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Apr20-09, 01:15 PM
P: 20
Q = vA (flow = velocity times area)
v = Q / A
Q = 60 L / min = 1 L / s = 0.001 m^3/s
A (cylinder) = pi * 0.30m * 0.015m = 0.01413 m^2
V = Q / A = 0.071 m /s = 0.25 kph

This is around the rim of the mouthpiece, where the velocity is the highest. Around the inside of the mouth itself the area is pi * r^2 = 0.07 m^2, which is higher so the velocity is lower. Outside of the lip the velocity is also lower. Water is coming in from all directions so the effective area crossing their paths is higher.

Q2. No idea. It depends how fine the particles are, and you'll have to look up some obscure handbook to find the equations. You're probably best off just trying and seeing. But with the low velocity from Q1, you probably won't get much suction at all. Best to lower the mouthpiece a bit and/or make it less wide. A friend of mine had a small aquarium with a small hose with about a meter of pressure (gravity powered) doing the sucking to pick up loose debris. That translates to a velocity of around 4.5 m/s (h = v^2 / 2g, g = 9.8 m/s), so maybe you could use that as a starting point. Just don't exceed a few meters of suction head (also = v^2 / 2g) or you'll get pump cavitation. Depending on suction shape the actual suction head is often half of this or less, but it'll help you ballpark it at least.

Q3. It would be the exact same size. The static head loss is equal to the water surface at the discharge minus the water surface at the suction, which are both the same regardless of the setup. That's assuming you're discharging back into the seabed somewhere else. If you're discharging 2 meters above the seabed into open air, then there's 2 meters you could save simply by running a hose to discharge under water. There's still no need to move the pump.

If it helps you conceptually, think of it this way: Being under 7 meters of water hurts as much at the discharge as it helps at the suction.

Btw, with only a meter or so of head (all from velocity in Q2), you don't need a very powerful pump regardless. 0.001 m^3/s (flow) * 9.8m^2/s (gravity) * 1m (head) * 1000kg/m^3 (water density)= 9.8 Watts. That's at 100% efficiency, actual pumps might be 40-80% efficient. The power will be more if the head is higher than 1 meter. A lot of pumps are rated at 3 meters or more, so you might not be able to find such a small / low pressure pump.