(10%) Superheated steam at 10 MPa and 480?C leaves the steam generator of a vapo
ID: 2991506 • Letter: #
Question
(10%) Superheated steam at 10 MPa and 480?C leaves the steam generator of a vaporpower plant. The power plant is operated by a Rankine cycle (may not be ideal). The
pressure at the exit of the turbine is 10 kPa, and the turbine operates adiabatically. The
mass flow rate of the fluid is 80 kg/s. The power output in the turbine is 65 MW.
Assume both the pump and turbine are adiabatic and the isentropic efficiency of the
pump is 100%. Neglect potential and kinetic energy changes.
a) Calculate the isentropic efficiency of the turbine. [ = 65%]
b) If the specific enthalpy at the exit of the pump is 203.2 kJ/kg, what is the heat transfer
rate in the boiler? [249.45 MW] Calculate the thermal efficiency of the cycle. [=26%]
Neglect the power to run the pump.
c) What is the maximum power output of the turbine if the turbine is operated
reversibly? [= 100 MW] What is the maximum thermal efficiency of the cycle? [=40%]
Neglect the power to run the pump.
Explanation / Answer
At T1 = 480 C and P1 = 10 Mpa, fom steam table : h1 = 3322.9 KJ/Kg and s1 = 6.53 KJ/Kg.K
now, s2 = s1 = 6.53 KJ/Kg as 1-2 is adiabatic
At P2 = 10 Kpa and s2 = 6.53 Kj/Kg, from steam table :
h2 = 2067.87 KJ/Kg
ideal work = m.( h1 - h2) = 80 x ( 3322.9 - 2067.87) = 100.4 MW
actual work = 65 MW
a) ientropic efficiency = 65/100.4 = 0.647 = 64.7 % = 65 %
b) exit of pump = h4 = 203.2 KJ/Kg
so, heat addition = m.(h1 - h4) = 80 x (3322.9 - 203.2) = 249.576 MW
c) thermal efficency of pump = We/Q = 65/249.576 = 0.26 = 26%
d) max power output = m.(h1 - h2) = 80 x ( 3322.9 - 2067.87) = 100.4 MW
e) max efficiency = 100.4/249.576 = 0.4 = 40%
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