# Otto Cycle Question

#### The Otto Cycle

You have a “standard” Otto cycle. At the start of the adiabatic compression $P=100 kPa$ and $T=25 C$. If the efficiency of this engine is $eta=0.48$ and the gas is air, please find T and P of the air at the end of the compression. Please specify what is the property and what are the assumptions you make.

By Luc1992 (Own work) [CC BY-SA 4.0], via Wikimedia Commons

#### Working principle

The Otto cycle is made from two adiabatic (isentropic) transformations and two constant volume transformations. Intake air is initially compressed isentropically (without heat exchange), thus its temperature increases drastically from outside ambient temperature, up to fuel ingnition temperature. When the fuel ignite a constant volume transformation happens quickly, so that the pressure increases. This is followed by an isentropic (adiabatic) relaxation until the initial volume of the cylinder. A fourth constant volume transformation happens, so that the system arrives in its initial condition.

The assumption that we make is that cold-air intake is used. Otherwise heat capacity could vary and fuel ingnition could happen before minimum volume.

#### Answer

Property: for air the isentropic expansion factor $gamma =Cp/Cv = (7/2)R / [(5/2)R] =7/5 =1.4$

The thermal efficiency of the above described Otto cycle is

$eta = 1-1/[r^{gamma-1}]$ (From Wiki)

where $r =V1/V2$ is the compression ratio.

$0.48 =1- 1/r^{0.4}$

$(1/r)^{0.4} = 0.52$

$r =5.128$

For the isentropic process 1-2 (compression process) we have

$P1*V1^{gamma} =P2*V2^{gamma}$

$P1/P2 =(V2/V1)^{gamma} = (1/r)^{gamma} =(1/5.128)^{1.4} =0.1014$

$P2 =P1/0.1014 =100*10^3/0.1014 =986.19*10^3 Pa =986.19 kPa$

The same for temperatures we have

$T1*V1^{gamma-1} = T2*V2{(gamma-1}$

$T1/T2 = (V2/V1){gamma-1} = (1/r)^{gamma-1} =(1/5.128)^{0.4} =0.52$

T2 =T1/0.52 =(25+273) /0.52 =573.08 K = 300 Celsius deg.

(T is absolute temperature hence it is in Kelvin deg.)

Assumption: cold air