Transformer Primary Current

[timm333]

A question about using transformer in opposite direction. For example a 500 kVA 13.8/4.16 kV transformer to be used as 4.16/13.8 kV. Originally the current on the 13.8 kV side was 20.9 A. But when used in the opposite direction, the current on the original 13.8 kV side would become 69.4 A (3.3 times more). As the winding on this side is not designed for this higher current, so we should keep the new load 3.3 times less than 500 kVA. Should we now use a fuse in series with the upstream relay in order to keep the let through current 3.3 times less? Thanks for help.

 

[Electric-Light]

A question about using transformer in opposite direction. For example a 500 kVA 13.8/4.16 kV transformer to be used as 4.16/13.8 kV. Originally the current on the 13.8 kV side was 20.9 A. But when used in the opposite direction, the current on the original 13.8 kV side would become 69.4 A (3.3 times more). As the winding on this side is not designed for this higher current, so we should keep the new load 3.3 times less than 500 kVA. Should we now use a fuse in series with the upstream relay in order to keep the let through current 3.3 times less? Thanks for help.

What is the direction of power flow and what is the energy source and what are you sinking energy into?
What's the power factor and is it leading or lagging?

 

[Ingenieur]

A question about using transformer in opposite direction. For example a 500 kVA 13.8/4.16 kV transformer to be used as 4.16/13.8 kV. Originally the current on the 13.8 kV side was 20.9 A. But when used in the opposite direction, the current on the original 13.8 kV side would become 69.4 A (3.3 times more). As the winding on this side is not designed for this higher current, so we should keep the new load 3.3 times less than 500 kVA. Should we now use a fuse in series with the upstream relay in order to keep the let through current 3.3 times less? Thanks for help.

the currents remain the same since S or the kva remains the same

if the power source is the hi side and load is the low, you step down voltage and step up current

if the power source is the low side and load is the hi, you step up voltage and step down current

the voltages and currents stay the same as rated on each side
either way at rated kva: the 13.8 side will see 20.9 A and the 4.16 side will see 69.5, regardless which is primary or secondary

you base protection on the actual current

 

[ron]

Originally the current on the 13.8 kV side was 20.9 A. But when used in the opposite direction, the current on the original 13.8 kV side would become 69.4 A (3.3 times more).

timm333

This could only be the case if the load was more than 500kVA. And as you mention, the winding would overload.

Calcs are based on the load (as mentioned by previous responses), which for that transformer shouldn't be higher than 500kVA (other than short overloads that may happen).

The 13.8kV side should never be more than 20.9A if the load is no more than 500kVA

 

[timm333]

Sorry did not provide good explanation. The actual voltages are different but did not want to write the actual voltages here so just used 13.8 and 4.16 kV for the purpose of understanding. Here is the situation: suppose we have 500 kVA transformer, 13.8 kV supply coming from utility, the transformer steps it down to 4.16 kV and feeds to 4.16 kV motor loads. Now let’s says that instead of 13.8 kV supply, we have a supply of 1.3 kV but still we want to run the 4.16 kV loads. (I know 1.3 kV is not a standard voltage, but just using is for the purpose of understanding).


So we use the transformer in the opposite direction and step up the 1.3 kV to 4.16 kV in order to run the 4.16 kV loads. The original 13.8 kV side is now used as 4.16 kV, and the original 4.16 kV side is now used as 1.3 kV. Even though the kVA would be the same, but the voltage now on both sides is lower than original rating, so the current would go up (about 3.3 times up) in both wingdings. As the wingdings are not designed for this higher current, so we should keep the new load 3.3 times less than 500 kVA. Should we now use a fuse in series with the upstream relay in order to keep the let through current 3.3 times less? Thanks.

 

[GoldDigger]

Trying to run a power transformer at one third of its design voltage and in reverse is, IMHO, a misguided exercise.
The current would have to be 3.3 times the design current to deliver full power.

With 3.3 times the current, the power lost in the transformer would be 1about ten times the design power, regardless of the voltage applied.
By my calculations you would have to reduce the power rating by a factor of 3.3 to get the transformer operating at normal temperature rise.
Combined with the reduced overall voltage the transformer should be OK.
But at that voltage, the magnetizing current will be in a totally different range relative to core hysteresis, etc. and although that is unlikely to be a problem it does not feel right to me.

