Moving a Large Volume of Water

I don't know the answer, but how high is the water on the inlet?

The water would be 12' deep. The inlet would be submerged 6' or so under the water surface.

You want to know, "how much additional water will be moved?" by the 120 GPM inflow?

I don't suppose that the answer, 120 GPM additional water would be moved (every minute), is what you want.

If I understand the underlying question, this would require a a very complex set of calculations or computer simulations.

The rate-of-change-of-volume would be:

(120 gal /1 min ) x ( 1 Ft³ / 7.481 gal ) =16.04 Ft³ / min

For a rectangular ''Canal'' that is L Ft long, its' volume would be:

(12 Ft-deep) x (120 Ft-wide )x ( L Ft-long ) = 1,500 x L Ft²

Thus, the water-volume would increase by:

( 16.04 Ft3 / min) / ( 1,500 x L FT² ) = ( 0.0107/L ) vertical feet-per-min

=> a purely vertical water-level-change-rate of (0.1283/L) inches / min.,

or, slightly more than L times one-eighth-of-an-inch-per-min.,

in the absense of all other inputs.

Thank you for calculations.

Let me correct myself by re-writing part of the last sentence as:

''or, slightly more than ( 1 / L (feet)) times one-eighth-of-a-vertical-inch-per-min., ''

Once again, haste made waste, for which I do apologize to you.

I don't know the answer, but the more it rains, the more interesting these questions become.

Looking at it as an electrical engineer, the pump imparts velocity to the water which spreads out similar to a magnetic field, i.e., nonending flow lines. So my answer is "all the water in the canal gets some velocity from the pump."

I would have to agree...the water has a velocity of 1.02 inches per second at the pipe outlet....so the effect is minimal...

This situation is like using a mixing eductor. IIRC, the entrained flow can be about 4x the pressurized flow.

However, a 24" dia. pump outlet should give way more than 120 gpm. The description needs more detail.

As Tornado said, a 24" dia. pump should output way more than 120 GPM. Are you sure its not 120 GPS?

24" is roughly the diameter of a barrel, and 120 gallons is just over 2 barrels. That's a really slow pump!

The amount of additional water moved will depend highly on the size and shape of the surroundings of the pump. a simple straight tube in a 150' wide canal will be very inefficient.

Yes. One needs the pressure/flowrate performance curve of the pump. The curve needs to be overlaid on the performance curve of the system it is pumping through, which can be determined by measuring site conditions. Those two curves intersect at the Operating Point, which can then be used to evaluate the proposed arrangement to determine compatibility and suitability.

A little arithmetic will then yield answers to this particular question and a number of others.

All pump manufacturers publish performance curves as part of their technical data on pumps.

Perry, "The Chemical Engineer's Handbook", any edition, would be a useful reference work to have on the side of the desk, perhaps?

Correction on pump out put.... it was 24" outlet, 90 gallons /sec. Specs were phoned into me for different pumps that I don't understand anyway... What size or design of pump would it take to make even a slow current ( .5 mph > 1 mph) …. in a 150' w x 12' straight canal.... guesses are acceptable...I need a rough idea only...…….

Which sort of <...gallons...>, please, as there are several?

Assuming plug flow, 90 US gallons/sec into a canal 150ft by 12ft cross section would produce a velocity a little over ¹/₁₆in/sec, which is likely to be swamped by other effects.

...so in order to provide the required maximum velocity it would take nearly 19 of those pumps.

...or a pump that could deliver more volume of water? How big of pump in discharge diameter and/or Gallons per second would you think?

You haven't said so specifically, but I get the impression that your goal is to get the water in the canal flowing, or flowing faster. I'm no hydraulic engineer, but I suspect that the more efficient method would be to construct a weir, and pump the water over the weir. Gravity will then be working with you to keep the water flowing. If the water isn't flowing fast enough, then the canal either has too low a slope, or has an obstruction or silting downstream. Pumping the water over the weir will effectively increase the slope.

