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More on heating
By SmoggyTurnip, Section Wind Posted on Mon Mar 06, 2006 at 03:24:41 PM MST
MPPT - is it really worth it?

There is allot of discussion about wind turbines for use in heating lately. I am happy about this because that is exactly what I am doing. Before I started the project I thought that some sort of MPPT control would be a great benefit but now I am not so sure. The following graphs are the results of my calculations and I am making allot of assumptions here. I have not yet completed my gennie and in the design process I found that I had to "guess" about allot of things when it comes to the behavior of blades and wind. The graphs on the right hand side in the following pics are my guess for the power curves for 5.5 meter blades with a design TSR of 7. Based on my limited research (mostly this board) I assumed that the nonloaded TSR would be double that of the design TSR as that seems to be what most people have found. In order that the calculations would be easy I gave the power curves the exact shape of a parabola. From all of the power curves I have ever seen (not many) this approximation seems reasonable to me. Shown on the power curve graphs are also the generator power curves. The red line is the power in the stator, the black line is the power in the load and the other line is the sum or the 2. The first graph is for a load of 24.5 ohms on each phase. The second graph is for a load or 44 ohms on each phase. Now the left hand graphs show the actual power output by taking the intersection of the gennie total power output and the power available at the blades. Again the red line is the power in the stator, The black line is the power in the load. Since I assumed a parabola for the power curves it was not too difficult to set up an equation to give the resistance value for a particular power output. Doing a bit of calculus - taking the derivative and setting to zero I was able to calculate the load resistance that maximized the power output for each wind velocity. That resistance value is the curved black line that opens to the right. So for example in a 28 mph wind the best resistance value is around 20 ohms and in a 12mph wind the best load is just above 40 ohms. The bright green line is the power that would be produced in the load when using the best resistance value. i.e. that would be the power in the load if the ideal MPPT circuit was used. The lighter green line is the power that would be used in the stator using the best load. The first set of graphs was for the situation when a 24.5 ohm load was used. This is the ideal load for the higher wind speed. The second set of graphs assumed a 44 ohm load this is the best resistance value at around 12 mph wind. I was very surprised to see that by picking a load that maximized the output power at 28 mph the output power was very close to being maximized at all lower wind speeds as well. It seems that by lowering the resistance the speed of the turbine is reduced bringing it closer to the ideal load but more than 95% of the extra load on the blades comes from within the stator. You can see from the first graph that the green line (MPPT output) is almost an exact copy ofthe black line (fixed load maximized for 28mph). To me this says that MPPT is not worth the effort. What do you think?

More on heating | 26 comments (26 topical)

Re: More on heating (3.00 / 0) (#1)
by Nando on Mon Mar 06, 2006 at 09:39:38 AM MST

Unhappily I can not follow your graphs since I do not see any colors in my PC.

To really get MPPT going either with the wind mill or the computer, one needs to properly set the system parameters, which I could not see well in your dissertation.

Take the standard formula for the wind mill versus wind velocity and generate the wind power curves.

Setup the generator to produce, let's say @ peak RPM a high voltage at least 5 times the battery voltage.

Define the internal resistance of the wind mill windings and calculate a % loss in the generator that is around 10 % the output energy.

Plot both curves or at least the output power curve with a load resistance that varies following the output power curve.

Then replace the variable load with a battery bank, the same you had for the 5 times the output voltage of the wind mill.

Using the generator parameters with a battery load, generate the output power curve.

Compare both curves for efficiency power transfer - that is the MPPT gain.

Good for you that you can handle high level math. -- congratulations.

TSR does not enter in the calculations -- TSR is just an adaptation to be able to match the wind to the generator.

Name each graph to be able to follow it, sometimes one insert several graphs and they appear in different order as one expected.

Make the traces with different styles for easy viewing, if possible.

I could tell the load variation, but I prefer that you find it when the proper parameters are implemented.

Output power versus wind velocity or the equivalent RPM of the generator.

