# Water flow rate through a slot at surface

## Main Question or Discussion Point

I've been searching all over in aquarium related forums for an answer to this questions and haven't been able to find a definitive answer.

Background:

Many saltwater reef aquariums utilize a filtration system located in a sub-tank (called a sump) beneath or remote of the display tank. Water drains from the display tank into the sump, goes through whatever filtration system is in place, and it pumped back into the display tank.

The water level in the display tank is usually maintained by an "overflow box" which is simply a box inside the tank that is sealed in place, and there are holes drilled and bulkheads in place so that the water that flows over the top of this box then finds it's way into the drain plumbing. So if the power is lost, the water level in the tank only drops to the lowest point at the top of the overflow box. If you google "reef ready aquarium" you will see exactly what I am referring to.

Most of these overflow towers have "teeth" along the top edge to prevent snails, fish, etc from escaping the tank and getting stuck in the plumbing system. I have been unable to find a reference of any type that definitively shows what flow rate can be expected through such teeth in various configurations.

So, here is my question:

What is the flow rate that can be achieved through a 1/4" wide slot?

Being an engineer myself, I understand that at different water levels, there will be different flow rates. I am interested to know what the variance is as well. So let's clarify the question:

I have a slot that is 1/4" wide and 2 inches tall. The bottom of the slot is square (generally they are rounded, but for simplicity, let's say they're squared off). The water level on one side of the slot is raised such that it is maintained at a constant level equal to 1/2" above the bottom of the slot. The water level on the other side of the slot is significantly lower (this is not relevant, because as long as the water level is not equal to or higher than the bottom of the slot, the water will exit the slot at the same rate).

What is the flow rate through this slot?

What is the flow rate if the water level is raised to 3/4"?

1"?

1.25"?

etc.

I'm sure there is an equation that covers this, that's what I'm looking for. I figure that there has to be some component of it that takes into account viscosity and turbulence or something. For instance, if the slot was 1/8" wide the flow rate would likely be less than half of the 1/4", and conversely, if the slot was 1/2" wide the flow rate would likely be greater than doubled.

So if anyone can help me out or at least point me in the right direction, it would be much appreciated!

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Well, I am no fluid dynamics kind of guy, but I think that if you 'plug' the slot, first, and keep that water level...you should be able to calculate the profile of the pressure along the height of the slot...this is the pressure that would drive the water when you 'unplug' the slot...I presume that this pressure is simply due to the weight of the water itself and hence, it is zero at the top of the water and increases with depth.

Then, going through the slot may be consider an orifice, since there is no pipe before of after the slot...although given the symmetry of the problem (if I understood it correct) maybe it could be said that there is a pipe before but not one after the slot, in which case, the loss coefficient is one of an exit...which I think it is 1.0 velocity head.

You can calculate the hydraulic diameter of the slit for the various heights that you describe...hyd. dia. is 4 times the cross sectional area divided by the wet perimeter.

Then, you can try applying flow equations...except that your pressure is not uniform, so I am not how you can go about it; in reality, there will be stuff going on, like the water along the bottom of the slot moving faster than at the top and dragging more water at the top than warrants due to its pressure....and I am just babbling here...

hope this helps.

actually, I take the exit thing back...it is not an exit because the opening of the 'exit' is not the same cross section as the 'pipe' upstream. So, I think it needs to be treated as some kind of orifice...

I have consulted one of out MEs and he pointed me to a book that talks about flows through orifices and weirs, but it was really no help. The one formula that it had looked like it would work, then there was a not saying it was only valid if the L dimension was at least 2x the H dimension. Since I have L=0.25 and H=0.5 or more, that didn't work.

The weir calculation was for non-orifice flow, so that's no help either.

Really, there's got to be someone out there that can figure this one out.

I have talked to a couple of tank manufacturers and have been told generally how they figure flow, but still have not been able to verify anything they've told me. One of them said that if they build an overflow box with 8 slots, each 1/4" wide, and each 1" tall, then run them at 50% (1/2 of the way up the slot) that they get 400 gallons per hour. So by that, I would figure 50 GPH for a 1/4" wide slot with 1/2" of water running into it, but I have no way to verify this.

