
2
m − K
= 0.001383 ( )
W
W
Hence, U
o
= 723 ( )
2
m-K
流
Q = U o A o (LMTD)
Hence A o = Q / [U o (LMTD)] = 78300 (W) / [723 (W/m 2 -K) x17.38(K)]
= = 6.22 (m 2 )
度 = A o / [ 數 x ]
=6.22/(74 x 3.14 x 0.016)
= 1.68 (m)

∆T 11 0 C
Q = 78.3 kW T B Q = 78.3 kW
T water
T A
=45 0 C 1/h o
F o
(d o
/2k)ln(d o
/d i
) (d o
/d i
)(F i
) (d o
/d i
)(1/h i
)
略
Q = h o A ∆T = h o A (T A –T B )
度
∆T = Q / (h o A)
= 78300 (W) / [6.22 (m 2 ) x 1171 (W/m-K)]
= 10.8 0 C
: ∆T = 0.2 0 C

例 4: 冷 (shell-and-tube type)
------ Oil on Tube Side
Tube passes (n) 數
Water: one shell pass
Oil: n tube passes
Water, T 1
= 35 0 C
Water, T 2
= 40 0 C
L
.
20
Q = m
• oil coil∆Toil
= [( )(L/s) × (0.866)(kg/L)](2000)(J/kg - K)(55−
50)
60
= 2887 (W) = 2.887 (kW)
ρ oil = 866 (kg/m3)
c w = 4180 (J/kg-K)
oil flowrate = 20 (L/min)
Oil,
t 2
=50 0 C
Oil
t 1
=55 0 C

.
•
m
w
=
Q/(c
w
∆T
w
) = 2887 (W) / [(4180)(J/kg - K)(40 − 35)(K)]
3
= 0.138 (kg/s) = 0.000138 (m /s) = 0.138 (L/s) = 8.288 (L/min)
U i = 180 (W/m 2 -K)
(i.e. laminar flow)
For counterflow arrangement :
(55 − 35) − (50 − 40)
LMTD =
ln[(55−
35)/(50 − 40)]
=
0
14.4 C
p
t
2
=
T
1
− t
− t
1
1
From Fig.,
=
50
35
−
−
55
= 0.25
55
R
F = 0.99 (correction factor)
;
T1
=
t
2
− T
− t
2
1
=
35
50
−
−
40
55
= 1
Q = U i A i F (LMTD),
A i = Q / [U i F (LMTD)] = (2887) / [(180)(0.99)(14.4)] = 1.13 (m 2 )
Hence, A i = 1.13 (m 2 )

For oil :
k = 0.14 (W/m-K), ν = 9.6 x 10 -5 (m 2 -s), ρ = 866 (kg/m 3 )
c p = 2 (kJ/kg-K) , Pr = 1200 , µ = 0.0831 (N-s/m 2 )
• = d i =0.02 (m) ; d o = 0.026 (m)
laminar flow design
A i = (tube length) (π d i ) ; nL = 1.13 / [(3.14)(0.02)] = 18 (m)
若 L = 1(m) , tube passes = n = 18
U i :
1 m (for one tube pass)
tube pass
度 度
Re =
Gz
−1
•
4moil
πd
iµ
=
(x/di)
=
Re Pr
(4)(0.2887)
(3.14)(0.02)(0.0831)
=
(1/0.02)
(221)(1200)
=
=
221
0.00018
laminar flow
Nu m = 26.0

略 !
略 !
略 !
1
U
i
=
d
(
d
i
o
)(
1
h
o
)
+
d
(
d
i
o
)F
o
+
di
(
2k
c
d
)ln(
d
o
i
)
+
F
i
+
1
h
i
!
流 度 h 0 > > h i 略 !
h m = Nu m k / d i = (26)(0.14) / (0.02) = 182 (W/m 2 -K)
= h i
U i ≈ h i !

