Why is Rated Voltage Used for Full Load Voltage Regulation Calculations?

[Jawad]

Hi,
I want to calculate voltage regulation of a transfromer. I have learned that voltage regulation is (Vnl-Vfl)/Vfl.
While solving some problems for practice I came across multiple questions. In some of them where the load is given I understand how to solve them. I just find the No load voltage and then the voltage at load when the load is connected.
But in the problems where it is written that find the voltage regulation at full load and "LOAD" is not specified there the writer find voltage regulation like this:
1. Find the voltage at the secondary of the transformer by adding the voltage drop at the Rp and Xp and voltage at output.
2. Uses rated voltage in place of Vfl.
3. Finds the Voltage Regulation.

I can't understand why is the writer doing it. Spent my whole day trying to understand it. Please Help.

 

[Babadag]

Actually rated voltage as defined on IEC 60076-1 CH. 3.4.3:
“rated voltage of a winding (Ur) the voltage assigned to be applied, or developed at NO-LOAD, between the terminals of an untapped winding, or of a tapped winding connected on the principal tapping. For a three-phase winding it is the voltage between line terminals[PHASE-TO-PHASE].
I did not find an ANSI/IEEE a clear definition. However I think the rated secondary voltage has to be no-load voltage.
Now, the full load voltage at transformer secondary terminals it is the primary rated voltage considered constant [that means independent of load] reduced by primary to secondary number of turns ratio [this is approximate the ratio of primary to secondary voltage rated -at no-load] minus the transformer full load voltage drop.
Actually Vfl=Vnl-VD then Vnl-Vfl=VD.
See IEEE 141 CH.3.11 Calculation of voltage drops.
The exact formula:
VD=Vp+I*R*COSFI+I*X*SINFI-SQRT(V^2-(I*X*COSFI-I*R*SINFI)^2)
Approximate formula: I*R*COSFI+I*X*SINFI
From the short-circuit voltage[vk%] and R/X ratio we can calculate R and X .
Let's say a transformer of 1000 kVA 13.8/0.48 kV vk%=5.75% and X/R [See ANSI C37.010] 6.5 cos(fi)=0.9 [=pf]. sin(fi)=sqrt(1-0.9^2)= 0.43589
From Zk^2=R^2+X^2 and X/R=6.5 we get X=0.013094 and R=0.002014 ohm
I=1000[kVA]/SQRT(3)/0.48[kV]=1202.81A then
VD=1202.81*(0.002014*.9+0.013094*.9)=9.045V.
We shall multiply by sqrt(3) in order to find the phase-to-phase voltage drop and finally we get 15.67 V[3.26%].

 

[Babadag]

Sorry. I think I skipped Zk=V^2/S*vk%/100 [V=0.48kV,S=1 MVA]
Zk=0.48^2/1*5.75/100=0.013248