And I'm an engineer, so I want to know!!

There is a Handbook of Hydraulic Resistance that I have seen around the office....it is by Idelchik, you can find it in Amazon

Here's the closest thing I've been able to come up with

Q = 2/3 C L H sqrt(2 g H) (forgive me I don't know how to make it look pretty)

C=0.62, g=32.174 ft/s^2. Changing to inches makes g=386.1 in/sec^2

This reduces now to

Q = 11.49 L H^1.5 which is in^3/s

Converting units to GPH: multiply by 3600 s/hr and 7.48gal/ft^3 divide by 1728 ft^3/in^3

Q(gph) = 179.1 L H^1.5

Now plug in L=0.25, H=0.5, Q=15.83 GPH.

I got a figure from a tank builder of 40 GPH for a 1" tall 1/4" wide slot with the water at 1". So let's plug that number in and Q = 179.1 * 0.25 = 44.75. Looks like he's using the same formula...

So, how does my math look? Do I have it all right?

I have no idea...I am electrical engineer! Then again I have found myself doing fluid flow calculations here and there and that's when I go borrow Idelchik.

Fluid dynamics seems such an empirical field, though, who is to say whether an equation is correct or not? If it works, it works...

Then again, that formula seems simple enough...notice that the term sqrt(2gH) is nothing else than the instantaneous speed of an object that has been falling for a distance H...I guess it has something to do with the potential energy it started from, in the first place...

Well I'm an EE too. So we're in the same boat...

Do you mind explaining why you need to know? Are you designing one yourself?
www.guarriello.net/overflow.htm[/URL] might help.

It's a closed loop system, so I don't know why you'd need to worry about the flow through the slots. That overflow box should be filled with water as well, no? So you'll maintain suction as long as the tank fluid level remains at the slots. Am I missing something?

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You don't want the water to go over the top of the slots and for a rimless tank with an overflow box 1/4" below the rim if you want the water level not to exceed a certain level in relation to the top of the tank, nailing down the necessary depth of the slot to maintain that with a given flow rate is critical.

People sometimes run thousands of gallons per hour through these type of systems which can easily top an overflow.

I do a little acrylic fabrication and I made an overflow box the other day from someone had 42 3/4" tall teeth and he's running 1100 GPH through it on a set such as is described above (rimless, etc) and the water will go up to within about 1/2" of the top of the tank.

Also, this is not considered a closed-loop system in the frame of reference of the aquarium realm. A true closed-loop system has the intake well below the water line.

But the reason I need to know is so that if anyone else asks me to build one for them, I'll know exactly how many teeth to make and how deep for the flow rate throughput and maximum tank water level they desire.

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Fluid dynamics seems such an empirical field, though, who is to say whether an equation is correct or not? If it works, it works...
That is just because you haven't gotten into the nitty-gritty of fluid mechanics. It is most certainly not that way for everything.

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At any rate, you most certainly can find a good rough estimate of the flow rate through the slot for a given height. I will start by stating this is assuming an inviscid fluid (which can be corrected later). Even keeping it inviscid will give you a good rough idea.

The fluid below the slot doesn't matter for the flow rate, so we assume the coordinate system originates at the bottom of the slot, which we call $y=0$. The pressure at any given height when the fluid rises to $y=h$ is given by:

$$p = \rho g (h-y) + p_{atm}$$

Now, assuming inviscid flow, we can use Bernoulli's equation, which is (neglecting height terms since we are just looking at the flow through that constant-height area):

$$p_1 + \frac{1}{2} \rho v_1^2 = p_2 + \frac{1}{2} \rho v_2^2$$

Letting station 1 be on the water side of the slot and station 2 be on the empty side, we would be able to simplify to the following:

$$\rho g (h-y)+p_{atm} = p_{atm} + \frac{1}{2} \rho v_2^2$$

Rearranging:

$$v_2 = \sqrt{2g (h-y)}$$

Now, that give the velocity immediately at the exit of the slot at a given fluid height. You can then use that to find volumetric flow rate. The volumetric flow rate through an arbitrarily small area of the slot is:

