Panel board Load Calculation

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i.qasim

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Location
Saudi Arabia
Dear Engineers,

I have to panel board schedules in front of me, both designed by well known engineering firms.
as shown in images.

my question is.. which on is correct... OR where am I stumbling...
[in one schedule, they have calculated total load by taking average of the single phase loads on each phase. in other, they have added the single phase loads on each phase :/ ]

pb - Copy.jpg pb.jpg

Lets say I have 18 lights of single phase load 10W each.
I divide it 6 lights per phase in a panel board,
what is the load on this panel board:
60W? as each phase has 60W load.
or 180W?

and what load will I use in the formula to find current for CB and cable selection
"P=uder_the_root 3*V*I*power_factor"

thank you very much.
 
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GoldDigger

Moderator
Staff member
Location
Placerville, CA, USA
Occupation
Retired PV System Designer
The size of the panel )bus capacity and main breaker size) would be determined by the actual current on the most heavily loaded phase wire.
The size of a transformer feeding the panel would be determined by the total kVA as long as the loads are closely balanced.

The sqrt(3) factor would only come into play when starting from the wattage or VA of line to line loads.
 

kwired

Electron manager
Location
NE Nebraska
Dear Engineers,

I have to panel board schedules in front of me, both designed by well known engineering firms.
as shown in images.

my question is.. which on is correct... OR where am I stumbling...
[in one schedule, they have calculated total load by taking average of the single phase loads on each phase. in other, they have added the single phase loads on each phase :/ ]

View attachment 13984 View attachment 13985

Lets say I have 18 lights of single phase load 10W each.
I divide it 6 lights per phase in a panel board,
what is the load on this panel board:
60W? as each phase has 60W load.
or 180W?



and what load will I use in the formula to find current for CB and cable selection
"P=uder_the_root 3*V*I*power_factor"

thank you very much.

Watts is watts it will not change.

Your example with the 18 ten watt lights is 180 watts, period. It may be split across the different parts of a multiwire source which along with applied voltage will determine how many amps will flow from each part of the source.

So lets assume 120 volts and 18 ten watt lamps:

Put them all on a single 120 volt two wire source: you have 180 watts / 120 = 1.5 amps

Put them on a three wire 120/240 source and evenly split them across the neutral you still have 180 watts /240 = .75 amps per line and no load on neutral.

Put them on a four wire 208/120 source and evenly split them across the neutral you still have 180 watts / 208/1.732 = just under .50 amps per line and no load on neutral.

Start to unbalance the load you will get variances in what percentage of the source is carrying how much capacity, but if still the same 180 watts is being supplied you are still providing 180 watts. If your source were only rated 180 watts then you can't put all 180 on just one coil or phase of a multiwire source as this would overload portions of that source.
 

GoldDigger

Moderator
Staff member
Location
Placerville, CA, USA
Occupation
Retired PV System Designer
Just FYI, in the three phase example the current is 180 divided by 120, then divided by three giving exactly .5 A.
Your calculation with 208 and sqrt(3) does not give exactly .5 because 208 is not exactly 120 sqrt(3).
 

kwired

Electron manager
Location
NE Nebraska
Just FYI, in the three phase example the current is 180 divided by 120, then divided by three giving exactly .5 A.
Your calculation with 208 and sqrt(3) does not give exactly .5 because 208 is not exactly 120 sqrt(3).
After some thought that does make sense.

If line to neutral voltage is exactly 120.00 then line to line should be more like 207.8460969082652752232935609807....... that is as many digits as is displayed on the calculator I used and my method would have come even closer to .5 had I used that figure.
 

kwired

Electron manager
Location
NE Nebraska
can I have an NEC reference for all this!!
NEC primarily does load calculations like being asked about in Volt-Amps.

To figure out current per phase you need to know some basic electrical theory to convert that from the VA, NEC does not discuss theory.

The example used in the OP has 180 VA of load. It may or may not be divided across multiple segments of the source supplying it, but is still 180 VA even if divided across three phases of a source.
 

charlie b

Moderator
Staff member
Location
Lockport, IL
Occupation
Semi-Retired Electrical Engineer
I have to panel board schedules in front of me, both designed by well known engineering firms. as shown in images. my question is.. which on is correct... OR where am I stumbling...
The loads should never be added in terms of amps. It can be right, and it can be wrong, and it is not easy to tell the difference. So I would say the second schedule should be carefully checked, before it is used. The math should always be done in units of power (VA or KVA). Conversion to amps should be the final step. This greatly reduces the risk of getting the wrong answer.

 
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