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.