Derive the efficiency of air standard Otto cycle.

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written 8.2 years ago by

ak.amitkhare.ak • 560

Fig 1.3 (a) and (b) shows the theoretical p-V diagram and T-s diagrams of this cycle respectively.

• The point 1 represents that cylinder is full of air with volume V1 pressure P1 and absolute temperature T1.

• Line 1-2 represents the adiabatic compression of air due to which p1, V1 and T1 change to p2, V2 and T2, respectively.

• Line 2-3 shows the supply of heat to the air at constant volume so that p2 and T2 change to p3 and T3. (V3 being the same as V2).

• Line 3-4 represents the adiabatic expansion of the air. During expansion p3, V3 and T3 change to a final value of p4. V4 or V1 and T4 respectively.

• Line 4-1 shows the rejection of heat by air at constant volume till original state (point 1) reaches. Consider 1 kg of air (working substance):

Heat supplied at constant volume = Cv(T3-T2).

Heat rejected at constant volume = Cv(T4-T1).

But, Work done = Heat supplied – Heat rejected

    $$= Cv(T3-T2) - Cv(T4-T1)$$

$Efficiency = \frac{{Work{\text{ done}}}}{{Heat{\text{ supplied}}}} = \frac{{{C_v}({T_3} - {T_2}) - {C_v}({T_4} - {T_1})}}{{{C_v}({T_3} - {T_2})}}$

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

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