Now that can't be done if there is boat traffic in the canal, without making a lock.

In a canal that wide, pushing some of the water faster is just going to cause reverse flow at other points away from the pump, unless something is done to block that reverse flow.

Ahhh…. That means the water will move away from the pump , but may circle back on the other side negating the flow. Some one warned me of that at the outset. Hmmm this may not be a viable answer.

A weir is out because there will be boat traffic occasionally.

Well, with all said, I might have to forget about creating an artificial current at this point. The powers in charge are concerned about too big of units, too much diesel or electricity and too much expense on upkeep... They cant keep their Gumbo and eat it too!

thank you all for the comments...……...

<unsubscribes>

If I vaguely understand, you are considering using a submerged pump to increase the velocity of a body of moving water. This would mean a differential head across the pump of close to zero. A very expensive place on the curve to run a pump and the motor would have to be sized accordingly (big). The effect would be negligible.

If you were to try this, you would need an impeller with a high specific speed - similar to a ship's propeller. If you increase the velocity of the water downstream the pump you have to ensure that you can supply the volume upstream.

If you were to try this, you would need an impeller with a high specific speed - similar to a ship's propeller. If you increase the velocity of the water downstream the pump you have to ensure that you can supply the volume upstream.

One would ordinarily select a larger pump from the proposed supplier of the one in question. Until one runs out of options, as it were.

The additional water in the <...canal...> has got to go somewhere, so the full picture of what is going on has not yet been revealed to the forum.

The economically useful range for pipework is usually taken in the region of 1-3m/s.

So for 19 x 90 USgal/sec the pipework would then ideally be no larger than 2.9m/9.5ft and no smaller than 1.5m/5ft diameter.

gallons of water

He is asking whether they are US, Imperial, or some other kind of gallon. Obviously, you are talking about US gallons.

Maybe, though it's best to be sure. Other countries use other measuring systems. One mustn't assume that the installation is in any particular country until advised.

• A mix-up in units caused the loss of the Mars Climate Orbiter mission.

1 mph = 1.47 fps

150' x 12' = 1800 sq ft ... 1 cubic foot of water equals 7.48 gallons of water .....

1800 x 7.48 = 13,464 gallons = 1800 cu ft

So you need a pump that produces significantly more than 90 gps...The current flow need not directly involve the entire body of water, what you need is a channel flow effect,where the middle of the canal reaches the .5 mph goal...So if we took the middle section of the canal, say 25% , would be around 3400 gallons per sec....You would need about 14 such 24" pipes with a velocity of 10' per sec...each...my back of the envelope calculation....

OK. That's going to make this project too cumbersome and expensive. It was a thought of mine, but I wanted to check it out here first. The original idea was to keep water moving gently, reduce any stagnation. Its way out of my area of expertise now.

thank you all again.

You could go another direction...

Yes....we may have to look at that. That , are the Diffuser system, with the bubbles.

Be careful putting too much air into the water. The loss of water density can cause boats passing through the area to sink.

What specifically is the concern with stagnation? Mosquitoes? Biofilms and other growies? Smells?

Perhaps these folks could provide some direction:

http://www.current-systems.com/

Have you researched eductor nozzles?

Eductor nozzles for fluid mixing and agitation

A horizontal side entry mixing impeller will get the water moving, similar to mixing chests in large industrial plants, such as paper mills. These are typically horizontal turbines, mixer manufacturers such as Philadelphia or Lightnin have the empirical data collected to let you know if you can get this velocity in your canal section

Some kind of propeller device (think along lines of outboard motor) might work better for this situation.

OK... again.…. these new ideas seem worth a try....I'll contact the sites you all mentioned....I can see the "propeller" thing shoving water.

A difference in levels also helps in <...shoving water...>.

There's more to this post than has currently been disclosed.

Of course, the <...canal...> has to be open at the other end and at overflow weir level for there to be a <...velocity...> in it.

