Register to reply 
Microhydro Project: Calculation Help! 
Share this thread: 
#1
Dec211, 01:09 PM

P: 9

Hi all,
Basically I need help working out how much energy we can potentially produce. We have a waterfall 50m high that has a constant year round flow. We would take off water into a 6" vertical steel pipe at the top and run it to the bottom into a turbine. I don't know the friction of the pipe but can I work this out? Solutions that find the kinetic energy of the water at the bottom is fine as I dont know the conversion rate of a typical microhydro turbine either. I do know that:



#2
Dec211, 02:14 PM

P: 820

What's your flow?
You have everything you need except that...we can't determine that, that has to do with your system design and the water system it is a part of. 


#3
Dec211, 02:28 PM

P: 688

Yes, you could Darcy–Weisbach equation:
h = f [itex]\frac{L}{D}[/itex] [itex]\frac{v^{2}}{2}[/itex] h = head loss = 50 m f = Darcy friction factor ≈ 0.015 for fully developed turbulent flow in 6" pipe per Crane TP 410 L = length of pipe ≈ 50 (should include equivalent length of other fittings and end effects) D = pipe ID (depends on pipe schedule number) = 6.065 inch for standard wall v = velocity of fluid in pipe. Solve for velocity, v. Then the volume flow is: Q = v [itex]A_{x}[/itex] [itex]A_{x}[/itex] = pipe cross sectional area = ( [itex]\pi[/itex] / 4) [itex]D^{2}[/itex] 


#4
Dec211, 03:04 PM

P: 9

Microhydro Project: Calculation Help!
So 50 = 0.015 x 50/0.154051 (inches to m) x v^2/2(9.8) gives me a solution of v = 14.19 m/s and so Q = v x pi x r^2 which is 14.19 x 0.0182 which gives me a flow rate of 0.26 m^3/s. Am I correct?



#5
Dec211, 03:33 PM

P: 9

And if I plug it into my other formula I get P = ( 260 (l/sec) * 50 * 1 )/102 and therefore I sould have a max generation capacity of 127.45kW



#6
Dec211, 07:18 PM

P: 688

mattpbarry, your numbers are good if there are no mechanical losses.
Include the [itex]e_{max}[/itex] efficiency in your above worksheet for a more realistic value. 


#7
Dec211, 10:12 PM

P: 9

Ok so if I use and emax of 60% then I get a max of 76.4kW. What is the Darcy friction factor for an 8" pipe?



#8
Dec311, 12:40 PM

P: 9

I wrote a program to calculate the efficiency of various pipe diameters. For a 100kW system which pipe diameter would you choose? Again I dont know what the rivers total flow is but its about 8m wide, 1m deep and fairly rapid.



#9
Dec311, 02:22 PM

P: 688

If you would like to calculate it, I recommend the Swamee–Jain correlation (see http://en.wikipedia.org/wiki/Darcy_f...actor_formulae). Or you can look it up on the Moody Chart. 


#10
Dec311, 03:00 PM

P: 688

Basically, when I have to "size" engineering equipment, I keep this point in mind: * Design equipment no bigger or heavier than it needs to be. In this case, the bigger pipe makes more power. But, it costs more, it is heavier (which makes the structural design more difficult). So I would establish and specify how much power you need and then choose the smallest pipe that meets this. 


