Calculation Method of Unbalanced Panel

[7JLAman4]

I have a 3-Ph 120VAC Delta connected power panel. I was given a method for calculating/estimating the panel feeder cable 3-Ph load ampacity. My question is, how accurate is this method below?

The 3 phase load was calculated as follows; for an unbalanced load on a panel, add the two highest phases together and multiply by 0.866 to get the 3 phase load.

In my given case, I have a single load on the panel, 4.4kW 1-Ph 120VAC load connected to phases AB which is claimed to draw 46A. Using the above method, the 3-Ph load is calculated/estimated at:

(46.0(AB) + 0.0(BC)) x 0.866 = 40.0A.

Has anyone used this this method before and how accurate is it? What could the 0.866 refer to? I understand that this is not the demand load used for calculating the required feeder size.

 

[VTer]

No.1. How can you have a 3-phase 120VAC Delta connected power panel? Do you mean 208V/120Y 3ph, 4W? Please verify.

No.2. The method you mentioned above is not very accurate, and seems more complicated than it should be. It is probably some rule of thumb which I have not heard before. Best way to get the feeder size or the size of panel is to add all the VA that will be supplied by your panel. Take the VA and use VA=sqrt3*V*I to solve for your “I” which will help you determine your feeder size. Depending on your application you can add a growth factor to your VA. As far as load balancing, try to spread out your loads evenly across the 3 phases. Make a excel spreadsheet, if you don’t have one already, to help you balance loads.

Hope this helps.


"Throughout space there is energy. Is this energy static or kinetic! If static our hopes are in vain; if kinetic — and this we know it is, for certain — then it is a mere question of time when men will succeed in attaching their machinery to the very wheelwork of nature". – Nikola Tesla

 

[DanDel]

If you're looking for cable ampacity, the load in kW is not needed, you just need the amperage on each phase.
This amperage should be based on the load in amps plus 25%, assuming that the O/C device is not already present. The cable size must match or exceed the amperage of the O/C device.
If you're load is 46A per phase, then the cable and O/C device should be sized for 46A x 1.25 = 58A or greater.

 

[7JLAman4]

VTer:
A1: The power panel is fed from a transformer bank consisting of (3)10kVA 480/120VAC single phase transformers connected to provide 3-Ph 3-Wire delta power to the panel.

A2: In this case I agree with your approach for estimating assuming that the loads are closely balanced. If the loads are unbalanced or any particular phase does not have any loads connected, the resultant value could be lower than required. This method works fine for determining the required rating of the panel supply transformer.

The method I have used for determining the panel demand load for sizing the feeder cable and circuit protection has been as follows:

for each phase; multiply largest load by 125% + remaining loads + 50% of spares. The largest of the three multiplied by sqrt(3) determines the panel demand load.

 

[7anoter4]

I wander why for a single phase load you need 3 phases. If there are 3 single phase transformer then for one transformer phase A is as "live" at single phase and phase B is as "neutral" then IA=-IB. So, if you don't expect any developing in the future 1.25*46=57.5 will be the cable current capacity for both conductors.
If the 3 transformers will be connected in delta at first and only 3 conductors are connected from the transformers to panel then, if there is no load on the remaining phases- the phase BC and CA- the current through transformers will be divided in two parallel ways: one through AB and the second through BC and CA in series. But the current carried by conductors will be still the same.
If in the future all 3 transformers will be loaded-with the same load- the total load will be 3*5520=16560 VA. Then the current per phase will be: 15560/sqrt (3)/120=74.9A

 

[waross]

The load on a panel for the purposes of ampacity is taken as the maximum load on any one phase.
The load on an unbalanced panel will approach the load on the heaviest phase PLUS 86.6% of the load on the heaviest phase.
OR, 186.6% of the load on the heaviest phase.
This is often used for panels with moderate unbalances.
It it provides an accurate limit, but not an accurate current. For sizing panels and cables the limit is often more important.
The true sum of currents from two phases must be derived with vector addition. 86.6% is only accurate when the currents are equal, but does provide an accurate limit.

Bill
--------------------
"Why not the best?"
Jimmy Carter

 

[rbulsara]

Jlamann:

Lets start with the basics. I am not sure who is "claiming" 46A for a 4.4kW load at 120V.

4.4 kW at 120V will draw 4400/120=36.67A. The poor heater does not know where the 120V is coming from.(If the heater is drawing 46A then it is rated more than 4.4kW.)


If you have only one heater connected on your delta source across A-B. The line A and B each will see 36.67A. Line C will see zero amps. So the ampacity of the conductor needs to be at least 36.7A in this case with only one 4.4kw load. (not accounting for safety and code margin).

The 0.87 or .866 (cosine of 30 degrees) factor comes in while calculating kW load, which I am not sure why you need that, but here is the calc to check the math.

