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Double Pipe Heat Exchanger Project #3 Calculations
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Transcript of Double Pipe Heat Exchanger Project #3 Calculations
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5/26/2018 Double Pipe Heat Exchanger Project #3 Calculations
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Design Requirements: Properties of Kerosene (In
1) Heats 2.5 kg/s of kerosene from 25 C to 35 C m.k= 2.50
2) Schedule 40 Steel Pipe .k = 730
3) Water at 90 C available to heat the kerosene kf.k = 0.132
4) Maximum pressure drop in each pipe is 70 kPa v.k = 5.47945E-07
.k = 0.0004
Schedule 40 Steel Pipe: Cp.k = 2470
Acceptable Pipe Sizes Pr.k = 7.48
ND.t = 2 ND.a = 3 .k = 7.32073E-08
ID.t = 0.05252 m ID.a = 0.07792 m t1 = 25
OD.t = 0.06034 m OD.a = 0.0889 m t2 = 35
A.t = 0.002166 m A.a = 0.004769 m Rd= 0.0002
Hydraulic Diameter, Dh= 0.01758 m .k = 7.32073E-08
Effective Diameter, De = 0.040282 m Pr.k = 7.48
Assumptions:
1) The double pipe heat exchanger is operating in counterflow.
2) The properties of Kerosense are taken from problem 43 on page 443 in the textbook
3)
q=(m.k)(Cp.k)(t2-t1)
q = 61750 J/s
Economic Velocity Range for Kerosene: 1.4 m/s 2.8 m/s
Mass Flow Rates for Velocity Range: 2.214 kg/s 4.428 kg/s
Economic Velocity Range for Water: 1.4 m/s 2.8 m/s
Flow Areas
At= 0.002166403 m
Aa= 0.001908997 m
Fluid Velocities
Vt= 1.58 m/s
Va= 2.20 m/s
Reynolds Numbers
Ret= 151518.4
Rea= 269362.4
Assuming we are operating under the assumption that there are no heat losses inside the heat exchanger, is
warmer fluid must be equal to the heat gained by the cooler fluid?
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Nusselt Numbers
Nut= 717.42
Nua= 630.55
Convection Coefficients
hi= 1803.112 W/m*K
ha= 10550.395 W/m*K
ht= 1569.430 W/m*K
Exchanger Coefficient
U0= 1366.20 W/m*K
Outlet Temperature Calculations
R = 0.350
A0= 1.148 m L = 6.06 m
Ecounter= 0.848T2= 86.50 C T2=
86.50 C
t2= 35 C
Log Mean Temperature Difference
LMTD = 58.19 C t1= 55 C LMTD =
t2= 61.50426 C
Heat Balance
qw= 61750
qc= 61750
q =
U = 924.3
Friction Factors
= 0.000046 m
/IDt= 0.000875857
/Dh= 0.00261661
Re.a = 117556.231
ft= 0.021
fa= 0.027
Pressure Drop:
pt= 2219.89 Pa 2.2 kPa
pa= 24630.23 Pa 24.6 kPa
Effectiveness
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N = 0.254
C = 0.350
E = 0.216
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Dh
er Tube): Properties of Water (Annular Tube):
kg/s m.w= 4.20 kg/s p.max = 70000 Pa
kg/m .w = 1000 kg/m L = 6.056641 m
W/m*K kf.w = 0.674 W/m*K L = 0.001
m/s v.w = 3.29E-07 m/s
N*s/m .w = 0.000329 N*s/m
J/kg*K Cp.w = 4206 J/kg*K C = 0.349574
Pr.w = 2.05
m/s .w = 1.66E-07 m/s
C T1 = 90 C
C T2 = 86.50 C
m*K/W Rd= 0.00015 m*K/W
m/s .w = 1.6E-07 m/s
Pr.w = 2.05
it possible to assume that the that the heat lost by the
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58.19 C
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Design Requirements: Properties of Kerosene (In
1) Heats 2.5 kg/s of kerosene from 25 C to 35 C m.k= 2.50
2) Schedule 40 Steel Pipe .k = 730
3) Water at 90 C available to heat the kerosene kf.k = 0.132
4) Maximum pressure drop in each pipe is 70 kPa v.k = 5.47945E-07
.k = 0.0004
Schedule 40 Steel Pipe: Cp.k = 2470
Acceptable Pipe Sizes Pr.k = 7.48
ND.t = 1.25 ND.a = 2 .k = 7.32073E-08
ID.t = 0.03504 m ID.a = 0.05252 m t1 = 25
OD.t = 0.04216 m OD.a = 0.06034 m t2 = 35
A.t = 0.0009643 m A.a = 0.002166 m Rd= 0.0002
Hydraulic Diameter, Dh= 0.01036 m .k = 7.32073E-08
Effective Diameter, De = 0.023266 m Pr.k = 7.48
Assumptions:
1) The double pipe heat exchanger is operating in counterflow.
