Air Standard Cycles Questions and Answers

Answer: B

The compression ratio is more than the expansion ratio.

In a diesel cycle, heat is added at constant pressure while the piston moves out, so combustion continues to the cut-off point. With compression ratio \(r = V_{1}/V_{2}\) and cut-off ratio \(\rho = V_{3}/V_{2} \gt 1\), the expansion ratio is

\[ r_{e} = \frac{V_{4}}{V_{3}} = \frac{r}{\rho} \lt r \]

So expansion starts from a larger volume and the expansion ratio is smaller. They are equal only in the Otto cycle, where heat is added at constant volume. 'Variable' is wrong because the inequality always holds.

Answer: C

The order is \(\eta_{diesel} \gt \eta_{dual} \gt \eta_{otto}\).

When the maximum pressure and temperature are fixed (the practical limit of the engine) along with the heat rejected, the Otto cycle needs the lowest compression ratio to stay within the pressure limit, and the Diesel cycle can use the highest. On a T-s diagram the Diesel cycle then receives the most heat for the same heat rejected, so it is the most efficient, the Otto cycle the least, and the dual cycle lies between.

This is the reverse of the comparison at the same compression ratio, where the Otto cycle is the most efficient, which is why options ranking Otto first are wrong here.

Answer: A

The order is \(\eta_{otto} \gt \eta_{dual} \gt \eta_{diesel}\).

At the same compression ratio and the same heat rejection, the Otto cycle adds all its heat at constant volume, at the highest average temperature, while the Diesel cycle adds heat at constant pressure during expansion, at a lower average temperature. The dual cycle adds part of its heat each way, so its efficiency lies between them.

The higher the mean temperature of heat addition for the same heat rejected, the higher the efficiency, which gives Otto, then dual, then Diesel. Any order putting the Diesel or dual cycle first reverses this.

Answer: C

The indicated mean effective pressure is 5.250 bar.

Mean effective pressure is the work per cycle divided by the swept volume. With compression ratio r, the total volume is \(rV_c\), so the swept volume is \((r - 1)V_c\):

\[ \begin{aligned} V_s &= (5.5 - 1)V_c = 4.5V_c \\ p_m &= \frac{23.625 \times 10^5\, V_c}{4.5\, V_c} \\ &= 5.25 \times 10^5\ \text{Pa} = 5.25\ \text{bar} \end{aligned} \]

Dividing by the total volume, \(5.5V_c\), gives the 4.295 bar option, and multiplying by 4.5 instead of dividing gives 106.3 bar. The clearance volume cancels, which is why the answer can be found without knowing it.