Where is this in the code?

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Dana114

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This is a question on a quiz I took. I cannot find calculations. Where is this in the code book. My supervisor is stumped on this one also. So far we have had no luck.


A single-phase, 5 hp motor is located 110 ft from a panelboard. The nameplate indicates that the voltage is 115/230 and the FLA is 52/26A. What size conductor is required if the motor windings are connected in parallel and operate at 115V? Note: Apply the NEC recommended voltage-drop limits.
 
Dana114 said:
This is a question on a quiz I took. I cannot find calculations. Where is this in the code book. My supervisor is stumped on this one also. So far we have had no luck.


A single-phase, 5 hp motor is located 110 ft from a panelboard. The nameplate indicates that the voltage is 115/230 and the FLA is 52/26A. What size conductor is required if the motor windings are connected in parallel and operate at 115V? Note: Apply the NEC recommended voltage-drop limits.
Click to expand...
Motor information is in 430 and conductor ampacity is in table 310.16.
To start forget about the voltage requirements. Calculate the conductor size
and OC device for the motor. Then you can calculate the VD using the selected conductor. If the VD is to high you can select a larger conductor.
Put your numbers up and lets see the results.
 
5hp single phase 110 ft from panel 115 FLA 52 amp (230 FLA 26A)

1)
look up FLC
table 430.248 FLC = 56A

conductor size
430 part II
single motor
430.22 assume continuous duty
not less than 125% (56A) =70A

2) look up wire (long way)
assuming this is new motor, all contacts 75 degree and THHN Cu (otherwise you need to use 60 degree column)
310.16
4 AWG

3) check Vdrop
Vdrop=2kid/Circmils (single phase) look up k and circ mils chap 9 table 8
? should I use the FLC here or the FLA of 52A ?? using FLC
=2 * 12.9 * 70 * 110/ 41740
=4.75volts

4.75/115 = 4.1%

toohigh (needs to be 3%
so, check next size or do the 1 shot calculation
(changing to one-shot calculation)
Vdrop of 3% on 115 volts = .03 * 115 =3.45v

Circmils of wire at 3% drop = 2kid/Vdrop

=2 * 12.9 * 70 *110 /3.45
=57582

look up for >57582 in table 8
calls for #2 AWG

hmmm, i woulda thought that #3 would work
checking #3 voltage drop
Vdrop = 2kid/circmils
=2*12.9*70*110/52620
=3.77 volts
3.77/115=3.2% , haha, the formula was right


(my numbers were from 2005 codebook)
I would appreciate it if someone would check me - I am not sure whether to use FLC or FLA amps when doing the voltage drop calc ? (if FLA, then I recalc it cause it would be lower size wire)
 
thanks don

ok, so fixing calculation

1)
ok, so lets calc circmils again
Circmils of wire at 3% drop = 2kid/Vdrop

=2 * 12.9 * 52 *110 /3.45
=42776

look up table 8 chap 9
answer is #3 wire
 
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ok, i just looked back at my notes and according to my notes from mike holt's motors CD you are supposed to use the FLC to calculate the conductor ampacities (430.6) (which means the first answer was right )(?)
 
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You use the table ampacities to find the code minimum size conductor. The code does not address how to calculate the voltage drop because, except for some rare cases, it is not required by the code. There is no reason to use anything higher than the actual load current to calculate the voltage drop.
 
Dana114 said:
...A single-phase, 5 hp motor is located 110 ft from a panelboard. The nameplate indicates that the voltage is 115/230 and the FLA is 52/26A. What size conductor is required if the motor windings are connected in parallel and operate at 115V? Note: Apply the NEC recommended voltage-drop limits.
Click to expand...
Assuming copper cond's, steel C. (worst case), 0.8 PF, & 120V nom. source:

1) 52A x 1.25 = 65A, hence min. req. cond. by amperage = #4 AWG (70A @ 60 deg. C.

2) VD = 2(110 ft.)(sqrt[0.31^2 + 0.060^2] ohms)(52A) / 1000 ft. = 3.61V.
Therefore %VD = 100%(3.61V / 120V) = 3.01%.

The voltage at the motor will be 120V - 3.61V = 116.4V.
Hence, the current should be slightly less than 52A: 52A (115V /116.4V) = 51.4A, meaning the voltage drop should decrease to just below 3% (< max. code recommended).

So, it would be #4 AWG.
 
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