Question

NOTE: This is a multi-part question. Once an answer is submitted, you will be unable to return to this part. The gas-turbine cycle of a combined gas-steam power plant has a pressure ratio of 8. Air enters the compressor at 290 K and the turbine at 1400 K. The combustion gases leaving the gas turbine are used to heat the steam at 15 MPa to 450°C in a heat exchanger. The combustion gases leave the heat exchanger at 247°C. Steam expands in a high-pressure turbine to a pressure of 3 MPa and is reheated in the combustion chamber to 500°C before it expands in a low-pressure turbine to 10 kPa. The mass flow rate of steam is 12 kg/s. Assume isentropic efficiencies of 100 percent for the pump, 85 percent for the compressor, and 90 percent for the gas and steam turbines. Determine the mass flow rate of air in the gas-turbine cycle. Use steam tables and the table containing the ideal-gas properties of air. (You must provide an answer before moving on to the next part.) The mass flow rate of air in the gas-turbine cycle is D k g/s.

Answer 1

Consider Gas turbine Plant: First gas turbine Plant of us drate diagram T-S in Brayton cycle es As we know, isentropic Given T, = 290k As we show, T. 1954. It is lisen topics T = 290 (8) { Yair = 14 It is Process r = 5.25.32K it takes more Efficiency it But work due to than compressor ideal given! [.: Compresor 8. ] (-) a Ti TZ = 290 + 525.32-290 0.85 T2 = 566.846

Similarly between 3& 4 1400 & Ty = 772.86K But actually due to turbine efficiency it Produces less work than expected given . (nb) = 09 (73 Tu) : Vereine 9077 Tzn = 1400 – 0.9 (1400 -772-86) TU = 835:58 835.57k S per given air from state at Ty till Temperature = 247 °C = 520k heat is extracted by Heat Exchanger and used for heating Steam in steam Power plant inplate co Heat Extracted by Heat Exchanger = sair Cair (T4 - 520) Extracted my 0,= ma (1005) (838:57–520) ☆ * 317.15 KM 9 = 317.15 m

Now, Tas diagram of steam Power Plant IS MRa, 4500 13ooc WE a and steam tables using mollier chart a) At 15 MPa, 450°c h = 3155 kJ/k96 s = 6.14 ks K is isentropic Prosess b) As statt i to state co S,= S2 = 6:14 & At 3 MPa, Se 6.14 ha = 2780 kg isobaric heat addition c) State a to state 3 . At 3MPa, 500c h₂=3455 KB/S Sz = 7.23 kolik

d) Expansion state 3 to state 4 is isentropic At lokpa, bu Su= S3 = 7-23 1 . At lokpa, Su = 7.23' hu= 2290.5 Kg e) state u to states is isobaric (isothermal heat rejection . Ab lokpa he = 417.5 kJ/kg vos = Vp = 0.00104 Kereta Son (Pom posed) f) State 6 enthally calculated from following formula as it is isentropic Compression (Pump hochz = V (AP) ho = 417.5+ 0.00106 (15x103 -10) ha = 43309 KJ Heat Extracted from to charge state heat exchanger is used from states to state I h-ho = 3155 - 433.09 = 2721-11 KJ

As per given in steam = 1219/5 Q = 12 ( 2721.91) 14 = 32-663 MW In heat exchanger Heat Gas- turbine Power plant get balanced between and steam Power Plant 317.15 ma 232-663 x 103 ma = 102-99kg/5 ma ~ 103kg/s

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