Question

A heat exchanger is to heat water (cp = 4.18 kJ/kg·°C) from 25°C to 60°C at a rate of 0.5 kg/s. The heating is to be accomplished by geothermal water (cp = 4.31 kJ/kg·°C) available at 140°C at a mass flow rate of 0.3 kg/s. Determine the rate of heat transfer in the heat exchanger and the exit tempera

Answer 1

73.15 kW, 196.6°C

Step-by-step explanation:

Energy in - Energy out = change in energy

`E_(in)-E_(out)=\Delta E\\\\\Delta E=0\\\\E_(in)-E_(out)=0\\\\E_(in)=E_(out)\\\\Q_(in)+\dot mh_1=\dot mh_2\\\\Q_(in)=\dot mh_2-\dot mh_1\\\\Q_(in)=\dot m c_p(T_2-T_1)`

The rate of heat transfer to cold water is given as:

`Q_(in)=\dot m c_p(T_2-T_1)\\\\\dot m =0.5\ kg/s,c_p=4.18\ kJ/kg.^oC, T_2=60^oC,T_1=25^oC\\\\Q_(in)=(0.5 *4.18)(60-25)=73.15\ kW`

For the geothermal water:

`Q_(in)=\dot m c_p(T_2-T_1)\\\\\dot m =0.3\ kg/s,c_p=4.31\ kJ/kg.^oC, ,T_1=140^oC\\\\T_2=(Q)/(\dot m c_p) +T_1=(73.15)/(0.3*4.31)+140=196.6\\ \\T_2=196.6^oC`