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.
1 – 2
2 – 3
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= cv (T3 – T2)
q R =heat rejection =cv (T4 – T1)
efficiency ɳ otto = 1 = q R / q s = 1 – cu (T4 – T1) / cu ( T3 –T2) = 1 – T4 –T1 / T3 – T2
r = compression ratio = u1 / u2
Expansion ratio = u4 u3 = u1 / u2 = r
For process 1-2
T3 = T4 r γ-1
Substituting the values of T2 and T3 in eqn. (10.1) we get
ɳ otto = 1 – T4 – T1 / T4 r γ-1 – T1 r γ-1 = 1 ( T4 –T1) / r γ-1 ( T4 – T1)
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.
v2 = v3 = V c = 1
p3 p2 = p4 / p1 = a
P2 /p1 = ( u1 / u2) γ and p3 / p4 = (u4 / u3)γ
P2 / p1 = r γ and p3 / 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
Pm = w net / v s
Pm = 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- T1 / T3 ) . Comparing this expression with the efficiency of the otto cycle i.e., ( 1- T4 / T3) , we see that the efficiency of otto cycle is lower as T4 > T1. This is obvious also because in otto cycle all heat is not supplied at the highest temperature of the cycle i.e., T3 and not rejected at the lowest temperature of the cycle i.e., T4 as it is done in case of the Carnot cycle.