208/1 and 480/1 load calcs?

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The answer is 9.3A. (assuming pf=1.0 and symmetrical components are not invoked due to an unbalanced system, thanks Rattus:rolleyes: )
Click to expand...


So let me get this straight. I have a feeder to a 3 phase 208Y/120 volt panel that contains one 2 pole CB connected to phases A & B. The only connected load is rated at 2228 watts at 208 volts. So you're telling me that if I take current readings on the feeder my current reading will be less than 10.7 amps on the A & B phases?
 
Infinity,

In the specific example of a _single_ 2228W 208V resistive load A to B, as the _only_ load on the feeder, the measured current would be 10.7A, as rattus noted, and I think that you agree with.

This is, however, an unbalanced system, and even though we've specified a resistive load, the current on A and B will be out of phase with the respective line to neutral voltages on A and B.

If we had _three_ of these heaters, and placed them in a balanced fashion on our feeder, then the total current on each feeder leg would be 18.6A, with each leg feeding half the VA of two loads.

I believe that for the case of calculating the over-all load on a feeder, with many diverse loads, one can reasonably make the _approximation_ that the overall load is balanced, in which case we can _approximate_ the incremental current on each leg caused by the 2228W load. If we approximate the system as balanced, then the approximate additional current on legs A and B caused by the 2228W load will be 9.3A.

-Jon
 
Watts is watts!

Watts is watts!

kingpb,

Watts is watts no matter how you slice it........

The load is 2228Watts.
Each leg of the single phase load is 120V to ground and this single phase load has two 120V legs.
The total wattage of the load is 2228 / 120 = 18.57Amps. Since there are two legs of 120V, each would carry 50% of the total amperage, or 9.285Amps.

Now if we have 3 single phase loads at 208V and each one draws 2228Watts, the total watts would be 6684Watts.
This would be a balanced load of 2 of the A phase, 2 of the B phase and 2 of the C phase.
That means there is 6684Watts divided evenly between 3 phases of 120V which equals 2228Watts per phase.
That means 2228Watts divided by 120V equals 18.57Amps on each phase. Each phase is used only once per breaker-load so each 2 pole breaker would use one half of the 18.57Amps per leg. Half of 18.57Amps is 9.285Amps.
That means each 2 pole breaker supplies each load with 9.285Amps two times for a total of 2228Watts.

Or, 2228W x 3 = 6684W / 208V x sqrt of 3 (360)= 18.57A per single phase load / 2 = 9.285A per leg.


........right?:mrgreen:
 
Sparky,

I read the original post to mean a line to line load of 2228 Watts. Then the applied voltage is 208V which yields an actual current of 10.7A. In this case the current lags Vphase for a PF of 0.867.

If these were line to neutral loads the PFs would be 1.0, and I think you would not need a 2P breaker.
 
kingpb said:
..because of the differences in a 208Y/120V system versus a 120/240V ..the line-to-line voltage is not double the line-neutral voltage, and thusly you have to use the phase load, divided by the line-to-neutral voltage.
Click to expand...
This approach was new to me, but I made it click better by visualizing it this way.

Using 2 phases only.
1) With given total load = 2228Watts
2) Must get phase load = 1114Watts 1/2 of load
3) Must use phase volts = 120Volts
4) Divide per phase amps = 9.283A

I can remember this nice trick for a long time, if it works for 3-phase loads?

Say an odd 3-phase motor.
1) With given total load = 2228Watts
2) Must get phase load = 742.67Watts or 1/3 of load
3) Must use phase volts = 120Volts
4) Divide per phase amps = 6.19A

If this is right, then I like it and won't forget it.
 
Ramsy,
You neglected the power factor in your first example. Throw in a PF of 0.867; then the current is 10.7A. Yes, there are phase differences between the current and the two phase voltages even if the line to line load is resistive.

There can be no argument that the current through the (resistive) line to line load is,

I = P/V = 2228W/208V = 10.7A

The load is balanced in your second example, therefore you can treat it like a wye configuration (as you did) even if it is wired in a delta.
 
ramsy said:
This approach was new to me, but I made it click better by visualizing it this way.

Using 2 phases only.
1) With given total load = 2228Watts
2) Must get phase load = 1114Watts 1/2 of load
3) Must use phase volts = 120Volts
4) Divide per phase amps = 9.283A

I can remember this nice trick for a long time, if it works for 3-phase loads?

