120VAC vs 240VAC: Energy cost comparison.

[PeterJ]

No, I'm not trying to steal from my electric company, I just want to understand this. I've been told by many instructors over the years that 240VAC is cheaper to operate than 120VAC. So here is the hypothetical:

It takes a consistent 2400W to perform the required heating at a U.S. residence. I'm presenting 3 scenarios used to achieve the same 2400W of heating in simple resistive circuits:

#1 - A single 2400W 240VAC heater is used.

#2 - Two 1200W 120VAC heaters are used, both drawing from the same phase.

#3 - Two 1200W 120VAC heaters are used, each drawing from opposite phases.


Scenario 1 draws 10A, and scenario #2, and #3 each draw 20A.

The Question: How does the electrical power meter differentiate between scenario #1, #2, and #3?

Thanks,
PeterJ

 

[augie47]

In terms of energy costs, the difference would be infinitesimal. You are consuming and being charged for wattage, hence kilowatt hours.
In theoretical terms if you looked at the entire system there might be some losses from resistance of wiring, etc., but in terms of load (wattage), no difference.

 

[GoldDigger]

The Question: How does the electrical power meter differentiate between scenario #1, #2, and #3?

Thanks,
PeterJ

Basically, the power meter looks at the current in each of the opposite phase 120 volt wires, and adds up the total (multiplying by the voltage (120 or more accurately 240/2) along the way.) That gives watts, which you are billed for.
The actual details are a bit more complicated than that, in that the meter also has to take into account the phase angle between the voltage and the current, but the above description is close enough to answer your concern.

 

[iwire]

Watts are watts regardless of the voltage and you are charged per watt.

 

[jumper]

Watts are watts regardless of the voltage and you are charged per watt.

That is what I have always learned. The energy consumption is the same.

 

[Sahib]

In scenario#1 and scenario#3, there is no current in neutral. So there is significant energy saving in both of the cases if long lengths circuits are present.

 

[iwire]

In scenario#1 and scenario#3, there is no current in neutral. So there is significant energy saving in both of the cases if long lengths circuits are present.

That depends entirely on the length of the circuits and the wire sizes used.

With a little engineering the 120 volt circuits can be more efficient than the 240 circuits.

 

[Sahib]

With a little engineering the 120 volt circuits can be more efficient than the 240 circuits.

Really?!

 

[macmikeman]

Define "costs" first. 240 volt circuitry uses less copper than equal consumption 120 volt stuff. Branch circuit wiring can be less expensive as well. However there will be $7.75 more in circuit breaker expense for the two pole than the single pole breaker up to 50 amps.

 

[iwire]

Really?!

As far as KWH cost to the customer, yes without a doubt.

 

[iwire]

Define "costs" first. 240 volt circuitry uses less copper than equal consumption 120 volt stuff. Branch circuit wiring can be less expensive as well. However there will be $7.75 more in circuit breaker expense for the two pole than the single pole breaker up to 50 amps.

:thumbsup:

I agree with you, we need to be on the same page as far as costs.

So far my comments have been based only on KWH usage.

 

[Sahib]

Define "costs" first. 240 volt circuitry uses less copper than equal consumption 120 volt stuff. Branch circuit wiring can be less expensive as well. However there will be $7.75 more in circuit breaker expense for the two pole than the single pole breaker up to 50 amps.

What happens at the break-even point and beyond?

 

[PeterJ]

Gentlemen,

Please address the question: How does the electrical power meter differentiate between scenario #1, #2, and #3?

I know of no way to measure watts in this scenario. I think the wheel in the meter is running on current, and the KW amount is factored into the design, but I do not know this.

Thanks,
PeterJ

 

[PeterJ]

GoldDigger,

Thank you for your answer. I don't understand how the meter can differentiate between Scenario #1, and Scenario #3.

I did come across this information by "Geezer" on another site:

"The traditional meter usually has a metal disc that rotates as power is consumed. The disc is actually part of an electric motor that rotates at a speed that is proportional to the product of the voltage and the current being supplied. If there is voltage but no current flow, it does not rotate, and if there happened to be a current but no voltage (very unlikely), it would not rotate either.

