Question

Consider a diesel-fired steam power plant that produces 400 MW of electric power. The turbine inlet conditions of 5 MPa and 500°C and a condenser pressure of 25 kPa. The state of the water at the inlet of the pump is saturated liquid. The steam has quality of (x = 0.88) at the inlet of the condenser. Water from a nearby river is used to cool the condenser. To prevent thermal pollution, the cooling water is not allowed to experience more than 8°C as it flows through the condenser. The diesel has a heating value (energy released when the fuel is burned) of 40,100 kJ/kg. Assuming that 80 percent of this energy is transferred to the steam in the boiler and that the electric generator has an efficiency of 95 percent, determine (a) the overall plant efficiency (the ratio of net electric power output to the energy input as fuel), (b) the required rate of diesel supply and (c) the minimum mass flow rate of the cooling water from the river.

Answer 1

Given: W ̇=400MW, Q=40100 kJ/kg, η_generator=95%=0.95,η_Boiler=80%=0.80
Turbine inlet condition
P=5MPa,T_3=500℃
Using Properties table for steam
T_3>T_Saturation (=264℃), Hence state is superheated
h_3g=3433.8kJ/kg,s_3g=6.976kJ/(kg-K)
Turbine outlet or inlet to condenser condition
P=25 kPa=0.25 MPa,x=0.88
h_4f=271.9 kJ/kg, h_4fg=2618.2 kJ/kg
h_4=h_4f+xh_4fg
h_4=271.9+0.88(2618.2)
h_4=2575.92 kJ/kg
Condensor outlet or inlet to pump condition
P=25 kPa,Staturation state (Given)
〖h_f=h〗_1=271.9 kJ/kg, v_f=0.001020 m^3/kg
Work done by pump,W_p=-∫▒〖v_f dP〗
W_p=0.001020 (5000-25)=5 kJ/kg
Pump outlet or inlet to boiler condition
h_2=h_1+W_p
h_2=271.9 +5
h_2=276.9
Heat supplied to the boiler
〖Q=(h〗_3g-h_2)=3433.8-276.9=3156.9
Work done by turbine
W_T=h_3g-h_4=3433.8-2575.92=857.88
Cycle efficiency
η_cycle=(W_T-W_P)/Q
η_cycle=(857.88-5)/(3156.9)
η_cycle=27%
Overall efficiency,(η)= η_cycle×η_generator× η_Boiler
η=0.27×0.95×0.80
η=20.52%
Diesel supply ((m_f ) ̇)

(m_f ) ̇=(power generated(W ̇))/ηQ
(m_f ) ̇=(400×1000)/((0.2052)(40100))
(m_f ) ̇=48.61 kg/s
Mass flow rate of cooling water ((m_w ) ̇)
Heat rejected by gases = heat taken by water
Temperature at state 4, T_4=65℃ (using steam table)
(m_f ) ̇(h_4-h_1)= (m_w ) ̇c_P (T_4-T_w)
(m_w ) ̇=(48.61(2575.92-271.9) )/(4.18×(65-8))
(m_w ) ̇=470 kg/s