real generator efficiency??

[scorman]

I am doing some benchtesting of a model and came to realize that what I measure is NOT necessarily what I think I am measuring ...or is it?

Basic question: can you define what is meant by generator efficiency of up to 85% ??

Some of the home brew axials I presume are a bit less then those made with commercial tolerances and laminated cores. I am disussing resistance heater load such as a calrod in a hot water tank,  not battery charging.

But how is that efficiency really measured or calculated?

I have been going over the aspect of the internal windings being in series with the load resistor. BUT, we measure the voltage across the load resistor to calculate power produced.

As a simplistic example, if the output voltage is 120v for an Rl = 10.6 ohms, then  1.35KW output is recorded     P=v^2/Rl ohms

If I choose a 5.5KW 230v calrod element I get 10.6 ohms (convenient, huh?)

BUT, what about the internal windings of the generator ? how does it fit into the equation? There is power consumed by the windings inside the generator, which is why they heat up.... just another resistance heating element (more complicated but to make a point)

In my benchtop experiment with 12inch rotor, I have a 12v PM Pittman DC servo whose Rt of the windings is 2 ohms. I can get from 2500rpm down to near zero = 5v max

I can apply NL, 10 ohms, 2.5, or 0.5 ohms load Rl while the rotor is spinning

When open circuit ie NL, the back EMF of the generator is what is measured and is linear with the rpm.

LINK

note that at any rpm, the voltage measured across Rl is proportional to the factor (Rl/Rl+Rt)

IF load is 2 ohms, and I measured 1v across the Rl, then the voltage divider effect means that 1 v is also dropped across the windings Rt and  1/2 of the turbines power is wasted in the generator and the efficiency is 50% ...but I can use a 0.5 ohm load as well, and now 80% of the power is in the generator, and if Rl=10 ohms then 83% of the power is across the load and 17% inside the generator ( but not providing much load at all).

Total power produce by the turbine is Pload x (Rl+Rt/Rl)

Back to real turbine situation:

Does the 85% efficiency number mean that for a load RL = 10.6 ohms, that the generator must be built with an internal loss of  15% of 1.35KW = 202 watts and the winding Rt is now defined by 202=v^2/Rt  and if Rt= 1.0ohm by coil construction parameters ( # windings/wire size), then the internal volage drop is only  14.2v ....   Am I missing a basic relationship here?

BTW, w/r to another thread, a simple result of the benchtesting is that full load = 1/2 rpm of NL ...so IF you are at TSR=7 at full load, and the generator circuit gets zapped, that turbine is spinning at 420mph tip speed at 30mph WS ..watch out!

Stew Corman from sunny Endicott

scorman@stny.rr.com

I removed the link you had in the intro. Instructions clearly state "No HTML in intro"

[Flux]

Why confusion over efficiency? seems a simple concept, it is the %of eenrgy out over that put in.

If you stick to metric units, if you put 100W of mechanical energy in then you will get 100W of electrical energy out. If it is 85% efficient than you will only get 85W out.

Mechanical power in is calculated from torque and speed of the driving source ( prime mover). With a conventional alternator you will have direct mechanical losses from friction and windage that you can't avoid. If it is an iron cored machine then you will have iron loss in the form of eddy and hysteresis loss which you also can not avoid.

As soon as you connect a load then you also have a copper loss that is load dependent.

If you are not prepared to measure true energy in at the mechanical source, then you can approximate these losses if you have sufficient details and enough understanding.

You will probably not be in a very good position to other than guess at mechanical and iron loss. You can make a better estimate of the copper loss but life is not that simple especially with rectifier loads

For air gap machines and heating loads you should be able to figure the copper loss fairly accurately.

As a rough approximation the efficiency is directly related to the load resistance and the winding resistance. For high efficiency the machine internal impedance should be a small fraction of the load resistance. Once you start looking at iron cored machines other factors cloud the issue but air gap machines play the game fairly well.

