help with calculations please

[snake]

hi friends,i am about to buy power resistors to be used as dump loads for my 12v wind turbine.i am estimating to get approximately 10amps/H,about 120vdc.i am having a lot of 10 of these resistors rated at 0.1ohm 3%,25watts each resistor.

i am having difficulties in performing the calculations to get the total power that these resistors will dump.

please,can someone help me with these calculations.

thanks

[ghurd]

I have a feeling you are looking at something very incorrectly.

A 12V system will be dumping at about 14.4V.  The 120V it says on the motor does not count.

Ohm's and Watt's Laws:

At 14.4V / 0.1 ohms = 144A

At 144A x 144A x 0.1 ohms = 2074W

The only way to make any use of those resistors is to series them all together for 1 ohm.

They will dump 14.4A.

And 14.4A x 14.4A x 1 ohm = 207W for all 10, or 20.7W each.

G-

[imsmooth]

First calculate how much power you need to dump.

You can connect your resistors in series or parallel, depending on what current you want to draw in order to satisfy how much power you need to draw from the turbine.  Let's say all your resistors have the same R to make things easier.  Let's say your voltage is 14v and each resistor is 1 ohm rated at 10w.

If you have 10 of them:

10 * 1ohm = 10ohms total resistance.

Current through system is 14v/10ohms = 1.4A

Power as heat = 14v * 1.4A = 19.6watts

You should use a safety factor to make sure your dump load does not overheat.  A factor of 2 is a good start, so your total wattage of resistors should be at least 40w

total wattage your resistors can handle = 10w * 10 = 200w so you are ok.

What if you connected them in parallel?

Total resistance = 1/R1 + 1/R2 + ... = 1/Rt

So, 10/1 = 1/Rt ---> Rt = 1/10 ohm

Total current in system is 14v/(1/10ohm) = 140A

Total wattage of resistors needed = 14v * 140A = 1960w

Wattage need for each resistor = (140A/10)^2 * 1 = 196w

Wattage rating of each resistor = 10w

Your 100w bank of resistors will not work here.

You need to know what resistance you want for your load to keep your turbine from spinning too fast.  While you would have had an acceptable dump load with regard to the 1.4A of current it would handle, the power it would draw from the turbine would have been insufficient to keep it under control.  If you need to draw 300w of power from the turbine you would have make a bank of resistors that will draw enough current to meet this need.

[imsmooth]

Edit: 10w * 10 = 100w, not 200w

[ghurd]

It sounds like you have a misunderstanding of typical dump load controllers.

'how much power you need to draw from the turbine' and 'what resistance you want for your load to keep your turbine from spinning too fast' are the type of statements which would indicate something is not clearly understood.

G-

[TomW]

Goods points. Glen.

It is common knowledge that you Do Not dump turbine power but that you dump battery power.

Seems to be a common misconception that gets regurgitated all the time here.

I draw so much fire when I make these kind of points that I have more or less stopped doing it.

Just tossing you some backup. Probably still draw fire but, hey its part of the fun I guess.

Personally, I think accuracy is important in this field.

I won't even touch the math which also had errors IMHO.

Now back to the welder and hammer. I understand those much better than people.

Tom

[snake]

i tried to do the calculations which the first guy posted,it seem that the calculations are good.if i connect all those resistors in series,i will give me 250 watts and 1 ohm.the average amps i get from my wind gen is 10amps at 14.4V lets say  15v which will make a total of 150watts,right?

so with the calculations:

V=15

R=1

V=IR

15V=I*1

I=15A

so,the resistors in series will be able to dump a total of 15A at 15V

P=IV

P=15*15

P=225watts

right?if i am wrong,please correct me.

which make the resistors effective to dump the power

[imsmooth]

How do you dump battery power if you do not have batteries, but are grid-tied?  What is being dumped then?

[TomW]

jonathan;

This whole direct grid tie thing is pretty much beyond most DIY folks so I am not sure why it even gets discussed.

Seems a no brainer from here to simply brake a grid tied turbine rather than dump but I don't find it worthy of the time to discuss myself.

To the point Glen said " typical dump load controllers".

Typical units here use a battery.

Some can always find a way to change the point but I will leave it at that.

Despite my hope to the contrary I still drew fire with an honest assessment.

As far as the calculations go most folks need to learn a few very simple formulas and laws WRT electricity. It starts with E[volts]=I[amps] times R[resistance] Nearly every other relationship derives from that.  

Oh watts is volts times amps or amps squared times resistance.

Tom

[ghurd]

Correct.  

Be warned about 20W in a 25W resistor.

If they are on very long, they get hot enough to give fingers blisters.

Mine seem to increase in temperature very linearly at 1F degree per second over most of the common operating conditions (10C to 70C) in 22C ambient.

G-

[imsmooth]

Ok, I won't bring up gt systems in the future.  I must be missing the I^2 * R thing.  If you are refering to the equation (140A/10)^2 * 1 this is

(140A/10 resistors) = 14A/per resistor seen in parallel; each resistor is 1 ohm

(14A)^2 * 1 ohm

What did I miss?

[snake]

thanks for all your advice friends

[rossw]

While the most common dump controllers do indeed seem to dump battery output, likely because this is convenient, easy to understand and one controller will do a multitude of sources, I don't (personally) believe it to be the best solution.

My own wind turbine, for instance - while there is a battery voltage sensing circuit (as in your typical dumpload controller), and it has hysteresis (like many of the typical dumpload controllers do), the actual "dump" part dumps the 2KW turbine into a 3-phase dumpload at the AC SIDE.

This has several advantages as I see it.

