I2R loss in common neutral circuit

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peteo

peteo

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A recent posting here made it clear that shared neutrals is an important subject. Rather than discuss the safety or installation requirements, it would be valuable to find a more meaningful way to express energy savings than ?it saves electricity.? This subject is not addressed in any of the Cal Energy Commission or Berkeley Lab papers I?ve seen.:confused:

Looking at statistical data such as http://www.eia.doe.gov/emeu/cbecs/cbecs2003/detailed_tables_2003/2003set8/2003pdf/b44.pdf and using even simple payback methods is too much to deal with for this. No ?typical? branch circuit could be found. Instead, it was decided to solve someone else?s sample for a benchmark value. The 120/240 common neutral circuit in article 225 was able to be solved, for ?how much current would it take, to save 60 watts.? This ignores any harmonic effects, derating, cost of copper, cable fill, etc. It is contended that a MWBC setup will fall between the homerun and common neutral values. Anyone can understand that paying for a 60W light bulb should be avoided.

Please understand what the intention is here, not to compare life cycle or installation costs, but just to point out that ?doing it this way will make your meter move less for the same load.? This is intended to be an intuitively obvious comparison of competing alternatives.


Questions:
1) ?The energy used heating cables is 60W when the circuits carry 9.4 amps.?
2) Assumption of 12 AWG wire at 0.85 pf; simple DC resistance formula (conveniently, Table 9 gives 1.7 ohm/1000?). Formulas used and numbers arrived at
3) Assumption that cables heating in the wall will be reflected at customers electric meter.
4) Drawings used to show the options.
5) A better statement, easier to remember or more useful or accurate, whatever.


Calculation for 8 ? 50 foot 12 gauge conductors equivalent to a 60W I2R loss:

Step 1 (8*50?)(1.7/1000?) => 6.8 ohms
Step 2 I2R = W => I = sqrt(60W/0.68ohms) = sqrt(88.2) => I = 9.4A



PS if the above is correct, the heating is 100W at 12.1 Amps and 150W at 14.85 Amps. In the 60W case the above translates into a significant energy savings, 60W on an 1134 W circuit or 5%. Free electricity? Why I?m still skeptical about this?
 
I think that your math is a bit wrong, but maybe I am misunderstanding it. If I understand what you are trying to calculate, then you've calculated losses that are much larger than reality.

First, I think that your 6.8 ohm result is a typo. In the next line you use 0.68 ohms, which is the resistance of 400 feet of 12ga conductor.

In your first equation you use 8*50 to get the 400 feet, and in your diagrams you show 8 circuits, which suggests that your home run length is only 50 feet, not 400 feet. The resistance of 50 feet of 12ga conductor is only 0.085 ohms.

Finally, in a circuit you have loss on both sides of the circuit, so depending upon the configuration you may need to double the resistance.

1) For a standard 'home run' circuit, 200 feet in length, you have a resistance of 0.68 ohms. As you correctly calculate, the current needed to give 60W of loss is 9.4A. The voltage drop to the load would be 6.4V, the power delivered to the load would be 1070W, with 60W lost as heat in the wire.

2) Lets change this circuit to a MWBC. Now to get the loss of 60W, we again need 9.4A, but now the voltage drop is versus 240V, and the power delivered is versus 240V. For the same 60W lost in the wires, we deliver 2196W to the load.

3) Now let us consider your _8_ circuits. These are 50 foot long circuits, and _each_ carries 9.4A
3A) Home-runs only. The circuit is 50 feet in length, 100 foot round trip. Resistance is 0.17 ohms. Voltage drop is 1.6V. Loss in each circuit is 15W, so with 8 circuits the total loss is 120W. Power delivered to the load is 1113W, efficiency of 98.7%

3B) MWBC. Each circuit is 50 feet in length, 100 foot round trip. Because these are balanced MWBC there is no current flow, and thus no loss in the neutral. Loss in each circuit _pair_ is 15W, total loss over 8 circuits (4 MWBC) is 60W. Voltage drop is 1.6V. Loss in each pair of circuits is 15W, power delivered to the load is 2241W per pair of circuits, efficiency is 99.3%. _However_ if you turn off half the circuits (all of the red halves), then the total loss will remain 60W and efficiency will drop to 98.7%

3C) Shared neutral circuit. Again there is no current flow in the neutral, and thus no loss in the neutral. Total loss is 60W. However if you turn off all of the red circuits, you get a different result. The loss on the 'hots' is still 30W, but the loss on the neutral is different. Now there is 4*9.4A = 37.6A flowing in the neutral. Because of the requirements of a shared neutral circuit, the neutral needs an ampacity of 80A, and is thus a #4 conductor (ignore derating for the moment). The resistance of #4 is 0.3 ohms/1000 feet, so the loss on the neutral is 21W. So in the shared neutral circuit, the total loss with half the circuits off would be 51W.

