How to predict the volume of a tank of water at any time? (fluid motion)

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SUMMARY

The discussion focuses on predicting the volume of a cylindrical tank filled with liquid as it drains through a pipe using fluid dynamics principles. Key equations mentioned include the Hagen–Poiseuille equation for viscous flow and Pascal's law for pressure changes due to fluid height. The relationship between volumetric flow rate (Q) and height change (Δh) is established, leading to the formulation Q = k Δh, where k incorporates factors like fluid density and pipe dimensions. The conversation emphasizes the need for a comprehensive equation that accounts for viscosity and pipe length in the context of a vertical tank drain problem.

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  • Understanding of fluid dynamics principles, specifically Hagen–Poiseuille equation
  • Knowledge of Pascal's law and its application in fluid mechanics
  • Familiarity with volumetric flow rate calculations
  • Basic concepts of pressure drop in fluid systems
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Zaxo3000
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We have a cylindrical tank of known dimensions. Attached at the bottom is a pipe perpendicular to the tank of known dimensions. We fill the tank with a liquid and let it drain out through the pipe. What formula will predict the volume of the tank at any time t? I believe the Hagen–Poiseuille equation (involves the viscosity of the liquid) and Pascal's law (involves the density of the liquid) are involved, but i cannot derive an equation for the volume at any time. All working and any explanation where ever necessary would be appreciated.

Heres my attempt at it:
ΔP = 8μLQ / πr^4

ΔP is the pressure drop
L is the length of pipe
μ is the dynamic viscosity
Q is the volumetric flow rate
r is the radius

ΔP = ρgΔh

g is acceleration due to gravity
ρ is the fluid density
Δh is the height of fluid

ΔP = 8μLQ / πr^4
ΔP = ρgΔh
Volume of tank = πr^2h
8μLQ / πr^4 = ρgΔh
Q = ρgΔh ✕ πr^4 / 8μL

This is essentially equal to Q = k Δh where k is = ρg ✕ πr^4 / 8μL
Im fairly sure is is redundant however as of course the rate of flow is proportional to the change in height of water as when the Δh is multiplied by the area of the circular face of the tank it will give the volume of the cylinder of water which has left through the pipe during the Δh. There has to be an equation for the relationship I am looking for as the results of the test will be consistent.
 
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Is you tank horizontal or vertical? This will govern the method to apply.
 
edgepflow said:
Is you tank horizontal or vertical? This will govern the method to apply.
Its a vertical standing tank, on its circular face.
 
edgepflow said:
This is just the classic tank drain problem. All my best books are at work but try

http://www.lmnoeng.com/Tank/TankTime.htm
Thanks edgepflow for the link. However that websites seems to be focused more on a Torricelli's law situation, which was created for holes. It does appear to take into account an oriface discharge constant but only relates it to the type of hole. As a large part of my experiment will be testing different viscosity and lengths of the pipe, a formula (such as the Hagen–Poiseuille equation) which takes these both into account is something that should be part of the answer.
 
Zaxo3000 said:
Thanks edgepflow for the link. However that websites seems to be focused more on a Torricelli's law situation, which was created for holes. It does appear to take into account an oriface discharge constant but only relates it to the type of hole. As a large part of my experiment will be testing different viscosity and lengths of the pipe, a formula (such as the Hagen–Poiseuille equation) which takes these both into account is something that should be part of the answer.
I think you can modify this treatment to include your other effects as follows:

DP = rho * g * h = KT * rho * v^2/2 + f (L/D) * rho * v^2/2

K = sum of all resistance coefficients (including opening)
L = length of pipe
and the rest are the usual meening.

Note that you can capture your viscosity effects with correlations for f that depend on Reynold's Number.

So solve this above equation for Q as a function h.

Then, as was done in the link: Integrate Q(h) = -A dh/dt.
 

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