Prop-to-Alternator Mis-Match, an Extreme Case

[SparWeb]

Matt,
Your idea still interests me.  It would look ugly, but...

You can get extra chord and twist by adding something like this:





Something like a strake made of sheet metal that extends the leading edge.
I arbitrarily ended the "strake" at 1/2 span so that it's not spinning too fast (but still quite fast).
It would need to be attached very securely (not easy to do through fiberglass) and sealed around the edges.
Probably filled with foam to keep the surface from bulging in and out.
Really hard to predict how effective it would be.  I still think it's a neat idea.
Thank you

[MattM]

Trapezoidal extension were underwhelming.  I'd leave them as simple rectangles.  The simple shape flattens under pressure, making it self adjusting.  The wider the less flex.

[Adriaan Kragten]

Matt,
Your idea still interests me.  It would look ugly, but...

You can get extra chord and twist by adding something like this:

(Attachment Link)

(Attachment Link)

Something like a strake made of sheet metal that extends the leading edge.
I arbitrarily ended the "strake" at 1/2 span so that it's not spinning too fast (but still quite fast).
It would need to be attached very securely (not easy to do through fiberglass) and sealed around the edges.
Probably filled with foam to keep the surface from bulging in and out.
Really hard to predict how effective it would be.  I still think it's a neat idea.
Thank you

In my new note: "Comparing starting torques of constant chord and tapered blades" I have taken a tapered and twisted blade and divided the blade into five sections of each the same length. I have calculated the contribution of each section to the total starting torque and I found that the contribution of the inner section is only about 15 %. This was the case for a blade with a very wide inner blade section with very large blade angles. The reason is that the average radius of this section is rather small. So if the starting torque of a certain rotor at a certain wind speed is really too low for a certain generator, increasing of the chord and blade angle of the inner blade section won't help a lot.

There are only two correct ways to solve the problem.
 
1) If the rotor diameter is maintained, one has to design a new rotor with a lower design tip speed ratio. This new rotor will get wider blades with larger blade angles and the starting torque coefficient of this rotor will increase about quadratic to one divided by the decrease of the design tip speed ratio. Adding more of the same blades gives a higher starting torque coefficient but if the solidity increases, the design tip speed ratio must decrease and this means that you need larger blade angles. You don't get larger blade angles if you simply add more of the same blades. Another problem might be that the Reynolds value decreases for more blades running at a lower tip speed ratio so it is better to take three blades with a wide chord than five blades with a small chord.

2) Increase of the rotor diameter results in increase of the torque level with a factor R^3 and this is also the case for the starting torque coefficient if the tip speed ratio is maintained. Adding spacers in between the hub and the blade root of existing blades to increase the rotor diameter makes that the blade angles will be wrong especially at the blade root. Small spacers might be allowed but one should check the blade geometry using the aerodynamic theory as given in KD 35 to find out what spacer length is allowed without getting much too large blade angles at the blade root.

[wbuffetjr1]

We measured line to line.

[MattM]

If I wanted best performance from an extension I'd find a guy with a computerized Autobrake and go for mounting out on the outer section blade with a more complicated shape like this....
But I am 1,500 miles away from the old shop and its in new hands these days.  Otherwise I'd have loved to send you free samples.  The 'ogee' shape looks odd but it was discovered by pure accident while trying out different more conventional shapes. More bends in metal give it more stiffness at a perpendicular angle which is why my brother and I originally stumbled across the shape.

I like both shapes, though.  During experiments I could cram 24 of the simple blade shapes into a 12" diameter rotor plate.  It was tough to squeeze 8 into the same rotor plate using the more complicated shape due to their inflexible design.  I was using them supported only by the rotor and fatigue of sheet metal became an issue just outside the rotor plate due to blade twist. 

Blades that aren't allowed to twist last indefinitely.  The heaviest blades from 16 gauge had the best lifespan of just over a year for 30 inch blades when unsupported.  We experimented with up to ten foot long and one foot wide blades, but the forces at that point were scary so we scaled down to 6-7 foot diameter targets.  We were leaving them free spinning through 50+ mph winds.  if you are supporting the blade at the mounting point then fatigue isn't going to be an issue.

Unsupported metal blades eventually crack and either bend or fall off.  I never had a single supported blade wear out.  Interesting that the ones that fail always failed in low winds and never in high ones.  Metal loves tension and failure seems to come when the tension is removed.  People here on the forum were saying the sheet metal blades were going to fly off but it never happened.  Most of the time the blade simply deformed and it could no longer rotate due to the imbalance.  Every unsupported blade that fell off merely landed at the tower base.  I was kind of disappointed.

[SparWeb]

All good info Adriaan, thanks.
...
Cutin Speed = 39.6 VAC * 100 RPM / 18 VAC = 220RPM

I doubt this calculation, at least if the given voltage at 100 rpm is the DC voltage after the rectifier. The open DC voltage is proportional to the rotational speed. So if the open DC voltage is 18 V at 100 rpm and if it is assumed that the battery voltage including the voltage drop over the rectifier diodes is 55 V, it means that the required rotational speed for an open DC voltage of 55 V is 100 * 55 / 18 = 305.6 rpm.

