Heat Exchanger Area Comparison for Parallel and Counter Flow

A heat exchanger heats 25,000 kg/hr of water entering at 30^oC while cooling 20,000 kg/hr of water from 100^oC to 80^oC. Determine the area necessary for (i) Parallel flow arrangement
(ii) Counter flow arrangement.
Overall heat transfer coefficient may be assumed as 1,600 W/m²K.

Calculations:

Heat of heat transferred by the cooling water = Q = mC_pΔT = 20000 x 4.184 x (100 - 80) = 1673600 kJ/hr = 464.89 kW

This will be the amount of heat transferred to the water that is getting heated up. Therefore, temperature change is = 1673600 / (25000 x 4.184) = 16^oC

i.e., outlet temperature = 30 + 16 = 46^oC

With these data, the following temperature-length diagram is drawn.

(i) Parallel flow arrangement:

ΔT_o = 100 - 30 = 70^oC

ΔT_L = 80 - 46 = 34^oC

LMTD = ΔT_ln = (ΔT_o - ΔT_L) / ln (ΔT_o/ΔT_L)

= (70 - 34) / ln (70/34) = 49.85^oC

A = Q / (UΔT_ln) = 464.89 x 10³ / (1600 x 49.85) = 5.8286 m²

Area required for parallel flow = 5.8286 m²

(ii) Counter flow arrangement:

ΔT_o = 80 - 30 = 50^oC

ΔT_L = 100 - 46 = 54^oC

LMTD = ΔT_ln = (ΔT_o - ΔT_L) / ln (ΔT_o/ΔT_L)

= (50 - 54) / ln (50/54) = 51.97^oC

A = Q / (UΔT_ln) = 464.89 x 10³ / (1600 x 51.97) = 5.5908 m²

Area required for counter flow = 5.5908 m²