Consider a cylinder of a gasoline engine at the beginning of the compression cycle, during which a fuel/air mixture (for our purposes mostly composed of nitrogen and oxygen, i.e. an ideal gas of diatomic molecules) at 300 K and 1 bar is compressed down to one-tenth volume (compression ratio of 10:1). Assume that the compression is rapid so no heat exchange occurs with the environment. Calculate the pressure and the temperature of the compressed gas. In a diesel engine the compression ratios are typically much higher; redo the same calculation with the compression ratio of 20:1.

This system experiments an adiabatic compression, as such compression happens with no heat interaction between the piston-cylinder device and surroundings. This is a particular case of polytropic process, in which there is no entropy generation according to the Second Law of Thermodynamics.

The compression process is represented by the following formulas:

[tex]\frac{p_{2}}{p_{1}} = \left(\frac{V_{1}}{V_{2}} \right) ^{\gamma}[/tex] (Eq. 1)

[tex]\frac{T_{2}}{T_{1}} = \left(\frac{V_{1}}{V_{2}} \right)^{\gamma - 1}[/tex] (Eq. 2)

Where:

[tex]p_{1}[/tex], [tex]p_{2}[/tex] - Initial and final pressures, measured in bar.

[tex]T_{1}[/tex], [tex]T_{2}[/tex] - Initial and final temperatures, measured in Kelvins.

[tex]V_{1}[/tex], [tex]V_{2}[/tex] - Initial and final volumes, measured in cubic meters.

[tex]\gamma[/tex] - Specific heat ratio of air, dimensionless.

From Theory of Diesel and Otto Cycles we know that compression ratio is defined as:

[tex]r_{c} = \frac{V_{1}}{V_{2}}[/tex] (Eq. 3)

And (Eqs. 1, 2) can be rewritten as follows:

[tex]\frac{p_{2}}{p_{1}} = r_{c}^{\gamma}[/tex] (Eq. 1b)

[tex]\frac{T_{2}}{T_{1}} = r_{c}^{\gamma - 1}[/tex] (Eq. 2b)

Then, we clear final pressure and pressure in each expression and calculate them for each case:

[tex]p_{2} = p_{1}\cdot r_{c}^{\gamma}[/tex]

[tex]T_{2} = T_{1}\cdot r_{c}^{\gamma-1}[/tex]

(i) [tex]r_{c} = 10[/tex], [tex]p_{1} = 1\,bar[/tex], [tex]T_{1} = 300\,K[/tex], [tex]\gamma = 1.4[/tex]

[tex]p_{2} = (1\,bar)\cdot 10^{1.4}[/tex]

[tex]p_{2} = 25.119\,bar[/tex]

[tex]T_{2} = (300\,K)\cdot 10^{0.4}[/tex]

[tex]T_{2} = 753.566\,K[/tex]

Final pressure and temperature are 25.119 bar and 753.566 K.

(ii) [tex]r_{c} = 20[/tex], [tex]p_{1} = 1\,bar[/tex], [tex]T_{1} = 300\,K[/tex], [tex]\gamma = 1.4[/tex]

[tex]p_{2} = (1\,bar)\cdot 20^{1.4}[/tex]

[tex]p_{2} = 66.289\,bar[/tex]

[tex]T_{2} = (300\,K)\cdot 20^{0.4}[/tex]

[tex]T_{2} = 994.336\,K[/tex]

Final pressure and temperature are 66.289 bar and 994.336 K.