Consider a regenerativereheat vapor power cycle utilizing st

Consider a regenerative-reheat vapor power cycle utilizing steam as the working fluid. Steam leaves the steam generator and enters the turbine at 4 MPa, 400!C. Steam is extracted from the first turbine at 400 kPa and fed into an open feedwater heater which operates at 400 kP a. The steam exiting the open heater is saturated liquid at 400 kPa. After expansion to 150 kPa in the first turbine, the steam is reheated to 320!C before entering the second turbine, where it expands to the condenser pressure of 10 kP a. Steam leaves the condenser as saturated liquid. The isentropic e!ciency of all turbines and pumps is 90%. Determine (a) net work generated in the cycle per unit mass of steam entering the first turbine, (b) thermal e!ciency of the cycle.

Solution

a) From steam table :

properties of steam entering turbine 1 :

h₁ = 3213.6 kJ/kg, (corresponding to pressure 4MPa, temperature 400^oC)

properties of steam leaving turbine 1 :

h₂\' = 2737.6 kJ/kg (corresponding to saturation pressure 400kPa)

properties of steam entering turbine 2:

h3 = 3127.2 kJ/kg (corresponding to pressure 150kPa, temperature 320^oC)

properties of steam leaving turbine 2:

h₄\' = 2584.7 kJ/kg (corresponding to saturation pressure 10kPa)

properties at inlet of pump :

h₅ = 191.8 kJ/kg

properties at outlet of pump :

h₆\' = 1087.4 kJ/kg

efficiency of Turbine ₁ = 0.9 = (h₁ - h₂\')/(h₁ - h₂)

Therefore h₂ = 2684.711 kJ/kg

efficiency of Turbine ₂ = 0.9 = (h₃ - h₄\')/(h₃ - h₄)

Therefore h₄ = 2524.422 kJ/kg

Turbine work = W_t= (h₁ - h₂) + (h₃ - h₄) = 1131.667 kJ/kg

efficiency of pump = 0.9 = (h₆ - h₅)/(h₆\' - h₅)

Therefore h₆ = 996.84 kJ/kg

Pump work = W_p= (h₆ - h₅) = 806.04 kJ/kg

Net work done =W_t -W_p = 325.627 kJ/kg

b) Heat supplied = Q = (h₁ - h₆)+(h₃ - h₂) = 2659.249 kJ/kg

Thermal efficiency of cycle = (Net work)/(Heat supplied) = 12.24%