For the type of graphs you have, it would be better to place one by one and a bit bigger for easy reading.

You will discover that MPPT is WORTH the effort, as a fellow in UK found out when he placed his wind mill to harvest MPPT energy for heating, I have been assisting him for several months now, with about 900+ KW-hr/month of harvested energy to keep is farm house warm.

Nando

Re: More on heating (3.00 / 0) (#3)
by SmoggyTurnip on Mon Mar 06, 2006 at 10:23:25 AM MST

Hi Nando, I am sorry you don't have a color monitor. Thanks for the response. Here is my reply: Nando: To really get MPPT going either with the wind mill or the computer, one needs to properly set the system parameters, which I could not see well in your dissertation. Smoggy: I have no idea what you are talking about here. What are the "system parameters" I why do I have to set them. Nando: Take the standard formula for the wind mill versus wind velocity and generate the wind power curves. Smoggy: Again I don't know what you are talking about. What is the standard formula you mention? Nando: Setup the generator to produce, let's say @ peak RPM a high voltage at least 5 times the battery voltage. Smoggy: There is no battery in the system I am talking about. It is strictly resistive heating. No need for DC - Conversion just adds loses. Nando: Define the internal resistance of the wind mill windings and calculate a % loss in the generator that is around 10 % the output energy. Plot both curves or at least the output power curve with a load resistance that varies following the output power curve. Smoggy: Thats pretty much what the graphs shows. Nando: Then replace the variable load with a battery bank, the same you had for the 5 times the output voltage of the wind mill. Smoggy: No batteries here. Nando: Using the generator parameters with a battery load, generate the output power curve. Compare both curves for efficiency power transfer - that is the MPPT gain. Smoggy: Pretty much what I did except for resistive load instead of battery. Nando: Good for you that you can handle high level math. -- congratulations. Smoggy: Thanks. Nando: TSR does not enter in the calculations -- TSR is just an adaptation to be able to match the wind to the generator. Smoggy: Matching the wind to the generator/load is what it is all about. TSR matters greatly here. Nando: Name each graph to be able to follow it, sometimes one insert several graphs and they appear in different order as one expected. Make the traces with different styles for easy viewing, if possible. Smoggy: I did the best I could with the graphs, I wish I could get them to show up better. Nando: I could tell the load variation, but I prefer that you find it when the proper parameters are implemented. Smoggy: I guess I just don't know what you mean here. Nando: You will discover that MPPT is WORTH the effort, as a fellow in UK found out when he placed his wind mill to harvest MPPT energy for heating, I have been assisting him for several months now, with about 900+ KW-hr/month of harvested energy to keep is farm house warm. Smoggy: Do you have any data to show how much extra power he is getting because of the MPPT. Maybe some nice graphs properly placed and explained :) Anything at all to convince me. When increasing the load to the generator the generator always becomes less efficient and if the blades are operating above design TSR they will become more efficient. What I am tring to show is that the added efficiency of bringing the blade rpm into a more efficient TSR is almost completely lost to the added inefficiency of the generator. I know that there are people who will disagree but I would like them to show me some data or reasoning that would support it. Just to say that you have added MPPT to a system and an improvement was made is not that convincing as I have no idea what the before and after pictures actually were.
The sooner you start the longer it takes.
[ Parent ]

Re: More on heating (3.00 / 0) (#11)
by Nando on Mon Mar 06, 2006 at 02:25:44 PM MST

The problem is not the monitor but me, my eyes.

IT SEEMS if I read right that you have a fixed load on each phase, RIGHT ?

If so, MPPT will not be measured

Read again my message, you need to use each measured voltage at each air velocity point, get 10 % of the generated voltage and with it and the (phase winding resistance * 1.73), calculate the current, using this current with the 90 % voltage gives you the output energy and this energy represents a load ( heater ) with a resistive value and the next wind velocity you do the same to obtain another resistive value specific to that new wind velocity point and so ON, so the resistance LOAD is NOT FIXED it is VARIABLE following the wind energy.

