An ideal Otto cycle has a compression ratio of 10.5, takes in air at 9

We have an ideal Otto cycle with a compression ratio of:
${\displaystyle r=10.5}$
Air is injected in to the engine at the next pressure and temperature:
${\displaystyle p_1=90kPa}$
${\displaystyle T_1=40°C\approx 313K}$
The power of the cycle is:
${\displaystyle W=90kW}$
We are to determine the coefficient of efficiency and the rate of heat intake.
The specific heat ratio for air we get from table A-2 and it’s value is:
${\displaystyle k=1.1}$
The coefficient of thermal efficiency for an Otto cycle with constant specific heats is:
${\displaystyle \eta =1-\frac{1}{r^{\kappa -1}}}$
${\displaystyle =1-r^{1-\kappa }}$
${\displaystyle =1-{10.5}^{1-1.4}}$
${\displaystyle \Rightarrow \eta =0.61}$
The rate of heat input is: ${\displaystyle \eta =\frac{\dot{W}}{\dot{Q_{\in }}}}$
${\displaystyle \Rightarrow Q_{in}=\frac{\dot{W}}{\eta }}$
${\displaystyle =\frac{90kW}{0.61}}$
${\displaystyle \Rightarrow Q_{in}=147.541kW}$

Step 1
Otto cycles have following process:
1-2 -isentropic compression process
2-3-constnat volume heat addition
3-4-isentropic expansion process
4-1 constant volume heat rejection.
Write the pressure and volume relation for isentropic process:
${\displaystyle P_2{V}_2^{\gamma }=P_1{V}_1^{\gamma }\dots ..\left(1\right)}$
Use the ideal gas equation in Equation (1). ${\displaystyle \frac{P_2{V}_2}{T_2}=\frac{P_1{V}_1}{T_1}\dots ..\left(2\right)}$
Step 2
Use the equation (1) and (2).
${\displaystyle T_1{V}_1^{\gamma -1}=T_2{V}_2^{\gamma -1}}$
${\displaystyle T_2=\frac{T_1{V}_1^{\gamma -1}}{V_2^{\gamma -1}}}$
${\displaystyle =T_1{\left(\frac{V_1}{V_2}\right)}^{\gamma -1}}$
${\displaystyle T_2=T_1{\left(r\right)}^{\gamma -1}}$
Substitute the known values in the above Equation. ${\displaystyle T_2=\left(313\right){\left(10.5\right)}^{1.4-1}}$
${\displaystyle 801.715K}$
Step 3
Write the expression for the thermal efficiency:
${\displaystyle {\eta }_{th}=1-\frac{T_1}{T_2}}$
${\displaystyle {\eta }_{th}=1-\frac{313}{801.715}}$
${\displaystyle =1-0.39}$
${\displaystyle =0.61}$
Thus, the thermal efficiency of the cycle is 0.61.
Step 4
Calculate the heat added using the thermal efficiency:
${\displaystyle W_{nct}={\eta }_{han}\times Q_{\in }}$
${\displaystyle Q_{in}=\frac{W_{et}}{{\eta }_{the}}}$
${\displaystyle Q_{in}=\frac{90kW}{0.61}}$
${\displaystyle Q_{in}=147.54kW}$
Thus, the heat added to the cycle is 147.54 kW.