Three Phase Transformer Voltage Drop Calculation

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jamie300993

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Hi, If possible, I would like someone to clarify whether or not I am correct in my calculation of the voltage drop for a three phase Dyn Transformer?

The transformer in question is a 3MVA 11kV/415V Dyn Transformer which has the following p.u. R & X values:

R = 0.00355pu = 0.355%
X = 0.0901pu = 9.01%

If I want to calculate the secondary voltage drop, and the load power factor is 0.85 with a Primary transformer current of 55A, would the following be correct;

IPrimary (Phase) = 55A

Secondary Phase current = Line Current (Secondary is star connected ) = 55*(11000/415) = 1457.83A

Converting X & R to ohms:

R(ohms) = (10*%R*kVsecondary2)/kVA = (10*0.3555*0.4152)/3000 = 0.0002038?

X(ohms) = (10*%X*kVsecondary2)/kVA = (10*9.01*0.4152)/3000 = 0.0051725?

? = cos^-1(0.85) = 31.788?

Volt Drop = Sqrt(3) * 1457.83 * (0.0002038*0.85 + 0.0051725*sin(31.788)) = 7.32V

Does this mean that the voltage to the load will be 415-7.32 = 407.68V?? This is the bit I would like confirming if the remainder of the calculations happen to be correct?


Many Thanks in advance for any help :)
 
jamie300993 said:
Volt Drop = Sqrt(3) * 1457.83 * (0.0002038*0.85 + 0.0051725*sin(31.788)) = 7.32V

Does this mean that the voltage to the load will be 415-7.32 = 407.68V?? This is the bit I would like confirming if the remainder of the calculations happen to be correct?


Many Thanks in advance for any help :)
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In my opinion it is correct. However X/R =25.38 it seems to me too high [usually has to be 10].:happyyes:
 
ron said:
I'm not sure if the smiley face was because the "usually has to be 10" was a joke, but as you know, close to the generation plant might be very high x/r (15 to 30)
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I think he was referring to the xfmr itself which has nothing to do with the X/R of where its used.
 
Jamie,

I believe you overlooked the fact that with a primary current of 55A the Xfmr is operating at about 1/3 of capacity. This requires modification of the Resistive and Reactive voltage-drops accordingly! A second point is that the above impedance related V-Drops are not linear, primarily due to loading losses!

A simpler, but somewhat less accurate method is called the Voltage-Regulation Calculation or V-Reg! It relates a transformer?s secondary-voltage for a specific loading to its rated-secondary voltage at full-load. It ignores Xfmr-losses, and presumes V-drop is proportional to Loading:

V-Reg (%) = A + B2/200, where:
A = 100 x Ld x {(Rt x Pf) + Xt x Sqrt (1-Pf?)}
B = 100 x Ld x {(Xt x Pf) - Rt x Sqrt (1-Pf?)}
Rt, Xt, = Xfmr Imp components in per-unit (0.00355 & 0.0901, respectively).
Ld = Xfmr Loading in per-unit (55/157 = 0.35).
Pf = Load Power-factor in per-unit, 0.85.

Then Voltage-Drop (V-drop, in Volts) is:
V-Drop = (V-Reg/100) x Xfmr?s Rated Sec?y Voltage! Using the V-Reg method with your parameters; Rt =0.00355, Xt=0.0901, Ld=0.35, Pf=0.85, Vrated=415, results in a V-Drp of 1.99 %, or 8.3V!

However, Julius Right?s comment about Xt/Rt ratio is correct. Also, the Xt value influences V-Drp much more than Rt. So, if a more conventional Xt for your Xfmr's design is used, say 10%, then the corresponding V-Drp is 1.15 % and 4.8 Volts!

Regards, Phil Corso
 
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IEEE 141 Voltage Drop.jpg
The calculation presented by Phil Corse it is of course more accurate. You can see this theory in IEEE 141 ch.3.11 Calculation of voltage drops.
Approximate voltage drop=I*R*cos(fi)+I*X*sin(fi).
For exact calculations, the following formula may be used:
actual voltage drop=es+I*R*cos(fi)+I*X*sin(fi)-sqrt(es^2-(I*X*cos(fi)-I*R*sin(fi))^2)
I don't think the transformer reactance will change much at 1/3 of rated current with respect the rated current load if the voltage stays unchanged.
But the resistance, depending on temperature, will change as the source of heat-the losses-will decrease.
If we will take the laminate losses as 10% of total losses at rated load and the copper losses will decrease with square of current -that means will be 1/9 of 90% =10% the new total losses will be only 20% of full load losses. Let?s say the winding average temperature at full load is 90oC in environmental temperature of 30oC.The difference of 60oC will drop to 20/100*60=12oC and the winding average temperature will be 42 oC. The resistance will be R90*(42+234.5)/(90+234.5)=85%R90.
Since the resistance is not an important parameter in voltage regulation calculation the temperature drop could be neglected.
 
Sorry, Phil Corso. I did not intend to change your name! I apologize for my mistake.
It seems to me it is the time to buy new glasses.:ashamed1:
 
Hi, sorry for not commenting on this thread for a while but I have had other problems to contend with which have taken priority. I have managed to get hold of a copy of the IEEE 141-1993 standard and as previously stated have found that the approximate voltage drop can be calculated by;

I*R*cos(fi) + I*X*sin(fi) ,

For a three phase system, am I correct in thinking I would simply need to multiply the value of this equation by sqrt(3) to get line voltage? If I was to use the voltage drop in terms of the secondary voltage, would the obtained value simply be deducted from the rated secondary voltage?

For the exact voltage drop equation (written in the second to last post above this post), eS is referred to as the sending end voltage (bus voltage). If this were applied to a transformer, would this simply be the rated secondary voltage?

Many thanks


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It was my mistake, sorry:ashamed1:. I forgot to mention es=VL??_L/sqrt(3) where VL_L it is phase-to-phase voltage -primary or secondary considering Vsec=Vprim*nsec/nprim nsec=number of secondary winding turns and nprim respectively of primary winding.Since rated voltages of a transformer are: primary the supply rated and secondary it is no-load voltage- then the ratio of these voltages it is very close to ratio of turn numbers.
In order to recalculate the result- between phase voltage- one has to multiply er with sqrt(3),of course.
 
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