$$dQ = v_2 \;dx \;dy$$

Integrating over a slot that is $W$ in width, we get:

$$Q = \int_0^h \int_{-\frac{W}{2}}^{\frac{W}{2}} v_2\;dx\;dy = W \int_0^h v_2\;dy$$

$$Q = W\sqrt{2g} \int_0^h \sqrt{(h-y)}\;dy$$

$$Q = W\sqrt{2g} \left[ -\frac{2}{3}(h-y)^{\frac{3}{2}} \right]_0^h$$

$$Q = \frac{2}{3}h^{\frac{3}{2}}W\sqrt{2g}$$

That will give you a pretty good estimate of your flow rate, just make sure your units are consistent. This is definitely going to be off by a small factor as a result of the fact that water most certainly is viscous, but it is a decent approximation.

One conclusion you can make is that neglecting the viscous effects, the flow rate depends linearly on the width, meaning twice the width gives you twice the flow rate. It depends on height of the water to the $\frac{3}{2}$ power, so doubling the height gives you about 2.82 times the flow rate.

So basically, that's the same equation that I posted up above

Q = 2/3 C L H sqrt(2 g H)
Except the C is missing from yours. I got that from a mechanical engineering handbook, and was stated as the equation was the theoretical discharge over a weir of length L. It went further to explain that the actual discharge resulted in the value of C=0.62 which is the discharge coefficient. This reduces the equation to 3.33 * L * H^1.5 when L and H are in feet and G is in ft/sec^2. I converted everything to inches and GPH and came up with Q = 179 L H^1.5 so except for the coefficient C, which probably accounts for viscosity of water, I think you just confirmed my equation!

The only other thing to be considered I believe is the issue of surface tension. Obviously if you take extreme cases such as a 1/8" slot compared to a 24" slot, the surface tension is going to get in the way much more the smaller the slot is.

The H^1.5 factor accounts completely for the higher flow factor of 2.82 (which I also came up with, glad to get that confirmed), and since 99.9% of the time all I care about is a 1/4" wide slot, I think I have my answer.

I also have a situation where I have a system (that is currently being built) that has a 21" wide end with a 16" wide and 1.5" deep notch cut out of it, leading to an external overflow box where all the drain plumbing is located. This is a true weir and was designed for a high-flow system. I had this system built so that it could handle 3000 GPH over the weir, and the manufacturer told me that the water would be about halfway up the 1.5" notch. Solving the arrived upon equation for Q=3000 with C=0.62 and L=16, H actually works out to just over 1", not 3/4". However I wonder if in this situation the equation does not take into account the absence of surface tension.

I wonder if this is the case, because if I remove the C=0.62 factor and use Q=3000, L=16 then H works out to exactly 0.75.

So maybe the answer here is that the viscosity effectively increases as the slot narrows. It just so happens that C equals approximately 0.62 for a 1/4" wide slot, and that is only (so far) verified with anecdotal evidence (which is reliable)

So the value of the discharge coefficient C is my only question at this point.

Gold Member
The effect of viscosity will definitely increase as the slot narrows, so that very well may be your culprit. It would lead to a slightly lower discharge coefficient and thus a greater height to maintain that flow rate.

I would imagine that the effect of surface tension would be quite minimal.

What about salinity? Marine aquariums typically run at a salinity between 29 and 35 ppt. I'm guessing that would have a relatively minor effect when compared to the issue of the slot width.

You are picking nits, I think. When you calculate the fluid levels and flows, you are getting an estimate that is disregarding viscous effects, the roughness of the edges of the cuts, the non-uniformity of the cut dimensions, the waves/surges in the water, etc. Your salinity isn't going to matter much for your calculations, I expect.

Gold Member
Salinity will never affect it... at all. It only affects density, which canceled out early on.

Gotcha. I had read in that ME text that hard corners vs rounded on the intake affected flow rate also. I think many miss this as almost all tanks I've seen with an overflow are cut with a router (CNC or by hand w/template). I would have to read up on how that would affect the flow rates, but again, such a change would drop the water level so insignificantly, so why bother...