Flux

[scorman]

Flux,

" if you put 100W of mechanical energy in then you will get 100W of electrical energy out. If it is 85% efficient than you will only get 85W out."

The question I was posing was where is "OUT"?

It has to be useable power

You missed a salient point in my experimental data ...depending upon the relationship of the load resistor to the internal winding resistance, yes, you can have 100w of mechanical energy, but you don't want the generator to be using the 80% bulk of that energy to heat the atmosphere !

BUT, what is meant when a generator is spec'd at 85% efficiency ?? 85% of what?

how is it applied to the real world ?

"As a rough approximation the efficiency is directly related to the load resistance and the winding resistance. For high efficiency the machine internal impedance should be a small fraction of the load resistance."

The statement I agree with ...question is how to apply this knowledge ...the 85% generator does NOT know what resistor it is being attached to OR is that part of the spec??

Stew

[Flux]

Sorry , I thought it was obvious that output was the power delivered to the load.

Anything that doesn't appear in the load is loss. None of the energy lost in the windings, core loss or bearing loss can be used in the load. The only possible exception would be a heating system where you could reclaim the internal losses as heat. This would be nothing other than a special case of an eddy current brake.

At 85% efficiency your 100W of mechanical energy would deliver 85W to the load, where the rest goes is not relevant.

Flux

[Nando]

Can I put my comments here?.

Efficiency of a generator is one thing, efficiency of the system is another thing.

Generator to Load:

You have the internal resistance of the generator (= Rg) that for a 3 phase runs 1.5 times the phase resistance ( some indicate 1.3 times).

The inductance of the generator that may not be a factor in the calculations

The resistance of the cabling from the generator to the load.

The equivalent resistance of the conversion from AC to DC if such is the case

The resistance of the load.

Assuming that the generator sees : cabling and load resistance = Rl

(No AC/DC conversion)

Efficiency would be = Rl/ ( Rg+Rl)

Lets say that the generator phase resistance = 1 Ohm or 1.5 ohms Rg

and the load = 15 ohms

eff= 15/ (15+1.5) = 15/16.5 = 90.9090909 %

Let's assume that the generator is producing 1 Wk = 1000 watt

P = V^2/ R

V= Square root ( P * R) = SqR( 1000 * 16.5) = 128 Volts

Where 128 * 1.5/ 16.5 = 11.6363 volts are for the generator internal resistance

128 *15/ 16.5 = 116.3636 are for the load

Power wise :

Power Generator = 11.6363^2/ 1,5 = 90.26898 watts

power Load = 116.3636^2/ 15 = 902.6991 watt

total power = 90.26898 + 902.6991 = 992.968 watts due to round offs = 1000 watts.

We now need to determinee the wind mill overal Input power which is another subject.

Nando

[Nando]

Can you detail the chart you have included.

It seems that the generator does have a very low internal resistance, better than 100 times less than the lowest load.

Nando

[Flux]

I have had another look at this. I can't see where the graph in the link relates to the other example and I still can't see what is worrying you.

"Back to real turbine situation:

Does the 85% efficiency number mean that for a load RL = 10.6 ohms, that the generator must be built with an internal loss of  15% of 1.35KW = 202 watts and the winding Rt is now defined by 202=v^2/Rt  and if Rt= 1.0ohm by coil construction parameters ( # windings/wire size), then the internal volage drop is only  14.2v ....   Am I missing a basic relationship here?"

Yes this is right, I can't see where this relates to the graph.

Are you reluctant to accept that the internal drop needs to be so low. You are right that at 50% you loose half the power in the generator so you seem to be understanding the thing ok.

I don't follow your last statement at all but you are right that there will be a large increase in prop speed if you loose the load, this is a peculiarity of the prop not of the generator.

