O.T.: Efficiency for resistive heating of 120V AC vs 240V?

Taking into consideration the North American residential standards of 120V AC and 240V AC, you often see it stated that 240V electrical resistance heaters are "more efficient" than 120V.

I understand that it takes fewer amps to deliver a given wattage when the voltage is higher, but the electric company charges by the watt, not the amp.

So in what way, and by what general amount (in terms of what the homeowner pays or in other terms) is a 1500-watt 240 volt electric baseboard heater more efficient than a 1500-watt 120V heater?

lbenson said:

Taking into consideration the North American residential standards of 120V AC and 240V AC, you often see it stated that 240V electrical resistance heaters are "more efficient" than 120V.

I understand that it takes fewer amps to deliver a given wattage when the voltage is higher, but the electric company charges by the watt, not the amp.

So in what way, and by what general amount (in terms of what the homeowner pays or in other terms) is a 1500-watt 240 volt electric baseboard heater more efficient than a 1500-watt 120V heater?

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More efficient means less loss.
What are the losses? It's the power lost in the cables going to the heater. ie heat not where you want it.
The loss is proportional to the square of the current times the resistance of the cable.
Because it's a square law, the losses go up quickly as current increases.

Efficiency is probably meant in terms of ability to produce heat versus power applied.

Could it be that it's not P=IV which is important, but P=I2R or P=V2/R ?

Added : Probably what BeanieBots is also saying, though I was thinking in terms of the heating element resistance rather than losses. And probably why I couldn't get the maths to work to support the claim that higher voltage, lower current is better.

Ok, thanks. I can understand that with losses proportional to amps squared, lower amps in the 240V circuit means less loss to heat delivered other than to the heating element, so you do pay more per kw of heat with 120V elements because of losses not in those elements.

Now, any kind of quantification for, say, 100 feet, round trip, of #12 wire to a 1500-watt heater for a 240V circuit and a 120V circuit?

On the face of it I'd go along with BB's theory. I often glaze over when people say 2kW heater A is more efficient than 2kW heater B.

"Could it be that it's not P=IV which is important, but P=I2R or P=V2/R ?"
- they're merely rearrangements of the same equation (by substitution of Ohm's law). And , naturally, it effects things whether or not you include the the cable parameters in the terms.
You leccy meter just measures the watts going through it. It doesn't know what's (haha) on the end and doesn't compensate for cable losses.

hippy said:

Efficiency is probably meant in terms of ability to produce heat versus power applied.

.

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When I read this, I had a WTF moment. I am sure you did not mean that 120V 1kw heater produces different amount of heat to 240V 1kw???? ;-)

The efficiency is mainly to do with line losses getting the power to your house. I know you don't pay until it goes through your meter but the losses have to be paid for so are built in to the pricing.

As an aside, many years ago, I read that a filament globe only produces around 5 - 10% light energy as is contained in the coal burnt to produce the power. Don't quote me on the figure!

fred

I wouldn't deny that everyone is paying for line loses outside their homes, but it's based on the meter reading.

What I'm trying to get at is the difference between the meter reading when supplying a nominal 1500-watt electrical resistance heater at 120V AC for a given length of wiring from the meter (or more practically, from the breaker panel), and the meter reading when supplying a 1500-watt 240V heater, same wiring.

First thing is to get the concepts right. Energy is what you pay for from the power company and the units usually used are watt-hours - that is, energy is power times time so the Energy company (basically) integrates power use over time to figure out energy consumption in the billing interval.

Power is the rate at which energy is transferred from one place to another. In electrical terms, that's volts times amps. You can use Ohm's law to express this in different ways based on the load (resistance) in the circuit. Very useful is that power is resistance times amps squared because voltage is easily manipulated while resistance is a parameter of what you are trying to do. (P=EI, E=IR, P= (IR)I = I^2 R)

As for the meter reading - since the power - volts times amps - remains constant, doubling voltage halves the amps. If your house is wired to code, wire size (resistance) is set so as to keep voltage losses below 3% or so at maximum circuit rating determined by the circuit breaker on that circuit. If you want to calculate the actual wire losses, http://www.powerstream.com/Wire_Size.htm is one of the better calculator pages out there, I think. - for instance, a typical 20 amp 120v household circuit is specified in code to be 12 gauge wire which has 2.2% voltage drop on a 40' run - 2.6v drop over 80' of wire for 20a is 52 watts lost in the line for a 1400 watt effective load or 0.3% of the load power is lost in the wiring.

