is higher voltage always more efficient?

[jeff262]

just curious if this holds true for everything? i knoiw that most of my brushless setups are but i was curious if this holds true for things like your garage heaters where you can use 110v, 240v, will the higher voltages be more efficient? i have been told to use higher voltages but i don't under stand why a higher voltage would be more efficient since i takes the same amps to make a btu doesn't it?
thanks jeff

[village_idiot]

Higher voltage uses fewer amps to do the same work. The meter on your power is a glorified amp meter, so running the heater at 220volts draws about half the amps as running it at 110 volts.

watts=amps*volts

work can be expressed as watts or many other things (like horsepower).

[andrewm1973]

Jeff,

P=VI
V=IR
so P=I^2R

P = power
V = voltage
I = current
R = resistance

Now P=I^2R is the one we are interested in.

That is Power is the Square of current times the resistance.

In your motor there are at least TWO types of work being done. One is the spinning of the prop/wheels - the work you WANT done. The other is heating of the air around it - the work you DONT want done but which the electricity has to do. There are other minor ones you don't want either like work being put into making noise from bearings and stuff like that but for this example we are ignoring them.

Anyways - all the work you don't want done is called LOSS.

The loss that is caused by the resistance is called I2R loss.

The biggest contributor to the size of I2R loss is I (cause its sqaured). If you can reduce I you can reduce parasitic heating.

Because power is also expressed as P = VI you can get the same amount of power out by increasing V and reducing I.

Doing this you have gotten rid of a percentage of your I2R loss the motor will have less LOSS as heat in the windings and you can smile :D

NOW - in heating your garage. The entire thing is a resistor. All the energy in the form of electricity is being converted to heat. Heat is not the enemy here. Heat is the reason for using the electricity. I2R loss is no longer a concern. (this is assuming the "garage heater" is not a reverse cycle airconditioner)

The reason for using 220V over 110V would be purely for size. I imagine you can only pull 1/2 the current from a 110V plug as you can a 220V plug. Over here on the other side of Kansas with all the witches and munchkins we use 240V and can pull a max of 10Amp from a single socket. That is 2400watts. I would guess that 10Amps or maybe 15amps would be your max on 110V too. If it is 10Amps then the max size your heater coudl be on 110V is 1100W. If you used a 220W plug then you could have 2200W.

Unless the electricity companies meters are very crappy - I can't imagine you would get charged any less for using the 220V option instead of the 110V option.

If you want better efficiency get a reverse cycle heater (assuming you dont live where it snows)

[AndyOne]

Efficiency is expressed by the amount of power lost in the system. Using a 100W bulb at 220V the current draw would be 0.454A and at 110V the current draw would be 0.909A. The bulb in itself would have similar efficiency but the power loss in the cable feeding it would be proportional to the square of the current it's carrying. So if the cable resistance is 1Ohm the power loss is 0.206W at 220V and 0.826W at 110V

from P=I^2R.

I hope this helps.

Andy.

[MatC]

Garage heater: 240 volts is used to minimise the size of the copper cable needed. It results in slightly lower losses (which is why power companies move power around using large voltages and low currents), but not much difference in terms of efficiency.

Motors: not sure there is actually any difference in efficiency with higher voltages? Anyone know better?

[vintage1]

Motors themselves do not show any real efficiency gains with higher voltages if they are wound to take them correctly

That is, at similar RPM two motors - one on 3s wound with three turns, one on 6s wound with six turns, will behave identically.

However the gains are to be had in the wiring and speed controllers, which can be of lighter gauge wire, and/or use smaller and lighter transistors.

Over here we use 240V AC for house circuitry, and can pull 13A from any socket. Houses are fused at 60A or 100A total, and things like power showers and cookers with special cable can pull 35A, or 45A using fixed wiring.

Local distribution is at 11KV..a sensible compromise between cable size and insulation requirements.

Trunks are IIRC 45KV, 132Kv and 270KV depending on the size.

Trains running of a third fat rail use 630V DC IIRC and need substations along the track to feed them.

Overhead train wiring uses something like 100KV IIRC - can't remember. Its good at breaking through ice and means less heavy gauge wire is needed.

[MatC]

Thanks vintage1 (can I call you V1 for short? ;) ) ...
So are the higher voltage cordless drills you see (they've gone from 7.2 to 24v in the time I can remember) just marketing?

[vintage1]

yes and no.

(Our cordless are up to 14.4v and more).

Yes, in the sense that voltage has no impact on power in absolute terms, but no in the sense that (as Astro found out) a generally safe level of current from a sub C cell is about 20A, and so if you want decent cell life and reliability more power needs more cells.

If you could still look at Astro's old Cobalt range, you would find the sport motors were all rated at 25A-35A..the more power was gotten from more cells. The competition motors were rated higher..for short duration high power.

[pilotpete2]

Overhead train wiring uses something like 100KV IIRC - can't remember. Its good at breaking through ice and means less heavy gauge wire is needed.
vintage1,
A fellow club member is a retired driver for Sl-Amtrak. a while back running a southbound diesel out of Boston he was cruising at 100MPH on the electrified main Northeast corridor, there was a sagging cattenary wire just after a little dip in the tracks that caused just a little bounce at just the wrong place :o , the air horn just over the cab hit the wire :eek: Knocked out service for hours :rolleyes:
Pete

[vintage1]

vintage1,
A fellow club member is a retired driver for Sl-Amtrak. a while back running a southbound diesel out of Boston he was cruising at 100MPH on the electrified main Northeast corridor, there was a sagging cattenary wire just after a little dip in the tracks that caused just a little bounce at just the wrong place :o , the air horn just over the cab hit the wire :eek: Knocked out service for hours :rolleyes:
Pete

Over here it is usually a day.

On some of the busiest commuting lines in the world :D

[pilotpete2]

And here I thought you guys were running 300mph mag-lev trains :)
Pete

[Norman Adlam]

And here I thought you guys were running 300mph mag-lev trains :)
Pete

Nearly right! 30mph Mag-less trains! :p

Cheers,

[Ron van Sommeren]

And here I thought you guys were running 300mph mag-lev trains :) PeteGermany (+ export to China) and Japan.

Vriendelijke groeten ;) Ron

[mtucker]

Controllers can sometimes be less efficient at higher voltages. For example, a lot of wide range DC power supplies will be their least efficient at their lowest Vin rating and their highest Vin rating with best efficiency being in the middle of the Vin range. I would think the same thing could apply to motor controllers.

At lower voltages it is as people have explained... high current and the I-squared losses eating in to the efficiency.
At higher voltages though, some of the switches connected to the positive of your voltage source (normally MOSFETs) take a longer time to turn on (because the gates of the MOSFETs have to be driven with a lot of voltage to get them fully turned on). It takes time to change the voltage on the gates (the control pin). Switching losses is the name for the power loss of the transistor turning from completely off to hard on.

Matt