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Heat Transfer Rate from Effectiveness-NTU Method

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Water enters a parallel flow double-pipe heat exchanger at 15^{o}C, flowing at the rate of 1200 kg/hr. It is heated by oil (C_{p} =2000 J/kg.K), flowing at the rate of 500 kg/hr from an inlet temperature of 90^{o}C. For an area of 1 m^{2} and an overall heat transfer coefficient of 1,200 W/m^{2}.K determine the total heat transfer and the outlet temperatures of water and oil.

Calculations:

Since the outlet temperature of the fluids are not given, the problem can be best solved by ε-NTU method.

C_{min} = (500/3600) x 2000 = 277.78 W/^{o}C = C_{h}

C_{max} = (1200/3600) x 4184 = 1394.67 W/^{o}C = C_{c}

C = C_{min}/C_{max} = 277.78 / 1394.67 = 0.2

N = NTU = UA / C_{min} = 1200 x 1 / 277.78 = 4.32

For parallel flow, relation between effectiveness (ε) and NTU (N) is given by,

ε = {1 - exp[ -N(1 + C)]} / (1 + C) = {1 - exp[ - 4.32 x (1 + 0.2)]} / (1 + 0.2)

= 0.8287

Total heat transfer rate Q = εC_{min}(T_{h,in} - T_{c,in}) = 0.8287 x 277.78 x (90 - 15) = 17265 W

T_{h,out} = T_{h,in} - Q/C_{h} = 90 - 17265/277.78 = 27.85^{o}C

Outlet temperature of oil = 27.85^{o}C

T_{c,out} = T_{c,in} + Q/C_{c} = 15 + 17265/1394.67 = 27.38^{o}C

Outlet temperature of water = 27.38^{o}C

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Last Modified on: 04-Feb-2022

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