# Formula to describe flow over thin weir

I have thin weir, really just a sheet of metal with a smooth and straight upper edge. I want to know how much water will flow over this weir, when the water level on the higher side is a few cm above the edge.
I'm interested in the special case when the water on the lower side i below the level of the weir:

• What formula describes this flow?
• What are important caveats?
• Flow rate in an open system depends on the pressure behind it. You need to know a lot more about the supply system. Mar 17 '16 at 11:38
• Is the water on the left side a static amount or is it constantly being replenished? That is the main parameter that is lacking.
– hazzey
Mar 17 '16 at 16:57
• The size/strength of the weir is irrelevant so long as it doesn't collapse. The flow rate, for an infinite reservoir, depends only on the drag due to atmosphere and the weir's edge -- and friction within the reservoir which restricts the speed at which the water higher than the edge can move across the remainder of the water Mar 17 '16 at 17:49

According to Engineering Toolbox, the flow rate over a weir is given by:

$$Q=\frac{2}{3}C_d b \sqrt{2g}\, h^\frac{3}{2}$$

where $Q$ is the flow rate, $C_d$ is an empirical discharge coefficient, $b$ is the width of the weir, $g$ is gravitational acceleration, and $h$ is the head of fluid above the weir. That same page has a few other formulas for various weir shapes. The wild card here is certainly the discharge coefficient; it will depend on a lot of parameters. This page has a chart for determining the coefficient based on several dimensions of the weir and surrounding channel.

Carlton is mistaken. Look at the exponent carefully. It should be 3/2, and not 2/3, this will obviously make a huge difference.

In another answer, I give an explanation for why this equation is the case:

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 neglect the C term which is experimentally derived.