Time taken for pressure equalization between two tanks

[Raghav Seetharamu]

Homework Statement

How to calculate time taken (rough approximation) for pressure equalization between two tanks.
Both the tanks have same fluid (Air).

Homework Equations

Rough equilibrium pressure can be achieved by using equations Ptot = (P1V1+P2V2)/Vtot.
Thought of using exponential equation P=Ae^(Bt) for calculating time required for pressure equalization.
A and B are constants.
P=A1e^(-Bt) for tank with decreasing pressure;
P=A2e^(Bt) for tank with increasing pressure;

Exponential constant B, is assumed to be same for the tanks. Am I correct with this assumption?

The Attempt at a Solution

P1 = 1000mbar; P2 = 0.1mbar, V1=6m3, V2 = 1m3,
Ptot = 860 mbar

P=1000*e^(-Bt) for tank with decreasing pressure ;
P=0.1*e^(Bt) for tank with increasing pressure;

How to obtain the exponential constant and hence the time taken for pressure equalization using these equations?

 

[BvU]

Hello Rahghav, $\qquad$ $\qquad$ !

To settle this somewhat, you need more information: a certain amount of material has to be transported from one vessel to the other. The rate depends on the pressure difference and on the resistance in that transport: is it a thin and very long tube or a big short pipe ?

 

[Raghav Seetharamu]

Hello Rahghav, $\qquad$ $\qquad$ !

To settle this somewhat, you need more information: a certain amount of material has to be transported from one vessel to the other. The rate depends on the pressure difference and on the resistance in that transport: is it a thin and very long tube or a big short pipe ?

Thanks for reverting:)
They are connected through relatively big and short pipes.

 

[haruspex]

Suppose the pressures at time t are P₁(t), P_pipe(t), P₂(t). A= cross-sectional area of pipe, L=length.
Mass velocity in pipe = v(t).
Can you write some equations for how the pressure differences result in acceleration of air into, through and out of the pipe, and for how the rate of flow affects P₁ and P₂?
Bear in mind that density depends on pressure. You should probably assume adiabatic compression/decompression.