I'm trying to calculate the time for the liquid in two connected reservoirs (open to the air) to equilibrate. The tricky part is the connecting region, which consists of a porous medium. I understand that without the porous medium, I can apply Torricelli's law to account for the change of pressure head over time. However, Torricelli's law doesn't apply to flow through porous medium. I also understand that I can apply Darcy approximation for flow through a porous medium. But Darcy law doesn't take into account for the change in pressure term in the equation.

What I have tried:

1) I tried derive the solution by combining Torricelli's law and Darcy law. But the solution turned out to be weird where I get t = Constant* ln(ΔH) where the Constant consists of area of reservoir, cross sectional area of opening, permeability of the porous medium, length, viscosity, density, gravitational force. So this method failed.

2) I tried looking for unsteady flow solution for Darcy equation. Turned out that there are a few ways such as VANS approximation, Virtual Mass approximation and Volume-average energy equation. To be honest, I failed to follow through the whole process of derivation and I also realized that the solution which the paper provided may not be what I have been looking for. The unsteady state flow that they describe is the initial fluctuation of velocity profile but not a change due to decreasing pressure head.

My question is, how can I proceed to calculate the time for the two reservoirs to equilibrate?

Known parameters: - Initial ΔH - Permeability of the porous column (assumed to be a constant throughout) - Area of reservoir - Cross-sectional area of connecting part - Viscosity - Density - Length

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Calculation as suggested by Chester:

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The problem is with ln(0) and also if my ΔH is actually smaller than 1, then it will result in a negative value.

*Sorry that I uploaded this as a picture as I am not sure how to insert equations and special characters here.

  • $\begingroup$ Darcy's Law does take into account the pressure difference across the porous plug, which is $\rho g \Delta H$. Have you tried using that alone? ie Forget Torricelli's Law. If you are still having difficulty, please show your calculation. $\endgroup$ Aug 4, 2016 at 18:18
  • $\begingroup$ After giving it a second thought, I think you may be right that I can directly do an integration with Darcy's equation. This is also suggested by Chester. I am going to try that out. Thanks! $\endgroup$
    – SHL
    Aug 5, 2016 at 1:35

1 Answer 1


The pressure difference across the porous plug is given by Darcy's law:

$$P_L-P_R=\frac{\mu}{k}vL\tag{1}$$where L is the length of the plug, k is the permeabiity of the plug, v is the superficial velocity through the plug, and $\mu$ is the fluid viscosity. The volumetric flow rate through the plug is given by $vA$, where A is the area of the plug. So, $$A_c\frac{d(\Delta H)}{dt}=-2vA\tag{2}$$where $A_c$ is the cross sectional area of each of the two tanks. If we neglect kinetic energy effects, then $$P_L-P_R=\rho g\Delta H\tag{3}$$If you combine these equations, you obtain a differential equation for the time variation of $\Delta H$ involving only the areas, the permeability, the viscosity, the density, g, and the length of the plug.

  • $\begingroup$ Turned out what you have proposed is very similar to what I have done in my approach 1. I have a few doubts here though. First, in equation (2), where does the "2" come from? Second, I still get t= Constant* ln(ΔH) as my solution which doesn;t make sense to me though. I will show my calculation in the question later. Thanks for your help! $\endgroup$
    – SHL
    Aug 5, 2016 at 1:52
  • $\begingroup$ Sorry about your doubts. I have practical experience in flow through porous media. How can I help alleviate your doubts? $\endgroup$ Aug 5, 2016 at 1:54
  • $\begingroup$ @SHL : The factor of 2 arises because a decrease in height by 1m on one side causes an increase in height of 1m on the other side, hence the difference in height decreases by 2m. $\endgroup$ Aug 5, 2016 at 2:10
  • $\begingroup$ @Chester Miller : I have uploaded the calculation steps and I couldn't get a reasonable solution. Is there anything that I have done wrong? Thanks. $\endgroup$
    – SHL
    Aug 5, 2016 at 2:41
  • $\begingroup$ @ Sammy Gerbil: Thanks for explaining that to me, that makes a lot of sense! $\endgroup$
    – SHL
    Aug 5, 2016 at 2:43

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