Determine the mean effective pressure of an ideal Otto Cycle that uses air as the working fluid; its state at the beginning of the compression is 14 psia and 60oF; its temperature at the end of the combustion is 1500oF; and its compression ratio is 9. Use constant specific heats at room temperature.

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Answer:

31.282 psia

Explanation:

Calculate the mean effective pressure of an ideal Otto cycle

Given data:

compression ratio = 9

inlet pressure ( P1 ) = 14 psia

Initial temperature ( T1 ) = 60°F = 520 R

Final/maximum temp ( T3 ) = 1500°F = 1960 R

using the constant specific heats at room temperature

attached below is the detailed solution

Ver imagen batolisis

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;

c_v = 0.171 btu/lbm.R

c_p = 0.24 btu/lbm.R

R = 0.3704 btu/lbm.R

ratio of specific heats; k = 1.4

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

  • Formula for heat entering and heat exiting are;

Q_in = c_v(T₃ - T₂)

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

  • Net work done is given by;

W_net = Q_in - Q_out

W_net = 121.02 - 50.251

W_net = 70.769 btu/lbm

Formula for the mean effective pressure is;

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

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