A steam power plant operates on a simple ideal Rankine cycle

A steam power plant operates on a simple ideal Rankine cycle. The turbine receives steam at 698.15K and 4100 kPa, while the discharge stem is at 40 kPa. The mass flow rate of steam is 8.5 kg/s. Determine the work output of turbine. [At 4100kPa; H = 3272.3 kJ/kg; S = 6.8450 kJ/kgK At 40kPa; the corresponding V = 0.001022m3/kg and H_sat vap = 2636.9 kJ/kg; H_sat liq = 317.65 kJ/kg; S_sat vap = 7.6709 kJ/kgK; S_sat liq = 1.0216 kJ/kgK [Provide a schematic diagram and full detailed calculation for then full credit] Answer: The net-work: (a) -1993.6 kW; (b) -7748.6 kW; (c)-1094.6 kW

Solution

Net work output= Work of turbine - Work of pump

h1=3272.3 kJ/kg

h2=2636.3kJ/kg

h3=317.65 kJ/kg

W pump=v1 ( p2- p1)

   =.001022(4100-40 ) =4.149 kJ/kg

h4= h3+ 4.149 = 321.799 kJ/kg

Turbine work = h1-h2 = 636 kJ/kg

Wnet = Turbine work - pump work = 631.8 kJ/ kg

For mass for rate 8.5 kg/s

631.8 kJ/ kg * 8.5 kg/s =5370.7 kJ/s ( kW)