Compare that energy with a fixed resistor value Like equal to the generator winding resistance.

If done properly, the MPPT may represent 85+ % power harvesting versus yours fixed value of around 50 %

With MPPT you may dissipate 10 % in the generator and 90 % in the external load

With a fixed load the efficiency is defined by the load value and the generator resistive value --

There is another loss that is the bad gap rotor-to-stator but that can be left for
later discussions.

TSR if the generator and load are MPPT matched the TSR is "FIXED" and that is just a ratio of wind velocity and tip speed for proper generator rotational needs to produce the needed power under the wind velocity and the cube power law.

The project in UK does not have data to be able to show what you want; it just shows the power harvesting versus wind velocity with certain problems due this to the wind mill delay to respond to the wind, so many measurements show very low TSR and some VERY HIGH, like 12 to 15 -- the data needs purging of many points to have a logical wind mill response.
I have inserted a European sold Chinese generator that in past days I wrote an explanation of how to obtain at least 85 % efficiency versus the around 50 % on the chart

Nando

[ Parent ]

Re: More on heating (3.00 / 0) (#13)
by SmoggyTurnip on Tue Mar 07, 2006 at 07:08:58 AM MST

Hi Nando,

This is what you said:
Read again my message, you need to use each measured voltage at each air velocity point, get 10 % of the generated voltage and with it and the (phase winding resistance * 1.73), calculate the current, using this current with the 90 % voltage gives you the output energy and this energy represents a load ( heater ) with a resistive value and the next wind velocity you do the same to obtain another resistive value specific to that new wind velocity point and so ON, so the resistance LOAD is NOT FIXED it is VARIABLE following the wind energy.

Yes this is what MPPT is.
This is what the geen lines on the left hand graphs are showing.
Since you and others are having trouble with the colors
I will try to explain the graphs verbially.

Graph #1 (Top left)

Bottom axis is wind velocity (0 to 28 mph).
Left axis is Power in watts.
Right axis is Resistance.

There are 5 curves on this graph. (it may only look like 4
because two of the curves are almost identical.

There are 2 curves that increas with wind velocity and
end up at the top right hand corner of the graph. They
are almost identical. (One is green and one is black for
the people who can see color) These curves are to be
read from the left vertical axis representing power.
The first curve is the power generated in the FIXED load
of 24.5 ohms. The second curve is the power generated in
a load that is changed using MPPT. The resistive value
of the MPPT load is shown by the curve that decreases with
wind velocity. So for example on the graph at 6 mph wind
the MPPT load is 60 ohms and the MPPT resistance at 28 mph
is around 20 ohms. There are 2 other curves that start
at 0 and increase with wind velocity. The one that ends
up at the lowest point at 28 mph (Red for those that get
color) is the power in the stator ( to be read from left
vertical axis) with the fixed load. The other is the
power in the stator when using the MPPT load.

Now if you pick an arbitray wind speed say 15 mph and
look at the power curves you will see that the power
in the load is nearly the same for the fixed load of 24.5
ohms and the MPPT load of about 30 ohms.

In fact at any wind speed the power in the load is
almost the same for the MPPT load and the fixed load.

Now I am pretty certain that my math is correct as I
have been working on this for months now and have checked
rechecked and rechecked it in many different ways, mostly
because I was so surprised by the results.

So if the math is right but the result is wrong then there
must be something wrong with my power curves, which I admit
are just a creation of my own to make the math easier, but
I think that the general shape is somewhat realalistic, and
if it is anywhere close to realistic then the effect that I
am pointing out must show up to some extent when using MPPT.

The effect can be summed up by saying that some of the gains
in bringing the blades into a more efficient rpm are lost in
both the lower rpm of the generator and the lower efficiency.

