Hydraulic flow rate

Given the height of the reservoir and the diameter of the pipe,
how do you calculate the flow rate of the water?

Flame not, lest ye be phlegmed.

Depends on how strong your pump is and what type it is.

<b>P75 @ 90 Mhz</b>... because I need the speed.

This is a question for a chem. engineer!

Basically, the answer is that it depends.

Given an initial height of water, the water will exit the pipe with some initial velocity Vo.

Without any further info, you need to make some assumptions. Assume no mechanical work is done on the fluid, and that friction losses on the walls of the resevoir and friction losses from the sudden expansion as the water exits the pipe are negligible.

Also assume that the pressure inside the tank is atmospheric, and that the water is incompressible.

Finally, assume that the water level at the top of the tank is not changing, at least as we calculate initial velocity. This is the weakest assumption, but if the pipe diameter is small compared to the resevoir we are probably OK.

Now: Get out your copy of Unit Operations of Chemical Engineering (McCabe, Smith and Harriott 2001) and look up the bernoulli equation (eqn 4.62)

Define the top of the water in the resevoir to be plane 'a' and the bottom of the exit pipe to be plane 'b'. Then:

pa/p +gZa + aVa^2/2 = pb/p +gzB +aVb^2/2 + hf

where:
pa is pressure at plane a
p is density of water (should be rho, but I don't want to type greek)
pb is pressure at plane b
hf is all friction losses
a is alpha, a velocity correction factor
g is gravity, about 9.8 m/s^2
Za is the height of plane a
Zb is the height of plane b
Va is the velocity at plane a
Vb is the velocity at plane b

Now: Pa = Pb = atmospheric pressure, so the terms cancel
We assume hf is 0, no friction losses
Va is zero (water level of tank not changing)
define Zb to be 0
Assume that a = 1 (this corresponds to turbulent flow. If the flow is laminar, the velocity correction term will be closer to 2, but this seems beyond the scope of the problem)

So: We are left with

gZa= = Vb^2 / 2

You know g to be 9.8 m/s^2
Plug in Za, the height of the water in the resevoir
multiply by 2
Take the square root to find Vb, the initial exit velocity of the water from the pipe.
Then: multiply the velocity by the cross sectional area of the pipe (pi * diameter^2 / 4) to get volumetric flow rate.

Now you have volumetric flow rate, which is what you were looking for!

Don't you wish <b>you</b> were a Chem. E?

<b>1.4 Ghz AMD T-Bird underclocked to 1 Ghz...just to be safe!</b>

Man your smart!



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Wow... I didn't learned that yet on school.

<b>P75 @ 90 Mhz</b>... because I need the speed.