Respuesta :
Solution :
The volume of the rigid cylinder = [tex]$100 \ cm^3 = 100 \times 10^{-6} \ m^3$[/tex]
Initial pressure inside the cylinder, [tex]$P_i = 800 \ kPa$[/tex]
Initial temperature inside the cylinder, [tex]$T_i = 35^\circ C= 308 \ K$[/tex]
Final temperature inside the cylinder, [tex]$T_f = 5^\circ C= 278 \ K$[/tex]
Final pressure inside the cylinder, [tex]$P_f = 100 \ kPa$[/tex]
Area of the hole, A = [tex]$0.3 \ mm^2 = 3 \times 10^{-7} \ m^2$[/tex]
Velocity of the air through the hole, V = 100 m/s
The final pressure and the temperature inside the cylinder will be the condition same as the ambient conditions.
At initial state, from the equation of state,
PV = mRT, where R = 287 J/kg-K for air
[tex]$800 \times 10^3 \times100 \times10^{-6} = m_1 \times 287 \times 308$[/tex]
∴ [tex]$m_1=9.1 \times 10^{-4} \ kg$[/tex]
Since the exit condition does not change with time, we have ,
At ambient condition, [tex]$P_f = 100 \kPa$[/tex] and [tex]$T_f= 278 \ K$[/tex].
Therefore, we can find the density of the air
[tex]$P=\rho R T$[/tex]
[tex]$\rho = \frac{P}{RT}$[/tex]
[tex]$=\frac{100 \times 10^3}{287 \times 278}$[/tex]
[tex]$= 1.25 \ kg/m^3$[/tex]
Mass flow rate of air from the cylinder = [tex]$\dot m$[/tex]
[tex]$\dot m$[/tex] can be written as [tex]$\dot m$[/tex] [tex]$=\rho_{f} \times A \times v$[/tex]
[tex]$\dot m$[/tex] = [tex]$1.25 \times 3 \times 10^{-7} \times 100$[/tex]
[tex]$\dot m$[/tex] = [tex]$3.75 \times 10^{-5}$[/tex] kg/s
Mass escaped from the cylinder in 5 seconds
[tex]$m=3.75 \times 10^{-5} \times 5$[/tex]
[tex]$= 1.875 \times 10^{-4} \ kg$[/tex]
Mass of air remaining in the cylinder after 5 seconds :
[tex]$m_2 = m_1 - m$[/tex]
[tex]$m_2 = 9.1 \times 10^{-4} - 1.875 \times 10^{-4}$[/tex]
[tex]$m_2 = 7.225 \times 10^{-4} \ kg$[/tex]
= 0.7225 grams