#11
Dec311, 03:51 PM

Sci Advisor
HW Helper
PF Gold
P: 2,903

A simple formula for calculating power potential is given on Wikipedia here:
http://en.wikipedia.org/wiki/Hydropower where P is Power in kilowatts, h is height in meters, r is flow rate in cubic meters per second, g is acceleration due to gravity of 9.8 m/s2, and k is a coefficient of efficiency ranging from 0 to 1. (Edit: I suspect this is the same equation you have. I didn't check.) There's another reference here that looks pretty reasonable and has the same equation. http://practicalaction.org/docs/tech...ydro_power.pdf Note that the DarcyWeisbach equation is for determining irreversible pressure loss through a pipe. So if you calculate a head loss of 50 meters through the pipe, and you only have 50 meters of head pressure, the power that you could potentially get out of it is zero. A well designed system will have a very small pressure loss through the inlet piping to your turbine because that irreversible pressure loss means you get that much less power out of your turbine. The friction factor you're using, 0.015 is about right. It will change depending on Reynolds number (ie: velocity and other factors, but primarily velocity). But it looks like you're calculating how much water would come out given you've completely used up all the head pressure. I'd suggest keeping the pressure loss down to around 10% or less of the available head pressure, or select a larger pipe or smaller turbine. For a 6" pipe, I'd suggest a flow of around 0.08 m^{3}/s assuming a purely vertical pipe, 50 meters long. If your pipe isn't vertical, you need to take into account the actual length of pipe and accept that your flow rate will also need to be decreased to offset the added pipe length. Assuming a verticle pipe however, gives you an outlet pressure of about 45 meters, so put the 0.08 m^{3}/s and 45 m of head into the equation above to determine power. I'd suggest an efficiency of around 0.6 to 0.8. That should give you a ballpark estimate of how much power is available assuming you will be using a 6" pipe. If your stream has a higher flow capacity than 0.08 m^{3}/s and you want more power out of it, you can up the size of the pipe and get more flow. 


#12
Dec311, 11:50 PM

P: 688

So going back to our basic formula of the form: P = h r g k, I hope we can help the original poster adjust their code to remain general. The idea I am thinking about is how to figure out r in a manner that could be easily coded. If there were no friction, we would get ideal flow: v = √2gh. Now as you mentioned, try to limit pressure loss to 10%. So we could use 90% of the 50 m to figure out the velocity (i.e. 45 m). And the friction loss is limited to 5 m. Now, going back to the DarcyWeisbach equation: hL = f (L/D) [itex]\rho[/itex] [itex]v^{2}[/itex] / 2 We could solve for the diameter, D, to limit the head loss to 10% (5m in this case). Let me know if you agree. 


#13
Dec411, 08:07 AM

Sci Advisor
HW Helper
PF Gold
P: 2,903

Hi edgeflow,
Personally, I prefer to use a spreadsheet and would do it slightly differently. I would have as input the pipe size and length, flow rate, head available, efficiency, etc. The output is power available after flowing through the pipe, and power available given no frictional loss. Knowing the percentage of power we don't have because we lose it to frictional losses helps to design the system more efficiently. If a spreadsheet was set up to calculate  head loss through the pipe  power available at the end of the pipe  power available assuming no frictional losses you can then goal seak on the output and have it change the input to match whatever output you desire. So you could goal seak on either diameter, length or flow rate to determine the output. I guess there's a few ways to do this, just comes down to what is easiest and most efficient for the OP. 


#14
Dec1311, 08:17 AM

P: 6

Hi edgeflow,
Its been a long time since I figured this stuff out... but It strikes me that solving for max flow through the pipe is not necessary the correct approach. (Please forgive me if this was already dealt with) Would you not have a nozzle at the end of your penstock that would minimize head loss, while maximizing velocity at nozzle (which is all that matters) prior to hitting your pelton wheel (insert turbine of your choice here). Thus you will need to balance nozzle diameter and pipe diameter until you arrive at your best return on the dollar. Once you have the formulas figured out, this should be easily solved with a spreadsheet (whatif). I have seen in several real world situations where they started with a large pipe and dropped to smaller and smaller diameter pipes as the pressure built up allowing for faster velocity as a method of cost savings while getting much better results than just sticking with the smaller pipe the whole way. 


Register to reply 
Related Discussions  
Rainfall Driven Micro HydroGenerating System  Introductory Physics Homework  1  
Rainfall driven micro hydro system  Introductory Physics Homework  1  
Micro Hydro Design Help  Engineering Systems & Design  2  
Micro Processor project making a Gun Chronograph  Electrical Engineering  4 