Question

A natural gas turbine is used to generate 5MW of power using air as the working fluid. The turbine operates at a mass flowrate of mair=10kg/sec the inlet conditions of Pmax=8MPa and Tmax=627C, and exhaust conditions of P2a=300kPa and T2a. The outside air temperature is at Tenv=25C and Qlose=300kW is lost to the environment. Its isentropic efficiency is nt. Use R=Cp-Cy, Cpve=1.06kJ/kgK, Cav=.773kJ/kgK, k=1.4 and Show work for credit!! a) (3pts) What is the isentropic efficiency, nt. of the turbine? b) (3pts) What is the entropy change of the air for this process? c) (2pts) What is the generated entropy, Sgen, for this process? d) (2pts) If the turbine operates on a cycle between Tmax and Tmin=T2a what is highest efficiency that this power plant could achieve? kJ/kgk kJ/kg. K Temperature, Air 250 300 350 400 450 500 550 600 650 700 750 800 900 1000 1.003 1.005 1.008 1.013 1.020 1.029 1.040 1.051 1.063 1.075 1.087 1.099 1.121 1.142 0.716 0.718 0.721 0.726 0.733 0.742 0.753 0.764 0.776 0.788 0.800 0.812 0.834 0.855

Answer 1

This question is from a gas turbine, in which power produced by the turbine is given so exhaust temperature can easily find out, after that just basic formulas of a gas turbine are used.

/ 45 4 - Solution Giuen Data + - Power geneanted boy TA - turbino(P)= 5 MW Wooking fluid = Alio / mais flow gate (ñ) = 10 kg /sue. S • Inlet conditions of turbine = 1. Piram = P = 8 MPa Tman. = Ti = 627 °c Exhaust conditions ; - Taa = Tea Paa = BookPa Outside aio tempegature = 25 Heat lost (Qgast) = BookW Idents opic efficiency of turbine = my

1 Use - R = Cp - Cv Cang. = 1.06 kJ /rg k Evang. = 0.773 kJ/kg k KE 1.4 ; Analy.sis : (a) To Find - Isentoopic efficiency of Turbins (9) P=5mW Using eq. t Q =oH+W COH= Change in Enthalpy. AH = Reat-'W [" OH = 300 - 5000 OH = - 4700 kW N OH = ŕ cp (Tan - T.) = -4700 Taa -T; = =443.396 Toa = Ti - 443:396 = 183.6°C Taa = 183.6°C

If turbine expangion process is Tsentropic , then we can write → K -1 (5). (*)* 5 = 0.391 Tas = 900 x 0.391 Tas = 351.9k = 79.240 ^, = 7 - To 697 – 18306 627 - 79.24 = 0.8094 OR 80.94% y Answer

- (b) To find & Entropy change of air Analysis for this process costobling • Entropy change foo ideal gas in flow process • is given by - E E AS er min ( Ep lon T -R len E 4S.10 (1.06 km 5.606 - 0.287.len ) 4.5.10 60-7197 0.400) Kokek-> Answers (0) To find – En toepy generated (Sgen) s jon = A Scoaline + Asusending E AS susunding = elements - ago Kereta E Szasounding = 1.079 KW E Ifwe, ourdina

: Syam. = (2.31+ 1,079) Kue Sgom. = 3.289 KW K Answer To find : Highest efficiency of power plant when turbine operates on a cycle between Tmar. and Tmin - Too ( Phan rawatan = 1 - hart./ Power blant = rp = pressure Ration = max. K -1 . . Tmen. I Mp = 10.75 Imen. / 1 pouces Hand = 1 - (10.75)0-2857 0.4926 OR 49.26% (max. Power plant = Ansuar