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

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Ingenieur

Senior Member
Location
Earth
Not at 50Hz.


READ

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
 

Besoeker

Senior Member
Location
UK
no problem
you can see from the curves the X (or L inductance) of the transformer increases with size
I think that is incorrect


This from the spreadsheet I generated and used for many years.
The W is omega but I'm too indolent to do the search and replace.


Transformer Regulation

Transformer
Rating (kVA) 1500 2279 Max amps
Voltage (V) 380 219 Vph
Impedance (%) 0.05
Resistance (W) 1.16E-05
Reactance (W) 4.67E-05 0.148616764

Nominal Load
kW (kW) 300
pf pu 0.2
kVA (kVA) 1501
Current (A) 2280
phi (rad) 1.3694
Z (W) 0.09621861
R (W) 0.019243722
X (W) 0.094274599 4.898979486 0.0882

Total
R (W) 1.93E-02
X (W) 9.43E-02 4.90E+00
Z (W) 0.096266666
Actual current (A) 2279 0
pc FLC (%) 100%
Actual kVA (kVA) 1500

Output
V 219
Regulation 0.05%

You will see that the X is 0.094 ohms.
Change the rating to 2500kVA and the X drops to 0.056 ohms.
Larger transformer, lower value of X.

Whether you accept this or not is somewhat irrelevant. Over many years I have had to provide guaranteed (and tested) NNT efficiencies at the bid stage for the drive systems we manufactured and supplied. Many of these went into the water industry. One of the things about that industry is that, as a rule, the water is just there. There is obviously capital cost for installation of pumps, pipework, and drives. But energy consumption is a very significant part of running costs.

That's a little background on why the insistence for guaranteed NNT efficiencies. It puts you between a rock and a hard place - give too low the NNT, even by a fraction of a percent, and a more optimistic competitor quoting a better figure may win the contract. Too high, fail to meet it and get clobbered with swingeing financial penalties.

That makes you very careful about the accuracy of your calculations. Measure in twice, cut it once comes to mind. But, over the many years, we always met guaranteed figures. Since I was the person doing those calculations I must have been getting something right.
 

Ingenieur

Senior Member
Location
Earth
I think that is incorrect


This from the spreadsheet I generated and used for many years.
The W is omega but I'm too indolent to do the search and replace.


Transformer Regulation

Transformer
Rating (kVA) 1500 2279 Max amps
Voltage (V) 380 219 Vph
Impedance (%) 0.05
Resistance (W) 1.16E-05
Reactance (W) 4.67E-05 0.148616764

Nominal Load
kW (kW) 300
pf pu 0.2
kVA (kVA) 1501
Current (A) 2280
phi (rad) 1.3694
Z (W) 0.09621861
R (W) 0.019243722
X (W) 0.094274599 4.898979486 0.0882

Total
R (W) 1.93E-02
X (W) 9.43E-02 4.90E+00
Z (W) 0.096266666
Actual current (A) 2279 0
pc FLC (%) 100%
Actual kVA (kVA) 1500

Output
V 219
Regulation 0.05%

You will see that the X is 0.094 ohms.
Change the rating to 2500kVA and the X drops to 0.056 ohms.
Larger transformer, lower value of X.

Whether you accept this or not is somewhat irrelevant. Over many years I have had to provide guaranteed (and tested) NNT efficiencies at the bid stage for the drive systems we manufactured and supplied. Many of these went into the water industry. One of the things about that industry is that, as a rule, the water is just there. There is obviously capital cost for installation of pumps, pipework, and drives. But energy consumption is a very significant part of running costs.

That's a little background on why the insistence for guaranteed NNT efficiencies. It puts you between a rock and a hard place - give too low the NNT, even by a fraction of a percent, and a more optimistic competitor quoting a better figure may win the contract. Too high, fail to meet it and get clobbered with swingeing financial penalties.

