Question

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.

Answer 1

Answer:

(i) Final pressure and temperature are 25.119 bar and 753.566 K, (ii) Final pressure and temperature are 66.289 bar and 994.336 K.

Explanation:

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:

$\frac{p_{2}}{p_{1}} = \left(\frac{V_{1}}{V_{2}} \right) ^{\gamma}$ (Eq. 1)

$\frac{T_{2}}{T_{1}} = \left(\frac{V_{1}}{V_{2}} \right)^{\gamma - 1}$ (Eq. 2)

Where:

$p_{1}$, $p_{2}$ - Initial and final pressures, measured in bar.

$T_{1}$, $T_{2}$ - Initial and final temperatures, measured in Kelvins.

$V_{1}$, $V_{2}$ - Initial and final volumes, measured in cubic meters.

$\gamma$ - Specific heat ratio of air, dimensionless.

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

$r_{c} = \frac{V_{1}}{V_{2}}$ (Eq. 3)

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

$\frac{p_{2}}{p_{1}} = r_{c}^{\gamma}$ (Eq. 1b)

$\frac{T_{2}}{T_{1}} = r_{c}^{\gamma - 1}$ (Eq. 2b)

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

$p_{2} = p_{1}\cdot r_{c}^{\gamma}$

$T_{2} = T_{1}\cdot r_{c}^{\gamma-1}$

(i) $r_{c} = 10$, $p_{1} = 1\,bar$, $T_{1} = 300\,K$, $\gamma = 1.4$

$p_{2} = (1\,bar)\cdot 10^{1.4}$

$p_{2} = 25.119\,bar$

$T_{2} = (300\,K)\cdot 10^{0.4}$

$T_{2} = 753.566\,K$

Final pressure and temperature are 25.119 bar and 753.566 K.

(ii) $r_{c} = 20$, $p_{1} = 1\,bar$, $T_{1} = 300\,K$, $\gamma = 1.4$

$p_{2} = (1\,bar)\cdot 20^{1.4}$

$p_{2} = 66.289\,bar$

$T_{2} = (300\,K)\cdot 20^{0.4}$

$T_{2} = 994.336\,K$

Final pressure and temperature are 66.289 bar and 994.336 K.