If it isn't, then all one will get from <...shoving water...> is a rise in level.

That would be an impeller, since it is stationary and the water moves.

You probably already knew that.

Eductors have no moving parts and take no additional energy to increase the velocity of the fluid stream.

hA hA I call anything with blades on it a prop...…...

"... That would be an impeller, since it is stationary and the water moves. ...."

.

Stationary with respect to what frame of reference? Isn't the water usually going to be in motion wrt the blades of said prop/imp when operating?

Consider the aeration method used at waste treatment plants,consisting of compressed air piping installed in the bottom of the containment pond and air forced through to create the bubbles and oxygenate the water,thus preventing stagnation and the water going septic.Much more efficient than trying to move tons of water.

for mosquito control,there is a biological agent called Bacillus thuringiensis serotype israelensis (Bti) that kill only the mosquito larvae;harmless to everything else.

Perhaps a nozzle using a higher pressure/lower flow ,as in a fire hose would produce enough turbulence to prevent stagnation..

A handy calculator for determining nozzle size/velocity/flow is linked below:

http://irrigation.wsu.edu/Content/Calculators/Sprinkler/Nozzle-Requirements.php

What you are trying to create is a flume where the surface of the water flows at 1mph (0.447m/sec) in a 12ft (3.6576m) wide channel that is 150ft (45.72m) long. Moving the surface water will prevent stagnation. What happens to the other 9ft 9 inches has very little effect on stagnation so why would you want to expend energy moving it? A rough empirical formula for calculating the velocity of water in a flume is Elliots Open Ditch Formula. V (m/sec) = 0.3√mh where m is the hydraulic radius in meters and h is the channel gradient in m/km or mm/m. If you were trying to move the whole volume of the canal the hydraulic radius would be 3.75 but as you only need the move the surface the hydraulic radius is approx 0.25. This formula makes some assumptions about the Mannings n number which should be valid for a man made canal and works for calculating when moving all the water in a flume, but a large error will be created in moving only the surface water, however the wide tolerance ½-1mph in your specification should compensate for the error. To create the gradient you need to suck water out of one end of the canal reducing it's depth and put it back into the other end increasing the depth by an equal amount, and this doubling effect means you only need to increase the depth by ½ the height to achieve the required head. If you remove 1.22m³/sec (44ft³/sec) and pump it to the other end you will create a head of 150mm (6 inches) which is what you need to get an average flow rate of 1mph. Note that the flow rate in the center of the channel will be faster (roughly double) the flow rate 18 inches from the edges. This is where the slope and roughness of the walls, the size, positioning and shape of the inlet pipe, any obstructions, wind and wave action etc. cause variables that are incalculable.

Good morning folks. It is now 8a.m. and I have just looked at the calculations I did at 2a.m. this morning after a long day driving back from holiday. I dropped a decimal point so you only need to pump 4.4ft³/sec not 44ft³/sec as posted. My mistake, my apologies. That converts to approx 33 (US) gallons/second.

"What you are trying to create is a flume where the surface of the water flows at 1mph (0.447m/sec) in a 12ft (3.6576m) wide channel that is 150ft (45.72m) long"

NO! The 12 ft.dimension is the depth, and the 150 ft. is the width. We've never been given any information regarding the length of the canal, but it is probably safe to assume that it is much greater than 150 ft.

Thus, regardless of the time of day or night, your calculations don't apply.

Netmaker: I just reviewed all of your posts in this thread, and you have never stated your real objective.

From your Original Post, I assumed that you wanted to make water flow through the canal, but then you say there is no current. If there is truly no current, then both ends of the canal must be at the same elevation, with no specific inlet and outlet. If that is the case, then no amount of pumping will cause any current other than a recirculating current.

Is your real objective mosquito control, prevention of biological film formation, or something else entirely?

No engineer can solve a problem unless he/she knows what the problem really is!