Question

An ideal Otto cycle has a compression ratio of \(10.5,\) takes in air at \(90 \mathrm{kPa}\) and \(40^{\circ} \mathrm{C},\) and is repeated 2500 times per minute.

Short answer

Question: Determine the efficiency, work output, and mean effective pressure (MEP) of an Otto cycle with a compression ratio of 10.5, intake air pressure of 90 kPa, intake air temperature of 40°C, and 2500 cycles per minute. Solution: 1. Efficiency of the Otto cycle is approximately 0.578 or 57.8%. 2. Work output is given by the expression: \(w_{out} = 0.718\,\text{kj/kg K}(T_3 - 603.94\,\text{K})\) 3. Mean effective pressure (MEP) can be calculated as: \(MEP = \frac{0.718\,\text{kj/kg K}(T_3 - 603.94\,\text{K})}{V_1(1 - \frac{1}{10.5})}\)

Step-by-step solution

Calculate the efficiency of the Otto cycle

The efficiency of an ideal Otto cycle is given by the formula: $$ \eta_{Otto} = 1 - \frac{1}{r^{(\gamma - 1)}} $$ where \(\eta_{Otto}\) is the efficiency, \(r\) is the compression ratio, and \(\gamma\) is the heat capacity ratio. For air, we can assume \(\gamma = 1.4\). Using the given compression ratio of 10.5, we can calculate the efficiency: $$ \eta_{Otto} = 1 - \frac{1}{10.5^{(1.4 - 1)}} \approx 0.578 $$

Determine the initial and final state properties for the working fluid

We are given the initial pressure (\(P_1 = 90\,\text{kPa}\)) and temperature (\(T_1 = 40^\circ\text{C} = 313.15\,\text{K}\)) of the air intake. We can calculate the final pressure after the isentropic compression process by using the relation: $$ \frac{P_2}{P_1} = \left(\frac{V_1}{V_2}\right)^\gamma = r^\gamma $$ Substituting the values, we get: $$ P_2 = P_1 \times r^\gamma = 90\,\text{kPa} \times 10.5^{1.4} \approx 3,\!189.86\,\text{kPa} $$

Calculate the heat input and work output for each process in the cycle

Now, we will use the first law of thermodynamics for a closed system to calculate the heat input and work output for each process in the cycle. Heat input during the constant volume process (\(q_{in}\)) is given by: $$ q_{in} = C_{v}(T_3 - T_2) $$ where \(C_{v}\) is the specific heat at constant volume. For air, we can assume \(C_{v} = 0.718\,\text{kj/kg K}\). To calculate the heat input, we need to determine the temperature at the end of the isentropic compression process (\(T_2\)). We can use the relation: $$ \frac{T_2}{T_1} = \left(\frac{V_1}{V_2}\right)^{\gamma - 1} = r^{\gamma - 1} $$ Substituting the values, we get: $$ T_2 = T_1 \times r^{\gamma - 1} = 313.15\,\text{K} \times 10.5^{0.4} \approx 603.94\,\text{K} $$ Now, we can calculate the heat input (\(q_{in}\)): $$ q_{in} = C_{v}(T_3 - T_2) = 0.718\,\text{kj/kg K}(T_3 - 603.94\,\text{K}) $$ For an ideal Otto cycle, the heat input is equal to the work output. Therefore, the work output (\(w_{out}\)) is given by: $$ w_{out} = q_{in} = 0.718\,\text{kj/kg K}(T_3 - 603.94\,\text{K}) $$

Calculate the mean effective pressure (MEP) of the cycle

The mean effective pressure (MEP) is the hypothetical pressure that, if constant throughout the expansion process, would result in the same net work output as in the actual cycle. It is defined as: $$ MEP = \frac{w_{net}}{V_1 - V_2} $$ For an ideal Otto cycle, the work output is equal to the net work done (\(w_{net}\)), and we can rewrite the MEP expression as: $$ MEP = \frac{w_{out}}{V_1 - V_2} = \frac{0.718\,\text{kj/kg K}(T_3 - 603.94\,\text{K})}{V_1 - V_2} $$ We can use the compression ratio to find the volume ratio: $$ \frac{V_1}{V_2} = r = 10.5 $$ So the MEP becomes: $$ MEP = \frac{0.718\,\text{kj/kg K}(T_3 - 603.94\,\text{K})}{V_1(1 - \frac{1}{10.5})} $$ The cycle is repeated 2500 times per minute. Using the calculated values, one can estimate the power and efficiency of the cycle.