Question

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.

Answer 1

Answer:

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.