Respuesta :
Answer:
The heat exchanger is 29.26 kJ/s.
The exit temperature of geothermal water is 117.37°C.
Explanation:
Given that,
Specific heat [tex]c_{p}=4.18\ kJ/kg°C[/tex]
Initial temperature = 25°C
Final temperature = 60°C
Specific heat by geothermal [tex]c_{p}=4.31\ kJ/kg°C[/tex]
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
[tex]Q=\dot{m}c_{p}\Delta T[/tex]
Put the value into the formula
[tex]Q=0.2\times4.18\times(60-25)[/tex]
[tex]Q=29.26\ kJ/s[/tex]
We need to calculate the exit temperature of geothermal water
Heat transferred to feed water= Heat transferred by geothermal water
[tex]0.2\times4.18\times(60-25)=0.3\times4.31\times(140-T_{exit})[/tex]
[tex]T_{exit}=\dfrac{29.26-181.02}{1.293}[/tex]
[tex]T_{exit}=117.37^{\circ}C[/tex]
Hence, The rate of heat exchanger is 29.26 kJ/s.
The exit temperature of geothermal water is 117.37°C.
