correct amount of circuits

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normbac

Senior Member
72 light fixtures ballast 280 watts #10 cu wire

72 x 280 watts = 20,160 x125% = 25,200 *277V = 90.97A * 30 amp circuits = 3.03
Question is does the .03 mean that I will need a fourth circuit or is rounding down permissible I am trying to use three 30a rated switches to control 3 groups of 24 lights

TIA & Merry Xmas
 
Y

yired29

Senior Member
normbac said:
72 light fixtures ballast 280 watts #10 cu wire

72 x 280 watts = 20,160 x125% = 25,200 *277V = 90.97A * 30 amp circuits = 3.03
Question is does the .03 mean that I will need a fourth circuit or is rounding down permissible I am trying to use three 30a rated switches to control 3 groups of 24 lights

TIA & Merry Xmas
Click to expand...

IMHO You would need to have 4 OCPD you can't have a fraction of a OCPD. There are no provisions for rounding the number of OCPD like there is for ampere 220.5 (B).
 
S

satcom

Senior Member
normbac said:
72 light fixtures ballast 280 watts #10 cu wire

72 x 280 watts = 20,160 x125% = 25,200 *277V = 90.97A * 30 amp circuits = 3.03
Question is does the .03 mean that I will need a fourth circuit or is rounding down permissible I am trying to use three 30a rated switches to control 3 groups of 24 lights

TIA & Merry Xmas
Click to expand...

You may have more then the .03 to worry about with that type of load. Are they electronic ballast?
 
Y

yired29

Senior Member
normbac said:
72 light fixtures ballast 280 watts #10 cu wire

72 x 280 watts = 20,160 x125% = 25,200 *277V = 90.97A * 30 amp circuits = 3.03
Question is does the .03 mean that I will need a fourth circuit or is rounding down permissible I am trying to use three 30a rated switches to control 3 groups of 24 lights

TIA & Merry Xmas
Click to expand...

But if we do the math like this I believe it would work.

72 / 3 = 24
24 x 280 = 6720
6720 / 277 = 24.26
24.26 x 1.25 = 30.325
We could round 30.325 down to 30 amps.
This way there could be only three circuits. One MWBC

If I missed something please correct me.
 
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yired29

Senior Member
normbac said:
I decided to go with four circuits but what are you referring to on name plate is that actual bulb wattage which would be 54w x 6
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54 x 6 = 324watts not 280watts.

The ballast amperage rating is not necessarly the same as the bulb wattage.

Ballast,Electronic,F17T8 Lamp
Item # 3G817
Fluorescent Electronic Ballast, Lamp F17T8, Number of Lamps 2, Input Voltage 277 Volts, Line Current 0.15 Amp, Lamp Length 24 Inches

2- 17 watt lamps = 34 watts
34 / 277 = .122 amps

Ballast says .15 amps
 
N

normbac

Senior Member
yired29 said:
54 x 6 = 324watts not 280watts.

The ballast amperage rating is not necessarly the same as the bulb wattage.

Ballast,Electronic,F17T8 Lamp
Item # 3G817
Fluorescent Electronic Ballast, Lamp F17T8, Number of Lamps 2, Input Voltage 277 Volts, Line Current 0.15 Amp, Lamp Length 24 Inches

2- 17 watt lamps = 34 watts
34 / 277 = .122 amps

Ballast says .15 amps
Click to expand...
Always thought suppose to use ballast wattage? they are T5 HO 6 lamp
 
Dennis Alwon

Dennis Alwon

Moderator
Staff member
Location
Chapel Hill, NC
Occupation
Retired Electrical Contractor
Norm I think you can run 3 circuits. Each fixture is 280 watts at 277 volts is virtually 1 amp apiece. On a 30 amp circuit you can have 24 amps for continuous load. 24 *3= 72.
 
Dennis Alwon

Dennis Alwon

Moderator
Staff member
Location
Chapel Hill, NC
Occupation
Retired Electrical Contractor
Also look at 220.5(B)

220.5 Calculations.
(A) Voltages. Unless other voltages are specified, for purposes of calculating branch-circuit and feeder loads, nominal system voltages of 120, 120/240, 208Y/120, 240, 347, 480Y/277, 480, 600Y/347, and 600 volts shall be used.
(B) Fractions of an Ampere. Where calculations result in a fraction of an ampere that is less than 0.5, such fractions shall be permitted to be dropped.
Click to expand...
 
Y

yired29

Senior Member
Dennis Alwon said:
Did you read article 220.5(B)? It clearly states that you can esp. with an 80% factor already in the equation.
Click to expand...

There are two different subjects here. You can round down for amperage 220.5 (B) but not for the number of circuits.

If you had a calculation that required anything over a whole number when it comes to circuit calculations you need to use the next whole number.

If you had a calculation that required anything over .49999 amps when it comes to amperage calculations you need to use the next whole number if less than .5 you can round down.


I don't think I'm missing something.
 
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