• For a cable of 6 mm² and installation method B2, the maximum current is 30 A x 0.8 =
24 A, which is insufficient. For a cable of 10 mm², the maximum allowed current is 50 A
x 0.8 = 40 A. So 2 parallel cables of 3 x 10 mm² +10 mm² are sufficient.
In case of installation method C, the maximum current of a 6 mm² cable at 40 °C is 36
A x 0.8 = 28.8 A. 2 parallel cables of 3 x 6 + 6 mm² will be sufficient.
• Install 32 A fuses on each cable.
Cable sizing according UL/cUL
Calculation method according UL 508A, table 28.1 column 5: allowable ampacities of insulated
copper conductors (75 °C (167 °F)).
Maximum allowed current in function of the wire size (NFPA70 table 310.16)
AWG or kcmil
Maximum current
14
20 A
12
25 A
10
35A
8
50 A
6
65 A
4
85 A
3
100 A
2
115 A
1
130 A
Correction factors
Ambient temperature
Correction factor
21-25 °C (70-77 °F)
1.05
26-30 °C (78-86 °F)
1.00
31-35 °C (87-95 °F)
0.94
36-40 °C (96-104 °F)
0.88
Calculation method for UL:
• Single supply cables (3 1 PE - configuration (1)):
• Add 25 % to the total current (I
tot
from the tables) (see UL 508A 28.3.2: "Ampacity shall
have 125 % of the full load current").
• Install the prescribed maximum fuse on each cable
• Parallel supply cable (2 x 3 2 PE - configuration (2)):
• Add 25 % to the total current (I
tot
from the tables) and divide by 2
• Multiply the ampacity of the cables with 0.8 (see UL 508A table 28.1 continued)
• Install fuses of half the size of the recommended maximum fuse size on each cable.
• When using 2 x 3 phase + 2 PE as in (3):
• Add 25 % to the total current (I
tot
from the tables) and divide by √3
• Multiply the ampacity of the cables with 0.8 (see UL 508A table 28.1 continued)
• Fuse size: the recommended maximum fuse size divided by √3 on each cable.
• Size PE cable:
• For supply cables up to AWG8: same size as the supply cables
Instruction book
2920 7140 52
101
Содержание SF 15+
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