A steam power plant operates on a simple ideal Rankine cycle between the pressur
ID: 508615 • Letter: A
Question
A steam power plant operates on a simple ideal Rankine cycle between the pressure limits 3 MPa and 50 kPa. The temperature of the steam at the turbine inlet is 400 degree C, and the mass flow rate of steam through the cycle is 60 kg/s. Determine: a) The thermal efficiency of this cycle. b) The net power output of the power plant. 3. Refrigerant-134a eneters the compressor of a refrigerator as superheated vapor at 0.14 MPa and -10C at a rate of 0.05 kg/s and leaves at 0.80 MPa and 50C. The refrigerant is cooled in the condenser to 26C and 0.72 MPa and is throttled to 0.15 MPa. Disregarding any heat transfer and pressure drops in the connecting lines between the components, determine: a) The rate of heat removal from the refrigerated space b) The power input to the compressor c) The coefficient of performance of this refrigerator The refrigerant enthalpies at various states are given in the table above.Explanation / Answer
State 1, at 50 kPa, h1 = hf = 335.01 kJ/kg (refer to the steam tables)
vf = 0.001029 m3/kg
wp,in = vf (P2 - P1)
wp,in = 0.001029 ( 3000-50) (1 kJ / 1 kPa . m3)
wp,in = 3.036 kJ/kg
h2 = h1 + wp,in
h2 = 335.01 + 3.036
h2 = 338.046 kJ/kg
State 3, at 3 MPa, h3 = 2994.3 kJ/kg and s3 = 6.5412 kJ/kg.K (refer to the steam tables)
State 4, at 50 kPa, s3 = s4
Therefore, x4 = (s4 - sf) / sfg = (6.5412 - 1.0756) / 6.5355
x4 = 0.836
h4 = hf + (x4 . hfg)
h4 = 335.01 + (0.836 X 2308)
h4 = 2264.498 kJ/kg
qin = h3 - h2 = 2656.254
qout = h4 - h1 = 1929.488
wnet = qin - qout = 726.766 kJ/kg
net power output = mass flow rate X wnet
net power output = 60 X 726.766
net power output = 43605.96 = 43.606 MW
thermal efficiency = 1 - (qout / qin)
thermal efficiency = 1 - 0.726
thermal efficiency = 0.273 = 27.3 %
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