Formula Question

[bkludecke]

OK, I was brushing up on formulas etc and am confused. My book shows that to convert single phase amps to three phase amps we multiply by .58.

Say I've got 12 perfectly balanced 20 amp loads on a 120/240 1 phase panelboard. Now we want to put these same loads on a new 120/208 3 phase panel board. I come up with 120 amps per phase on the single phase system and 80 amps per phase on the 3 phase system. But when I use the .58 multiplier I get 69 amps.

Where did I go wrong?

 

[Besoeker]

OK, I was brushing up on formulas etc and am confused. My book shows that to convert single phase amps to three phase amps we multiply by .58.

Say I've got 12 perfectly balanced 20 amp loads on a 120/240 1 phase panelboard. Now we want to put these same loads on a new 120/208 3 phase panel board. I come up with 120 amps per phase on the single phase system and 80 amps per phase on the 3 phase system. But when I use the .58 multiplier I get 69 amps.

Where did I go wrong?

Perhaps by making it too complicated?

Your 12 20A loads are single phase loads. That doesn't change.

In the first case they are split with six on each side of the 120-0-120 supply hence 120A each side.
In the second for balance you'd have four on each phase thus 80A in each.

 

[bkludecke]

OK, but why is the answer different when I use the formula?

 

[jim dungar]

OK, but why is the answer different when I use the formula?

The .58 multiplier is for converting single phase VA to three phase VA. Remember, there is also a voltage change.

120A x 240V = 28,800VA single phase
28,800VA x .58 = 16,704VA three phase
16,704VA/208v = 80A three phase.

 

[gar]

090317-1010 EST

bkludecke:

A formula without a knowledge of the basis of the formula can produce results that are unrelated to the real question.

For example:

P = V*I is valid for average power for steady state DC or steady state AC with a resistive load. Here the capital letters for the variables imply RMS values. That means averaged over some time period.

p = v*i is valid under any conditions for the instantaneous power as calculated from the instantaneous values. If you integrate this equation over time and divide by time, then you will get the same result as P = V*I for the limitations I specified above. If the resistive load and steady state conditions are removed and when those conditions do not exist, then P = V*I is not valid for average power, but P = integral of v*i over a time period relative to time and divided by that time period is valid.

.

 

[bkludecke]

Ok good. Now let's say that instead of 12, 20amp 120 volt loads on the single phase panel, I've got 6 20 amp 240 volt loads. Now I take those same loads and put them on a 3 phase 240 volt delta panel. I get 80 amps per phase with the 240 volt delta system. Now the multipier doesn't work.:-?

 

[Smart $]

Ok good. Now let's say that instead of 12, 20amp 120 volt loads on the single phase panel, I've got 6 20 amp 240 volt loads. Now I take those same loads and put them on a 3 phase 240 volt delta panel. I get 80 amps per phase with the 240 volt delta system. Now the multipier doesn't work.:-?

6 ? 20A = 120A "1?"
120A ? .577 = 69.3A "3?"​

Proof...

6 ? 20A ? 240V "1?" = 28,800VA
28,800VA ? (240V "3?" ? √3) = 69.3A​

FWIW, 0.577 is the decimal approximation of 1 ? √3...

120A ? √3 = 69.3A​

 

[bkludecke]

6 ? 20A = 120A
120A ? .577 = 69.3A
​

Proof...

6 ? 20A ? 240V "1?" = 28,800VA
28,800VA ? (240V "3?" ? √3) = 69.3A
​

FWIW, 0.577 is approx. equal to 1 ? √3...

120A ? √3 = 69.3A
​

Right. But when I plot the loads on a panel schedule I get 80 amps per phase. Where am I going wrong?

 

[jim dungar]

Right. But when I plot the loads on a panel schedule I get 80 amps per phase. Where am I going wrong?

You are forgetting the voltage change and the resultant relationship of the currents.

It appears that you want a single multiplier to use to go between all of these possible combinations you have described.

Given (12) 20A @ 120V loads = 28.8kVA

The different line currents and rounded kVA are:
Single phase 2-wire 240V = 120A loads connected (2) in series, then line-line = 28.8kVA
Single phase 3-wire 120/240V = 120A loads connected line-neutral = 28.8kVA
Three phase 3-wire 240V delta = 69A loads connected (2) in series, then line-line = 28.8kVA
Three phase 4-wire 208Y/120 = 80A loads connected line-neutral = 28.8kVA

 

[Smart $]

Right. But when I plot the loads on a panel schedule I get 80 amps per phase. Where am I going wrong?

