50kw impluse turbine

westbuteast

has anyone here ever bought a 50 kw pelton turbine or installed a micro hydro generator supplying 400 v 3 phase power at 50 Hz?

Coles

How much head and flow does the resource have?

westbuteast

Coles wrote: »

How much head and flow does the resource have?

I have 206 meters of head and a flow rate of .035 meters cubed a second. I have estimated that there would be 20 bar at the end of nozzle hitting the turbine blades.

Coles

westbuteast wrote: »

I have 206 meters of head and a flow rate of .035 meters cubed a second. I have estimated that there would be 20 bar at the end of nozzle hitting the turbine blades.

Wow! That's a lot of pressure. Have you figured out what length the penstock would need to be?

westbuteast

Coles wrote: »

westbuteast wrote: »

I have 206 meters of head and a flow rate of .035 meters cubed a second. I have estimated that there would be 20 bar at the end of nozzle hitting the turbine blades.

Wow! That's a lot of pressure. Have you figured out what length the penstock would need to be?

I'm looking at 900 meters to the power house. This is a survey I'm doin more of a visability stuby. Have a 150 mm pipe picked and that should give me 46 kw with all losses taken into account friction factors and all.

Darragh3206

I can recommend some manufactures of micro hydro turbines that I have dealt with in the past and had good experiences with if you like. I will pm you my information. Hydropower is my main business.

Coles

westbuteast wrote: »

I'm looking at 900 meters to the power house. This is a survey I'm doin more of a visability stuby. Have a 150 mm pipe picked and that should give me 46 kw with all losses taken into account friction factors and all.

How have you estimated the available flow at 35L/s?

I'm developing a site for a client at present in Wicklow that will use a Crossflow turbine, 65L/s and a head of 14m, which will have a single phase 6Kw capacity. There's nothing different about the 3 phase production, but you will need to check the proximity of the electricity lines or there could be a significant cost involved. Many turbines actually produce 3 phase and then convert it to single phase before exporting the production.

The most expensive element of your project will be the penstock, but with a 1:5 gradient it sounds like it should be feasible. Certainly worth investigating.

PM sent.

mchammer

westbuteast wrote: »

I have 206 meters of head and a flow rate of .035 meters cubed a second. I have estimated that there would be 20 bar at the end of nozzle hitting the turbine blades.

hello - can I ask how you calculate the pressure at the nozzle? have you ever gone any further with the project? thanks am looking into a similar project in time..

Coles

mchammer wrote: »

hello - can I ask how you calculate the pressure at the nozzle? have you ever gone any further with the project? thanks am looking into a similar project in time..

The vertical height of water (the head) will give an approximate measure of pressure, but a certain amount of pressure will be lost due to friction within the penstock when the water is flowing. The amount of friction loss will depend on the penstock material, diameter, water velocity. Also pressure will be lost as the water flow negotiates bends, valves, junctions etc..

mchammer

Coles wrote: »

The vertical height of water (the head) will give an approximate measure of pressure, but a certain amount of pressure will be lost due to friction within the penstock when the water is flowing.

Thanks for that Coles.. I found an good guide to hydro eventually in between all the crap websites..

http://www.canyonhydro.com/guide/HydroGuide7.html

Basically 1 psi = 2.31 vertical feet so roughly if I have 200ft of head i can expect a theoretical static pressure of 86.5psi. If the nozzle area is .5 of a sq inch the theoretical pressure should double i.e. 173 psi.. at 85% efficiency due to head loss I should still theoretically get 130psi or thereabouts assuming a continuous flow...

am i right or way off?
cheers
mc

Coles

mchammer wrote: »

Thanks for that Coles.. I found an good guide to hydro eventually in between all the crap websites..

http://www.canyonhydro.com/guide/HydroGuide7.html

Basically 1 psi = 2.31 vertical feet so roughly if I have 200ft of head i can expect a theoretical static pressure of 86.5psi. If the nozzle area is .5 of a sq inch the theoretical pressure should double i.e. 173 psi.. at 85% efficiency due to head loss I should still theoretically get 130psi or thereabouts assuming a continuous flow...

am i right or way off?
cheers
mc

Almost, except for the last bit. You can't increase the pressure at the nozzle beyond the static pressure (there is one exception to this rule but it usually means you are about to experience a very expensive mechanical failure)

I'd advise that you stick to metric units rather than that other stuff, but let's have a quick look at pressure...

The pressure is derived from the vertical height of water (the 'static head of water') above the turbine.

