A well-insulated heat exchanger is to heat water (cp 5 4.18 kJ/kg?8C) from 25 to 608C at a rate of 0.50 kg/s. The heating is to be accomplished by geothermal water (cp 5 4.31 kJ/kg?8C) available at 1408C at a mass flow rate of 0.75 kg/s. Determine (a) the rate of heat transfer and (b) the rate of entropy generation in the heat exchanger.

a) The rate of heat transfer in the heat exchanger = 73.15 kJ/s = 73.15 kW

b) The rate of entropy generation = 0.050 kJ/s°C = 50 J/s°C = 50 W/K

Explanation:

The rate of heat transfer in the heat exchanger is equal to the rate of heat transfer required to change the temperature of one of the fluids involved in the heat exchange.

Rate of heat transfer required to raise the temperature of the water from 25°C to 60°C is given as mCΔT

where m = mass flowrate of the water = 0.5 kg/s

C = specific heat capacity of water = 4.18 kJ/kg°C

ΔT = change in temperature = 60 - 25 = 35°C

Rate of heat transfer = 0.5 × 4.18 × 35 = 73.15 kJ/s = 73.15 kW

b) The rate of entropy generation is given as the sum of entropy generation for the normal water and the geothermal water.

Normally, rate of entropy generation is given as mC In (T₂/T₁)

For the normal water

Rate of entropy generation = mC In (T₂/T₁)

m = mass flowrate = 0.5 kg/s

C = specific heat capacity of water = 4.18 kJ/kg°C

T₂ = final temperature of the water = 60°C = 333.15 K

T₁ = initial temperature of the water = 25°C = 298.15 K

Rate of entropy generation = 0.5 × 4.18 × In(333.15/298.15) = 0.232 kJ/s.°C

For the geothermal water

Rate of entropy generation = mC In (T₂/T₁)

m = mass flowrate = 0.75 kg/s

C = specific heat capacity of water = 4.31 kJ/kg°C

T₁ = initial temperature of the geothermal water = 140°C = 413.15 K

T₂ = final temperature of the geothermal water = ?

To find the final temperature, we note that the rate of heat transfer to the normal water is the same as rate of heat loss by the geothermal water

Rate of heat loss by the geothermal water = -73.15 kJ/s (minus because of rhe direction of heat loss)

-73.15 = mCΔT = 0.75 × 4.31 × (T₂ - 140)

(T₂ - 140) = -73.15/(0.75×4.31) = -22.63

T₂ = 140 - 22.63 = 117.37°C = 390.52 K

Rate of entropy generation = 0.75× 4.31 × In (390.52/413.15) = -0.182 kJ/s°C

Total rate of entropy generation = 0.232 - 0.182 = 0.050 kJ/s°C = 50 J/s°C = 50 W/K

Hope this Helps!!!

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