| New Reply |
Micro-hydro Project: Calculation Help! |
Share Thread |
| Dec2-11, 01:09 PM | #1 |
|
|
Micro-hydro Project: Calculation Help!
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 micro-hydro turbine either. I do know that: Code:
Pmax = Qmax*Hmax*emax / K
Where:
Pmax=Maximum Power Available (kW)
Qmax=Flow (Volume/time)
Hmax=Head (Vertical drop in m)
emax=Efficiency of the turbine (use a value of 1 for max power available)
K=Unit conversion factor (see table below for some common values)
For Q measured in K is equal to
ft3/min 708 (ft4)/(min*kW)
ft3/sec 11.8 (ft4)/(sec*kW)
l/sec 102 (l*ft)/(sec*kW)
gal/min 5302 (gal*ft)/(min*kW)
|
| Dec2-11, 02:14 PM | #2 |
|
|
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. |
| Dec2-11, 02:28 PM | #3 |
|
|
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] |
| Dec2-11, 03:04 PM | #4 |
|
|
Micro-hydro 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?
|
| Dec2-11, 03:33 PM | #5 |
|
|
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
|
| Dec2-11, 07:18 PM | #6 |
|
|
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. |
| Dec2-11, 10:12 PM | #7 |
|
|
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?
|
| Dec3-11, 12:40 PM | #8 |
|
|
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.
Code:
Diameter: 5" Friction Factor: 0.018 Flow Rate: 148.96765539522767 l/sec 10% 7.30233604878567 kW 20% 14.60467209757134 kW 30% 21.90700814635701 kW 40% 29.20934419514268 kW 50% 36.511680243928346 kW 60% 43.81401629271402 kW 70% 51.11635234149969 kW 80% 58.41868839028536 kW 90% 65.72102443907103 kW 100% 73.02336048785669 kW Diameter: 6" Friction Factor: 0.015000000000000001 Flow Rate: 257.4161085229533 l/sec 10% 12.618436692301634 kW 20% 25.236873384603268 kW 30% 37.8553100769049 kW 40% 50.473746769206535 kW 50% 63.09218346150817 kW 60% 75.7106201538098 kW 70% 88.32905684611144 kW 80% 100.94749353841307 kW 90% 113.5659302307147 kW 100% 126.18436692301634 kW Diameter: 7" Friction Factor: 0.012857142857142857 Flow Rate: 408.7672464045047 l/sec 10% 20.037610117867878 kW 20% 40.075220235735756 kW 30% 60.11283035360363 kW 40% 80.15044047147151 kW 50% 100.18805058933938 kW 60% 120.22566070720725 kW 70% 140.26327082507513 kW 80% 160.30088094294302 kW 90% 180.3384910608109 kW 100% 200.37610117867877 kW Diameter: 8" Friction Factor: 0.01125 Flow Rate: 610.1715164988525 l/sec 10% 29.910368455826106 kW 20% 59.82073691165221 kW 30% 89.7311053674783 kW 40% 119.64147382330442 kW 50% 149.5518422791305 kW 60% 179.4622107349566 kW 70% 209.37257919078272 kW 80% 239.28294764660885 kW 90% 269.1933161024349 kW 100% 299.103684558261 kW Diameter: 9" Friction Factor: 0.01 Flow Rate: 868.7793662649676 l/sec 10% 42.58722383651803 kW 20% 85.17444767303606 kW 30% 127.76167150955406 kW 40% 170.3488953460721 kW 50% 212.93611918259012 kW 60% 255.52334301910813 kW 70% 298.11056685562613 kW 80% 340.6977906921442 kW 90% 383.2850145286622 kW 100% 425.87223836518024 kW Diameter: 10" Friction Factor: 0.009 Flow Rate: 1191.7412431618213 l/sec 10% 58.41868839028536 kW 20% 116.83737678057072 kW 30% 175.2560651708561 kW 40% 233.67475356114144 kW 50% 292.09344195142677 kW 60% 350.5121303417122 kW 70% 408.93081873199753 kW 80% 467.3495071222829 kW 90% 525.7681955125682 kW 100% 584.1868839028535 kW Diameter: 11" Friction Factor: 0.008181818181818182 Flow Rate: 1586.2075946483837 l/sec 10% 77.75527424746979 kW 20% 155.51054849493957 kW 30% 233.26582274240937 kW 40% 311.02109698987914 kW 50% 388.77637123734894 kW 60% 466.53164548481874 kW 70% 544.2869197322885 kW 80% 622.0421939797583 kW 90% 699.7974682272281 kW 100% 777.5527424746979 kW Diameter: 12" Friction Factor: 0.007500000000000001 Flow Rate: 2059.3288681836266 l/sec 10% 100.94749353841307 kW 20% 201.89498707682614 kW 30% 302.8424806152392 kW 40% 403.7899741536523 kW 50% 504.73746769206537 kW 60% 605.6849612304784 kW 70% 706.6324547688915 kW 80% 807.5799483073046 kW 90% 908.5274418457176 kW 100% 1009.4749353841307 kW Diameter: 13" Friction Factor: 0.006923076923076922 Flow Rate: 2618.2555112265213 l/sec 10% 128.34585839345692 kW 20% 256.69171678691384 kW 30% 385.03757518037077 kW 40% 513.3834335738277 kW 50% 641.7292919672847 kW 60% 770.0751503607415 kW 70% 898.4210087541985 kW 80% 1026.7668671476554 kW 90% 1155.1127255411125 kW 100% 1283.4585839345693 kW Diameter: 14" Friction Factor: 0.0064285714285714285 Flow Rate: 3270.1379712360376 l/sec 10% 160.30088094294302 kW 20% 320.60176188588605 kW 30% 480.902642828829 kW 40% 641.2035237717721 kW 50% 801.5044047147151 kW 60% 961.805285657658 kW 70% 1122.106166600601 kW 80% 1282.4070475435442 kW 90% 1442.7079284864872 kW 100% 1603.0088094294301 kW Diameter: 15" Friction Factor: 0.005999999999999999 Flow Rate: 4022.1266956711474 l/sec 10% 197.16307331721313 kW 20% 394.32614663442627 kW 30% 591.4892199516393 kW 40% 788.6522932688525 kW 50% 985.8153665860655 kW 60% 1182.9784399032785 kW 70% 1380.1415132204916 kW 80% 1577.304586537705 kW 90% 1774.4676598549179 kW 100% 1971.630733172131 kW |
| Dec3-11, 02:22 PM | #9 |
|
|
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. |
|
| Dec3-11, 03:00 PM | #10 |
|
|
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. |
|
| Dec3-11, 03:51 PM | #11 |
|
|
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 Darcy-Weisbach 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 m3/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 m3/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 m3/s and you want more power out of it, you can up the size of the pipe and get more flow. |
| Dec3-11, 11:50 PM | #12 |
|
|
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 Darcy-Weisbach 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. |
|
| Dec4-11, 08:07 AM | #13 |
|
|
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. |
|
| Dec13-11, 08:17 AM | #14 |
|
|
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 (what-if). 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. |
| New Reply |
Similar discussions for: Micro-hydro Project: Calculation Help!
|
||||
| Thread | Forum | Replies | ||
| Rainfall Driven Micro Hydro-Generating 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 | ||
Quote by mattpbarry
