Bernoulli three reservoir with unknown elevation

[Edmund68]

Hi,

I have a tank with 2 compartments which are isolated from one another. Each compartment have a discharge outlet pipe (Q1, Q2) which combined together into a single discharge pipe (Q3) somewhere downstream.

The friction factor, length and diameter of the piping are known.

With the total flow (Q3) from the 2 compartments given, I would like to know if it is possible to determine the elevation of the water in each of the compartment. Q1 and Q2 is unknown.

I understand the standard solution for three reservoir problem is an iterative process which begins with guessing the total head at the junction where all the pipe meets, and then with known elevation calculate the flow of each pipe. However over here, the elevation is unknown.

Appreciate if you can share your thoughts. Thank you!

 

[quark]

When two tanks or isolated compartments are discharging in to a common header, without external force, the level should get balanced.if the tanks are not underground, I am missing something.

 

[LittleInch]

Is this a steady state analysis or transient?

How is the water level maintained or are you staying at the same level until it runs out?

I think there are a few too many variables and you need to start with some sort of fixed item, be it level of one tank, flow or pressure. Up to you which one you fix.

If it's transient you don't stand a chance.

Remember - More details = better answers
Also: If you get a response it's polite to respond to it.

 

[GeoEnvGuy]

I assume this is a school question. The answer is q1+q2=q3

Pipes of known diameter, length, and friction.

Solve for the flows in the pipe, also stated it is a tank with 2 isolated compartments so they are at the same elevation.

 

[Edmund68]

Yes I am assuming both compartments to be the same level now to solve the problem. Total head HT available at the common header can be easily solved. By assuming Z2=Z3 and hence Hf2=Hf3, I can do trail and error to find the v2 and v3 which will give me Z2 and Z3 eventually. But is this method of assuming both compartments to be same level correct?

I know by connecting both compartments to a common header, the level will equalised if the there is no flow. But I would like to know the level if flow is present.

This is a steady state problem. The water from both compartment is downstream of a weir, hence there is no control of level in the compartment. However, I can keep the flow constant via pump upstream of the weir.

The ultimate aim is to find the maximum flow allowed before the weir becomes submerged.

 

[quark]

In the absence of unequal and continuous infeed to both the compartments, if Z2>Z3, then Z2 compartment path becomes least resistant and will flow more till levels equals. Since, piping length from Z2 is longer, Z2 can be slightly higher but can be ignored. The head drop for both the tanks till common point in the header should be, ideally, equal.

 

[hydrae]

This will require trail and error until your error is small enough
tricky with the Darcy-Wabash (sp) equations

I would put this problem into a hydraulic modeling software (such as EPAnet which is free.)
you have h1, h2, h3, d1, d2, d3, L1, L2, L3, f1, f2, f3
also known q3 = q2+q1

you need q1, q2, q3 which will give you V1, V2, V3

f1, f2 and f3 need to be converted to C1, C2, C3 values for the Hazen Williams equation

Note this solution method is only valid for water with Reynolds numbers greater than 4000

Hydrae

 

[LittleInch]

I'm still not getting this or what item is fixed or variable.

"The water from both compartment is downstream of a weir, hence there is no control of level in the compartment. However, I can keep the flow constant via pump upstream of the weir."

If you can't control the level how can you control the flow?

Where is this pump?

If you sketch out the WHOLE system we might be able to figure something out...


Remember - More details = better answers
Also: If you get a response it's polite to respond to it.

 

[Edmund68]

Hello all, thank you for your replies.

As mentioned above, friction factor, length and diameter of the piping are known. v1 is known as well.

I want to solve for v2,v3,z2,z3.

If z2 is not equal to z3, I don't think I can use trail and error method. But since they are connected to a common header, will z2 and z3 going to equalised eventually?