Steam enters a turbine at 600 psia, 900°F, and 250 ft/s. It expands through a pr
ID: 1940531 • Letter: S
Question
Steam enters a turbine at 600 psia, 900°F, and 250 ft/s. It expands through a pressure ratio of 10:1 (i.e. to 60 psia) and leaves the turbine at a velocity of 500 ft/s.a) If the process is adiabatic and internally reversible, calculate the work output per pound of steam. Be sure to account for changes in kinetic energy across the turbine.
b) Now consider the case where the internal irreversibilities reduce the work output to 85 percent of the reversible case. For the same pressure ratio, find the final temperature and final specific entropy.
Explanation / Answer
(a) Q = 0 (since adiabatic)
For energy balance: Q-W = (h2-h1) + 0.5 (V22 - V12) where h = specific Enthalpy, V = Velocity, Q = Heat tran = sfer, W = Work done.
P1 = 600 psia = 41.37 bar, P2 = 60 psia = 4.14 bar
T1 = 900 oF = 755.4 K = 482 oC.
Looking at saturated liquid-vapour tables for water we notice that for 600 psia pressure, saturation temperature is 486.3 oF. Since T1 is higher than this temperature, state 1 is supereated. Looking into superheated steam tables against P1 and T1 we get, h1 = 1462.9 Btu/lb, s1 = 1.6766 Btu/lb.oR, v1 = 1.302 ft3/lb.
Since process is adiabatic and internally reversible, entropy will remain unchanged. Hence, s2 = s1.
From steam tables again, At P2 = 60 psi, sg = 1.6443 Btu/lb.oR. This is lower than s2. Hence, steam is still superheated at state 2. Looking in tables against P2, s2 we get, T2 = 340 oF, h2 = 1203 Btu/lb, v2 = 7.71 ft3/lb.
Putting values in energy equation, we get 0-W = (1203-1462.9) + 0.5(500^2 - 250^2)/25037 = - 259.9 + 3.744 = - 256.156 Btu/lb (Here relation 1 Btu/lb = 25037 ft2/s2 has been used.)
Hence, W = 256.156 Btu/lb.
(b) W = 0.85 * 256.156 = 217.73 Btu/lb.
Assuming polytropic process pvn = constant.
W = P1v1 ((P2/P1)(n-1)/n-1)/(1-n)
By putting values and adjusting for units, we get, n = 1.237.
Now, T2/T1 = (P2/P1)(n-1)/n
By putting values, we get T2 = 415 oF.
Looking in superheated tables against P2, T2 we get, s2 = 1.7216 Btu/lb.oR.