That makes you very careful about the accuracy of your calculations. Measure in twice, cut it once comes to mind. But, over the many years, we always met guaranteed figures. Since I was the person doing those calculations I must have been getting something right.


look at the curves
MVA X/R
5 4
10 15
50 30
100 38
200 45
 
Last edited:

Besoeker

Senior Member
Location
UK
Nope.
Look at the calculations I provided.
Where are they at fault?
The consequences of them being wrong would, as I have already explained, been crippling for our business.

Here are some more........

POWER (kW) 2244
Nmin (rpm) 500
Nmax (rpm) 1489
Range (rpm) 989
OC Rotor Volts (Vac) 1628
Recovery V (Vdc) 1460
Recovery V (Vac) 1460
Supply Voltage (Vac) 11000
Tx Ratio (pu) 0
Rotor Nom Amps (A) 835
Calc'd Rotor Amps (A) 835
Number of fans 4
Increment (rpm) 124

Performance Data 1082 947 811 675 539 404 268 134 0
Speed (rpm) 500 624 747 871 995 1118 1242 1365 1489
Rotor Volts (V) 1081 946 811 676 541 405 270 135 0
Chopper Ratio (pu) 1.00 0.88 0.75 0.63 0.50 0.38 0.25 0.13 0.00
Rotor Amps pu (pu) 1.00 1.00 1.00 1.00 1.00 1.00 1.00 1.00 1.00
Rotor Amps (A) 94 146 210 286 372 471 580 702 835
Rotor I (per equiv cct) (A) 102 153 214 287 372 469 577 699 835
LV Recovery Amps (A) 102 133 161 180 186 176 144 87 0
HV Recovery Amps (A) 14 18 21 24 25 23 19 12 0

Supply Harmonics

5th Harmonic (A) 2.58 3.37 4.05 4.53 4.69 4.43 3.64 2.20 0.00
7th Harmonic (A) 1.76 2.30 2.77 3.10 3.21 3.03 2.49 1.51 0.00

11th Harmonic (A) 1.08 1.42 1.71 1.91 1.98 1.87 1.53 0.93 0.00
13th Harmonic (A) 0.88 1.15 1.39 1.55 1.61 1.52 1.25 0.75 0.00

17th Harmonic (A) 0.68 0.89 1.07 1.19 1.23 1.17 0.96 0.58 0.00
19th Harmonic (A) 0.61 0.80 0.96 1.07 1.11 1.05 0.86 0.52 0.00

Losses
Chopper Amps (Adc) 125 187 262 352 456 574 707 856 1022
Max Chpr Amps (Adc) 1022
Chpr Chk Res (mohm) 11
Inverter Amps (Adc) 125 163 197 220 228 215 177 107 0
Max Invr Amps (Adc) 228
Invr Chk Res (mohm) 48
Max Rec reqired (kVA) 471
Rect/Ch loss (W) 675 1009 1417 1900 2461 3099 3819 4624 5520
DCC1 loss (W) 165 369 728 1310 2196 3484 5291 7758 11054
Inverter loss (W) 675 883 1063 1188 1230 1162 955 578 0
DCC2 loss (W) 752 1285 1862 2326 2495 2227 1503 551 0
Tx Fe loss (W) 3294 3294 3294 3294 3294 3294 3294 3294 3294
Tx Cu loss (W) 1418 2423 3511 4386 4705 4199 2834 1039 0
Fans (W) 500 500 500 500 500 500 500 500 500

Total Losses (kW) 7.5 9.8 12.4 14.9 16.9 18.0 18.2 18.3 20.4
Output (kW) 85 165 284 449 669 950 1301 1730 2244
Shaft Power (kW) 85 165 284 449 669 950 1301 1730 2244
Speed (rpm) 500 624 747 871 995 1118 1242 1365 1489
Torque (pu) 0.11 0.18 0.25 0.34 0.45 0.56 0.70 0.84 1.00

