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The compression ratio in an Otto cycle is a crucial parameter defined as the ratio of the maximum to the minimum volume in the cylinder. Mathematically, it is expressed as: \[ r = \frac{V_1}{V_2} \] Here, \( V_1 \) is the initial volume before compression and \( V_2 \) is the volume after compression. In this particular problem, the compression ratio is given as 7.5. The higher the compression ratio, the more the air-fuel mixture is compressed, leading to higher efficiency and power output. However, it can also result in higher engine temperatures and potential knocking issues. Understanding the role of compression ratio helps in analyzing engine performance and optimizing fuel efficiency.

An isentropic process is a thermodynamic process in which entropy remains constant. In the context of the Otto cycle, both the compression and expansion processes are ideally assumed to be isentropic. For an isentropic compression, the relationship between temperature and volume is given by: \[ \frac{T_2}{T_1} = \left( \frac{V_1}{V_2} \right)^{k-1} \] Here, \(T_1\) and \(T_2\) are the initial and final temperatures, and \(k\) (or \(\gamma\)) is the specific heat ratio, typically 1.4 for air. Similarly, for isentropic expansion, the relation is: \[ \frac{T_4}{T_3} = \left( \frac{V_3}{V_4} \right)^{k-1} \] Where \(T_3\) and \(T_4\) are the temperatures before and after expansion. These relationships are essential for calculating state properties in cycles and are derived from the principles of thermodynamics.

Thermal efficiency of an Otto cycle is a measure of how well the cycle converts heat into work. It is defined as the ratio of the net work output to the heat input. For the Otto cycle, thermal efficiency can be expressed as: \[ \eta = 1 - \frac{q_{out}}{q_{in}} \] In the given problem, heat rejection \(q_{out}\) and heat addition \(q_{in}\) are computed using the specific heat at constant volume: \[ q_{out} = m \times c_v \times (T_4 - T_1) \] \[ q_{in} = m \times c_v \times (T_3 - T_2) \] With efficiencies calculated as: \[ \eta = 1 - \frac{0.114}{0.281} = 0.595 \] Meaning approximately 59.5 percent of the input energy is converted to useful work. Higher thermal efficiency indicates a more effective engine in converting fuel energy into mechanical work.

Mean effective pressure (MEP) is a representative measure of the engine's power output. It is defined as the average pressure which, if acted upon the pistons, would produce the same work as the cycle. MEP is calculated by the formula: \[ MEP = \frac{W_{net}}{V_1 - V_2} \] In this problem, \(W_{net}\) is the net work output and \(V_1\) and \(V_2\) are the initial and final volumes during compression. With computed values: \[ V_1 = 0.00206 \text{ m}^3 \] \[ V_2 = \frac{V_1}{r} = \frac{0.00206}{7.5} = 0.000275 \text{ m}^3 \] \[ MEP = \frac{0.167}{0.00206 - 0.000275} = 91.45 \text{ kPa} \] The MEP provides an insight into the efficiency and performance of an engine, helping compare different engines regardless of their size.