# Pipe Flow Calculation

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1. Dec 14, 2016

### George Steel

I used to have a slide chart that gave gpm through a level section of pipe (1/2" - 6") by measuring how long the exit stream was when it had dropped 4". Imagine a framing square with the blade on the pipe and the tongue hanging down in the stream. Slide the square out until the stream is hitting the 4" mark on the tongue - read length from end of pipe on the blade of the square. The slide chart would give the gpm based on various pipe sizes. Has anyone seen one of these charts? Mine was from 30 years ago. Maybe by a water pump company? Can anyone give the equation behind this method?

Thanks

2. Dec 14, 2016

### JBA

I have not seen this type of instrument but it is a unique and simple solution based upon the following:

The flow volume through a pipe Q = v (pipe flow velocity) x A ( pipe I.D. area), and by having selected the pipe size on the slide chart therefore inside diameter of the pipe it is then only necessary to determine the fluid flow velocity to determine the flow capacity.

Note: For convenience, this analysis will be done in the units of inches and seconds that can then easily be converted to gpm (gallons per min).

The time for the flow stream to drop the specified 4" is t = sqrt(2 x h / g), where g (the acceleration due to gravity) = 2.68 inches/second^2, and h = 4 inches:
therefore, t = sqrt(2 x 4 / 2.68) = 1.728 sec;

v (velocity of the stream through the pipe) = s (the horizontal distance from the end of the pipe) / t .

Now, all that is required is the measured horizontal stream distance s to solve the equation: Q (inches^3/sec) = s (in.) / 1.728 (sec) x A (in^2);

and, after units and in^3/gal conversion: Q (gpm) =.15 x s (in.) x A (in^2)

However, the A (area) used in the chart for a given pipe size may also contain an adjustment for flow friction; and, only a comparison of the formula result to the chart result will determine if that is true.

3. Dec 14, 2016

### jack action

The principle is based on the physics of trajectories, for which the basic equation is:
$$y = -\frac{g\sec^2\theta}{2v_0^2}+x\tan\theta$$
Where:
$y$ is the vertical location (m);
$x$ is the horizontal location (m);
$\theta$ is the angle of elevation;
$v_0$ is the initial velocity (m/s);
$g$ is the acceleration due to gravity (9.81 m/s²).

In your case, $\theta$ = 0 (so $\sec^2$0 = 1 and $\tan$0 = 0), $y$ = -4 in (=-0.1016 m; note the negative sign).

Isolating $v_0$:
$$v_0 = \sqrt{\frac{-g}{2y}}x = \sqrt{\frac{-9.81}{2 (-0.1016)}}x$$
$$v_0 = 6.95x$$
To get the volumetric pipe flow (m³/s), you multiply the initial flow velocity by the pipe area ($A =\frac{\pi}{4}d^2$; m²):
$$VF = v_0A = 6.95x\frac{\pi}{4}d^2 = 5.457d^2x$$
Using gpm for $VF$ and inches for $d$ and $x$, you get:
$$VF = 1.4174d^2x$$
So, for example, if it takes 6" for the flow to drop 4", the volumetric flow for a 1" pipe is:
$$VF = 1.4174(1)^2(6) = 8.5 gpm$$

I'm slow to type, so @JBA already answered, but my answer doesn't match his. I've been looking for the error and the problem is the 2.68 in his first equation. It should be 386.1, which is the correct acceleration due to gravity in in/s².

Last edited: Dec 15, 2016
4. Dec 15, 2016

### JBA

jack action, thank you. As soon as I saw your comment I realized I had divided when I should have multiplied. All of that work totally wasted by a stupid conversion error.

I have now rerun the calculation using the correct value for g = 386.16 in/sec^2 (based upon 32.18 ft/sec^2) and my answer is now equal to yours at VF = 1.4173 x (1)^2 x (6) = 8.504 gpm.

5. Dec 15, 2016

### George Steel

Thanks to you both.. Excellent description.