8-A) The ideal (internally reversible, frictionless) components of a steam power
ID: 1862656 • Letter: 8
Question
8-A)
The ideal (internally reversible, frictionless) components of a steam power-plant are shown. Superheated vapor at 4000 kPa (40 bars) & 500 degree C flows steadily into an adiabatic turbine at (1), which is connected to a condenser at (2). Saturated liquid exits from the condenser at 20 kPa, entering the adiabatic pump at (3). High-pressure liquid leaves the pump at 4000 kPa (40 bars), flowing into the boiler at (4). Neglect changes in kinetic and potential energy. The steam mass flow rate is 50 kg/s. Assume source TH = Tmax.cycle and cold sink Tc =Tmin.cycie. For the write-up clearly sketch the components, carefully draw the T-s diagram and organize a state table. Find the turbine & pump power, W t.out. W p.in (MW) Find the heat transfer rates, Q iN.hot, Q out.cold (MJ/s), & entropy creation rates, 2 sigma 3, 4 sigma 1 (MJ/K-s) Find Rankine cycle efficiency vs. the reversible limit, eta th = W CYCLE/Q IN,HOT = %Explanation / Answer
From steam table, at P1 = 4000 kPa and T1 = 500 deg C, we get x1 = superheated vapor, h1 = 3450 kJ/kg, s1 = 7.09 kJ/kg-K
From steam table, at P2 = 20 kPa and s2 = s1 = 7.09 kJ/kg-K, we get x2 = 0.884, h2 = 2340 kJ/kg, T2 = 60.1 deg C
From steam table, at P3 = 20 kPa and x3 = 0 (sat. liquid), we get T3 = 60.1 deg C, h3 = 251 kJ/kg, s3 = 0.832 kJ/kg-K
From steam table, at P4 = 4000 kPa and s4 = s3 = 0.832 kJ/kg-K, we get x4 = subcooled liquid, h4 = 255 kJ/kg, T4 = 60.2 deg C
Work done in turbine = m*(h1 - h2) = 50*(3450 - 2340) = 55500 kW
Work done on pump = m*(h4 - h3) = 50*(255 - 251) = 200 kW
Net work = 55500 - 200 = 55300 kW
Heat input = m*(h1 - h4) = 50*(3450 - 255) = 159750 kW
Efficiency = 55300 / 159750 = 0.346 or 34.6 %
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