Load summary

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kencoel

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I have a 3 phase panel. The A phase reads at 78.84 the B phase reads 61.19 and the C phase reads 50.14 amps. Does this mean that the load at the time of the reading is the three combined for a total load amps of 190.17 ?
 
kencoel said:
I have a 3 phase panel. The A phase reads at 78.84 the B phase reads 61.19 and the C phase reads 50.14 amps. Does this mean that the load at the time of the reading is the three combined for a total load amps of 190.17 ?
Click to expand...

No.
Adding the currents like that is a bit meaningless.
 
Load Calculating

Load Calculating

(...continued from a previous post.) I am trying to determine how to figure the full load amps of an existing commercial kitchen panel. I have read each leg of each phase during peak hours and added in some new load/amps for some kitchen equipment. I need to submit this information for a permit. The A phase reads 78.84 amps, the B phase reads 61.19 amps and the C phase reads 50.14 all included with the new equipment turned on. How do you determine what the Total load amps for this panel would be? Thanks.
 
kencoel said:
Does this mean that the load at the time of the reading is the three combined for a total load amps of 190.17 ?
Click to expand...

No. All you care about is the amps on each "phase conductor:. The only time you would add them is when you have several conductors that combine into a single conductor (i.e. circuit 1 + circuit 2 pigtailed into a single feeder).

It does not matter if this is three-phase or single-phase, "total load amps" created by adding amps from different phase conductors is a meaningless number.
 
kencoel said:
I have a 3 phase panel. The A phase reads at 78.84 the B phase reads 61.19 and the C phase reads 50.14 amps. Does this mean that the load at the time of the reading is the three combined for a total load amps of 190.17 ?
Click to expand...

No. The max your panel is drawing is 78.84 amps.
 
drbond24 said:
I predict that a lecture from charlie b is fast approaching. :D
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You are too kind. :D

I'll keep it simple. Read posts #3 and #7.

Oh, and if you divide your 190.17 by 3, and get about 63.4, you can use this to get a reasonable (if no precisely accurate) idea of the load on you panel.

(Edited to correct post numbers, after having combined two threads into one.)
 
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NOTE: The originator started a second thread, rather than post additional comments on this thread, withoujt actually addressing a separate issue. I have combined the two threads into one, since we do not allow duplicate threads.

Please continue the discussion here.
 
090115-1322 EST

Adding the magnitude of the three currents is not necessarily meaningless.

Suppose the entire load is resistive, it is a Y system, and the neutral is connected.

Then as an approximation --- the total load power assuming a balanced source voltage is
Power total = Vline to neutral * (Iline1 + Iline2 + Iline3).
Make it a delta load load and it is still
Power total = Vline to line * (Iline1 + Iline2 + Iline3) / Sq-root of 3.

.
 
gar said:
090115-1322 EST

Adding the magnitude of the three currents is not necessarily meaningless.

Suppose the entire load is resistive, it is a Y system, and the neutral is connected.
Click to expand...
Gar -
(hard to get the tone of voice right with my keyboard - picture a non-antagonistic Ryan penguin slap comming:rolleyes::rolleyes:)
You are way too sharp for this comment. I'm going to attribute the comment to a tongue-in cheek. (I know, "Smile when you say that, stranger" I am:D)

cf
 
gar said:
Adding the magnitude of the three currents is not necessarily meaningless.
Click to expand...
I said "total load amps" is a meaningless number.

gar said:
Suppose the entire load is resistive, it is a Y system, and the neutral is connected.
Click to expand...
Now you are changing the OP. But my post is still valid.
jim dungar said:
The only time you would add them is when you have several conductors that combine into a single conductor
Click to expand...

At the neutral point of a wye, you combining several conductors into one.
 
gar said:
090115-1322 est

adding the magnitude of the three currents is not necessarily meaningless.

Suppose the entire load is resistive, it is a y system, and the neutral is connected.

Then as an approximation --- the total load power assuming a balanced source voltage is
power total = vline to neutral * (iline1 + iline2 + iline3).
Make it a delta load load and it is still
power total = vline to line * (iline1 + iline2 + iline3) / sq-root of 3.

.
Click to expand...

p= sqrt(3)*{[(v1*i1)+(v2*i2)+(v3*i3)]/3}
 
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