c) A geothermal power plant utilizes steam produced by natural means underground
Steam wells are drilled to tap this steam supply which is available at 4.5 bar and 175°C.
The steam leaves the turbine at 100 mm Hg absolute pressure. The turbine isentropic
efficiency is 0.75. Calculate the efficiency of the plant. If the unit produces 12.5 MW.
what is the steam flow rate?
Answers
Answer:
At 4 bar at 5 bar 150°C 200°C 152°C 200°C h = 2752.8 h = 2860.5 h = 2748.7 h = 2855.4 s = 6.9299 s = 7.1706 s = 6.8213 s = 7.0592 ∴at 4 bar 175°C at 5 bar, 175°C h = 12752.8(2860.52752.8)2+−h = 1751522748.7(2855.42748.7)200152⎛⎞−+−⎜⎟−⎝⎠= 2806.7 kJ/kg = 2800 kJ/kg s= 6.9299 + 1(7.17066.9299)2−s= 236.8213(7.05926.8213)48+−= 7.0503 kJ/kg – K = 6.9353 ∴ at 4.5 bar 175°C 1h= 2806.728002+= 2803.4 kJ/kg 1s= 7.05036.93532+= 6.9928 kJ/kg – K Pressure 100 mm Hg = 33100m(13.610 ) kg /m1000××× 9.81 m/s2 = 0.13342 bar = 13.342 kPa Here also entropy 6.9928 kJ/kg – K So from S. T. At 10 kPa at 15 kPa hf= 191.83 fs= 0.6493 fs= 0.7549 hf= 225.94 hfg = 2392.8 gs= 8.1502 gs= 8.0085 hfg = 2373.1 ∴at 13.342 kPa [Interpolation] fs= 1513.3420.64931510⎛⎞−+⎜⎟−⎝⎠(0.7549 – 0.6493) = 0.68432 kJ/kg – K gs= 1513.3428.15021510⎛⎞−+⎜⎟−⎝⎠(8.0085 – 8.1502) = 8.1032 kJ/kg – K ∴If dryness fraction is x then 6.9928 = 0.68432 + x (8.1032 – 0.68432) ∴x = 0.85033 ∴At 13.342 kPa
Answer:
42% , 5.92kg/s
Explanation: