Fluid flow application for Pelton Turbine issue
Fluid flow application for Pelton Turbine issue
(OP)
Hello,
I'm currently undergoing a MicroHydro project where I need to call on my learnings from fluid mechanics to optimise my system. I'm becoming very unsure of my working and so would like some feedback on my logic.
I need the minimum figures to produce 1kwh from a pelton wheel turbine - generator set up with the following parameters:
1. Open dam of water 7m above the turbine.
2. Water flowing through a 50mm inside diameter pipe
3. Pelton wheel has a pitch diameter of 500mm
4. Losses are negligible (until I'm confident with my initial workings)
5. The turbine runs at 80% efficiecny
6. The generator runs at 80% efficiency
The initial jet velocity, derived from Bernoullis principle:
Vi= √2gh= √(2*9.81*7) = 11.72 m/s
The optimal wheel speed:
u = Vi/2 = 11.72/2 = 5.86 m/s
The RPM of the wheel:
RPM=u/(π*Dpitch )*60= 5.86/(π*0.5)*60=223.82 RPM
Volumetric Flow Rate of the water:
Q=Anozzle*Vi=(〖0.025〗^2*π)*11.72= 0.023m^3/s or 23 l/s
Power from the turbine:
PTurbine=μ*p*g*h*Q=0.8*1*9.81*7*0.023=1263.53Watts
Power from the generator:
PGenerator=PTurbine*0.8=1263.52*0.8=1010.8224 Watts
∴ With a head of 7m,a pitch diameter of 500mm and a nozzle diameter of 50mm, the required output is satified.
NOTE: These workings are displayed in the attachments in an easier to understand form.
Now here are my questions:
1. In the derivation of bernoullis, the pressure/(density*gravity) is constant on both sides of the Steady Flow Energy Equation. This is because the dam at the top is open (so atmospheric pressure) and the water exiting the pipe at the bottom is exposed to the atmosphere (so atmospheric pressure). Therefore the two pressure constants can be removed from the SFEE? Is that correct?
2. The volumetric flow rate seems incredibly unrealistic to me, do you agree?
3. I feel there are errors in the logic somewhere but I'm unsure where, can you think of anything?
What I decided to do was create a spreadsheet based on my workings in the above example to show the differences of values with varying user input (shown in yellow).
http:/ /i33.photo bucket.com /albums/d5 2/Serene_b eing/pelto ntable.jpg
From the outlined information, can anyone see any obvious flaws in my logic. My knowledge of fluid dynamics is poor as it was not covered much in my degree.
Thanks for any info.
DFC101
I'm currently undergoing a MicroHydro project where I need to call on my learnings from fluid mechanics to optimise my system. I'm becoming very unsure of my working and so would like some feedback on my logic.
I need the minimum figures to produce 1kwh from a pelton wheel turbine - generator set up with the following parameters:
1. Open dam of water 7m above the turbine.
2. Water flowing through a 50mm inside diameter pipe
3. Pelton wheel has a pitch diameter of 500mm
4. Losses are negligible (until I'm confident with my initial workings)
5. The turbine runs at 80% efficiecny
6. The generator runs at 80% efficiency
The initial jet velocity, derived from Bernoullis principle:
Vi= √2gh= √(2*9.81*7) = 11.72 m/s
The optimal wheel speed:
u = Vi/2 = 11.72/2 = 5.86 m/s
The RPM of the wheel:
RPM=u/(π*Dpitch )*60= 5.86/(π*0.5)*60=223.82 RPM
Volumetric Flow Rate of the water:
Q=Anozzle*Vi=(〖0.025〗^2*π)*11.72= 0.023m^3/s or 23 l/s
Power from the turbine:
PTurbine=μ*p*g*h*Q=0.8*1*9.81*7*0.023=1263.53Watts
Power from the generator:
PGenerator=PTurbine*0.8=1263.52*0.8=1010.8224 Watts
∴ With a head of 7m,a pitch diameter of 500mm and a nozzle diameter of 50mm, the required output is satified.
NOTE: These workings are displayed in the attachments in an easier to understand form.
Now here are my questions:
1. In the derivation of bernoullis, the pressure/(density*gravity) is constant on both sides of the Steady Flow Energy Equation. This is because the dam at the top is open (so atmospheric pressure) and the water exiting the pipe at the bottom is exposed to the atmosphere (so atmospheric pressure). Therefore the two pressure constants can be removed from the SFEE? Is that correct?
2. The volumetric flow rate seems incredibly unrealistic to me, do you agree?
3. I feel there are errors in the logic somewhere but I'm unsure where, can you think of anything?
What I decided to do was create a spreadsheet based on my workings in the above example to show the differences of values with varying user input (shown in yellow).
http:/
From the outlined information, can anyone see any obvious flaws in my logic. My knowledge of fluid dynamics is poor as it was not covered much in my degree.
Thanks for any info.
DFC101





RE: Fluid flow application for Pelton Turbine issue
The numbers appear correct, aside from a possible discharge coefficient loss at the dam to pipe entrance, and the 23 l/s is the theoretical maximum steady state velocity you could reach. The other numbers appear reasonable, although I didn't check any of them.
BTW, we give extra points for posting the spreadsheet. Saves us a lot of typing just to check someone else's numbers.
//virtualpipeline.spaces.live.com
RE: Fluid flow application for Pelton Turbine issue
I understand what you mean, a 7m head of water is tremendous I guess when you look at it in the manner which you described.
I will being to incorporate losses into my system now that I feel more confident in my workings.
Attached here is the spreadhsheet I created
RE: Fluid flow application for Pelton Turbine issue
I still very well remember a rogue wave that came over the top of the Galveston jetty that I was fishing off of and washed my A out into the channel. That was only 3 m.
Thanks for the sheet.
//virtualpipeline.spaces.live.com
RE: Fluid flow application for Pelton Turbine issue
RE: Fluid flow application for Pelton Turbine issue
http://www.eng-tips.com/viewthread.cfm?qid=283070
Please note that you are talking about a Pelton turbine. therefore the turbine is located in atmosphere and you have to calculated the turbine output based on the jet speed not the efficiency.RO.g.h.Q formula. Please take a look to some related academic books for turbine calculations.