13
$\begingroup$

I am a newbie in water system design but I am currently faced with the exact situation below on my land, and I need to know whether gravity alone is sufficient in order to fill Tank 2 from Tank 1, as I already experienced backflow.

Please have a look at this image:

enter image description here

Note: let's not worry about how Tank 1 is being filled as it probably does not matter - I just made sure I never used a pipe diameter smaller than 0.5" between Tank 1 and Tank 2.

As you can see:

  • My water source is the overflow of Tank 1

  • I have put a 1" diameter pipe in the first 'downhill' section of my path

  • The second, longest section of the path is a pipe of 0.5" diameter

  • The system is 'powered' through gravity alone

Questions

  • Is there any missing variable (eg path length?) on my drawing in order to resolve this system? (What are the important variables?)

  • The flow at the source (the overflow of Tank 1) can be nearly 0. How does this variable influence the system? (Can a larger flow at the source help going up the 'uphill' sections between the tanks?)

  • Is the 'head' variable important at all?

  • I used a bigger pipe for the first section in an attempt to 'make more weight' and create sufficient pressure to go up the small uphill that follows. Should I limit the use of that bigger pipe strictly to the downhill section, and does it actually make sense to have used a bigger pipe at all?

$\endgroup$
  • 2
    $\begingroup$ Your "Head" variable is certainly the most important. You have to take it the greater possible. I think it would be preferable to avoid ascending sections. After all, old roman aqueduct were all build with a little, but always negative, slope. $\endgroup$ – Trimok Jun 30 '13 at 17:37
  • $\begingroup$ Yes, in fact my current understanding about the head variable is: "Since the flow and pressure are nearly 0 at the source - the source being the overflow of Tank 1 - the fluid dynamics is such that the downhill pipe must first fill up with water before the pressure is sufficient to push the water in the following uphill sections. So in this specific situation the "head" (ie the pressure build-up) only exists conditionally. Correct? $\endgroup$ – John Doisneau Jun 30 '13 at 22:30
  • $\begingroup$ One more note: unfortunately I cannot really play with the head variable. It is already set, and I know that it is quite small even though I did not measure it. I know it is there though, because in one of my tests I could actually get water going out at Tank 2. $\endgroup$ – John Doisneau Jun 30 '13 at 22:33
  • 3
    $\begingroup$ If the input were guaranteed to be under water for all times, and if you started out the pipe without significant air pockets, then you'd have a working siphon, which would work even if parts of the pipe were above the input level by up to 10 meters. But perhaps you require more robustness. $\endgroup$ – user10851 Jul 2 '13 at 3:55
  • 1
    $\begingroup$ BTW, it is even possible to make water flow to a higher tank without using a pump, although only a fraction of it can be raised, the rest being dumped lower down. See homework problem 15 in ch. 11 of this book: lightandmatter.com/lm $\endgroup$ – Ben Crowell Jul 26 '13 at 13:56
8
$\begingroup$

You want to make sure your pipe is sized such that the flow, Q, out of tank 1 is greater than the flow into tank 1.

Or, at some level, you want to know the maximum flow, Q, that the system will allow. You can solve this kind of problem using the Bernoulli Equation for conservation of energy with the Darcy Weisbach equation to account for frictional losses equation:

$\frac{P_1}{\gamma}+z_1+\frac{v_1^2}{2 g}=\frac{P_2}{\gamma}+z_2+\frac{v_2^2}{2 g}+f\frac{L}{D}\frac{v^2}{2g}$

If we assume both tanks are under atmospheric pressure, the fluid in both tanks is not moving, and we neglect minor losses from the bends in the pipe and the change in pipe diameter then the equation reduces to:

$z_1-z_2=f\frac{L}{D}\frac{v^2}{2g}$

Where the term $z_1-z_2$ is what you're calling head. Actually, because both tanks are full, it does not matter if the pipes are at the top or botom of the tanks, As long as both pipes are submerged the change in head (from free surface to free surface) is what's important. so the variables we have are:

  • $head$: the difference in water surface elevations
  • $f$:a friction factor that depends on what the pipe is made of, and how fast the fluid is moving in it,
  • $L$: the pipe length,
  • $D$: the pipe diameter,
  • $v$: the fluid velocity,
  • $g$: is gravity, which is a constant

Your system is a little more complicated because we have two pipe diameters, so the equation should look like, $z_1-z_2=f_A\frac{L_A}{D_A}\frac{v^2}{2g}+f_B\frac{L_B}{D_B}\frac{v^2}{2g}$, although i'm going to focus on the concept here, so i'm going to stick with the simpler equation.

to find the friction factor, $f$, a moody diagram is typically used... although because you're having backflow at times, i'm going to assume your flow is low enough that we can estimate the friction factor as:

$f=\frac{64}{Re}=\frac{64 \nu}{v*D}$ where $\nu$ is the kinematic viscosity of water.

