An air receiver of volume 5.5 m³ contains air at 16 bar and 42°C. A valve is opened and some air is allowed to blow out to atmosphere. The pressure of the air in the receiver drops rapidly to 12 bar when the valve is then closed.

Calculate the mass of air which has left the receiver.

Question

An air receiver of volume 5.5 m³ contains air at 16 bar and 42°C. A valve is opened and some air is allowed to blow out to atmosphere. The pressure of the air in the receiver drops rapidly to 12 bar when the valve is then closed.

Calculate the mass of air which has left the receiver.

Accepted Answer

Initial volume of air, [latex]V_1=5.5[/latex] m³

Initial pressure of air, [latex]p_1=16[/latex] bar

Initial temperature of air, [latex]T_1=42+273=315[/latex] K

Final volume of air, [latex]V_2=V_1=5.5[/latex] m³

Final pressure of air, [latex]p_2=12[/latex] bar

Mass of air which left the receiver :

Mass of air in the initial condition,

[latex]m_1=\frac{p_1 V_1}{R T_1}=\frac{16 imes 10^5 imes 5.5}{\left(0.287 imes 10^3 ight) imes 315}=97.34[/latex] kg.

Assuming that the mass in the receiver undergoes a reversible adiabatic process, then

[latex]\frac{T_2}{T_1}=\left(\frac{p_2}{p_1} ight)^{\frac{\gamma-1}{\gamma}}[/latex]

[latex]\frac{T_2}{315}=\left(\frac{12}{16} ight)^{\frac{14-1}{14}}=\left(\frac{12}{16} ight)^{0.286} \quad ext { or } \quad T_2=315 imes\left(\frac{12}{16} ight)^{0.286}=290[/latex] K

Now mass of air in the receiver in final condition,

[latex]m_2=\frac{p_2 V_2}{R T_2}=\frac{12 imes 10^5 imes 5.5}{\left(0.287 imes 10^3 ight) imes 290}=79.3[/latex] kg.

∴ Mass of air which left the receiver,

[latex]m=m_1-m_2=97.34-79.3= 1 8 . 0 4[/latex] kg.

Question 4.64: An air receiver of volume 5.5 m³ contains air at 16 bar and ...

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