Yes, that would work, but I think it is a bad idea. If it means using an existing surplus transformer instead of purchasing the correct one (possibly on a long lead time), it may still be justified.

ETA: Heavily edited because I screwed up the first time...

 

[gar]

170119-1431 EST

timm333:

You have a 500 kVA transformer with a 13.8 kV primary.

Then you want to use the transformer with that primary (now being used as a secondary) having a voltage output of 4.16 kV.

This means that you now have to view the transformer as a 500/3.32 = 150 kVA transformer to avoid overheating said winding. For normal life of a transformer you can not run more current thru a transformer winding than what that winding was designed for.

.

 

[timm333]

Thanks. I think the required short circuit rating of the incoming cable from utility would be the same as long as the load does not exceed 150 kVA. How will reversing the transformer affect the protection of the incoming feeder?

 

[Ingenieur]

Thanks. I think the required short circuit rating of the incoming cable from utility would be the same as long as the load does not exceed 150 kVA. How will reversing the transformer affect the protection of the incoming feeder?

new kva = 4.16/13.8 x 500 = 150 kva
or
1.3/4.16 x 150 = 156 kva
use the lower 150 kva

input current = 67 A, base primary protection on this
output = 21 A, secondary protection on this

to calc fault current determine the actual Z in ohms and I fault = 4160/Z

 

[timm333]

The actual Z would be the Z of cable and transformer combined. Will the Z of transformer be the same as mentioned on the transformer datasheet (as long as the load does not exceed 150 KVA)?

 

[Electric-Light]

I'm not getting what he's trying to do. Resisting back on an aircraft carrier being lowered with a 4.16kV crane with motor running at synchronous speed and sending power back into the grid?

 

[Ingenieur]

The actual Z would be the Z of cable and transformer combined. Will the Z of transformer be the same as mentioned on the transformer datasheet (as long as the load does not exceed 150 KVA)?

it's the same
based on the base Z (derived from rated kva and v) x pu%/100 = Z Ohms

another way
i fault = kva/(1.732 x v sec x pu Z)
Z Ohms = v sec / i fault

add the wire Z
i fault = v sec/(Z fmr + Z wire)
infinite bus no utility Z

 

[GoldDigger]

I'm not getting what he's trying to do. Resisting back on an aircraft carrier being lowered with a 4.16kV crane with motor running at synchronous speed and sending power back into the grid?

It seems to me that he has an existing 13.8kV feed going to a 4.16kV crane.
All of a sudden he no longer has that 13.8 feed available, but gets a 1.38kV feed instead (all numbers are made up examples though).
He wants to keep the same 4.16kV crane working at the same full load watts, so he turns the transformer around. (3.3 being very roughly sqrt(10))

Which reminds me that sqrt(3) approches 2 for large values of 3. :angel:

 

[timm333]

If the transformer has impedance of 6%, then:

I,fault = 500/(1.732 x 4.16 x 0.06)=1.16 kA

Now should it be: Z,ohm= 4.16/1.16=3.59 Ohm
Or should it be: Z,ohm=4.16/(1.732 x 1.16)=2.07 Ohm ?

Can we usually ignore the utility impedance for cable sizing?

Also is this cable sizing method valid for all cables (LV, MV, and HV)?

Thanks

 

[Ingenieur]

If the transformer has impedance of 6%, then:

I,fault = 500/(1.732 x 4.16 x 0.06)=1.16 kA

Now should it be: Z,ohm= 4.16/1.16=3.59 Ohm
Or should it be: Z,ohm=4.16/(1.732 x 1.16)=2.07 Ohm ?

Can we usually ignore the utility impedance for cable sizing?

Also is this cable sizing method valid for all cables (LV, MV, and HV)?

Thanks

line to line fault (which is typically larger than l-g so should be used for sizing) is the 3.59 Ohm

the 2.07 is your single line equivalent Z (mult by 1.732 to get phase)
Z base = (4160)^2 / 500000 = 34.61 Ohm
Z act = 0.06 x 34.61 = 2.077 Ohm (single line equiv) or 3.597 phase (call it 3.6)

this is not used for sizing the cable but determining a fault current magnitude
the utility impedance will lower the fault current so it can usually be ignored unless you need a lower fault current so you can use lower rated equipment if the infinite bus puts you on the border of 2 sizes, typically does not change things much

if you want to estimate line Z influence:
assume 3% Vdrop (Vd) (can be 2% for short runs, whatever you feel is appropriate)
Vll = phase V
I full load Ifl = kva/(1.732 x Vll)

plugging and crunching
Vd = 0.03 x Vll = Ifl x Z (If)
Z = 0.03 x Vll^2 x 1.732 / kva = 0.03 x 4160^2 x 1.732 / 50000 = 1.798 Ohm (call it 1.8)

new fault current
= 4160/(3.6 + 1.8) = 0.77 kA (reduced from 1.16 kA by considering the conductors)

