3
$\begingroup$

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.

$\endgroup$
8
  • $\begingroup$ 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
    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
    Jul 24, 2014 at 19:34
  • 1
    $\begingroup$ 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
    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) $\endgroup$ 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
    Jul 24, 2014 at 19:48

4 Answers 4

3
$\begingroup$

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.

$\endgroup$
4
  • $\begingroup$ 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
    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. $\endgroup$ 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
    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. $\endgroup$ Jul 25, 2014 at 0:37
2
$\begingroup$

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.
$\endgroup$
6
  • $\begingroup$ 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
    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. $\endgroup$ 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
    Jul 24, 2014 at 21:15
  • 1
    $\begingroup$ @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. $\endgroup$ 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
    Jul 25, 2014 at 0:53
0
$\begingroup$

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.

$\endgroup$
0
$\begingroup$

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

$\endgroup$
5
  • 1
    $\begingroup$ 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... $\endgroup$ 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
    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
    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
    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
    Jul 25, 2014 at 0:13

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.