208V 1phase loads

[jumper]

Don't feel bad I have no idea what either of you are saying. :grin:

Complex numbers and angular velocity(phase angle in time), lotsa fun.

j is the square root of -1. Imaginary numbers can be a hard concept to grasp until you have played with them and I still get confused alot of the time.

 

[mull982]

Here is the math - hope it helps.

There are six 3kW, 208V heaters connected balanced between the phases, so there is a 6kW load between A-B, B-C and C-A. The load current is 6000W/208V=28.85A. The three load currents are 120? out of phase with each other, so lets say Iab=28.85<0, Ibc=28.85<-120, Ica=28.85<-240.

The current flowing on Bus A will be Ia=Iab-Ica (likewise, Ib=Ibc-Iab & Ic=Ica-Ibc.)

Converting: Iab=28.85<0 = 28.85 + j0
.................Ibc=28.85<-120 = -14.43-j24.98
.................Ica=28.85<-240 = -14.43+j24.98

So calculating for current on Bus A: Ia=Iab-Ica =>
.................Ia= (28.85 + j0) - (-14.43+j24.98)
.................Ia= 43.28 - j24.98

Converting back: Ia=50<-30

The current on Bus A (Ia) has a magnitude of 50 (and is 30? separated from Iab)

This is exactly how I approached this problem. You could obviously all divide the kVA up betwwn phases so phase A would have 6kVA total on it and then using 6kVA/120V =50A. I believe someone mentioned this in an earlier post howver.

So now after seeing how the balanced panel calc works out it has me back to the OP's unbalanced situation wondering the difference between the two methods of calculations he first presented. Using vector math or simply adding up the currents from the 6 heaters all between phases A & B we come up with 86A. However is we split each of the loads between phases A&B with 3kVA each and then add up a sinle phase we get 9kVA / 120v = 75A.

So why doesn't dividing the load between phases and adding the total load on each phase work for this case as it does when the loads are balanced across all 3 phases? Does this method only work for balanced 3 phase panels?

 

[david luchini]

This is exactly how I approached this problem. You could obviously all divide the kVA up betwwn phases so phase A would have 6kVA total on it and then using 6kVA/120V =50A. I believe someone mentioned this in an earlier post howver.

I would not approach the problem as you have stated it (in red) above. There is not 6kVA on phase A, and on B and on C. There is 6kVA connected between A and B, 6kVA between B and C and 6kVA between C and A

So now after seeing how the balanced panel calc works out it has me back to the OP's unbalanced situation wondering the difference between the two methods of calculations he first presented. Using vector math or simply adding up the currents from the 6 heaters all between phases A & B we come up with 86A. However is we split each of the loads between phases A&B with 3kVA each and then add up a sinle phase we get 9kVA / 120v = 75A.

So why doesn't dividing the load between phases and adding the total load on each phase work for this case as it does when the loads are balanced across all 3 phases? Does this method only work for balanced 3 phase panels?

It doesn't work to split the loads between A & B and divide by 120V, because none of the load is connected to neutral. If you had 6 heaters which were rated 3kVA each at 120V, and you connected three heaters from A-N and three heaters from B-N, then you would would get 75A on phase A and B, and 75A on the neutral. Since none of the loads are neutral connected in the OP, you cannot use 120V in the calculation.

 

[Little Bill]

Obviously that was my opinion originally but when an EE jumps in and says otherwise I have to wonder. Now we have 2 EE's arguing it out. That makes it not simple.

Dennis,
I gave you the right answer 1 week and 4000 calculations ago!:grin:

 

[augie47]

Obviously that was my opinion originally but when an EE jumps in and says otherwise I have to wonder. Now we have 2 EE's arguing it out. That makes it not simple.

That's what EEs do best

 

[mull982]

It doesn't work to split the loads between A & B and divide by 120V, because none of the load is connected to neutral. If you had 6 heaters which were rated 3kVA each at 120V, and you connected three heaters from A-N and three heaters from B-N, then you would would get 75A on phase A and B, and 75A on the neutral. Since none of the loads are neutral connected in the OP, you cannot use 120V in the calculation.

