Transformer short circuit impedance in relation to 50Hz vs 60Hz

[mbrooke]

How does transformer impedance vary from 50Hz to 60Hz and visa versa? If you take say a 480 volt delta 208Y 100Kva 10%Z 60Hz transformer and place it on a 400 volt 50Hz supply by how much will the impedance drop?


My understanding is that the impedance will go down substantially for two reasons: 1. The current will stay the same but the reduced voltage required to prevent magnetic saturation will lower the kva, thus when calculated the short circuit per kva will be higher. 2. X=2x3.14xfxL (if I am applying this equation correctly) dictates that reactive impedance goes down with frequency, so the equivalent magnetic circuit when drawn out of a transformer would theoretically present a lower impedance and in turn more fault current on the secondary. Would this correct? And how would one go about calculating it correctly?


Second, if I order a 100Kva 50Hz transformer, would it automatically have a lower impedance because their is 20% more iron right off the bat?

 

[Ingenieur]

The PU Z is based on a specific voltage
let's use a 1 ph 100 kva to simplify
480:120
fla 833.33

Put a v source on the primary
short the sec
ramp voltage up until sec i = fla
assume this happens at 48 v
PU Z = 48/480 ~ 10%

at 400 v rating it would be 48/400 = 12%
but transformer kva must be derated assuming sec i is the limiting factor
At 400 v S = 100 x 833.33 = 83.33 kva


the base Z is derived from S = v^2 / Z or Z = v^2 / S
480 2.3 Ohm
400 1.92 Ohm

actual Z
480 2.3 x 0.1 = 0.23 Ohm
400 1.92 x 0.12 = 0.23 Ohm



if tested at 50 vs 60 Hz it may be a bit lower since Z = R + j X
and X = j 2 Pi freq L
so if X goes down by 50/60 Z will go down
if you know the X/R ratio you can estimate by how much

 

[mbrooke]

This is done with a variac being turned up and a short circuited secondary until rated current is reached? And yes rated current on the secondary is the limiting factor here.

 

[Ingenieur]

This is done with a variac being turned up and a short circuited secondary until rated current is reached? And yes rated current on the secondary is the limiting factor here.

Yes, a big one depending on kva lol

typical X/R for a xfmr like this is 0.7 range
so Z might drop 5% or 0.95 x 0.23 = 0.219, neglible

another take
v prim 48
v sec 12
i sec 833.33
Z sec = v sec / i sec = 12 / 833.33 = 0.0144 Ohm
reflected to prim 4^2 x 0.0144 = 0.23 Ohm

 

[gar]

160814-2303 EDT

mbrooke:

Ingenieur has given you information to answer your question.

To word it differently I submit the following"

1. Assumptions ---
--- A specific transformer is to be evaluated under two different excitation conditions.
--- The transformer equivalent circuit is invariant for these two excitation conditions.
--- The transformer simply looks like a series circuit of a resistor and an inductor.
--- The resistance remains constant vs frequency.
--- The inductive reactance varies in proportion to freuency.
--- Short circuit conditions means the load is a series circuit of resistance and inductance = to the transformer impedance.

2. The transformer impedance, if only inductive, would be proportional to frequency. But with R and Xl both approximately the same at 60 Hz, then Z won't drop as rapidly as frequency does. At very low frequencies Z becomes approximately R.

3. Since your test drops the source voltage in proportion to frequency, and since Z does not drop this rapidly this means the short circuit current will be slightly less at 50 Hz compared to 60 Hz. If you only had inductive reactance, then short circuit current would have remained constant.

.

 

[Julius Right]

I agree with Inginieur.
If S=100 kVA and 480 V the rated current will be 100/0.48/sqrt(3)=120.28 A
The short-circuit impedance of 100 kVA 480V 10% at 60 Hz is:
0.48^2/0.1*0.1=0.2304 ohm [0.48 kV,0.1 MVA,10%]
If x/r=2 then X=0.2304/SQRT(1+1/4)=0.20608 ohm.
and R=0.20608/2=0.10304 ohm.
In my opinion, the rated current stays the same at 400V 50Hz.
then at 400 V 50Hz and 120.28 A S50hz=sqrt(3)*400*120.28/1000=83.33 kVA.
If we consider the leakage inductance[L] as constant [even the main magnetic flow
will be larger ,a bit, and then this inductance could be less but still negligible].
then the reactance will be now 0.20608*50/60=0.17173 ohm.
The resistance will decrease, a bit, due to reduced skin and proximity effect. However,
it is still negligible since the conductor diameter is small.
So the impedance will be Z50Hz=sqrt(0.171733^2+0.103038^2)=0.20027 ohm.
If Zn=0.4^2/0.08333=1.92 then vk%=0.20027/1.92*100=10.4%

 

[gar]

160815-2000 EDT

Using Ingenieur's X/R = 0.7 the impedance drops about 5.2% from 60 to 50 Hz as he indicated. By mbrooke's definition of the problem the source voltage drops by about 16.7%.