MPPT would not be worth the effort if these gains and losses
were equal, which is why I am asking for someone to offer me
some data or logic to support the idea of MPPT.

Saying it is so does not make it so.

I don't have the time to explain the other graphs right now,
maybe later.

Have a great day.
The sooner you start the longer it takes.
[ Parent ]

Re: More on heating (3.00 / 0) (#2)
by Flux on Mon Mar 06, 2006 at 09:52:57 AM MST

I don't think you need mppt but I do think that you will have to make a much better attempt at it than using constant resistance.

I, like most other people probably, have no true idea of the prop power curve shape, but one thing I have noticed is that in low winds the thing is happy to run above design tsr to some extent. Tsr 7 may run up to 8 in low winds. In high winds I find that things work better below design tsr. Although it will still produce useful power at tsr 8 it would be better at 6.

The big issue for me is not the power output but noise. from the look of your curves it seems as though you will be up at about 10 in high winds. For all the effect it has it might as well be running away light, the noise will be frantic.

If you set the load in high wind to optimum tsr you will stall in the lower winds.
I just don't think that you can even cover a 2:1 wind speed range with fixed resistance and have half decent results without serious noise in high wind.

If the prop tracked at constant tsr then you would have four times increase in power with a double in wind speed. The power goes up twice this so you are down in runaway. Added to this is the tendency for the prop tsr to want to fall and I think you are way off mark.

Without some attempt at mppt battery charging is inefficient but quiet and pleasant, most of us can live with that for the type of load in use. For heating you need the decent efficiency to make it worthwhile and if you insist on a fixed load then you will have a lot of cold days or sleepless nights.
Flux

Re: More on heating (3.00 / 0) (#4)
by kitno455 on Mon Mar 06, 2006 at 10:41:14 AM MST

but flux- i think he is suggesting exactly that- change the resistance of the load to match the power output of the gen? or am i mis-reading this whole thing.

smoggy- dont set your posting to 'html format', it is eating all your linefeeds.

allan

[ Parent ]

Re: More on heating (3.00 / 0) (#6)
by SmoggyTurnip on Mon Mar 06, 2006 at 10:49:07 AM MST

Hey thanks, I was wondering
what was happening to my line feeds.
The sooner you start the longer it takes.
[ Parent ]

Re: More on heating (3.00 / 0) (#7)
by SmoggyTurnip on Mon Mar 06, 2006 at 10:54:30 AM MST

Oh and yes I guess you must be missreading. I am suggesting that a properly selected fixed resistive load would be almost as good as MPPT. Of coarse this is not based on any experience - only on calculations which are in turn based on some pretty loose assumptions about the power curves for turbine blades. Flux is saying that a fixed resistance that works well in high winds will stall the blades in low winds. I just don't understand why.
The sooner you start the longer it takes.
[ Parent ]

Re: More on heating (3.00 / 0) (#22)
by Nando on Thu Mar 09, 2006 at 06:49:56 AM MST

SmoggyTurnip:

Are you now aware why a fixed value resistor does not allow the wind mill generator to operate at low wind velocities?.

Can you calculate the torque the mill generates and, as well, the energy the generator produces and the resistor's current effects on the generated torque ?.

STALLING OF THE GENERATOR will occur,

Have you done a new set of curves with the proper power setting reference to demonstrate the difference between MPPT and fixed resistance loads FOR FINAL DEFINITION of MPPT BENEFITS.

Nando

[ Parent ]

Re: More on heating (3.00 / 0) (#23)
by SmoggyTurnip on Thu Mar 09, 2006 at 08:07:36 AM MST

Nando:
Are you now aware why a fixed value resistor
does not allow the wind mill generator to operate
at low wind velocities?.

Smoggy:
You have a good point here Nando.
Before I go any farther let me explain the graph
In the top right hand corner.

The parabolas which are opening down are the
power curves for an imaginary set of blades. The
represent the power delivered to the shaft for a
given RPM.

Each one of those parabolas is for
the same blade set but a different wind speed.

