Question

power plant

3- A combined power plant consisting of a simple open cycle gas turbine unit, using air as the working fluid (neglect fuel mass flow rate), and a steam turbine is to be designed such that the heat rejected at the gas turbine unit is to be utilized to produce steam at the generator for the steam turbine. The exhaust air leaving the generator is at (200 C°). Draw a simple sketch for the plant and the (T-s) diagram and calculate the mass flow rate of steam per (kg/s) of airflow, the total power output per (kg/s) of airflow and the overall thermal efficiency of the plant. Assume ideal cycles. Neglect the feed pump work of Rankine cycle. The data of each cycle are as follows:- Gas Turbine Cycle Steam Turbine Cycle Pressure Minimum 1 0.07 (bar) Maximum 5 Temperature Minimum 20 (C) Maximum 830 300 30

Answer 1

2 T-S DIAGRAM ht sbar GAS TURBINE 35 CYCLE. 30bar STEAM CYCLE 0.07bas IS 45 S NOTE:- The state points in gas turbine cycle is represented using integers (i.e., 1,2,3,4) and state Points in the Steam cycle are represented using integers followed by Suffix 8 (1.e., 15, 25, 35, 45). GAS TURBINE CYCLE:- The minimum Pressure = P, = Iban The minimum temperature= T = 20°c = 293K. The Pressure at exit of compressos = P2 = 5bar. The maximum temperature in cycle = Tz=8301 = 1103k

We know that, T = ($) T3 (op) I= Ta. 1-4-1 of To 1103 T = 464.059K 293 Tu Tu = 696.4176k Temperature at the exit of compaesson = T2 =464-OSIK Temperature at the exit turbine Tu 696.4176k. NOTE :-Given that, the exhaust air leaving the gas turbine is utilized to produce steam at the generator. Det, Te Exit temperature of gas (air) in the generator Te=200°c = 473. ISK.) let, T; → Inlet temperature of air in the generator. Ti = Tu = 696041766 STEAM CYCLE: Minimum Pressure in the cycle = Pis = Pas = 0.07 bar 7KPA since the cycle is assumed to be ideal, The inlet to the Pump is saturated liquid. .: Enthalpy at inlet to the Pump = his = heapsat=ZkPa - his = 163.32 KJ/kg. since, the pump work is to be neglected (: given in statement .. has = his = 163-32kJ/kg

The Pressure in the generator = Pa Pag- 3obar - 3 mpa, The maximum temperature in steam cycle = Tac 300°c from super heated steam tables, = has at P3s = 3MPa and T3s 300°C, Enthalpy of steam = = 2994. IS KI kg Entropy of steam = Sys 6.56 KJ since the cycle is ideal, the turbine is isentsopic. ..Isus = 53s = 6.54 KJ /kg K. KJ/KgK, from saturated steam tables, at Psat = 0.07bas Sf = 0.559 KJ/KgK Sfg = 7.7178 kJ/kgK, we know that, So Sft aus Sfg 77.4967 ng 5 6.54 = 0.559 + TL5.707178 ) Pus= 0.77496 Rus = 0.77496 at Psat = 0.07bas, he = 163.32 I 3 hag = 2409.6 KI Kg. y has = hqtxos hag = 163.32 + (0.77496 X 2409•6) = 2030 66 KJ/Kg. "Enthally of steam at exit of turbine = h = 2030.6KJ kg "AS

ie, Energy lost by air Applying energy balance across generator, ie, max cea (T; - Te) = mg (has- has). ► maxl.cos (696.6/76-473-15) - s (2994-15- 163.32 Ms ma -0.07926 .. mass flow sate of steam per(Kg/s) & air flow = 0.07926 s - mass flow rate of steam Pes[kg/s) of air flow = 0.07926 Kg sec Net work produced by gas turbine per kg of air = way - We = [1h3 - bu) - (ha-hi] = cp[ta-Tu) - (T2-T,]] = 1.005 (1103-696-41) - (464.05 - 293)] 236.717 KJ Kg. Net Wosk Produced by steam cycle per kg 0 steam has hus 2994.15 - 2030.66 = 963.49 KJ Kg .. Total Power output = max Wome + (mis x Women Steam tusbine Cycle = 1x236.717) + (0.079 26 X 963.49) - 313.0839kW.

The total Power output per (K9/s) 4 air flow = 313-0839kW Finding overall thermal efficiency: - Total Power output 313.0834 KW 313.839 nou IXI-005 (103 Rate of heat supplied Maxcoal T3 - Tz) 466-059) -0.4875 = 48.75% loverall = 48.75% overall thermal efficiency of Plant