Understanding the Otto Cycle Heat Addition

This problem involves calculating the specific heat added during the combustion process in an Otto cycle. We are given the compression ratio, initial conditions (pressure and temperature), and the maximum temperature reached after heat addition. We need to find the amount of heat added per unit mass (specific heat addition).

Given Data:

• Compression Ratio, $r_c = \frac{V_1}{V_2} = 9$
• Initial Pressure, $P_1 = 100$ kPa
• Initial Temperature, $T_1 = 10^\circ \text{C}$
• Maximum Temperature, $T_3 = 2500$ K

Assumptions:

We assume the working fluid is air, and we use the standard properties for air:

• Specific heat ratio, $\gamma = 1.4$
• Specific heat at constant volume, $c_v = 0.7175$ kJ/(kg·K)

Step-by-Step Calculation

1. Convert Initial Temperature to Kelvin:

   The initial temperature $T_1$ must be in Kelvin for thermodynamic calculations.

   Initial Temperature, $T_1 = 10^\circ \text{C} = 10 + 273.15 = 283.15$ K

2. Calculate Temperature after Isentropic Compression ($T_2$):

   In an Otto cycle, the compression process (1-2) is assumed to be isentropic (adiabatic and reversible). The relationship between temperature and volume during an isentropic process is given by:

   $$ \frac{T_2}{T_1} = \left(\frac{V_1}{V_2}\right)^{\gamma-1} $$

   Substituting the values:

   $$ T_2 = T_1 \times (r_c)^{\gamma-1} $$

   $$ T_2 = 283.15 \, \text{K} \times (9)^{1.4 - 1} $$

   $$ T_2 = 283.15 \, \text{K} \times (9)^{0.4} $$

   Calculating $(9)^{0.4}$:

   $$ (9)^{0.4} \approx 2.4118 $$

   $$ T_2 \approx 283.15 \, \text{K} \times 2.4118 $$

   $$ T_2 \approx 683.0 \, \text{K} $$

3. Calculate Specific Heat Addition ($q_{in}$):

   The heat addition process (2-3) in an Otto cycle occurs at constant volume. The specific heat added is calculated using the change in temperature and the specific heat at constant volume ($c_v$):

   $$ q_{in} = c_v (T_3 - T_2) $$

   Substituting the values:

   $$ q_{in} = 0.7175 \, \frac{\text{kJ}}{\text{kg} \cdot \text{K}} \times (2500 \, \text{K} - 683.0 \, \text{K}) $$

   $$ q_{in} = 0.7175 \, \frac{\text{kJ}}{\text{kg} \cdot \text{K}} \times 1817.0 \, \text{K} $$

   $$ q_{in} \approx 1304.5 \, \frac{\text{kJ}}{\text{kg}} $$

4. Convert Result to MJ/kg:

   Since $1$ MJ = $1000$ kJ, we convert kJ/kg to MJ/kg:

   $$ q_{in} \approx \frac{1304.5}{1000} \, \frac{\text{MJ}}{\text{kg}} $$

   $$ q_{in} \approx 1.3045 \, \text{MJ/kg} $$

Conclusion

The calculated specific heat addition by combustion is approximately $1.3045$ MJ/kg. This value closely matches one of the provided options.