Question

TASK 3
a) A single cylinder petrol engine has a volume compression ratio of 8:1, takes in a mixture of fuel and air at a temperature of 250°C and its pressure is 101 kPa. If the pressure at the end of the compression stroke is 1.5 MPa what will be its final temperature?
b) A compressed air storage cylinder has a volume of 0.5 m3 and contains air at an absolute pressure of 1.8 MPa and temperature 20°C. A quantity of the air is released during which the temperature of the remaining air falls to 15°C and the pressure to 1.5 MPa. Calculate the mass of the air released. The characteristic gas constant for air is 287 Jkg−1K−1.

Answer 1

Answer on Question #59710-Engineering-Material Science Engineering

a) A single cylinder petrol engine has a volume compression ratio of 8:1, takes in a mixture of fuel and air at a temperature of $250^{\circ}\mathrm{C}$ and its pressure is $101\mathrm{kPa}$. If the pressure at the end of the compression stroke is $1.5\mathrm{MPa}$ what will be its final temperature?

Solution

$$\frac {p _ {1} V _ {1}}{T _ {1}} = \frac {p _ {2} V _ {2}}{T _ {2}}$$ $$T _ {2} = T _ {1} \frac {p _ {2} V _ {2}}{p _ {1} V _ {1}} = (250 + 273.15) \frac {1.5 \cdot 10^{6}}{101 \cdot 10^{3}} \frac {1}{8} = 971.19 \mathrm{K} = (250 - 273.15)^{\circ} \mathrm{C} = 698^{\circ} \mathrm{C}$$

b) A compressed air storage cylinder has a volume of 0.5 m³ and contains air at an absolute pressure of 1.8 MPa and temperature $20^{\circ}\mathrm{C}$. A quantity of the air is released during which the temperature of the remaining air falls to $15^{\circ}\mathrm{C}$ and the pressure to 1.5 MPa. Calculate the mass of the air released. The characteristic gas constant for air is 287 Jkg⁻¹K⁻¹.

Solution

$$pV = nRT$$

I'll eventually use the molecular weight of air (28.7 g/mol); this will provide essentially the same info as the "characteristic gas constant".

In the initial state,

$$(1.8 \cdot 10^{6} \mathrm{Pa})(0.5 \mathrm{m}^{3}) = n_{1} \left(8.314 \frac{\mathrm{J}}{(\mathrm{mol} \mathrm{K})}\right) (293.15 \mathrm{K})$$

In the final state,

$$(1.5 \cdot 10^{6} \mathrm{Pa})(0.5 \mathrm{m}^{3}) = n_{2} \left(8.314 \frac{\mathrm{J}}{(\mathrm{mol} \mathrm{K})}\right) (288.15 \mathrm{K})$$ $$n_{1} - n_{2} = \left[ \frac{0.5 \mathrm{m}^{3}}{\left(8.314 \frac{\mathrm{J}}{(\mathrm{mol} \mathrm{K})}\right)} \right] \left( \frac{1.8 \cdot 10^{6} \mathrm{Pa}}{293.15} \mathrm{K} - \frac{1.5 \cdot 10^{6} \mathrm{Pa}}{288.15} \mathrm{K} \right) = 56.2 \mathrm{moles}$$ $$m = M n = 56.2 \mathrm{mol} \left(28.7 \frac{\mathrm{g}}{\mathrm{mol}}\right) = 1.61 \mathrm{kg}$$

https://www.AssignmentExpert.com