To determine the mean effective pressure (MEP) of an air-standard Otto cycle, we need to first calculate the net work done by the cycle and the displacement volume per unit mass. The mean effective pressure is a crucial performance parameter for internal combustion engines.

Otto Cycle Process Overview

An air-standard Otto cycle consists of four internally reversible processes:

• Isentropic Compression (1-2): Air is compressed without heat transfer.
• Constant Volume Heat Addition (2-3): Heat is added to the air at constant volume.
• Isentropic Expansion (3-4): Air expands without heat transfer, doing work.
• Constant Volume Heat Rejection (4-1): Heat is rejected from the air at constant volume, returning it to the initial state.

Given Parameters of the Otto Cycle

Let's list the given values for the Otto cycle problem:

• Compression ratio, \(r = 10\)
• Pressure at the beginning of compression, \(P_1 = 100 \text{ kPa}\)
• Temperature at the beginning of compression, \(T_1 = 27 \text{ }^\circ\text{C}\)
• Heat added at constant volume, \(Q_{in} = 1500 \text{ kJ/kg}\)
• Heat rejected at constant volume, \(Q_{out} = 700 \text{ kJ/kg}\)
• Specific gas constant for air, \(R = 0.287 \text{ kJ/kg-K}\)

Temperature Conversion and Net Work Calculation

First, convert the initial temperature from Celsius to Kelvin:

\(T_1 = 27 \text{ }^\circ\text{C} + 273.15 = 300.15 \text{ K}\)

(For simplicity in such problems, it's common to use \(273\), so \(27 + 273 = 300 \text{ K}\) is also acceptable.)

The net work done per unit mass in the Otto cycle is the difference between the heat added and the heat rejected:

\(W_{net} = Q_{in} - Q_{out}\)

Substitute the given values:

\(W_{net} = 1500 \text{ kJ/kg} - 700 \text{ kJ/kg}\)

\(W_{net} = 800 \text{ kJ/kg}\)

Specific Volume Calculation

To calculate the mean effective pressure, we need the specific volume at the beginning (\(v_1\)) and end (\(v_2\)) of the compression stroke. We can find \(v_1\) using the ideal gas equation of state:

\(P_1 v_1 = R T_1\)

So, the specific volume at state 1 is:

\(v_1 = \frac{R T_1}{P_1}\)

Substitute the values:

\(v_1 = \frac{(0.287 \text{ kJ/kg-K}) \times (300 \text{ K})}{100 \text{ kPa}}\)

\(v_1 = \frac{86.1 \text{ kJ/kg}}{100 \text{ kPa}}\)

Since \(1 \text{ kJ} = 1 \text{ kPa} \cdot \text{m}^3\), the units simplify to \( \text{m}^3\text{/kg}\):

\(v_1 = 0.861 \text{ m}^3\text{/kg}\)

The compression ratio \(r\) is defined as the ratio of the maximum to minimum volume in the cycle, which is \(v_1 / v_2\):

\(r = \frac{v_1}{v_2}\)

From this, we can find \(v_2\):

\(v_2 = \frac{v_1}{r}\)

Substitute the values:

\(v_2 = \frac{0.861 \text{ m}^3\text{/kg}}{10}\)

\(v_2 = 0.0861 \text{ m}^3\text{/kg}\)

Mean Effective Pressure (MEP) Calculation

The mean effective pressure (MEP) is defined as the net work per cycle divided by the displacement volume. In terms of specific values (per unit mass), it is:

\(MEP = \frac{W_{net}}{v_1 - v_2}\)

Substitute the calculated values for \(W_{net}\), \(v_1\), and \(v_2\):

\(MEP = \frac{800 \text{ kJ/kg}}{(0.861 - 0.0861) \text{ m}^3\text{/kg}}\)

\(MEP = \frac{800 \text{ kJ/kg}}{0.7749 \text{ m}^3\text{/kg}}\)

Again, recalling that \(1 \text{ kJ} = 1 \text{ kPa} \cdot \text{m}^3\), the units become kPa:

\(MEP \approx 1032.38 \text{ kPa}\)

Rounding to the nearest whole number, the mean effective pressure of the cycle is approximately 1032 kPa.

The calculated mean effective pressure is 1032 kPa.