# Answer to Question #83233 in Mechanical Engineering for Jha Devarsh

air at 90 kPa and 40oC, and is repeated 2500 times per minute. Using

constant specific heats at room temperature, determine the thermal

efficiency of this cycle and the rate of heat input if the cycle is to

produce 90 kW of power.

The thermal efficiency of an ideal Otto cycle is given by:

η=1-r^(1-k), (1)

where r = 8 – the compression ratio,

k = 1.4 – the adiabatic index for the ideal air.

Substitute into (1):

η=1-〖10.5〗^(1-1.4)=0.610.

The power P of the cycle is given by the ratio:

P=W/τ, (2)

where W is the work of the cycle,

is the time needed to complete the cycle once, which equals to:

τ=60/2500=0.024 s.

From (3) we can find the work of the cycle as follow:

W=Pτ,

W=90∙0.024=2.16 kJ.

The thermal efficiency of a cycle is given by:

η=W/Q_in , (3)

where Qin is the heat input for the cycle.

From (3) we can find the heat input Qin to the cycle as follow:

Q_in=W/η, (4)

Q_in=2.16/0.61=3.54 kJ.

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