
Concept explainers
(a)
The maximum temperature in the cycle.
(a)

Answer to Problem 153P
The maximum temperature in the cycle is
Explanation of Solution
Draw the
Refer to Table A-2b, obtain the properties of air at 1000 K.
Express the compression ratio.
Here, clearance volume is
Express the total volume of the engine at the beginning of compression process (state 1).
Process 1-2: Isentropic compression
Calculate the temperature at state 2.
Here, temperature at state 1 is
Calculate the pressure at state 2.
Here, pressure at state 1 is
Process 2-x and x-3: Constant-volume and constant pressure heat addition processes
Calculate the temperature at x state.
Here, pressure at x state is
Calculate the heat addition to the process 2-x.
Here, constant volume specific heat is
Calculate the heat addition to the process x-3.
Here, constant pressure specific heat is
Conclusion:
Substitute 16 for r and 1.8 L for
The value of
Substitute
Substitute 343 K for
Substitute 95 kPa for
Substitute 3859 kPa for
Substitute
Substitute
Thus, the maximum temperature in the cycle is
(b)
The net work output.
The thermal efficiency.
(b)

Answer to Problem 153P
The net work output is
The thermal efficiency is
Explanation of Solution
Express the total heat addition to the process.
Calculate the volume at state 3.
Here, volume at state x is
Process 3-4: Isentropic expansion
Calculate the temperature at state 4.
Here, volume at state 3 and 4 are
Calculate the pressure at state 4.
Here, pressure at state 3 and 4 are
Process 4-1: Constant volume heat rejection
Calculate the heat rejection.
Calculate the net work output.
Calculate the thermal efficiency.
Conclusion:
Substitute 702.6 kJ/kg for
Substitute
Substitute 2308 K for
Substitute 7500 kPa for
Substitute
Substitute 569.3 kJ/kg for
Thus, the net work output is
Substitute 835.8 kJ/kg for
Thus, the thermal efficiency is
(c)
The mean effective pressure.
(c)

Answer to Problem 153P
The mean effective pressure is
Explanation of Solution
Calculate the mass.
Calculate the mean effective pressure.
Conclusion:
Substitute 95 kPa for
Substitute 0.001853 kg for m,
Thus, the mean effective pressure is
(d)
The power for engine speed of 3500 rpm.
(d)

Answer to Problem 153P
The power for engine speed of 3500 rpm is
Explanation of Solution
Calculate the power for engine speed of 3500 rpm.
Here, engine speed is
Conclusion:
Substitute 2200 rev/min for
Thus, the power for engine speed of 3500 rpm is
(e)
The second law efficiency of the cycle.
The rate of exergy output with the exhaust gases.
(e)

Answer to Problem 153P
The second law efficiency of the cycle is
The rate of exergy output with the exhaust gases is
Explanation of Solution
Express the maximum thermal efficiency of the cycle.
Here, dead state temperature is
Express the second law efficiency of the cycle.
Calculate the rate of exergy of the exhaust gases.
Here, specific internal energy at state 4, dead state is
Conclusion:
Substitute
Substitute 0.8709 for
Thus, the second law efficiency of the cycle is
Substitute
Thus, the rate of exergy output with the exhaust gases is
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Chapter 9 Solutions
EBK THERMODYNAMICS: AN ENGINEERING APPR
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