5. Air with T = 300 K and P = 100 kPa enters a Brayton-cycle compressor inlet. T

Question

5. Air with T = 300 K and P = 100 kPa enters a Brayton-cycle compressor inlet. The combustion chamber adds 670 kJ/kg of air. Due to material considerations, the maximum gas temperature allowed is 1200 K. Assume ideal gas, a polytropic process, and the values in Table A.5 are valid. The polytropic equations summarized on page 295 (Chapter 6) are valid for open and closed systems for P, T, and V only. HOWEVER, the polytropic specific work equations are only valid for closed systems. You cannot use them to calculate compressor and turbine work. a. What is the maximum allowable compression ratio? b. For the maximum allowable compression ratio, what is the net work? c. What percentage of the turbine work is needed to drive the compressor? d. For the maximum allowable compression ratio, what is the cycle efficiency? e. SKETCH a TS diagram for this process, showing the cycle relative to the 2-phase region

Explanation / Answer

Combustion (h3) = h2 + qH

2w3 = 0

Tmax = T3 = 1200 K

h2 = h3 - qH

h2 = 1277.8 - 670 = 607.8

T2 will be around 600 K

Pr2 = 13.0923

T1 = 300 K

Pr1 = 1.1146

Ideal compression ratio = P2 / P1 = Pr2 / Pr1 = 11.75

Maximum allowable compression ratio = 11.75

Ideal expansion ratio (Pr4) = Pr3 / (P3/P4) = 191.174 / 11.75 = 16.27

So if we interpolate linearly we wil get T4 = 636 K   and   h4 = 645.7

wT = h3 - h4 = 1277.8 - 645.7 = 632.1

- wC = h2 - h1 = 607.8 - 300.47 = 307.3

Net work (wnet) = wT + wC = 632.1 - 307.3 = 324.8

% of turbine work needed to drive the compressor = (307.3 / 632.1) * 100 = 48.62 %

Cycle efficiency = wnet / qH = 324.8 / 670 = 0.485

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