 

[Ingenieur]

and for those who care, lol
why a l-l fault > l-g one (typically)

the proof involves symmetrical components, which I won't use it to derive the equations, but we need the sequence Z's (which the utility may provide)
usually Z1 = Z2 and Z0 = 2 x Z1 (or Z2), just a rule of thumb
Zf is the fault Z, usually considered 0 but seldom is (most faults are arcing or have some Z)
and Zf for a l-g is usually higher than l-l, makes sense, a wire to ground vs. a wire to wire

l-g
Vf = Vll / 1.732
If = 3 (Vll / 1.732) / (Z0 + Z1 + Z2 + 3Zf)
let Zf = 0, Z1 = Z2 and Z0 = 2 x Z1, and Vf = Vll / 1.732
If = 3 Vll / (1.732 x 4 Z1)
If = Vll / (2.32 Z1)

l-l
Vf = Vll
If = Vll / (Z1 + Z2 + Zf) (you can see already it will be larger already: no Z0, Vf vs 3Zf, and Vll vs Vll/1.732)
let Zf = 0, Z1 = Z2 and Z0 = 2 x Z1
If = Vll / (2 Z1)

l-g: If = Vll / (2.32 Z1)
l-l: If = Vll / (2 Z1)

all else being equal the l-l should be ~ 2.32/2 = 1.16 times larger than the l-g, even greater if Zf is considered or the system is not well grounded (high Z0)

 

[Electric-Light]

Available fault current of the transformer is obtained by extrapolation from the primary voltage required to bring the shunted secondary to induce the rated current rather than tryin' it out real scale.

Given 7200 to 240v transformer rated at 120kVA 500A. if the variable voltage transformer feeding the primary needed to be raised to 216V before the secondary pushed 500A through a short, its extrapolated that if you applied a short bar on secondary with the primary of 7200v applied, the short circuit current is 16.67kA. 500A @ 3%. 500/0.03 = 16.67kA

Rumor has it that real scale testing on something like that doesn't go the same as real scale test on a 720v to 24v 120VA 5A transformer.

Question: if you don't have a variac big enough to energize the primary to induce full rated secondary current, how much lower can you get away with and still have a reasonably accurate calculation with linear regression?

 

[Ingenieur]

Available fault current of the transformer is obtained by extrapolation from the primary voltage required to bring the shunted secondary to induce the rated current rather than tryin' it out real scale.

Given 7200 to 240v transformer rated at 120kVA 500A. if the variable voltage transformer feeding the primary needed to be raised to 216V before the secondary pushed 500A through a short, its extrapolated that if you applied a short bar on secondary with the primary of 7200v applied, the short circuit current is 16.67kA. 500A @ 3%. 500/0.03 = 16.67kA

Rumor has it that real scale testing on something like that doesn't go the same as real scale test on a 720v to 24v 120VA 5A transformer.

Question: if you don't have a variac big enough to energize the primary to induce full rated secondary current, how much lower can you get away with and still have a reasonably accurate calculation with linear regression?

the pu Z is determined as follows (kind of follows what you posted)
shunt sec
apply V to prim
increase V until sec I = rated
if a 7200 prim and V is raised to 360 (= I rated)
then pu Z = 360/7200 = 0.05 or 5%

it can be shown this is the PU z and since Z is linear with V and I
I fault = 7200/360 x I rated or 1/0.05 x I rated

 

[electrofelon]

the pu Z is determined as follows (kind of follows what you posted)
shunt sec
apply V to prim
increase V until sec I = rated
if a 7200 prim and V is raised to 360 (= I rated)
then pu Z = 360/7200 = 0.05 or 5%

it can be shown this is the PU z and since Z is linear with V and I
I fault = 7200/360 x I rated or 1/0.05 x I rated

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

 

[Electric-Light]

the pu Z is determined as follows (kind of follows what you posted)
shunt sec
apply V to prim
increase V until sec I = rated

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.