Which Watt Meter

[Wingman]

YEAH! Have fun Andy. Would have recommended the one I use but it's no longer available.

[gangster]

wattage. ?? No such measurement

[Gary Manuel]

Posted by PatMc on 20/01/2019 18:06:38:

Posted by Gary Manuel on 20/01/2019 17:37:18:

When choosing an electric setup, CURRENT and WATTAGE are both important. Exceed the maximum of either and you are asking for trouble.

A motor with a maximum current of 45A that runs nicely at (say) 40A on a 3 cell setup may overheat at 40A on a 4 cell setup if the maximum power rating is exceeded.

PS I use a HK Watt meter and can confirm that it works well.

Edited By Gary Manuel on 20/01/2019 17:41:21

Sorry Gary, that's not correct.
Exceeding the stated power (wattage) has no ill effect on a motor so long as the current is within the motor's limits.

I to use a HK Wattmeter, which IMO excellent. This one.

Actually "Wattmeter" seems to be a common misnomer. AFAICS all "wattmeters" for our use are in fact multimeters, in that they measure a range of parameters of which measuring amps is the most important & watts probably the least important.

Sorry to dwell on this, but maximum power is based on a motor's physical ability to dissipate power in the form of heat and is related to it's mass and cooling ability. It's maximum current rating is based on the conductor size within it's windings. I maintain that neither maximum should be exceeded.

I have exactly the same Watt meter as Pat and can recommend it.

Edited By Gary Manuel on 20/01/2019 20:56:21

[Trevor]

Personally, I find a clamp meter to be much more convenient than a watt meter, partly because I get involved in checking other people’s models and the clamp meter gets around any connector incompatibility issues. Look carefully at the specs though - many of the cheaper clamp meters don’t have a suitable dc current range.

Trevor

[Former Member]

[This posting has been removed]

[john stones 1 - Moderator]

I love lecky advice threads, they're very competitive.

[PatMc]

Posted by Gary Manuel on 20/01/2019 20:54:45:

 

Sorry to dwell on this, but maximum power is based on a motor's physical ability to dissipate power in the form of heat and is related to it's mass and cooling ability. It's maximum current rating is based on the conductor size within it's windings. I maintain that neither maximum should be exceeded.

Gary, only the wasted power is dissipated as heat, the greatest part is dissipated in driving the load [prop].

As I stated previously, heat generated in motor (i.e. wasted power) = current x (winding resistance)² watts

Note this figure is dependant on current regardless of overall power.

PS since the OP has achieved his objective I don't think there's any harm in expanding the subject.

Edited By PatMc on 20/01/2019 22:48:59

[Gary Manuel]

Posted by PatMc on 20/01/2019 22:44:35:

Posted by Gary Manuel on 20/01/2019 20:54:45:

 

Sorry to dwell on this, but maximum power is based on a motor's physical ability to dissipate power in the form of heat and is related to it's mass and cooling ability. It's maximum current rating is based on the conductor size within it's windings. I maintain that neither maximum should be exceeded.

Gary, only the wasted power is dissipated as heat, the greatest part is dissipated in driving the load [prop].

As I stated previously, heat generated in motor (i.e. wasted power) = current x (winding resistance)² watts

Note this figure is dependant on current regardless of overall power.

PS since the OP has achieved his objective I don't think there's any harm in expanding the subject.

Edited By PatMc on 20/01/2019 22:48:59

Pat - Your second line is incorrect.

It should be heat generated in motor (i.e. wasted power) = current² x (winding resistance) watts.

It is dependent on current squared, which is derived from W=IV, where V can be substituted for IR so that W = I²R. The power wasted is a function of both voltage and current.

 

Posted by john stones 1 on 20/01/2019 22:22:40:

I love lecky advice threads, they're very competitive.

I like to think that we are being informative rather than competitive.

Edited By Gary Manuel on 20/01/2019 23:37:56

[john stones 1 - Moderator]

I'm sure you would.

[Mike Blandford]

Posted by Gary Manuel on 20/01/2019 23:28:13:

It should be heat generated in motor (i.e. wasted power) = current² x (winding resistance) watts.

It is dependent on current squared, which is derived from W=IV, where V can be substituted for IR so that W = I²R. The power wasted is a function of both voltage and current.

Yes, but the voltage to be used is the voltage in the windings that is caused by the current passing through the winding resistance, NOT the voltage applied to the motor. Most of the applied voltage is opposed by the voltage developed in the windings (back emf) caused by the motor rotating. Using a higher applied voltage (e.g. 4 cells instead of 3), and keeping the current the same means the motor is rotating faster, while dissipating the same amount of heat (maybe a bit more due to higher mechanical losses).

