Question 9.4: The Ideal Diesel Cycle An ideal Diesel cycle with air as the...

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\cdot ft ^{3} / \mathrm{lbm} \cdot \mathrm{R}, and its other properties at room temperature are c_{p}=0.240  \mathrm{Btu} / \mathrm{lbm} \cdot \mathrm{R}, c_{v}=0.171  \mathrm{Btu} / \mathrm{lbm} \cdot \mathrm{R}, and k=1.4 (Table A-2Ea).

Analysis     The P-V diagram of the ideal Diesel cycle described is shown in Fig. 9–25. 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:

\begin{array}{l}V_{2}=\frac{V_{1}}{r}=\frac{117  \mathrm{in}^{3}}{18}=6.5  \mathrm{in}^{3} \\\\V_{3}=r_{c} V_{2}=(2)\left(6.5  \mathrm{in}^{3}\right)=13  \mathrm{in}^{3} \\\\V_{4}=V_{1}=117  \mathrm{in}^{3}\end{array}

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

T_{2}=T_{1}\left(\frac{V_{1}}{V_{2}}\right)^{k-1}=(540  \mathrm{R})(18)^{1.4-1}=1716  \mathbf{R}

P_{2}=P_{1}\left(\frac{V_{1}}{V_{2}}\right)^{k}=(14.7  \mathrm{psia})(18)^{1.4}=841  \mathrm{psia}

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

\begin{array}{c}P_{3}=P_{2}=841  \mathrm{psia} \\\frac{P_{2} \mathrm{~V}_{2}}{T_{2}}=\frac{P_{3} V_{3}}{T_{3}} \rightarrow T_{3}=T_{2}\left(\frac{V_{3}}{V_{2}}\right)=(1716  \mathrm{R})(2)=3432  \mathrm{R}\end{array}

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

\begin{array}{l}T_{4}=T_{3}\left(\frac{V_{3}}{V_{4}}\right)^{k-1}=(3432  \mathrm{R})\left(\frac{13 \mathrm{in}^{3}}{117  \mathrm{in}^{3}}\right)^{1.4-1}=1425  \mathrm{R} \\P_{4}=P_{3}\left(\frac{V_{3}}{V_{4}}\right)^{k}=(841  \mathrm{psia})\left(\frac{13  \mathrm{in}^{3}}{117  \mathrm{in}^{3}}\right)^{1.4}=38.8  \mathrm{psia}\end{array}

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

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

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

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

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

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

Thus,

W_{\text {net }}=Q_{\text {in }}-Q_{\text {out }}=2.051-0.754=1.297  \mathrm{Btu}

Then the thermal efficiency becomes

\eta_{\mathrm{th}}=\frac{W_{\text {net }}}{Q_{\text {in }}}=\frac{1.297  \mathrm{Btu}}{2.051   \mathrm{Btu}}=0.632 \text { or } 63.2 \%

The thermal efficiency of this Diesel cycle under the cold-air-standard assumptions could also be determined from Eq. 9-12.

\eta_{\text {th,Diesel }}=1-\frac{1}{r^{k-1}}\left[\frac{r_{c}^{k}-1}{k\left(r_{c}-1\right)}\right]          (9-12)

(c) The mean effective pressure is determined from its definition, Eq. 9-4:

\mathrm{MEP}=\frac{W_{\mathrm{net}}}{V_{\max }-V_{\min }}=\frac{w_{\mathrm{net}}}{v_{\max }-v_{\min }}        ( kPa )                                     (9-4)

\begin{aligned}\mathrm{MEP} &=\frac{W_{\text {net }}}{V_{\max }-V_{\min }}=\frac{W_{\text {net }}}{V_{1}-V_{2}}=\frac{1.297  \mathrm{Btu}}{(117-6.5)  \mathrm{in}^{3}}\left(\frac{778.17  \mathrm{lbf} \cdot \mathrm{ft}}{1  \mathrm{Btu}}\right)\left(\frac{12  \mathrm{in}}{1  \mathrm{ft}}\right) \\&=110  \mathrm{psia}\end{aligned}

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.