Question

please show work for all sub parts

A combined cycle gas turbine / vapor power plant uses the turbine exhaust as the energy source for the boiler. Each power system uses a single turbine. The gas power system is modeled as an ideal air-standard Brayton cycle. The vapor power system is modeled as an ideal Rankine cycle. Given specific operating conditions determine the temperature and pressure at each state, the rate of heat transfer in the boiler, the power output of each turbine, and the overall efficiency. |--Given Values-- Brayton Compressor Inlet: T1 (K) = 316 Brayton Compressor Inlet: P1 (kPa) = 138 Brayton Heat Exchanger Exit: T5 (K) = 523 Brayton: rp = 10.2 Brayton: rc = 2.7 Brayton: AV (m3/s) = 1.5 Rankine Turbine Inlet: T7 (C) = 560 Rankine Turbine Inlet: P7 (MPa) = 14 Rankine Condenser Inlet: P8 (kPa) = 250 a) Determine the specific enthalpy (kJ/kg) at the compressor inlet. Your Answer = 316.274 Correct! Exact Answer= 316.27 +/- 1.2E+00 b) Determine the specific enthalpy (kJ/kg) at the combustor inlet. Your Answer = 614.026 Correct! Exact Answer=614.04 +/- 2.3E+00 c) Determine the specific enthalpy (kJ/kg) at the gas turbine inlet. Your Answer = 1803.99 Correct! Exact Answer= 1804.05 +/- 6.1E+00 d) Determine the temperature (K) at the gas turbine inlet. Your Answer = 1637.982 Correct! Exact Answer= 1638.0 +/- 5.6E+00 e) Determine the pressure (kPa) at the gas turbine inlet. Your Answer = 1407.6 Correct! Exact Answer= 1407.6 +/- 4.0E+00 f) Determine the specific enthalpy (kJ/kg) at the gas turbine exit. Your Answer = 902.592 Incorrect g) Determine the specific enthalpy (kJ/kg) of the gas at the heat exchanger exit. Your Answer = 525.615 Correct! Exact Answer=526.74 +/- 1.7E+00 h) Datar

Answer 1

10.2 Bray Given :- Brayton compressor inlet (Ti)= 316 K Brayton inget compressor (P) = 138 kPa Brayton. heat exchanger exit (T5) = 523.5 ton (op) Brayton (rc) = 2.7 Brayton (AV) 1.5 m3/s Rankine turbine inlet (Ty ) = 560°c Rankine turbine inlet (Py = 14MPa condenser inlet Kpa 250 Rankine inlet @ specific enthalpy. at compressor his 316.27 kJ/kg T = 31SK b Sp. enthalpy @ com bustor inlet Ch2) have Pr. we At Ti = 316K 1.6629 we know Pra= Pri. P2 1-6629 X 10.2 Pr2 = 16.9612 Pr2 = Now at 16.9612 614.03 T2 = 606.67 K Specific enthalpy gas turbine iniet T3 = rex Tz 2.7 X 606.67 T3 = 1638 K At T3 16 38 K 13 = 1804.01 kJ/kg @ Temperature at the gas turbine inlet T3 = = 1638K e Pressure at the gas. turbine inlet Prz = 874.42 T3 = 1638K But P3= P2 and 10-2 » P2 = 138 x 10.2 Pi P2 = 1407.6kPa P2 Pa = 1407.6 kPa specific enthalpy at the gas turbine exit Pa P4 Pr4 874.42 10.2 P2 Pi P3 P4 Pr4 .. Pr4 = 85.73 Now h4 966.85 KJ/kg sp. enthalpy of the gas at the heat exchanger exit Ths 526.74 kJ XJ/kg @ T5 = 523 K mass How rate of in gas power system SAV = AV RTI ma 138 0.287 x 316 x 1.5 ma = 2. 282 kg/s iy Net power output of brayton cycle. Winet aña [cha-h4) - Cha-h)] 2.282 [ (1804.01 - 966.85) - C614.03 – 316-272] I wnet 1230.91 kW js Rate of heat transfer to combustor (&c) å ca ma Cha-ha) 2.282 % (1804.01 - 614.03) äc 2715.53 KW KY Rate of heat transfer to the water in boiler hg hf@250 kPa 535.37 kJ/kg Wpa vf @ 250 kPax (14000 250) 0.0010672% (14000 - 250) = 14.67 kJ/cg ho hz hgt wpz 550.04 kJ/kg 3486 kJ/kg @ 14mpa, 560°C 6.5941 kJ/kg IZ 575 ay P8 250kPa S8 - Sf Sfg 6.5941 -1.6072 7.05.27 -1.6072 58 S7 28 0.92 hga hft. ngx hfg 535:32 +0.9 2x 2181-5 h8= 2542.35 KJ/kg ms ina hu-hs hr-ho mg. 966.85 2.282 526.7Ų 550.04 3486 mis 0.34.21 kgls åb ins x Chy-h6) 0.3421 (3486-550.04) åb = Too 4.4 KW l] specific enthalpy at turobine inlet hy 2 3486 kJ/kg m] sp. entropy at turbine in let 57= 6.5941 kJ/kg k n] sp. enthalpy at condenser inlet h8 2542.35 KJ/kg o] specific enthalpy at pump inlet hg 535.37 KI/kg p] specític enthalpy at pump exit ho 550.04 kJ/kg q] mass flow rate of water ms = 0.3421 kg/s o] Net power developed ins [Chy-h8) - Cho-hg)] = 0.3421 x [(3486-2542.35) (556.04 -535.37) Wnet 928.98 KW S] Thermal efficiency of combined cycle + Whet, s 123091 + 928.98 2715.53 Whet, og Mth Qin 79.54 %.