110v verses 220v Load

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brian777

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The guys at work were all wondering how running a 220v airconditioner or any load for that matter would be cheaper per kilowatt hr than 110v. I know it is I just dont know how to explain it electrically. Can anyone help me on this?

Brian
 
With no other variables, to produce the same amount of work they would consume the same amout of power, E X I = W

Let's assume a hypothetical 120 v AC unit at 20 amps

120 x 20 = 2400 watts

now if we use the same unit at twice the voltage we would use half the current.

240 x 10 = 2400 watts

Roger
 
The simple answer is that there are fewer losses. More of the energy supplied by the power system goes into the room (in the form of cooler air), and less of the energy is lost in the wires (i.e., I^2R losses) or in the inefficiencies of the A/C equipment (e.g., the motor).
 
charlie b said:
The simple answer is that there are fewer losses. More of the energy supplied by the power system goes into the room (in the form of cooler air), and less of the energy is lost in the wires (i.e., I^2R losses) or in the inefficiencies of the A/C equipment (e.g., the motor).
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Energy wise that is true, but the energy savings may not be noticable much over the life of the unit.

One might think that the somewhat smaller size of a 220V motor versus a 110V motor might make a 220V A/C less expensive for the same BTU of cooling. This does not take into consideration that 110V units outsell 220V units by a lot, so the lower voltage units are less expensive.
 
Asking this is like asking 'Which weighs more, a pound of feathers or a pound of lead?'

The first order answer is that if you have a 1KW load designed for 120V, and you compare it to a 1KW load designed for 240V, then they cost the same to operate.

But then you find out that the pound of feathers takes up more space, and so it gets supplied in a heavier bag.

Delivering 1KW at 120V will in general be less efficient than delivering that same 1KW at 240V. This means that you have more losses in the wires, or you've had to pay for thicker wires. So you might have power transmission effects that make one voltage more efficient than the other.

But the thing that trumps this all is differences in the appliances. Two air conditioners of the same BTU rating might consume vastly different amounts of power. A high efficiency 120V unit will be cheaper to operate than a low efficiency 240V unit, compared at the job of cooling the same room.

-Jon
 
winnie said:
Asking this is like asking 'Which weighs more, a pound of feathers or a pound of lead?'
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But it is not the same as asking, "Which weighs more, a pound of feathers or a pound of gold?" The correct answer to that one is that a pound of feathers outweighs a pound of gold. :smile:

(Really! On the level! :smile: )
 
charlie b said:
But it is not the same as asking, "Which weighs more, a pound of feathers or a pound of gold?" The correct answer to that one is that a pound of feathers outweighs a pound of gold. :smile:

(Really! On the level! :smile: )
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True, but apples to oranges comparisons will get you nowhere.:rolleyes:
 
charlie b said:

But it is not the same as asking, "Which weighs more, a pound of feathers or a pound of gold?" The correct answer to that one is that a pound of feathers outweighs a pound of gold. :smile:

(Really! On the level! :smile: )
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Gold is measured in the troy system in which there are 12 troy ounces (each 480 grains) to a troy pound.

Feathers would be measured in the avoirdupois system that has 16 ounces (each 437.5 grains) to the pound.

So a pound of feathers weighs 7000 grains, and a pound of gold weighs 5760 grains.
 
brian777 said:
The guys at work were all wondering how running a 220v airconditioner or any load for that matter would be cheaper per kilowatt hr than 110v. I know it is I just dont know how to explain it electrically. Can anyone help me on this?

Brian
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You could say that volts are cheap while amps are expensive. As the others have said, either unit uses the same amount of power, but the higher voltage pulls a lower current. The higher the current the more losses any system experiences, so lower currents are more efficient. Switching one air conditioner from 110 to 220 isn't going to make much difference, but if an entire building were designed around this you would save some $$. Use 277 lighting instead of 120, 480 heat pumps instead of 208. It adds up.
 
charlie b said:
The simple answer is that there are fewer losses. More of the energy supplied by the power system goes into the room (in the form of cooler air), and less of the energy is lost in the wires (i.e., I^2R losses) or in the inefficiencies of the A/C equipment (e.g., the motor).
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You lost me, assuming I use larger conductors with the lower voltage the losses would be equal would they not? :-?
 
