There must be a pretty simple first-approximation formula, assuming standard gravity and water density, an infinite reservoir behind, no ripples, no turbulence, just constant smooth flow. If the water level in the reservoir is X inches higher than the top of the straight-line-horizontal weir, how many gallons per second are flowing over each foot of the weir’s length?
You are missing a couple of dimensions inc. velocity.
X inches above the weir by foot of weir describes an area i.e. square inches.
You are wanting a rate of volume i.e. cubic inches/time.
But apply the Francis formula:
Q = 2/3 C x L sqrt 2g x H^ 3/2
Q: Flow rate (in m³/s)
C: Discharge coefficient (typically between 0.60 and 0.65 for sharp-crested)
L: Width or length of the weir crest (in m)
g: Acceleration due to gravity (9.81 m/s²)
H: Height (head) of the water above the weir crest (in m)
Thanks for the link. So, if I picked the right graph, looks like 10 gal/sec per foot if the flow is 7 inches deep.
No, his dimensions are valid. The whole weir has a rate measured in volume per time (such as gallons per second). But of course it depends on how wide the weir is, so you need volume per time per width (such as gal/s per foot).
And when I was working it in my head, I got that same formula, so I’m glad to see my fluid dynamics isn’t as rusty as I thought.
I must be misreading something. Seems like the graph is saying 600 gal/min per foot of weir if the water is 7 inches deep, but that would mean the water’s flowing at 2.3 feet/sec, which has to be wrong?
It wouldn’t all be flowing at the same speed, but I get 1.87 m/s at the bottom edge, so that seems reasonable.
Do you think that number is too high or too low?
Does the flow rate really not depend on the liquid density or viscosity? So, water, honey, mercury, lava, it’s all the same?
Or, are the liquid parameters baked into the constant?
(I love the scientists who put two-thirds of a proportionality constant in an equation. Probably from integrating to get the height to three-halves power.)
Density cancels out. A more-dense fluid would need more force to push it out, but it would have that more force from the greater weight of the fluid above it.
Viscosity and other non-ideal-fluid parameters could in principle be relevant, but water is low enough viscosity that I suspect that for the depths we’re looking at here, it’s negligible.
The value of the whole proportionality constant is \frac{2\sqrt{2g}}{3}. That’s the value regardless of how you derive it (though yes, the easiest way involves an integral). If they put in any other value, it’d be wrong.
10 gallons is a box 1 foot by 7 inches, 2.3 feet long, so I read that last graph to be saying if the water is 7 inches deep over a weir that’s 1 ft long, it’s flowing at a gentle 2.3 ft/sec, average. If the water level is 7 inches higher than the top of the weir, it would actually be a torrent.
Oh, wait, I missed the C in the equation. That’d be, basically, a distillation into a single number of how non-ideal the fluid is. An ideal fluid would have that at exactly 1.