In an air-standard Brayton cycle, air enters the compressor at 1000 lbm/hr, 14.7

Question

In an air-standard Brayton cycle, air enters the compressor at 1000 lbm/hr, 14.7 psia. and 40degreeF. and it leaves the compressor at 70 psia. The inlet turbine temperature is 1440degreeF. Use the method of Table 16-1 to determine the following: If the compression process of the above cycle is accomplished in two equal stages (each pressurization ratio is (70 /14.7)) with intercooling back to 70degreeF in between, determine the efficiency and net work of the cycle. Complete the table below * Do not use Table 9-1 formulas. Use the method of Appendix I.

Explanation / Answer

At state 1, P1 = 14.7 psia, T1 = 40 deg F, From air standard tables h1 = 119.48 Btu/lb

At state 2, P2 = 70 psia, T2 = T1*(P2/P1)^(1-1/) = (40+460)*(70/14.7)^(1-1/1.4) = 781 R = 781-460 deg F = 321 deg F, From air std tables, h2 (at 321 deg F is) = 186.94 Btu/lb

At state 3, P3 = P2 = 70 psia, T3 = 1440 deg F. From air standard tables, h3 (at 1440 deg F is) = 477.09 Btu/lb

at state 4, P4 = P1 = 14.7 psia, T4 = T3*(P4/P3)^(1-1/) = (1440+460)*(14.7/70)^(1-1/1.4) = 1216 R = 1216-460 deg F = 756 deg F. From air std tables, h4 (at 756 deg F is) = 296.4 Btu/lb

Let intermediate state of compression before intercooling be 2' and after intercooling be 2''. We have

T2' = T1*(P2'/P1)^(1-1/) = (40+460)*(70/14.7)^(1-1/1.4) = 625 deg R = 625-460 deg F = 165 deg F

From table, h2' corresponding to 165 deg F is = 149.3 Btu/lb and h2'' corresponding to 70 deg F is h2'' = 126.8 Btu/lb.

Similary for second compression stage, T2 = (70+460)*(70/14.7)^(1-1/1.4) = 663 deg R = 663-460 deg F = 203 deg F

from table, h2 corresponding to 203 deg F is = 158 Btu/lb

Turbine work Wt = h3-h4 = 477.09-296.4 = 181.05 Btu/lb

Compressor work = (h2'-h1) + (h2 - h2'') = (149.3-119.48) + (158-126.8) = 61.02 Btu/lb

Net work = Wt - Wc = 181.05-61.02 = 120.03 Btu/lb

Net work = 1000 lb/hr*120.03 Btu/lb = 120030 Btu/hr

Heat input = h3-h2 = 477.09-158 = 319.09 Btu/lb

Efficiency = Net work/ Heat input = 120.03/319.09 = 0.376 = 37.6 %

State Specific enthalpy (Btu/lb) Pressure (psia) Temperature (deg F) 1 119.48 14.7 40 2 186.94 70 321 3 477.09 70 1440 4 296.4 14.7 756

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