1. It's much easier on the rectifiers
   
2. It's a more efficient control with less points of failure
   
3. It completely avoids the problem of "over-dumping" the batteries, as may
   
   

   happen for example, if you are only just getting too high on your battery

   voltages. If you have (say) 500W going into your batteries, they could get

   high enough to trip the dump - but if your turbine can make 2KW, then you

   probably need at least 2KW dump - which will be very tough on the batteries.

It's quite easy to achive battery monitoring with "wild-side" dumping, many of the dumpload controllers presented here over the last few years are readily adaptable to this method of control (especially those that merely operate a relay)

[ghurd]

Not sure I agree with that.

I would like to get your take on this,

#1- The rectifiers have been taking the current to the battery for a long time already.  It should not be abusing them if they are properly sized and cooled.

#2-  I don't understand what you mean by "It's a more efficient control".

As long as the battery voltage is maintained high and accurately, either seems fine to me?

#3- I believe dumping the max charging amps should not be hard on the battery, unless the battery is undersized for the system.

I mean, the battery on a 1000W windmill should be able to supply well over 1000W without trouble, or the battery is too small.  I have never been comfortable charging over C/10 (or believed it was efficient), but I am comfortable drawing C/5.

Your 2KW, making say 500W or 1KW, then dumps to the 2KW load.

Isn't that fairly violent?

My stuff (small) did not appreciate those conditions.  Anything to tame it down a step was in danger of running away unless the battery was left in parallel (meaning not a double throw relay deciding where all the current would be directed).

G-

[Airstream]

Was the term sought above C/20 if "over" was in relation to higher current?

Did you intend 5%, 10%, or 20% Amperage?

C/X = Battery Amp Hour Rating divided by X

C/20 is 5 percent of capacity...

C/10 is 10 percent of capacity...

C/5 is 20 percent of capacity...

20% of any medium or large bank is manly amperage!

[ghurd]

Yup,

C/10 amps.

Even with solar, C/10 max power amp PV gets the battery to regulation voltage pretty quickly, and a good part of the available power after that is simply not stored.  My take on that is either the charger is oversized for the system, or the battery is undersized.  For solar I like about C/15 for the type of things I do.

And C/5 amps for drain.  That's only about 250W out of a 100AH battery, or 500W out of a pair of golf cart batteries.  Not an unreasonable load, IMHO.  And not an unreasonable amount of load for the fractions of a second a dump load on the DC side.

G-

[rossw]

Agreed... but my feelings are this (I should say - this is based on a single-stage or "bang-bang" controller - all changes with progressive and/or PWM systems)

1. The diodes (when used with batteries) have a fairly short duty-cycle with reasonably long recovery period. The current flowing also is likely to be less than a "maximum capacity" dump is going to pull. Its unlikely the turbine will be making full power very often, whereas the dumpload needs to be of the maximum the turbine can make (in order to keep the thing under control). (It could be worse of course - it could be used as a brake in which case peak-currents will be VERY high)
   
2. How do I put this into words? Charging slowly at lets say, 10 amps, enough to let the batteries slowly soak up power and their voltage to rise slowly. Eventually, they'll get to the point at which you want to stop pouring power into them. With a load across the batteries thats going to pull 40 amps, you suddenly go from 10A charge to 30-40A discharge. I consider that less than optimal. If you dump at the "wild" side, you simply stop putting more charge into the batteries, so they go from 10A charge to very little or no charge. That has to be "kinder" to the batteries, surely?
   
3. The size of the battery bank I suppose, determines to a large extent how big your turbine can be. a 2KW turbine would be a problem on string of 7AH gel-cells! My own main bank is 24 x 2V/500AH. It's too small for my needs, but its all I've got. They are AGM, and are rated to charge at well in excess of C10, so taking 40A isn't a problem, but I also have nearly 2KW of PV. In theory, I could have close to 40A of PV and 40A or wind to deal with. One load across the batteries to "tame" them is a big ask. Fortunately with the PV, I can simply NOT pump amps in to the batteries (series controller) - because the PV won't overspeed if they run open. I'm happier with the dump for the turbine being on the raw side because it introduces no significant inductive spikes or sudden discharge current on my DC side.
   

[ghurd]

Understood.

And wow.  4KW for 24V 500AH.

G-

[rossw]

24 x 2V/500AH is 48V/500AH. (Or if S/P as 2 x 12 x 2V/500AH = 24V/1000AH)

This is probably why we should talk of our battery banks in watt-hours rather than amp-hours

(48V,500AH = 24KWH; 24V,1000AH = 24KWH)

[Flux]

This is way off the original question but does raise some interesting points.

Dumping on the input side of the rectifier has advantages and snags. It does prevent battery discharge if the controller fails. It works perfectly well with a fast pwm or proportional control but for bang bang operation then it presents problems in that it will slow the mill every time the dump comes in. If the dump is going to be capable of taking the full machine power in high wind the application of the dump will probably stop the machine in lower winds.

If you can stand this continual starting and stopping then it may be fine. The mechanical loading may be rather high.

If you do it in a couple of stages within close voltage set points then it will be better, as would bang bang directly on the battery be if done in stages.

The idea of pwm dump with another rectifier feeding the dump load ahead of the battery rectifier is a good method but I am not sure if the common controllers can be configured to do this without doing something to the circuitry. With pwm the controller will just divert sufficient to do exactly the same as it would if fed from the battery. It would reduce rectifier load on the battery rectifier and prevent any possible battery discharge with a failed controller.

The scheme that doesn't work is one which dumps the mill to resistors and disconnects the battery.

Flux