Answer:
(a) The rate of heat transfer is 238.26 kW.
(b) Rate of entropy generation is 0.063 kW/K.
Explanation:
(a) Taking the heat gained by the cold water into consideration,
[tex]Q_{in} =mc(T_{2} -T_{1})[/tex]
Q = (0.95 kg/s)(4.18 kJ/kg.⁰C)(70-10) = 238.26 kW.
(b) Entropy generation = entropy in cold water + entropy in hot water
[tex]S_{generated} = (m_{c} c_{c} (ln(\frac{T_{c2}}{T_{c1}})) + (m_{h} c_{h} (ln(\frac{T_{h2}}{T_{h1}}))[/tex]
But [tex]\frac{T_{h2}}{T_{h1}} = 1-\frac{Q_{h} }{m_{h}c_{h}T_{h1}}[/tex]
∴ [tex]S_{generated}[/tex] = (0.95)(4.18)(ln(343/283)) + (1.6)(4.19)(ln(1- (238.26/((1.6)(4.19)(358))))
[tex]S_{generated}[/tex] = 0.7636 + (- 0.7009) = 0.0627 kW/K.
The rate of heat transfer and the rate of entropy generation in the heat exchanger is mathematically given as
Q = 238.26 kW.
S= 0.0627 kW/K.
Question Parameter(s):
Cold water (cp = 4.18 kJ/kg⋅°C) leading to a shower enters a well-insulated
counterflow heat exchanger at 10°C at a rate of 0.95 kg/s
and is heated to 70°C by hot water (cp = 4.19 kJ/kg⋅°C) that enters at 85°C at a rate of 1.6 kg/s
Generally, the equation for the Heat gain is mathematically given as
Q =mc(T_{2} -T_{1})
Therefore
Q = (0.95)(4.18)(70-10)
Q = 238.26 kW.
In conclusion for Entropy generation
E = entropy in cold water + entropy in hot water
Therefore
[tex]S= (0.95)(4.18)(ln(343/283)) + (1.6)(4.19)(ln(1- (238.26/((1.6)(4.19)(358))))[/tex]
S= 0.7636 + (- 0.7009)
S= 0.0627 kW/K.
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