Vertical Water Flow

Alright, I need to know how to calculate the flow rate of water that is flowing from a 3-m high resovior through a 3-cm diameter hole on the bottom surface through a rounded pipe.

I have looked throughly through the chapter in my thermal-fuid sciences book, and know that the K[sub]L[/sub] value for a rounded pipe is 0.03, and that K[sub]L[/sub] = h[sub]L[/sub]/(V[sup]2[/sup]/2g)

This equation would be ideal to use, since V is the average flow rate. However, I do not know the head loss, h[sub]L[/sub]. I have another equation for determining the head loss, but that relies on more properties I don’t have, such as pressure loss, and the flow rate itself.

I cannot seem to find any way to get the flow rate knowing only the area, height, and loss coefficient. Any help would be appreciated.

How about this:

You express your minor head loss h[sub]lm[/sub] with the equation h[sub]lm[/sub]=K((V[sub]avg[/sub][sup]2[/sup])/2).

Using fliud statics you can calculate the hydrostatic force on an area with the diameter of your pipe at the proper distance with

F=Integral p dA, dp/dh=rho*g, and F=mass flow rate *V.

So pressure at the bottom surface of the reservoir is p = rhogh = 999kg/m[sup]3[/sup]*9.81m/s[sup]2[/sup]*3m = 29400.57Pa

Your area A is 7.068 X 10[sup]-6[/sup]m[sup]2[/sup], so F = p*A = .208 N

Since mass flow rate is equal to rho*V[sub]avg[/sub]A, F is also equal to rhoV[sub]avg[/sub][sup]2[/sup]*A. This gives you a mean velocity entering the pipe of V[sub]avg[/sub] = sqrt(p/rho) = 5.42m/s.

And then you use this with your head loss equation to calculate the head loss for the pipe, and thus loss in pressure and the flow rate after entering the pipe.

Or anyway, I’m pretty sure this is right.

No, that won’t work, because I don’t know how deep the resovoir is. It’s 3m off the ground, not 3m deep.

So your pipe is three meters long?

Because if you know the length of your pipe and the vertical displacement you can calculate the relative change in pressure and use that for your calculation of (major) head loss in the pipe.

The minor loss at the pipe’s entrance, I should think, can also be calculated with relative changes in pressure.

Flow out of an orifice in a tank of constant cross section is:

Q = c * A * square root of (2 * g * h *)

This little h is the height difference between the orifice and the water’s surface (static pressure).

I don’t think you can accurately predict the outflow without knowing the depth of the water in the tank.
equation from “Fluid Mechanics and Hydraulics”, Giles

The problem here, Firx, is that you haven’t stated your problem clearly enough. The equation you’re looking for is:

p[sub]1[/sub]/ñ + Ü[sub]1[/sub]V[sub]1[/sub][sup]2[/sup]/2 + gz[sub]1[/sub] = h[sub]lT[/sub] + p[sub]2[/sub]/ñ + Ü p[sub]2[/sub]V[sub]2[/sub] [sup]2[/sup]/2 + gz [sub]2[/sub]

What we need to know is:

  1. the elevation of the water surface and the outlet
  2. the opening type in the bottom of the tank (you need to account for losses here as well)
  3. the size of the reservoir

Since you haven’t specified this information, I’m going to make some assumptions.

If we’re discharging to the atmosphere, we can make p[sub]1[/sub] and p[sub]2[/sub] go away by assuming they’re equal. If the surface area of the reservoir is very large compared to the 3 cm pipe, we can assume V[sub]1[/sub] is 0. We set our datum at the outlet of the 3 cm pipe, so z[sub]2[/sub] is 0. Our equation now reduces to:

gz[sub]1[/sub] = h[sub]lT[/sub] + Ü[sub]2[/sub]V[sub]2[/sub][sup]2[/sup]/2

The term Ü[sub]2[/sub] is the kinetic energy coefficient. Since there’s a lot I don’t know about your pipe, I’m going to assume that it equals 1. I’m also going to assume a square edged inlet with a value of k=0.5. With the K value for pipe losses that you stated above, we have:

gz[sub]1[/sub] = 0.5V[sub]2[/sub][sup]2[/sup]/2 + 0.03V[sub]2[/sub]2/2 + V[sub]2[/sub][sup]2[/sup]/2

gz[sub]1[/sub] = 0.765V[sub]2[/sub][sup]2[/sup]

V[sub]2[/sub] = (1.307gz[sub]1[/sub])[sup]0.5[/sup]

If you know the height, z[sub]1[/sub], you can then solve.

Any more precision than that, we need to more about your situation.

