Turbine Design for fire-hose driven shaft

[Clay2]

I am working on a design for a water powered drive shaft. The water source would be from a 2.5 inch diameter hose, at approx 75-100 psi. The goal would be to get 55-115 RPM from the shaft and max torque in the smallest possible handheld device. That being said, it seems my best options for driving such a shaft would be a turbine of one of three designs: crossflow vs turgo vs pelton. What do you think is the best drive to get these results? A pitot-tube pressure of at least 344 kilopascals (kPa) (75 pounds per square inch (psi)), and a flow rate of at least 300 liters per minute (lpm) (80 gallons per minute (gpm)); So we are talking about .3m^3/m @ about 75 psi. So approx 0.005 m^3/s

 

[anorlunda]

Welcome to PF. That is a very powerful water stream you are talking about.

The pictures below are from https://en.wikipedia.org/wiki/Water_turbine

First, I recommend just talking about power and power efficiency rather than RPM and torque. It is much simpler and more to the point.

Second, the key parameter is the flow rate m³/s. You gave hose diameter and pressure. Do you know the flow rate? Once you know the flow rate and pressure, you can look it up on a curve such as this to see which kind of turbine is most applicable.

 

[Clay2]

A pitot-tube pressure of at least 344 kilopascals (kPa) (50 pounds per square inch (psi)), and a flow rate of at least 300 liters per minute (lpm) (80 gallons per minute (gpm)); So we are talking about .3m^3/m @ about 50 psi. So approx 0.005 m^3/s. Looks like 50 psi is about 36 meters of head.

 

[anorlunda]

A pitot-tube pressure of at least 344 kilopascals (kPa) (50 pounds per square inch (psi)), and a flow rate of at least 300 liters per minute (lpm) (80 gallons per minute (gpm)); So we are talking about .3m^3/m @ about 50 psi. So approx 0.005 m^3/s. Looks like 50 psi is about 36 meters of head.

That sounds modest. You said 2.5 inch hose and 100 PSI (I presume 100 at the nozzle.) Fire equipment catalogs say up to 700 gpm for similar numbers.

Let's say 0.04 m³/sec and 70 m head. In any case, that's 2 decades off-scale on that diagram. Too bad.

This reference may give you design data that can be used on such a small scale turbine.

Sayers, A. T. (1990). Hydraulic and Compressible Flow Turbomachines. Mcgraw Hill Book Co Ltd. ISBN 978-0-07-707219-3.

Edit: By the way, please confirm that this is just a design exercise and that you're not really going to build one. I ask because of safety concerns.

 

[The Fez]

From a practical sense, using a pump-as-turbine would make the most sense. Any off the shelf centrifugal pump can be hooked up "backwards" to create torque from the water. This type of application is being used to recapture energy from pressure reducing stations in domestic water systems.

 

[The Fez]

That being said, nothing in the range of pressure/flow you are talking about would be "handheld."

 

[Clay2]

That being said, nothing in the range of pressure/flow you are talking about would be "handheld."

Okay, yea, that is what I have gotten from some other forums as well like: https://www.eng-tips.com/viewthread.cfm?qid=449391.

PAT (pump as turbine). So, basically here is where the Physics comes in...

That pressure/flow, 75psi min and 80 gpm min is what is available at the hose/inlet to PAT. What I need out of the PAT is approx 336 in-lb torque or 40 Nm @ 55-115 RPM... can this be done?

 

[jrmichler]

The book Centrifugal and Axial Flow Pumps, by A.J. Stepanoff has a Figure 13.2 showing the performance of a double suction pump in all four quadrants (forward and reverse rotation, forward and reverse flow). I did not take the time to fully understand the figure, but it looks possible. You would have to work with a pump supplier to select a pump that would do the job. And, like @The Fez says, it would not be handheld.

Be aware that the low RPM would result in very low efficiency. You would probably need to add a gear reducer.

Let's check to see if it is theoretically possible:

1) Power available = GPM X PSI / 1714 = 80 X 75 / 1714 = 3.5 hp.
2) Power desired = 28 ft-lbs X 115 RPM / 5250 = 0.61 hp

You need only 0.61 / 3.5 = 18% efficiency to make it work. So it's theoretically possible.

 

[Clay2]

Great info,
Or approx a 5-1 reduction gear @ approx 80 something overall efficiency?

You need only 0.61 / 3.5 = 18% efficiency to make it work. So it's theoretically possible.[/QUOTE]

 

[The Fez]

A lot of the pumps you would be looking at work at 3450 rpm, so something like a 25:1 or greater reduction gear would be appropriate. Reduction gears don't affect efficiency like you are talking about, so you don't need 25x efficiency for a 25:1 reduction. However, gear reducers do have some loss of efficiency depending on the type you specify. For instance a 25:1 worm gear reducer has about a 85% efficiency, so you would need 0.18/0.85 or somewhat over 21% efficiency, which is doable. You can look up Power Transmission Efficiencies for more information.