g A rigid air cylinder with a volume of 100 cm3 is punctured with a hole having a crosssectional area of 0.3 mm2 . The original pressure and temperature of the air inside the cylinder are 800 kPa and 35 C. As the air leaves the hole in the cylinder, it reaches a pressure of 100 kPa and a temperature of 5 C. The velocity of the air as it escapes through the hole is 100 m/s. Calculate the original mass of air inside the tank and the mass in the tank 5 seconds after it is punctured, assuming the exit conditions of the air remain independent of time g

Solution :

The volume of the rigid cylinder =

Initial pressure inside the cylinder,

Initial temperature inside the cylinder,

Final temperature inside the cylinder,

Final pressure inside the cylinder,

Area of the hole, A =

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

∴  

Since the exit condition does not change with time, we have ,

At ambient condition, and .

Therefore, we can find the density of the air

Mass flow rate of air from the cylinder =

can be written as

=

= kg/s

Mass escaped from the cylinder in 5 seconds

Mass of air remaining in the cylinder after 5 seconds :

    = 0.7225 grams

50 inches should be the resultant vector

the gas collects in coils that then becomes a liquid that then flows through the coils inside the refrigerator