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Transformer Voltage Drop 3

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joozu6

Electrical
May 29, 2003
37
Can anyone give me an idea on how to calculate voltage drop through a transformer due to load?

Our system is 12470/7200 primary. For example I want to calculate the voltage drop through a 25 kVA 120/240 padmount transformer with a 4kVA load. The %Z on the nameplate is 1.9 and the R/X ratio is 0.9. I assumed a power factor of 90%. My calculations don't seem to come out right.

Thanks.
 
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For a 25kVA xfmr with a 1.9% impedance, the voltage will drop 1.9% when full load is pulled through the xfmr.
 
Yes, but reactance varies with the load, right?
 
Voltage drop through any impedance (line or transformer) can be approximated by VD = IRcosø + IXsinø. If you make I, R, and X in per unit of the transformer rating, the answer will be in pu.

Ipu = 4/25 = 0.16
X = Z·sin(atan(X/R)) = 0.019·sin(48°) = 0.0141
R = Z·cos(atan(X/R)) = 0.0127
cosø = pf = 0.9
sinø = sin(acos(0.9)) = 0.436

VD = 0.16·0.0141·0.9 + 0.16·0.0127·0.436 = 0.00292 or 0.292%

Notice that this is a little less than 4/25 times %Z. If the X/R ratio were larger, the difference would be greater.
 
That helps, jghrist. Thanks.

The only thing that perplexes me is Ipu being a ratio of the load kVA per rated kVA. Wouldn't Ipu be Vpu/Zpu?
 
I think I got it now. Current changes and voltage is constant so load kva/rated kva give Ipu.

Thanks.
 
I think the values of R and X are inverted in the final calculation, aren't?
The final value would be = 0.281%
 
Comment: VD = IRcosø + IXsinø is approximately valid when acos(.9)=25.8deg ~ atan(X/Rac)=48deg ???? These do not seem to be approximately equal.
VD=.16 x .0127 x .9 + .16 x .0141 x .436 = .281% as indicated in the previous posting
 
Suggestion: IEEE Std 141-1993 Red Book indicates on page 98:
actual voltage drop=
Es+IRcosF+IXsinF-sqrt[Es**2-(IXcosF-IRsinF)**2]=
1+.00281-sqrt[1-.16(.0127x.9-.0141x.436)]=.00323 or .323%
about 15% higher voltage drop
 
piterpol is correct about the inversion of R & X. :~/

The load phase angle ø and the transformer impedance angle atan(X/R) do not have to be approximately equal for the VD equation to be accurate. The equation is fairly accurate unless the angle between the sending end and receiving end voltages is large. This is the case if the load phase angle is nearly equal to the impedance angle, but it is also the case with small voltage drops at any angle. In the extreme case, with pf = 1 (0°) and X/R very large (90°), the equation will give 0% VD regardless of impedance or load. I·Z would have to be >0.1 pu for the actual VD to be >0.5% in this case.
 
Suggestion: Reference:
1. Wire and Cable, Power Plant Electrical Reference Series, Volume 4, Electric Power Research Institute, page 4-33 reads:
For a situation in which the load power factor angle (theta) is close to the cable impedance angle atan(X/Rac) the "exact Equation 4-12 for voltage drop" can be approximated by Equation 4-13:
Vs-Vl=Rac x I x cos(theta) + X x I x sin(theta)

 
Thanks, guys. That helped a lot.

I have two more questions concerning %impedance:

1. What's the difference between %IZ and %Z?
2. For a 120/240V transformer is the %Z on the nameplate for the 120 winding or across the 240 winding?

Charts of transformer voltage drops that I compared had different values (both had the same %PF), so i'm assuming the variable is the impedance.
 
joozu6,
1. I don't think I've ever seen %IZ.

2. The %Z is the same for both windings; that's why it is used rather than an ohmic value. It relates to the kVA rating of the transformer and can be calculated into ohms for each side of the transformer, using the 'base' and percent or 'per unit' values. You can search this forum for more info on how this works. I believe it has been explained in detail before several times.
 
Maybe %IZ means per unit I times %Z.

The nameplate %Z includes both the primary and secondary windings. It would apply to either the whole 240 volt winding if the two secondary windings are connected in series or to the two 120 volt windings connected in parallel (not the normal connection, but would be used for a two-wire service to use the full transformer capacity).
 
Thanks guys. Very much appreciated.
 
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