The curves do not represent any real blade because
I could not find any power curves for real blades, although
I have searched extensively on the internet. All I could
imaginary curves used for explaining blade theory.

The other curves on that graph represent the generator
electrical power at a given RPM for a fixed load.
the bottom curve is the power in the stator, the next
curve is the power in the fixed resistor, and the third
is the total electrical power (load plus stator). Now these curves
do not include the frictional power losses or any other losses.
to find the operating point at a given wind speed I just took
the intersection between the wind power curve and the total
electrical power curve to get the RPM. As you can see from
the graphs there is all wind power curves have an point of
intersection with the generator total power curve suggesting that
the blades will never stall. Since we both know that blades do stall
then we know that there is something wrong with the graphs. So
I will lay this argument to rest until I have my generator complete
and do some testing with it. I will say that I will not be convinced
until I see some data. I find it strange that you have been working
in this area for some time now but don't have any actual data to
show the benefits of MPPT but you are strongly convinced that
that the benefits are great.

Have a great day

.
The sooner you start the longer it takes.
[ Parent ]

Re: More on heating (3.00 / 0) (#24)
by Nando on Thu Mar 09, 2006 at 08:51:15 AM MST

The data I took several decades ago, I do not have with me, they are with the company that paid for all the work.

Unhappily one use a lot a math to prove a point to your own point of view, but for others that see the data, they can not either follow your reasoning or see the error in the display of that math calculations.

In your case, I would suggest to do a chart that gives a simpler and easier way to display your data, like the curve I attached in my past messages.

You do not need a SPECIFIC set of blades to prove the MPPT benefits, you only need a the wind air cube power curve, a generator, with known phase winding resistance, attached to a blade, let's say with a TSR of 6 or 7, which will give you the mill RPM, the generator volts/RPM, a defined top RPM, like 300 RPM and a peak voltage at that RPM.

You do not need to limit the generator peak power -- though I suggest a top RPM.

Curves:

Wind Power Vs RPM

Generator voltage Vs RPM

Assume 10 % dissipation to the generator & plot GEN POWER curve vs RPM

Plot POWER curve of output load Vs RPM

since you want to use FIXED resistance loads, do 2 values

Resistance 1 = Generator resistance

Resistance 2 = 2 or 3 times the generator resistance

If you care, calculate the MPPT resistance at the peak voltage you have set and use it as a FIXED VALUE RESISTANCE. ---

And plot them as above --- You indeed will discover the differences and if you calculate the torque vs RPM and the load effects on the generator to see if the generator may allow the wind mill to rotate under those load conditions, you may notice that the MPPT may be the one that allows the wind mill to rotate.

lastly, I spent sometime looking for an article that may assist you in your quest.

http://home.att.net/~africantech/ESME/windhns1/WindHns1.htm

Nando

[ Parent ]

Re: More on heating (3.00 / 0) (#25)
by SmoggyTurnip on Thu Mar 09, 2006 at 09:46:04 AM MST

Plot POWER curve of output load Vs RPM

HOW???
The sooner you start the longer it takes.
[ Parent ]

Re: More on heating (3.00 / 0) (#26)
by Nando on Thu Mar 09, 2006 at 09:57:56 AM MST

FROM PAST MESSAGE:

The problem is not the monitor but me, my eyes.
IT SEEMS if I read right that you have a fixed load on each phase, RIGHT ?

If so, MPPT will not be measured

Compare that energy with a fixed resistor value Like equal to the generator winding resistance

READ AGAIN:

http://www.fieldlines.com/comments/2005/6/16/16214/9124/3#3

Use a non defined wind mill with a certain AREA and TSR common to MPPT and the fixed resistors

Do you need more info ?

