Transformer Primary Current

[Electric-Light]

Say testing out 69K to 13.2k 5MVA that would have a FLA 219A. At 3% Z..

The test equipment has to push 2070v @ 42A to make the 219A happen.

If you can get away with a lower value like 20% of rated amp, you can get by with 414v x 8.4A.

Then, you're at a level the test equipment can get by with operating on 208-240v 15-20A level power source.

 

[Ingenieur]

So couldnt then the %Z be calculated from Any applied voltage, not just the one that results in rated secondary current

Sort of
if 180 gave 1/2 i rated
180/7200 = 2.5%
you could unitize it to 1 pu i by mult by 2
the std is to do it at rated i

 

[timm333]

Thanks for the explanation by symmetrical components. To me symmetrical-components are easier than per-unit. Per unit is confusing, especially when it solves the problem using single phase equations. For example when you used: If = 4160/(3.6+1.8), it is confusing as it seems like a single phase calculation.

Voltage drop would be more prominent at lower voltages than at the higher voltages. Can we ignore voltage drop for HV short cables of 145 kV?

Also can the cables from GIS breaker be sized same as normal cables?

For sizing of HV cables (like 145 kV underground cables), I already have the standards (IEC 60287-3-1 and IEC 60183). But I am not able to find a document which shows a solved example. If there is a document on internet which gives the step by step hand calculations of high voltage cable sizing, it will be very helpful. Thanks

 

[Ingenieur]

But if Isc Sec rated is accomplished at 100v, does Isc= 75% @ 75v? 50% @ 50v?
Can you do 5, 10, 15, 25% and linearize and get acceptably close match?

It's always convenient to be able to test within the limits of 120v 15A circuit which is essentially guaranteed available anywhere and the test equipment wouldn't be gigantic.

The current is not the sc current, it is the rated operating current
The sc i would be the sec i if the prim v were raise to the rared 7200

if i rated at 100 then pu Z = 100/7200 = 1.4%
i fault = 7200/100 x i rated = 1/0.014 x i rated
of course if V rated/operating is lower so will fault i since Z is ~ constant
but we talkinf in terms of transformer rated parameters

the test is no issue for mfgs making xx MVA xfmrs
They do BIL and SIL at xxx kv levels

 

[GoldDigger]

But if Isc Sec rated is accomplished at 100v, does Isc= 75% @ 75v? 50% @ 50v?
Can you do 5, 10, 15, 25% and linearize and get acceptably close match?

It's always convenient to be able to test within the limits of 120v 15A circuit which is essentially guaranteed available anywhere and the test equipment wouldn't be gigantic.

The secondary current should be proportional to primary voltage as long as you are not driving the core into saturation. And with the secondary shorted that should not happen as long as you stay below the rated input voltage on the primary side.
The primary side load current will be high because of the short, but the magnetizing current will stay well below the operating level.

mobile

 

[Ingenieur]

Thanks for the explanation by symmetrical components. To me symmetrical-components are easier than per-unit. Per unit is confusing, especially when it solves the problem using single phase equations. For example when you used: If = 4160/(3.6+1.8), it is confusing as it seems like a single phase calculation.

Voltage drop would be more prominent at lower voltages than at the higher voltages. Can we ignore voltage drop for HV short cables of 145 kV?

Also can the cables from GIS breaker be sized same as normal cables?

For sizing of HV cables (like 145 kV underground cables), I already have the standards (IEC 60287-3-1 and IEC 60183). But I am not able to find a document which shows a solved example. If there is a document on internet which gives the step by step hand calculations of high voltage cable sizing, it will be very helpful. Thanks

Actual both are used
sym component calcs are usually done in pu
it is a single ph calc, shorted line to line

that is the beauty of sym comp, they make the calcs simpe algebraic calcs
typically only for 1 ph

Vdrop calcs for the cable are the same except reactive components need considered
use the cable mfgs data

 

[timm333]

Yes symmetrical components is good technique. But I would think that the result should be multiplied/divided by 1.732 (or 3) to convert it back to three phase.

The only thing which I am not able to understand is that when we solve the problem by symmetrical components/per unit, then why at the end we don't convert the answer back to three phase?

