Timber, I think back to running a sluice in the creek. I had a Keene 52a. Now I would set this up with 1" per foot drop and about 1" of water over the riffle, Then I would slowly feed in material till my 3rd riffle had the slurry. I would stop and wait a second then repeat. So you saying that being we would feed less water across this sluice we would shrink the sluice size down to accommodate less water. Now I may be stupid, but that same scoop I just added for a 10" sluice wont process the same in a smaller sluice. It would overload the riffles. I would have to slow my feed rate of material down.
I understand that there is a excess of water in a dredge, I'm sure it by design. It's prob due to lack of classification, When your sucking up a 5" rock, you need a certain amount of water to push it down the sluice.
But you guys are talking numbers that are smaller, instead of 500gpm from the jet, 250gpm then instead of 1000gpm from the flare to the sluice your saying 750. Your saying the sluice would shrink to match the flow. I don't see how your having the same amount of material processing across your sluice more efficiently. I see less material being fed because of the smaller sluice due to less water.
This goes back to what I am tryi8ing to get at is that water doesn't expand or contract under dredging conditions, Water coming out of the jet is flowing from a high pressure to a low pressure, causeing water in the hose to start moving toward a higher pressure zone ( called suction) If a hose (don't know any numbers, just pulling this out of my ass) will hold 1/2 cubic foot of water for a 6"x1' section of hose, you apply a force to cause a flow (venture) that hose will still only hold that volume of water. The hose doesn't expand nor contract and neither does the water.
Am I right so far? Tell me if I'm not!
Once flow is started the only thing that can change is the velocity, water doesn't expand/contract, neither does the hose. If you jet is only supplying 250gpm of flow lets say in 1 second that's 2 cups of water. Lets break the 1/2 cubic foot of water down to cups. That's 59.85 rounded. So being water cant compress under these conditions, we are adding 2 cups of water so we much take away 2 cups of water. Do we still not end up with 59.85 cups? We're talking in a ideal condition the system is airtight or underwater and only sucking in minute quantities of water to fill in air gaps. So the only variable here is velocity. You cant cram 1.2 gallons of water into a 1 gallon can. See where I'm going with this>? Your velocity will (increase
) due to more efficient push? (1 hand vrs 2 hands) Or is it going to decrease being you have a fixed volume of water and your adding and taking away. This is the part that I am unsure of, but logic tells me it will increase due to more efficient push. My point is I do not see how you are going to have less unless the velocity decreases because your volume can't change unless you create a "void"
I'm no scientist, hell I can only count to 21 (10 fingers/toes plus little me)(LOL) I just cant wrap my head around how you would end up with less water unless your velocity changed (slowed) and in that case you would have less suction and less material being sucked and that doesn't equal more efficiently
Once again, tell me if I'm wrong here.
