Question

A four-cylinder two-stroke 2.4 -L diesel engine that operates on an ideal Diesel cycle has a compression ratio of 22 and a cutoff ratio of \(1.8 .\) Air is at \(70^{\circ} \mathrm{C}\) and \(97 \mathrm{kPa}\) at the beginning of the compression process. Using the cold-airstandard assumptions, determine how much power the engine will deliver at \(3500 \mathrm{rpm}\)

Short answer

After calculating the work done during each process and the net work done per cycle, we can find the power delivered by the engine by multiplying the total work done with the number of cycles per second: \(P = W_{total} \cdot \frac{RPM}{60}\) By plugging in the actual values for the compression ratio, cutoff ratio, and RPM, the final answer for the power delivered by the engine can be obtained.

Step-by-step solution

Calculate the initial and final pressures and temperatures during the compression process

The air is assumed to be an ideal gas, so we can use the ideal gas law: \(PV=T\). The compression ratio (\(r_c\)) is given as 22, which means \(V_1 = 22V_2\). Using the relation for isentropic processes: \(P_1 V_1^k = P_2 V_2^k\) , we can find the final temperature and pressure: \(P_2 = \frac{P_1 V_1^k}{V_2^k} = \frac{P_1 (r_c)^k}{1^k}\) , where \(k=1.4\) \(P_2 = \frac{97(22)^{1.4}}{1^{1.4}} = 6099.61\) kPa \(T_1\) is given as \(70^{\circ}\mathrm{C}\), which is \(343.15\ \mathrm{K}\). \(T_2 = T_1 \frac{P_2}{P_1} \frac{V_2}{V_1} = 343.15 \frac{6099.61}{97}\frac{1}{22}\) \(T_2 = 967.30\ \mathrm{K}\)

Calculate the initial and final pressures and temperatures during the heat addition process

During the heat addition process (to the cutoff point), the volume ratio (\(r_c'\)) is given as 1.8. Therefore, \(V_3= 1.8V_2\). Since the process is at constant pressure, \(P_3=P_2\). We can find \(T_3\) using the ideal gas law: \(T_3 = T_2 \frac{V_3}{V_2} = 967.30 \cdot 1.8 = 1740.94\ \mathrm{K}\)

Calculate the initial and final pressures and temperatures during the expansion process

During the expansion process, the isentropic process relation applies: \(P_3V_3^k=P_4V_4^k\). Since the expansion is isentropic, we have \(V_1 = V_4\) and \(r_c = \frac{V_1}{V_2} = \frac{V_4}{V_3}\). We can find \(P_4\) using the isentropic process relation: \(P_4 = \frac{P_3 V_3^k}{V_4^k} = \frac{P_3 (\frac{V_1}{r_c})^k}{V_1^k} = \frac{P_3}{r_c^k}\) \(P_4 = \frac{6099.61}{(22)^{1.4}} = 451.73\) kPa \(T_4 = T_3 \frac{P_4}{P_3} \frac{V_4}{V_3} = 1740.94 \frac{451.73}{6099.61}\cdot\frac{22}{1.8}\) \(T_4 = 726.25\ \mathrm{K}\)

Calculate the work done during each process

The work done during each process can be found using the ideal gas relationships: \(W_{12} = c_v (T_1 - T_2)=-c_v (T_2 - T_1)\) \(W_{23} = P_2(V_3 - V_2) =P_2(1.8-1)V_2 \) \(W_{34} = c_v (T_3 - T_4)=-c_v (T_4 - T_3)\) \(W_{41} = 0\), since the work done is 0 during constant volume cooling The net work done per cycle: \(W_{net} = W_{12} + W_{23} + W_{34} + W_{41}\)

Calculate the net work done by the engine and the power delivered

The overall work done by the engine can be calculated by multiplying the net work done per cycle by the number of cylinders: \(W_{total} = 4W_{net}\) Now, we can find the power output by multiplying the total work done with the number of cycles per second: \(P = W_{total} \cdot \frac{RPM}{60}\) where RPM = Revolutions Per Minute = 3500. Final step is to plug-in the actual values of \(W_{12}\), \(W_{23}\), and \(W_{34}\), and then using the above formula to calculate the power. Once we calculate the power with the given values of compression ratio, cut off ratio and RPM, we'll get the final answer for power delivered by the engine.