Question

Problems 2 through 6 below are to be considered with reference to an Ideal Brayton Cycle with the following operating parameters: • Atmospheric conditions of 300K @ 100 kPa at the compressor inlet • 1.250K @ 900 kPa at the turbine inlet Be sure to read each problem carefully and utilize the specified assumptions. Use the air standard assumptions and treat specific heats as described in each problem. Also, assume isentropic mechanical energy transfers and isobaric thermal energy transfers throughout. For all problems, use the station numbers indicated in the accompanying figures to keep track of your calculations. 2. For the cycle described above, compute the following results: • Required heat input to the cycle, per unit mass, • Net work transferred out of the cycle, per unit mass • Cycle thermal efficiency Comp I Torb combustion ! Chamber Ideal Brayton Cycle Using the following sets of assumptions: a) Cold air standard assumptions (air standard assumptions with constant room temperature specific heats b) Air standard assumptions with average temperature properties determined by entering average temperatures for the process of interest into the Spreadsheet found under the useful links" tab found in your course website EGR 2320 Fall '19 Thermodynamics Test #1 Dr. Short -0.9. c) Cold air standard assumptions and the following mechanical efficiencies: 0.9 d) Air standard assumptions with both average temperature specific beats and the following mechanical efficiencies: Ne n u 0.90

Answer 1

Brayton cycle comp. combustion Twritine) I chamber F US Brayton cycle with turbine and compr efficiency. T= 3ook T3 = 1250k P, = 100k Pa = Py = Pus, P=900kPa = P2 = P25 I c = r = 0.9.8=14 ? Acc. to the cp: loos kulkg K. J Assumptions - Calculation of Temp at all points upe Tas (P3) Tas = 300(90) Tab = 562.03 x] > isentopic process 1.= 0.9 = Tas TI - Tp - 300=562-03-300 - T2 = 591.15k ] TRTI - Tus= 12.50 [Tus = 667.38* ] 0.9 Y ) Tз . /Р. 1 = 0.9 = T3 - Ty a T3 - Tu= 0.9 (1250-667-88) T4 : 725.64 x] T3-Tus ② Heat input to the cycle (As) (22-3) Per unit mais 0s = Cp (T3-Tz) ) Qs = 1:005( 1250-591.15] Is = 662.144 KJ / kg 2 Net work output (ulnet) per unit man Whet = WT- We - Cp (T3 - Tu) - Cp (T2-T ) m

Wnet/ in = 1.005 (1250 - 725.64) - 1.005( 591.15-300) Whet = 526.98 - 292.6 Wnet = 234.37 kJ / kg © Cycle Thermal Efficiency cy) = Wnet. => 1 To 234.37. X 100 Heat gupplied (as) 662.144 35.39%
I have solved all the three parts by taking the assumptions into consideration of specific heats and compressor and turbine efficiencies . I have written all the formulas that I have used so that you can understand the solution easily and also use them for future reference.

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