Engineering, 14.04.2020 22:21 angelespinosa521

Determine the mean effective pressure of an ideal Otto cycle that uses air as the working fluid if its state at the beginning of the compression is 14 psia and 60°F, its temperature at the end of the combustion is 1500°F, and its compression ratio is 9. Use constant specific heats at room temperature. The properties of air at room temperature are R = 0.3704 psia·ft3/lbm·R, cp = 0.240 Btu/lbm·R, cv = 0.171 Btu/lbm·R, and k = 1.4.

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Answer from: BrainGeek4848

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)

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

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

Read more about compression ratio at; link

Answer from: jaquezdaking4919

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 $\frac{w_{net}}{\alpha _1 - \alpha_1 }$

$= \frac{Heat\, Supplied - Rejected \, Heat}{\alpha_1 -\frac{\alpha_1 }{r} }$

= $\frac{q_{in} -q_{out}}{\alpha_1-\frac{\alpha_1 }{r} }$

$=\frac{P_1}{RT_1} \frac{r}{r-1} (c_v(T_3-T_2)-c_v(T_4 - T_1))$

$=\frac{P_1}{RT_1} \frac{c_v r}{r-1} (T_3(1-r^{1-k})+T_1(1 - r^{k-1}))$

= $=\frac{14\cdot 0.171 \cdot 9}{(9-1)\cdot 0.06855 \cdot 520} (1960(1-9^{1-1.4})+520(1-9^{1.4-1}))$

31.268 psi.

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