Nando

[ Parent ]

Re: More on heating (3.00 / 0) (#5)
by SmoggyTurnip on Mon Mar 06, 2006 at 10:47:37 AM MST

Hi Flux, Flux: The big issue for me is not the power output but noise. from the look of your curves it seems as though you will be up at about 10 in high winds. For all the effect it has it might as well be running away light, the noise will be frantic. Smoggy: Yes I agree that the graph shows a very high TSR in high winds but I plan on having the thing furl at about 24 mph. Flux: Without some attempt at mppt battery charging is inefficient but quiet and pleasant, most of us can live with that for the type of load in use. For heating you need the decent efficiency to make it worthwhile and if you insist on a fixed load then you will have a lot of cold days or sleepless nights. Flux Smoggy: You may be right here (except for the part about losing sleep). But I ask you, where is my reasoning wrong? What assumptions am I making that are not right.
The sooner you start the longer it takes.
[ Parent ]

Re: More on heating (3.00 / 0) (#8)
by Flux on Mon Mar 06, 2006 at 11:13:02 AM MST

Probably your reasoning is ok as long as you restrict yourself to a narrow band of wind speed. The narrower the band the better the track at fixed load.

You will have to remove the load at the low end, you can't just leave it there and you may find that you have to raise this from 12 to about 15 mph. There is not a lot of heat in low winds so you may not see much difference.

If you intend to furl at 24 mph then again you are restricting your maximum heat so over the speed range from 15 to 24 mph you may manage a track that is not significantly worse power wise than with the correct load match, if you like noise, high gyroscopic forces and all that goes with it then you should be ok.

Much depends on your wind area. If you have oversized the prop to produce useful power in lower winds and winds over 24 mph are infrequent then it may be near enough.

I would have thought that you would try to track from 12 to 30 mph for heating, heat is the lowest form of energy and you need lots of it to be much use. The low winds will only give a bit of water preheat or something.

I just don't see the issue about banging in a few more heaters as the wind picks up.

No need for calculus on the mppt, just a few steps of control to keep it near and keep things safe and quiet. A gear box in a car is not mppt but it does help the thing go along in a reasonable way.
Flux

[ Parent ]

Re: More on heating (3.00 / 0) (#9)
by SmoggyTurnip on Mon Mar 06, 2006 at 11:32:24 AM MST

Flux: I would have thought that you would try to track from 12 to 30 mph for heating, heat is the lowest form of energy and you need lots of it to be much use. The low winds will only give a bit of water preheat or something. Smoggy: If I want to go into higher winds then I would have to increase the load on the blades because as you said the blades would be going to tsr 10. To do this I would need to reduce the load resistance. This would have the effect of lowering the RPM and decreasing the efficiency of the alternator. This greatly increases the heat in the stator and I want to avoid stator burn out. On Hugh's pages describing the Nirvana machine he shows what can happen when the load resistance is too low. He also says on that page that he needs to make his alternator more efficient. The only way I can see to make an alternator more efficient is to increase the ratio of load resistance to internal resistance.
The sooner you start the longer it takes.
[ Parent ]

Re: More on heating (3.00 / 0) (#10)
by Flux on Mon Mar 06, 2006 at 11:47:57 AM MST

You have a point. If you have built the alternator you are stuck.

I have always said that the problem with making best use of mppt is that you end up with very large and expensive alternator.

There is no way round this, producing alternators to have the same efficiency at 200 rpm as a normal one at 2000 rpm doesn't come without a price tag.

You will indeed have to furl at a safe wind speed but I would be reluctant to let the speed go too high, the gain in efficiency is not going to let you furl at more than a few mph and with a nasty increase in noise.
Flux

[ Parent ]

Re: More on heating (3.00 / 0) (#12)
by erichtopp on Mon Mar 06, 2006 at 03:15:16 PM MST

Hummmm, I can't see any colors in the graphs at all. I'm not sure if it's my eyes or my monitor, or both my eyes and the monitor, or that there isn't any color in the graph, or that there is color in the graph but I don't see it. Right now I see 8 graphs on the monitor, for some strange reason there seems to be doubles of each graph. At this point I'm not sure if there is something different between the double graphs but then I've been looking at them for a while.

