Ampacity correction and adjustment factors on table 310.15(B)(3)(a)

[Lucaslight]

I'm a little confused about what our instructor has been telling us, let's say we have 15 H.P. 480V 3 phase motor with 21A F.L.A.'s after multiplying by 125% we get 27.25A for a #10 wire, now let's say we run conduit from the first motor to another identical motor, and then from there to another identical motor.
for the second conduit "E.M.T." We have to derate due to the fact that now we have 6 wires. He is saying that because 80% is the reciprocal of 1.25%=

(21 Amps)(1.25)=27.25> (27.25 Amps)(.80) = 21 amps

bringins us back down to 21 amps F.L.A.

but then on the 3rd conduit we would need to adjust again since we have 9 current carrying conductors. "Which I believe is 70%" In this case he told us to divide by the percentage which is what I've seen in different questions that have been posted here before.

(21 Amps)(1.25)=27.25 amps > (27.25 Amps) / (.70)=38.92 Amps

I was thinking the reciprocal thing does make sense but wouldn't you just follow the math and divide by .80 instead of multiplying? That would appropriately increase your wire size since you have more wires "4 to 6 wires for 80%" correction.

(21 Amps)(1.25)=27.25 Amps > (27.25 Amps) / (.80)=34.06 Amps

im confused... :?

 

[david luchini]

I was thinking the reciprocal thing does make sense but wouldn't you just follow the math and divide by .80 instead of multiplying? That would appropriately increase your wire size since you have more wires "4 to 6 wires for 80%" correction.

(21 Amps)(1.25)=26.25 Amps > (26.25 Amps) / (.80)=32.81 Amps

im confused... :?

You are correct, you would divide in both cases.

If your #10 wire is rated 90Deg, then #10 would be acceptable for all of the conduits...40A * 0.7 = 28A.

 

[Smart $]

Your gut feeling is right.

FLA ? 125% ? Correction factor ? Adjustment factor <= Table value.

Table value ? Correction factor ? Adjustment factor >= FLA ? 125%

 

[Lucaslight]

uhmm

uhmm

So in Table 310.15(B)(3)(a), the percentages are used to indicate us that for a certain number of wires,
"more than 3" we would have to multiply the percentages by the (F.L.A*1.25) in order to obtain the derated Amperage values.

Now I've read this online, and it contradict what I'm understanding...

this is telling me that by reducing the conductors by 70% the insulation rating will not overheat... am I not supposed to increase the wire by 70% in order to ease current flow? therefore decreasing the chance of overheating?

 

[Smart $]

So in Table 310.15(B)(3)(a), the percentages are used to indicate us that for a certain number of wires,
"more than 3" we would have to multiply the percentages by the (F.L.A*1.25) in order to obtain the derated Amperage values.

Now I've read this online, and it contradict what I'm understanding...

this is telling me that by reducing the conductors by 70% the insulation rating will not overheat... am I not supposed to increase the wire by 70% in order to ease current flow? therefore decreasing the chance of overheating?

Multiply the allowable ampacity (table value) by the percentage of Table 310.15(B)(3)(a), in effect reducing the ampacity of the conductor to keep its temperature from exceeding the insulation temperature rating. This adjusted ampacity must equal or exceed the required minimum circuit ampacity (e.g. 125% FLA).

 

[Lucaslight]

Soo...

Soo...

So the wire size decreases as the adjusted amperage decreases?

 

[david luchini]

So the wire size decreases as the adjusted amperage decreases?

No, you've still got it backwards.

Try FLA * 1.25 ? 0.8.

 

[Smart $]

So the wire size decreases as the adjusted amperage decreases?

The more a conductor is derated (i.e. adjustment percentage lowered), the larger the wire size has to be to provide the same level of ampacity.