Say an odd 3-phase motor.
1) With given total load = 2228Watts
2) Must get phase load = 742.67Watts or 1/3 of load
3) Must use phase volts = 120Volts
4) Divide per phase amps = 6.19A

If this is right, then I like it and won't forget it.
Click to expand...

Yes, Roger it works for three phase loads as well. Rattus is correct in needing to figure a power factor, for those loads requiring it.

i.e. if the 2228W had a pf of 0.93, you simply convert KW to KVA before dividing by the number of phases or poles.

In this case say it was a motor, and had a pf = 0.93, then it would be:

2228W/0.93 = 2396 VA (Total VA)

Then divide by 2 (2 phases/poles), and then divide by your line voltage 120V (for 120V system).

I = 9.98A (OK, 10A)
 
King,

I am trying to make the point that, in this example, PF must be considered even for a resistive load because the load current leads/lags the phase voltages by 30 degrees.

Then,

P = 2*Vphase*I*cos(30) = 2*120V*10.71A*0.867 = 2228W

I see no way to obtain 9.3A in this example, although this may be the accepted method for panel schedules.
 
rattus: "I am trying to make the point that, in this example, PF must be considered even for a resistive load because the load current leads/lags the phase voltages by 30 degrees."

Don't get me wrong, in theory I agree 100%, the angle between voltages and current play a real part in electricity. For panel calculations, I will adjust the rated load using the pf in the load determination, and thusly provide the VA rating on the schedule not the watts, avoiding any confusion.

In this discussion, I wanted to avoid to much engineering theory, simply because, as we have both been witness too, when we start talking angles, magnitudes, and phases, the topics can quickly escalate into a brew-ha-ha over minute details. Specifcally, remember the "light bulb is not a purely resistive load argument"?

I'll bet everybody agrees that the 3 phase formula looks like this:

Ptot = sqrt3 * Vll * Iph * cos(ang)

Now solve for current; when you do, you are actually dividing P by 3, (load on each phase), and converting, either the current or voltage (depending on delta or wye) to phase current or voltage. (formula works the same for either, you can all do the math). Anyway, point has been made.

Anyone up for a lively discussion on symmetrical components? Nothing like a little positive, negative, and zero sequence talk. Now that's engineering! :D
 
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kingpb said:
The answer is 9.3A. (assuming pf=1.0 and symmetrical components are not invoked due to an unbalanced system, thanks Rattus:rolleyes: )
Click to expand...


Sorry kingpb, but you are mistaken. For a single phase load connected between two phases, the current throught the load is the power divided by the line-line voltage. For the original example, 1228 amps/208 volts is 10.7 amps. That's ohms law, pure and simple (assuming a resistive load).

And if that is the only load on the panel, that same 10.7 amps must be the current flowing into 2 phases of the panel mains.


On the other hand, for my example of dividing the power by two, and putting half on each phase, you mentioned for a single load, the load current has to equal the input current. I agree with that. And you made it clear that it does not in my example.

The problem is somewhere in how I divided the loads into the two phases. I haven't figured out where I went wrong yet, but it is obvious that the input current has to be 10.7 amps, not 9.2 amps.

Steve
 
steve66 said:
Sorry kingpb, but you are mistaken. For a single phase load connected between two phases, the current throught the load is the power divided by the line-line voltage. For the original example, 1228 amps/208 volts is 10.7 amps. That's ohms law, pure and simple (assuming a resistive load).

And if that is the only load on the panel, that same 10.7 amps must be the current flowing into 2 phases of the panel mains.


On the other hand, for my example of dividing the power by two, and putting half on each phase, you mentioned for a single load, the load current has to equal the input current. I agree with that. And you made it clear that it does not in my example.

The problem is somewhere in how I divided the loads into the two phases. I haven't figured out where I went wrong yet, but it is obvious that the input current has to be 10.7 amps, not 9.2 amps.

Steve
Click to expand...

When you figure out your mistake you will see I was correct. Please review my previous post where I provided the power formula derivation. If you want a reference, go to Elements of Power System Analysis, by William D Stevenson, Jr. The formula for power is derived there.

As far as Ohms Law, E = I*R, has nothing to do with power.
 
steve66:

The issue is that you are getting two different numbers because they are both perfectly valid answers for slightly different questions. The current flowing through the heater is 10.7A. The increase in current flow that this heater causes on the feeder is approximately 9.3A

To see why this is the case, consider this single load as part of a _balanced_ set of loads on the same panel.

