The Otto cycle is the ideal prototype of most internal combustion engines in use today. The cycle was clearly described for the first time (1862) by a Frenchman, Beau de Rochas. Fourteen years later, Nicholas A. Otto (1832-1891), a German, independently invented the same cycle, and going further than Beau de Rochas, he built an engine to operate on it. The ideal cycle consists of two isentropic and two constant volume processes, 1-2-3-4 Fig. 1

It is customary to analyze the ideal cycle as though there were no suction and exhaust strokes 0-l and as though the working substance were air only Such an analysis is referred to as an air-standard analysis.

Even though the real engine exhausts its working substance we are Justified in assuming a constant volume cooling 4-1 for a standard of comparison; the work W is the same (friction of flow neglected ) whether the working substance is exhausted or cooled; the heat supplied QA is the same in each case; and consequently from the energy equation of a cycle, QA-QR = W, the heat rejected in each case must be the same. Assuming constant specific heats, we find (Fig. 1) 


	QA = wcv(T3 - T2) Btu,

	QR = wcv(Tl - T4) = -wcv(T4 - T1) Btu.


The net work W is SUM(Q),* so that


(a)	W = wcv(T3 - T2) - wcv(T4 - Tl) Btu.


* The work from the pV plane, Fig. 1, is (in Btu)


	W = (p2V2-p1V1)/(J(l-k))+(p4V4-p3V3)/(J(l-k)) = (wR(T2-Tl+T4-T3))/(J(l-k)).

	
Using cv = (R/J)/(k-1), we get


	W = -wcv(T2-Tl+T4-T3) = wcv(T3-T2+Tl-T4)


as in equation (a) the thermal efficiency of the Otto cycle is


	e = W/QA = (wcv(T3-T2)-wcv(T4-T1))/(wcv(T3-T2)


or


(b)	e = 1 - (T4-T1)/(t3-T2)
To simplify equation (b), use the TV relation for an isentropic process. Thus, T4/T3 = (V3/V2)k-1 and T1/T2 = (V2/V1)k-l; or, since V3 = V2 and V4 = V1, 
	T4 = T3(V3/V4)k-1 = T3(V2/V1)k-1, and T1 = T2(V2/V1)k-1.


Substituting these values of T4 and T1 into (b), we find


	e = 1 - (T3(V3/V4)k-1 - T2(V2/V1)k-1)/(T3-T2 = 1 - (V2/V1)k-1.
Let the adiabatic compression ratio Vl/V2 be represented by the symbol r then 
	e = 1 - 1/rk-1




Basic Functional Referance