Question

A heat exchanger is to heat water (cp = 4.18 kJ/kg·°C) from 25 to 60°C at a rate of 0.2 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 temperature of geothermal water.

Answer 1

Answer:

The heat exchanger is 29.26 kJ/s.

The exit temperature of geothermal water is 117.37°C.

Explanation:

Given that,

Specific heat $c_{p}=4.18\ kJ/kg°C$

Initial temperature = 25°C

Final temperature = 60°C

Specific heat by geothermal $c_{p}=4.31\ kJ/kg°C$

Temperature = 140°C

Mass flow rate = 0.3 kg/s

We need to calculate the rate of heat transfer in heat exchange

Using formula of  rate of heat transfer in heat exchanger

$Q=\dot{m}c_{p}\Delta T$

Put the value into the formula

$Q=0.2\times4.18\times(60-25)$

$Q=29.26\ kJ/s$

We need to calculate the exit temperature of geothermal water

Heat transferred to feed water= Heat transferred by geothermal water

$0.2\times4.18\times(60-25)=0.3\times4.31\times(140-T_{exit})$

$T_{exit}=\dfrac{29.26-181.02}{1.293}$

$T_{exit}=117.37^{\circ}C$

Hence, The rate of heat exchanger is 29.26 kJ/s.

The exit temperature of geothermal water is 117.37°C.