# Fill rate of tank for elevated vessel [closed] I'm trying to model a system that consists of a rectangular vessel (storage tank) that fills a cylindrical tank. I would like the compute the time it takes to fill a tank based on an initial vessel water height $H_\text{vw}$.

The vessel tank, seen on the left:

• is open to the atmosphere and filled with water
• is rectangular
• has a height of $H_\text{v}$ (constant)
• has an instantaneous water level of $H_\text{vw}$
• is always being filled by a constant flow $F_\text{f}$ whenever the water level - $H_\text{vw}$ is less than the tank height $H_\text{v}$. $F_\text{f}$ is in gallons per minute, but I can easily convert it to inches of water per minute.
• The bottom of the vessel tank sits $H_\text{ag}$ above the bottom of the fill tank.

The fill tank, to the right:

• is open to the atmosphere
• is cylindrical in shape with a diameter $d$
• has an instantaneous water level of $H_\text{tw}$
• has a maximum water height of $H_\text{t}$
• is connected to the vessel tank via a 2" flexible smooth walled hose.

I have a constraint to fill the tank within a certain amount of time $t_\text{max}$, for a given initial water level in the vessel $H_\text{vw}(0)$,g so I need to choose a height $H_\text{ag}$ that makes this possible.

After I solve for $H_\text{ag}$, I would like to know how long it takes to fill a tank for any given initial water level $H_\text{vw}(0)$.

I would also like to model the flow rate $\text{gpm}(t)$ in gallons per minute during the fill process.

## closed as off-topic by Brian Moths, ZeroTheHero, Jon Custer, Yashas, John RennieMar 21 '17 at 8:08

This question appears to be off-topic. The users who voted to close gave this specific reason:

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• Given this is a physical problem is $F_{F}$ actually a binary flow? (ie constant while $H_{VW}<H_{v}$ and $0$ at $H_{VW}=H_{v}$) Or is there some system like the ballcock in a toilet where the flow decreases as the height increases? – Lio Elbammalf Mar 20 '17 at 22:54
• Thanks Lio. Ff is constant. it is two 2" municipal metered water valves (approximately 320 gpm combined). It will be cut-off when the vessel is full, but should run at full speed when the Vessel is not full. There maybe some "slowly shutting off", to keep from water hammer and hysterics at the very top, but I'm not worried about accounting for that yet. – Jason Mar 20 '17 at 23:00
• The whole point of this vessel is to decrease fill times (increase flow rate) during peak hours by having a sustained output flow rate >= to the 2 municipal water lines. – Jason Mar 20 '17 at 23:04
• I think the answer is no from, Torricelli's law, which seems like the best case possible. $$gpm = a*v$$ $$a=\pi\cdot (\frac{2"}{2})^2 = \pi in^2$$ (single 2 inch outlet) $$gpm(max)=av\cdot7.48\frac{g}{ft^3}\cdot60\frac{sec}{min}$$ $v=\sqrt{2gh}$ (Torricelli's law) $$v=\sqrt{2\cdot32.2\frac{ft}{sec^2}\cdot(8'+10')}=34\frac{ft}{sec}$$ $$gpm(max)=aV\cdot7.48\frac{g}{ft^3}\cdot60\frac{sec}{min}=\pi in^2\cdot1\frac{ft^2}{144in^2}\cdot34\frac{ft}{sec}\cdot60\frac{sec}{min}\cdot7.48\frac{g}{ft^3}=332 gpm$$ Does this sound right? – Jason Mar 20 '17 at 23:47
• I have retracted my close vote since you have now shown some effort, and I think there may be a conceptual question or two hidden within your comments. – sammy gerbil Mar 21 '17 at 0:10

Nevertheless you could get a good enough estimate by assuming that the water surface in the cylindrical tank remains at the mid level and calculate the head $h$ from this. Then use Torricelli's Law to get the particle flow speed $v$ based on this value of $h$, but use the cross-section area $A$ of the pipe to calculate volume flow rate $Av$ at the outlet level.