Calculation of the change in short circuit current results in about a 12% drop from 60 to 50 Hz.

.

 

[mbrooke]

Ok, let me digest all this. :thumbsup: But, Id still like to spice things up a bit. I tried to get the X/R ratios of a few small 480/208 units but was unable too, however I was able to get the X/R ratios of a few power transformers. One being 115kv/23kv 28/37.3/46.7MVA delta wye transformer on a 100MVA base: R=1.829 X=40.215.


Using this given X/R, how would the same transformer behave on 50Hz?

 

[gar]

160816-1016 EDT

mbrooke:

With the X/R ratio you have presented there won't be much change in the short circuit current because the impedance is largely inductive.

Let:
V60 = source voltage at 60 Hz
V50 = source voltage at 50 Hz
ISS60 = short circuit current at 60 hz
ISS50 = short circuit current at 50 Hz
ZINT60 = series internal impedance of transformer at 60 Hz
ZINT50 = series internal impedance of transformer at 50 Hz
R60 = resistance at 60 Hz
R50 = rewistance at 50 Hz
R60 = R50
X60 = inductive reactance at 60 Hz
X50 = inductive reactance at 50 Hz

X magnitude = R * 22.04 from mbrooke

ISS60 = V60/ZINT60
ISS50 = V50/ZINT50

V50 = V60*5/6 from mbrooke specification

X50 = X60*5/6 from equation X for an inductor

ISS50 = ( V60 * 5/6 ) / ZINT50
ISS60 = V60 / ZINT60

ISS50/ISS60 = ( 5/6 ) * ZINT60 / ZINT50

I have to leave for a short time. I will finish when I return. I have not proofread above.

.

 

[gar]

160816-1119 EDT

Continuing:

ZINT60 = ( R^2 + (22.04*R)^2 ) )^0.5 = R * ( 1 + 486 ) = 487 * R
ZINT50 = ( R^2 + (22.04*R*5/6)^2 )^0.5 = R * ( 1 + 405 ) = 406 * R

Thus
ZINT50 / ZINT60 = 0.834
for comparison
5/6 = 0.834

So
ISS50/ISS60 = ( 5/6 ) * ZINT60 / ZINT50 = 0.834 / 0.834 = 1

.

 

[Ingenieur]

Ok, let me digest all this. :thumbsup: But, Id still like to spice things up a bit. I tried to get the X/R ratios of a few small 480/208 units but was unable too, however I was able to get the X/R ratios of a few power transformers. One being 115kv/23kv 28/37.3/46.7MVA delta wye transformer on a 100MVA base: R=1.829 X=40.215.


Using this given X/R, how would the same transformer behave on 50Hz?

since X/R = 22

the Z change would be = sqrt(1^2 + (50/60 22)^2) / sqrt(1^2 + 22^2) = 0.835 or ~17% less
as X/R gets larger, the X becomes more significant, ie approaches 50/60 = 0.833 (vs 0.835 above)
so once X/R > 10, you can assume Z is almost proportional with freq

that X/R is way too high
for your range <2 and most likely less than 1
some X/R tables http://arcadvisor.com/files/GET3550Fapp41-43.pdf

for example table 14
100 kva 1 phase
X 3.55
R 2.12
X/R 0.6

112.5 kva 3 ph
X/R 0.63

using 0.6 the Z change would be = sqrt(1^2 + (50/60 0.6)^2) / sqrt(1^2 + 0.6^2) = 0.96 or ~only 4% less

 

[Ingenieur]

since X/R = 22

the Z change would be = sqrt(1^2 + (50/60 22)^2) / sqrt(1^2 + 22^2) = 0.835 or ~17% less
as X/R gets larger, the X becomes more significant, ie approaches 50/60 = 0.833 (vs 0.835 above)
so once X/R > 10, you can assume Z is almost proportional with freq

that X/R is way too high
for your range <2 and most likely less than 1
some X/R tables http://arcadvisor.com/files/GET3550Fapp41-43.pdf

for example table 14
100 kva 1 phase
X 3.55
R 2.12
X/R 0.6

112.5 kva 3 ph
X/R 0.63

using 0.6 the Z change would be = sqrt(1^2 + (50/60 0.6)^2) / sqrt(1^2 + 0.6^2) = 0.96 or ~only 4% less

oops
X/R
100 kva 1 phase
X 3.55
R 2.12
X/R 1.7

112.5 kva 3 ph
X/R 1.6

using 1.6 the Z change would be = sqrt(1^2 + (50/60 1.6)^2) / sqrt(1^2 + 1.6^2) = 0.88 or ~only 12% less