Do please check my arithmetic - it shouldn't change the fact that line power losses in properly designed household distribution systems are down in the noise and pale in comparison to other issues and factors.

I'll bite...

If your meter reads kWhours (highly likely) then nothing - both "consume" 1500W, and if you leave them on for an hour, each will clock up 1.5KWh on the meeter

If your meter reads amps (unlikely) then the 120V AC meter will read about 12A, and the 240V AC meter will read about 6A

lbenson said:

I wouldn't deny that everyone is paying for line loses outside their homes, but it's based on the meter reading.

What I'm trying to get at is the difference between the meter reading when supplying a nominal 1500-watt electrical resistance heater at 120V AC for a given length of wiring from the meter (or more practically, from the breaker panel), and the meter reading when supplying a 1500-watt 240V heater, same wiring.

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I concur with Bryanl. The difference in your house is negligible. There is a difference as the resistance stays the same and current doubles. Your saving over winter would be in the order of cents if you use 240v.

Fred

You would have to consider the element resistance and the out and return resistance of the cable as as potential dividers.

Comparing 120V and 240V (230V) the element resistance would be nominally half for the lower voltage source and therefore the relative values of the cable resistance would be more significant. I2R losses being of more concern.

It's no accident that the National Grid in the UK, and doubtless elsewhere transmit electrical power over the country at tens of Kilo Volts to avoid losses due to cable heating.

One could argue that the lower voltage is actually more efficient because of the extra bonus heating from the supply cable . . .

In standard use the effects will be fairly minimal, so unless you have a long thin appliance lead, I shouldn't get too excited by it all.

Best just to turn your weather up by 1 degree C to achieve better economies in the power system.

To be pedantic MartinM57, you have assumed a perfect wire to the load and a perfect load. There will be a small loss in the practical wiring. This loss will be greater at 120V, so your room will be getting slightly less of the 1.5kw. ;-)

Fred

lbenson said:

I wouldn't deny that everyone is paying for line loses outside their homes, but it's based on the meter reading.

What I'm trying to get at is the difference between the meter reading when supplying a nominal 1500-watt electrical resistance heater at 120V AC for a given length of wiring from the meter (or more practically, from the breaker panel), and the meter reading when supplying a 1500-watt 240V heater, same wiring.

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As others have stated the higher voltage transmission will always be more efficient.

Let's make it easy, a 1200 watt heater.

For 240 VAC that would be 5 amps, 48 ohms
For 120 VAC that would be 10 amps, 12 ohms

The cable is stated to be the same in both cases so keeping it simple let's call it 5 ohms round trip

Total resistance for 240 VAC = 53 ohms
Total resistance for 120 VAC = 17 ohms

Current for a 240 VAC supply = 240/53 = 4.528 A, = 1086.79 watts seen by the meter, = 984.13 watts seen by the heater = 90.5% of metered power
Current for a 120 VAC supply = 120/17 = 7.059 A, = 847.06 watts seen by the meter, = 597.95 watts seen by the heater = 70.6% of metered power

Obviously 5 ohms is higher than the cable resistance would be in a real life situation (although if you have been used to UK ring mains and 240 volts you would be horrified by some US 120 volt domestic installations) but the mathematical relationship is good.

(Most US houses get a 240 volt supply which is from the 120 - 0 - 120 secondary of the local transformer. All regular wall outlets use one or other of the 120 volt legs. Some larger loads e.g. oven, cooktop, air conditioner are hard wired across the 240 volts. There is usually one 30 amp, 240 outlet that is used for the clothes dryer. The scenario of a plug in 1500 watt heater @ 240 volts is unlikely.)

freddagg said:

To be pedantic MartinM57, you have assumed a perfect wire to the load and a perfect load. There will be a small loss in the practical wiring. This loss will be greater at 120V, so your room will be getting slightly less of the 1.5kw. ;-)

Fred

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Yes, it is pedantic If OP can see any significant (interpret that as you will) difference in the kWH readings on the meter, I would be very surprised (or I would find that the wiring in the 120VAC case was getting pretty warm).

I had a variation on this discussion with an electrical engineer I once worked for. (It was a painful proposition.) I was to design and build an electric heater for a box that was to be out in the cold for a few hours. I selected a small 12V battery to power it, along with an appropriately sized resistor for a heater, and a simple temperature control circuit. My boss informed me that I must use a 24V battery to increase the efficiency, as the heater would then draw less current from the battery and put more of the energy into the heater resistor.

