Transformer Impedance

My question is can this be considered the overall impedance of the transformer. I am getting two values using two different methods. one is 0.0137 ohms and the other is 0.0073 ohms.

Answer to your question is NO.

How did you get the other values?

Thanks mate

I will take you through the cals

----> sqr(V)/S = (sqr(433)/1200000) * 5.09/100 = 0.00795 ohms

----> 5.09/100 * 433 = 22V. FLA is 1600Amps

Therefore 22V/1600A = 0.01375

The second calculation is for the short circuit impedance and the first one is using the base values. Which one is correct??

Thanks again..

hi guest,

can you explain the same method for 500KVA, 11/415V, 5%?

Use this example

Given:

Connection: D-Y (4160V/480V) Size = 300KVA Ipri = 79.92A Isec = 692.62A

PF = Cosq = 0.25 Sinq = Sin(Cos-1q) = 0.968

R = 1.7% = 0.017pu @ 300KVA base X = 4.96% = .0496pu @ 300KVA base

Using primary side:

Zact = Zpu x Zbase I = 79.92A Zbase= V2/Sbase = (4160)2 / 300KVA = 57.685W

Ract = (.017)( 57.685) = 0.98065W

Xact = (0.0496)(57.685) = 2.86119W

VD = IRCosq + IXSinq ÿFrom IEEE Red Book

VD = I[RCosq + XSinq]

VD = 79.92[(0.98065)( 0.25) + (2.86119)( 0.968)] = 240.85V

VD3q= VD x ÿ3 = 240.85V x ÿ3 = 417.16V

VD3q % = (417.16V / 4160V) x 100 = 10.03%

Connection: D-Y (4160V/480V) Size = 300KVA Ipri = 79.92A Isec = 692.62A

Is this correct or am I wrong ???

P = I x U x sqrt (3)

300 kVA = I x 4160 x 1.73 so Ipri becomes 41.7 Ampères ....

300 kVA = I x 480 x 1.73 so Isec becomes 361 Ampères ....

You are right.....even I doubted my calculating abilities...remaining part of reply by Wareagle also I couldn't get.

Nor did I understand the remaining part. .......

Perhaps Wareagle can explain it one more time ....

Or the one who rated this as a good answer ...

you can consider it as overall impedance of the transformer .

before starting calculation , you have to assume base MVA, V, I, Impedance.

in your case

transformer impedance = 0.0509 * (base MVA / 1.2 MVA )

I can't quite make out what you are saying.

Could it be that the lower impedance is the series loss impedance (effectively the series resistance) and the higher one is the impedance at short circuit?

If so, the reason they are different is that the true short-circuit impedance must include the transformer's leakage inductance, whereas the loss impedance relates to the losses when the transformer is loaded with a largely resistive load - and the main effect of the leakage inductance is to change the phase of the output rather than the Voltage.

I looked up the standard definition of winding impedance: it is the effective series impedance (= leakage-inductance plus transformed winding resistances) of the transformer.
% winding impedance is the winding impedance as a percentage of the impedance needed to generate nominal loading.

There will be two values for this - one viewed from the supply side, the other viewed from the load. Which you want depends on the application.

Having said this, the numbers in the header-posting are both wrong. I wouldn't care to disinter how you got there, but the correct formula is:
Z_winding = V_nom²/VA_rating x (%impedance/100)

So the winding impedance viewed from the 690-Volt side is specified to be 0.02-Ohms, and the impedance viewed from the 433-Volt side is specified to be 0.008 Ohms.

(BTW, the full-load currents should be 1739 and 2771 Amp rms respectively, and it all comes out in the wash)

Sample Voltage Drop Calculation Across Transformer

XF2-001:

Given:

Connection: D-Y (4160V/480V) Size = 300KVA Ipri = 79.92A Isec = 692.62A

PF = Cosq = 0.25 Sinq = Sin(Cos-1q) = 0.968

R = 1.7% = 0.017pu @ 300KVA base X = 4.96% = .0496pu @ 300KVA base

Using primary side:

Zact = Zpu x Zbase I = 79.92A Zbase= V2/Sbase = (4160)2 / 300KVA = 57.685W

Ract = (.017)( 57.685) = 0.98065W

Xact = (0.0496)(57.685) = 2.86119W

VD = IRCosq + IXSinq From IEEE Red Book

VD = I[RCosq + XSinq]

VD = 79.92[(0.98065)( 0.25) + (2.86119)( 0.968)] = 240.85V

VD3q= VD x √3 = 240.85V x √3 = 417.16V

VD3q % = (417.16V / 4160V) x 100 = 10.03%

Using secondary side:

Zact = Zpu x Zbase I = 692.62A Zbase= V2/Sbase = (480)2 / 300KVA = 0.768W

Ract = (.017)( 0.768) = 0.013056W

Xact = (0.0496)( 0.768) = 0.0380928W

VD = I[RCosq + XSinq] = 692.62[(0.013056)( 0.25) + (0.0380928)( 0.968)] = 27.80 V

VD3q= VD x ÿ3 = 27.80 x ÿ3 = 48.15 V

VD3q % = (48.15V / 480V) X 100 = 10.03%

This is a correction to the first post

Sample Voltage Drop Calculation Across Transformer

XF2-001:

Given:

Connection: D-Y (4160V/480V) Size = 300KVA Ipri = 79.92A Isec = 692.62A

PF = Cosq = 0.25 Sinq = Sin(Cos-1q) = 0.968

R = 1.7% = 0.017pu @ 300KVA base X = 4.96% = .0496pu @ 300KVA base

Using primary side:

Zact = Zpu x Zbase I = 79.92A Zbase= V2/Sbase = (4160)2 / 300KVA = 57.685W

Ract = (.017)( 57.685) = 0.98065W

Xact = (0.0496)(57.685) = 2.86119W

VD = IRCosq + IXSinq

VD = I[RCosq + XSinq]

VD = 79.92[(0.98065)( 0.25) + (2.86119)( 0.968)] = 240.85V

VD3q= VD x ÿ3 = 240.85V x ÿ3 = 417.16V

VD3q % = (417.16V / 4160V) x 100 = 10.03%

Using secondary side:

Zact = Zpu x Zbase I = 692.62A Zbase= V2/Sbase = (480)2 / 300KVA = 0.768W

Ract = (.017)( 0.768) = 0.013056W

Xact = (0.0496)( 0.768) = 0.0380928W

VD = I[RCosq + XSinq] = 692.62[(0.013056)( 0.25) + (0.0380928)( 0.968)] = 27.80 V

VD3q= VD x √3 = 27.80 x √3 = 48.15 V

VD3q % = (48.15V / 480V) X 100 = 10.03%