You're not. This is an inherent problem with summing VA of line-to-line loads on a panel schedule. The only way to get accurate line current is to use vector math (or an equivalent means).

 

[Besoeker]

OK, but why is the answer different when I use the formula?

The formula isn't relevant to the application.
You are not converting a single-phase load to a three-phase load.

 

[Besoeker]

Right. But when I plot the loads on a panel schedule I get 80 amps per phase. Where am I going wrong?

You aren't.

 

[bkludecke]

I'm starting to get it. No wait. If I take 6 20amp 240volt 1 phase loads and then put them on a 3 phase 240volt delta panelboard I think I have gone from 120 amps per line single phase to (essentially) 4 3 phase loads totaling 80 amps per line. At least that's how it appears on a panel schedule. Line to line voltage doesn't change. So in this example the .58 or 1.73 multipliers don't seem to apply.

Now I take my 6 20amp 240volt 1 phase loads totaling 120 amps and apply the .58 multiplier to get 3 phase current and I get 69 amps. I think I may have torn through the mass/time/space/beer intergalactic fabric.

This is what happens when an old geezer like me (last ac theory class was in 1974) tries to remember stuff.

 

[Smart $]

I'm starting to get it. No wait. If I take 6 20amp 240volt 1 phase loads and then put them on a 3 phase 240volt delta panelboard I think I have gone from 120 amps per line single phase to (essentially) 4 3 phase loads totaling 80 amps per line. At least that's how it appears on a panel schedule. Line to line voltage doesn't change. So in this example the .58 or 1.73 multipliers don't seem to apply.

Now I take my 6 20amp 240volt 1 phase loads totaling 120 amps and apply the .58 multiplier to get 3 phase current and I get 69 amps. I think I may have torn through the mass/time/space/beer intergalactic fabric.

This is what happens when an old geezer like me (last ac theory class was in 1974) tries to remember stuff.

Actually, connecting (6) 20A, 240V loads in balanced fashion to a three phase system would be line-current-wise the same as connecting two 14,400VA 3? loads, or just one 28,800VA 3? load...

28,800 ? (240 ? √3) = 69.3A​

...attributed per line for these 6 loads.

There is no easy way to show this on a standard/typical panel schedule... and it is even worse when there are unbalanced line-to-line loads!!!

I have at times considered making an Excel spreadsheet that does all the calculations and fills in the amounts as appropriate. I've concluded it will take some scripting for the spreadsheet to perform as intended and I suck at scripting :grin:

 

[rattus]

Am I missing something?

Am I missing something?

Seems to me we are comparing the two legs of a 120/240V service to the three legs of a 120/208V service. The load comprise 12 20A branch circuits. In the first case we have 120A in each of the two legs In the second case we have 80A in each of the 3 legs. The line to line voltages do not matter in this case--sqrt(3) does not apply.

As for the 0.58 factor, that formula does not apply. The correct factor is 0.67.

 

[gar]

090318-1029 EST

I concur with rattus.

If you have 6 equal resistive loads at 120 V that total to 28,800 W (VA), then when equally distributed on a 120-0-120 single phase source the hot line current magnitudes are I = 14,400/120 = 120 A, and neutral is 0.

Reorganize the loads so there are 28,800/3 = 9600 W per phase in a 3 phase Y 208/120 connection with each 9600 W load between a phase leg and neutral, then I = 9600/120 = 80 A.

Note in a Y load the line current is identical with the load current.


When the problem was changed to a delta load you also changed the 3 phase voltage to 240 line-to-line from 208 line-to-line. In this case when you redistribute the loads to get 9600 W per phase, then the current in the load on one side of the delta is 40 A, 9600/240 = 40. Now you perform the vector sum of the currents at one node between the single line feeding two sides of the delta and you have 2*(40*cos 30 deg) = 2*40*0.866 = 80*0.866 = 69.28 A. Take the ratio between 240 and 208 (this is not quite the correct ratio), the correct ratio is 240/207.84 = 1.1547, and multiply by 69.28 resulting in 1.1547*69.28 = 79.9976 = 80.0 A.

Because you scaled the line-to-line voltage up the line current was reduced.

If you had retained the same line-to-line voltage of 208 V, changed from a Y load to a delta load of the same power by changing the resistance values, then your line current would have remained at 80 A, and the current in one side of the delta would have been 40*1.1547 = 46.188 A, and 46.188*207.84 = 9599.71 W. This rounds to 9600 W.

.