The isentropic compression ratio of a new lawn mower Otto cycle gasoline engine is [latex]8.00[/latex] to [latex]1[/latex], and the inlet air temperature is [latex]T_3 = 70.0°\text{F}[/latex] at a pressure of [latex]p_3 = 14.7 \text{psia}[/latex]. Determine a. The air temperature at the end of the isentropic compression stroke [latex]T_{4s}[/latex]. b. The pressure at the end of the isentropic compression stroke before ignition occurs [latex]p_{4s}[/latex]. c. The Otto cold ASC thermal efficiency of this engine.

a. The isentropic compression ratio for an Otto cycle engine is defined as [latex]\text{CR} = \frac{v_3}{v_{4s}} = (\frac{T_3}{T_{4s}})^{\frac{1}{1−k} }[/latex] from which we have [latex]T_{4s} = \frac{T_3}{\text{CR}^{1-k}} = T_3 × \text{CR}^{k-1} = (70.0 + 459.67 \text{R}) (8.00)^{0.40} = 1220 \text{R}[/latex] b. For the Otto cycle, the isentropic pressure and compression ratios are related by [latex]\text{PR} = \text{CR}^{k}[/latex], where [latex]\text{PR} = p_{4s} / p_3[/latex] and [latex]\text{CR} = v_3/v_{4s}[/latex]. Then, [latex]p_{4s} = p_3 \text{CR}^{k} = (14.7 \text{psia}) (8.00)^{1.40} = 270. \text{psia}[/latex] c. Equation (13.30) gives the Otto cold ASC thermal efficiency as [latex](η_T)_{\substack{\text{Otto}\\\text{cold ASC}\\}} = 1 - \frac{T_3}{T_{4s}} = 1 - \text{PR}^{\frac{1-k}{k}} = 1 - \text{CR}^{1-k} = 1- (8.00)^{1−1.40} = 0.565 = 56.5\%[/latex]

Question 13.14: The isentropic compression ratio of a new lawn mower Otto cy...

Modern Engineering Thermodynamics

The isentropic compression ratio of a new lawn mower Otto cycle gasoline engine is $8.00$ to $1$ , and the inlet air temperature is $T_3 = 70.0°\text{F}$ at a pressure of $p_3 = 14.7 \text{psia}$ . Determine

a. The air temperature at the end of the isentropic compression stroke $T_{4s}$ .
b. The pressure at the end of the isentropic compression stroke before ignition occurs $p_{4s}$ .
c. The Otto cold ASC thermal efficiency of this engine.

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a. The isentropic compression ratio for an Otto cycle engine is defined as

$\text{CR} = \frac{v_3}{v_{4s}} = (\frac{T_3}{T_{4s}})^{\frac{1}{1−k} }$

from which we have

$T_{4s} = \frac{T_3}{\text{CR}^{1-k}} = T_3 × \text{CR}^{k-1} = (70.0 + 459.67 \text{R}) (8.00)^{0.40} = 1220 \text{R}$

b. For the Otto cycle, the isentropic pressure and compression ratios are related by $\text{PR} = \text{CR}^{k}$ , where $\text{PR} = p_{4s} / p_3$ and $\text{CR} = v_3/v_{4s}$ . Then,

$p_{4s} = p_3 \text{CR}^{k} = (14.7 \text{psia}) (8.00)^{1.40} = 270. \text{psia}$

c. Equation (13.30) gives the Otto cold ASC thermal efficiency as

$(η_T)_{\substack{\text{Otto}\\\text{cold ASC}\\}} = 1 – \frac{T_3}{T_{4s}} = 1 – \text{PR}^{\frac{1-k}{k}} = 1 – \text{CR}^{1-k} = 1- (8.00)^{1−1.40} = 0.565 = 56.5\%$