Concept explainers
(a)
Draw the
(a)
Answer to Problem 173RP
The
Explanation of Solution
Draw the
Thus, the
(b)
The expression for the back work ratio as a function of k and r.
(b)
Answer to Problem 173RP
The expression for the back work ratio as a function of k and r is
Explanation of Solution
Find the work of compression using the first law for process 1-2.
Here, heat interaction during the process 1-2 is
Write the expression of expansion work.
Here, gas constant is R, specific volume at state 2 and 3 is
Write the expression of back work ratio using the equations (I) and (II).
Here, temperature at state 1, 2, and 3 are
Conclusion:
Process 1-2: Isentropic
Calculate the ratio of
Here, pressure at state 1 and 2 is
Process 2-3: Constant pressure
Calculate the expression for
Here, volume at state 1 and 2 is
Process 3-1: Constant volume
Calculate the expression for
Substitute
Thus, the expression for the back work ratio as a function of k and r is
(c)
The expression for the cycle thermal efficiency as a function of k and r.
(c)
Answer to Problem 173RP
The expression for the cycle thermal efficiency as a function of k and r is
Explanation of Solution
Express out the heat addition and heat rejection in the process using first law to the closed system for processes 2-3 and 3-1.
Here, constant pressure specific heat is
Express the cycle thermal efficiency.
Conclusion:
Substitute
Substitute
Thus, the expression for the cycle thermal efficiency as a function of k and r is
(d)
The value of the back work ratio and thermal efficiency as r goes to unity.
(d)
Answer to Problem 173RP
The value of the back work ratio and thermal efficiency as r goes to unity is
Explanation of Solution
Recall the expression of back work ratio and apply the limits as r goes to infinity.
Thus, the value of the back work ratio as r goes to unity is
Recall the expression of cycle thermal efficiency and apply the limits as r goes to infinity.
Thus, the value of the thermal efficiency as r goes to unity is
From the results of cycle thermal efficiency and back work ratio values of 0 and 1, it shows that no expansion and net work can be done whether you add heat to the system when there is no compression
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Chapter 9 Solutions
Thermodynamics: An Engineering Approach
- An air-standard cycle is executed in a closed system with 0.5 kg of air and consists of the following three processes: 1-2 Isentropic compression from 100 kPa and 27C to 1 MPa 2-3 P = constant heat addition in the amount of 416 kJ 3-1 = 5 c1v + c2 heat rejection to initial state (c1 and c2 are constants) (a) Show the cycle on P-v and T-s diagrams. (b) Calculate the heat rejected. (c) Determine the thermal efficiency. Assume constant specific heats at room temperature.arrow_forwardA compression ratio of 8 is achieved by using an ideal air-standard Otto cycle engine. The working fluid has a pressure of 100 kPa and a temperature of 27°C at the start of the compression process, and the constant volume heat addition process supplies 800 kJ/kg of heat to the working fluid. What are the (a )the temperature, volume and pressure of the air at the end of each process (in K, m3 and kPa) (b) the net work output/cycle [kJ/kg], and (c) the thermal efficiency of this engine cyclearrow_forwardAn air-standard cycle is executed in a closed system and is composed of the following four processes: 1-2 Isentropic compression from 100 kPa and 27°C to 1 MPa 2-3 P = constant heat addition in amount of 2800 kJ/kg 3-4 V = constant heat rejection to 100 kPa P = constant heat rejection to initial state (a) Show the cycle on P-v and T-s diagrams. (b) Calculate the maximum temperature in the cycle. (c) Determine the thermal efficiency. Assume constant specific heats at room temperature.arrow_forward
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