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Answer:
The Entropy generated by the steam = 2.821 kJ/K
Explanation:
Total volume of container = 5m³
Heat transfer does not exist between system and surrounding, dQ = 0
At the first chamber, temperature of water at saturated liquid is 300°C
From the steam table:
Specific enthalpy of saturated liquid at 300°C ,
Specific internal energy of saturated liquid at 300°C, 1332.7 kJ/kg
For closed system, the first law of thermodynamics state that:
dQ = dw + dU..................(1)
work done for free expansion, dw =0
0 = 0 + dU
dU = 0 , i.e. U₁ = U₂
At the second chamber,
The final pressure, P₂ = 50 kPa
From the steam table, at P₂ = 50 kPa, = 340.49 kJ/kg
Let the dryness fraction at the second chamber = x
Since U₁ = U₂
1332.7 = 340.49 + x2140.7
Dryness fraction, x = 0.463
From steam table, the specific volume is,
V₂ = 5 m³
1.5 = 5/m₂
m₂ = 3.33 kg
At 300°C
From the steam table,
Therefore the entropy generated will be :
Entropy = mass* (S₂ - S₁)
Entropy = 3.33* (4.102 - 3.2548)
Entropy = 2.821 kJ/K