Water level rise in a tank, continuous flow in, high exit

[Diokhan]

Homework Statement

: See below paragraphs[/B]

Homework Equations

: I'm not sure on which equations I need.[/B]

The Attempt at a Solution

: I'm so sorry, this really isn't my strong point. Using the figures below, the tank surface is 113.184 sq in. The side pipe has a cross section area of 3.14 sq in. The water height in the tank needs to be 42" which is 1.0668 meters and I don't know what to do next. Please see paragraphs below.[/B]

I was asked to post this here and not in the technical section, however this isn't homework or coarse work. I'm 37 years old and have been out of school for 20 years. I am engineering a series of gravity driven bioreactors for aquaculture and I need help calculating what height the exit pipes need to be based on the following situation:

There is cylinder column of water in a tank of a known diameter. Water enters from above at a known flowrate. Water exits out the bottom through a pipe of a known diameter, and this exit pipe routes up the side of the tank to a known height so that as water enters the tank, it stays at a known height.

For some example numbers to plug in - the tank is 12" diameter and 48" tall, water enters from above at 30 gpm, exits through a 2" pipe. Water needs to remain 6" from the top of the tank. At what height does the exit pipe outlet need to be?

I really don't know how to solve this, though I'm sure that Bernoulli's equation comes into play. I've been able to find many examples where water just exits out a hole of a known diameter at the bottom while the tank fills, but not where water enters from the top while exits out the bottom, routes through a pipe, then exits at a known height.

Please show all work. I would like to turn this into an excel spreadsheet so that I can play with different flowrates, tank diameters, tank heights, exit pipe diameters, and the height of water from the top of the tank. I will always be solving for the height of the exit pipe.

THANK YOU ALL!

 

[berkeman]

Welcome to the PF.

So the water exiting the 2" pipe just flows out into open air, like into the next tank into the next bioreactor's top? Or is there some backpressure greater than 1atm?

If into open air, then you are asking what gauge pressure does it take to drive 30gmp through a 2" pipe, which will tell you how far above the pipe exit height the water in the big tank will rise, right?

 

[Diokhan]

Thanks for the response. Assume no back pressure in the scenario. I am more asking on what height I need the exit to be on the pipe to keep the larger tank at a certain water level for a certain incoming flowrate.

I do know that the higher the flowrate, the more the water level in the tank will rise to reach its happy spot (equilibrium's not the term, or is it?) I also know that a smaller exit pipe would affect this as well. As well as a larger tank with everything else the same would be less affected. Of course the easy way out is to put the exit halfway up with a gate valve to control main tank height, but I know there is a math solution to have the above diagram work the way intended. The point of the configuration is so the tank will not drain below the exit height when flow is cut off. The tube upward the exit is a siphon break.

The closest thing I could find are examples that solve this:

(Answer: (a) 8Q²/π²gDo⁴, (b) at 12.5 cm/s)

I know I could take solutions and rearrange formulas to solve for different parts, however in my diagram in post 1, the exit is near the top side of the tank and I don't know how this would change what formulas I would use to solve.

 

[berkeman]

I am more asking on what height I need the exit to be on the pipe to keep the larger tank at a certain water level for a certain incoming flowrate.

My intuition says that you won't be able to get 30 gpm through a 2" exit pipe with such low gauge pressure, but I could be wrong. I'll let @Chestermiller and others give you a better and more quantitative response...

 

[haruspex]

Bernoulli's equation comes into play.

Right.
If we ignore drag and assume smooth flow, we only have to consider the two levels at atmospheric pressure.
$P+\frac 12\rho v^2+\rho gh$ is constant.
Since the ambient pressure is atmospheric at both points, we can drop the P term.
Since the volume flow rate is the same at both, the linear velocities are inversely proportional to the cross-sectional areas. If we write A for the tank area, a for the pipe area, and V_in for the input volume flow rate:
$2g\Delta h=( \frac{V_{in}}a)^2-( \frac{V_{in}}A)^2$
For the example numbers you quote, I get a head of 4cm.

You show your final horizontal pipe as being the same diameter as the ascending pipe. That could be a problem. Inevitably, the level in the ascending pipe would rise above the top of the horizontal pipe, reducing the head difference between the two points at atmospheric pressure. If possible, make the ascending pipe the narrower of the two.

 

[Chestermiller]

I looked at the frictional pressure drop in the vertical section of the 2" pipe, and it is negligible. I see that haruspex just posted, and I get roughly the same answer that he does. The centerline of the horizontal section of 2" pipe should be about 4.4 cm (roughly 2") below the water level in the 12" pipe. This will give roughly a discharge rate of 30 gpm.

 

[Nidum]

Sometimes useful to do a sensitivity analysis in problems like this where the actual requirement is for depth of liquid in a tank to stay reasonably constant .

Basically do any likely variations in the incoming flow rate cause large or small variations in the depth of water in the tank ?

 

[Diokhan]

Thank you for giving me a fish with 4.4cm! I really appreciate all of you taking the time to calc this.

Now please teach me to fish. Please show me your work so I can see how this was derived, step by step. I'd like to be able to calculate these figures also as well as trying different figures for different sizes of tank diameter, height, and outlet pipe diameter, and flowrates.

I tried and got .11, so I obviously did something wrong. Instead of trying to type out my work, I took a photo that I hope you're able to see alright:

 

[haruspex]

Thank you for giving me a fish with 4.4cm! I really appreciate all of you taking the time to calc this.

Now please teach me to fish. Please show me your work so I can see how this was derived, step by step. I'd like to be able to calculate these figures also as well as trying different figures for different sizes of tank diameter, height, and outlet pipe diameter, and flowrates.

I tried and got .11, so I obviously did something wrong. Instead of trying to type out my work, I took a photo that I hope you're able to see alright:
View attachment 209186

The V_in in my post #5 was the input volume flow rate, so the units would be m³/s.

 

[Diokhan]

Thank you, haruspex, that was my error. I am also now able to get 4.4cm

You show your final horizontal pipe as being the same diameter as the ascending pipe. That could be a problem. Inevitably, the level in the ascending pipe would rise above the top of the horizontal pipe, reducing the head difference between the two points at atmospheric pressure. If possible, make the ascending pipe the narrower of the two.

Do you mean like this with a narrower vent? : For my own knowledge, if using the same diameter, why a reduction in head difference be a problem if level in output pipe were to equalize with the level in the main tank as long as flow in = flow out?

Basically do any likely variations in the incoming flow rate cause large or small variations in the depth of water in the tank ?

Flow should be fairly consistent. Likely only differences would be a lower input flow set point to where tank level rise is less of a concern than it is at higher flow.

 

[haruspex]

Do you mean like this with a narrower vent?

No. I wrote that I would make the horizontal pipe leading away from that point wider than the vertical pipe leading up to it from the tank.

 

[Diokhan]

Got it. The exit point (2") will connect to a larger 3" pipe to direct it where it ties into the rest of the system. Thanks for your time and help on this. I was also able to make a spreadsheet to try different variables to see what had an effect on what. I am going to mark this as solved. Thanks again.