Operating costs will be comparable, but 220 VAC motors are less likely to cause noticeable voltage drops when starting, assuming reasonable wiring capacity to begin with.
 
iwire said:
You lost me, assuming I use larger conductors with the lower voltage the losses would be equal would they not? :-?
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Not quite. The system with the higher voltage will have a lower current, and will therefore use a smaller (higher resistance) wire. That part you have correct.

But the power lost within the wires varies with the value of resistance, but also varies with the square of the current. So if the current is half, that will cut the power loss down to 25%. At the same time, since the current is half, the wire is smaller, and the resistance of the wire might be double. That will tend to double the power loss in the wire. The combined effect is (1/4) times (2), for a net of (1/2). Thus, with the higher voltage (all else remaining the same), you get a savings of half the I^2R power losses in the wires.

That said, I think this is going to be a small difference. I think the improved efficiencies you get from the higher voltage equipment will be more significant.
 
charlie b said:
Thus, with the higher voltage (all else remaining the same), you get a savings of half the I^2R power losses in the wires.
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Let's suppose you wire each of our two hypothetical circuits for the same % of voltage drop; that's the usual measuring stick, right?

How would that affect your conclusion? I think it still works. I've always been a higher-voltage-is-better kind of guy myself.
 
LarryFine said:
Let's suppose you wire each of our two hypothetical circuits for the same % of voltage drop; that's the usual measuring stick, right? How would that affect your conclusion? I think it still works.
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I must admit that I changed my opinion twice, while typing out my reply. Each time I thought it through, I came up with a different opinion. So I had better post this quickly, before I have time to change my opinion again. ;) :grin:

My opinion is that it will not work the same way, if you express your hypothetical question in terms of "same % voltage drop." Indeed, the power losses turn out to be worse for the higher voltage system. Please note, however, that the "usual measuring stick" is really not having the same %VD. Rather, we select conductor sizes based on load amps, and we check to make sure the VD is within acceptable limits. That will not always make the VD have the same percent value for the 220V and 110V situations.

Here is the basis for my conclusion. There are two ways of calculating the power loss in a wire. One is I^2R. The other is (V^2)/R. In the later expression, the "V" is not the system voltage, but rather the voltage drop along the wire. So if you propose to keep the same voltage drop percentage in a 110V system and in a 220V system, the actual "V" number (VD along the line) will be double for the 220V system. So when you compare the ?(V^2)/R? of the 220V system to the ?(V^2)/R? of the 110V system, you get a ?double the VD valued, squared? factor multiplied by a ?half the resistance value.? The result of ?(2 times 2) divided by 2? is that you have just doubled the amount of power losses in the wire.
 
Charlie, try figuring the losses using I*V.

If you design for the same % voltage drop, and maintain constant load KW, then the wiring loss will remain constant.

For the same load power, I_load * V_load is constant. As V_load increases, I_load decreases. (Note: this is not a real load characteristic, this is a design load, where we adjust things so that same power is consumed at different voltages.) Since we've made V_drop a constant proportion of V_supply, I_load is inversly proportional to V_load, V_drop * I_load remains constant.

-Jon
 
The 120 volt unit runs cheaper. When you go into Sears and buy the thing, the expert man there will answer any question you may have about whether it is necessary to run a dedicated line for it with this method" See the cord end? This will fit all your standard receptacle outlets maa'm, no need at all to call in an electrician. So therefore they are saving money..... for a while.;)
 
The simple answer is that your utility electric meter reads neither volts nor amps, but the product of the two which we call VA or Watts. If a load measures 1000 watts, you will pay for 1 kwh for each hour that that load is energized. For an air conditioners of equal wattage, a 240-volt unit will draw half the amperage of a 120-volt unit. Any savings will be due to this lower amperage. Lower amperage allows the electrician to install a smaller (less expensive conductor). Lower amperage will allow less "voltage drop" on a given conductor, so the unit is more likely to receive the optimum voltage for which it was designed. These savings are usually minuscule over the life of a unit. Much more savings may be realized by purchasing a unit that is properly sized for the application.
 
And too:

And too:

The plugs and receptacles tend to get hot with these plugin units, and that problem seems to be worse with 120V.

And, although you save a bit on wire with 240V, a two-pole breaker is required, so the 240V installation may cost a bit more on a short run.

I vote for 240V in any case.
 
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