Question

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

Answer 1

Answer:

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;

$W_{p_1 \rightarrow p_2}$ = $nRTln\frac{V_{p1}}{V_{p2}}$ =  $p_1 \cdot v_1\cdot ln\frac{V_{p1}}{V_{p2}}$ = 1.694 × 0.1 MPa ×$ln\frac{1.694}{0.039}$ =  638855.89 J

= 0.639 MJ.

Answer 2

Answer:

408 J/kg

662696.7 J/kg

Explanation:

Workdone by compression process in a pump can be isothermally expressed as :

$W = P_1V_1 \ In \frac{P_2}{P_1}$

For water at inlet pressure 1. 0 bar;

the specific values for :

saturated vapor $v_g = 1.694 m^2 kg$

saturated liquid $v_f = 0.001043 \ m^2 kg$

∴ we have

$P_1$ = 1.0 bar = 1. 0 × 10⁵ N/m²

$P_2$ = 50 bar = 50  × 10⁵ N/m²

For saturated water output from condenser work; we have

W = $P_1V_f \ In (\frac{P_2}{P_1})$

W = $(1.0*10^{-5})(0.001043) In ( \frac{50}{10} ) J/kg$

W = 408 J/kg

For saturated vapor output from condenser

Work required :

$W = P_1V_g \ In ( \frac{P_2}{P_1} )$

W = $1.0*10^5*1.694 \ In ( \frac{50}{1.0}) J/kg$

W = 662696.7 J/kg