I tried to explain to him that it really made no difference because all of the energy would be dissipated in the same insulated box in the form of heat, no matter what, and that a battery of twice the voltage would, (in order to fit in the space available,) have only half the current capacity. He never did grasp any of this and I found another employer as quickly as I could.

Put simply, if you're trying to generate heat electrically in a closed system, efficiency doesn't matter at all, because any losses in the system will become heat, which is what you're trying to get in any event, and power = current times voltage, no squares involved.

In the case of a 120V heater or a 220V heater, in the 220V system a slightly smaller amount of heat will be dissipated in the wiring going to the heater from the step-down transformer than will be dissipated in the wiring leading to the 120V heater. As has been noted earlier, the difference in the cost of heat where you want it will be pennies different.

Pongo finally said it. I was waiting to see if anyone would.

All USA residences are wired with 220/240 to the house for use by high current appliances / systems. There is no real difference between here and there as far as the higher voltage / higher current use goes. Here in the USA also have 120 Volt circuits for low current devices.

The difference might be with small 120 Volt space heaters, but those are always limited by the 15/20 amp fused circuits they are plugged into. These small heating appliances don't create enough voltage drop in the wiring to make a measurable / practical difference.

John West said:

I had a variation on this discussion with an electrical engineer I once worked for. (It was a painful proposition.) I was to design and build an electric heater for a box...

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I've had a similarly painful discussion with a design engineer for a heater system in a closed and well insulated box, but he wasn't as smart as you. He very accurately calculated the heat needed, converted that to volts/amps/resistance, but then wanted to ensure those parameters by having the resistance fed from a linear stabilized PSU included in the box!

Thanks for all of the comments, and before anyone worries, the question was not offered in order to guide any practical behavior--just to gain understanding.

So as I understand it, it is as BeanieBots said in the first reply--the issue of loss is not in the heater elements, but in the wiring. If, as in John West's example, you have a near-perfect environment, with negligible losses outside the heating elements, there would not be much difference. The chart on this site -- http://www.interfacebus.com/Reference_Cable_AWG_Sizes.html -- says AWG 12 gage wire has a resistance of 1.6 ohms per 1000 feet.

Thanks to bryanl for this summary of North American standards: "wire size (resistance) is set so as to keep voltage losses below 3% or so at maximum circuit rating determined by the circuit breaker on that circuit". And also, based on the calculator he linked to, on an 80-foot round trip there is a 1.36% voltage drop for 1500 watts, 120 volts, 12.5 amps, compared to a .34% voltage drop for 1500 watts, 240 volts, 6.25 amps.

So if I have understood these numbers correctly, there is about twice the loss with 120V compared to 240V for this example, but the difference between the two is less than one percent of total metered usage.

You have to admit that it was a very passion fuel subject which made a significant number of posts in quite a short period of time. So quite a bit learned about the feelings and views of the readership.

I probably ought to leave this alone but there are few subleties here that create the sort of misunderstandings that drive a physics teacher insane ...

The load resistance is irrelevant as power consumption in various parts of the circuit is at issue and the load resistance is doing the useful part of the energy expended. So the design of the load is simply a matter of matching it to the supply to get the desired job done.

The key is to figure out the power used in the rest of the circuit as that is not doing the desired work. Efficiency is the ratio of the power doing what you want the power to do compared to the power that's doing other things that can't be avoided.

And, yes, squares do get involved because of the available design parameters for the power distribution system. This is related to why circuit breakers are rated in amps yet your interest is in the power going to your load. The primarly limiting factor in electrical system power distribution is resistance. The relationship between power and resistance does involve a power of two exponent. Electrical power systems are fixed voltage, variable current where the wire size, which determines resistance losses in the circuit, is the primary cost factor. (note that similar concerns influence circuit design for the pins on the Picaxe.)

Yes, you can select a voltage for an electrical power delivery system to use but changing that voltage has its own losses. This is why AC is used for the grid, why cars are heading from 12v to 42v systems, why computers often run off 12v or 48v DC to move the reduction to volt levels needed by processors as local as possible, why household (US, anyway) use split phase systems, why many industrial services use 480 v 3 phase, and so on. Modern electronics is helping with the efficiencies of voltage conversion as you can see if you look at the 5 and 3 volt regulator options for our Picaxe boards. Heaters only need a resistance element to match desired power level to supply voltage. Most other appliances and devices have a much more complicated situation and have to do things like voltage changing, current limiting, and power conditioning to match the device to the available power source.

