An ideal Otto cycle has a compression ratio of 10.5, takes in air at 9

kursval7z 2021-11-25 Answered
An ideal Otto cycle has a compression ratio of 10.5, takes in air at 90 kPa and 408C, and is repeated 2500 times per minute. Using constant specific heats at room temperature, determine the thermal efficiency of this cycle and the rate of heat input if the cycle is to produce 90 kW of power.

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Expert Answer

tnie54
Answered 2021-11-26 Author has 7322 answers

We have an ideal Otto cycle with a compression ratio of:
\(\displaystyle{r}={10.5}\)
Air is injected in to the engine at the next pressure and temperature:
\(\displaystyle{p}_{{{1}}}={90}{k}{P}{a}\)
\(\displaystyle{T}_{{{1}}}={40}°{C}\approx{313}{K}\)
The power of the cycle is:
\(\displaystyle{W}={90}{k}{W}\)
We are to determine the coefficient of efficiency and the rate of heat intake.
The specific heat ratio for air we get from table A-2 and it’s value is:
\(\displaystyle{k}={1.1}\)
The coefficient of thermal efficiency for an Otto cycle with constant specific heats is:
\(\displaystyle\eta={1}-{\frac{{{1}}}{{{r}^{{\kappa-{1}}}}}}\)
\(\displaystyle={1}-{r}^{{{1}-\kappa}}\)
\(\displaystyle={1}-{10.5}^{{{1}-{1.4}}}\)
\(\displaystyle\Rightarrow\eta={0.61}\)
The rate of heat input is: \(\displaystyle\eta={\frac{{\dot{{{W}}}}}{{\dot{{{Q}_{{\in}}}}}}}\)
\(\displaystyle\Rightarrow{Q}_{{in}}={\frac{{\dot{{{W}}}}}{{\eta}}}\)
\(\displaystyle={\frac{{{90}{k}{W}}}{{{0.61}}}}\)
\(\displaystyle\Rightarrow{Q}_{{in}}={147.541}{k}{W}\)

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Daniel Williams
Answered 2021-11-27 Author has 9250 answers

Step 1
Otto cycles have following process:
1-2 -isentropic compression process
2-3-constnat volume heat addition
3-4-isentropic expansion process
4-1 constant volume heat rejection.
Write the pressure and volume relation for isentropic process:
\(\displaystyle{P}_{{{2}}}{{V}_{{{2}}}^{{\gamma}}}={P}_{{{1}}}{{V}_{{{1}}}^{{\gamma}}}\ldots..{\left({1}\right)}\)
Use the ideal gas equation in Equation (1). \(\displaystyle{\frac{{{P}_{{{2}}}{V}_{{{2}}}}}{{{T}_{{{2}}}}}}={\frac{{{P}_{{{1}}}{V}_{{{1}}}}}{{{T}_{{{1}}}}}}\ldots..{\left({2}\right)}\)
Step 2
Use the equation (1) and (2).
\(\displaystyle{T}_{{{1}}}{{V}_{{{1}}}^{{\gamma-{1}}}}={T}_{{{2}}}{{V}_{{{2}}}^{{\gamma-{1}}}}\)
\(\displaystyle{T}_{{{2}}}={\frac{{{T}_{{{1}}}{{V}_{{{1}}}^{{\gamma-{1}}}}}}{{{{V}_{{{2}}}^{{\gamma-{1}}}}}}}\)
\(\displaystyle={T}_{{{1}}}{\left({\frac{{{V}_{{{1}}}}}{{{V}_{{{2}}}}}}\right)}^{{\gamma-{1}}}\)
\(\displaystyle{T}_{{{2}}}={T}_{{{1}}}{\left({r}\right)}^{{\gamma-{1}}}\)
Substitute the known values in the above Equation. \(\displaystyle{T}_{{{2}}}={\left({313}\right)}{\left({10.5}\right)}^{{{1.4}-{1}}}\)
\(\displaystyle{801.715}{K}\)
Step 3
Write the expression for the thermal efficiency:
\(\displaystyle\eta_{{{t}{h}}}={1}-{\frac{{{T}_{{{1}}}}}{{{T}_{{{2}}}}}}\)
\(\displaystyle\eta_{{{t}{h}}}={1}-{\frac{{{313}}}{{{801.715}}}}\)
\(\displaystyle={1}-{0.39}\)
\(\displaystyle={0.61}\)
Thus, the thermal efficiency of the cycle is 0.61.
Step 4
Calculate the heat added using the thermal efficiency:
\(\displaystyle{W}_{{{n}{c}{t}}}=\eta_{{{h}{a}{n}}}\times{Q}_{{\in}}\)
\(\displaystyle{Q}_{{in}}={\frac{{{W}_{{{e}{t}}}}}{{\eta_{{{t}{h}{e}}}}}}\)
\(\displaystyle{Q}_{{in}}={\frac{{{90}{k}{W}}}{{{0.61}}}}\)
\(\displaystyle{Q}_{{in}}={147.54}{k}{W}\)
Thus, the heat added to the cycle is 147.54 kW.

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