PARALLELED FEEDERS
- Thread starter patl
- Start date
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- Location
- Illinois
- Occupation
- retired electrician
Re: PARALLELED FEEDERS
Yes. Each conductor produces heat and that is the reason for derating.
Don
Yes. Each conductor produces heat and that is the reason for derating.
Don
- Location
- Lockport, IL
- Occupation
- Semi-Retired Electrical Engineer
Re: PARALLELED FEEDERS
It's not intuitive, and may even seem contrary to intuition, but it is true nonetheless. Let?s look at heat generation (I*2R), but ignore de-rating factors for the moment.
Consider a pair of 1/0. The ampacity is 150 amps per conductor, or 300 amps total. But each conductor will generate heat separately from the other. The heating caused by current in the two wires is found by 2 times (150) times (150) times the resistance. Let?s call the result H1, so that H1 = 45,000 times R1.
Now consider a single 350. Its ampacity is 310 amps, but let us limit it to the 300 amps that we discussed above. The heating caused by current in the one wire is found by (300) times (300) times the resistance. Let?s call the result H2, so that H2 = 90,000 times R2.
Next, compare R1 and R2. From Table 8, the resistance of a 350 (R2 in this problem) is about 26% of the resistance of a single 1/0 (R1 in this problem). So if we substitute R2 = 0.26 times R1, we get H2 = 90,000 times (0.26) times R1, or H2 = 23,000 times R1.
Summary of results:
H1 = 45,000 times R1
H2 = 23,000 times R1
Conclusion, the heat generated by the pair of 1/0 cables is almost twice as much as the heat generated by a single 350.
It's not intuitive, and may even seem contrary to intuition, but it is true nonetheless. Let?s look at heat generation (I*2R), but ignore de-rating factors for the moment.
Consider a pair of 1/0. The ampacity is 150 amps per conductor, or 300 amps total. But each conductor will generate heat separately from the other. The heating caused by current in the two wires is found by 2 times (150) times (150) times the resistance. Let?s call the result H1, so that H1 = 45,000 times R1.
Now consider a single 350. Its ampacity is 310 amps, but let us limit it to the 300 amps that we discussed above. The heating caused by current in the one wire is found by (300) times (300) times the resistance. Let?s call the result H2, so that H2 = 90,000 times R2.
Next, compare R1 and R2. From Table 8, the resistance of a 350 (R2 in this problem) is about 26% of the resistance of a single 1/0 (R1 in this problem). So if we substitute R2 = 0.26 times R1, we get H2 = 90,000 times (0.26) times R1, or H2 = 23,000 times R1.
Summary of results:
H1 = 45,000 times R1
H2 = 23,000 times R1
Conclusion, the heat generated by the pair of 1/0 cables is almost twice as much as the heat generated by a single 350.
- Location
- Lockport, IL
- Occupation
- Semi-Retired Electrical Engineer
Re: PARALLELED FEEDERS
Suppose the run is 1000 feet long. Per Table 8, the resistance of 1/0 is 0.122 ohms per 1000 feet, so R1 is 0.122 ohms. The resistance of 350 is 0.0367 ohms per 1000 feet, so R2 is 0.0367 ohms.
H1 = 2 times I*2 times R1
H1 = 2 times (150)*2 times R1
H1 = 45,000 times R1
H1 = 45,000 times 0.122
H1 = 5,490
H2 = I*2 times R2
H2 = (300)*2 times R2
H2 = 90,000 times R2
H2 = 90,000 times 0.0367
H2 = 3,300
H1 > H2
QED
Suppose the run is 1000 feet long. Per Table 8, the resistance of 1/0 is 0.122 ohms per 1000 feet, so R1 is 0.122 ohms. The resistance of 350 is 0.0367 ohms per 1000 feet, so R2 is 0.0367 ohms.
H1 = 2 times I*2 times R1
H1 = 2 times (150)*2 times R1
H1 = 45,000 times R1
H1 = 45,000 times 0.122
H1 = 5,490
H2 = I*2 times R2
H2 = (300)*2 times R2
H2 = 90,000 times R2
H2 = 90,000 times 0.0367
H2 = 3,300
H1 > H2
QED
- Location
- Illinois
- Occupation
- retired electrician
Re: PARALLELED FEEDERS
If you use the DC resistance of 0.122 ohms/1000' for 1/0 and 0.0367 ohms per 1000' of 350 kcmil and use a load of 300 amps the 350 will produce 3.3 watts of heat per foot and each of the 1/0s will produce 2.745 watts of heat per foot. The two paralleled 1/0s will have a total of 5.49 watts per foot. This is 66% more heat for the paralleled 1/0s.
