Answer :
Answer:
a) Check the attached file
b) [tex]\phi = U_{2} - U_{1}[/tex]
c)Energy transfer by work, [tex]W = 545 kJ/kg[/tex]
Energy transfer by heat, E = 543 kJ/kg
d)The entropy production, [tex]\triangle S = 0.0015 kJ/kg-K[/tex]
The process is irreversible
Explanation:
Initial Pressure, P₁ = 10 bar
Initial temperature, T₁ = 360⁰ C
Final Pressure, P₂ = 1.5 bar
At P₁ = 10 bar and T₁ = 360°C
Getting the other parameters from the steam table:
V₁ = 0.28736 m³/kg, U₁ = 2892.05 kJ/kg, h₁ = 379.09 kJ/kg, S₁ = 7.3366 kJ/K
Using Boyle's law, to get V₂
P₁ V₁ = P₂V₂
10 * 0.28736 * = 1.5 * V₂
V₂ = 1.9157 m³/kg
At P₂ = 1.5 bar and V₂ = 1.9157 m³/kg
U₂ = 2890 kJ/Kg, h₂ = 3180 kJ/kg, S₂ = 8.2 kJ/K, T₂ = 351° C
a) Check the attached file for the P-V and T-S diagrams
b) Write the engineering model
The system described is a closed system
[tex]\phi = \triangle U + W\\\phi = U_{2} - U_{1}[/tex]
c) Energy transfer by work and heat
Energy transform by work
[tex]W = P_{1} V_{1}ln(\frac{P_{1} }{P_{2} } )\\ W = (10*0.287)ln(\frac{10}{1.5} )\\W = 545 kJ/kg[/tex]
Energy transform by heat
E = ( U₂ - U₁) + w
E = (2890 - 2892.05) + 545
E = 543 kJ
d)
[tex](\triangle S)_{123} = S_{2} - S_{1}\\(\triangle S)_{123} = 8.2 - 7.3366\\(\triangle S)_{123} = 0.8634 kJ/kg-K\\[/tex]
[tex]( \triangle S)_{b} = \frac{-q}{T_{b} }\\( \triangle S)_{b} = \frac{-543}{630 }\\( \triangle S)_{b} = -0.8619 kJ/kg-K[/tex]
[tex]\triangle S = (\triangle S)_{sys} + (\triangle S)_{b}\\\triangle S = 0.8634 - 0.8619\\\triangle S = 0.0015 kJ/kg-K[/tex]
The process is irreversible since ΔS > 0