To simplify things, assume that the only things on this panel are matched resistive loads; and that we have 3 of these 2228W 208V heaters. We balance the panel by connecting one heater from A to B, one from B to C, and one from C to A. The total load is now 6684W, and it is now a _balanced_ three phase resistive load. Each phase will now be supplying 18.6A. Each leg is supplying 2 heaters, and each leg is supplying 18.6A, so it is perfectly reasonable to say that each heater is adding 9.3A of load to each leg.

Examine what is happening at the phase A bus bar. 10.7A is flowing through the appropriate breaker to the A-B connected heater. 10.7A is also flowing through a different breaker to the A-C connected heater. But these two current flows are _not_ in phase with each other, and when you add them up, the total current flowing from the feeder into the phase A bus is only 18.6A. If the two heaters on each leg together result in 18.6A of current flow on the feeder, then it is totally reasonable to say that each adds 9.3A load to the feeder.

Thus the 9.3A number is an approximation that is only as valid as the assumption of over balance on the panel. If the panel is not reasonably balanced, then the load added to the feeder by this heater will be something greater than 9.3A.

-Jon
 
kingpb said:
When you figure out your mistake you will see I was correct. Please review my previous post where I provided the power formula derivation. If you want a reference, go to Elements of Power System Analysis, by William D Stevenson, Jr. The formula for power is derived there.

As far as Ohms Law, E = I*R, has nothing to do with power.
Click to expand...


Kingpb:

If you take 208 volts from a transformer without a centertap, and place it across a 2228 watt load what current do you get? Of course it is 2228/208 = 10.7 amps.

How could it be any different if we add a center tap that isn't connected to the load?

Do you see where you went wrong yet? Its P=I*E. Are you saying P is not = I * E?

Steve
 
winnie said:
steve66:

The issue is that you are getting two different numbers because they are both perfectly valid answers for slightly different questions. The current flowing through the heater is 10.7A. The increase in current flow that this heater causes on the feeder is approximately 9.3A

To see why this is the case, consider this single load as part of a _balanced_ set of loads on the same panel.

To simplify things, assume that the only things on this panel are matched resistive loads; and that we have 3 of these 2228W 208V heaters. We balance the panel by connecting one heater from A to B, one from B to C, and one from C to A. The total load is now 6684W, and it is now a _balanced_ three phase resistive load. Each phase will now be supplying 18.6A. Each leg is supplying 2 heaters, and each leg is supplying 18.6A, so it is perfectly reasonable to say that each heater is adding 9.3A of load to each leg.

Examine what is happening at the phase A bus bar. 10.7A is flowing through the appropriate breaker to the A-B connected heater. 10.7A is also flowing through a different breaker to the A-C connected heater. But these two current flows are _not_ in phase with each other, and when you add them up, the total current flowing from the feeder into the phase A bus is only 18.6A. If the two heaters on each leg together result in 18.6A of current flow on the feeder, then it is totally reasonable to say that each adds 9.3A load to the feeder.

Thus the 9.3A number is an approximation that is only as valid as the assumption of over balance on the panel. If the panel is not reasonably balanced, then the load added to the feeder by this heater will be something greater than 9.3A.

-Jon
Click to expand...

I didn't realize that dividing the power among more than one phase only works exactly for a balanced panel. And we generally been arguing over the correct solution to a single line-line load on a panel. Since that makes the panel unballanced, I shouldn't have divided the load into 2 separate 1114 watt loads.

Again, we are back to the fact that a single line-line load of 1228 Watts causes 10.7 amps to flow through the load, 10.7 amps to flow through each pole of the breaker, and 10.7 amps to flow into each of two main phases iinto the panel.

Now, if we have a panel that already has (2) of these loads connected, and we add a third load between the other two phases, then you have the correct answer in your last post. We still have 10.7 amps through the load, and 10.7 amps through each pole of the breaker, but it only adds 9.2 amps to the two phases of the panel main. The difference (as you and Rattus mentioned) is because on the panel main, the 10.7 amps adds to an already present current that is somewhat out of phase with the 10.7 amps.
 
kingpb said:
Steve,

Are you talking about a DC system, then yes, P=I*E

Otherwise, for ac it's Pph=Vph*I*cos(ang)
Click to expand...

P=I*E if we deal with rms voltages, and assume resistive loads. I think all along we have been assuming resistive loads. No need to add complications at least until we agree on the basics;)
 
The Answer hope this resolves any Issues

The Answer hope this resolves any Issues

Lets first take the calculation of a three phase balanced system with a unity power factor.

