Homework SourseSteam enters an adiabatic utilityclass power turbine at 1000
Steam enters an adiabatic utility-class power turbine at 1000 psia and 1100 F(temperature). At a pressure of 120 psia and 550 F, 20% of the mass flow rate of the system is extracted from the turbine for use in an industrial process heat exchanger. The remainder of the steam expands through the turbine and exhausts with a quality of 90%. The turbine is 4000hp.
a) determine the mass flow rate of the steam entering at 1000 psia and 1100F.
b) determine the entropy at the entrance, extraction and exit conditions.
c) sketch the process on the T-s diagram.
Steam enters an adiabatic utility-class power turbine at 1000 psia and 1100 F(temperature). At a pressure of 120 psia and 550 F, 20% of the mass flow rate of the system is extracted from the turbine for use in an industrial process heat exchanger. The remainder of the steam expands through the turbine and exhausts with a quality of 90%. The turbine is 4000hp.
a) determine the mass flow rate of the steam entering at 1000 psia and 1100F.
b) determine the entropy at the entrance, extraction and exit conditions.
c) sketch the process on the T-s diagram.
a) determine the mass flow rate of the steam entering at 1000 psia and 1100F.
b) determine the entropy at the entrance, extraction and exit conditions.
c) sketch the process on the T-s diagram.
Solution
a)
From steam properties at P1 = 1000 psia and T1 = 1100 F, we get h1 = 1560 Btu / lb, and entropy s1 = 1.69 Btu / lb-R
From steam properties at P2 = 120 psia and T2 = 550 F, we get h2 = 1300 Btu / lb, and entropy s2 = 1.71 Btu / lb-R
From steam properties at s3 = s2 = 1.71 Btu / lb-R and x3 = 0.9, we get T3 = 145 deg F, h3 = 1020 Btu / lb
Power developed = 4000 hp = 169629 Btu/min
Power = m*(h1 - h2) + (m - 0.2*m)*(h2 - h3)
169629 = m*(1560 - 1300) + 0.8*m*(1300 - 1020)
m = 350.47 lb / min
b)
As found above,
s1 = 1.69 Btu / lb-R
s2 = 1.71 Btu / lb-R
s3 = 1.71 Btu / lb-R
