Basic Mechanical Engineering (3110006)

This thermodynamic cycle is operated with constant volume heat addition and consists of two reversible adiabatic (isentropic) processes and two constant volume processes.

Isentropic Compression Process (1 – 2)

At point 1 cylinder is full of air with a volume ${\mathrm{V}}_1$ , pressure ${\mathrm{p}}_1$ and temperature ${\mathrm{T}}_1$.
Piston moves from BDC to TDC and an ideal gas (air) is compressed isentropically to state point 2 through compression ratio, $\mathrm{r} = \frac{{\mathrm{V}}_1}{{\mathrm{V}}_2}$.
The law of adiabatic process is, ${\mathrm{pV}}^{\mathrm{\gamma }}=\mathrm{C}$.
The relation for the adiabatic process 1-2,

$\frac{{\mathrm{T}}_2}{{\mathrm{T}}_1} = {\left(\frac{{\mathrm{V}}_1}{{\mathrm{V}}_2}\right)}^{\mathrm{\gamma }-1} = {\mathrm{r}}^{ \mathrm{\gamma }-1}$

$\Rightarrow {\mathrm{T}}_2 = {\mathrm{T}}_1 {\mathrm{r}}^{ \mathrm{\gamma }-1} \& {\mathrm{T}}_1 = \frac{{\mathrm{T}}_2}{{\mathrm{r}}^{ \mathrm{\gamma }-1}}$

Constant Volume Heat Addition Process (2 – 3)

Heat is added at constant volume from an external heat source. The pressure rises at constant volume.
The heat heat supplied during the process 2 – 3,

${\mathrm{Q}}_{\mathrm{s}} = {\mathrm{Q}}_{( 2-3 )} = {\mathrm{mC}}_{\mathrm{V}} ( {\mathrm{T}}_3 - {\mathrm{T}}_2 )$

Isentropic Expansion Process (3 – 4)

The increased high pressure exerts a greater amount of force on the piston and pushes it towards the BDC.
Expansion of working fluid takes place isentropically and work is done by the system.
The volume ratio $\frac{{\mathrm{V}}_4}{{\mathrm{V}}_3}$ is called isentropic expansion ratio.

The relation for adiabatic process (3 – 4),

$\frac{{\mathrm{T}}_3}{{\mathrm{T}}_4} = {\left(\frac{{\mathrm{V}}_3}{{\mathrm{V}}_4}\right)}^{\mathrm{\gamma } - 1} = {\left(\frac{{\mathrm{V}}_1}{{\mathrm{V}}_2}\right)}^{\mathrm{\gamma } - 1} = \mathrm{r}{ }^{\mathrm{\gamma } - 1} \left( \because {\mathrm{V}}_3={\mathrm{V}}_2 ,{\mathrm{V}}_4={\mathrm{V}}_1 \right)$ 

$\Rightarrow {\mathrm{T}}_4 = \frac{{\mathrm{T}}_3}{\mathrm{r}{ }^{\mathrm{\gamma } - 1}}$ 

Constant volume Heat Rejection Process (4 – 1)

Heat is rejected to the external sink at constant volume.
This process is so controlled that ultimately the working fluid comes to its initial state 1 and the cycle is repeated.
Heat rejected during process 4 – 1,  

${\mathrm{Q}}_{\mathrm{r}} = {\mathrm{Q}}_{4-1}={\mathrm{mC}}_{\mathrm{V}} ( {\mathrm{T}}_4 - {\mathrm{T}}_1 )$

Thermal Efficiency of an Otto Cycle is,

$\mathrm{\eta } = \frac{\mathrm{Networkdone}}{\mathrm{Heatsupplied}} = \frac{{\mathrm{W}}_{\mathrm{net}}}{{\mathrm{Q}}_{\mathrm{s}}}$ 

$\mathrm{\eta } = \frac{{\mathrm{C}}_{\mathrm{v}} ( {\mathrm{T}}_3 - {\mathrm{T}}_2 ) - {\mathrm{C}}_{\mathrm{V}} ( {\mathrm{T}}_4 - {\mathrm{T}}_1 )}{{\mathrm{C}}_{\mathrm{v}} ( {\mathrm{T}}_3 - {\mathrm{T}}_2 )}$ 

$\mathrm{\eta } = 1 - \frac{ {\mathrm{C}}_{\mathrm{V}} ( {\mathrm{T}}_4 - {\mathrm{T}}_1 )}{{\mathrm{C}}_{\mathrm{v}} ( {\mathrm{T}}_3 - {\mathrm{T}}_2 )}$ 

Substitute the value of ${\mathrm{T}}_4$ and ${\mathrm{T}}_1$ in above equation,

$\mathrm{\eta } = 1 - \frac{{\displaystyle \frac{{\mathrm{T}}_3}{ {\mathrm{r}}^{\mathrm{\gamma }-1} }} - {\displaystyle \frac{{\mathrm{T}}_2}{ {\mathrm{r}}^{\mathrm{\gamma }-1} }}}{{\mathrm{T}}_3 - {\mathrm{T}}_2}$ 

$\mathrm{\eta } = 1 - \frac{{\displaystyle 1}}{ {\mathrm{r}}^{ \mathrm{\gamma } - 1 }}$