Question

An air-standard Diesel cycle has a compression ratio of \(18.2 .\) Air is at \(120^{\circ} \mathrm{F}\) and 14.7 psia at the beginning of the compression process and at \(3200 \mathrm{R}\) at the end of the heataddition process. Accounting for the variation of specific heats with temperature, determine \((a)\) the cutoff ratio, \((b)\) the heat rejection per unit mass, and ( \(c\) ) the thermal efficiency.

Short answer

The cutoff ratio for the given air-standard Diesel cycle is approximately 1.842. To find the heat rejection per unit mass and the thermal efficiency, use the calculated values of \(W_{in}\), \(W_{out}\), \(Q_{in}\), and \(Q_{out}\) from step 4. The heat rejection per unit mass can be calculated as \(q_{out} = Q_{out}/m\), and the thermal efficiency can be calculated as \(\eta_{th} = 1 - \frac{Q_{out}}{Q_{in}}\).

Step-by-step solution

1. Initial and final states of each process

Given: - Compression ratio: \(r_c = 18.2\) - Initial state of compression process: \(T_1 = 120^{\circ} \mathrm{F}\) and \(P_1 = 14.7\) psia - Temperature at the end of the heat addition process: \(T_3 = 3200\,\mathrm{R}\) Let's convert the given temperature values to Rankine: \(T_1 = 120^{\circ}\mathrm{F} + 459.67\,\mathrm{R} = 579.67\,\mathrm{R}\) Now, we have the initial state of the compression process and the state at the end of the heat addition process. We will use the isentropic process equation to determine the other states.

2. Isentropic process calculation

For an isentropic process, we can use the following relationship between temperature and pressure: $$ \frac{T_2}{T_1} = \left(\frac{P_2}{P_1}\right)^{(\gamma-1)/\gamma} $$ where \(T_2\) is the temperature at the end of compression and \(P_2\) is the pressure at the end of compression. For air, take \(\gamma = 1.4\), so the relationship becomes: $$ \frac{T_2}{T_1} = \left(\frac{P_2}{P_1}\right)^{0.286} $$ We know that \(T_1 = 579.67\,\mathrm{R}\), \(P_1 = 14.7\,\mathrm{psia}\), and \(r_c = 18.2\). Using the compression ratio: $$ P_2 = P_1 \times r_c = 14.7\,\mathrm{psia} \times 18.2 = 267.54\,\mathrm{psia} $$ Now, using the isentropic process equation to find \(T_2\): $$ \frac{T_2}{579.67\,\mathrm{R}} = \left(\frac{267.54\,\mathrm{psia}}{14.7\,\mathrm{psia}}\right)^{0.286} $$ $$ T_2 = 579.67\,\mathrm{R} \times \left(\frac{267.54\,\mathrm{psia}}{14.7\,\mathrm{psia}}\right)^{0.286} \approx 1736.85\,\mathrm{R} $$

3. Cutoff ratio calculation

In the Diesel cycle, the cutoff ratio \(r_{co}\) is defined as: $$ r_{co} = \frac{T_3}{T_2} $$ Therefore, $$ r_{co} = \frac{3200\,\mathrm{R}}{1736.85\,\mathrm{R}} \approx 1.842 $$

4. Specific work input/output and heat transfer calculations

We can now calculate the specific work input/output (\(W_{in}\), \(W_{out}\)) and the heat transfer (\(Q_{in}\), \(Q_{out}\)) during the cycle. For the isentropic compression: $$ W_{in} = c_v (T_1 - T_2) = c_v (579.67\,\mathrm{R} - 1736.85\,\mathrm{R}) $$ For the heat addition process: $$ Q_{in} = c_p (T_3 - T_2) = c_p (3200\,\mathrm{R} - 1736.85\,\mathrm{R}) $$ For the expansion process: $$ W_{out} = Q_{in} - W_{in} $$ For the heat rejection process: $$ Q_{out} = W_{out} - W_{in} $$

5. Heat rejection and thermal efficiency calculations

Now, we can calculate the heat rejection and the thermal efficiency. Heat rejection per unit mass: $$ q_{out} = Q_{out} / m $$ Thermal efficiency (\(\eta_{th}\)) is the ratio of net work output to the heat input: $$ \eta_{th} = \frac{W_{out} - W_{in}}{Q_{in}} = \frac{Q_{in} - Q_{out}}{Q_{ }in} = 1 - \frac{Q_{out}}{Q_{in}} $$ Therefore, we have the following results: - Cutoff ratio: \(1.842\) - Heat rejection per unit mass: Calculate \(q_{out}\) using the values of \(Q_{in}\), \(Q_{out}\), and \(W_{in}\), \(W_{out}\) from step 4. - Thermal efficiency: Calculate \(\eta_{th}\) using the heat rejection per unit mass and the heat input from step 4.