Calculate water flow rate through orifice

Question

I'm not very good with fluid physics, and need some help. Imagine the following setup with water contained in-front of a wall with an opening on the bottom:

How do I calculate the water flow $Q$?. I have made some re-search and found I need to (partially) calculate the pressure across the opening (orifice). But I don't know the pressure on the back side of the orifice. Can this be solved in any way?

Note: I'm not saying "please give me the solution, I'm lazy". I want to figure it out myself. But since, in this case, I only found formulas involving calculating pressure drop, I canno't use them to solve the problem. Therefore I'm turning my face to you, to see if there's another way to solve this problem.

Update: The "tank" holding the water is actually a big lake, and the opening is how much the water gate have opened. I need to very precisely calculate how much water flows through the opening.

Answer 1

First assume that $h$ doesn't change very much because you have a large body of water (we can relax this condition later). Let's also assume that the hole is small compared to the depth ($d \ll h$) - we'll relax this too. For this case, the answer is straightforward, you'd use Bernoulli's equations and simply set the static pressure ($\rho g h$) equal to the dynamic pressure ($\frac{1}{2}\rho v^2$). Then you'd pull out $v$ and multiply it by the area $A$ of the hole to get $Q$, since $Q$ is the volumetric flow rate.

Now, let's relax the condition that $d \ll h$. Since the pressure at the hole varies with depth, the velocity will vary too. You can treat this like a calculus problem where you calculate the incremental change in velocity as a function of height. To calculate $Q$, you'd need to integrate $w \int v(x) \,\mathrm{d}x$ for $x = 0$ to $x = d$. Note $w$ would be the width of your hole into the page (assuming a square hole).

Once you obtain the expression above ($Q$ as a function of $h$), you could then relax the condition that $h$ be constant by noting that $h$ will depend on the volumetric flow rate and the geometry of the lake. Once you have $Q(h)$ from the previous step you can use that to calculate $h(t)$ and back-substitute that into your equation from the previous step.

Answer 2

I have used the Darcy Formula together with the following formulas for a quick numeric solution (only a few iterations needed)

1. $$h_f = \frac{\Delta P}{\rho g}$$
2. $$ f = {\rm Darcy}(Re)$$
3. $$ h_f = f\,\frac{L}{D}\,\left( \frac{v^2}{2 g} \right) $$
4. Solve above for $v$
5. $$ Re = \frac{\rho D\,v}{\mu} $$
6. Go to step 2 until $f$ converges to a value.

Answer 3

The NCEES: FE Reference Handbook has some good material on fluid flow through a submerged orifice in its fluid mechanics section. You can search for it online. NCEES will provide you with one free of charge.

Answer 4

From a Wolfram article we get the simplified Bernoulli equation:

$$Q = a c \sqrt{2 g h}$$

Where

• $Q$: the flow rate ($\mathrm{m^3/s}$)
• $a$: the area of the hole ($\mathrm{m^2}$)
• $c$: flow coefficient (dimensionless)
• $g$: the gravity acceleration ($\mathrm{m/s^2}$)
• $h$: the depth of the hole ($\mathrm{m}$)

That is valid for a small enough hole, but since your hole can be big, we have to use integral calculus. Moreover, I think that the flow coefficient can be set as 1 for a big hole. And the area of the hole can be calculated as the width of the hole times the height (assuming a square hole).

So $$\begin{align}\renewcommand{\intd}{\,\mathrm{d}} Q &= \int_{h-d}^h \sqrt{2 g y}\,w \intd y \\ &= \int_{h-d}^h w \sqrt{2 g} \sqrt{\vphantom{2}y} \intd y \\ &= w \sqrt{2 g} \int_{h-d}^h \sqrt{\vphantom{2}y} \intd y \\ &= w \sqrt{2 g} \left[\frac{2}{3} \sqrt{y^3}\right]_{h-d}^h \\ &= \frac{2}{3} w \sqrt{2 g} \left[\sqrt{y^3}\right]_{h-d}^h \\ &= \frac{2}{3} w \sqrt{2 g} \left(\sqrt{h^3} - \sqrt{(h-d)^3}\right) \end{align}$$