how can we know cable size for a given current?
if any one have catalogue plz mail to
Answers were Sorted based on User's Feedback
Answer / m.faisal siddiqui.
The selection of a cable size for a given value of current
can be done using the cables & tables provided by the cable
manufacturer. The selection of cable size also depends upon
the lenght of cable from source to load. The total voltage
drop should not exceed 2.5% of rated voltage. Following
factors shall be considered while selecting a cable for
1. Rated Current.
2. Capacity of Protective Circuit Breaker used.
3. Total Length of cable.
4. Total voltage drop.
5. Laying Method.i.e. underground,in conduit,in trays etc.
6 No. of cables.i.e. grouping factor.
7. Ambient Temperature of installed area.
Temperature Correction Factor to be applied.
8. Type of cable. i.e. singel core,multicore,PVC insulated,
Armoured, XLPE insulated etc.
|Is This Answer Correct ?||25 Yes||2 No|
Answer / shrikant muthineni
For an aluminium conductor we can load approximately 2.5
amperes and for a copper conductor aroundf 6 amps
|Is This Answer Correct ?||20 Yes||7 No|
Answer / vasu
Two things are very much common on the power system engineering field while doing a cable sizing, these are:
Cables are separately sized based on voltage levels as said above.
Formulas used for LT side:
Full load current = KW/(KV x sqrt3 x PF x η)
Voltage Drop =
sqrt 3*IFL*(R COS Ø + X SIN Ø)*L/(KV*10000*N.R.)
IFL = Full load current of the Load connected in Amps
KW = Rating of the Load connected
KV = System Voltage
R = Resistance of the cable in Ohm/Ph/KM
X = Reactance of the cable in Ohm/Ph/KM
L = Length of the cable in meters
N.R = No of Runs of the cable
cosØ = Power factor of the system
PF = Power factor
Voltage Drop =
sqrt 3*Ist*(R COS Ø + X SIN Ø)*L/(KV*10000*N.R.)
where Ist=starting current which can be 6 times the FLA of current.In technical term it is called as LRC(locked rotor current)
Full load current = KVA / KV x sqrt 3
Short circuit current :
A = Ish x sqrt(t)/ k
t = short circuit time in sec
k = constant
A = Conductor size in Sqmm
Ish = Short circuit current in Amps
Derating Factor :
Ambient temperature correction factor for cables laid in 0.9 Trays at 50 degree C = 0.9
Group derating factor:
Multi Core cables
Cables are laid with a spacing of one Diameter and routed in cable trays and trays in tiers by 300mm =0.9
Ageing factor= 0.95
Therefore the Total derating factor = 0.9 x 0.9 x 0.95 = 0.7695
Total Derating Factor= 0.77
The rest calculation for voltage drop are same as that of LT side
|Is This Answer Correct ?||8 Yes||3 No|
Answer / bijumone.g
roughly you can take it as 1.5 times of the conductor
size.eg. 300 sqmm can conduct 450 A approximately and
moreover it also depends on the application like
underground or overhead.
if you want in detail you can check it in finolex website.
|Is This Answer Correct ?||12 Yes||9 No|
Answer / manoj sharma
Thank U very much VASU
|Is This Answer Correct ?||6 Yes||3 No|
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