###
Heat Exchanger Area Comparison for Parallel and Counter Flow

Home ->
ChE Learning Resources
->
Solved Problems
->
Heat Transfer->

A heat exchanger heats 25,000 kg/hr of water entering at 30^{o}C while cooling 20,000 kg/hr of water from 100^{o}C to 80^{o}C. Determine the area necessary for (i) Parallel flow arrangement

(ii) Counter flow arrangement.

Overall heat transfer coefficient may be assumed as 1,600 W/m^{2}K.

Calculations:

Heat of heat transferred by the cooling water = Q = mC_{p}ΔT = 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^{o}C

i.e., outlet temperature = 30 + 16 = 46^{o}C

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

(i) Parallel flow arrangement:

ΔT_{o} = 100 - 30 = 70^{o}C

ΔT_{L} = 80 - 46 = 34^{o}C

LMTD = ΔT_{ln} = (ΔT_{o} - ΔT_{L}) / ln (ΔT_{o}/ΔT_{L})

= (70 - 34) / ln (70/34) = 49.85^{o}C

A = Q / (UΔT_{ln}) = 464.89 x 10^{3} / (1600 x 49.85) = 5.8286 m^{2}

Area required for parallel flow = 5.8286 m^{2}

(ii) Counter flow arrangement:

ΔT_{o} = 80 - 30 = 50^{o}C

ΔT_{L} = 100 - 46 = 54^{o}C

LMTD = ΔT_{ln} = (ΔT_{o} - ΔT_{L}) / ln (ΔT_{o}/ΔT_{L})

= (50 - 54) / ln (50/54) = 51.97^{o}C

A = Q / (UΔT_{ln}) = 464.89 x 10^{3} / (1600 x 51.97) = 5.5908 m^{2}

Area required for counter flow = 5.5908 m^{2}

[Index]

Last Modified on: 01-May-2024

Chemical Engineering Learning Resources - msubbu

e-mail: learn[AT]msubbu.academy

www.msubbu.in