**2 Answers**

written 8.3 years ago by |

Air standard Otto cycle on (a) P-v diagram (b) T-s diagram

Processes: -

0-1: a fresh mixture of fuel-air is drawn into the cylinder at constant pressure

1-2: isentropic compression

2-3: energy addition at constant volume

3-4: isentropic expansion4-1: combustion products leave the cylinder

1-0: the piston pushes out the remaining combustion products at constant pressure

Since the net work done in processes 0-1 and 1-0 is zero, for thermodynamic analysis, we consider the 1-2-3-4 only.

The thermal efficiency of the cycle is given by $$η = \frac{W_{net}}{Q_1} = \frac{Q_1 - Q_2}{Q_1}$$

where Q1 and Q2 denote the energy absorbed and rejected as heat respectively.

For a constant volume process,

$$Q = ΔU$$

If ‘m’ is the mass of the air which is undergoing the cyclic process,

$$ΔU = mC_vΔT$$

Energy is absorbed during the process 2-3 and energy is rejected during the process 4-1.

Hence,

$$Q_1 = U_3 - U_2 = mC_v(T_3 - T_2)$$ $$Q_2 = U_4 - U_1 = mC_v(T_4 - T_1)$$ $$η = 1 - \frac{T_4 - T_1}{T_3 - T_2}$$

For an ideal gas undergoing an isentropic process (process 1-2 and 3-4), $$Tv^{γ-1} = constant$$

Hence, $$\frac{T_1}{T_2} = (\frac{v_2}{v_1})^{γ-1}$$

and $$\frac{T_4}{T_3} = (\frac{v_3}{v_4})^{γ-1}$$

But $v_1 = v_4 and v_2 = v_3$.Hence we get,

$$\frac{T_1}{T_2} = \frac{T_4}{T_3} \ \ or \ \ \frac{T_1}{T_4} = \frac{T_2}{T_3}$$

$$1 - \frac{T_1}{T_4} = 1 - \frac{T_2}{T_3} \ \ or \ \ \frac{T_4 - T_1}{T_4} = \frac{T_3 - T_2}{T_3}$$

$$\frac{T_4 - T_1}{T_3 - T_2} = \frac{T_4}{T_3} = \frac{T_1}{T_2}$$

$$η = 1 - \frac{T_1}{T_2} = 1 - (\frac{v_2}{v_3})^{y-1} = 1 - (\frac{1}{γ_0})^{y-1}$$

Where the compression ration $r_0$ is defined as $r_0 = \frac{v_1}{v_2}$