1. The problem:
Finding the net head.
This website (http://www.oregon.gov/ENERGY/RENEW/H...ro_index.shtml), says that the gross or "static" head is the vertical distance between the top of penstock and the point at which the water is discharged from the turbine. It then says "Net head is gross head minus the pressure or head losses due to friction and turbulence in the penstock." I am confused about this statement. How do I go about finding net or "dynamic" head?? If I know the pressure at the top of the penstock, and the pressure at the bottom, how do I apply this information?
2. Relevant equations
I'm not sure if any of these are relevant. . .
P2-P1 = -y (Z2- Z1)
(The pressure difference between two altitudese. . )
dp= -y dz
(Change in pressure as it is related to change in weight. . .)
y= (density) (g)
3. The attempt at a solution
I'm very confused. . . I just need to understand the method of determining net head.
Thank you in advance!
Finding the net head. . .
Moderators: kgudger, bfinio, MadelineB, Moderators
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imatreyu
- Posts: 6
- Joined: Sun Jan 17, 2010 2:50 am
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- Project Question: Investigating the possibility of building a rain-water dependent turbine on Mt. Waialeale, Kauai
- Project Due Date: 1/20/10
- Project Status: I am conducting my experiment
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paulsdecarli
- Former Expert
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Re: Finding the net head. . .
I agree, that is confusing. The concept of net head is a way of accounting for the losses in the system. It is the head one would need if there were no losses in the system. The losses will depend on the diameter and length of the penstock....smaller diameter and longer mean more losses...losses are also velocity dependent.
In other words , one can't calculate the net head without putting in all the details. It can be fairly complicated for even an experienced hydraulic engineer.
I think you can see why a scale model might not have the same net head ratio as the real thing.
In other words , one can't calculate the net head without putting in all the details. It can be fairly complicated for even an experienced hydraulic engineer.
I think you can see why a scale model might not have the same net head ratio as the real thing.
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imatreyu
- Posts: 6
- Joined: Sun Jan 17, 2010 2:50 am
- Occupation: Student: 11th grade
- Project Question: Investigating the possibility of building a rain-water dependent turbine on Mt. Waialeale, Kauai
- Project Due Date: 1/20/10
- Project Status: I am conducting my experiment
Re: Finding the net head. . .
Mm.. .
Well, I was wondering if possibly you would know of a realistic ideal penstock diameter for something that must travel a fairly long distance (down the side of a mountain).
Well, I was wondering if possibly you would know of a realistic ideal penstock diameter for something that must travel a fairly long distance (down the side of a mountain).
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deleted-71827
- Former Expert
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- Project Question: Neuroregeneration
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Re: Finding the net head. . .
Hi!
I think that the penstock dimensions really depends on what particular distance or other measurements that you are dealing with. Here are a couple of links that might give you a little bit of insight into some very rough estimates of sizes-
http://smallhydro.com/tags/foundation/
http://www.malibuhydro.com/engin.htm
Hope this helps!
I think that the penstock dimensions really depends on what particular distance or other measurements that you are dealing with. Here are a couple of links that might give you a little bit of insight into some very rough estimates of sizes-
http://smallhydro.com/tags/foundation/
http://www.malibuhydro.com/engin.htm
Hope this helps!
"There is a single light of science, and to brighten it anywhere is to brighten it everywhere." -Isaac Asimov
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deleted-71588
- Former Expert
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- Joined: Mon Oct 16, 2006 11:47 am
Re: Finding the net head. . .
As an engineer, I would look at these kinds of problems a different way. What pressure and flow rate do you need at the bottom? What is the density and viscosity of the fluid? What is the static height? Calculate the zero flow static pressure and subtract the pressure need to determine how much pressure loss you can tolerate.
All engineering solutions involve a cost factor. You can always build structure to straighten out the path and you can always use larger diameter piping to achieve the solution but these come at a cost.
So what science project are you trying to do? What is your hypothesis? This will help us understand how the questions you have been asking are related to your science project so we can move this out of a general theoretical areana where things depend on lots of variables and some VERY complicated fluid dynamic modeling mathmatics with all sorts of undetermined boundary condintions to what you need to figure out for your project.
All engineering solutions involve a cost factor. You can always build structure to straighten out the path and you can always use larger diameter piping to achieve the solution but these come at a cost.
So what science project are you trying to do? What is your hypothesis? This will help us understand how the questions you have been asking are related to your science project so we can move this out of a general theoretical areana where things depend on lots of variables and some VERY complicated fluid dynamic modeling mathmatics with all sorts of undetermined boundary condintions to what you need to figure out for your project.
-Craig

