Steam enters a turbine operating at steady state at a pressure of 1000 lbf/in2 a
ID: 2991410 • Letter: S
Question
Steam enters a turbine operating at steady state at a pressure of 1000 lbf/in2 and a temperature of 1100°F, and exits at a pressure of 4.0 lbf/in2 as a saturated liquid-vapor mixture with a quality of 0.96. Stray heat transfer from the turbine to the surroundings occurs. The ambient temperature is 90°F. Measurements indicate that the magnitude of the rate of the stray heat transfer is 3% of the turbine power. The mass flow rate of water through the turbine is 10 lbm/s.(a) Calculate the turbine power (Btu/s).
(b) For an enlarged control volume that includes the turbine and enough of the
surroundings so that the boundary temperature is 90°F, calculate the rate of entropy
production (Btu/s-°R).
Please show work. Final answer for A) 4620 Btu/s; (B) 1.31 Btu/s·R
Explanation / Answer
At p1 = 1000 lbf/in2 and T1 = 1100 F, from superheated steam tables, enthalpy h1 = 1562.9 Btu/lb and entropy s1 = 1.6908 Btu/lb-R
At p2 = 4 lbf/in2 and x=0.96, enthalpy h2 = 120.89 + 0.96(1006.4) = 1087.034 Btu/lb
and entropy s2 = 0.2198 + 0.96(1.6426) = 1.7967 Btu/lb-R
(a) Theoretical Turbine power = m(h1-h2) = 10(1562.9-1087.034) = 4758.66 Btu/s
Heat transfer = 0.03*4758.66 = 142.76 Btu/s
Actual turbine power = theoretical turbine power - heat transfer
= 4758.66 - 142.76 = 4616 Btu/s 4620 Btu/s
(b) Entropy production due to expansion in turbine = m(s2-s1) = 10(1.7967-1.6908) = 1.059 Btu/s-R
Entropy generated due to heat transfer to surroundings = Heat transfer / (460+90) = 142.76/550 = 0.26 Btu/s-R
Total entropy generation = 1.059+0.26 = 1.319 Btu/s-R