Pressure (P) = density of fluid (p)* gravity (g) * height of fluid (h)

If there is 10m of head then static pressure at the turbine will be...

Pressure (P) = 1000 (kg/m3) * 9.81 (m2/s) * 10 (m)
Pressure (P) = 98100 Pascals = 98.1 kPascals
1 kPascal = .145 psi
so the Pressure at the turbine is about 14.4 psi

BUT what happens when the water is flowing? A certain amount of the 'static head pressure' is lost due to friction within the penstock. The actual amount of frictional losses within the Penstock depends on a range of factors but the most important issue is the velocity of water. If the water is allowed to flow too quickly then the frictional losses will make the scheme inefficient. If the flow is too slow then too much money has been wasted on a very expensive over sized penstock. It's a balancing act to size the penstock correctly but it's probably the most critical factor for a successful scheme.

Let's assume 10% frictional losses. Our 'Static Head' has now been reduced by 10% and is now called the 'Dynamic Head', the actual pressure at the turbine when the water is flowing. In this case 89 kPascals (13 psi), the equivalent to the static pressure of 9m of water.

Rather than looking at pressures it's far easier to look directly at the electrical production with the following formula...

Capacity kW = Head (m) * Flow (m3/s) * 9.81 (gravity, m2/s) * efficiency (%)

Probably easiest to use the Static Head measurement and to use a lower efficiency % to accommodate the frictional losses.

We have 10m of head and let's assume a flow of 200 Litres/sec (0.2m3/s), and a system efficiency of 70%. This system efficiency rate takes into account the 10% frictional losses, the mechanical losses within the turbine, the losses within the generator, and any transmission losses to the point of consumption, etc.

Capacity (kW) = 10 (m) * 0.2 (m3/s) * 9.81 (m2/s) * 70%
Capacity (kW) = 14kW

The amount of electricity that we can produce depends on how long we can keep the turbine running at full capacity. A 'Load Factor' of 60% would assume that the generator is running at the equivalent of full output for 60% of the time, and over a year it's annual production will be...

Production (kWHrs) = 14 (kW) * 365 (days) * 24 (hours) * 60% (Load Factor)
Production (kWHrs) = 73,600 kWHrs

I hope that helps a bit.

A helpful source.

mchammer

Coles wrote: »

I hope that helps a bit.

A helpful source.

much appreciated Coles - super response

Capt'n Midnight

mchammer wrote: »

http://www.canyonhydro.com/guide/HydroGuide7.html

Basically 1 psi = 2.31 vertical feet so roughly if I have 200ft of head i can expect a theoretical static pressure of 86.5psi. If the nozzle area is .5 of a sq inch the theoretical pressure should double i.e. 173 psi.. at 85% efficiency due to head loss I should still theoretically get 130psi or thereabouts assuming a continuous flow...

And that's why I love the metric system

Atmospheric pressure = 1 bar ~ 10m of water ~100,000 Newtons / m2

and scaling up dimensions is just soooo much easier

mchammer

Capt'n Midnight wrote: »

And that's why I love the metric system

Atmospheric pressure = 1 bar ~ 10m of water ~100,000 Newtons / m2

and scaling up dimensions is just soooo much easier

thanks for that also.. was out on site and took a small vid ... took fairly accurate gps too and the head is 80+ metres over about 210 metres... hard to work out flow but it can't be much more than around 5 litres a second? have access to 2 streams of roughly same flow so it might be a case of 2 intakes? anyone hazard a guess at flow rate? should have brought a bucket!

Coles

It's difficult to guess the flow (I'd say 3-5 ltrs/s), and it's a bit meaningless too. A single spot measurement won't tell you much about the long term flow conditions. In order to accurately build up a picture of the flow you would need to repeatedly take measurements and compare your results to a nearby gauged stream. Unfortunately this method is unlikely to prove useful as there are very few gauged upland streams, particularly ones with such small catchments (<1km2?), and it's unlikely that there would be suitable one close by.

It won't work to have multiple intakes, so it would be worth considering positioning your intake at the confluence of the two streams even though you lose some of your head.

I think you have a really excellent resource. Well worth considering.

Capt'n Midnight

you could use a V shaped notch in a weir to measure flow rate



http://www.engineeringtoolbox.com/weirs-flow-rate-d_592.html

http://ocw.usu.edu/biological_and_irrigation_engineering/Irrigation___Conveyance_Control_Systems/6300__Weirs_for_Flow_Measurement_Lecture_Notes.pdf