If the given voltage is the AC voltage in between two of the three wires of a winding which is connected in star, the AC phase voltage in between the star point and a wire end is a factor square root of 3 lower and so 10.39 V at 100 rpm. If the 3-phase winding is rectified in star, the DC voltage is a factor 2.339 higher (see KD 340 formula 13) and so 24.3 V at 100 rpm (if the voltage drop over the rectifier is neglected). So for a voltage of 55 V you need a rotational speed of 100 * 55 / 24.3 = 226.3 rpm. This is only a little higher than what you have calculated. I think that the difference is caused by the fact that you have neglected the factor 0.955 out of formula 13. The relation in between the DC voltage and the AC phase voltage including the voltage drop over the rectifier, is given by formula 14 of KD 340.

If 18 V is the AC phase voltage at 100 rpm, I find that the DC voltage = 18 * 1.7321 * 1.4142 * 0.955 - 1.4 = 40.71 V at 100 rpm.

I've been pondering this response since you posted it.  I really don't think it needs to be complicated.
One thing that was uncertain as you were typing was the exactly connection and measurement of the stator voltage
Since it's in star and the measurements were on the line ends, that's been cleared up and the highlighted part of your calculation is the one that applies.  But I can't make sense of the calculation.

I measure this result when I do the test in person:  https://youtu.be/WHzkbGx5wZk

In the test video, I measure 19.4 VAC and 27.7 VDC.  The conversion is about 1.4.
For a 48V system rather than 24, then I would see about 38.8 VAC and 55.4 VDC.

I would rather go with a conversion that represents what I really measure.  I have reason to trust my instruments.

I'm quite certain that with a 18V/100 RPM stator in star, the 48V cut-in speed will be about 220 RPM.

[Adriaan Kragten]

The point is what you call the phase voltage. In my definition of KD 340, I call the phase voltage the effective AC voltage Ueff of one phase and I give the formula for the effective DC voltage UDCeff if you know the phase voltage. If the winding is connected in star and if the star point is hidden somewhere inside the generator, you can only measure the AC voltage between two of the three phases. This voltage is a factor square root of 3 higher than the phase voltage. So if you use this voltage, the DC voltage will be a factor square root of 3 lower than when you would have measured the phase voltage. But even if you do so, you can't neglect the factor 0.955 because if you neglect this factor you get the peak value of the fluctuation of the DC voltage and not the effective DC voltage. If you take any book of electro-mechanics, you will find the same formula to transform AC into DC as I used in KD 340.

[JW]

Can someone specify this is not RMS voltage.

[Adriaan Kragten]

All good info Adriaan, thanks.
...
Cutin Speed = 39.6 VAC * 100 RPM / 18 VAC = 220RPM

I'm quite certain that with a 18V/100 RPM stator in star, the 48V cut-in speed will be about 220 RPM.

If you follow my formulas you find the following value of UDCeff for n = 220 rpm.

If the AC voltage in between two phases is 18 V at 100 rpm for a winding connected in star, the AC phase voltage is a factor square root of 3 = 1.7321 lower and so Ueff = 10.392 V at n = 100 rpm. Ueff at 220 rpm is a factor 220 / 100 = 2.2 higher. So Ueff  = 22.863 V AC at n = 220 rpm. If you substitute this value in formula 14 of KD 340 you find that UDCeff = 0.955 * 1.4142 * 1.7321 * 22.863 - 1.4 = 52.1 V. This is about the open voltage for a 48 V battery if it has just been charged. So it is right that the cut in wind speed is about 220 rpm and your measurements are in accordance to the theory.

[SparWeb]

Hi JW,
All of my measurements are V-RMS.  I've been using true-RMS meters ever since I found out that simpler meters are confused when the waveform is no a pure sine. 
I think I know what you're getting at.  The DC voltage clamps the peaks of the AC, which would be a sine-wave, except that the DC voltage flattens the tops.

Hi Adriaan,
OK that seems to be cleared up.  It was more convenient for me to simply take measurements in-situ.

[SparWeb]

Firstly, I want to thank Chad (WBuffetJR) for inviting me to look at this more closely.
I think this stuff is fun, and it's been a chance for me to think carefully about blade designs that I normally don't think about.  5 blades instead of 3.  TSR less than 4 rather than between 5 or 6.

I have gone back to the Blade Element Analysis and used it to work out blade designs to suit WBuffetJR's alternator.  I've worked through several iterations for comparison:

1   9 feet      3 blades   TSR 3.8
2   9 feet      5 blades   TSR 3.8
3   10 feet     3 blades   TSR 3.8
4   10 feet     3 blades   TSR 4.2
5   10 feet     5 blades   TSR 4.2


Here's an example of how the calculations went for each blade.  Simplified using the 3/4 span point for reference:

Tip Radius = 4.5 feet
3/4 Radius = 3.4 feet
Root chord = 8.0 inches = 0.67 ft
3/4R chord = 4.5 inches = 0.38 ft

Slope of the lift curve = 0.090 1/degree
angle of zero lift = -2 degrees
incidence of blade twist = 7 degrees

Now let's suppose the wind is 30 kph, and it's turning at 220 RPM.
This is a point on the graph below, so you can check my work.