Don
If you use the DC resistance of 0.122 ohms/1000' for 1/0 and 0.0367 ohms per 1000' of 350 kcmil and use a load of 300 amps the 350 will produce 3.3 watts of heat per foot and each of the 1/0s will produce 2.745 watts of heat per foot. The two paralleled 1/0s will have a total of 5.49 watts per foot. This is 66% more heat for the paralleled 1/0s.
Don
ryan_618
Senior Member
- Location
- Salt Lake City, Utah
Re: PARALLELED FEEDERS
You're outcome would be a non-listed assembly!
You're outcome would be a non-listed assembly!
Re: PARALLELED FEEDERS
Besides not being a listed assembly, it would would be a little less than a 4/0 and wouldn't be allowed to carry more than somewhere around 230 amps.
Roger
[ November 10, 2003, 05:14 PM: Message edited by: roger ]
Besides not being a listed assembly, it would would be a little less than a 4/0 and wouldn't be allowed to carry more than somewhere around 230 amps.
Roger
[ November 10, 2003, 05:14 PM: Message edited by: roger ]
- Location
- Illinois
- Occupation
- retired electrician
Re: PARALLELED FEEDERS
Pat,
The amount of heat produced is directly related to the cross sectional are of the conductors. Look at the area of two 1/0s and compare that to the area of the 350 kcmil.
If you use the cross sectional area of two 1/0s to make a single conductor, the results would be exactly the same as the ones posted by Charlie and myself in this thread.
Don
Pat,
The amount of heat produced is directly related to the cross sectional are of the conductors. Look at the area of two 1/0s and compare that to the area of the 350 kcmil.
If you use the cross sectional area of two 1/0s to make a single conductor, the results would be exactly the same as the ones posted by Charlie and myself in this thread.
Don
- Location
- Lockport, IL
- Occupation
- Semi-Retired Electrical Engineer
Re: PARALLELED FEEDERS
Looking at it another way (but still getting the same results): If you put two identical resistors in parallel, the net resistance is cut by half. Thus, if you put two 1/0 cables in parallel, you reduce the resistance in half. But the resistance of a single 350 is not half, but rather one quarter, of the resistance of a single 1/0. Thus, the single 350 has half the resistance of the parallel combination of a pair of 1/0 cables. Therefore, it would generate about half the heat.
By the way, I would be grateful if you were to turn off the CAPS LOCK feature, next time you post.
Looking at it another way (but still getting the same results): If you put two identical resistors in parallel, the net resistance is cut by half. Thus, if you put two 1/0 cables in parallel, you reduce the resistance in half. But the resistance of a single 350 is not half, but rather one quarter, of the resistance of a single 1/0. Thus, the single 350 has half the resistance of the parallel combination of a pair of 1/0 cables. Therefore, it would generate about half the heat.
By the way, I would be grateful if you were to turn off the CAPS LOCK feature, next time you post.
Re: PARALLELED FEEDERS
Relating to Don's last comment, notice that the
area of the 2 1/0 = 0.190 sq in x 2 = 0.218 and the area of the 350 = 0.364 sq in, a 67% advantage
and yet the current capacity is almost the same for both circuits.
The AC/DC resistance ratio for 1/0 = 1 where the
AC/DC resistance ratio for 350 = 1.08. In reality the heat generated in the 350 may be 8% higher that caculated. The ability to dissapate this heat is limited to a great extent by the insulation resistance. In table 310.13 the insulation thickness for 1/0 conductors using THHN in 50 mils and for 350 the insulation thickness is 60 mils. This point is made just to show 2 reasons why the parallel smaller conductor
may have the same current capacity as a much larger conductor.
Relating to Don's last comment, notice that the
area of the 2 1/0 = 0.190 sq in x 2 = 0.218 and the area of the 350 = 0.364 sq in, a 67% advantage
and yet the current capacity is almost the same for both circuits.
The AC/DC resistance ratio for 1/0 = 1 where the
AC/DC resistance ratio for 350 = 1.08. In reality the heat generated in the 350 may be 8% higher that caculated. The ability to dissapate this heat is limited to a great extent by the insulation resistance. In table 310.13 the insulation thickness for 1/0 conductors using THHN in 50 mils and for 350 the insulation thickness is 60 mils. This point is made just to show 2 reasons why the parallel smaller conductor
may have the same current capacity as a much larger conductor.
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