Micro Hydro Turbine, pressure and head, efficiency 1

[madassteve]

Hi,

I have a question regarding the calculation of turbine efficiency.

If there is a certain head available at a hydro site, say 20m then the pressure will be approx 2 bar. However the pressure at the inlet of the turbine will be lower than this when the turbine is running becuase of the flow not being entirely restricted, say 1 bar.

So when calculating the amount of power in the water to find the turbine efficiency, should the potential head of 20m be used or 10m (the head equivalent of 1 bar)?

Cheers

 

[BigInch]

Depends on how you are getting the water to the turbine, which usually has at least some frictional flow losses associated with the piping, ducting, directional changes and valving, etc. encountered along the way. If you have a water pool at an elevation 20 m above the turbine but 200 m away, you'll need a combination of open channel, pipe or ducting of some sort equalling at least 201 meters. Whatever you have left of the 20 m head, after subtracting the sum of all flow losses, is available as inlet head at the turbine. If those losses sum to 10 meters, then you have 20-10 = 10 m inlet head. It could be more or less. The larger diameter and the shorter in length the duct is, the more power you will have available at the turbine inlet. To know the power that you can extract from that remaining head using the turbine, you must know the turbine's efficiency in converting that inlet head to actual output shaft power and the efficiency of the alternater, or whatever other mechanical equipment you are using to (presumedly) make electricity with that shaft power. You will need to multiply the inlet head by the turbine efficiency, the gear box (if any) efficiency and the alternator/generator efficiency. After that, you may have additional wire losses to account for, say if you have 100 meters of wire to your final point of use (at the light bulb, etc.).

**********************
"Pumping accounts for 20% of the world’s energy used by electric motors and 25-50% of the total electrical energy usage in certain industrial facilities."-DOE statistic (Note: Make that 99% for pipeline companies)

http://virtualpipeline.spaces.live.com/

 

[madassteve]

Ok thanks, that is helpful but I think I phrased exactly what I was trying to ask quite poorly.

Say for example if you have a substantially large 20m high storage tank, and at the bottom of the drop you have a ball valve connected to the end of the penstock pipe with a pressure gauge one side, and open to atmosphere on the other side, when the ball valve is shut the pressure will read appxox 2 bar. If the ball valve is then opened the pressure will obviously drop to a lower value, is the head still constant at 20m, and just the pressure is lower?

 

[Zapster]

Is your question: should you use total head (velocity head + static head) at the turbine inlet verses static head? I think that it should be total head.

 

[BigInch]

Effectively you almost have the full 20 m head, but technically there is a bit of a loss of head as the water moves from points around the outlet through the tank wall and accelerates from zero velocity to enter the pipe nipple from the tank to the valve. That can be calculated by using what's called the tank nozzle discharge coefficient "Cd". Its normally about 0.65 to 0.75 of the velocity head, if the pipe exiting the tank is flush mounted with the inside of the tank.
This page shows it,

http://www.lmnoeng.com/TankDischarge.htm

Now, valve losses, which you may have some, if its not a full port, fully opened ball valve.

There can also be some losses associated with the water freely jetting out of the tank, but you might reasonably ignore those, if you had a turbine attached right there.



**********************
"Pumping accounts for 20% of the world’s energy used by electric motors and 25-50% of the total electrical energy usage in certain industrial facilities."-DOE statistic (Note: Make that 99% for pipeline companies)

http://virtualpipeline.spaces.live.com/