For unbalanced loads, you need to add the loads in each three phases separately. Now in order to do per phase calcs you need to use per phase voltage. For a three phase system per phase voltage is V(LL)/1.732=69.28V in this case. It does not matter whether actual source is wye or delta connected for this purpose.

Now since your system is delta, the 36.67A is in "phase" with the line to line voltage of 120V, which is 30 degree out of phase with phase voltage of 69.28V.

So the real (watts) load in the Phase A will be 36.67A*69.28V*cos(30)=36.67*69.28*0.866 or 2200W. Likewise, Phase B will also have 2200W and that in Phase C will be zero as there is no current in C. The total of three phase load is 2200+2200+0=4400 which what you had connected to begin with. The conductors between the trasfomer bank and the panel would read 36.67, 36.67 and 0 amps.

 

[7JLAman4]

rbulsara-
I do not believe that I had mentioned anything about a heater! The load is not a motor, not heater rather power supply or rack of supplies. My original inquiry was regarding the method for determining the load of the feeder neglecting any spares or demand factors to determine required circuit protection or cable size. I completely agree with your mehtod and calculations. I was more concerned with a similar panel that fed multiple unbalanced single phase loads and the proper method for determining required circuit protection and cable size and if the method mentioned in the OP could be applied for a single 1-Ph load on a 3-Ph panel.

In my industry, when selecting the size of panel feeders supplied from other panels (not transformers) the assumption is that all loads are considered continuous and therefore the feeder breakers are rated for 100% plus an additional 25% of the largest load (for three phase loads). When the panel contains an unbalanced mix of 3-Ph and 1-Ph loads, the CB is sized based as above, but with regards to each phase or line current. Determine the phase with the largest cumilative load, then add 25% of largest plus 50% of provided spares.

 

[davidbeach]

Continuous is 125% of all loads, not 100% of all plus 25% of the largest.

It is also penny wise and pound foolish to feed panels at less than the panel rating. Sure it costs a bit more up front but the first time you don't have to re-pull the feeder to increase capacity you have paid for all the over sized feeders.

 

[rbulsara]

jlamann:

My bad. I assumed a heater since you mentioned 4.4kW and no kVA. That still does not change the calcs.

You also asked where does 0.87 comes from and hence my explanation. It has nothing to do with codes or demand factor. What you had in your original post made no sense to me.

For any kind of unbalanced load, you still calculate total load in each phase and the conductor ampacity must accomodate the highest current value in any of the conductors. Plus 25% more for "all continuous load" required by applicable Code (such as NEC) and any room for growth you want.

Only time 125% of largest load comes in play when there are motors loads of different sizes on a feeder.

 

[7anoter4]

I agree with davidbeach.
If this panel would be the only panel supplied from the transformers I should take into account all the possible supplied power [3*10=30 kVA] and the maximum steady state current would be : 30000/sqrt(3)/120=144.3 A .So I'll take this value for cable dimensioning.

 

[waross]

Where does 86.6% come from?
Loads on different phases, line to line loads and line to neutral loads are at different phase angles.
Vector addition must be used to calculate the actual sum of these currents.
When equal loads on different phases are added vectorily, the resultant current is "root 3" or 1.73 times the individual currents. 1.73 / 2 = .866 or 86.6%. When three equal loads are connected to a three phase source, the current drawn by each load will be 86.6% of the line current.


Bill
--------------------
"Why not the best?"
Jimmy Carter

 

[7JLAman4]

davidbeach-
7anoter4-

The following is an abbreviated version from an IEEE Std ., which is the basis for our calculations. When sizing the feeder cables, spares are included when determining the panel demand load which is the basis for sizing the circuit protection and feeder cable size.

If this panel was to be fed by a transformer, the panel feeder or secondary of the transformer would be sized for 100% of the transformer so that the feeder cable would not need to be replaced at a later date.

"Feeder and branch circuit conductors should be sized for 100% of their connected load.

Conductors for lighting, communications, and electronics circuits should be sized for the total connected load. The connected load should include 50% of the rating of spare circuits on switchboards or load centers and the average active circuit load for the spare circuits on distribution panels.

Conductors supplying an individual motor should have a continuous current-carrying capacity equal to 125% of the motor nameplate rating. Conductors supplying more than one motor should have a continuous current-carrying capacity equal to 125% of the largest motor plus the sum of the nameplate ratings of all other motors supplied, including 50% of the rating of spare switches on the distribution unit."

 

[davidbeach]

Continuous loads and motors are not the same thing, even if the word "continuous" happens to be used in the motor description. You want to follow those "recommendations" (context could be really helpful here, but I'm not going to try to figure out which standard) feel free to do so; I wouldn't. As I said, panels should generally be fed at their rating, not their present loading with some assumption about the future. Ran into undersized feeders way too many times to ever want to deliberately do that to a client.