To determine The net power output by the power plant operating on a simple Brayton cycle. Explanation Show the simple Brayton cycle, with air as the working fluid on the T − s diagram. For the given Brayton cycle with air as the working fluid T i , h i , P r i and P i are the temperature at i t h state, specific enthalpy at i t h state, relative pressure at i t h state, and pressure at i t h state respectively. Write the expression of the pressure ratio for the Brayton cycle ( r p ) . r p = P 2 P 1 (I) Write the expression of heat rejected from the power plant ( Q ˙ out ) . Q ˙ out = m ˙ ( h 4 − h 1 ) (II)) Here, the mass flow rate of air is m ˙ . Write the expression of relation between the relative pressure and actual pressure for the process 1-2. P r 2 = P 2 P 1 P r 1 (III) Write the expression of power given to the compressor ( W ˙ C,in ) . W ˙ C,in = m ˙ ( h 2 s − h 1 ) η C (IV) Here, the efficiency of the compressor is η C . Write the expression of power developed by the turbine ( W ˙ T,out ) . W ˙ T,out = m ˙ ( h 3 − h 4 ) (V) Write the expression of net power output of the power plant ( W ˙ net,out ) . W ˙ net,out = W ˙ T,out − W ˙ C,in (VI) Conclusion: Substitute 120 psia for P 2 and 15 psia for P 1 in Equation (I). r p = 120 psia 15 psia = 8 Refer Table A-17E given in the textbook to find the properties of air at 2,000 R ( T 3 ) and 1,200 R ( T 4 ) , which gives 504.71 Btu / lbm for h 3 , and 291.30 Btu / lbm for h 4 . Rearrange Equation (I) and find h 1 . Substitute 40 lbm / s for m ˙ , 291.30 Btu / lbm for h 4 and 6 , 400 Btu / s for Q ˙ out in Equation (II). h 1 = h 4 − Q ˙ out m ˙ = 291.30 Btu / lbm − 6 , 400 Btu / s 40 lbm / s = 131

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Chapter 9.12, Problem 91P

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The net power output by the power plant operating on a simple Brayton cycle.

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Chapter 9.12, Problem 89P

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THERMODYNAMICS(SI UNITS,INTL.ED)EBOOK>I

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