Re: More on heating (3.00 / 0) (#14)
by Nando on Tue Mar 07, 2006 at 09:13:00 AM MST

Please try to separate your writing for easier reading and understanding.

I had to display your curves in two different displays.

You have two sets of curves, one is MPPT_question.jpg and second MMPPT_question2.jpg call them MPPT & MPPT2 for easier writing.

MPPT has the power output equal (green & black) and generator dissipation almost equal.
MPPT2 has the power with a delta value in the outputs and in the generators

To be able to see what you have I will need to see the mathematics and how you are applying them.

Normally, one calculates the total power then calculate the generator dissipation and the produce output power is the Ptot - Pgen = Pout.

One may assume the wrong steps, though the math is right; MPPT seems to have such problem.

Power distribution is Pout = Pgen * [Rload /( Rload+Rgen)]

So see the difference in Rload for both cases and clearly the MPPT high ratio does generate HIGHER OUTPUT POWER.

Please explain the second graph curves in detail

One may assume the wrong resistance value in the NO MPPT loading, the wind mill has to have the capability to produce power under that FIXED load -- at low wind velocities.

MPPT has been showing for decades to be the highest producer of harvested energy.

Nando

Re: More on heating (3.00 / 0) (#15)
by SmoggyTurnip on Tue Mar 07, 2006 at 10:29:00 AM MST

Hi aging Nando.

You say:

"
Power distribution is Pout = Pgen * [Rload /( Rload+Rgen)]

So see the difference in Rload for both cases and clearly the MPPT high ratio does generate HIGHER OUTPUT POWER.
"

I agree with your formula but you are i don't agree with your second statement.

You are assumming that Pgen is the same for
both the fixed resistance case and the MPPT case.
This is not true because the RPM is lower for the MPPT case.
The two changing variables (Pgen and Rload) are competing in that equation
- one is increasing the result while the other is decreasing it.

It is my contention that
the final result does not vary enough
to make changing Rload worth the effort.
It also has the big disatvantage that
almost all the extra power is going into the stator.

I will try to explain the other graphs when I have time.
Can you please esplain what your graph is trying to show?
I am not sure why your graph shows lower efficiency for higher rpm.
Is your graph for battery charging or resistive heating.

When you say that MPPT has been shown
to be better for decades, are you talking
about battery charging or resistive heating
or both? And how much better?
And where was it shown so that I might read about it?

I am not saying you are wrong - I would just like to
see something to support your idea or for someone
to point out the error in my reasoning.

I am almost hoping I am wrong because I think it would be
fun to build an MPPT system but I don't want to go to all that
effort if it only gives me a few extra watts.

How much extra power can one expect?
The sooner you start the longer it takes.
[ Parent ]

Re: More on heating (3.00 / 0) (#16)
by Nando on Tue Mar 07, 2006 at 11:07:48 AM MST

Well, if you want to addapt every step to make your results as you want then change them. YOUR result will be the value you desire and not the real resultant value.