Efficiencies
ISK efficiency (%) 90.8% 93.6% 95.2% 96.3% 97.2% 97.9% 98.4% 98.8% 99.0%
Motor efficiency (%) 91.3% 93.3% 94.5% 95.3% 95.7% 96.0% 96.1% 96.2% 96.1%
Mechanical output (kW) 85.0 164.9 283.6 449.0 668.6 950.2 1301.5 1730.2 2244.0
Rotor Power (kW) 190.1 248.5 299.8 336.1 349.6 332.3 276.4 174.7 20.4
Stator Power (kW) 301.4 443.2 617.4 824.1 1063.5 1335.7 1641.2 1980.9 2356.6
ISK Output (kW) 182.6 238.7 287.4 321.2 332.7 314.3 258.2 156.3 0.0
ISK Losses (kW) 7.5 9.8 12.4 14.9 16.9 18.0 18.2 18.3 20.4
Motor Losses (kW) 26.3 29.8 34.0 39.1 45.4 53.3 63.3 76.0 92.2
Total Losses (kW) 33.8 39.6 46.4 54.0 62.2 71.3 81.5 94.4 112.6
Efficiency (%) 71.6% 80.6% 86.0% 89.3% 91.5% 93.0% 94.1% 94.8% 95.2%


A fan in a cement works.....
Real application, actual figures, tested efficiencies...........and proven eneergy savings.
Not quite a cube law application - inlet air temperature makes a big difference to torque required with the cold air after a process outage.

I'm sure you will have heard of the major company where they are installed and in operation.

Please understand that this is not a pisssing contest. I'm simply showing some data that I have had to provide in real life applications so that you might understand a little of what I was involved in and the basis for my comments. And improve..........

And I think my big dog still likes me even though I don't feed him.
He loves children - one for breakfast, one for dinner.

IMG_1917_zpssy8ys0rw.jpg


Handsome isn't he?
 

Ingenieur

Senior Member
Location
Earth
Xbase = V^2 / S
X actual in ohms = Xpu x V^2 / S
assume V and freq constant and case 1 and 2, ratio the equation
L1/L2 = X1pu / X2pu x S2 / S1
rearrange
L2 = L1 x X2pu / X1pu x S1 / S2
assume the Xxpu are equal
and S2 = 2 x S1 ie xfmr 2 S twice as large as xfmr 1 S
so L2 = L1/2 basically inversely proportional to S if Xpu constant

but in general the Xpu ratio increases faster than the S1/S2 ratio so L increases with S
but depending on xfrm type and quality it could remain constant or decrease

but what we really care about is X/R and it always increase with S for a given type/quality of xfmr
 
Last edited:

Besoeker

Senior Member
Location
UK
but what we really care about is X/R and it always increase with S for a given type/quality of xfmr

That is usually the case but let me remind you of your claim:

"you can see from the curves the X (or L inductance) of the transformer increases with size"

That's about X, not about X/R.
Might I kindly suggest that you be consistent about your claims? Or at at least try to be.
 

Ingenieur

Senior Member
Location
Earth
That is usually the case but let me remind you of your claim:

"you can see from the curves the X (or L inductance) of the transformer increases with size"

That's about X, not about X/R.
Might I kindly suggest that you be consistent about your claims? Or at at least try to be.

I am
see my post above and copied below, you do alot of selective editing
as consistent as your 'I think' lol
you can always find an exception to the rule
can I kindly ask you to lose the boner for me and stop stalking?
you've been proven wrong everytime
isn't it getting old?

Xbase = V^2 / S
X actual in ohms = Xpu x V^2 / S
assume V and freq constant and case 1 and 2, ratio the equation
L1/L2 = X1pu / X2pu x S2 / S1
rearrange
L2 = L1 x X2pu / X1pu x S1 / S2
assume the Xxpu are equal
and S2 = 2 x S1 ie xfmr 2 S twice as large as xfmr 1 S
so L2 = L1/2 basically inversely proportional to S if Xpu constant

but in general the Xpu ratio increases faster than the S1/S2 ratio so L increases with S
but depending on xfrm type and quality it could remain constant or decrease

but what we really care about is X/R and it always increase with S for a given type/quality of xfmr
 
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