Answer:
The Entropy generated = 2.821 kJ/K
Step-by-step explanation:
Total volume of container = 5m³
Heat transfer between system and surrounding= 0. dQ = 0
1st chamber, temperature of water at saturated liquid is 300°C
From the steam table:
Specific enthalpy of saturated liquid at 300°C is given as h_{f} = 1344.8 kJ/kg
Specific internal energy of saturated liquid at 300°C is given as U_{f1} = 1332.7 kJ/kg
the first law of thermodynamics state that (for a closed system) :
dQ = dw + dU..................(1)
work done for free expansion=
dw =00 = 0 + dU
dU = 0 .
that is, U₁ = U₂
At the second chamber,
The final pressure P₂ is given as = 50 kPa
From the steam table, at P₂ = 50 kPa, U_{f2} = 340.49 kJ/kg
(U_{fg} )_{2} = 2142.7 kJ/kg
Let the dryness fraction at the second chamber = x
U_{2} = U_{f2} + U_{fg2}
U_{2} = 340.49 + x2140.7
Since U₁ = U₂
1332.7 = 340.49 + x2140.7
Dryness fraction; x = 0.463
From the steam table, the specific volume is:
u_{f2} = 0.00103 m^{3} /kg\\
u_{2} = u_{f2} + xu_{fg2}
u_{2} = 0.00103 + 0.463(3.2393)
u_{2} = 1.5 m^{3} /kg
u_{2} = v2/ m2
V₂ = 5 m³
1.5 = 5/m₂
m₂ = 3.33 kg
At 300°C S_{1} = S_{f} = 3.2548kJ/kg-k
S_{2} = S_{f2} + xS_{fg2}
From the steam table,
S_{f2} = 1.0912 kJ/kg-k\\S_{fg2} = 6.5019 kJ/kg-k\\S_{2} = 1.0912 + 0.463(6.5019)\\S_{2} = 4.102 kJ/kg-k
Therefore the entropy generated will be :
Entropy = mass* (S₂ - S₁)
Entropy = 3.33* (4.102 - 3.2548)
Entropy = 2.821 kJ/K