Analyzing the Diesel Cycle
At the beginning of the compression process of an air-standard Diesel cycle operating with a compression ratio of 18, the temperature is 300 K and the pressure is 0.1 MPa. The cutoff ratio for the cycle is 2. Determine (a) the temperature and pressure at the end of each process of the cycle, (b) the thermal efficiency, (c) the mean effective pressure, in MPa.
Known An air-standard Diesel cycle is executed with specified conditions at the beginning of the compression stroke. The compression and cutoff ratios are given.
Find Determine the temperature and pressure at the end of each process, the thermal efficiency, and mean effective pressure.
Schematic and Given Data:
Engineering Model
1. The air in the piston–cylinder assembly is the closed system.
2. The compression and expansion 3–4 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
a. The analysis begins by determining properties at each principal state of the cycle. With T_{1}=300 K, Table A-22 gives u_{1}=214.07 kJ/kg and v_{ r 1}=621.2. For the isentropic compression process 1–2
v_{ r 2}=\frac{V_{2}}{V_{1}} v_{ r 1}=\frac{v_{ r 1}}{r}=\frac{621.2}{18}=34.51
Interpolating in Table A-22, we get T_{2}=898.3 K \text { and } h_{2}=930.98 kJ/kg. With the ideal gas equation of state
p_{2}=p_{1} \frac{T_{2}}{T_{1}} \frac{V_{1}}{V_{2}}=(0.1)\left(\frac{898.3}{300}\right)(18)=5.39 MPa
The pressure at state 2 can be evaluated alternatively using the isentropic relationship, p_{2}=p_{1}\left(p_{ r 2} / p_{ r 1}\right).
Since Process 2–3 occurs at constant pressure, the ideal gas equation of state gives
T_{3}=\frac{V_{3}}{V_{2}} T_{2}
Introducing the cutoff ratio, r_{ c }=V_{3} / V_{2}
T_{3}=r_{ c } T_{2}=2(898.3)=1796.6 K
From Table A-22, h_{3}=1999.1 kJ / kg \text { and } v_{ r 3}=3.97.
For the isentropic expansion process 3–4
v_{ r 4}=\frac{V_{4}}{V_{3}} v_{ r 3}=\frac{V_{4}}{V_{2}} \frac{V_{2}}{V_{3}} v_{ r 3}
Introducing V_{4}=V_{1}, the compression ratio r, and the cutoff ratio r_{ c }, we have
v_{ r 4}=\frac{r}{r_{ c }} v_{ r 3}=\frac{18}{2}(3.97)=35.73
Interpolating in Table A-22 with v_{ r 4}, \text { we get } u_{4}=664.3 kJ/kg and T_{4}=887.7 K. The pressure at state 4 can be found using the isentropic relationship p_{4}=p_{3}\left(p_{ r 4} / p_{ r 3}\right) or the ideal gas equation of state applied at states 1 and 4. With V_{4}=V_{1}, the ideal gas equation of state gives
p_{4}=p_{1} \frac{T_{4}}{T_{1}}=(0.1 MPa )\left(\frac{887.7 K }{300 K }\right)=0.3 MPa
b. The thermal efficiency is found using
\eta=1-\frac{Q_{41} / m}{Q_{23} / m}=1-\frac{u_{4}-u_{1}}{h_{3}-h_{2}}
1 =1-\frac{664.3-214.07}{1999.1-930.98}=0.578(57.8 \%)
c. The mean effective pressure written in terms of specific volumes 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
\frac{W_{\text {cycle }}}{m}=\frac{Q_{23}}{m}-\frac{Q_{41}}{m}=\left(h_{3}-h_{2}\right)-\left(u_{4}-u_{1}\right)
=(1999.1-930.98)-(664.3-214.07)
= 617.9 kJ/kg
The specific volume at state 1 is
v_{1}=\frac{(\bar{R} / M) T_{1}}{p_{1}}=\frac{\left(\frac{8314}{28.97} \frac{ N \cdot m }{ kg \cdot K }\right)(300 K )}{10^{5} N / m ^{2}}=0.861 m ^{3} / kg
Inserting values
\text { mep }=\frac{617.9 kJ / kg }{0.861(1-1 / 18) m ^{3} / kg }\left|\frac{10^{3} N \cdot m }{1 kJ } \right| \left|\frac{1 MPa }{10^{6} N / m ^{2}}\right|
= 0.76 MPa
1 This solution uses the air tables, which account explicitly for the variation of the specific heats with temperature. Note that Eq. 9.13 based on the assumption of constant specific heats has not been used to determine the thermal efficiency. The cold air-standard solution of this example is left as an exercise.
\eta=1-\frac{1}{r^{k-1}}\left[\frac{r_{ c }^{k}-1}{k\left(r_{ c }-1\right)}\right] (cold air-standard basis) (9.13)
Skills Developed
Ability to…
• sketch the Diesel cycle p–υ 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
If the mass of air is 0.0123 kg, what is the displacement volume, in L? Ans. 10 L.