Mike

Edited By Mike Blandford on 21/01/2019 00:36:14

[PatMc]

Posted by Gary Manuel on 20/01/2019 23:28:13:

Posted by PatMc on 20/01/2019 22:44:35:

Gary, only the wasted power is dissipated as heat, the greatest part is dissipated in driving the load [prop].

As I stated previously, heat generated in motor (i.e. wasted power) = current x (winding resistance)² watts

Note this figure is dependant on current regardless of overall power.

PS since the OP has achieved his objective I don't think there's any harm in expanding the subject.

 

Pat - Your second line is incorrect.

It should be heat generated in motor (i.e. wasted power) = current² x (winding resistance) watts.

It is dependent on current squared, which is derived from W=IV, where V can be substituted for IR so that W = I²R. The power wasted is a function of both voltage and current.

 

Oops, my slip up.

You're quite right Gary, I put it the wrong way round, as you say power = I²R

However this doesn't change my point - with any given current the heat generating power will remain the same independant of the applied voltage.

As you stated V = IR but note in this case V = the applied voltage minus the back emf.

If the applied voltage is increased the motor rpm will increase by the same ratio, this in turn will increase the back emf by the same ratio & at the any given current the voltage across the windings will be the same.

 

 

Edited By PatMc on 21/01/2019 00:46:02

[gangster]

So in a nutshell Buy a Wattmeter as they are called for this purpose, the cheap 3 button ones are working well for most of us. Do not exceed the max current. Buy a few sizes of prop as well around the chosen range .Prop size is probably as significant as anything else in this business

Edited By gangster on 21/01/2019 09:13:04

[Attilio Rausse]

Posted by gangster on 21/01/2019 08:57:46:

So in a nutshell Buy a Wattmeter as they are called for this purpose, the cheap 3 button ones are working well for most of us. Do not exceed the max current. Buy a few sizes of prop as well around the chosen range .Prop size is probably as significant as anything else in this business

Edited By gangster on 21/01/2019 09:13:04

[Robin Etherton]

Ditto for Trevor’s comments.

Go for a clamp enter every time.

If you are not completely au fait with what you are doing it’s much safer.

Ask me how I know

[Robin Etherton]

Clamp meter oops

[Gary Manuel]

Posted by PatMc on 21/01/2019 00:45:04:

Oops, my slip up.

You're quite right Gary, I put it the wrong way round, as you say power = I²R

However this doesn't change my point - with any given current the heat generating power will remain the same independant of the applied voltage.

As you stated V = IR but note in this case V = the applied voltage minus the back emf.

If the applied voltage is increased the motor rpm will increase by the same ratio, this in turn will increase the back emf by the same ratio & at the any given current the voltage across the windings will be the same.

What you are not recognising Pat, is that "R" is not a fixed value. The "Resistance" value is actually made up of the copper losses (impedance) and iron losses (reactance) plus friction losses. Copper losses are proportional to current and behave exactly as you have described, but iron losses are proportional to rotational speed. This effectively increases the value of "R" in these over-simplified formulae.

For a given current (adjusted by prop selection) with a higher applied voltage, the effective value of "R" will be greater, which results in greater losses in the form of heat. This is why a maximum wattage is quoted for a motor and this value should not be exceeded.

[Dave Hess]

It's true that current is the biggest cause of heat and measuring the current gives a good idea of whether you'll have a problem with over-heating as long as you know the motor's current rating, but that's not the whole story. Many of the rules and formulae given above are over-simplified. They're correct for a resistor, but not for a motor because the coils are inductors. It's the impedence that resists the applied voltage and the impedence is the sum of the resistance and the reactance. The reactance changes with the motor's speed, which is why the current changes with speed, and it's also why the efficiency is dependent on speed.

It's very important that people understand the relationship between speed and efficiency because speed not only affects the way the motor heats up but also the rate at which you waste your battery during flight.

If you watch this guy's videos, you can get a much better idea of what's going on. In some of his earlier ones, he explains how the graphs work because there's a lot on them and it'll probably take a while to figure them out. The coloured zones are the efficiency zones with blue as good and red as bad. The black line that ramps down from top left to bottom right represents full throttle for each propeller, which is where we need to look for over-heating. If you get stuck, I can probably explain it. Once you have them figured, they're dead simple, and you should be enlightened:

Edited By Dave Hess on 21/01/2019 12:20:37

[Geoff S]

Posted by Robin Etherton on 21/01/2019 09:32:22:

Ditto for Trevor’s comments.

Go for a clamp meter every time.

If you are not completely au fait with what you are doing it’s much safer.

Ask me how I know

I have a clamp meter but don't use it as much as my watt meters. One reason it gets used less frequently than it might is that it's quite big and difficult to get it where it's needed - another is that I forget I've got it!

One thing to be aware of is that the cheaper ones are ac only so, before purchase, ensure it measures dc as well as ac current.