 

[mbrooke]

since X/R = 22

the Z change would be = sqrt(1^2 + (50/60 22)^2) / sqrt(1^2 + 22^2) = 0.835 or ~17% less
as X/R gets larger, the X becomes more significant, ie approaches 50/60 = 0.833 (vs 0.835 above)
so once X/R > 10, you can assume Z is almost proportional with freq

that X/R is way too high
for your range <2 and most likely less than 1
some X/R tables http://arcadvisor.com/files/GET3550Fapp41-43.pdf

for example table 14
100 kva 1 phase
X 3.55
R 2.12
X/R 0.6

112.5 kva 3 ph
X/R 0.63

using 0.6 the Z change would be = sqrt(1^2 + (50/60 0.6)^2) / sqrt(1^2 + 0.6^2) = 0.96 or ~only 4% less

Why do you say my X/R is to high? :blink: Here is another one to a (newer) 40/50/60MVA unit:


R=1.155 X= 39.42


Delta wye, 115kv to 23kv.

 

[gar]

160817-0927 EDT

Some curves and information:
http://arcadvisor.com/files/GET3550Fapp41-43.pdf
http://www.powerstudies.com/sites/www.powerstudies.com/files/ImportanceofX-over-RRatios.pdf
https://www.powellind.com/sites/downloads/ProductAssets/01.4TB.047 X-R Ratio.pdf
http://grouper.ieee.org/groups/tran...performance/TF_C57_12_00/XoverRdiscussion.pdf

.

 

[Ingenieur]

Why do you say my X/R is to high? :blink: Here is another one to a (newer) 40/50/60MVA unit:


R=1.155 X= 39.42


Delta wye, 115kv to 23kv.

that is for a large HV power transformer 50 MVA KV range

your original question referenced a 0.100 MVA 0.480 kV to 0.208 kV range xfmr
BIG difference in X/R

the link I posted has info for transformers from >100 MVA down to <50 kva
http://arcadvisor.com/files/GET3550Fapp41-43.pdf
table 8B
50 MVA X/R 30
0.1 MVA ~2

table 14 dry type qht
3 phase 75 kva
X 1.91
R 2.27
X/R 0.84

 

[mbrooke]

that is for a large HV power transformer 50 MVA KV range

your original question referenced a 0.100 MVA 0.480 kV to 0.208 kV range xfmr
BIG difference in X/R

the link I posted has info for transformers from >100 MVA down to <50 kva
http://arcadvisor.com/files/GET3550Fapp41-43.pdf
table 8B
50 MVA X/R 30
0.1 MVA ~2

table 14 dry type qht
3 phase 75 kva
X 1.91
R 2.27
X/R 0.84

transformers
I know, but I could not find the X/R ratios to small dry types. I was able to get the ratios to some large power transformers which is a score in my book Its easier for me to understand on something that size. But my apologies about it.

 

[Ingenieur]

transformers
I know, but I could not find the X/R ratios to small dry types. I was able to get the ratios to some large power transformers which is a score in my book Its easier for me to understand on something that size. But my apologies about it.

no problem
you can see from the curves the X (or L inductance) of the transformer increases with size

on smaller xfrms you can essentially ignore it
on larger it approaches the 50/60 ratio
example for 60 to 50 Hz

 

[mbrooke]

no problem
you can see from the curves the X (or L inductance) of the transformer increases with size

on smaller xfrms you can essentially ignore it
on larger it approaches the 50/60 ratio
example for 60 to 50 Hz

What is the 50/60 ratio?


And this might be obvious, but do paralleled units exhibit more R, less X when analyzed in power systems then one larger unit? For example, a 115-34.5kv substation, one with a single 100MVA unit and another with 4 paralleled 25MVA units? I've seen in some parts of the world where instead of using two 600MVA 345/115 auto transformers like we do here in the USA, they will use say 6 200MVA units because it does something with transient stability/consequences of a weaker system (I think).

 

[mbrooke]

And on the side... a 100MVA 115/34.5kv Ygr autotransfomer would have a lower X/R ratio than a 100MVA 115/34.5kv Delta wye unit?

 

[Julius Right]

According to ANSI C37.010 the X/R ratio has to be around 2. See:
http://arcadvisor.com/files/GET3550Fapp41-43.pdf