In a Counter flow heat exchanger design, fluid rates and specific heats were chosen in such a manner that heat capacities of both the fluids are same. A hot fluid enters at 100°C and leaves at 60°C. The cold fluid enters heat exchanger at 40°C. The mean temperature difference between the two fluids is

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ISRO Refrigeration and Air Conditioning 2018 Official

Option 3 : 20°C

ISRO Scientist ME 2020 Paper

1652

80 Questions
240 Marks
90 Mins

__Concept:__

The total heat transfer rate in the heat exchanger is given by

Q = U × A × θm

Where U = overall heat transfer coefficient in W/m2-K,

A = effective surface area of heat exchanger in m2,

θm = log mean temperature difference \(\left( {LMTD} \right) = \frac{{{\bf{\Delta }}{T_i} - {\bf{\Delta }}{T_e}}}{{{\bf{ln}}\left( {\frac{{{\bf{\Delta }}{T_i}}}{{{\bf{\Delta }}{T_e}}}} \right)}}\)

Also, Q = Cph (Thi-The) = Cpc (Tce-Tci)

i.e. heat lost by hot fluid = heat gain by the cold fluid

where Cph and Cpc are heat capacity of hot and cold fluid respectively in kJ/K

Since the heat capacity for hot and cold fluids are the same and it is a case of counter flow so the LMTD will be the difference in temperature of either side.

__Calculation:__

__Given:__

T_{hi} = 100°C, T_{he} = 60°C, Tci = 40°C

Q = Cph (Thi-The) = Cpc (Tce-Tci)

Thi-The = Tce-Tci

100 - 60 = T_{ce} - 40

Tce = 80°C

ΔTi = Thi – Tce = 100 - 80 = 20°C

ΔTe = The - Tci = 60 - 40 = 20°C

As ΔTi and ΔTe both are equal so the heat exchanger is a balanced type counter flow heat exchanger so, the mean temperature difference for this case is 20 °C.