A power cycle under development has two options for a condenser. One outputs saturated water and one outputs saturated vapor. The condenser outlet provides fluid to a pump. The pump’s inlet pressure is 1 bar and the required outlet pressure is 50 bar. If the pump does not change the internal energy of the fluid (∆u = 0), determine the work per unit mass required for the liquid and the vapor

The work per unit mass required for the liquid is 4.9 MJ

The work per unit mass required for the vapor is 0.639 MJ.

Explanation:

Per unit mass of vapor, we have

p₁ = 1 bar = 0.1 MPa

p₂ = 50 bar = 5 MPa

mass of saturated water = 1 kg

Volume of saturated water = 1 m³

Since the water is incompressible, we have

Pump work = v×(p₂ - p₁) =  1 m³ × (5 MPa - 1 MPa) = 4.9 MJ

For the steam, we have. 1 kg steam at 1 bar we have the specific volume v₁ = 1.694 m³/kg also

1 kg steam at 50 bar has a specific volume v₂ = 0.039 m³/kg

Since the internal energy is constant, that is ΔU = 0, then, the temperature can be assumed to be constant. Therefore, we have

Work done in moving from p₁ to p₂ is given by;

= =   = 1.694 × 0.1 MPa × =  638855.89 J

= 0.639 MJ.

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