Micro-hydro Project: Calculation Help

[mattpbarry]

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 don't know the conversion rate of a typical micro-hydro turbine either.

I do know that:

P[SUB]max[/SUB] = Q[SUB]max[/SUB]*H[SUB]max[/SUB]*e[SUB]max[/SUB][SIZE="5"] / K[/SIZE]


Where:

    P[SUB]max[/SUB]=Maximum Power Available (kW)
    Q[SUB]max[/SUB]=Flow (Volume/time)
    H[SUB]max[/SUB]=Head (Vertical drop in m)
    e[SUB]max[/SUB]=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)

If that helps. Thanks in advance. If you need any more data I'll try find it.

 

[Travis_King]

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.

 

[edgepflow]

Yes, you could Darcy–Weisbach equation:

h = f $\frac{L}{D}$ $\frac{v^{2}}{2}$

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 $A_{x}$

$A_{x}$ = pipe cross sectional area = ( $\pi$ / 4) $D^{2}$

 

[mattpbarry]

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?

 

[mattpbarry]

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

 

[edgepflow]

mattpbarry, your numbers are good if there are no mechanical losses.

Include the $e_{max}$ efficiency in your above worksheet for a more realistic value.

 

[mattpbarry]

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?

 

[mattpbarry]

I wrote a program to calculate the efficiency of various pipe diameters. For a 100kW system which pipe diameter would you choose? Again I don't know what the rivers total flow is but its about 8m wide, 1m deep and fairly rapid.

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

 

[edgepflow]

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?

Crane Technical Paper 410 has a nice table of Darcy Friction factors for turbulent flow in steel pipes. Unfortunately, I keep this book at work or I would look it up for you.

If you would like to calculate it, I recommend the Swamee–Jain correlation (see http://en.wikipedia.org/wiki/Darcy_friction_factor_formulae).

Or you can look it up on the Moody Chart.

 

[edgepflow]

I wrote a program to calculate the efficiency of various pipe diameters. For a 100kW system which pipe diameter would you choose? Again I don't know what the rivers total flow is but its about 8m wide, 1m deep and fairly rapid.

Your code can help you make a real world decision.

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.

 

[Q_Goest]

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/technical_information_service/micro_hydro_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.

 

[edgepflow]

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/technical_information_service/micro_hydro_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.

You bring up a good point about a 50m of friction head loss leaving no energy available for doing work. This idea started bugging me lately.

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) $\rho$ $v^{2}$ / 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.

 

[Q_Goest]

Hi edgeflow,

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.

I agree - the next step here is how best to code this. You could solve for diameter as you suggest. That means you would need to input some percentage of head loss expected through the pipe just as you say.

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.

 

[flyboy_1234]

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.