Pφ = Vφ* Iφ* cos(θ) and we know that PLL = 3 * (Vφ* Iφ* cos(θ)

and since we have a unity power factor of cos(θ) = 1

we get

PLL = 3 * (Vφ* Iφ)

and since

Vφ = VLL/sqrt(3)

we get

PLL = 3 * (VLL/sqrt(3) * Iφ)

then skipping a few math steps we get

Iφ = PLL/(sqrt(3)*VLL)

Now lets take our current example a 208Y/120V 3 phase panel feeding a single phase 208V 2228W load.


Assuming a balance load and unity power factor (ideal case),


Pφ = Vφ* Iφ* cos(θ)

and we know that since we have a 2 pole breaker we will have 2 lines so we get

PLL = 2 * (Vφ* Iφ* cos(θ))

and since we have a unity power factor of cos(θ) = 1

we get

PLL = 2 * (Vφ* Iφ)

and since

Vφ = VLL/sqrt(3)

we get

PLL = 2 * (VLL/sqrt(3) * Iφ)

then getting rid of the radical in the denominator we have

Iφ = ((2*sqrt(3))/3) * PLL * VLL)

or

Iφ = 1.15* (PLL * VLL)

now pluging in the numbers

Iφ = 1.15 * 2228 * 208

which equals

Iφ = 9.3 Amps
 
C_CLEV said:
Now lets take our current example a 208Y/120V 3 phase panel feeding a single phase 208V 2228W load.

Assuming a balance load and unity power factor (ideal case),
Click to expand...

If I assume that 1 = 0, then I can perform lots of nifty mathematical tricks. *grin*

Your calculations are _correct_ under the assumption of a balanced load and unity power factor.

But a _single_ phase load connected to a _three_ phase panel _cannot_ be a balanced load.

The assumptions do not fit the example, and therefor the equations do not fit the example.

-Jon
 
C_CLEV:

I think you are making the same mistake that Kingpb is. Dividing the power by two, and assuming the phases are ballanced may be a common way to approximate the main current into a panel, it is not the correct way to calculate the current in a load, or the main current into a panel with only one single phase load.

Steve
 
OK, all this is forcing me to delve into symmetrical components, which I did not want to do. I was hoping someone else would step up and provide the details. But, it appears inqusitive minds want to know the answer to this riddle.

BTW: For all those that said the current through the load is 10.7A are correct, you can all sleep better tonite.

This scenario has stumped many an engineer, designer, and electrician. Some repsonses were getting close, and I thought Steve and Winnie were going to nail it, my hats off to their effort.:)

I'll give the answer, then explain were it comes from - For a 2228W load connected across 2 phases of a 3 phase 208Y/120V panelboard (and no other load connected) the amount of current that will be seen on the Phase A, and Phase B feeder is: drum roll............................................ 6.2A.

I know, seems impossible, but that is the magic of what we call unbalanced systems, and the reason symmetrical components exist. For anybody that wants to see printed verification, go to the Handbook of Electric Power Calculations, by H. Wayne Beaty, third edition, pg 13.9.

The unbalanced 3-ph circuit total power is the sum of the powers represented by the separate positive, negative, and zero phase sequence products. The input voltages are balanced so the negative and zero sequence voltages become zero, and we are left with the following:

Ptot = 3Vph1*Iph1*cos(ang1)

Becasue the way the angles work out using symetrical components, cos(ang1) = 1, so, calculating current says: 2228/(3*120) = 6.2A.

Ths has been confirmed using ETAP program, I recommend everyone that has the ability to check this using a program to do so.

Now, for the crux of the issue, if you were to put another 1114W load on Phase C, thereby blalancing the system, the current on each of the feeders goes to 9.3A. Again, I can confirm this using ETAP.

Conclusion: Although you will have 10.7A across the load, you will never see more current on the panel then as if the entire panel is balanced, (i.e. 9.3A versus 6.2A.) This is why for panel sizing I always try to teach people to use a three phase balanced system, and when showing the VA on the panel you need to treat it as balanced, (i.e. divide load by poles and divide by voltage) even though the load current, for sizing the branch circuit may have a higher current at the load. I dont' beleive that many panel schedules treat unbalanced load properly, I know ETAP does.

Plus, it would be a disaster trying to recalculate unbalanced load current on the panel feeder, every time you added a circuit to a panel. Especially for guys in the field. It's easier, and actually more conservative, to treat it as a balanced load.

It is now beer 30, Cheers!!!!!
 
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