Geoff

[Peter Beeney]

Post 1

May I just add another little ‘joule’ to the collection of gems here, as I invariably look at this in a slightly different way. First, making a simple statement to myself, - generally when a current flows through a conductor with a resistance* it creates heat; it also creates a magnetic field surrounding the conductor; the two conditions are independent of each other, the heat does not generally affect the magnetic field.

Second simple statement to self, - power is defined by the rate at which energy is used, or energy divided by time.

I always consider the mechanical output from the motor as being separate from the electrical input power inasmuch that the mechanical power, as measured by a dynamometer at the prop shaft, is the direct result of the interaction between two magnetic fields, it’s actually a deflecting action, whereas the electrical power, as measured by a watt meter at the battery is the result of the current flowing through the resistance of the conductor and simply creating heat. The magnetic field is incidental, it’s always there but it comes for free in this particular case.

So because the heat produced is the main concern I’m now thinking that the value of the resistance is all important, it looks as though lowering the resistance will reduce the heat produced. Using the little formula again, I squared R, sorry about more arithmetic, if we consider a motor with a resistance of 0.100 ohms and a current flowing through of 10 amps then the the power expended as heat will be 10 x 10 x 0.1, which equals 10 watts of heat. If the resistance is lowered to 0.010 ohms the the result is 1 watt of heat and going down one more notch to 0.001 it will of course be only 0.1 watt. That’s just one hundred milliwatts. But as the amps are constant the magnetic field strength will not have changed, so the power expended by the motor shaft remains the same throughout.

For me this resistance, or lack of it, is a bit of a Holy Grail as far as motors are concerned; and indeed, the resistance of whole power train needs to be as low as possible.

With regard to the little power conundrum, if I stay within the motor’s parameters set by the manufacturer then I might just stay out of trouble. But that’s not always guaranteed either. However, if I start increasing the cell count then that might soon become noticeable, and I think this is what happens when I do. It’s an electrical principle that if we increase the voltage across a given resistance the current increases by the same amount and as a consequence the watts, (dissipated as heat), increase by a larger amount. 5 volts across a 1 ohm resistance = 5 amps, therefore 5V x 5A = 25 watts. Double the voltage to 10 then 10V x 10A = 100 watts, a x4 increase. But the magnetic field around the coils is created by the current flow only, not the voltage, so that can only increase by up to a maximum of 2 times, thus the torque can only increase up to 2 times, but the heat dissipated is now rising greater in proportion to the increase in mechanical power. I have to reduce the prop size to lighten the load to allow the motor to turn faster. If, however, I change to a larger prop, increasing the load, the motor proportionally reduces the rpm, the back emf reduces again in proportion, more current flows through the coils, and from the above statement the heat dissipation increases greater in relation to the decrease in back emf. It might soon start overheating.

* If some conductors are supercooled they lose all their resistance, now I reckon an infinitely large current would flow, driven by an infinitely small voltage; but I guess the supply unit is always going to have some resistance, so that wouldn’t work, also there comes a saturation point where the magnetic field will not increase any further. But it has it’s advantages, I believe the coils for the electromagnets at the Large Hadron Collider at Cern are supercooled; I’ve mentioned this before in similar threads long ago. Did you see recently that they’ve mooted an even bigger one, 4 times the diameter and 10 times the power, as I recall. I’d think they might even have to have a specially lengthened cheque book to accommodate all the zeros to pay the monthly power bill.

[Peter Beeney]

Post 2

And in the same vein I notice that a Danish company, (I think), has developed a supercooled generator for wind turbines at sea. The size is reduced but the output is greater. But it still has to rely on the wind blowing I guess. I can’t remember much info. although interestingly there was no mention of the running costs, I guess this soaks up some (a lot?) of that increased output. Although I’m sure it’s possible they’ve developed better conductors and better (cheaper?) ways of keeping it supercooled by now. I have read some snippets about renewables and it does seem that the maintenance and running costs are very expensive indeed.

I too use a clip-on power meter for current flow checks, but in close collaboration with a rev counter. In fact I invariably reach for the tacho first; certainly to always check the unloaded kV first; from there I can often guesstimate near enough as to what is going on. Occasionally, in marginal cases, I’d use a contact thermometer to look for any possible hotspots.

Drifting downwind and slightly off course now, but I’ve also noticed that the University of Stockholm have given NiMHs a bit of a dusting off - they’re now NiH2. In short they can make them live a very long life indeed. Very useful as massive renewable energy storage load-levelling/sharing lumps etc. Also automotive use perhaps, if they can ever get the various manufactures to agree to supply them in the first instance. A similar move to small lithiums would be nice…

As it happens I have a nicad pack in a rechargeable torch, which is now about forty years old; …and still going strong! It’s a slightly daunting thought sometimes that this now stands a very good chance of actually seeing me off…

PB