Description
Manufacturer | ABB |
Brand | ABB |
Series | module |
Part Number | SS832 |
Product Type | module |
Quality | 100% New Original |
Stock | In stock |
Delivery time | 1-3 days after Payment |
After-sales Service | Have |
Warranty | 1 year |
Shipping term | DHL / FEDEX/ EMS /UPS/TNT/EMS |
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Protection of feeders
1.2 Installation and dimensioning of cables
Table 14: Correction factors for soil thermal resistivities other than
2.5 Km/W
For correction factor k2
’’:
• for cables laid directly in the ground or if there are not other conductors within
the same duct, the value of k2
’’ is 1;
• if several conductors of similar sizes are present in the same duct (for the
meaning of “group of similar conductors”, see the paragraphs above), k2
’’ is
obtained from the first row of Table 5;
• if the conductors are not of similar size, the correction factor is calculated by
using this formula:
where:
n is the number of circuits in the duct.
Correction factor k3
Soil thermal resistivity influences the heat dissipation of the cable. Soil with low
thermal resistivity facilitates heat dissipation, whereas soil with high thermal
resistivity limits heat dissipation. IEC 60364-5-52 states as reference value for
the soil thermal resistivity 2.5 Km/W
To summarize:
Use this procedure to determine the cross section of the cable:
1. from Table 10, determine the correction factor k1
according to the insulation
material and the ground temperature;
2. use Table 11, Table 12, Table 13 or the formula for groups of non-similar cables to determine the correction factor k2
according to the distance between
cables or ducts;
3. from Table 14 determine factor k3
corresponding to the soil thermal resistivity;
4. calculate the value of the current I’b by dividing the load current Ib (or the
rated current of the protective device) by the product of the correction factors
calculated:
5. from Tables 15-16, determine the cross section of the cable with I0 ≥ I’b, according to the method of installation, the insulation and conductive material
and the number of live conductors;
6. the actual cable current carrying capacity is calculated by