Maximizing Water Flow: Solving the Pump Effect Challenge in Shallow Waters

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SUMMARY

This discussion addresses the challenges of maximizing water flow using a suction hose (100m, Ø 5cm) connected to a pump (60 l/min) for dredging purposes. Key calculations reveal that the water velocity at the mouthpiece is approximately 0.071 m/s, which may be insufficient for effective suction of loose particles from the seabed. The discussion emphasizes the importance of adjusting the mouthpiece's position and size to enhance suction efficiency. Additionally, it highlights that the pump's power requirements are minimal, with an estimated need of around 9.8 Watts at 100% efficiency, depending on the suction head and pressure losses.

PREREQUISITES
  • Understanding of fluid dynamics principles, specifically flow rate and velocity calculations.
  • Familiarity with suction pump specifications and performance metrics.
  • Knowledge of pressure drop calculations in long hoses.
  • Experience with experimental methods for optimizing suction equipment.
NEXT STEPS
  • Research "Fluid dynamics flow rate calculations" to deepen understanding of velocity and area relationships.
  • Explore "Suction pump efficiency" to learn about optimizing pump performance for specific applications.
  • Investigate "Pressure drop in hoses" using online calculators to assess impacts on suction performance.
  • Experiment with "Mouthpiece design for dredging" to determine optimal configurations for particle pickup.
USEFUL FOR

This discussion is beneficial for marine engineers, environmental scientists, and anyone involved in sediment dredging or water flow optimization in shallow waters.

TorMcOst
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Dear all,

I have worked with a challenge for two full days now without managing to solve it. I would be MOST grateful if anyone could help me sort out my confusion concerning the problem.

I have a suction hose (100m, Ø 5cm) that is connected to a larger mouthpiece in the end (Ø 30cm). The suction end of the hose is submerged to 7 m draught and the other end are connected to a pump (sucking 60 l/min) in an assisting vessel (2m above surface area). The mouthpiece will be placed 1,5cm from the seabed and are ment to suck up all loose particles on the seabed.

Q1. What will the speed of the water passing between the seabed and the sucking mouthpiece be?

Q2. Would a 60 l/min suction pump be sufficient in order to pump up all loose particles if the sucking mouthpiece were moved over the seabed with 0,7 m/s (still 1,5cm distance between seabed and mouthpiece)?

Q3. How much smaller pump would be needed if the pump was placed right behind the sucking mouthpiece on the seabed?

Thank you very much in advance!

Yours Faithfully,

Tor
 
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Q1.
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.
 
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Hi Tor-
The previous answer is very comprehensive, I just want to add that there are some online flow and pressure drop calculators such as this one:
http://www.pipeflowcalculations.com/pressuredrop/index.htm
The calculator indicates that the pressure drop in your 100 m hose is about 0.09 bar (0.09 atm) so there is no problem in the sucking limit of the pump. The Reynolds number in the hose is 1900, which is acceptable. Sounds like you are dredging for gold in the bottom of one of the rivers in California. The choice of mouthpiece will have to be determined by experimentation. You will need a certain minimum flow velocity to loosten the silt you are trying to pick up. Good luck.
Bob S
 
Last edited by a moderator:
Ah, I didn't notice the 100m length, that'll add a little pressure loss. For some reason I thought it was a short hose. To convert back and forth between meters of water and other forms of pressure, just find the pressure of a column of water X meters high. For example, 1 meter high of water weighs 9800 N per square meter, which is 9800 Pa or 9.8 kPa or roughly 0.098 atm. So from Bob's calculation that's another meter of pressure for your pump, for 2 meters or about 0.2 atm (rounded) or about 20 kPa total assuming you use 1 meter of pressure for suction (> 4.5 m/s velocity, see my other post above).
 

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