An engine working on air standard Otto cycle is supplied with air at 0.1 MPa and 35°C. The compression ratio is 8. The heat supplied is 500kJ/kg. Property data for air: c_p = 1.005 kJ/kg K, c_v = 0.718 kJ/kg K, R = 0.287 kJ/kg K. The maximum temperature (in K) of the cycle is _________ (correct to one decimal place).

Correct Answer: 1403.9

To determine the maximum temperature of the Otto cycle, we analyze the different states of the cycle shown in the P-V diagram:
State 1 represents the start of the isentropic compression process (1-2), and State 3 represents the state after constant volume heat addition (2-3), where the temperature and pressure reach their maximum values ($T_{\text{max}}=T_3$).

1. Given Parameter Values:
Initial pressure: $P_1=0.1\text{MPa}$
Initial temperature: $T_1=35°\text{C}=35+273=308\text{K}$
Compression ratio: $r=\frac{V_1}{V_2}$
Heat supplied: $Q_{\text{s}}=500\text{kJ/kg}$
Specific heat at constant pressure: $c_p=1.005\text{kJ/kg K}$
Specific heat at constant volume: $c_v=0.718\text{kJ/kg K}$

2. Calculate the Ratio of Specific Heats ($\gamma$):

$\gamma =\frac{c_p}{c_v}=\frac{1.005}{0.718}\approx 1.40$

3. Isentropic Compression Process (1-2):
Using the relation between temperature and volume for an isentropic process:

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

Substituting the given values:

$T_2=308\times {\left(8\right)}^{1.40-1}=308\times 8^{0.4}$

$T_2\approx 308\times 2.2974=707.6\text{K}$

4. Constant Volume Heat Addition Process (2-3):
The heat supplied is given by the relation:

$Q_{\text{s}}=c_v\left(T_3-T_2\right)$

Substituting the values:

$500=0.718\times \left(T_3-707.6\right)$

Solving for $T_3$:

$T_3-707.6=\frac{500}{0.718}\approx 696.38\text{K}$

$T_3=707.6+696.38=1403.98\text{K}\approx 1403.9\text{K}$

Thus, the maximum temperature of the cycle is 1403.9 K.