220 RPM = 23 radian/sec
30 kph = 27.3 ft/sec

The tip speed ratio is:

TSR = (4.5 ft) * (23 rad/sec) / (27.3 ft/sec) = 3.8
This TSR matches the design TSR of the blades.

The angle of attack of the blade:

AoA = arctan (1/TSR) = arctan (1/3.8 ) = 14.7 degrees

At the 3/4 radius point:
AoA = arctan [1/(TSR*3/4)] = arctan [1/3.8*3/4)] = 19.3 degrees

Account for the "inflow factor" on the angle of attack

AoA = arctan [1/TSR*3/4*(1+0.33)/2] = 13.2 degrees

This gets the coefficents of lift and drag at the point 3/4 out the radius:

CL = 0.090 1/degree * (13.2 + 2.0 - 7.0) = 0.74
CD = 0.020 + (13.2 + 2.0 - 7.0)^2 / 2.1 = 0.030

We also need the density of air now.  This is where the altitude is a penalty:
rho, sea level = 0.002378 slug / cubic foot
rho, 10,000 ft = 0.001800 slug / cubic foot

This information can give the forces acting at this point on the radius:

The local airspeed at this point on the radius is:

V,local,3/4 = 3.4 feet * 23 radian/sec = 78 ft/sec

dL = (0.001800 slug / cubic foot) / 2 * (78 ft/sec)^2 * (3.4 ft) * (0.38 ft) * (0.74)
dL = 5.1 pounds/ft

dD = (0.001800 slug / cubic foot) / 2 * (78 ft/sec)^2 * (3.4 ft) * (0.38 ft) * (0.030)
dD = 0.20 pounds/ft


The lift creates useful torque, while the drag mostly resists it.  The forces are applied at angles to the axis of rotation. Some trigonometry is needed to find the torque.

Torque, lift = 3 blades * dL * R * 0.75 * sin(AoA) = 3 * 5.1 Lb/ft * 4.5 ft * sin(13.2deg) = 15.6 Lb*ft
Torque, drag = 3 blades * dD * R * 0.75 * cos(AoA) = 3 * 0.2 Lb/ft * 4.5 ft * cos(13.2deg) =  2.7 Lb*ft

Net Torque = 15.6 - 2.7 = 12.9 Lb*ft = 17.4 N*m

Power = Torque * speed = 17. N*m * 23 radian/sec = 402 Watt

[SparWeb]

The previous calculations found a point where there is 30 kph (20 mph) wind blowing, making the blades turn at 220 RPM and they will produce 400 Watts of mechanical power at the shaft. 

To check the match with the alternator, we can look at what the alternator needs to make it turn at 220 RPM.  I worked that out before, and got about 100 Watts.

So if there's 400 Watts available, but only 100W being used, then the blades will accelerate to a higher speed.  In a 30 kph wind the 9' / TSR=3.8 blades will settle at a speed about 300 RPM.  They seem to be matching up better with this configuration of blade.

With this as my starting point, I plotted the rest of the points and I arrived at this graph:

[SparWeb]

Compared to the blades that WBuffetJR has now, it's getting close, but not yet right.
At the point I illustrated before, the match is possible, but at higher speeds, the blades go into a pretty deep stall. 
The power curve for the alternator seems to follow the sloping line for the 50 kph blade line, so it's hard to say where they'll balance but what it means is that the blade needs a very high angle of attack and will probably be stalled.

We had some discussion of adding blades, so here's how those curves change with 5 blades, keeping all other things equal.

[SparWeb]

That works even better. 
Another way to improve the match is to keep 3 blades, and increase the diameter to 10 feet.
That results in this:

[SparWeb]

That seems to hit the sweet spot.  The blade power curves have also shifted up a lot, offering much more mechanical power to the alternator for any given wind speed.  My first pass gave about 1.5 kW in a 40 kph (24 mph) wind, but I have later found a blade configuration that provides more than 2.3 KW at the shaft.

Going forward with this design, I can predict the output power curve that this can produce:

[SparWeb]

Going back to Hugh Piggott's blade shape spreadsheet:

[SparWeb]

I've made some changes to Hugh's spreadsheet.  It normally just plots a theoretical "best" chord, but the TSR=3.8 is so low, the theoretical chord is very very wide - almost 20 inches.
The work I did above shows that a 12 inch chord will be fine, so I put in a modification to the spreadsheet to allow the chord to be limited.

Even with that modification, the blades are pretty hard to make.  The solid laminated blank will have to measure 12" X 9" X 60".  The blades I carved this summer only measured 9" X 6" X 60".  Each of my blades weighed almost 10 pounds, and with the hub the whole assembly was about 35 pounds.  The blade set for Chad's machine may be close to 50 pounds.

This is a challenge to deal with, and I need to ask WBuffetJR if he thinks the axle shaft from his generator can handle it.