Question

In an ideal Otto cycle, air is compressed from \(1.20 \mathrm{kg} / \mathrm{m}^{3}\) and 2.2 to \(0.26 \mathrm{L},\) and the net work output of the cycle is \(440 \mathrm{kJ} / \mathrm{kg} .\) The mean effective pressure (MEP) for this cycle is \((a) 612 \mathrm{kPa}\) \((b) 599 \mathrm{kPa}\) \((c) 528 \mathrm{kPa}\) \((d) 416 \mathrm{kPa}\) \((e) 367 \mathrm{kPa}\)

Short answer

Answer: The mean effective pressure (MEP) of this Otto cycle is approximately 528 kPa.

Step-by-step solution

Calculate the initial and final volume of the air

We are given the initial density of the air \(\rho_i=1.20 \mathrm{kg/m^3}\) and we have to compress it to a final volume 0.26 L. The initial volume can be calculated using the ideal gas law: \(PV=mRT\). Here, \(P\) is pressure, \(V\) is volume, \(m\) is mass, \(R\) is the specific gas constant, and \(T\) is temperature. Since we are given the initial conditions, we can assume a mass of 1 kg for the air. Considering air to be an ideal gas, we can use the specific gas constant for air, \(R=287 \mathrm{J/kg K}\). We are given the initial pressure \(P_i=2.2\times10^5 \mathrm{Pa}\). We can now calculate the initial volume \(V_i\) as: \[V_i = \frac{m}{\rho_i} = \frac{1 \mathrm{kg}}{1.20 \mathrm{kg/m^3}} \approx 0.833 \mathrm{m^3}\] Since we are given the final volume \(V_f = 0.26 \times 10^{-3} \mathrm{m^3}\) (converting L to m³), we can now proceed to calculate the work done during compression and expansion.

Calculate the work done during compression and expansion

For an ideal Otto cycle, work done during the compression process \((W_C)\) and expansion process \((W_E)\) can be calculated using the following equations: \(W_C = mR(T_2-T_1)/(1 - k)\) and \(W_E = mR(T_4-T_3)/(1-k)\) where \(T_1, T_2, T_3, T_4\) are the temperatures at points 1, 2, 3, 4 (respectively), \(m\) is mass, \(R\) is the specific gas constant, and \(k\) is the specific heat ratio (for air, \(k=1.4\)). We have to find \(T_2 - T_1\) and \(T_4 - T_3\). We can use the relation: \(\frac{T_2}{T_1} = \frac{P_2}{P_1}(\frac{V_1}{V_2})\) and \(\frac{T_4}{T_3} = \frac{P_1}{P_2}(\frac{V_1}{V_2})\) Since we are not given temperatures, we can ignore the temperature terms and find the ratio of work done \((W_C)/(W_E)\): \[\frac{W_C}{W_E} = \frac{T_2-T_1}{T_4-T_3} \] To find \(W_C\) and \(W_E\), we are given the net work output of the cycle \((W_{net})\) as 440 kJ/kg. Since \(W_{net} = W_E - W_C\), we can find the ratio of work done: \[\frac{W_C}{W_E} = 1 - \frac{W_{net}}{W_E} \Longrightarrow W_E = \frac{W_{net}}{1-\frac{W_C}{W_E}}\]

Calculate the Mean Effective Pressure (MEP)

Mean Effective Pressure (MEP) is defined as the hypothetical constant pressure that, if maintained throughout the entire cycle, would result in the same amount of net work output. The formula for MEP is: \[MEP = \frac{W_{net}}{V_\text{displaced}}\] We know that \(W_{net}=440\,\text{kJ/kg}\). To find \(V_\text{displaced}\) (the difference in volume), we simply subtract the final volume from the initial volume: \[V_\text{displaced}=V_i-V_f \approx 0.833\,\text{m}^3 - 0.26 \times 10^{-3}\,\text{m}^3\] Now, we can find the MEP: \[MEP = \frac{440\times10^3\,\text{J/kg}}{0.833\,\text{m}^3 - 0.26 \times 10^{-3}\,\text{m}^3} \approx 528000\,\text{Pa} = 528\,\text{kPa}\] So the mean effective pressure (MEP) for this Otto cycle is approximately 528 kPa, which corresponds to the option (c).

Key concepts

Engine Thermodynamics

Understanding the principles of engine thermodynamics is crucial for comprehending how energy is converted into work in engines like the ideal Otto cycle. The Otto cycle is a thermodynamic process typical for internal combustion engines and involves four distinct stages: intake, compression, power, and exhaust. During the intake, air (and usually fuel) is drawn into the engine. Compression then increases the pressure and temperature of the mixture, followed by ignition, which causes the power stroke as the gases expand. Finally, the exhaust stroke expels the burnt gases, and the process starts over.

The cycle is categorized as an ideal process, meaning it assumes no energy losses or inefficiencies. Real engines deviate from this ideal due to friction, heat loss, and other factors. However, examining the ideal cycle provides significant insight into engine operation, paving the way for real-world applications and improvements. Engine thermodynamics not only dictates the operation of these phases but also relies on principles such as the conservation of energy and the ideal gas law to predict the performance and efficiency of the engine.

Mean Effective Pressure

Mean Effective Pressure (MEP) is a critical term in the evaluation of an engine's performance. It represents the average pressure that acts on the pistons of an engine that would produce the same amount of work during one cycle as that produced under varying pressures. In simpler terms, it is a helpful way to summarize the net output of an engine in terms of a single, constant pressure value that equates to the same amount of work done over the entire cycle.

MEP is important as it provides an indication of an engine's capability to do work, regardless of its size. It allows for the comparison of engines on a common basis. For instance, when an Otto cycle operates, the MEP can be used to evaluate the effectiveness of the compression and power strokes by equating the net work done to a hypothetical constant pressure that could achieve the same result. The calculation of MEP, as seen in the exercise, considers the work done (energy output) and the volume displacement (space the piston travels), making it a comprehensive measure of engine efficiency.

Ideal Gas Law

The ideal gas law is a fundamental equation that is invaluable in fields such as engine thermodynamics. It relates the pressure (P), volume (V), temperature (T), and amount of substance (n) of an ideal gas through the equation:
\[PV = nRT\]
where R represents the gas constant, specific to the gas in question. In our engine context, it's common to calculate mass-based processes, using the mass (m) and specific gas constant (R), where R is tailored for units of mass rather than amount of substance, thus giving us the modified form:\[PV = mRT\]This law presupposes the gas being analyzed behaves ideally, which means its molecules do not attract or repel each other and take up a negligible space relative to the container. While no real gas is perfectly ideal, many behave ideally under common conditions, allowing this simplification to be useful for calculating changes in state during an engine cycle.

In the problem at hand, the ideal gas law was initially employed to calculate the volume of the air before compression, indicating its underlying importance in determining states from known quantities. It also permits the direct correlation between thermodynamic state variables, essential for analyzing changes during processes such as those within the Otto cycle.