Answer / zahid ali

2E + 1 KV = 67 KV DC Hi pot
67 * .707 = 47 KV AC Hi pot

Answer / bapi

AS HI POT TEST IS CONDUCTED IN DC SO APPROX. 1.5 TIMES OF
AC SUPPLY VOLTAGE IS APPLIED FOR 1 MINUTE.
FOR 33 kV CABLE APPROX. 50kV FOR 1 MINUTE WILL BE APPLIED.
FOR TESTING OF "R" PHASE WE HAVE TO SHORT "Y" AND "B" PHASE
ALONG WITH ARMOUR & IT SHOULD BE EARTHED. SIMILARLY FOR Y &
B PHASE.
AFTER EACH PHASE TESTING IT SHOULD BE DISCHARGED THROUGH
DISCHARGE RODS.

Answer / s.jeganath

(2E+1)1.414=dc
2E+1 =AC

(2*33000+1)1.414=DC
67000*1.414=DC
94738v=DC (it is a 100%)
but apply in 80% only it is standard for 33kv cable
APPly voltage dc=94738*0.8=75790V
(75.790KV DC)

Answer / mathiraja

2E+1KV=67KV FOR AC VOLTS
(2E+1KV)*1.414=94.73KV FOR DC VOLTS
ACCORDING TO IEC60950 STANDARDS.

Answer / manikandan n

For DC Hipot test the applied voltage calculating formula is equal to (2E+1000)*1.414*(80/100)
E = system voltage
1.414 = For DC RMS conversion.
80% = test conduct in site at second time.

Answer / ram sundar

For DC HIPOT = 1.414*(2E+1)
E=33/1.73=19.07
DC Hipot = 1.414*(2*19.07+1)
55.34kv approx 56Kv

Answer / sunil choudhary

FOR TESTING 6.6KV CABLE HIPOT TESTING.HOW MUCH KV WE HAVE
TO
APPLY .GIVE IN DETAILS.

sunil choudhary

Answer / yousaf

D.C. voltage equal to 3 x U0 shall be applied for 15 minutes. The standard is 1EC5840.

57KV FOR 15 MIN.

Answer / jim sreedharan

For 33 KV cables we do three tests.

1. Vlf Test: We apply 10 kv dc between graphite and armour
at a low frequency (normally between 0.02 hz to 0.05 hz)
for 1 hour.

2. Insulation Resistance :- We apply 5 kv dc for 1 min

3. High Voltage ac Test:- We apply 57.1 kv for 30 mins
(33/1.732)x 3 = 57.1 kv

God Bless

Answer / jayasanker

33KV CABLE
FORMULA:2E+1*1.414
2*33+1*1.414 = 94KV DC
IF BUS BAR 2*33+1 = 67KV AC

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