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#19114  12/20/02 12:18 PM
Motor branch circuit calculation


I thought it might be fun to pose a motor branchcircuit 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 Squirrelcage induction Service factor 1.15 continuous duty 480 volt
Motor #2: 40HP 3PH code letter F Squirrelcage 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 THWN2, 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 shortcircuit/ground fault protection.
Calculate the minimum size rigid conduit needed for these two circuits.
Extra points for showing your work.



#19115  12/21/02 12:03 AM
Re: Motor branch circuit calculation


Matt  are you trying to get us to do your homework for you?



#19116  12/21/02 12:08 AM
Re: Motor branch circuit calculation

Member

Joined: Aug 2002
Posts: 42
Fla.


So you ran the 2 motor feeds in the same conduit??



#19117  12/21/02 04:24 PM
Re: Motor branch circuit calculation


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 12212002).]



#19118  12/21/02 07:39 PM
Re: Motor branch circuit calculation


Matt: Thought I would take a try at your motor problem. I am using a 1999 NEC code book.
Motor #1 FLA (table 430150)34 amps Wire size #6 thwn2 Correction factors 110F .87 (table 31016) Adjustment factor more than 3 wires in raceway 80% 31015b2a Continous load 125% 43022a
Fuse size 60amp 430152 34x175% round up to nearest fuse size 43052c1 exc.1
Motor #2 FLA 52 amps (table 430150) Wire size #4 thwn2 Correction factors 110F .87 (table 31016) Adjustment factor more than 3 wires in raceway 80% 31015b2a Continous load 125% 43022a
Fuse size 100 amp 430152 52x175% round up to nearest fuse size 43052c1 exc. 1
Minimum size RMC 1 1/14" 3#4 THWN2 .0824x3=.2472 3#6 THWN2 .0507x3=.1521 Chapter 9 table 5
.2472+.1521=.3993 1 1/4 RMC Chapter 9 table 4
Let me know if I am correct.
[This message has been edited by fla sparkey (edited 12212002).]
[This message has been edited by fla sparkey (edited 12212002).]



#19119  12/23/02 10:19 AM
Re: Motor branch circuit calculation


Fla Sparky, You're right on the money!
Here is my break down;
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 THWN2, 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 THWN2
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 THWN2.
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 THWN2, 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 11/4" RMC has a 40% capacity of 0.610, a 1" only 0.355. Our RMC size is 11/4".




