Determine the mean effective pressure of an ideal Otto cycle that uses air as the

The mean effective pressure of the ideal Otto Cycle that uses air as the working fluid is; 31.28 psia

We are given;

Initial pressure; P₁ = 14 psia

Initial temperature; T₁ = 60°F = 520 R

Maximum temperature; T₃ = 1500°F = 1960 R

Compression ratio = 9

From tables, the specific heat capacities and constants are;

Let us first calculate the initial volume from the formula;

V₁ = RT₁/P₁

V₁ = (0.3704 * 520)/14

V₁ = 13.7577 ft³/lb.m

Compression ratio is; V₁/V₂ = 9

Thus;

V₂ = V₁/9

V₂ = 13.7577/9

V₂ = 1.5286 ft³/lb.m

To get the temperature T₂, we will use the formula;

T₂ = T₁(V₁/V₂)^(k - 1)

T₂ = 520(9)^(1.4 - 1)

T₂ = 1252.277 R

Similarly;

T₄ = T₃/9

T₄ = 1960/(9^(1.4 - 1))

T₄ = 813.87 R

Q_in = c_v(T₃ - T₂)

Thus;

Q_in = 0.171(1960 - 1252.277)

Q_in = 121.02 btu/lbm

Q_out = c_v(T₄ - T₁)

Q_out = 0.171(813.87 - 520)

Q_out = 50.251 btu/lbm

W_net = Q_in - Q_out

W_net = 121.02 - 50.251

W_net = 70.769 btu/lbm

mean effective pressure = W_net/(V₁ - V₂)

mean effective pressure = 70.769/(13.7577 - 1.5286)

mean effective pressure = 5.7869 btu.ft³

Converting to psia gives;

Mean effective pressure = 31.28 psia

Read more about compression ratio at; link

The mean effective pressure of the Otto cycle is  31.268 psi

Explanation:

The mean effective pressure is obtained by dividing the work done in the working stroke process of the cycle by the volume of the stroke ar the stroke volume.

In the Otto cycle, therefore, we are to apply an expression for the work done and the volume of the Otto cycle stroke to derive the value of the mean effective pressure as follows.

Here we have the mean effective pressure given by

MEP

=

=

 31.268 psi.

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