The net work output (kJ kg–1) of the cycle is

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Direction: In a steam power plant operating on the Rankine cycle, steam enters the turbine at 4 MPa, 350°C and exists at a pressure of 15 kPa. Then it enters the condenser and exists as saturated water. Next, a pump feeds back the water to the boiler. The adiabatic efficiency of the turbine is 90%. The thermodynamic states of water and steam are given in the table.

h is specific enthalpy, s is specific entropy and v the specific volume; subscript f and g denote saturated liquid state and saturated vapour state.

The net work output (kJ kg^–1) of the cycle is

1. 498
2. 775
3. 860
4. 957

Correct Option: A

Given: Rankine cycle on T-S diagram
P₃ = 4 MPa
T₃ = 350°C
η_t = 0.9
h₃ =3092.5 kJ/kg
s₃ = s₄ = 6.5821 kJ/kgk
h₁ = 225.94 kJ/kg = h_f
h_g = 2599.1 kJ/kg

h_fg = h_g – h_f = 2373.16 kJ/kg
s₁ = s_f = 0.7549 kJ/kgK
s_g = 8.0085 kJ/kgK
s_fg = 8.0085 – 0.7949 = 7.2536 kJ/kg K
Also, s₃ = s₄ = 6.5821
∴ s₄ = s₁ + x s_fg
= 0.7549 + x (7.2536) = 6.5821
⇒ x = 0.8
h₄ = h₁ + xh_fg
= 225.94 + (0.8 × 2375.16) = 2132.4 kJ/kg
Net work output of the cycle
= η_t (h₃ – h₄)
= 0.9 (3092.5 – 2132.4)
= 864 kJ/kg
≈ 860 kJ/kg
Heat supplied to the cycle = h₃ – h₂
Since we do not have information regarding h₂, therefore neglect compressor work and take
h₂ = h₁
∴ Q_s = 3092.5 – 2866 kJ/kg = 2863 kJ/kg