Respuesta :
Answer:
a) [tex]\mu=3.3[/tex]
b) [tex]Q=1440.7KJ/Kmol[/tex]
c) [tex]W=1872.9KJ/Kmol[/tex]
Explanation:
From the question we are told that:
Initial Temperature [tex]T_1=325k[/tex]
initial Pressure [tex]P_1=2 bar[/tex]
Final Pressure [tex]P_2=4 bar[/tex]
iso-thermal expansion [tex]T_2=250k[/tex]
a)
Generally the equation for Coefficient of performance is mathematically given by
[tex]\mu=\frac{T_2}{T_1-T_2}[/tex]
[tex]\mu=\frac{250}{325-250}[/tex]
[tex]\mu=3.3[/tex]
b)
Generally the equation for Heat Expansion is mathematically given by
[tex]Q=RT_2 In(\frac{P_2}{P_1})[/tex]
Where
R=Gas constant
[tex]R=8.314462618[/tex]
Therefore
[tex]Q=8.314462618*250 In(\frac{4}{2})[/tex]
[tex]Q=1440.7KJ/Kmol[/tex]
c)
Generally the equation for work input is mathematically given by
[tex]W=RT_1 In(\frac{P_2}{P_1})[/tex]
[tex]W=8.314462618*250 In(\frac{4}{2})[/tex]
[tex]W=1872.9KJ/Kmol[/tex]
The coefficient of performance is 3.33, the heat transfer in the isothermal expansion is 1440.71kJ/K.mol and the work input is calculated as 1872.92kJ/K.mol
Given Data:
- T1 = 325K
- P1 = 2 bar
- P2 = 4 bar
- T2 = 250K
Isothermal expansion occurs at 250K.
a) The coefficient of performance
This is calculated as
COP =[tex]\frac{T_2}{T_1-T_2}=\frac{250}{325-250} =3.33[/tex]
b) Heat Transfer in isothermal expansion
[tex]Q = RT_2In(\frac{p_2}{p_1})[/tex]
Therefore; In isothermal process du = 0
R = 8.314 AkJ/K.mol
Q = 8.314 * 250 In(4/2)
Q = 1440.71kJ/K.mol
c) Work Input
W[tex]_i_n[/tex]=[tex]RT_1In(\frac{p_2}{p_1})\\W_i_n=8.314*325In(4/2)\\W_i_n=1872.92kJ/K.mol[/tex]
The work input is 1872.92kJ/K.mol
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