The one constant in the equation(s) is the minimum required circuit ampacity (MRCA) [e.g. 125% FLA]

1-3 conductors...no adjustment
4?6 80% (allowable ampacity must be not less than 1/80% or 125% MRCA)
7?9 70% (allowable ampacity must be not less than 1/70% or 143% MRCA)
10?20 50% (allowable ampacity must be not less than 1/50% or 200% MRCA)
21?30 45% (allowable ampacity must be not less than 1/45% or 222% MRCA)
31?40 40% (allowable ampacity must be not less than 1/40% or 250% MRCA)
41 and above 35% (allowable ampacity must be not less than 1/35% or 286% MRCA)

Again, allowable ampacity is the table value for the conductor type and conditions of use. None of the above take into consideration any other 'adjustment', such as correction for ambient temperature per 310.15(B)(2).

 

[Lucaslight]

The more a conductor is derated (i.e. adjustment percentage lowered), the larger the wire size has to be to provide the same level of ampacity.

The one constant in the equation(s) is the minimum required circuit ampacity (MRCA) [e.g. 125% FLA]

.

oh ok... so I was right the first time, the reciprocal part that our instructor mentioned seemed to keep throwing me off :slaphead:... thanks a lot.
I was also watching some of Mike Holt's videos on youtube about the subject but now it makes a lot of sense, thanks a lot.

 

[kwired]

I'm a little confused about what our instructor has been telling us, let's say we have 15 H.P. 480V 3 phase motor with 21A F.L.A.'s after multiplying by 125% we get 27.25A for a #10 wire, now let's say we run conduit from the first motor to another identical motor, and then from there to another identical motor.
for the second conduit "E.M.T." We have to derate due to the fact that now we have 6 wires. He is saying that because 80% is the reciprocal of 1.25%=

(21 Amps)(1.25)=27.25> (27.25 Amps)(.80) = 21 amps

bringins us back down to 21 amps F.L.A.

but then on the 3rd conduit we would need to adjust again since we have 9 current carrying conductors. "Which I believe is 70%" In this case he told us to divide by the percentage which is what I've seen in different questions that have been posted here before.

(21 Amps)(1.25)=27.25 amps > (27.25 Amps) / (.70)=38.92 Amps

I was thinking the reciprocal thing does make sense but wouldn't you just follow the math and divide by .80 instead of multiplying? That would appropriately increase your wire size since you have more wires "4 to 6 wires for 80%" correction.

(21 Amps)(1.25)=27.25 Amps > (27.25 Amps) / (.80)=34.06 Amps

im confused... :?

Minimum conductor ampacity must be 125% of motor full load value taken from tables at end of art 430 - I think you got that down. So you have a 21 amp motor times 125% = 26.25 (assuming your 27.25 was a typing error).

This means you must have a minimum (likely at 75C) conductor ampacity of 26.25 amps no matter what adjustments end up being.

Now come adjustments - but we can apply 90C insulation ampacity values when doing adjustments. There are two ways to do this one is to figure we need a 10 AWG based on no adjustments and then derate the 10 AWG to either 80%, 70% or whatever applies, and if that results in too little ampacity left over - increase conductor size and do it all again.

OR if you multiply the minimum ampacity needed by the reciprocal of the adjustment factor - you end up with a final minimum ampacity and select your conductor instead of repeating calculations until you find the right conductor size.

So to find minimum ampacity in one set of calculations, start with minimum necessary ampacity which was 26.25
If there are 4-6 conductors requiring 80% adjustment the reciprocal is 1.25 - so 26.25 x 1.25 = 32.81 (remember you are selecting from 90 deg column for this value) or for 70% for 7-9 conductors the reciprocal is 1.43, 26.25 x 1.43 = 37.54.

Reciprocal is 1 divided by the number you are "reciprocating"

1/.80 = 1.25
1/.70 = 1.43

you also get same result by dividing by the adjustment factor like your instructor told you.

so if the base were 100 we can either divide by .80 and the result is 1.25 or multiply by 1.25 and the result is still 1.25.

try a few different values at different percentages and you will see, is easier to see when you use a base of 100 in your trials. One of the biggest things that maybe will throw a person off is getting the decimal in the right place - remember when you drop the percent sign you need to move the decimal two places to the left to be able to directly use that number in calculations.