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

Answer:

73.15 kW, 196.6°C

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:

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