Mean Nusselt Number for Thermally and
Hydrodynamically Developing Laminar Flow
Constant Wall Temperature – Circular Tube

Pressure Drop (∆p, 力 降 )
Re = 221 ; laminar flow ;
; f = 64/Re = 0.29
V = volumetric flowrate / cross sectional area
= 0.000138 (m 3 /s) / [π(0.02) 2 /4](m 2 )
= 0.44 (m/s)
∆p
2
ρV
L'
= [ ][ f ]
2 D
3
2
866(kg/m ) × [0.44(m/s)]
=
[(0.29)(18 / 0.02)]
2
2
= 21828 (kg/m -s ) = 21828(pa) = 21.828 (kpa)

Pressure Drop – friction plus flow acceleration
Re / (L’/D) = 0.245
∆p
=
4 c
f app
ρV
(
2
2
L'
)( )
D
=
f
ρV
(
2
2
L'
) ( )
D
≈
64 ρV
( )(
Re 2
2
L'
)( )
D

例 5: 冷 (shell-and-tube type)
------- Oil on Shell Side
冷
L / 9
L
T 2 =50 0 C
t 1 =30 0 C
t 2 =34 0 C
T 1 =80 0 C
0.15 m
3/8” steel pipe (40 schedule)
wall thickness = 3.2 mm
d i = 10.7 mm ; d o = 17.1 mm
S
S
T
D
= 25 (mm) ;
= 23.6 (mm)
S
L
= 20 (mm)
S D
S T
S L

•
moil
= 12 (kg/min) = 0.2 (kg/s)
= 13.02
(L/min) = 2.17 × 10
-4
3
(m /s)
量 ≡ Q
Q moil
cp,oil(T1
− T2
)
= 10800 (W)
=
0.2(kg/s) × 1800(J/kg
− K) × (80 − 50)(K)
For oil:
ν
oil
=
= • K)
流 理 :
2×
10
-5
(m
2
/s)
;
Pr
oil
= 100
;
k
oil
=
0.125 (W/m - K)
ρ oil
=
920 (kg/m
3
)
;
c
p,oil
= 1800 (kJ/kg -
For water :
ν
w
ρ w
= 6.5×
10
-7
(m
= 1000 (kg/m
3
2
)
/s)
;
;
c
Pr
p,w
w
= 4 ;
k
w
= 4180 (kJ/kg - K)
= 0.65 (W/m - K)

(i) Water Side (tube side) :
•
mwater
Q
10800 (W)
= = = 0.646 (kg/s)
c (t − t ) 4(K) × 4180 (J/kg-K)
p,w 2 1
•
•
mwater
−4 3
water volumetric flowrate ( V ) = = 6.46×
10 (m /s)
ρ w
•
V / 11 (tubes)
flow velocity in a tube (U
i
) = = 0.653 (m/s)
2
( πd i
/4)
Ud
i i
Re = = 10756 (turbulent flow)
ν
0.8 0.4
Dittus-Boelter Equation: NuD
= 0.023ReD
Pr = 67.2
k
w
Hence, h
w
= ( )Nu
D
= 4085 (W/m2
−K)
di

行 度 L = 3.6 (m)
baffle 離 L / 9 = 0.4 (m)
(ii) Oil Side (shell side) : Tube bank
•
m
U
U
Re
From Table :
Nu
h
o
oil
∞
max
D
= ρU
=
=
m
0.0036 (m/s)
=
D,max
=
k
= (
d
(
S
=
U
ν
F C Re
oil
o
∞
)
A
•
oil/(ρoil
T
ST
− d
max
Nu
oil
A)
d
)
U
9.8
C = 1.04;
m
D,max
D
o
o
=
=
0.2(kg/s)/[(920)(kg/m
∞
Pr
=
n
(
;
m
Pr
Pr
)
= 88.5 (W/m
0.0115 (m/s)
=
w
0.4;
1/4
略
2
n
= 12.1
- K)
=
3
0.36;
) × (0.15)(m) × 0.4(m)]
F
=
0.89