2) The properties of Kerosense are taken from problem 43 on page 443 in the textbook
3)
q=(m.k)(Cp.k)(t2-t1)
q = 61750 J/s
Economic Velocity Range for Kerosene: 1.4 m/s 2.8 m/s
Mass Flow Rates for Velocity Range: 0.986 kg/s 1.971 kg/s
Economic Velocity Range for Water: 1.4 m/s 2.8 m/s
Flow Areas
At= 0.000964313 m
Aa= 0.000770385 m
Fluid Velocities
Vt= 3.55 m/s
Va= 2.20 m/s
Reynolds Numbers
Ret= 227104.7
Rea= 155576.6
Assuming we are operating under the assumption that there are no heat losses inside the heat exchanger, is
warmer fluid must be equal to the heat gained by the cooler fluid?
-
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Nusselt Numbers
Nut= 991.70
Nua= 406.45
Convection Coefficients
hi= 3735.870 W/m*K
ha= 11774.630 W/m*K
ht= 3104.954 W/m*K
Exchanger Coefficient
U0= 2457.04 W/m*K
Outlet Temperature Calculations
R = 0.866
A0= 0.840 m L = 6.34 m
Ecounter= 0.956T2= 81.34 C T2=
81.34 C
t2= 35 C
Log Mean Temperature Difference
LMTD = 55.67 C t1= 55 C LMTD =
t2= 56.33762 C
Heat Balance
qw= 61750 J/s
qc= 61750 J/s
q = 61750 J/s
U = 1321.0
Friction Factors
= 0.000046 m
/IDt= 0.001312785
/Dh= 0.004440154
Re.a = 69276.59574
ft= 0.022
fa= 0.031
Pressure Drop:
pt= 18499.86 Pa 18.5 kPa
pa= 48393.20 Pa 48.4 kPa
Effectiveness
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N = 0.334
C = 0.866
E = 0.255
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Dh
er Tube): Properties of Water (Annular Tube):
kg/s m.w= 1.69 kg/s p.max = 70000 Pa
kg/m .w = 1000 kg/m L = 6.339982 m
W/m*K kf.w = 0.674 W/m*K L = 0.001
m/s v.w = 3.29E-07 m/s
N*s/m .w = 0.000329 N*s/m
J/kg*K Cp.w = 4206 J/kg*K C = 0.866238
Pr.w = 2.05
m/s .w = 1.66E-07 m/s
C T1 = 90 C
C T2 = 81.34 C
m*K/W Rd= 0.00015 m*K/W
m/s .w = 1.6E-07 m/s
Pr.w = 2.05
it possible to assume that the that the heat lost by the
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55.67 C
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Property 3" x 2" units 1 2" x 1.25" units 2
IDp 0.05252 m 0.03504 m
ODp 0.06034 m 0.04216 m
IDa 0.07792 m 0.05252 m
L 6.06 m 6.34 m
t1 25 C 25 C
t2 35 C 35 C
T1 90 C 90 C
T2 86.50 C 81.34 C
U0 1366.20 W/m*K 2457.04 W/m*K
A0 1.148 m2
0.840 m2
pp 2.22 kPa 18.50 kPa
pa 24.63 kPa 48.39 kPa
E 0.216 0.255
N 0.254 0.334
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Nominal Diameter OD (cm) ID (cm) Area (cm) OD (m) ID (m) Area (m)
0.125 1.029 0.683 0.3664 0.01029 0.00683 0.00003664
0.25 1.372 0.924 0.6706 0.01372 0.00924 0.00006706
0.375 1.714 1.252 1.233 0.01714 0.01252 0.0001233
0.5 2.134 1.58 1.961 0.02134 0.0158 0.0001961
0.75 2.667 2.093 3.441 0.02667 0.02093 0.00034411 3.34 2.664 5.574 0.0334 0.02664 0.0005574
1.25 4.216 3.504 9.643 0.04216 0.03504 0.0009643
1.5 4.826 4.09 13.13 0.04826 0.0409 0.001313
2 6.034 5.252 21.66 0.06034 0.05252 0.002166
2.5 7.303 6.271 30.89 0.07303 0.06271 0.003089
3 8.89 7.792 47.69 0.0889 0.07792 0.004769
3.5 10.16 9.012 63.79 0.1016 0.09012 0.006379
4 11.43 10.23 82.19 0.1143 0.1023 0.008219
5 14.13 12.82 129.1 0.1413 0.1282 0.01291
6 16.83 15.41 186.5 0.1683 0.1541 0.01865
8 21.91 20.27 322.7 0.2191 0.2027 0.03227
10 27.31 25.46 509.1 0.2731 0.2546 0.05091
12 32.39 30.33 722.5 0.3239 0.3033 0.07225