Question

An outboard motor for a boat is cooled by lake water at \(15.0^{\circ} \mathrm{C}\) and has a compression ratio of \(10.0 .\) Assume that the air is a diatomic gas. a) Calculate the efficiency of the engine's Otto cycle. b) Using your answer to part (a) and the fact that the efficiency of the Carnot cycle is greater than that of the Otto cycle, estimate the maximum temperature of the engine.

Short answer

Question: Estimate the efficiency of an Otto cycle engine with a compression ratio of 10 and the maximum temperature of the engine, given the lake water temperature is 15.0°C. Answer: The efficiency of the Otto cycle engine is approximately 33.93%, and the maximum temperature of the engine is approximately 459.7 K.

Step-by-step solution

Find the Efficiency of the Otto Cycle

To find the efficiency of the Otto cycle, we need to calculate the adiabatic index \(\gamma\), where \(\gamma=\frac{C_p}{C_v}\). For a diatomic gas, the adiabatic index \(\gamma=\frac{7}{5}=1.4\). Now we will use the efficiency formula for the Otto cycle: Efficiency (\(\eta\)) = \(1-\frac{1}{r^{\gamma-1}}\), where \(r\) is the compression ratio. Substitute the given values: \(r = 10\) and \(\gamma=1.4\) Efficiency (\(\eta\)) = \(1-\frac{1}{10^{(1.4-1)}}\) Calculate the efficiency: Efficiency (\(\eta\)) = \(1-\frac{1}{10^{0.4}}\) = \(1-0.6607\) = \(0.3393\) or \(33.93\%\)

Estimate the Maximum Temperature of the Engine

To estimate the maximum temperature of the engine, we will use the efficiency of the Otto cycle and the efficiency of the Carnot cycle. The efficiency of the Carnot cycle is given by: Efficiency (\(\eta_C\)) = \(1-\frac{T_C}{T_H}\), where \(T_C\) is the temperature of the cold reservoir (lake water with \(15.0^\circ\)C), and \(T_H\) is the temperature of the hot reservoir (the maximum temperature of the engine). First, convert the given lake water temperature from Celsius to Kelvin: \(T_C=15+273.15=288.15 K\) Since the efficiency of the Carnot cycle is greater than the efficiency of the Otto cycle, let \(\eta_C > \eta\). Set a lower bound for the Carnot efficiency, let's say \(\eta_C = 1.1\eta\). \(\eta_C = 1-\frac{T_C}{T_H}\) Find \(T_H\): \(1.1\eta = 1-\frac{288.15}{T_H}\) Solve for \(T_H\): \(T_H = \frac{288.15}{1-1.1\eta}\) = \(\frac{288.15}{1-1.1(0.3393)}\) = \(\frac{288.15}{0.6267}\) = \(459.7 K\) The estimated maximum temperature of the engine is approximately 459.7 K.