An air-standard dual cycle has a compression ratio of 8.6 and displacement of Va = 2.2L.At the beginning of compression, p; - 95 kPa, and T: - 290 K. The heat addition is 5 kJ, with one quarter added at constant volume and the rest added at constant pressure. Determine: a) each of the unknown temperatures at the various states, in K. b) the net work of the cycle, in kJ. c) the power developed at 3000 cycles per minute, in kW. d) the thermal efficiency. e) the mean effective pressure, in kPa.

Elements Of Electromagnetics
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ISBN:9780190698614
Author:Sadiku, Matthew N. O.
Publisher:Sadiku, Matthew N. O.
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An air-standard dual cycle has a compression ratio of 8.6 and displacement of V3= 2.2 L. At the beginning of compression, p; - 95 kPa,
and T1- 290 K. The heat addition is 5 kJ, with one quarter added at constant volume and the rest added at constant pressure.
Determine:
a) each of the unknown temperatures at the various states, in K.
b) the net work of the cycle, in kJ.
c) the power developed at 3000 cycles per minute, in kW.
d) the thermal efficiency.
e) the mean effective pressure, in kPa.
Step 1
Your Answer
Correct Answer (Used)
Determine each of the unknown temperatures at the various states, in K.
T2 =
670.0
K
T3 =
1202
K
%3D
T4 =
2291
K
Ts =
1454
K
Step 2
X Your answer is incorrect.
Determine the net work of the cycle, in kJ.
Weycle =
2.61
kJ
Transcribed Image Text:An air-standard dual cycle has a compression ratio of 8.6 and displacement of V3= 2.2 L. At the beginning of compression, p; - 95 kPa, and T1- 290 K. The heat addition is 5 kJ, with one quarter added at constant volume and the rest added at constant pressure. Determine: a) each of the unknown temperatures at the various states, in K. b) the net work of the cycle, in kJ. c) the power developed at 3000 cycles per minute, in kW. d) the thermal efficiency. e) the mean effective pressure, in kPa. Step 1 Your Answer Correct Answer (Used) Determine each of the unknown temperatures at the various states, in K. T2 = 670.0 K T3 = 1202 K %3D T4 = 2291 K Ts = 1454 K Step 2 X Your answer is incorrect. Determine the net work of the cycle, in kJ. Weycle = 2.61 kJ
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