PU transformer Z calculation

[Ingenieur]

I was sent data by a transformer manufacturer
since this is used in an XP piece of equipment we do fault calcs to determine let thru energy based on fusing

transformer
12470/240/1
15 kva
they state PU Z 5.12%
prim test V 638.8 V
prim test I 1.2 A
so 638.8/12470 = 5.12%, X/R = 4.11/3.06
doesn't sound right when compared to standard tables, they state 1.8% for this xfrm

they also give the actual measured R for the prim and sec:
prim 128.641 Ohm
sec 0.02572 Ohm

when I calculate the PU act/base for the prim or sec I get 1.9% (close to the standard tables)
eg, Z prim = (128.641 + (12470/240)^2 x 0.02572) / (12470^2/15000) x 100 = 1.91%

what do you guys come up with based on the measured R?
big difference in I fault 5.12 vs 1.9
I have an email into the mfg

 

[steve66]

If a manufacturer told me the %Z was 5.12, I would go with it. (Actually I would go with 90% of it to account for a 10% tolerance.)

I you are using the tables I'm thinking of, they are just a rough estimate. The one I have says anywhere from 1.2% to 6%, and 5 falls in that range.

I'm not understanding your equation to calculation the %Z with the resistances. Wouldn't the winding inductances also figure into that equation?

 

[GoldDigger]

If a manufacturer told me the %Z was 5.12, I would go with it. (Actually I would go with 90% of it to account for a 10% tolerance.)

I you are using the tables I'm thinking of, they are just a rough estimate. The one I have says anywhere from 1.2% to 6%, and 5 falls in that range.

I'm not understanding your equation to calculation the %Z with the resistances. Wouldn't the winding inductances also figure into that equation?

The winding inductances would indeed figure in, as long as you properly account for the mutual inductance and the corresponding reduction of the effective inductances because of the load on the secondary.

 

[Ingenieur]

If a manufacturer told me the %Z was 5.12, I would go with it. (Actually I would go with 90% of it to account for a 10% tolerance.)

I you are using the tables I'm thinking of, they are just a rough estimate. The one I have says anywhere from 1.2% to 6%, and 5 falls in that range.

I'm not understanding your equation to calculation the %Z with the resistances. Wouldn't the winding inductances also figure into that equation?

He actually gave me the measured Z of each: prim and sec (it is Z, measured with 60 Hz)
my tables are actually very detailed
they give a number for 12470/240/1 15 kva 1.8%, no range

the total Z looking into a shorted sec from the prim is
Act = prim Z + a^2 sec x sec Z
Z base = V^2 / va
pu Z = act / base x 100

he said the 5% seems too high
he thinks he gave me the incorrect v and i data for a larger 3 ph (which the pu 5.12 was derived)
and the proper actual measure Z data

 

[Ingenieur]

here's the chart

 

[Bugman1400]

here's the chart

If this is a std pole pig, I am having a tough time accepting an imp of 5%. That high of an imp is typically reserved for the big boy xfmrs.

 

[Ingenieur]

If this is a std pole pig, I am having a tough time accepting an imp of 5%. That high of an imp is typically reserved for the big boy xfmrs.

my thoughts exactly
but not pole mntd, in a piece of equipment, but same construction
we see 5-10 MVA 3 ph in that range

 

[Julius Right]

I don't know how the ratio X/R is stated but 4.11/3.06=1.343 it seems to me too small. If X/R=4.77/3.04=1.57 then 5.12% it is correct.
R=R1+R2*(12.47/0.249)^2=285.93 ohm
X=1.57*285.93=448.65 ohm
Z=SQRT(285.93^2+448.65^2)=532.02 ohm
cosfi=R/Z=285.93/532.02=0.5374 [Secondary is short-circuited when tested]. sinfi=sqrt(1-cosfi^2)=0.8433.
Voltage drop= I*(R*cosfi+X*sinfi)=1.2*(285.93*0.5374+448.65.3*0.8433)= 638.4064 V.

 

[Julius Right]

I forgot to mention: in my opinion the resistance measurement was conducting at 20oC but the insulation class is B [130 oC]. So the resistance temperature factor is kT= (234.5+130)/ (234.5+20) =1.4322 for copper conductor and (228+130)/ (228+20) =1.4435 for AL conductor.
Then total R=1.4435*(128.641+0.02572*(12470/240) ^2) =285.93 ohm.:ashamed1:

 

[Julius Right]

At a second glance, even 4.11% reactance and 3.06% resistance it could be o.k.
R=(12470^2/15000)*3.06/100=317.22 ohm
In this case the measuring and rated temperature has to be different [let's say 180-class H-and 24.3 oC when measured].
X=(12470^2/15000)*4.11/100=426.07 ohm
Z=sqrt(R^2+X^2)=531.19 ohm
cosfi=0.5972 ; sinfi=0.802
DV=1.2*(317.22*0.5972+426.07*0.802)=637.38 V=5.11%

 

[NewtonLaw]

