Question

An ideal Diesel cycle has a maximum cycle temperature of \(2300^{\circ} \mathrm{F}\) and a cutoff ratio of \(1.4 .\) The state of the air at the beginning of the compression is \(P_{1}=14.4\) psia and \(T_{1}=50^{\circ} \mathrm{F}\). This cycle is executed in a four-stroke, eightcylinder engine with a cylinder bore of 4 in and a piston stroke of 4 in. The minimum volume enclosed in the cylinder is 4.5 percent of the maximum cylinder volume. Determine the power produced by this engine when it is operated at 1800 rpm. Use constant specific heats at room temperature.

Short answer

Using the information given for the diesel engine and applying the formulas for the ideal Diesel cycle, we have calculated the power output of the engine. We first computed the specific heats and determined the volumetric efficiency of the engine. Then, we calculated the mass of air in the cylinder and the net specific work output. Finally, with the number of cycles per second and the net specific work, we found the power output of the engine to be 5.47 kW when operated at 1800 rpm.

Step-by-step solution

Calculate the specific heats at constant pressure and constant volume

Given that we need to use constant specific heats at room temperature, we will assume the values of specific heats for air. We have: \(c_p = 1.005 \, \frac{kJ}{kg \cdot K}\) (specific heat at constant pressure) \(c_v = 0.718 \, \frac{kJ}{kg \cdot K}\) (specific heat at constant volume)

Determine the volumetric efficiency

We have the cutoff ratio \(r_c = 1.4\). Using the formulas for an ideal Diesel cycle, the volumetric efficiency (\(\eta_v\)) is given by: \(\eta_v = 1 - \frac{1}{r}\) where \(r\) is the compression ratio. Given that the minimum volume enclosed in the cylinder is 4.5% of the maximum cylinder volume, we have: \(r = \frac{V_{max}}{V_{min}} = \frac{100}{4.5} = 22.22\) Now we can calculate the volumetric efficiency \(\eta_v\): \(\eta_v = 1 - \frac{1}{22.22} = 0.955\)

Calculate the mass of air in the cylinder

Given the cylinder bore of 4 inches and the piston stroke of 4 inches, we can calculate the cylinder volume: \(V_{max} = \pi \left(\frac{4}{2}\right)^2 \cdot 4 = 50.27 \, in^3\) (cubic inches) Convert the given pressures and temperatures into SI units: \(P_1 = 14.4 \, psia = 99.3 \, kPa\) \(T_1 = 50^\circ F = 10^\circ C = 283.15 \, K\) Now, let's find the minimum volume enclosed in the cylinder, \(V_{min}\): \(V_{min} = 0.045 \cdot V_{max} = 0.045 \cdot 50.27 \, in^3 = 2.263 \, in^3\) Convert \(V_{min}\) to SI units (cubic meters): \(V_{min} = 2.263 \, in^3 \cdot 1.639 \times 10^{-5} \, \frac{m^3}{in^3} = 3.708 \times 10^{-5} \, m^3\) Using the ideal gas law, \(PV = mRT\), we can find the mass of air in the cylinder (\(m\)): \(m = \frac{P_1 V_{min}}{RT_1} = \frac{99.3 \times 10^3 \cdot 3.708 \times 10^{-5}}{287 \cdot 283.15} = 3.086 \times 10^{-3} \, kg\)

Calculate the net specific work and power output

Using the Diesel cycle efficiency formula, we have: \(\eta_{th}=1-\frac{1}{r^{(\gamma -1)}}\frac{r_c^{\gamma} -1}{\gamma (r_c -1)}\) where \(\gamma = \frac{c_p}{c_v}\) Now, find \(\gamma\): \(\gamma = \frac{1.005}{0.718} = 1.4\) Calculate the thermal efficiency (\(\eta_{th}\)): \(\eta_{th}=1-\frac{1}{22.22^{(1.4 -1)}}\frac{1.4^{1.4} -1}{1.4 (1.4 -1)} = 0.627\) Next, calculate the net specific work output \(w_{net}\): \(w_{net} = \eta_{th} Q_{in} = \eta_{th} c_v (T_2 - T_1)\) In order to calculate \(T_2\), we use the compression process (\(\gamma = \frac{c_p}{c_v}\)): \(T_2 = T_1 \cdot r^{\gamma - 1} = 283.15 \cdot 22.22^{1.4 - 1} = 813.96 \, K\) Now, we can find \(w_{net}\): \(w_{net} = 0.627 \cdot 0.718 (813.96 - 283.15) = 236.93 \, \frac{kJ}{kg}\) Finally, calculate the power output \(P_{out}\): \(P_{out} = m \cdot w_{net} \cdot n\) where \(n\) is the number of cycles per second (which can be derived from the given 1800 rpm): \(n = \frac{1800}{60} \cdot \frac{1}{4} = 7.5 \, \frac{cycles}{s}\) The power output is then: \(P_{out} = 3.086 \times 10^{-3} \cdot 236.93 \cdot 7.5 = 5.47 \, kW\) Therefore, the power produced by the engine when it is operated at 1800 rpm is 5.47 kW.