If you get into line losses for meter reading concerns, you are in the ballpark where you have to consider how much of the line between the meter and the heater is in the desired heating space (its heat losses are helping the heater!), the amount that the heater load resistance changes with temperature as it heats up, losses incurred in regulating temperature, connection losses, - good brainstorming project to make a list of these things. Don't forget metrology concerns, either.

@bryanl

What's the source for that 3% max voltage loss number that you quoted? I know that none of the houses I have owned in the US have met that.

lbenson said:

I wouldn't deny that everyone is paying for line loses outside their homes, but it's based on the meter reading.

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It certainly is!

The cost to generate your power might be of the order of 5 to 10 cents per kWh at the power station (which covers labour and power station profit.
But you pay around 3 to 4 times that based upon your household meter reading.
The difference is to cover the network operators labour costs and the losses in the power lines from the power station to your house.
The transmission and distribution lines have losses (the I2R losses BB flagged) and even transformers have losses of the order of a few percent.
That is why there are transmission lines at 132 kV, 330 kV and even 1,000 kV to reduce the current to move a given amount of energy around and thus reduce losses.
Volt drop is another reason for transmitting at higher voltages otherwise far bigger conductors needed that will be many times larger than needed for the greater current alone.

freddagg said:

hippy said:

Efficiency is probably meant in terms of ability to produce heat versus power applied.

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When I read this, I had a WTF moment. I am sure you did not mean that 120V 1kw heater produces different amount of heat to 240V 1kw???? ;-)

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It was rather clumsily stated.

As others have noted it is overcoming power loss which is the issue so it is "efficiency" in terms of generated power, transferred to the house, then to the heater which is the measure. It's not that the heater itself is more efficient with a higher voltage but that an electrical transmission system as a whole is more efficient using a higher voltage.

The losses in a national grid transmission system scale down to within the house, but the difference within a house, as also noted, will be far less significant. None the less they do exist as Pongo illustrates.

My "power applied" should have been "power generated" or "available at source". That is when measured at source, not at the point it is input into the actual heater.

bryanl said:

I probably ought to leave this alone but there are few subleties here that create the sort of misunderstandings that drive a physics teacher insane ...

And, yes, squares do get involved because of the available design parameters for the power distribution system.

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I stand by my earlier comment, as it was clearly stated to be in regards to the closed system I described, not the power distribution grid. No amount of mathematics or electronic wizardry will make a closed electrical heating system any more or less efficient, as it is a 100% efficient system no matter what components are used or how they are employed. Any energy losses in the electrical system anywhere will become heat at 100% efficiency. The laws of thermodynamics apply.

Line losses in the house will also be in the form of heat, which is what you want. So no difference whatsoever in the house, though a tiny bit more of the heat might be in the house wiring in a 120v system compared to a 230v system, but it still heats the house. In the distribution network it is another matter, there, high volts (or, more to the point, low amps) is by far the best. Which is why they use 440kV on main routes, in the UK at least.

The efficiency of 240V over 120V in a household type installation has to do with the reduced wire size from the breaker panel to the heating appliance. This is why large resistive heating loads such as cooking stoves and clothes dryers are wired at 240V. While they could be run on 120V the wire size would have to be larger (if you are following safety codes) and therefore more expensive.

Otherwise it is all the same...

Oh wow I can't resist... what about the radiated electromagnetic energy which escapes the house? have ya'll got shielded twisted wires? [inside a mu-metal steel box...] Miniscule but not zero and perhaps enough energy to run the incredible PICAXE chip!

This energy would cause heating by induction in the pipes (assuming metal) and hence produce further heat in the house. But not all will do so some may be wasted.

What about the capacitive losses? The wires are in close proximity so there are losses. These losses are real though small. There has been a strange phenomena of flashing compact flouros due it the capacitance of the switch wire passing enough current over time to enable the CF to flash.

Fred

premelec said:

Oh wow I can't resist... what about the radiated electromagnetic energy which escapes the house? have ya'll got shielded twisted wires? [inside a mu-metal steel box...] Miniscule but not zero and perhaps enough energy to run the incredible PICAXE chip!

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So we should all be living in Faraday cages.... huh??

However, picking up stray power for a PICAXE to run off.... interesting idea. Might work, then again it might not. Can also experiment with antennae and Germanium diodes to rectify. Need to live close to a transmitter though.