 

[timm333]

Single conductor HV cables have a metallic sheath, and a voltage is induced in this sheath. For example 100 V/m can be induced in the sheath of a 145 kV cable under normal operating conditions. Would this induced voltage in the sheath affect the voltage drop in the cable? Thanks

 

[NewtonLaw]

High Voltage Cable Losses

High Voltage Cable Losses

Single conductor HV cables have a metallic sheath, and a voltage is induced in this sheath. For example 100 V/m can be induced in the sheath of a 145 kV cable under normal operating conditions. Would this induced voltage in the sheath affect the voltage drop in the cable? Thanks

Usually this has a minimal affect on the voltage drop of the cable power conductor. It does affect the capacity in that the eddy currents that can be set up due to the power (and thus thermal heating) developed in the sheath. This power is on the order of 2 to 5% of the power loss of the main conductor. The values is derived from an old empirical formula which looked at the radius to the sheath and current flow through the main conductor. There were more but I have to look them up in my text books to find it. The thermal addition will reduce the overall cable rating accordingly.

 

[timm333]

Thanks, actually in addition to the metallic sheath, there are two semiconductor screens (shields) as well: one above the conductor and the second above the insulation. Do these two screens (shields) would affect the power capacity (and/or voltage drop) of the cable?

 

[NewtonLaw]

Thanks, actually in addition to the metallic sheath, there are two semiconductor screens (shields) as well: one above the conductor and the second above the insulation. Do these two screens (shields) would affect the power capacity (and/or voltage drop) of the cable?

The semiconductor shields are used for electric field mitigation between voids on the surface of the conductors themselves or on the surface of the insulation. Voids allow partial discharge to occur under some operating conditions. The semicoductive shields help to eliminate these stresses. Hope this helps.

 

[timm333]

Thanks. Sorry a couple of more questions. Cross bonding of underground high voltage cables is easy when the number of segments is exactly divisible by 3. For example if the cable length is 1500 m and the length of cable reel on one drum is 500 m, then two joints would be needed and so there will be total 3 segments, and so the cross binding would be easy.

However if the length of the cable is 1700 m and the length of cable reel on one drum is 500 m, then 3 joints would be required and so there will be 4 segments now (instead of 3). How to do the cross binding when there are 4 sections?


Also, when we have to size a HV cable for 130% load factor, will it be the same thing as 130% full load current?

 

[NewtonLaw]

Cable Cross Bonding Concern

Cable Cross Bonding Concern

Thanks. Sorry a couple of more questions. Cross bonding of underground high voltage cables is easy when the number of segments is exactly divisible by 3. For example if the cable length is 1500 m and the length of cable reel on one drum is 500 m, then two joints would be needed and so there will be total 3 segments, and so the cross binding would be easy.

However if the length of the cable is 1700 m and the length of cable reel on one drum is 500 m, then 3 joints would be required and so there will be 4 segments now (instead of 3). How to do the cross binding when there are 4 sections?


Also, when we have to size a HV cable for 130% load factor, will it be the same thing as 130% full load current?

From my experience, the three cross bonds will be sufficient. As to the load factor reaching 130%, you must also supply the total amount of time the cable will be at the 130% load factor. In essence you are saying you will over load the cable to 130% of capacity. This may OK if the 130% load factor lasts no longer than a few minutes followed by some time period that allows the cable to dissipate the heat caused by the overload. There is a maximum overload for every cable however and to be sure you can achieve satisfactory operation over the life of the cable, you should speak to the cable manufacturer about exactly what loading you can place on the cable without damaging the insulation. If you were using Okonite cable, there technical support personnel would gladly help to answer your questions on this. Hope this helps.

 

[STEVE312623]

OCPD 150 KVA 240V 3 PHASE TRANSFORMER?

OCPD 150 KVA 240V 3 PHASE TRANSFORMER?

Calculations show that on a 3- phase 150KVA 240 transformer full load amps are 361 amps and protection recommended on the primary side is 600 amps. This being said , would I be correct to believe that a 600 amp bus plug must be used since the power is coming from a bus duct?

Any assistance or direction will greatly appreciated!