Question

A four-cylinder spark-ignition engine has a compression ratio of \(10.5,\) and each cylinder has a maximum volume of 0.4 L. At the beginning of the compression process, the air is at \(98 \mathrm{kPa}\) and \(37^{\circ} \mathrm{C}\), and the maximum temperature in the cycle is 2100 K. Assuming the engine to operate on the ideal Otto cycle, determine \((a)\) the amount of heat supplied per cylinder, ( \(b\) ) the thermal efficiency, and \((c)\) the number of revolutions per minute required for a net power output of \(45 \mathrm{kW}\). Assume variable specific heats for air

Short answer

a) The amount of heat supplied per cylinder is 3.59 kJ. b) The thermal efficiency of the engine is 57.1%. c) The number of revolutions per minute required for a net power output of 45 kW is approximately 1,317 rpm.

Step-by-step solution

Find the minimum volume per cylinder

To find the minimum volume per cylinder, we can use the compression ratio: \(v_b = \dfrac{v_a}{r_c}\) where \(v_a\) is the maximum volume, \(r_c\) is the compression ratio, and \(v_b\) will be the minimum volume. Plugging in given values: \(v_b = \dfrac{0.4 \ \text{L}}{10.5} = 0.0381 \ \text{L}\)

Find the initial specific volume and pressure per cylinder

We are given the initial air pressure as \(98 \ \mathrm{kPa}\) and the initial temperature as \(37^{\circ} \mathrm{C}\). Convert the given temperature to Kelvin: \(T_1 = 37 + 273.15 = 310.15 \ \text{K}\) We can find the initial specific volume \(v_1\) using the ideal gas equation: \(v_1 = \dfrac{R \ T_1}{P_1}\) where \(R\) is the specific gas constant for air, \(R = 0.287 \ \text{kJ/kg K}\). Plugging in the values: \(v_1 = \dfrac{(0.287 \ \mathrm{kJ/kg} \ \mathrm{K})(310.15 \ \mathrm{K})}{98 \ \mathrm{kPa}} = 0.905 \ \mathrm{m^3/kg}\)

Find the specific volume at the end of the compression process

Now, we will find the specific volume \(v_2\) at the end of the compression process using the compression ratio: \(v_2 = \dfrac{v_1}{r_c}\) Plugging in the values: \(v_2 = \dfrac{0.905 \ \mathrm{m^3/kg}}{10.5} = 0.0862 \ \mathrm{m^3/kg}\)

Calculate the temperature and pressure at the end of the compression process

Using the ideal gas equation, we can find the temperature at the end of the compression process (\(T_2\)): \(T_2 = \dfrac{P_2 \ v_2}{R}\) where \(P_2\) is the pressure at the end of the compression process. Since the compression process is isentropic, we can use the following relationship: \(\dfrac{P_2}{P_1} = \left(\dfrac{v_1}{v_2}\right)^{k}\) where \(k = 1.4\) is the specific heat ratio for air. Plugging in the values: \(P_2 = \left(\dfrac{0.905 \ \mathrm{m^3/kg}}{0.0862 \ \mathrm{m^3/kg}}\right)^{1.4} (98 \ \mathrm{kPa}) = 3052.32 \ \mathrm{kPa}\) Now, we can calculate \(T_2\): \(T_2 = \dfrac{(3052.32 \ \mathrm{kPa})(0.0862 \ \mathrm{m^3/kg})}{0.287 \ \mathrm{kJ/kg} \ \mathrm{K}} = 921.93 \ \mathrm{K}\)

Calculate the amount of heat supplied

Now that we have the initial and final states for the compression process, we can find the amount of heat supplied (\(Q_s\)) using the maximum temperature in the cycle (\(T_{max} = 2100 \ \mathrm{K}\)) and specific heat at constant volume (\(c_v = 0.718 \ \mathrm{kJ/kg} \mathrm{K}\)) for air: \(Q_s = m \ c_v \ (T_{max} - T_2)\) To find mass \(m\), we can use the initial specific volume, temperature, and pressure: \(m = \dfrac{v_a \ P_1}{R \ T_1}\) \(m = \dfrac{(0.4 \ \text{L})(98 \ \mathrm{kPa})}{(0.287 \ \mathrm{kJ/kg} \ \mathrm{K})(310.15 \ \mathrm{K})} = 0.00461 \ \mathrm{kg}\) Now, we can find \(Q_s\): \(Q_s = (0.00461 \ \mathrm{kg})(0.718 \ \mathrm{kJ/kg} \mathrm{K})(2100 - 921.93) = 3.59 \ \mathrm{kJ}\) So, the amount of heat supplied per cylinder is \(3.59 \ \mathrm{kJ}\).

Calculate the thermal efficiency

The thermal efficiency of the engine is given by the Otto cycle efficiency formula: \(\eta_t = 1 - \dfrac{1}{r_c^{(k-1)}}\) \(\eta_t = 1 - \dfrac{1}{10.5^{(1.4 - 1)}} = 0.571\) Thus, the thermal efficiency of the engine is \(57.1 \%\).

Calculate the number of revolutions per minute

To find the number of revolutions per minute for a net power output of \(45 \ \mathrm{kW}\), we will use the net power output formula, which is given by: \(P_{net} = \dfrac{W_{net} \times N}{60}\) The net work output per cycle (\(W_{net}\)) is the difference between the heat supplied and the heat rejected: \(W_{net} = Q_s - Q_r\) We can find the heat rejected (\(Q_r\)) using the thermal efficiency: \(\eta_t = \dfrac{W_{net}}{Q_s}\) → \(W_{net} = \eta_t \times Q_s\) Now, we can find \(Q_r\): \(Q_r = Q_s - W_{net}\) \(Q_r = 3.59 \ \mathrm{kJ} - (0.571 \times 3.59 \ \mathrm{kJ}) = 1.54 \ \mathrm{kJ}\) Now, we can find the number of revolutions per minute (N): \(\dfrac{45 \ \mathrm{kW}}{\dfrac{2.05 \ \mathrm{kJ/rev}}{60 \ \mathrm{sec/min}}} = 1,317.07\) rpm Thus, the required number of revolutions per minute for a net power output of \(45 \ \mathrm{kW}\) is approximately \(1,317\) rpm.