Pelton turbine power calculation 1

[21121956]

Hello everybody:

In a general manner, the power delivered by a hydraulic turbine is:
P = ?t * ?g * ?tr * 9,81 * Qd * Hn [kW]
Where: ?t = turbine efficiency.
?g = generator efficiency.
?tr = transformer efficiency.
Qd = design flow, [m³/s].
Hn = Net head, [m].

Where: Hn = H gross – ? hf
H gross = Gross head, [m].
hf = Head losses, [m].

Being for a Pelton turbine:

? hf = h friction losses + h minor losses + h manifold losses + h jets loss

I have designed a spreadsheet to calculate that power, taking into account all the variables indicated above.

I have tested the accuracy of my spreadsheet against the calculations achieved during the design of small hydro power plants that have been running since several years ago and, for my surprise, I have found that for the calculations of ? hf, they have NOT taken h jets loss, that is a loss of around 6% of the Gross Head.

My question is: why is it possible that for the power calculation, sometimes h jets loss is taken into consideration and, sometimes it is left aside?

For a particular calculation, P = 10000 kW without h jets loss, and P results to be 9328 kW if h jets loss is added to the equation.

I will appreciate very much your feedback on this matter. Thanks.

 

[BigInch]

Jet losses would normally be included in the turbine efficiency. Mechanical effciency, I think what your "Power delivered by the turbine" says, would need to include additional efficiency factors for the power absorbed by the rotor/power supplied by the water, and other mechanical losses.

Possibly your unaccounted for losses are from not including those additional terms.

You might want to have a look at this, where it shows that the difference in jet and bucket speed would need to be included to describe the turbine's efficiency.

http://www.engr.mun.ca/~yuri/Courses/FluidsI/Pelton Turbine.pdf

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[21121956]

Hello everybody:

Thanks BigInch for your reply.

I think, I understand perfectly what you are talking about and, at the same time, I am aware of the subject exposed in your submitted document.

Here is, in brief, the data for one of the Projects:

Reservoir Water Level, max 948 m.a.s.l.
Reservoir Water Level, min 948 m.a.s.l.
Tailrace Water Level, max 500.2 m.a.s.l.
Tailrace Water Level, min 500 m.a.s.l.
Gross Head, 448 m

Low pressure conduction pipe diameter 1 m
Low pressure conduction pipe length 1156 m
Steel Penstock diameter 1 m
Steel Penstock length 1508 m

Flow rate, max 2.74 m³/s

Penstock:

Mean velocity 3.49 m/s
Trashrack losses 0.70 m
Losses at pipe entrance 0.02 m
Losses in bends and elbows 0.78 m
Friction losses in the pipe 9.53 m
Inlet valve losses 0.06 m
Manifold losses 0.20 m
Losses at pipe reductions 0.25 m
Nozzles (jets) losses 26.48 m
Total friction losses 38.03 m

Net Head for the Project 409.97 m

Turbine efficiency = 0.9
Generator efficiency =0.95
Transformer efficiency = 0.99

Power House:

Installed Capacity 9328 kW
Number and type of Units 2 Pelton horizontal

I really beg your pardon for the out of alignment of some part of the text of this post.

As you can see, the losses in the jets are of some important magnitude compared with the rest of losses.

 

[BigInch]

Then why don't you think the jet loss term is not already part of the turbine eff loss that results in an efficiency of 0.90 ?

With an eff of 0.90, 10% is lost in the turbine, of which the jet loss would represent appx 5% of the total power you would calculate if you assumed it was already included in the turbine efficiency number, so the jet loss would account for 1/2 the total losses of the turbine.

"I am sure it can be done. I've seen it on the internet." BigInch's favorite client.

"Being GREEN isn't easy." Kermit

http://www.youtube.com/watch?v=hpiIWMWWVco

http://virtualpipeline.spaces.live.com

 

[21121956]

Hello everybody:

Jet velocity, V1 = Cv * (2*9.81*Hn)^0,5 [1]
Cv = velocity coefficient, 0,97 ÷ 0,99
Velocity of the buckets (runner), u = 0,46 * V1

Tangential velocity at entrance of bucket:
Vt1 = V1 = Vr1 + u [2]
Vr1 = Relative velocity at entrance.

Tangential velocity at exit of bucket:
Vt2 = u – Vr2 * cos ?2 [3]
Vr2 = Relative velocity at exit, Vr2 = k * Vr1
?2 = Outlet angle of the jet, typically 165°.
k = Bucket coefficient, 0,85 ÷ 0,92

The Euler equation:
Energy transfer/mass = u*Vt1 – u*Vt2 [4]
Substituting [2] and [3] in [4]:

P = (?*Q)*u*[Vr1+u – (u - Vr2*cos ?2)] [5]
= (?*Q)*u*[(V1-u) + (k*(V1-u)*cos ?2)]
= (?*Q)*u*(V1-u)(1+k*cos ?2)

Efficiency of runner = Power / Kinetic Energy of the jet

= (?*Q)*u*(V1-u)(1+k*cos ?2) / (?*Q)*V1²/2 [6]

In the equation [1] the absolute velocity V1 is calculated with the net head, (Hn = Hgross – hf) and hf includes (or should include) the nozzles loss; then, at the moment to make the calculation of the runner efficiency, equation [6], the losses in the nozzles have been already taken into account.

In other words, the hjets occur just in the nozzle and not at the moment of Energy transfer from the jet (some distance appart of the nozzle) to the runner.

Once again, thank you for your comments.

 

[21121956]

Hello everybody:

Hi BigInch, I made a technical query to Voith, the german hydro turbine manufacturer and, their experts told me, among other things the following:

" The losses in the distributor line and the nozzles are included in the turbine efficiency, because the measurement of the hydraulic power in model and in prototype is done upstream the first bifurcation of the distributor line. Therefore a separate term for the nozzle efficiency is not necessary and also not considered by the relevant international standards (IEC)".

I think I am obliged to share this information with the people of the Forum, especially with you.

Thanks and one star for you.

 

[BigInch]

Thank you. Yes, I thought it was a question of the most logical place to draw the control volume for the machine.

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"Being GREEN isn't easy." Kermit

http://www.youtube.com/watch?v=hpiIWMWWVco

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