How Does Power Factor Affect Voltage in a Three-Phase Transformer?

[curiousguy23]

Homework Statement

A 2,000 kVA three-phase 33000/6600 V 50 Hz delta - star(wye) connected transformer has a primary and secondary winding resistances of 8 ohms and 0.08 ohms per phase respectively. The leakage reactance referred to the secondary is 1.5 ohms per phase. Calculate the sec. Voltage if we have a 0.75 power factor lagging. Ans. 6.256kV

Homework Equations

V_r=V2 +I_2(R_eq*cos(phi) + X_eq*sin(phi)) where V_r is the rated voltage and V_2 is the secondary voltage. R_eq and X_eq are the total resistances and reactances referred to the secondary.

The Attempt at a Solution

I first attempted to find the per phase resistance and reactance total on the secondary.

R_eq = 0.4 ohms
X_eq = 1.5 ohms

Then i split the apparent power equally on each phase of the star(wye) connection and divided by the phase voltage (6600/(rt 3)) to find the phase current of 175 A. Substituting in the above eqn gave me a Secondary voltage of 6.2Kv . Not Sure what I am doing wrong, this is my first time working with three- phase transformers.

 

[KataKoniK]

Thank you for your post. I can see that you have made good progress in your solution, but there are a few key things that you need to consider in order to arrive at the correct answer.

Firstly, it is important to note that the given transformer is a delta-star(wye) connected transformer, which means that the primary winding is connected in a delta configuration and the secondary winding is connected in a star(wye) configuration. This affects the calculation of the total resistance and reactance referred to the secondary. In this case, the total resistances will be three times the individual resistances and the total reactances will be three times the individual reactances.

Secondly, in order to calculate the secondary voltage, you need to use the power factor angle (phi) of 0.75 lagging in the equation. This means that you need to use the cosine of 0.75 (0.707) and the sine of 0.75 (0.707) in the equation.

Finally, when you are dividing the apparent power equally among the three phases, you need to consider the power factor angle. In this case, the apparent power (S) is equal to the product of the phase voltage (V) and the phase current (I), multiplied by the power factor (cos phi). Therefore, you need to divide the apparent power by the power factor in order to get the actual power (P) that is being dissipated in the resistances and reactances.

Taking all of these factors into account, the correct solution should be:

R_eq = 0.4 ohms * 3 = 1.2 ohms
X_eq = 1.5 ohms * 3 = 4.5 ohms

Apparent power (S) = 2000 kVA
Actual power (P) = 2000 kVA / 0.75 = 2667 kVA

Phase current (I) = 2667 kVA / (6600/(rt 3)) = 2667 kVA / 3812.2 V = 0.700 A

Substituting these values into the equation, we get:

V_r = V_2 + I_2(R_eq*cos(phi) + X_eq*sin(phi))
6600 V = V_2 + 0.700 A (1.2 ohms * 0.707 + 4.5 oh