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I want formula to calculate cable size as per load given in
kw & amp.I searched many sites but didn't right answer.Plz
reply me asap.
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Answer Posted / amit kumar tripathi
The Power Cable size shall be selected on the basis of
(a) Full load of Current Carrying Capacity
(b) Fault current Rating
(c) Steady state voltage drop
(a) Full load of Current Carrying Capacity
Transformer TRF-1 Rating (P) = 1600 KVA
Rated System Voltage (V) = 0.4 KV
Minimum size is required for full load current =P/_/3 x V
=2309.47Amps
Overall derating factor of multicore LV cables =0.660
From Cable catalogue, 1C x 400 Sqmm cable is selected and
Current rating of the cable in air = 680Amps
Therefore 6 cables Run is selected= 4080 Amps
Hence selected cable size is adequate
(b) Fault Current Rating criterion
Since feeder is protected by circuit breakekr, cable size is decided based on the bus of Fault current the transformer feeder .
The required size of cable for fault current is given by
formula =Isc. _/t
K
where,
Isc - Short circuit current in KA = 50KA
t - Fault Clearing time in seconds =1Sec
K - XLPE insulation factor = 0.094
Therefore required cable size = 531.91Sq mm
Hence, Actual cable size choose 6 Run - 1C x 400 Sq mm is adequate
(C) Steady state voltage drop
Ir - Full load current of motor = 2309.47Amp
Rated System Voltage (V)= 400V
Cos Ø = 0.85
Sin Ø = 0.53
L - length of the cable =5
N - No of cable runs per feeder =6 Run
R - Resistance of 1C x 400 Sq mm cable =0.104 Ohm / km
X - Reactance of 1C x 400 Sq mm cable =0.07 ohm / km
The voltage drop is given by formula =
√3 x Ir x L/1000 (R cosø1 + X sinø1)/N X V
Therefore voltage drop =0.1044 %
Voltage drop is less than permissible value of 6%
| Is This Answer Correct ? | 4 Yes | 4 No |
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