Flux

[scorman]

The graph was for a small 12v Pittman PM DC motor used as a generator for the experiment. It was calibrated in a lathe at the various speeds

It appears that with small physical size and corresponding low total wattage capability, that the windings at 2 ohms prevents this "system" from delivering

most of the 12 inch turbine power to the load resistor and instead, in order to load down the blades, most of the power was absorbed within the motor windings. That is fine , since total turbine power can easily be determined.

I was concerned that maybe I was looking at things wrongly when scaling up to a full size system, or perhaps it was possible to choose a wrong combo of generator/load to blades, other than the obvious rpm specs.

Can anyone give me numbers on these 1->2KW axial home builts for what the winding resistance actually measures? Are they really in the 1->2 ohm range?

I have a large DC brush motor with the following specs:

1725rpm, 23 amp at 125v = 2.8KW   R=V^2/P = 5.5ohms ....

so this would be only 66% electrically efficient with a 10.6ohm load resistor

so looking for motor/generators of similar power, lower voltage and higher amps (lower Rg) is goodness?

a 1200rpm/230v/10amp DC servo rated at 3HP has Rg= 23ohms ...NG!

thanks to those who have contributed here,

Stew

[Flux]

I think I am with you now. The toy motor in the graph would need a load of about 20 ohms to be reasonably efficient. Things are influenced by the voltage as well as the winding resistance.

"I have a large DC brush motor with the following specs:

1725rpm, 23 amp at 125v = 2.8KW   R=V^2/P = 5.5ohms ....

so this would be only 66% electrically efficient with a 10.6ohm load resistor"

No, you don't know its internal resistance. The rating will be the current level that the manufacturer decided it could handle safely at its rated voltage. The efficiency will have been chosen to be high enough that it will not burn out at rated current ( with dc it may also be limited by brush and commutator limitations as well as winding heating)

You would need to run it at constant speed, measure open circuit volts and then measure output volts into a known load to determine its internal resistance.

Now there is another big issue that is going to throw everything into chaos, you ask the typical resistance of a typical 1-2kW axial unit, that will depend on the system voltage. Two ohms is very high for a typical machine and at 12v you need to get down to a small fraction of an ohm.

It doesn't end here, the normal machines are matched by having a significant winding loss ( or line loss) and at the full load point you will not expect an efficiency of much over 50%. If you aim for high alternator efficiency the speed rises very little from cut in and the blades stall at winds just above cut in.

I am not sure what your application is but as you seem to be considering resistive loads you may be thinking of heating. If that is so and you want a decent alternator efficiency you will e forced to use a load controller that will change the effective load resistance with wind speed to keep the prop tracking wind speed. A fixed load resistor will either cause stall or run away.

Direct battery charging alternators can be 85% or more efficient at small loads but the efficiency falls constantly with wind speed. If you want to be up in the 85% region over the whole range you will not be able to directly connect it to a fixed voltage battery and achieve a workable overall efficiency. You have to compromise the electrical efficiency to get useful power from the prop over a range of wind speed.

All good fun, it all interacts and nothing is simple.

Flux

[scorman]

Thanks Flux for the new information....this is what I was looking for

"No, you don't know its internal resistance."

That was my problem ...thought I did...I was taking a too simplistic view

"The rating will be the current level that the manufacturer decided it could handle safely at its rated voltage. "

I was under the misunderstanding, that I could apply the simple P=v^2/R to the nameplate rating to determine the internal winding resistance of the motor/generator.

This is obviously a wrong assumption as I am typically Ohm's Law challenged <grin>.

So, w/o having the specific manufacturer's data on a specific motor, w/o bench testing it as you described, there is no way to know it's winding resistance just from nameplate specs

As an example ..I have considered this past eBay servo unit as a viable choice (except for the 3200rpm rating):

http://cgi.ebay.com/Allen-Bradley-AB-AC-Servo-Motor-Model-1326AB-Series-C_W0QQitemZ320099425343QQihZ
011QQcategoryZ78195QQssPageNameZWDVWQQrdZ1QQcmdZViewItem#addimg2

http://ebay.sunnkinglister.com/jivison/R324336003.JPG

at the AB site, they show a similar unit "Terminal Resistance ohms (L-L) at 25°C = 1.9 ohms " this would be the real winding resistance per each phase?

