What is the lowest attainable pressure? (Mass Balance/Control Volume Question)?

A tank of Volume = 1m^3 contains air that is maintained at a constant temperature. The absolute pressure in the tank Po= 1 bar. Air is evacuated at a constant rate of Q=uA=0.001 m^3/s. The mass flowrate of a leak, in a tank, is c(po-p), where c=0.00001m^3/s is a constant, ';po'; is the density of the ambient air, and ';p'; is the density of the air inside the tank. The initial density of the air inside the tank is also po. What is the lowest attainable pressure, P inside the tank.








lower case 'p' is density


capital 'P' is pressureWhat is the lowest attainable pressure? (Mass Balance/Control Volume Question)?
Assume the air is an ideal gas


P∙V = n∙R∙T


with n = m/M


%26lt;=%26gt;


P = (m/V) ∙ (R/M) ∙ T


%26lt;=%26gt;


P = ρ ∙ R' ∙ T


where


R' specific gas constant of air


R' = R/M = 8.314472J(molK / 0.028.97 kg/mol = 287J/kgK





Assuming isothermal conditions pressure inside the tank and density are proportional. So can solve the problem by finding minimum density.


Minimum density can be found from mass balance of the tank.


dm/dt = Min - Mout


assuming well mixed tank, density is everywhere the same and the leaving air stream has the same density. Hence:


dm/dt = c∙(ρ₀ - ρ) - Q∙ρ


expressing mass in tank in terms of density and volume:


d(V∙ρ)/dt = c∙(ρ₀ - ρ) - Q∙ρ


because the volume of the tank remains unchanged


V ∙ dρ/dt = c∙(ρ₀ - ρ) - Q∙ρ





At the min mum density the derivative equals zero. The physical meaning of this relation, that inlet and pullet stream are identical.


dρ/dt = 0


%26lt;=%26gt;


c∙(ρ₀ - ρ_min) - Q∙ρ_min = 0


%26lt;=%26gt;


ρ_min = c∙ρ₀ / (c + Q) = ρ₀ / (1 + Q/c)





therefore


p_min = ρ_min∙R'∙T


= (ρ₀∙R'∙T) / (1 + Q/c)


= P₀ / (1 + Q/c)


= 1atm / (1 + 0.001m³/s / 0.00001m³/s)


= 0.0099atm

bloom-