This makes the equation:

$z_1-z_2=\frac{64 \nu}{v*D}\frac{L}{D}\frac{v_A^2}{2g}$

solving for the flow rate $Q$ ($Q=V A$) gives:

$Q=\frac{\pi D^4 g (z_1-z_2)}{128 L \nu}$

So you can see that the maximum flow is strongly influenced by pipe diameter! The bigger the pipe, the more flow from tank 1 to tank 2 will be possible. Additionally, contracting from 1" pipe to 0.5" pipe is restricting the flow even more, do to energy losses at the connection. Shortening the length of pipe used will also help you, but not nearly as much as increasing the pipe diameter. Doubling the pipe diameter will give you 16 times as much flow! The change in elevation of the free-sufrace of both tanks (head) is for sure important, but if you can't change it, that's ok. Finally, the pipe material is an important factor at higher flow rates, but maybe not for your current situation at a low flow rate.

$\endgroup$
  • $\begingroup$ This was very thoughtful, thank you very much for these formulas applied to this particular situation! However like I mentioned above I believe we cannot consider that "both pipes are submerged" since my input is the overflow of Tank 1 and the flow is sometimes nearly 0 (and only part of the input pipe diameter is "under water"). Does the system then become too complicated to lay down its equations? $\endgroup$ – John Doisneau Jul 4 '13 at 22:36
  • 1
    $\begingroup$ @JohnDoisneau it is easiest to assume both are submerged, and is probably o.k. for your purposes. In reality the pipe could be operating in one of a couple states: (1) it is partially full the whole way (would be an open-channel flow problem instead of a pipe flow problem) (2) it is partially full part of the way and full the rest of the way (a combo problem). at any rate: Having the straightest shortest pipe line (with most constant slope possible), and largest feasible pipe diameter will maximize the flow between the two $\endgroup$ – mrsoltys Jul 7 '13 at 1:31
  • $\begingroup$ Thank you for the advice mrsoltys. I will accept your answer as it provided an in-depth understanding of my situation. $\endgroup$ – John Doisneau Jul 7 '13 at 23:16
2
$\begingroup$
  1. The main variables you're missing are those relating to energy losses to friction, such as the total surface area of the path. Why not use a straight pipe than a winding one?

  2. Of course the "head" is important, this is the potential difference driving the motion. If it were zero between Tank 1 and any point in the path, there would be no flow.

  3. Larger pipes are better.

$\endgroup$
  • $\begingroup$ Thank you dimension10. Both tanks must be full of water. I have continued reading about fluid dynamics applied to gravity-based water distribution systems and also found out that I probably experienced backflow because of air blocks in the several land ondulations leading to Tank 2. Since my head variable is small, to ensure my system works all the time, I may transform it into a pressure-based one, switching the water source point to the bottom of Tank 1, and using an automated valve at Tank 2! $\endgroup$ – John Doisneau Jun 30 '13 at 22:39
  • $\begingroup$ @JohnDoisneau: Instead of making it a pressure-based pump, why don't you increase the "head"? Also, you could but the ending of the pipe at Tank 2 at its base, to increase the value of the "head". $\endgroup$ – Abhimanyu Pallavi Sudhir Jul 1 '13 at 3:25
  • 2
    $\begingroup$ comment to point 2: head is the difference between the higher water surface and the lower one, provided there's a continous, filled pipe between them. It does not matter wether the outflow tank 1 is connectetd at the top or bottom, the head will be the same (as long as the pipe is filled) $\endgroup$ – mart Jul 2 '13 at 14:17
  • $\begingroup$ And given that the input is not "under water" how does it change the system's behaviour? $\endgroup$ – John Doisneau Jul 4 '13 at 22:32
  • $\begingroup$ No, I just mean the worst case would be to have only one drop of water "fill" the input every minute... $\endgroup$ – John Doisneau Jul 5 '13 at 9:15

protected by rob Aug 4 '16 at 16:21

Thank you for your interest in this question. Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site (the association bonus does not count).

Would you like to answer one of these unanswered questions instead?

Not the answer you're looking for? Browse other questions tagged or ask your own question.