-Jon
 
Peteo, I like to be a little more simplistc in showing this. I know the 4 circuits in the text are pretty unrealistic in a dwelling but it shows the point.


Hypothetical situation; A home has (4) 20 amp circuits (loaded to 15 amps each) wired as two wire circuits. We use #12 conductors, at 100?.


#12 = 1.98 ohms per K? x .100 = .198

I^2 R heat loss

15 x 15 x .198 x 8 = 356.4 watts

Now if we took the (4) circuits and installed them as multi wire branch circuits, we would reduce the current carrying conductors to 4. (remember hypothetically 15 amps so each grounded conductor is truly neutral.)

15 x 15 x .198 x 4 = 178.2 watts.

Now if we give this a time frame per day, times 365 days, times umpteen billion nationwide circuits, we have helped reduce the impact on the resources it takes to produce electricity.

Here's a graphic from Ed MacLaren

3wire.JPG


Roger
 
I am not an engineer and most of the math on your post is beyond me. I may be over-simplifying but I am not sure that the I2R loss for a typical circuit can be easily calculated. For example if the load is resistance the reduced voltage will result in a reduced wattage output from the equipment, so the total wattage will be close to the same figure that would result in a circuit with less voltage drop. Also meters do not read neutral loads, they calculate the amperage on the ungrounded conductors, so unless the shared neutral results in less wattage on these conductors the rotation of the meter will be unaffected. This is interestig and I would like to understand more. Please let me know if I am missing the obvious.
 
How come the impedances of load 1 and load 2 changed in diagram B?

These need to be constant in both examples.

Do the math, the answer may surprise you...
 
The impedances of the loads didn't change. There was simply a power increase due to decreased voltage drop before reaching the loads. That's exactly his point.
 
Are you sure?

(The result will still show more watts being consumed by the loads.)

It will also show more overall watts consumed.
Not same "total power consumed"

Most importantly the percent of "wasted power" drops in B
 
bcorbin said:
The impedances of the loads didn't change.
Click to expand...
Yes they did. Engy is right.

In "Diagram A," a current of 40 amps flows through "Load #1," and the voltage across that load is 112 volts. Ohm's law tells us that the resistance of the load is 112/40, or 2.8 ohms. In "Diagram B," a current of 40 amps flows through the same "Load #1," and the voltage across that load is 118 volts. Ohm's law tells us that the resistance of the load is 118/40, or 2.95 ohms.

Obviously, "Load #1" is different in the two diagrams. Without doing any math, you can see that "Load #2" is also different in the two diagrams. By inspection, we see that the same current gives a different voltage drop, so the two resistances cannot be the same.
 
Yes, the impedences changed, but only by about 5%. Not enough to worry about compared to the fact that the losses were almost cut in half.

Looking at it another way, in both cases, the power delivered to the load is constant. 12000 watts. So in one case, we lose 1000 watts to get 12KW to the load, in the other case we only loose 560 watts.

However, I have one other small issue. This diagram would apply to a feeder. It wouldn't apply to multiwire receptacle circuits. (I think a receptacle circuit is what the original poster was talking about. I don't know of anyone who uses separate neutrals for a feeder:) ).

Still, I think it gives you the general idea that a MWBC creates less power loss. The numbers would just be smaller for a 20A receptacle circuit.

Steve
 
I get numbers similar to Roger's. It is suprisingly high to me. Consider a single 100', branch circuit with 15A flowing on #12 wire. From my Square D calculator, at 95% PF, the voltage drop is .37 volts per amp per 100'. That gives a .37*15A = 5.55 V drop.

5.55V * 15A = 83 watts lost on the wiring.

If we make this circuit part of a 240/120V MWBC with 15 amps flowing on the other circuit, we save half this power. (All the current cancels out on the neutral, eliminating approx. half the voltage drop).

So we basically get a 40W savings per 100', 15A circuit.

Notice this would be lower for a 2 phase 208/120V MWBC, since there would still be some neutral current flowing.

Steve
 
Without paying particular attention to any one post, it looks like the dynamics of the circuitry is being left out sometimes. Here's what I'm getting at:

If, in someone's post above, they show a 40-amp load at the end of a circuit, are they saying that, at 120 volts, the load would be 40 amps, or with the voltage drop applied, the resulting current is 40 amps?

As Charlie said, the two resistances can't be the same. However, they must be, since we're discussing a given load impedance as a constant, and not the load current; the current is dependent on the total circuit impedance.

If you compare the numbers in Roger's pair of diagrams, he shows the total power consumption, as well as each load's current, as being the same. This is not correct: the only constants are the resistances (using the loads as simple resistances).