Then why do you see 3 phase panel schedules with the kVA's of loads split up between the phases and then a total kVA for each phase calculated at the bottom. This total phase kVA is then used for determining the current on that phase.

An example is a 3 phase panel schedule that has 3 phase loads split between 3 phases and single phase L-L loads split between 2 phases.

 

[skeshesh]

Because what you are calculating in a panel schedule is total load on the panel not each branch circuit. Another important use of the panel schedule is to balance the loads. Ideally, when you have a final panel schedule, the phases should be close to balanced. This is why its important to keep in mind when dealing with very unbalanced phases to calculate the current on the high let to ensure that the panel rating is not exceeded.

 

[CelsoSr]

Evenly dsitributed....

Evenly dsitributed....

So what would the load be if all six heaters were distributed evenly across all three phases?

If evenly distributed you get 2x3000 watts per phase. So 6000/208=28.8 amps. Minimum 30amps/3poles breaker

 

[jmccamish]

there are 1500 va per phase per heater. 6 heaters is 7500 va per phase. 7500/120=62.5 amps
70 amp breaker. assuming you dont balance the loads and put them all on the same two phases.

 

[CelsoSr]

Evenly dsitributed....

Evenly dsitributed....

If evenly distributed you get 2x3000 watts per phase. So 6000/208=28.8 amps. Minimum 30amps/3poles breaker

Changing my calc.:
There are four 'legs' x 14amps in each line=56a. So 60amps/3p breaker

 

[Dennis Alwon]

If evenly distributed you get 2x3000 watts per phase. So 6000/208=28.8 amps. Minimum 30amps/3poles breaker

Six thousand is one each phase --Should be 18000/ 360 = 50 OR 6000/120 = 50. I think I will tiptoe out of here quietly. It seems no one is reading thru the whole thread. We have had about 5 different answers.

 

[mull982]

Because what you are calculating in a panel schedule is total load on the panel not each branch circuit. Another important use of the panel schedule is to balance the loads. Ideally, when you have a final panel schedule, the phases should be close to balanced. This is why its important to keep in mind when dealing with very unbalanced phases to calculate the current on the high let to ensure that the panel rating is not exceeded.

So the method of dividing loads up on each phase and then calculating a current for each phase given the total kVA on that phase only works for a balanced panel?

 

[mull982]

Six thousand is one each phase --Should be 18000/ 360 = 50 OR 6000/120 = 50. I think I will tiptoe out of here quietly. It seems no one is reading thru the whole thread. We have had about 5 different answers.

I agree with this answer of 50A as you have shown.

Using 6kVA/120 also shows that in a balanced panel you can calculate each phase based off of L-G voltage such as 120v.

 

[david luchini]

there are 1500 va per phase per heater. 6 heaters is 7500 va per phase. 7500/120=62.5 amps
70 amp breaker. assuming you dont balance the loads and put them all on the same two phases.

I would think that 6 heaters at 1500 va per heater would give you 9000 va per phase,
which would give you 9000/120=75 amps, which would be the wrong answer for the 6 heaters on two phases.

6x3000VA/208V=86.5 Amps for the heaters on the same two phases.

 

[al hildenbrand]

We have had about 5 different answers.

I'll happily admit that my answer was wrong. . . so 4 different answers.

 

[Bjenks]

Most of these simple programs are not engineering programs, but done by someone who is using basic NEC mathmatics. In your option of 75A it is because the programmer assumes 180 degree phase shift between the two 120V phases or they are only giving you the real current and not including the imaginary current. This inadvertantly made the voltage phase to phase of 240V. Thus 18,000/240= 75A. If you did phasor mathmatics of 18kVA between A and B then the math would go as follows assuming a 30 degree phase shift in voltage in reference to 0: phase current is in phase with the voltage, thus 18kVA/208=86A with 30 degree phase shift. Then to find the real current it would be 86*cos 30 = 74.5A (looks familiar?) and the imaginary at 43A. Now to find the total current you find the Square Root of 74.5 sq + 43 sq which equals 86A. Which is the same as were we started.

Keep in mind putting loads from phase to neutral is diffent than putting it phase to phase. Unless of course you don't understand the math of the NEC code.