# Help in designing water turbine for given wattage

We need to design an archimedes screw turbines to power a pump of 80 watts.

Any tips to design this?

I have tried a way, but I think it's very wrong.

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Let's consider pumping 1 liter (1 kg) per second up h=1 meter. mgh = 9.81 kg-m2/sec2=9.81 Newton-meters = 9.81 joules.

So 1 liter per second up 1 meter is 9.81 watts.

Let's consider pumping 1 liter (1 kg) per second up h=1 meter. mgh = 9.81 kg-m2/sec2=9.81 Newton-meters = 9.81 joules.

So 1 liter per second up 1 meter is 9.81 watts.
So you're saying, 9.81Sin(63) would give the energy gotten from water being flowing down a tube at 63 degrees? Or is that wrong?

All I am saying is that, at 100% efficiency, 9.81 watts is needed to lift 1 liter of water per second up vertically 1 meter. Or 98.1 watts to lift 1 liter/sec up 10 meters. Now. you must describe in detail your engineering model.

All I am saying is that, at 100% efficiency, 9.81 watts is needed to lift 1 liter of water per second up vertically 1 meter. Or 98.1 watts to lift 1 liter/sec up 10 meters. Now. you must describe in detail your engineering model.
The task is to design an archimedes screw turbine to power an 80 watt pump.

I can assume flow rates, head, etc.

The turbine will be at 63 degrees, so I need to find out the size of the turbine.

russ_watters
Mentor
No, you don't need to find the size of the turbine yet. First you need to find the height and flow rate of the water supply. The size of the turbine is a function of the flow rate, but the wattage is a function of the height and flow rate. So the higher the drop, the smaller the turbine can be.

No, you don't need to find the size of the turbine yet. First you need to find the height and flow rate of the water supply. The size of the turbine is a function of the flow rate, but the wattage is a function of the height and flow rate. So the higher the drop, the smaller the turbine can be.
Ok, I can assume and make up heads, so assuming a head of 2m.

What's next?

I can assume flow rates too if need be.

I found this formula

"Power = Head X Flow X 7"

So I can fill in the head to get a flow rate?

Or this? This is the one I used at first but I got a veeeeery small area, for a 20m3/hr flow rate and 1.75 head difference.

Would this be correct?

Ph = q ρ g h / (3.6 106) (1)

where

Ph = power (kW)

q = flow capacity (m3/h)

ρ = density of fluid (kg/m3)

g = gravity (9.81 m/s2)

You need to understand the equations you are using. For example, 20 m3 is 20,000 liters. That is a big number. Using the formula I derived in post #2, pumping 1 liter/sec up 2 meters would require 2 x 9.81 ≈ 20 watts. What is the efficiency of your pumping system? Why don't you discuss its design? What is the "bucket" size and pitch on your Archimedes screw? What is its rpm? what is the gearing on your motor?

You need to understand the equations you are using. For example, 20 m3 is 20,000 liters. That is a big number. Using the formula I derived in post #2, pumping 1 liter/sec up 2 meters would require 2 x 9.81 ≈ 20 watts. What is the efficiency of your pumping system? Why don't you discuss its design? What is the "bucket" size and pitch on your Archimedes screw? What is its rpm? what is the gearing on your motor?
sorry, the q is in m3/hr, so it would be .0055 l^3 / second.

The aim is to design the screw, work out the forces and design the thickness of the shaft etc.

Maybe this is the wrong forum.

The efficiency of an archimedes screw is 84%.

The main aim is to design a system that aerates a septic tank.

usually the pumps are 80 watts, so we've decided to power the pump by using hydro power.