This cycle is named after Dr. A.N.OTTO, a German scientist (in 1876).

The cycle that may be used to approximate the operation of S.l engine is the Otto cycle. Fig. 10.3 shows the Otto cycle on P- v and T – s diagram. Let 1 kg mass of ideal gas circulate in the cycle.

Process
1 – 2 2 – 3 3 -4 4- 1 |
Description
Reversible adiabatic compression. Constant volume heat addition. Reversible adiabatic expansion. Constant volume heat rejection. |

**10.3.1 Derivation of Air Standard Efficiency of Otto-cycle**

q s = heat addition= c_{v} (T_{3} – T_{2})

q _{R} =heat rejection =c_{v} (T_{4} – T_{1})

efficiency ɳ _{otto} = 1 = q _{R} / q _{s} = 1 – c_{u} (T_{4} – T_{1}) / c_{u} ( T_{3} –T_{2}) = 1 – T_{4} –T_{1} / T_{3} – T_{2 }

r = compression ratio = u_{1} / u_{2}

Expansion ratio = u_{4} u_{3} = u_{1} / u_{2} = r

For process 1-2

T^{3} = T4 r ^{γ-1}

Substituting the values of T2 and T3 in eqn. (10.1) we get

ɳ _{otto }= 1 – T_{4} – T_{1} / T_{4} r ^{γ-1} – T_{1} r ^{γ-1} = 1 ( T_{4} –T_{1}) / r ^{γ-1} ( T_{4} – T_{1})

ɳ _{otto }= 1 – 1 / r ^{γ-1}

**10.3.2 Derivation of Mean Effective Pressure of Otto Cycle**

Let the clearance volume be unity and the ratio of compression and expansion each equal

to r and we consider one kg of air as working fluid.

v_{2} = v_{3} = V c = 1

p_{3} p_{2 }= p_{4 }/ p_{1} = a

P_{2} /p_{1} = ( u_{1} / u_{2}) γ and p_{3} / p_{4} = (u_{4} / u_{3})^{γ}

P_{2} / p_{1} = r ^{γ} and p_{3} / p4 = r ^{γ}

Net work done = area under the diagram

= (area under isentropic 3 – 4)- (area under isentropic 1- 2)

= 1 / γ -1 [ p4 r ^{γ} – p4 r ^{γ }– p1 r)]

= 1 / γ -1 [ a r ^{γ-1} – r ( r ^{γ-1} -1)]

= p1 r ( r ^{γ-1-1}) ( a-1) / γ-1

W _{net} = p1 r (a-1) (r ^{γ-1} -1) / γ-1

P_{m} = w _{net} / v s

P_{m} = p1 r (a-1) ( r ^{γ-1}) (r ^{γ-1} -1) / γ-1) ( r-1)

**Note :** The Carnot efficiency based upon the maximum temperature of this cycle is equal

to ( 1- T_{1} / T_{3} ) . Comparing this expression with the efficiency of the otto cycle i.e., ( 1- T_{4} / T_{3}) , we see that the efficiency of otto cycle is lower as T_{4} > T_{1}. This is obvious also because in otto cycle all heat is not supplied at the highest temperature of the cycle i.e., T_{3} and not rejected at the lowest temperature of the cycle i.e., T_{4} as it is done in case of the Carnot cycle.