%Z of a Transformer

%Z of a Transformer

I was sent data by a transformer manufacturer
since this is used in an XP piece of equipment we do fault calcs to determine let thru energy based on fusing

transformer
12470/240/1
15 kva
they state PU Z 5.12%
prim test V 638.8 V
prim test I 1.2 A
so 638.8/12470 = 5.12%, X/R = 4.11/3.06
doesn't sound right when compared to standard tables, they state 1.8% for this xfrm

they also give the actual measured R for the prim and sec:
prim 128.641 Ohm
sec 0.02572 Ohm

when I calculate the PU act/base for the prim or sec I get 1.9% (close to the standard tables)
eg, Z prim = (128.641 + (12470/240)^2 x 0.02572) / (12470^2/15000) x 100 = 1.91%

what do you guys come up with based on the measured R?
big difference in I fault 5.12 vs 1.9
I have an email into the mfg

Based on the test data you were given:

Z-Actual = 638.8 volts / 1.2 amps = 532.33 ohms on the primary side.

Total R referred to the primary side = 128.64 +(0.02572*(12470/240)^2) = 128.64+69.44 = 198.08

Xl = squart root or (Z^2-R^2) = square root of (532.33^2 - (198.08^2) = 494.11 ohms

X/R ratio = 494.11/198.08 = 2.49 (not 1.34 that you list above. Were these numbers given to you also?)

%Z of a transformer is based on the kVA rating used as the Base kVA and the voltage as the Base Voltage from these two numbers you may calculate the Base Impedance.

Z-base = (V-base)^2 / kVA-base = 12470^2/15000 = 10,366.73

Z% = (Z-actual/Z-base)*100 = (532.33/10366.73)*100 = 5.05% or approximately 5.1%. The error is in rounding numbers I think and is just another way to %Z that is exactly equal to what was given above for (638.8V/12470V)*100 = 5.21%

Seems to me the data you were give is correct. Hope this helps.

 

[Ingenieur]

Based on the test data you were given:

Z-Actual = 638.8 volts / 1.2 amps = 532.33 ohms on the primary side.

Total R referred to the primary side = 128.64 +(0.02572*(12470/240)^2) = 128.64+69.44 = 198.08

Xl = squart root or (Z^2-R^2) = square root of (532.33^2 - (198.08^2) = 494.11 ohms

X/R ratio = 494.11/198.08 = 2.49 (not 1.34 that you list above. Were these numbers given to you also?)

%Z of a transformer is based on the kVA rating used as the Base kVA and the voltage as the Base Voltage from these two numbers you may calculate the Base Impedance.

Z-base = (V-base)^2 / kVA-base = 12470^2/15000 = 10,366.73

Z% = (Z-actual/Z-base)*100 = (532.33/10366.73)*100 = 5.05% or approximately 5.1%. The error is in rounding numbers I think and is just another way to %Z that is exactly equal to what was given above for (638.8V/12470V)*100 = 5.21%

Seems to me the data you were give is correct. Hope this helps.

that was the x/r given
seems too low
they said the data was incorrect, was for a 3 ph
they have not gotten the new data to me

thanks
I will post it when I get it

 

[Ingenieur]

Using their x/r 1.3431
pu 3.2%
still seems high and x/r too low

 

[NewtonLaw]

Question?

Question?

Using their x/r 1.3431
pu 3.2%
still seems high and x/r too low

Would you show me how you got the "pu 3.2%" value? Is this a calculated value for the %Z of the transformer?

 

[Ingenieur]

Would you show me how you got the "pu 3.2%" value? Is this a calculated value for the %Z of the transformer?

equivalent R = 128.641 + (12470/240)^2 x 0.02572 = 198.076 (ref to primary)
X = 4.11/3.06 x 198.076 = 266.044
Z = sqrt(198.076^2 + 266.044^2) = 331.683

Z base = 12470^2/15000 = 10,366.73
pu = 331.683 / 10,366.73 = 0.03199 ~ 3.2%

the x/r 1.343 seems too low and pu Z of 3.2%, too high, for a 1 ph, 15 kva xfmr

 

[Ingenieur]

if you ignore their R measurements
Prim 128.641
Sec 0.02572

X/R = 4.11/3.06 assumed in pu%
Base 10366.73
R 317
X 426
Z = sqrt(317^2 + 426^2) = 531

531/10366.73 x 100 = 5.12%

 

[NewtonLaw]

There given x/r ratio is questionable

There given x/r ratio is questionable

if you ignore their R measurements
Prim 128.641
Sec 0.02572

X/R = 4.11/3.06 assumed in pu%
Base 10366.73
R 317
X 426
Z = sqrt(317^2 + 426^2) = 531

531/10366.73 x 100 = 5.12%

Since they give you the short circuit test via the primary where the primary full load is given 1.2 with a voltage of 638.8 volts you have the given magnitude then for Z actual,

so |z| = 638.8 / 1.203 = 531.1 Ohms

The they give you the measured resistance of the primary and the secondary, 128.641 ohms and 0.02572 ohms. Reflect the secondary to the primary by the square of the turns ratio and total primary R then is 198.08 ohms. These are measured values.

From these the reactance value = square root of (531.1^2 - 198.08^2) = 492.78 ohms and then the x/r = 492.78/198.08 = 2.49 from the test data supplied.


Did the manufacturer ever give you updated or corrected values?