1
U
o
=
1
h
o
+
1
=
88.5
+
= 0.0113
F
do
(
2k
0.0002
0.0002
d
)ln(
d
2
m − K
= 0.01227 ( )
W
W
Hence, Uo
= 82 ( )
2
m - K
o
+
+
c
+
o
i
)
0.0171 17.1
[ ]ln( )
(2)(40) 10.7
+
+
0.0001
d
(
d
+
o
i
)F
i
+
+
0.00028
d
(
d
o
i
)(
1
h
i
)
17.1
( )(0.000176)
10.7
+
0.000391
+
17.1
( )(
10.7
1
4085
)
A o = π x d o x L x (22) = 1.182 L
LMTD =
(80 − 34) − (50 − 30)
80 − 34
ln( )
50 − 30
=
31.2
0
C

:
R
=
T
t
1
2
−T
− t
2
1
=
80
34
−
−
50
30
Correction Factor (F')
=
=
7.5
0.98
;
P
=
t
T
2
1
−
−
t
t
1
1
=
34 -30
80 -30
=
0.08
Q = F’ U o A o (LMTD)
= (0.98)(82)(1.182 L)(31.2)
Hence, L = 3.64 (m) 來 不 !
= A o
= π x d o x L x (22) = 1.182 L = 4.31 (m 2 )

例 6: 冷 (shell-and-tube type)
------- 例 5 更 (3/4” steel pipe)
冷
L / 9
L
T 2 =50 0 C
t 1 =30 0 C
t 2 =34 0 C
T 1 =80 0 C
0.20 m
3/4” steel pipe (40 schedule)
wall thickness = 2.9 mm
d i = 20.9 mm ; d o = 26.7 mm
S L
S
S
T
D
= 32 (mm)
= 34 (mm)
;
S
L
=
30 (mm)
S D
S T

•
moil
= 12 (kg/min) = 0.2 (kg/s)
= 13.02
(L/min) = 2.17 × 10
-4
3
(m /s)
量 ≡ Q
Q moil
cp,oil(T1
− T2
)
= 10800 (W)
=
0.2(kg/s) × 1800(J/kg
− K) × (80 − 50)(K)
For oil:
ν
oil
=
= • K)
流 理 :
2×
10
-5
(m
2
/s)
;
Pr
oil
= 100
;
k
oil
=
0.125 (W/m - K)
ρ oil
=
920 (kg/m
3
)
;
c
p,oil
= 1800 (kJ/kg -
For water :
ν
w
ρ w
= 6.5×
10
-7
(m
= 1000 (kg/m
3
2
)
/s)
;
;
c
Pr
p,w
w
= 4 ;
k
w
= 4180 (kJ/kg - K)
= 0.65 (W/m - K)

(i) Water Side (tube side) :
Q
10800 (W)
•
mwater
= = = 0.646 (kg/s)
c
p,w
(t
2
− t
1) 4(K) × 4180 (J/kg-K)
• •
−4 3
water volumetric flowrate ( V ) = m water/ ρw
= 6.46×
10 (m /s)
•
V / 15 (tubes)
flow velocity in a tube (U
i) = = 0.126 (m/s)
2
( πd i
/4)
Ud
i i
Re = = 4040 (transitional flow)
ν
Gnielinski Equation: f ≈ 0.04 (smooth pipe)
(f/8)(Re −1000)Pr D 2/3
Nu
D
= [1 + ( ) ] = 25.6
1/2 2/3
1+ 12.7(f/8) (Pr −1) L
>>
Hence, h
k
= ( )Nu = 798 (W/m − K) = h
2/3
Usually L D, hence (D/L) can be neglected
w
w D 2 i
di
!

