I thought it might be fun to pose a motor branch-circuit calculation here, just to see how many different answers we come up with, and to see the different methods people use. Give it a whack if you have the time.

Lets say we have a rigid (RMC) conduit containing two 3PH motor circuits.

Motor #1: 25HP 3PH Code letter F Squirrel-cage induction Service factor 1.15 continuous duty 480 volt

Motor #2: 40HP 3PH code letter F Squirrel-cage induction Service factor 1.15 continuous duty 480 volt

The conduit is short, so we won't worry about voltage drop, but it does run through a boiler room where the ambient temperature is 110 degrees F. We are using THWN-2, 90 degree rated wire. Equipment temperature ratings are undetermined. The rigid conduit will be used as the equipment ground.

Calculate the minimum conductor size for each motor.

Calculate the maximum size dual element time delay fuse size for the short-circuit/ground fault protection.

Calculate the minimum size rigid conduit needed for these two circuits.

Yep, two motor circuits in the same conduit, and no, I'm not trying to get you to do my homework LOL. Just thought it would be interesting. I'll wait for a couple answers, and then post mine.

[This message has been edited by Matt M (edited 12-21-2002).]

Re: Motor branch circuit calculation#19118 12/21/0206:39 PM12/21/0206:39 PM

Motor #1: Step 1: 25HP = 34 amps from table 430.150 article 110.14C1a applies, so we know we must use the 60 degree column of table 310.16.

Step 2: 34 amps X 125% for continuous duty = 42.5 amps. A look in table 310.16 in the 60 degree column, shows that a #6 fits the bill. We'll jot it down for now.

Correction factors for adjacent conductors, and ambient temperature correction are a seperate calculation, and are calculated using 100% of the FLA, not the 125%.

Step 3: We need a correction factor of 80% for 6 adjacent current carrying conductors. 34 amps divided by .8 = 42.5 amps.

Step 4: We will be using the THWN-2, 90 degree rated wire, therefore we see an ambient temperature correction factor of .87 from the bottom of table 310.16. 42.5 amps divided by .87 = 48.85 amps.

Step 5: Looking in the 90 degree column of 310.16, our selected conductor size for 48.85 amps is a #8. Compare this conductor to the one selected in steps one and two, and the larger gets the nod.

Conductor size for motor #1 is #6 THWN-2

A look in table 430.52 shows that we need a dual element TD fuse of no larger than 175% of full load amps. 34 amps X 1.75 = 59.5. Round up to the nearest size gives us a 60 amp fuse.

Motor #2: Step 1: 52 FLA (table 430.150)

Step 2: 125% X 52 = 65 amps for continuous duty. Table 310.16 60 degree column shows that we need a #4 conductor. Jot it down for later comparison.

Step 3: 52 amps divided by .8 for adjacent current carrying conductor correction = 65 amps.

Step 4: 65 amps divided by .87 for ambient temperature correction = 74.7 amps. A trip to the 90 degree column of 310.16 shows a #6 conductor.

Step 5: Comparing the conductor selected in step 5 with the one determined in step 2 gives the nod to the larger #4 conductor.

Conductor size for motor #2 = #4 THWN-2.

52 FLA X 1.75 = 91 amps. Round up to the next standard size gives us a 100 amp DETD fuse.

A trip to chapter 9 table 5 shows an area of .0507 sq. inches for a #6 THWN-2, X 3 gives us .1521 sq. inches.

A single #4 THWN = .0824 sq. inches, X 3 gives us .2472 sq. inches.

.1521 + .2472 = .3993 sq. inches.

A trip to chapter 9 table 4 shows a 1-1/4" RMC has a 40% capacity of 0.610, a 1" only 0.355. Our RMC size is 1-1/4".

Re: Motor branch circuit calculation#19120 12/23/0212:44 PM12/23/0212:44 PM

I calculated out problem last night but my Internet Server was down so I was unable to post. Just to let you know I came up with same answers. That was fun Matt. Thanks for the post. Merry X-Mas to all!