Answer / rajan

U0 * 4 = 33/ 1.732

= 19.05 * 4
= 76KV

Answer / dhanraj

For 33 Kv cable DC hi pot test can be done like (2*E+1)1.414
here E=Voltage , 1.414 is the root 2 value.
Now (2*33+1)=67Kv
Then 67*1.414=94.738 Kv this is 100% of cable test.
In manufacturing factory they will test up to 80% of the voltage in that 100% (94.73*0.8)=75.79 Kv.
But in consumer side they will test up to 80% of voltage that injected in the factory voltage (75.79*0.8)=60.64 Kv.
Now 60.64 Kv is apply for 3 mins maximum.

Answer / michael aliganga

CABLE FACTORY AC TEST VOLTAGE =2E+1kV= 2X35+1= 71kV
DC HIPOT TEST:
CABLE INSTALLATION TEST DURING AND AFTER = ET= EF X 1.8 X 0.8
ET = 71 X 1.8 X 0.8 = 102kV, ICEA DC FIELD TEST VOLTAGE
(ICEA S-97-682) IS 100 kV FOR 35 KV SHIELDED POWER CABLE.

EF=CABLE MANUFACTURERS RECOMMENDED FACTORY AC TEST VOLTAGE
IF NOT AVAIABLE USE 2E+1KV
1.8 AC TO DC CONVERSION FACTOR (QUITE COMMON)
0.8 CONVERSION FACTOR FOR DERATING FROM FACTORY TESTING TO
ACCEPTANCE TESTING.

Answer / manohar

Hipot test conduct for 33kV cable is 1.5 times of rating
voltage in DC(49kV approx will be apply for 1 minute.)

Answer / martin thomas

Iec code
Ac= (2 x e) 1
Dc={(2 x e) 1} x 1.414
1.414=sqrt (2)
E= rated voltage

Aramco std

Dc =4 x u
U= phase voltage

Answer / b.gusain

Uo X 4 as per ICE

Answer / jagan

33kV cable hipot Test
33*2+1=67
67/√3=38.68 for single cable
38.68*√2=57.8 for diode dc convert RMS
57.8*0.8=46.3 for dielectric % reduce by year
Finally apply new cable 46.3KV /phase

Answer / chandan gupta

75.790 kv dc

Answer / a.shiva krishna

AC Hipot test voltage = Nominal input voltage * 2 + 1000
DC Hipot test voltage = AC Hipot test voltage * 1.414
for 33kv
ac hipot v/g = 33000*2+1000 = 67kv
dc hipot v/g = 67000 * 1.414 = 95kv

Answer / raveendra

(2E+1)=67 kV FOR AC HIPOT ( 1 minute test)
67*1.7 = 113.9 DC HIPOT ( i minute test)

for aged cables generally we will apply 65% of the above

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