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Air enters a gas turbine compressor at a temperature of 300 K and a pressure of 100 kPa, and exits at a temperature of 660 K and a pressure of 950 kPa. Before entering the turbine, there is a heat transfer of 750 kJ/kg to the air. a) Wk=? b) η=? (It will be assumed that the cycle works according to the ideal Brayton cycle and the specific heats do not change with temperature. k=1.4)

          Air enters a gas turbine compressor at a temperature of 300 K and a pressure of 100 kPa, and exits at a temperature of 660 K and a pressure of 950 kPa. Before entering the turbine, there is a heat transfer of 750 kJ/kg to the air. a) Wk=? b) η=? (It will be assumed that the cycle works according to the ideal Brayton cycle and the specific heats do not change with temperature. k=1.4)
        
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University Physics with Modern Physics
University Physics with Modern Physics
Hugh D. Young 14th Edition
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Air enters a gas turbine compressor at a temperature of 300 K and a pressure of 100 kPa, and exits at a temperature of 660 K and a pressure of 950 kPa. Before entering the turbine, there is a heat transfer of 750 kJ/kg to the air. a) Wk=? b) η=? (It will be assumed that the cycle works according to the ideal Brayton cycle and the specific heats do not change with temperature. k=1.4)
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Transcript

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00:01 Determine the power delivered by this plant, assuming constant -specific heats at room temperature.
00:07 Assuming constant -specific heats, t2s is equal to t1 times pressure 1, pressure 2 over pressure 1, to the power of k minus 1 divided by k.
00:20 So we plug in our 290 times 8 to the power of 0 .4 divided by 1 .4 to get 525 .3 kelvin, t4s is then equal to our 1 ,100 kelvin times 1 8 to the power of .4 divided by 1 .4, which is 607 .2 kelvin.
00:44 Our efficiency is then equal to 1 minus t4 minus t1 divided by t3 minus t2.
00:54 So we plug in 1 minus 607 .2 minus 290 divided by 1100.
01:02 Minus 525 .3 to get .48.
01:08 Our net work, our net power, is equal to our thermal efficiency times q .n.
01:15 Our 0 .448 times 35 ,000 kilowatts.
01:24 So we get 15 ,680 kilowatts for part a.
01:30 Now for part b, assuming variable -specific heats, at temperature warmth, which is 290 kelvin, we get h1 is equal to 290 .16 kilojoules per kilogram.
01:44 P at r1 is 1 .211...
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