The motor drives a train of gears that count the number of revolutions of the disc.

The only slightly tricky bit is that the disc reacts to the components of voltage and current that have the same phase relationship, and therefore it is measuring the actual power delivered. It really is recording energy delivered, usually in kilowatt-hours.

If, for some reason, the voltage drops and the lights go dim, the motor does slow down, so you are not paying for power that was not delivered."

I still, really, don't understand it.

PeterJ

 

[PeterJ]

Gentlemen,

Please address the op: The Question: How does the electrical power meter differentiate between scenario #1, #2, and #3?

Thanks,
PeterJ

 

[robbietan]

GoldDigger,

Thank you for your answer. I don't understand how the meter can differentiate between Scenario #1, and Scenario #3.

I did come across this information by "Geezer" on another site:

"The traditional meter usually has a metal disc that rotates as power is consumed. The disc is actually part of an electric motor that rotates at a speed that is proportional to the product of the voltage and the current being supplied. If there is voltage but no current flow, it does not rotate, and if there happened to be a current but no voltage (very unlikely), it would not rotate either.

The motor drives a train of gears that count the number of revolutions of the disc.

The only slightly tricky bit is that the disc reacts to the components of voltage and current that have the same phase relationship, and therefore it is measuring the actual power delivered. It really is recording energy delivered, usually in kilowatt-hours.

If, for some reason, the voltage drops and the lights go dim, the motor does slow down, so you are not paying for power that was not delivered."

I still, really, don't understand it.

PeterJ

during a voltage sag, the meter will still be able to record the consumption. however like all meters, there is a level of accuracy involved. for current transformers, there is a set range where they measure with 99.99% accuracy, go beyond this range and the accuracy falls off. same is true with potential transformers

 

[GoldDigger]

Gentlemen,

Please address the op: The Question: How does the electrical power meter differentiate between scenario #1, #2, and #3?

Thanks,
PeterJ

In case 1, there is 10A flowing in L1 and in L2.
In case 2, there is 20A flowing only in L1 and nothing in L2 (the return current is all on the neutral, N, which does not go through the meter.)
In case 3, there is 10A flowing in L1 and 10A flowing in L2, with no current in N.

Can you see that if the meter measures separately the L1 current and the L2 current it can distinguish between case 2 and the others (although it does not actually need to)?
Can you see that it cannot distinguish between case 1 and case 3, and will read the same amount of power used, which is exactly what it should do?

Formula: Power = (I_L1 + I_L2) x 120. Or to make it closer to the way it actually works: Power = (I_L1 + I_L2) x (240/2)

 

[robbietan]

Gentlemen,

Please address the op: The Question: How does the electrical power meter differentiate between scenario #1, #2, and #3?

Thanks,
PeterJ

being in a country where the voltage is 220V line to line, i would say that there is no difference here in terms of kilowatt hr consumption. used to be some meters have only one current transformer located on the "hot" leg. what some people would do is try to reverse the meter so that the CT will be measuring the neutral leg. and get some "savings". however newer meters, especially the electronic meters that is now standard, have both legs equipped with CT. and can accurately measure power even in the presence of high voltage distortion from harmonics

 

[PeterJ]

If I had to guess, I would expect that the meter wheel is driven by current from each scenario and the rate is cumulative according to the current measured from all three circuits. Thus, the energy cost for 240V and 120V applications are essentially the same.

But I don't KNOW this.

PeterJ

 

[PeterJ]

In case 1, there is 10A flowing in L1 and in L2.
In case 2, there is 20A flowing only in L1 and nothing in L2 (the return current is all on the neutral, N, which does not go through the meter.)
In case 3, there is 10A flowing in L1 and 10A flowing in L2, with no current in N.

Can you see that if the meter measures separately the L1 current and the L2 current it can distinguish between case 2 and the others (although it does not actually need to)?
Can you see that it cannot distinguish between case 1 and case 3, and will read the same amount of power used, which is exactly what it should do?

Formula: Power = (I_L1 + I_L2) x 120. Or to make it closer to the way it actually works: Power = (I_L1 + I_L2) x (240/2)

THANK YOU! This makes sense.

PeterJ