行 度 L = 1.35 (m)
baffle 離 L / 9 = 0.15 (m)
(ii) Oil Side (shell side) : Tube bank
•
moil
U
U
Re
From Table :
Nu
h
o
∞
max
D
= ρU
=
=
m
0.00725 (m/s)
=
D,max
=
k
= (
d
(
S
=
U
ν
F C Re
oil
o
∞
)
A
•
oil/(ρoil
T
ST
− d
max
Nu
oil
A)
d
)
U
58
C = 1.04;
m
D,max
D
o
o
=
=
0.2(kg/s)/[(920)(kg/m
∞
Pr
=
n
;
(
略
m
Pr
Pr
= 123 (W/m
0.0438 (m/s)
=
w
0.4;
)
2
1/4
- K)
n
=
= 26.1
3
0.36;
) × (0.15)(m) × 0.2(m)]
F
=
0.94

1
U
o
=
1
h
o
1
=
123
+
+
F
o
= 0.00815
do
(
2k
0.0002
0.0002
d
)ln(
d
2
m − K
= 0.01025 ( )
W
W
Hence, Uo
= 98 ( )
2
m - K
+
+
c
+
o
i
)
0.0267 26.7
[ ]ln( )
(2)(40) 20.9
+
+
(8×
10
d
(
d
-5
)
o
i
)F
+
i
+
+
d
(
d
o
0.00022
i
)(
1
h
i
)
26.7
( )(0.000176)
20.9
+
0.0016
+
26.7
( )(
20.9
1
798
)
A o = π x d o x L x (30) = 2.52 L
LMTD =
(80 − 34) − (50 − 30)
80 − 34
ln( )
50 − 30
=
31.2
0
C

:
80 − 50
R = = 7.5 ;
34 − 30
Correction Factor (F')
34 -30
P =
80 -30
= 0.98
=
0.08
Q = F’ U o A o (LMTD)
= (0.98)(98)(2.52 L)(31.2)
Hence, L = 1.44 (m)
來 1.35 m
不 !
= A o
= p x d o x L x (30) = 2.52 L = 3.63 (m 2 )

例 7: 冷 Single-Pass
L / 15
冷
T 2 = 8 0 C
R-22 冷
(x=100%)
t 1 =5 0 C
t 2 =5 0 C
T 1 =16 0 C
冷
L
R-22
(x=0.21)
冷
= 98 (kW)
: d o = 19 (mm) ; d i = 17 (mm)
wall thickness = 1.0 (mm) ; k c = 390 W/m-K

0.29 m
68
S
S
S
T
L
D
= 27 (mm)
= 25 (mm)
= 28.4 (mm)
S D
S L
S T
: refrigerant (R-22, 5 0 C): k f = 0.1 (W/m-K), µ f = 0.00023 (pa-s)
k g = 0.00945 (W/m-K), µ g = 0.0000122 (pa-s)
h f = 205.899 (kJ/kg) h g = 407.143 (kJ/kg)
water(10 0 C): ρ w
= 1000 (kg/m 3 ), µ w
= 0.001307 (pa-s),
k w
= 0.58 (W/m-K), c p,w =4194 (J/kg-K), Pr =9.45
F o = fouling factor (water to copper)=0.000176 (m 2 -K/W)

Refrigerant:
•
Q = m ref [(1−x)h fg
]
Q
98(kW)
•
m ref = = = 0.616 (kg/s)
(1−x)h fg
(1−0.21)(407.143 −205.899)(kJ/kg)
•
4(m
ref
/ tube number) 4(0.616/68)(kg/s)
Re = = = 2952
π d
i
µ
w
π (0.017)(m)(0.00023)(pa-s)
Gnielinski Equation:
f ≈ 0.04 (smooth pipe) ; Pr
in
= (0.21) (0.92)+(0.79)(2.49) =2.16
Pr =Pr =0.92 ; Pr =(Pr +Pr )/2=1.54 ;
out g m in out
k
in
=(0.21)(0.00945)+(0.79)(0.1)= 0.081 (W/m-K) ;
k = k = 0.00945 (W/m-K) ; k =(k +k )/2=0.0452
out
g
m in out
(f/8)(Re −1000)Pr D
= + =
1+ 12.7(f/8) (Pr −1) L
2/3
Nu
D
[1 ( ) ] 11.6
1/2 2/3
>>
k
= = − = hc
2/3
Usually L D, hence (D/L) can be neglected !
m
2
Hence, h ( )Nu
D
30.7 (W/m K)
di

