Heat Transfer Rate from Effectiveness-NTU Method

Water enters a parallel flow double-pipe heat exchanger at 15^oC, 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^oC. For an area of 1 m² and an overall heat transfer coefficient of 1,200 W/m².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/^oC = C_h

C_max = (1200/3600) x 4184 = 1394.67 W/^oC = 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^oC

Outlet temperature of oil = 27.85^oC

T_c,out = T_c,in + Q/C_c = 15 + 17265/1394.67 = 27.38^oC

Outlet temperature of water = 27.38^oC