Load calc

[RichB]

Good morning all,

Here is another test question and the 3 ways it is worked
Number 1 is directly from the test and numbers 2 and 3 are the two ways I worked it--cuz I am having a senior moment and just can't remember or find it in code.
Number one does not take into account the # of CCA, and here is my senior moment--I can't remember if we do or not with parallel runs. As the question stands I would read it to mean in conduit and to do a complete calculation.--As always TIA folks!

1. For a service with 9 parallel runs of XHHW 750 kcmil copper per phase, all run in an ambient temperature of 100? - 102? F., what is the total ampacity to each phase? Answer: ______________________­­____

Answer: 3762 amps [Ref: 750 kcmil wire is rated at 475 amps. (9)(475)(0.88) = 3762 amps]

2.
Example #1

9 sets of 750 KCML CU

Below using 75 degree column in both conductor and adjustment

750 KCML CU = 475 amps @ 75 degree column, x .70 conductor fill = 332.5 amps x .88 correction factor 75 degree column = 292.2 x 9 sets = 2633.4 amps



3.
Example #2

9 sets of 750 KCML CU

Below using 90 degree column in both conductor and adjustment

750 KCML CU = 535 amps @ 90 degree column, x .70 conductor fill = 374.5 amps x .91 correction factor 90 degree column = 340.7 x 9 sets = 3067.1

 

[lielec11]

The question is very vague and/or poorly worded. I guess you have to assume they're not running more than 3 current carrying conductors in any conduit?

 

[charlie b]

XHHW has a 90C rating available. So I would use the 90C column for both the initial ampacity and the temperature correction factor. The question did not state that more than three current-carrying conductors would be installed in the same raceway, so I would not apply that adjustment factor.

• (9) times (535 amps) times (0.91 correction factor from the 90C column) gives you 4,382 amps.

 

[GoldDigger]

XHHW has a 90C rating available. So I would use the 90C column for both the initial ampacity and the temperature correction factor. The question did not state that more than three current-carrying conductors would be installed in the same raceway, so I would not apply that adjustment factor.

• (9) times (535 amps) times (0.91 correction factor from the 90C column) gives you 4,382 amps.

But nothing in the question mentioned the temperature rating of the terminations.....

 

[david luchini]

I would assume for a "service" they figured either exposed to weather or underground (wet location) so the 75 deg rating of the XHHW is used.

The calculation looks right to me.

 

[Julius Right]

I agree with Example #1 but since all 3*9=27 conductors are in the same raceway instead of 0.7 then 0.45 adjustment factor has to be employed. So the total current will be 9*475*0.88*.45 = 1692.9 A/phase.
You could run the cables in an open top ventilated cable tray grouping cable in 3*1 or 4*1 cables per group and according to 392.17(2)(d) keeping 2.15 times cable overall diameter clearance.
Then the ampacity will be as per Table 310.15(B)(20) 638 A for 75 oC insulated 750 MCM copper conductor.

 

[jimdavis]

Where does it state that all 27 phase conductors are contained in the same raceway? Did I miss something? Also when an exam question does not give any information regarding termination temperature ratings, you should always follow 110.14(C)(1).