h
b
(
1
∆T
e
1 ⎪⎧
= ( ) ⎨
5 ⎪⎩
q
⎛
)
⎜
⎜
µ
⎝
f
w
=
( 0.00023)
µ
f
g(ρ
i
fg
− ρ
σ
(201244)
= (0.2)(49143)(0.683) =
i
fg
f
g(ρ
g
f
− ρ
σ
) ⎡ c
⎢
⎢⎣
C
6712
p,f
sf
i
g
)
∆T
fg
pr
e
n
f
(W/m
⎛ c
⎜
⎝ C
i fg = 201244 (J/kg) ; µ f = 0.00023 (pa-s) ; σ = 10.8 x 10 -3 (N/m )
Pr f = 2.49 ; c p,f = 1170 (J/kg-K) ; ρ f - ρ g = 1266 – 25 = 1241(kg/m 3 )
g = 9.81 (m/s 2 ) ; ∆T e ≈ 5 (K) ;
C sf ≈ 0.007 ; n ≈ 1.7
=
⎤
⎥
⎥⎦
h 6712
p,f
sf
i
∆T
fg
− K)
Pr
e
n
f
⎞
⎟
⎠
3
≡
h
b
∆T
(9.81)(1241) ⎡ (1170)(5)
−3
10.8×
10
⎢
⎣(0.007)(201244)(2.49)
b n 1/n 2 1/2 2
h
i
= h
c[1 + ( ) ] = (30.7)[1 + ( ) ] ≈ 6712 (W/m −K)
hc
30.7
2
3
⎞
⎟
⎟
⎠
e
1.7
⎤
⎥
⎦
3
⎪⎫
⎬
⎪⎭

Water:
• •
Q= mw
c
p,w
(T1 − T
2) = m w (kg/s) × 4194(J/kg − K) × (16 − 8)(K) = 98000 (W)
•
-3 3
m w = 2.921 (kg/s) = 2.921 × 10 (m /s) = 2.921 (L/s) = 175.3 (L/min)
行 度 L = 2.7 (m) baffle
離 L / 15 = 0.18 (m)
•
w
m
= ρU A
∞
•
3
U∞
= m w/(ρwA) = 2.921(kg/s)/[(1000)(kg/m ) × (0.29)(m) × 0.18(m)]
= 0.056 (m/s)
S
U d U d
= = = = =
T
max o max o
U
max
( ) U
∞
0.189 (m/s) ; ReD,max
2747
ST − do ν
w
µ
w
/ ρw
From Table: C = 0.35( S T
/S
L) = 0.378; m = 0.6; n = 0.36; F = 0.97
Pr
m n 1/4
NuD
= F C Re
D,max
Pr ( ) = (0.97)(0.378)(115.7)(2.24) = 95
Prw
k
w
2
h
o
= ( ) Nu
D
= 2901 (W/m -K)
do

1 1 d d d d 1
= + F + ( )ln( ) + ( )F + ( )( )
U h 2k d d d h
o o o o
o
i
o o c i i i i
1 0.019 19 19 1
= + 0.000176 + [ ]ln( ) + 0 + ( )( )
2901 (2)(390) 17 17 6712
= + + × + +
-6
0.0003447 0.000176 (2.7 10 ) 0 0.0001665
2
m − K
= 0.0006899 ( )
W
W
Hence, U
o
= 1450 ( )
2
m-K
LMTD =
(16 − 5) − (8 − 5)
16 − 5
ln( )
8 − 5
=
6.16
0
C

Q = U o A o (LMTD)
Hence A o = Q / [U o (LMTD)]
= 98000 (W) / [1450 (W/m 2 -K) x 6.16 (K)]
= = 10.97 (m 2 )
度 = A o / [ 數 x ]
= 10.97 /[(68)(3.14)(0.019)]
= 2.704 (m)
來 2.7 m !
度
度