One needs to set a group of equal initial values to notice the difference.
~~~~~~~~~~~~~~~~
In this analisis BOTH generators NON-MPPT and MPPT MUST and need to have same total power @ same RPM and SAME Rgen values TO ATTAIN THE PROPER ANSWER !!

As a matter of fact a MPPT system may produce higher voltage and the Rload/Rgen ration may be higher for higher % output.

Then the comparison will be real and non discriminatory and non selective.

I finally see why you are getting those results.

A MPPT wind Mill system is supperior as energy supplier and it does not matter if is for heating or for electric power harvesting -- the system needs to be designed MPPT from the beginning -- you are thinking about the standard 1/2 power wind mills ( 1/2 power to the battery and 1/2 power to the generator -- [ in principle not neessarily in the same ratio]-- cut in affeccts this ratio, as well as, gen build.

A real charting would be:

Get the generator Volts/RPM, Wind Power/RPM, plot [POWER * Rload / (Rgen+Rload)] for both cases plot efficiency for both cases, plot POWERgen(dissipation) for both cases.

You will get almost the graph I attached to my past message with additional curves for the MPPT case and the POWERgens.

LIKE I SAID THE MATH may be right, but the application wrong.

I did not generate the graph in my past message, I took it from a web page in Europe, and it is RIGHT with the efficiency SINCE the generator is absorving a lot of ITS generated power

[ Parent ]

Re: More on heating (3.00 / 0) (#18)
by SmoggyTurnip on Tue Mar 07, 2006 at 11:21:51 AM MST

Nando:
One needs to set a group of equal initial values to notice the difference.

Smoggy:
So has someone done this?
The sooner you start the longer it takes.
[ Parent ]

Re: More on heating (3.00 / 0) (#21)
by Nando on Tue Mar 07, 2006 at 11:54:04 AM MST

YES, ALWAYS in the industrial age, the way to properly show behaviors, changes and/or improvements and so ON.

Always, a reference is needed.

Nando

[ Parent ]

Re: More on heating (3.00 / 0) (#17)
by Nando on Tue Mar 07, 2006 at 11:14:18 AM MST

Well, if you want to adapt every step to make your results as you want then change them. YOUR result will be the value you desire and not the real resultant value.
One needs to set a group of equal initial values to notice the difference.
~~~~~~~~~~~~~~~~
In this analysis BOTH generators NON-MPPT and MPPT MUST and need to have same total power @ same RPM and SAME Rgen values TO ATTAIN THE PROPER ANSWER !!

As a matter of fact a MPPT system may produce higher voltage and the Rload/Rgen ration may be higher for higher % output.

Then the comparison will be real and non discriminatory and non selective.

I finally see why you are getting those results.

A MPPT wind Mill system is superior as energy supplier and it does not matter if is for heating or for electric power harvesting -- the system needs to be designed MPPT from the beginning -- you are thinking about the standard 1/2 power wind mills ( 1/2 power to the battery and 1/2 power to the generator -- [ in principle not necessarily in the same ratio]-- cut in affects this ratio, as well as, gen build.

A real charting would be:

Get the generator Volts/RPM, Wind Power/RPM, plot [POWER * Rload / (Rgen+Rload)] for both cases plot efficiency for both cases, plot POWERgen(dissipation) for both cases.

You will get almost the graph I attached to my past message with additional curves for the MPPT case and the POWERgens.

LIKE I SAID THE MATH may be right, but the application wrong.

I did not generate the graph in my past message, I took it from a web page in Europe, and it is RIGHT with the efficiency SINCE the generator is absorbing a lot of ITS generated power.

I AM SOOOOOOO SORRY, THE MESSAGE WENT OUT BY ERROR, I WAS GOING TO CHECK IT FOR ERRORS and clicked the wrong place -- I type fast and make mistakes that I need to correct ALWAYS.

Nando

[ Parent ]

Re: More on heating (3.00 / 0) (#19)
by SmoggyTurnip on Tue Mar 07, 2006 at 11:25:09 AM MST

No problem Nando - it happens.

Nando - I guess I am not sure what the point of your graph is.

Please tell me how is it relivant to the MPPT discussion.

What am I missing?
The sooner you start the longer it takes.
[ Parent ]

Re: More on heating (3.00 / 0) (#20)
by Nando on Tue Mar 07, 2006 at 11:51:02 AM MST

SOMETIME back I wrote a message where you were quasi-participant:

http://www.fieldlines.com/comments/2005/6/16/16214/9124/3#3

There, I indicated with the graph how too obtain much higher output power from the same generator.

I included it, in this thread, as a reference if you cared to search for it, since you did not, I got the info, which is in the above URL for you to examine.

Nando

[ Parent ]

More on heating | 26 comments (26 topical)

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