Time to Drain An Atmospheric Tank

AI Thread Summary
To determine the time required to drain an atmospheric tank, the discussion centers on using a first-order ordinary differential equation (ODE) that incorporates factors such as the tank's cross-sectional area, the orifice area, and the discharge coefficient. The equation accounts for the decreasing head pressure as the water level drops, which affects the flow rate. The derivation is based on the Bernoulli equation and considers the impact of frictional losses in a piped outlet. Adjustments for frictional losses are suggested to be included in the final head calculation, as these losses vary with the changing flow during drainage. Understanding these dynamics is crucial for accurately modeling the draining process.
bigNate
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I am a little rusty on my differential equations as I haven't seen them in a few years! I am looking for some direction on a question on a first order ODE. I am trying to find the time required to drain a tank that is open to atmosphere. Any help would be appreciated.

Thanks!
 
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It depends on the volume (mass) and volumetric (mass) flow rate.

Open to the atmosphere simply means that the surface of the liquid or the gas is at 1 atm of pressure. For liquids, the pressure will increase with depth, which could affect mass flow rate. The drain opening will also influence flow rate.
 
I understand that the head pressure is going to decrease as the water level goes down which will affect the velocity of water leaving the tank...I am trying to develop an equation to get the time required to drain the tank.
 
This is directly from a PE exam review book:

The general ODE covering the set up is:
-A\frac{dH}{dt}=A_oC_o\sqrt{(2g \Delta H)}

Integrating gives:

t=\frac{A\sqrt{\frac{2}{g}}}{A_o C_o}\left[H_i^{.5} - H_f^{.5}\right]

Where:
A= the cross sectional area of the tank
A_o= area of the flow orifice
g= acceleration due to gravity
C_o= orifice discharge coefficient (usually about 0.6)
H_i= Initial tank head in ft.
H_f= Final head

Give that a go and see how that does. Granted, if you have a very viscous fluid or the viscosity changes over time all bets are off since the Reynolds number is a function of that which then effects your discharge coefficient.
 
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Thanks. I assume this was derived from using the Bernoulli equation and an unsteady state material balance? I didn't get the same term on the right hand side...do you have the derivation as well?
 
What is the geometry of the tank? Spherical? Cylindrical? If cylindrical, Vertical or horizontal?
 
It is a rectangular concrete tank
 
bigNate said:
Thanks. I assume this was derived from using the Bernoulli equation and an unsteady state material balance? I didn't get the same term on the right hand side...do you have the derivation as well?
It is definitely a product of the Bernoulli equation. It actually looks more like a plain square edge orifice flow equation.
 
What would the formula be taking into account frictional losses due to pipe length and fittings etc. for a piped outlet (open to atmos)?
 
  • #10
krugan said:
What would the formula be taking into account frictional losses due to pipe length and fittings etc. for a piped outlet (open to atmos)?
You can include those losses into the final head (which is what that variable is there for). If you notice, the flow is going to be proportional to the delta head number. If you increase the outlet head, the flow decreases.
 
  • #11
Fred,

Since the flow is constantly changing as it's draining the frictional losses will also change. How do you determine what frictional losses to use in the final head?
 
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