Trough is 1.87 meters long and 0.5 meter wide and 0.22 meter deep. Water only flows over the 1.87 meter long side and from both sides (so x2). The water overflow is 0.012 meters tall (ruler positioned on edge of trough and water hits 1.2 cm up ruler, this is similar to depth of a waterfall at the ledge). It is consistent across trough length.

What is the flow rate of water over the edge?

I modeled water molecules like balls thrown upward and being decelerated by gravity.

$$\frac{2 \cdot 0.012}{t^2} = 9.81$$

Assuming final velocity of rising water molecule is zero, then initial velocity of 0.485 m/s can be calculated (as it leaves trough edge traveling upward). [sqrt (2*gravity*height)]

The average is therefore half of that at 0.243 m/s, since we are averaging with zero.

The cross sectional area over which water flows is 1.87 m * 0.012 m and multiplying that by 0.243 m/s gives 0.00545 m^3/s which converts to 86.4 gpm. There are two sides of the trough so multiply by 2 to get 172.8 gpm.

I derived this naively based on physics knowledge, so please tell me where I'm wrong regarding a simple model of the system.

  • 1
    $\begingroup$ Have you read the wiki page and followed the links? - en.wikipedia.org/wiki/Weir#Flow_measurement $\endgroup$
    – KalleMP
    Jul 20 '18 at 21:43
  • 1
    $\begingroup$ I found my solution. My work using acceleration is equivalent to sqrt (2*g*h) as it also gets you to velocity and can be derived from Bernoulli's equation. What I was missing was that I needed to integrate this over the vertical height, as pressure varies. Integrating in regards to h results in 2/3 * sqrt(2*g)*h^(3/2). This is the standard equation used in basic engineering models. Unfortunately, engineering texts rarely explain things from basic principles. My initial insight was gained by considering many thin slices of the weir space and modeling that in excel (a rough integration). $\endgroup$
    – SMy
    Jul 24 '18 at 12:28
  • $\begingroup$ related: engineering.stackexchange.com/questions/8001/… $\endgroup$
    – mart
    Jul 25 '18 at 13:46

At each location along the vertical distance $h$, the velocity is given by $\sqrt {2gh}$, where $g$ is the acceleration due to gravity, and this velocity of flow varies as $h$ changes along the vertical. So, you must integrate over the vertical distance since the relationship is nonlinear and you can't simply take the flow rate at the center. The first integral of the equation above is $\dfrac{2}{3}\sqrt{2g}\cdot h^{3/2}$. I could find the equation but not the derivation online (surprisingly), so this is what I did on my own.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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