Question

Problem 1 0/5 points (0%) Consider a combined gas-steam power plant that has a net power output of 500 MW. The pressure ratio of the gas-turbine cycle is 12. Air enters the compressor at 308 K and the turbine at 1300 K. The combustion gases leaving the gas turbine are used to heat the steam at 9 MPa to 360°C in a heat exchanger. The combustion gases leave the heat exchanger at 460 K. An open feedwater heater incorporated with the steam cycle operates at a pressure of 0.5 MPa. The condenser pressure is 29.7 kPa. Assume all the compression and expansion processes to be isentropic. {kJ/kg) (1) The net work of the gas cycle per kg of combustion gases. (2) The net work of the steam cycle per kg of steam. 6{kJ/kg} (3) The mass flow rate ratio of air to steam. 6{kg air/kg steam} (4) The total network output of combined cycle per kg of combustion gases. (5) The required rate of heat input to the combustion chamber. (6) The thermal efficiency of the combined cycle. {%} {KJ/kg} {W} Now assume isentropic efficiencies of 100 % for the pump, 76 % for the compressor, and 83 % for the gas turbines and 88 % for steam turbine. (7) The thermal efficiency of the combined cycle. 0 %}

Answer 1

Solution : Given : Net power output = 500 MW. Relative pressiue ratio = 12. Temp of air entering compressor -308 K emp. of air entenne turbine = 1300k Temp of steam = 360 C & Pressure of steam = ampa Temp. of combustion gases leaving heat exchanger 460k! Pressive inside seedwater heated = 0.5MPa. - Condenser pressure-29 Ikla. -70% Mat = 83% nst -88% np = 100% enthalpy of air at 308, K = 308 kJ/ug = ho Relative pressure, Prag 21.427... for isentropic process 8.9. Pra aro 12 » Pag = 17.12 Enthalpy value for this pressure using interpolation: has = 607 + 1712-163 0 - 607) 21.9.-16.3 - 614.76 lesleg Actual enthalpy due to compressor na has hg - 614.76 -308 ha ha ha-80-8 1 hq - 308 306.76 - 403.63 - ho~711: 63 kilug Now enthalpy of air at 13ook hion 1396 krug, Relative pressme, Pr=33)

Enthalpy value for this podeure using interpolation Whisbet 28.8.-21.9 (713-660) his - 660+ 27.58-21.91 has a 660 + 43.628 2 703.628 KJ /eg. Actual enthalpy due to turbine n - ho-hi - 1396-ho a hor his 1396- 703.628 0.83 - 1396-ho 692.37 7 h = 821.33 kilig. Properties of water at p, 29. 7upa Sp. enthalpy, h = h = 288.85 kJ/kg. Sp. volume, &, = = 0.001022 m3/ug. Work done between condenser a feed water Kpump = 3, (P2P) - -0.00102321500-297) - 0148 & Wpung - Ma chi - 0.49 h - 28.8.85 7 hq - 23.9.33 kJ/kg Properties at P₂ = 500 kla, sp. enthalpy , hz - h=640.81 kJ lug sp. volume = V = 0.00 109 m3/kg Work done between heater f boiler Wpump - 0.0010919-015) x103 = 9,265 & Wpump = (hu- 640·81) = 9.265 hy = 640.075 kJ/leg

Properties at ampa f 360°C Specific enthalpy & sp. entropy are: he - 2982.6 kilug S = 6.0942 ulugk Entropy at 65 & 7s. Sbs = 6:0942 og S7 = 6:0942 kugk Properties at 500kla & calculating doyness fraction h - 640.81 kilug ng a 2107.5 kilug Sf = 1.8621 les lugk, Segn 4.9571 kr/ugk S69 - Set x6s (Sgg" 6.0942 - 1.8621 + xo (4.9571) -> Kos - 0.853 chos = 640.81 +0.853 (2107.5) - 243815 kJ/kg Actual enthalpy due to efficien hy the . hg- hos - 0.88 - 2992.6-ho - 2992.6-243815 7 ho- 2504.99 kJ lig. Properties at 29.7 kPa to calculate dryness fraction: h=288.85 kilkg, hgga 2335.3 weg S = 019429 kJugl , Saga 6.826 kr/kgk Sze. S t Ras (88) 6.0942 = 0.9429 + x78 (6.826) 7 X78 = 0.7546 a has = 288.85 +0.7546 (2335.3) = 2051.06 Klug I of turbine

her has » 0.88 = 2992.6-hz 2992.6-2051.06 > hy=2164.04 kJ/kg At Ti - 460k 112 - 462 kJ/kg. Using steady state energy low equation mg (he-hu) a mais ch, -h12) 821.33-462 mas Mair s hy 2992.6-640.075 - mg 0.1527 ③ . Mass flow rate ratio of air to steam = 6.548 Am msteam - nie-hoz, 821:33-462 2 Mais Mais - 6.540 team extracted. (1-4) hq tyho = 1x hz = (1-4 289.33+ y 2504.99) - 640.81 > 289.33 - 289.33y + 2504.99 y = 640.81 24.66y= 351.48 » y = 0.1586 Calculating the turbine work: We a he-het(1-4) (he-ha) =299206-2504.99 + 101186) (2304,99-216404 Wy = 774.48 kilig (2) Network of the steam cycle Whetsteam = 774:18 - (1-001536) 0.48 -9,265 = 764.5 kilig Am

(1) Net work of the gas cycle = (horhu (ha-ha (1396 - 921:33) --'/711.63-308) & nedgan) 171.04 kilug. (W Net work output What a Wherfgen) + Winer (steam) e n 171.04 + 1764.5) & Wneta 287.79 kJ lug. Mass flow rate, soox 1000 173737 kg/l Ang Mair 287.79 6 Required heat input to combustion chamber Q = mais (ho- ha) - 1737.37 (1396 - 711.63) » Q = 1189.003 MW 1 (6) Thermal efficiency of the cycle n. 500 - 42.05% 2- 1189.003 (7) Thermal efficiency of combined cycle 2 1 X 0.76 x 0.83 x 0.88 = 55.51% Ary

Gascycle Steam cycle Temperature - Entropy diagrama
Summary:(1)Net work of gas cycle=171.04kj/kg.(2)Net work of steam cycle=764.5kj/kg.(3)Mass flow rate ratio of air to steam=6.548.(4)Total net work output=287.79kj/kg.(5)Required heat input=1189.003MW.(6)Thermal Efficiency of cycle=42.05%.(7) Combined thermal efficiency=55.51%.