Analyzing the Dual Cycle
At the beginning of the compression process of an air-standard dual cycle with a compression ratio of 18, the temperature is 300 K and the pressure is 0.1 MPa. The pressure ratio for the constant volume part of the heating process is 1.5:1. The volume ratio for the constant pressure part of the heating process is 1.2:1. Determine (a) the thermal efficiency and (b) the mean effective pressure, in MPa.
Known An air-standard dual cycle is executed in a piston–cylinder assembly. Conditions are known at the beginning of the compression process, and necessary volume and pressure ratios are specified.
Find Determine the thermal efficiency and the mep, in MPa.
Schematic and Given Data:
Engineering Model
1. The air in the piston–cylinder assembly is the closed system.
2. The compression and expansion 4–5 are adiabatic.
3. All processes are internally reversible.
4. The air is modeled as an ideal gas.
5. Kinetic and potential energy effects are negligible.
Analysis The analysis begins by determining properties at each principal state of the cycle. States 1 and 2 are the same as in Example 9.2, so u_{1}=214.07 kJ / kg , T_{2}=898.3 K , u_{2}=673.2 kJ / kg. Since Process 2–3 occurs at constant volume, the ideal gas equation of state reduces to give
T_{3}=\frac{p_{3}}{p_{2}} T_{2}=(1.5)(898.3)=1347.5 K
Interpolating in Table A-22, we get h_{3}=1452.6 kJ / kg \text { and } u_{3}=1065.8 kJ/kg.
Since Process 3–4 occurs at constant pressure, the ideal gas equation of state reduces to give
T_{4}=\frac{V_{4}}{V_{3}} T_{3}=(1.2)(1347.5)=1617 K
From Table A-22, h_{4}=1778.3 kJ / kg \text { and } v_{ r 4}=5.609.
Process 4–5 is an isentropic expansion, so
v_{ r 5}=v_{ r 4} \frac{V_{5}}{V_{4}}
The volume ratio V_{5} / V_{4} required by this equation can be expressed as
\frac{V_{5}}{V_{4}}=\frac{V_{5}}{V_{3}} \frac{V_{3}}{V_{4}}
With V_{5}=V_{1}, V_{2}=V_{3}, and given volume ratios
\frac{V_{5}}{V_{4}}=\frac{V_{1}}{V_{2}} \frac{V_{3}}{V_{4}}=18\left(\frac{1}{1.2}\right)=15
Inserting this in the above expression for v_{ r 5}
v_{ r 5}=(5.609)(15)=84.135
Interpolating in Table A-22, we get u_{5}=475.96 kJ/kg.
a. The thermal efficiency is
\eta=1-\frac{Q_{51} / m}{\left(Q_{23} / m+Q_{34} / m\right)}=1-\frac{\left(u_{5}-u_{1}\right)}{\left(u_{3}-u_{2}\right)+\left(h_{4}-h_{3}\right)}
=1-\frac{(475.96-214.07)}{(1065.8-673.2)+(1778.3-1452.6)}
= 0.635(63.5%)
b. The mean effective pressure is
\text { mep }=\frac{W_{\text {cycle }} / m}{v_{1}-v_{2}}=\frac{W_{\text {cycle }} / m}{v_{1}(1-1 / r)}
The net work of the cycle equals the net heat added, so
\text { mep }=\frac{\left(u_{3}-u_{2}\right)+\left(h_{4}-h_{3}\right)-\left(u_{5}-u_{1}\right)}{v_{1}(1-1 / r)}
The specific volume at state 1 is evaluated in Example 9.2 as v_{1}=0.861 m ^{3} / kg. Inserting values into the above expression for mep
[(1065.8-673.2)+(1778.3-1452.6)-(475.96-214.07)]
\text { mep }=\frac{\times\left(\frac{ kJ }{ kg }\right)\left|\frac{10^{3} N \cdot m }{1 kJ } \| \frac{1 MPa }{10^{6} N / m ^{2}}\right|}{0.861(1-1 / 18) m ^{3} / kg }
= 0.56 MPa
Skills Developed
Ability to…
• sketch the dual cycle p–v and T–s diagrams.
• evaluate temperatures and pressures at each principal state and retrieve necessary property data.
• calculate the thermal efficiency and mean effective pressure.
Quick Quiz
Evaluate the total heat addition and the net work of the cycle, each in kJ per kg of air. Ans. Q_{ in } / m=718 kJ / kg , W_{\text {cycle }} / m=456 kJ/kg.
