shrikeerthi said:
Hai,
I have the following situation: I have a closed container with a certain gas at a certain temperature(Tg) and pressure(pg). Now I open the container. The gas will escape through the opening to the atmosphere in order to create a pressure balance. There will be no significant increse in the pressure in the atmosphere but there will be noticible pressrue drop in the container. Does anyone know the formula to calculate the change in pressure over time in the container? Thanks in advance for any kind of help.
With Regards,
sk
Let's employ the conservation equations:(NOTE: variables without any subindex mean conditions inside the container.)
1) Continuity:
\dot m=-V\frac{d \rho}{dt} is the mass flow.
2) Energy:
\frac{d(\rho e V)}{dt}+ \dot m (e + \frac{U^2}{2} + \frac{P}{\rho})=0
which means that the time change of internal energy per unit mass (e) is due to the outflow of stagnation enthalpy, because:
h_o=e + \frac{U^2}{2} + \frac{P}{\rho}=c_p T is the stagnation enthalpy of the reservoir.
\frac{d(\rho c_v T V)}{dt}-V\frac{d\rho}{dt}c_p T=0
\frac{1}{\gamma-1}\frac{d(PV)}{dt}-V\frac{d\rho}{dt}c_p T=0
\frac{1}{\gamma-1}\frac{dP}{dt}-\frac{\gamma}{\gamma-1}\frac{P}{\rho}\frac{d\rho}{dt}=0
and we come to a surprise checking that the last equation is:
\frac{dP}{dt}=\frac{\gamma P}{\rho}\frac{d\rho}{dt}
which is the equation of an isentropic line. Remind I have written the energy equation neglecting terms of heat flux (which at Re_D Pr >>1 are in fact negligible. Integrating the last equation we have:
\frac{P}{\rho^\gamma}= \frac{P_o}{\rho_o^\gamma} which is a constant value and depends on initial values. This equation doesn't provide a value of P(t) yet.
The evolution of P(t) has a strong dependence on the orifice shape. For simplicity, we will assume it is a convergent nozzle, with subsonic flow everywhere. So that, the mass flow is not constant, and will depend on the inner conditions. The mass flow through a convergent nozzle can be written as:
\dot m(t)=-V\frac{d \rho}{dt}=\rho_s U_s A_s where "s" are the conditions just at the outlet section of the nozzle. Pay attention to this change of variables:
\dot m(t)=\rho a A_s \frac{\rho_s}{\rho}\frac{U_s}{a_s}\frac{a_s}{a}
The variable a is the Sound Speed. The Sound Speed is different inside the container (a) and just at the outlet (a_s) because the gas has been accelerated and thus its static temperature has changed along the streamline that goes from the inner part (at rest) to the outlet (at movement with velocity Us).
By means of the definition of total enthalpy:
h_o=h+\frac{U^2}{2}
one can calculate those fractions as a function of the Mach Number at the outlet section: M=\frac{U_s}{a_s}. In particular, the pressure evolution from inwards to the outlet can be written as:
\Big(\frac{P}{P_a}\Big)^{\frac{\gamma-1}{\gamma}}=1+\frac{\gamma-1}{2} M^2 where P_a is the ambient pressure.
Substituting this in the mass flow last equation:
\dot m(t)=-V\frac{d \rho}{dt}=\rho a A_s \Big(\frac{P}{P_a}\Big)^{-\frac{\gamma+1}{2\gamma}}\Big[\Big(\frac{P}{P_a}\Big)^{\frac{\gamma-1}{\gamma}}-1\Big]^{1/2} \Big(\frac{2}{\gamma-1}\Big)^{1/2}
So that, this last equation with the isentropic line will give you the evolution P(t), because there are two equations for two unknowns. Remind that \rho a=\sqrt{\gamma P \rho}.
You have checked that calculating the evolution of the pressure inside a container is not a trivial task. Energy and Continuity of mass considerations must be employed. Also, the shape of the nozzle is very important, due to the elipticity of the subsonic problem the information of boundaries must be propagated inside the container. Without the equation of the nozzle mass flow, you'll never reach a P(t) solution.
I hope that it will help you.
