- Location
- New Jersey
- Occupation
- Journeyman Electrician
An engineer gave me this chart for derating and the sizing of conductors feeding equipment in a machine room. Does Note d seem correct? IMO it is, providing there is not more than one outlet on the circuit. Any comments?
Ampacity: THHN-90 Degree Column (Table 310.16)
#12 30a
#10 40a
#8 55a
#6 75a
#4 95a
#2 130a
Derating Factors: Current carrying conductors (Table 310.15 (B)(2)(a))
4-6 80%
7-9 70%
10-20 50%
21-30 45%
31-40 40%
Standard Overcurrent Devices: (Article 240.6A)
15,20,25,30,35,40,45,50,60
Maximum Overcurrent Protection After Derating: (Article 240.4(B)(1)(2)(3))
Number of Current Carrying Conductors: Conductor Size Maximum Overcurrent Device
10-20 #6 THHN= 75a(50%)=37.5a- 40a max
21-30 #6 THHN= 75a(45%)=33.75a- 35a max
31-40 #6 THHN= 75a(40%)=30a- 30a max
10-20 #8 THHN= 55a(50%)=27.5a- 30a max
21-30 #8 THHN= 55a(45%)=24.5a- 25a max
31-40 #8 THHN= 55a(40%)=22a- 25a max
10-20 #10 THHN= 40a(50%)=20a- 20a max
21-30 #10 THHN= 40a(45%)=18a- 20a max
31-40 #10 THHN= 40a(40%)=16a- 20a max
10-20 #12 THHN= 30a(50%)=15a- 15a max
21-30 #12 THHN= 30a(45%)=13.5a- 15a max
31-40 #12 THHN= 30a(40%)=12a- 15a max
Notes: To derate multiple current carrying conductors in a raceway serving a single outlet
a) Find the conductor ampacity in Table [310.16]
b) Find the derating factor from Table [310.15(B)(2)(a)] based on the number of current carrying conductors in the raceway.
c) Multiply the ampacity by the derating factor and find the derated conductor value.
d) If the value doesn?t conform to a standard value in [240.6A], then round up to the next higher standard value.
[ January 12, 2006, 05:23 AM: Message edited by: infinity ]
Ampacity: THHN-90 Degree Column (Table 310.16)
#12 30a
#10 40a
#8 55a
#6 75a
#4 95a
#2 130a
Derating Factors: Current carrying conductors (Table 310.15 (B)(2)(a))
4-6 80%
7-9 70%
10-20 50%
21-30 45%
31-40 40%
Standard Overcurrent Devices: (Article 240.6A)
15,20,25,30,35,40,45,50,60
Maximum Overcurrent Protection After Derating: (Article 240.4(B)(1)(2)(3))
Number of Current Carrying Conductors: Conductor Size Maximum Overcurrent Device
10-20 #6 THHN= 75a(50%)=37.5a- 40a max
21-30 #6 THHN= 75a(45%)=33.75a- 35a max
31-40 #6 THHN= 75a(40%)=30a- 30a max
10-20 #8 THHN= 55a(50%)=27.5a- 30a max
21-30 #8 THHN= 55a(45%)=24.5a- 25a max
31-40 #8 THHN= 55a(40%)=22a- 25a max
10-20 #10 THHN= 40a(50%)=20a- 20a max
21-30 #10 THHN= 40a(45%)=18a- 20a max
31-40 #10 THHN= 40a(40%)=16a- 20a max
10-20 #12 THHN= 30a(50%)=15a- 15a max
21-30 #12 THHN= 30a(45%)=13.5a- 15a max
31-40 #12 THHN= 30a(40%)=12a- 15a max
Notes: To derate multiple current carrying conductors in a raceway serving a single outlet
a) Find the conductor ampacity in Table [310.16]
b) Find the derating factor from Table [310.15(B)(2)(a)] based on the number of current carrying conductors in the raceway.
c) Multiply the ampacity by the derating factor and find the derated conductor value.
d) If the value doesn?t conform to a standard value in [240.6A], then round up to the next higher standard value.
[ January 12, 2006, 05:23 AM: Message edited by: infinity ]
