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Common data for 1 & 2 Air enters the compressor
of an ideal air standard Brayton cycle at 100 kPa, 200C and the pressure ratio across the compressor is 12:1.Flow rate of air is 10 kg/s and maximum temperature in cycle is 11000degreeC. Compressor Power input is:-
a) 6.1 MW
b) 3.03 MW
c) 6.2 MW
d) None of these
Correct answer is option 'B'. Can you explain this answer?
Verified Answer
Common data for 1 & 2 Air enters the compressor... more of an idea...
P2/P1 = 12.
then T2= (12)^{0..4/1.4} × 293 = 595.94K.
Then in ideal brayton cycle work done by compressor is =(h2- h1)= m×Cp× (T2-T1)= 10×1.005× (595.94-293) = 3044.54 KW = 3.04 MW
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Most Upvoted Answer
Common data for 1 & 2 Air enters the compressor... more of an idea...
P2/P1 = 12. then T2= (12)^{0..4/1.4} × 293 = 595.94K. then in ideal brayton cycle work done by compressor is =(h2- h1)= m×Cp× (T2-T1)= 10×1.005× (595.94-293) = 3044.54 KW = 3.04 MW . So option D is correct
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Community Answer
Common data for 1 & 2 Air enters the compressor... more of an idea...
P2/P1 = 12.
then T2= (12)^{0..4/1.4} × 293 = 595.94K.
Then in ideal brayton cycle work done by compressor is =(h2- h1)= m×Cp× (T2-T1)= 10×1.005× (595.94-293) = 3044.54 KW = 3.04 MW
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Common data for 1 & 2 Air enters the compressor... more of an ideal air standard Brayton cycle at 100 kPa, 200C and the pressure ratio across the compressor is 12:1.Flow rate of air is 10 kg/s and maximum temperature in cycle is 11000degreeC. Compressor Power input is:-a) 6.1 MWb) 3.03 MWc) 6.2 MWd) None of theseCorrect answer is option 'B'. Can you explain this answer?
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