Question

ANSWER PLEASE

PROBLEM #2 (68 pts.) Steam at 200 psia, 600°F (State 1) enters an adiabatic turbine cperating at steady state with a mass flow rate of 16.5 lbm/min and exits at 15 psia, 240°F (State 2), Neglecting the changes in kinetic and potential energies and assuming the surroundings to be at T.- 537 R. Determine: a) the actual power output, in Btu/min. b) the reversible (or maximum) power output, in Btu/min. c) the rate of exergy destruction, in Btu/min. d) the second-law efficiency of the turbine. e) the isentropic efficiency of the turbine.

Answer 1

Enthalby and entropy at state 1 hi= 1322.31 Btu/ebm 2 At 600°F and zoopsia, S, = 1.68 B741ebmºR) S forom steam table Enthalby and entro by at state 2 hz = 1163.9 Btulebm 7 At 240°F and is bsia, 5,= 177 Btu/rebm •R) I foom steam table Enthalby and entropy at To= 537R ho= 45.4 Btu/ebm ? Saturated liquid So = 0.09 B tullbmºR) I values foom steam table We have mass flow rate, in= 16.5 lb m/min (a) Actual Power Out but is Wout = m ( ho-h2) = 16.5(1322.31 – 1163.9) → Wout = 2613.765 Btu/min (b) Reversible (or maximum ) power out but wser,ou: = [Chi-ha) - To (31-sa)] Waev,out = 16-5[(1322.31– 1163-9)- 534 (168–172)] Wreviout = 3411.21 Btu/min)

(6) Rate of Xdestroyed exergy destruction = Woer,out - Wout = 3411.21 - 2613.765 troyed → Xdestayed = 797.445. Btu/min d) The second law efficiency of the turbine no = Wouy = 2613.765 - Woer, out 3411.21 = MI = 76.624/ e when steam expands isen too bically in tobtine then the entropy at the exit of turbine - is equal to the inkt entropy. For Isentropic expansion, enthalby at the exit of the tusbine is given as r At 15 bsia and has= 1100.26 Btul&bm 7 | Sx = 1.68 Btu/(IbmºR).] - from steam table = isentoo ble efficiency of the twobine is given as na = hinha - 1322-31- 1163.9 hir has 1322:31 - 1100.26 R+=71-34%