There's no way the current for load #1 can be the same in both cases when the wattage is different; that implies not using the same load in both cases. It should be using the same load resistances, wire resistances, and supply voltages for both circuit arrangements.

If this were done, the total power consumption for A and B would be different; they would not both be 12KW. The proper way to solve this is to use the impedance of the load, calculated by using the current-at-what-voltage numbers.

For example, if load #1 was given to be 40 amps at 120 volts, we get (120/40) 3 ohms. Using this figure, and adding the two .1-ohm conductors, we can calculate the current to be (120/3.2) 37.5 amps. That's load #1's current in A.

Using that, voltage drop across the load is (37.5*3) 112.5 volts, and along each conductor is (37.5*0.1) 3.75 volts, for a total "loss" drop of (2*3.75) 7.5 volts. Total power is 4.5 KW. As a check, add that to the load, and we get 120 volts.

Circuit B is done the same way:

If it's 60 amps at 120 volts, it is 2 ohms, circuit current is 54.5 amps (rounded), load voltage is 109 volts, and conductor voltage drop is 10.9 volts. Total power is 6.54 KW (at 120 volts; the total voltage is 119.9 because of rounding earlier.)

Note that the total power for both circuits is not 12 KW; it's 11.04 KW. The second diagram I can do also, but not until later.
 
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Okay, now I have time for round two: Diagram B.

Once again, the constants are the supply voltages, the load resistances, and the wires betwixt the supply and the loads. The two loops are still calculated independently. The only real difference is that we add the currents in the neutral algebraically, meaning that we'll be adding a positive and a negative; the result being the imbalance.

Load #1 is 37.5 amps, and load #2 is 54.5 amps. If you look at just the neutral's current, the result is (54.5 - 37.5) 17 amps. The drop on the neutral is (17*0.1) 1.7 volts. The power losses are looked at for each resistance. For example, this neutral's loss would be (17*1.7) 28.9 watts. What are the other power losses, and how much is waste?

Back to Diagram A, load #1's power consumption is 4.5 KW: 4218.75 across the load, and 140.625 across each conductor. Total: 4500 watts. Remember this. For Diagram A, load #2's power consumption is 6.54 KW: 5,9405 watts across the load, and 297.025 watts across each conductor. Total: 6534.55; round up to 6540.

Now, let's use Diagram B and combine the neutrals. The voltage drops for the two hots and for the two loads are the same. What changes are the neutral's numbers. With a current of only 17 amps, the power loss is only 28.9 watts, where the two separate neutrals in Diagram A have a combined loss of (140.625 + 297.025) 437.65 watts.

For Diagram B, the total power consumption is found by adding the power of each load, one of each hot, and the shared neutral. Where does that leave us? Let's recap: The total power consumption of Diagram A is 11.04KW, while the total consumption of Diagram B is 11.01655, for a difference of 234.5 watts saved using the shared neutral.


Whew! I hope my numbers are correct.
 
Boy Larry, you sure went to a lot of trouble to agree with Ed's point. :) Now let's say that for what ever reason someone or some automatic means adjusted the loads and the loads currents remain the same regardless of voltage dropped across the loads, (yes the loads resistances would have to change) it ultimately shows the MWBC being the more efficient circuit which is what Ed's drawings are showing.


BTW, has anyone here heard from Ed? I hope he is doing well.



Roger
 
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Of course, a constant-current supply would adjust its voltage output to attempt to maintain the set current level. A varying load impedance to suit a constant power level is not a realistic condition.

My point of view was from a real-world, how-do-we-figure-this-out, practical angle. Way back when I was going to night school, we were discussing the cause and effect of voltage drop on a long extension cord.

I explained that loads don't "draw" current, they allow it to pass through. (Motors excluded for this example) We had to start with the load's current at normal voltage first, to determine its impedance.

Then the conuctor impedance can be added to the load impedance, to figure the actual current. Only then can a realistic voltage drop and power loss be calculated. But you guys already knew that.
 
Larry, I think that you bring up a good point about ignoring the dynamics of the load when considering voltage drop. At the same time, I think that it is a very good approximation technique to simply assume a constant load current; it is simply important that one be aware that the result is only approximate.

With any reasonable voltage drop, the voltage applied to the load will vary by less than 10%, so the current will similarly vary by less than 10%. Thus in reasonable conditions the error in the voltage drop calculation will be a few percent if you assume constant current. The error introduced by presuming constant current is probably smaller than the error introduced by not considering the temperature change of resistance of the conductors when calculating voltage drop.

I'd also like to note that with the existence of switching power supplies, constant power _loads_ on the electrical system are reasonably common.

-Jon
 
I guess I made a common mistake in my first glance. I knew the thrust of the argument, saw it verified (I knew it to be true), and missed the detail of the impedances changing slightly. Lesson learned.
 
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