From your previous posting:

"You have the internal resistance of the generator (= Rg) that for a 3 phase runs 1.5 times the phase resistance ( some indicate 1.3 times)."

Is this irrelevant if I use each phase to it's own 3 individual resistance heater elements?

Thanks again for your patience and expertise,

Stew

[Flux]

I don't know much about the modern synchronous ac servo motors, they will obviously generate in some form but they are iron cored and will not obey the simple ways of the air gap machines, at some point reactance is going to enter the equation. The phase resistance will give you the best case if you take it as internal impedance but the reactive component may limit you to well below the resistive value.

For 3 phase ac into a resistive load the behaviour will not be the same as for the rectified case and the 1.3 factor will not apply. The simplest way may be to work on phase resistance ( 1/2 the resistance between terminals) and do your calculations on a per phase basis. The 3 phase power will be 3 times the power in a single phase.

I would imagine that that servo motor would produce very little as a wind generator at a realistic speed. You are limited to 3.3A per phase and the voltage at a sensible speed would be low ( probably similar in output to a dc tape drive motor ( Amtec type thing).

Flux

[finnsawyer]

Some general thoughts here.  We assume you have a balanced three phase alternator and that its properties are linear at a given RPM.  You hang equal value resistors across the three phases with no common connections between resistors.  Then you can model any of the phases as seen at the load resistors looking back by a voltage source in series with a resistor, the values of which are assumed unknown, and which are independent of load.  So, you have two unknowns.  If you use two different values of the load resistors, you can simply by measuring the load voltages and calculating the load currents in the two cases determine values for the unknown voltage and resistance.  There are two caveats.  The RPM can not be allowed to change with load and the currents and voltages must stay in phase (no inductive or reactive effects).  That unknown voltage by the way is the open circuit voltage that is supposedly linear with RPM.  Knowing that relationship would allow you to compensate for changes in RPM due to changes in loading.  The advantage of this technique is that is allows you to determine the total resistance per phase looking back from the load.  It does require the alternator to behave as a linear device, an assumption that can be tested by doing the tests for a greater number of load resistances.  In that case the alternator voltage stays constant, but the alternator or system resistance varies in value (becomes non linear).  I am assuming no diodes are in the circuit at this point.

[Nando]

Flux:

You said:

For 3 phase ac into a resistive load the behavior will not be the same as for the rectified case and the 1.3 factor will not apply.

You are right if the connections is a four wire 3 phase, where the common is brought out.

If 3 phase with loads in between phases, the 1.5 times resistor value applies and this can be seen simply by calculating the voltages values of the three phases with one set at 90 degrees and doing a simple Kirkoff calculations.

Nando

[Flux]

Nando

I think we are looking at a different issue here.

With the air gap alternators star connected and charging a battery with a rectifier, to calculate the output current We take the difference between the emf and the battery voltage and divide by the effective internal resistance of the alternator.

Most people seem to take the measured resistance between 2 lines ( 2 x resistance per phase). This seems logical as the 3 phase rectifier conducts through 2 diodes at any instant and the current is carried at any time through 2 of the 3 coils.

This gives a rather optimistic result and It seems to work as though the effective resistance is 1.3 times the measured dc resistance ( there is negligible inductance to introduce a reactive component). This was the 1.3 factor I was referring to.

When calculating losses the actual resistance seems to be the value to use. Don't ask me to explain this factor, it just seems to give good results.

Flux

[Nando]

Servo motors have an inductance that it lower if compared to an equivalent regular motor.

The Ke ( Back EMF constant) varies dramatically with the manufacturer types.

They are quite useful for power generation, if the RPM is there, good or even excellent for small Hydro systems.

I have been looking at them for Hydro use.

Nando