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:

Fluid image

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.

  • $\begingroup$ If the hole is small enough for the pressure on it can be considered constant ($ d \ll h$), then the pressure is simply the fuild density times $h$ times gravity. $\endgroup$
    – rodrigo
    Commented Jul 24, 2014 at 19:29
  • $\begingroup$ Ah, unfortunately that's not the case. d varies and can almost be equal to h in some cases. $\endgroup$
    – Eric
    Commented Jul 24, 2014 at 19:34
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    $\begingroup$ I disagree with the close-votes on this question. This is an example of "what tools/physics do I need to solve this problem?", which we all agreed was on-topic in the meta thread $\endgroup$
    – Jim
    Commented Jul 24, 2014 at 19:42
  • $\begingroup$ Is the tank large enough such that $h$ stays constant? (not necessarily that $d\ll h$, it can be a very wide tank) $\endgroup$ Commented Jul 24, 2014 at 19:46
  • $\begingroup$ The tank is actually a big lake, and this is the water gate at the power plant. I need to very precisely calculate the water flow rate depending on how much the hatch have opened (the distance $d$) $\endgroup$
    – Eric
    Commented Jul 24, 2014 at 19:48

4 Answers 4


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.

  • $\begingroup$ Please excuse me since I haven't done calculus in 4 years. I have some questions: $v(x)=\sqrt{2gx}$ which I integrate as you described, right? But then $Q(h)$ as the result, what do you mean by calculating $h(t)$? Thank you in advance. $\endgroup$
    – Eric
    Commented Jul 24, 2014 at 23:27
  • $\begingroup$ @Eric Once you obtain Q(h), if h is not constant in time, you need to derive h as a function of time. So, if the lake has initial width, length and height, corresponding to an initial volume V, that volume will decrease at a rate of Q. So you can calculate, just from geometry, the change in height as a function of Q, which is itself a function of height. It's a self-consistent solution. $\endgroup$ Commented Jul 24, 2014 at 23:58
  • $\begingroup$ Ah right, but since the lake is quite large I can assume h is constant all time then. I have integrated the formula, which turns out much like rodrigo's solution; but it gives me wrong values (I have a diagram with approximate values). Am I missing out someting? I can give you the solution if you want. Thank you so much for your help, really! $\endgroup$
    – Eric
    Commented Jul 25, 2014 at 0:06
  • $\begingroup$ @Eric you should post these details in your original question. Also, I'm neglecting viscosity - if that is an assumption you're allowed to make on your homework assignment then these equations (Rodrigos) should get you in the right ballpark. $\endgroup$ Commented Jul 25, 2014 at 0:37

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.
  • $\begingroup$ At step 1, you have $\Delta{P}$, but I don't have that (as stated in the question), or did I misunderstand your initial step? $\endgroup$
    – Eric
    Commented Jul 24, 2014 at 21:05
  • $\begingroup$ Actually $\Delta P = \rho g h$ so $h_f = h$ actually. This is the driver (motive force) of the motion. What concerns me is the $L$ which the the length of the orifice. I am not sure it is defined here. $\endgroup$ Commented Jul 24, 2014 at 21:11
  • $\begingroup$ By length, do you mean length "into the paper"? Asuming the hole is square, what is $D$ then? The diameter I guess, but how does that apply to a square hole? Also, what is the initial $Re$? $\endgroup$
    – Eric
    Commented Jul 24, 2014 at 21:15
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    $\begingroup$ @Eric L is the length of your pipe - this applies only if theres a pipe attached to the orifice (not into the page). L is along the direction of flow. The $Re$ is the reynolds number, which takes into account viscosity. This captures the same physics as the $c$ coefficient in Rodrigo's equations. $\endgroup$ Commented Jul 25, 2014 at 0:47
  • $\begingroup$ Ah right, got you. I'm feeling a bit stupid no;, but in the first iteration how do I calculate the friction constant $f$ if I don't know the $Re$ value until step 5)? There must be an initial Re value, no? $\endgroup$
    – Eric
    Commented Jul 25, 2014 at 0:53

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.


From a Wolfram article we get the simplified Bernoulli equation:

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


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

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    $\begingroup$ Try to be mindful of the OP's request: 'I'm not saying "please give me the solution, I'm lazy'. I want to figure it out myself." Let's aim to please... $\endgroup$ Commented Jul 24, 2014 at 20:06
  • $\begingroup$ @user3814483: Ok... I'm not so sure my equations are correct, or even appropriate to the problem... just developing the idea... $\endgroup$
    – rodrigo
    Commented Jul 24, 2014 at 20:13
  • $\begingroup$ No this is actually fine, it's more complicated than I thought, thank you. I tried it but it gave me not quite the right numbers, but I'll play with it a bit. Thank you very much @rodrigo $\endgroup$
    – Eric
    Commented Jul 24, 2014 at 20:16
  • $\begingroup$ @Eric: Actually, I would expect the real value to be quite lower than my predictions. One, because real $c$ would be lower; Two, because the water coming out from the top of the hole will obstruct the flow from the botton of the hole... Unless there is a pipe, of course, but then things are even more complicated. $\endgroup$
    – rodrigo
    Commented Jul 24, 2014 at 20:26
  • $\begingroup$ Ah got you. Thank you very much for your help! I have a given flow coefficient actually, but values are still too high. I guess that is related to the 2) you mentioned. $\endgroup$
    – Eric
    Commented Jul 25, 2014 at 0:13

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