–Patch, who’s wondering how many formatting errors he made typing this

My mistake. bouv started this thread, not Firx.


I would love to give you those information, but that’s not supplied in the problem. All I’m given is that it is water, the opening at the bottom of the tank is 3m above the surface of the earth (not, that’s not the depth of the resovoir, just how high up it is. I am also not given the size of the resovoir in any way, shape or form. I do know the opening type, as I said earlier it is a ewll-rounded pipe, that has a K[sub]L[/sub] value of 0.03. Since I had to hand this in today, this is what I did:

I used the following equation,

P[sub]1[/sub]/ñg + V[sub]1[/sub][sup]2[/sup]/2g + z[sub]1[/sub] = P[sub]2[/sub]/ñg + V[sub]2[/sub][sup]2[/sup]/2g + z[sub]2[/sub] + h[sub]L[/sub]

Where ñ is the density (it’s suppossed to be a rho, but to me it looks like an n with a squigly line over it.) I assumed P[sub]1[/sub] and P[sub]2[/sub] were open to the atmosphere and equal, so those terms canel out, and that V[sub]1[/sub] = V[sub]2[/sub], so they cancel out as well. Z[sub]1[/sub] was 3m, and z[sub]2[/sub] was 0m. The trouble was with h[sub]L[/sub]. I could find the minor head loss in terms of V, which would then be great and I could find V, but I couldn’t incorperate the major loss at all, since it relies on f, which relies on the reynolds number, which relies on V itself, and other factors that are nmot known. So, since the height was only 3m, I assumed that the major head loss was neglible, and only minor head loss occured, which resulted in the equation:

h[sub]L[/sub] = K[sub]L[/sub]*(V[sup]2[/sup]/2g)

This results in h[sub]L[/sub] = 0.052V[sup]2[/sup]m. I then set that equal to my height of 3m, and solve for V. I got V = ~7.75 m/s.

Ah. Homework. :slight_smile:

If they don’t state information about the reservoir, then it’s safe to assume that it’s damn big (V[sub]1[/sub] = 0). You can also assume that the height is 3m from inlet to outlet.

Not having the problem in front of me doesn’t help, but some thoughts:

  1. You’re correct about P[sub]1[/sub] and P[sub]2[/sub]

  2. You’re incorrect about V[sub]1[/sub] = V[sub]2[/sub]. You don’t have the necessary information to make such a judgement, and given that you are flowing from a tank to a pipe makes this extremely unlikely.

If A[sub]1[/sub] is very very large compared to A[sub]2[/sub], then it is safe to assume that V[sub]1[/sub] << V[sub]2[/sub], and then V[sub]1[/sub] may be considered equal to zero. V[sub]1[/sub] then drops out of the equation. The reverse is also true. However, you can’t drop them both out in this problem.


K values are available for minor losses - their availability depends upon what sources you have. My old fluids book has a listing of Ks for a variety of situations, and that’s where I got the K=0.5 value in my prior post.

If no other information was given to make judgements about minor losses, then it’s fair to make the asusmption you did.

Hmm… you might want to check your math here. If K[sub]L[/sub]=0.03, then K[sub]L[/sub]/(2g) => 0.03/(29.81) =>0.0015. As we’re about to see, though, you did get the correct final equation. I suspect that the V[sub]2[/sub][sup]2[/sup]/2g crept back into your equation even though by your method it dropped out earlier.

The correct equation for you at this point should be:

z[sub]1[/sub] = V[sub]2[/sub][sup]2[/sup]/2g + K[sub]L[/sub]*(V[sup]2[/sup]/2g)

Substituting we have:

3m = V[sub]2[/sub][sup]2[/sup]/(29.81m/s[sup]2[/sup]) + 0.03V[sub]2[/sub][sup]2[/sup]/(2*9.81m/s[sup]2[/sup])

3m = 0.052s[sup]2[/sup]/m*V[sub]2[/sub][sup]2[/sup]

or V[sub]2[/sub] = 7.55 m/s

Since we were looking at the same formula going into the answer, I suspect that your 7.75 was a typo. You finally got the right answer, but your assumptions were incorrect. I hope you learned something from us that helps you in the future. :slight_smile:


My K[sub]L[/sub] was 1.03, because my book states that when water exists, there is a K[sub]L[/sub] of 1. But seeing as I already turned in this problem, it’s a mot point. Well, at least until the test rolls around…But hey, I got an A on the last test, and for a class on fluid dynamics and thermodynamics, reputed to be one of thr bitchiest classes in the engineering curriculem, I think that’s pretty good.

Any chance you misread the problem? I can see not telling how high the tank is off the ground, but I can’t see not telling you the height of water above the outlet.