The Ideal Diesel Cycle

An ideal Diesel cycle with air as the working fluid has a compression ratio of 18 and a cutoff ratio of 2. At the beginning of the compression process, the working fluid is at 14.7 psia, 80°F, and 117 in³. Utilizing the cold-air-standard assumptions, determine (a) the temperature and pressure of air a the end of each process, (b) the net work output and the thermal efficiency, and (c) the mean effective pressure.

Question

The Ideal Diesel Cycle

An ideal Diesel cycle with air as the working fluid has a compression ratio of 18 and a cutoff ratio of 2. At the beginning of the compression process, the working fluid is at 14.7 psia, 80°F, and 117 in³. Utilizing the cold-air-standard assumptions, determine (a) the temperature and pressure of air a the end of each process, (b) the net work output and the thermal efficiency, and (c) the mean effective pressure.

Accepted Answer

An ideal Diesel cycle is considered. The temperature and pressure at the end of each process, the net work output, the thermal efficiency, and the mean effective pressure are to be determined.
Assumptions 1 The cold-air-standard assumptions are applicable and thus air can be assumed to have constant specific heats at room temperature.
2 Kinetic and potential energy changes are negligible.
Properties The gas constant of air is R = 0.3704 psia · ft³/lbm · R and its other properties at room temperature are [latex] c_{p}=0.240 ext { Btu/lbm } \cdot R , c_{v}= [/latex]0.171 Btu/lbm · R, and k = 1.4 (Table A–2Ea).
Analysis The P-V diagram of the ideal Diesel cycle described is shown in Fig. 9–24. We note that the air contained in the cylinder forms a closed system.
(a) The temperature and pressure values at the end of each process can be determined by utilizing the ideal-gas isentropic relations for processes 1-2 and 3-4. But first we determine the volumes at the end of each process from the definitions of the compression ratio and the cutoff ratio:

[latex] \begin{aligned} &V_{2}=\frac{V_{1}}{r}=\frac{117 in ^{3}}{18}=6.5 in ^{3} \ &V_{3}=r_{c} V_{2}=(2)\left(6.5 in ^{3} ight)=13 in ^{3} \ &V_{4}=V_{1}=117 in ^{3} \end{aligned}[/latex]

Process 1-2 (isentropic compression of an ideal gas, constant specific heats):

[latex] \begin{aligned} T_{2} &=T_{1}\left(\frac{V_{1}} {V_{2}} ight)^{k-1}=(540 R )(18)^{1.4-1}=1716 R \ P_{2} &=P_{1}\left(\frac{V_{1}}{V_{2}} ight)^{k}=(14.7 psia )(18)^{1.4}=841 psia \end{aligned} [/latex]

Process 2-3 (constant-pressure heat addition to an ideal gas):

[latex] \begin{aligned} P_{3} &=P_{2}=841 psia \ \frac{P_{2} V_{2}}{T_{2}} &=\frac{P_{3} V_{3}}{T_{3}} ightarrow T_{3}=T_{2}\left(\frac{V_{3}}{V_{2}} ight)=(1716 R )(2)=3432 R \end{aligned} [/latex]

Process 3-4 (isentropic expansion of an ideal gas, constant specific heats):

[latex] \begin{aligned} &T_{4}=T_{3}\left(\frac{V_{3}}{V_{4}} ight)^{k-1}=(3432 R )\left(\frac{13 in ^{3}}{117 in ^{3}} ight)^{1.4-1}=1425 R \ &P_{4}=P_{3}\left(\frac{V_{3}}{V_{4}} ight)^{k}=(841 psia )\left(\frac{13 in ^{3}}{117 in ^{3}} ight)^{1.4}=38.8 psia \end{aligned} [/latex]

(b) The net work for a cycle is equivalent to the net heat transfer. But first we find the mass of air:

[latex] m=\frac{P_{1} V_{1}}{R T_{1}}=\frac{(14.7 psia )\left(117 in ^{3} ight)}{\left(0.3704 psia \cdot ft ^{3} / lbm \cdot R ight)(540 R )}\left(\frac{1 ft ^{3}}{1728 in ^{3}} ight)=0.00498 lbm [/latex]

Process 2-3 is a constant-pressure heat-addition process, for which the boundary work and Δu terms can be combined into Δh. Thus,

[latex] \begin{aligned} Q_{ ext {in }} &=m\left(h_{3}-h_{2} ight)=m c_{p}\left(T_{3}-T_{2} ight) \ &=(0.00498 lbm )(0.240 Btu / lbm \cdot R )[(3432-1716) R \ &=2.051 Btu \end{aligned} [/latex]

Process 4-1 is a constant-volume heat-rejection process (it involves no work interactions), and the amount of heat rejected is

[latex] \begin{aligned} Q_{ ext {out }} &=m\left(u_{4}-u_{1} ight)=m c_{v}\left(T_{4}-T_{1} ight) \ &=(0.00498 lbm )(0.171 Btu / lbm \cdot R )[(1425-540) R \ &=0.754 Btu \end{aligned} [/latex]

Thus,

[latex] W_{ ext {net }}=Q_{ ext {in }}-Q_{ ext {out }}=2.051-0.754=1.297 Btu [/latex]

Then the thermal efficiency becomes

[latex] \eta_{ th }=\frac{W_{ ext {net }}}{Q_{ ext {in }}}=\frac{1.297 Btu }{2.051 Btu }=0.632 ext { or } 63.2 \% [/latex]

The thermal efficiency of this Diesel cycle under the cold-air-standard assumptions could also be determined from Eq. 9–12.
(c) The mean effective pressure is determined from its definition, Eq. 9–4:

[latex] \eta_{ ext {th, Diesel }}=1-\frac{1}{r^{k-1}}\left[\frac{r_{c}^{k}-1}{k\left(r_{c}-1 ight)} ight] [/latex] (9–12)

[latex] MEP =\frac{W_{ ext {net }}}{V_{\max }-V_{\min }}=\frac{w_{ ext {net }}}{v_{\max }-v_{\min }} [/latex] (9–4)

[latex] \begin{aligned} MEP &=\frac{W_{ ext {net }}}{V_{\max }-V_{\min }}=\frac{W_{ ext {net }}}{V_{1}-V_{2}}=\frac{1.297 Btu }{(117-6.5) in ^{3}}\left(\frac{778.17 lbf \cdot ft }{1 Btu } ight)\left(\frac{12 in .}{1 ft } ight) \ &= 1 1 0 ext { psia } \end{aligned} [/latex]

Discussion Note that a constant pressure of 110 psia during the power stroke would produce the same net work output as the entire Diesel cycle.

Question 9.3: The Ideal Diesel Cycle An ideal Diesel cycle with air as the...

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