Answer / amit kumar tripathi

PERCENTAGE OF VOLTAGE DROP IS EQUAL TO
% V DROP = 1.732*Ir*L/1000 ( R COSØ + X SIN Ø ) / V * N
WHERE
Ir = FULL LOAD CURRENT
L = LENGTH OF THE CABLE
R = RESISTENCE OF THE CABLE
X = REACTENCE OF THE CABLE
COSØ = 0.8
SINØ = 0.53
V = VOLTAGE
N = NO. OF RUN CABLE
MAXIMUM % VOLTAGE DROP CONSIDERED 6% As per IE Rule & it cable size designs is Economical.

If Power rating & required cable length is increase , so your cable size is also increase.

Answer / muhammad usama qadir

you can calculate cable size by this link
http://www.solar-wind.co.uk/cable-sizing-DC-cables.html

Answer / tariqusman

if u want to find the cable size for a specific load. so u
find the load current. and just divid by 4, similsrly u
find the accurate size of cable for specific load. for
example. a load of 50kw, ist u find the load current, and
then divide by 4.
P = 50kw or 50,000watt
V = 440 volt
I = ?
p = 1.7320*440*I*0.8
I = p/1.7320*440*.8
I = 50,000/598.4
I = 83 Ampere

now u divid the load current by 4

83/4 = 20 sq.mm

its simple if u find the cbale size...

Answer / s.venkatraman

hai am also no exact answer
may be
eg 5HP motor select cable size 2.5sqmm cu cable, r 4 sqmm
Al cable.
work out
5Hp motor taken 90% full load current 7 amps.
that current * 5 times = 35 Amps (motor not run in full load
its my experiance)
2.5sqmm cu cable full load withstand current 36A
4sqmm Al cable full load with stand current 34A
so we choice Cu r Al cable

Answer / sukhvinder singh

Kv*load/volt


for single phase use 230 volt& for three phase use 440 volt
Kv =1000

Answer / 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%

Answer / er. naushad ashraf

Dear See 1 eg. of Cable Designing

From - 5 MVA power T/f LV side

To - 11 kV LA’s

• RMS Symmetrical Short Ckt current = 17.39 kA (As per Tender spec.)

• Fault Clearing Time = 1 sec

• Min. Cable Size = (17.39x1000xsq root 1)/94= 185 sqmm.

• Min. Cable Size = 1-3C 185 sqmm.

• Cable Length = 0.06 Km. (Assumed).

• Ifl = (5000x1000)/(1.734x11000x0.9x0.95) = 307 A.

• Derating Factor = Ca X Cg = .88x1 = 0.88

• Cable sized considered = 3C X 185 sqmm. (XLPE+Al).

• Cable Resistance (Ohm/Km) = 0.21

• Cable Reactance (Ohm/Km) = 0.087

• Cable Ampacity = 330 A.

• Cable Derated Current = 330x0.88 = 290.4 A.

• Min. No. of cable required = 307/290 = 1.0586 = 1 No.

• No. of Cable Runs Selected = 1 No.

• % Voltage Drop = (sqrt 3x307x0.06(0.21x0.9+0.087x0.43)x100)/(1x11000)
= 0.066 % (Acceptable)

Answer / tariqusman

sorry i remind you that if u determind the above value of
cable size, for example i find 20 sq. mm cble for 83 ampere
load . so u use the 25 sq. mm cable.

sorry for 20 sq mm. there no have 20 sq.

Answer / md. ekram ali

Ye sab galt jawab de rahe hai koi v correct answer nahi diya

Answer / naushad ashraf

As we Now that,

Area= (IscXrtT)/K, Factor for PVC or XLPE,

As we know that Derating Factor = AgXAaXAsXAf

and I= P/(rt3XVXCos#)
and I=DfXI

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