Transformer with PU question

[iwire]

a power transformer rated at 10KVA with with 440//220V. On the primary side, it has a per unit impedance of 0.01 + j0.1.

If a 10kVA load with a 0.8 pf leading connected to the secondary side, what is the voltage across secondary side?

The answer :

Load impedance is 100% of angle 36.87 (from PF of 0.8) which on a PU impedance of 0.8+j0.6.

V (secondary side) pu = 1/(0.8+j0.6) = 0.934

so the measured voltage on secondary will be 220 X 0.934 = 205.5 V

how they get 0.8+j0.6?

 

[iahim]

I can't follow their solution, but here is how I would solve it:

Vbase= 220V, Sbase=10kVA

Line current, I = 1<+36.87 pu (because load is leading pf)

Since the transformer impedance is in pu, it can be shifted to the secondary side as is, so voltage drop across the transformer impedance is (1<36.87) x (0.01+j0.1) = 0.1<121.159.

Vsecondary = Vload + Vdrop = 1 + 0.1<121.159 = 0.952<5.16

So Vsecondary = 209.46V

 

[iahim]

a power transformer rated at 10KVA with with 440//220V. On the primary side, it has a per unit impedance of 0.01 + j0.1.

If a 10kVA load with a 0.8 pf leading connected to the secondary side, what is the voltage across secondary side?

The answer :

Load impedance is 100% of angle 36.87 (from PF of 0.8) which on a PU impedance of 0.8+j0.6.

V (secondary side) pu = 1/(0.8+j0.6) = 0.934

so the measured voltage on secondary will be 220 X 0.934 = 205.5 V

how they get 0.8+j0.6?

If the load has leading power factor, the reactance of the load has to be capacitive, so the impedance cannot be 0.8+j0.6. Their answer is wrong! It should be 0.8-j0.6.

 

[msteiner]

a power transformer rated at 10KVA with with 440//220V. On the primary side, it has a per unit impedance of 0.01 + j0.1.

If a 10kVA load with a 0.8 pf leading connected to the secondary side, what is the voltage across secondary side?

The answer :

Load impedance is 100% of angle 36.87 (from PF of 0.8) which on a PU impedance of 0.8+j0.6.

V (secondary side) pu = 1/(0.8+j0.6) = 0.934

so the measured voltage on secondary will be 220 X 0.934 = 205.5 V

how they get 0.8+j0.6?

If the load has leading power factor, the reactance of the load has to be capacitive, so the impedance cannot be 0.8+j0.6. Their answer is wrong! It should be 0.8-j0.6.

iahim, are you sure about your last statement? My understanding is that a leading power factor (capacitive) has a positive power factor angle when expressed in polar form, and a positive imaginary component when expressed in rectangular form.

 

[iahim]

We should be taking a break from this, at least until the results are released.

I am sure that the reactance for a capacitive load is negative and for an inductive load is positive. Just read the problem: the transformer's impedance is 0.01+j0.1. Obviously the transformer is inductive, not capacitive.

 

[msteiner]

I'm taking the exam for the first time in April, so this is my ramping up time

Is it correct to say that current and impedance have opposite polarities, so the current to a capacitive load has a positive angle with respect to voltage, and the impedance is negative?

 

[iahim]

I'm taking the exam for the first time in April, so this is my ramping up time

Is it correct to say that current and impedance have opposite polarities, so the current to a capacitive load has a positive angle with respect to voltage, and the impedance is negative?

In that case, no break for you. :laugh: It's good that you started early.

I'm not sure I understand your question. The impedance for a capacitive load is not necessarily negative, just the imaginary component.

 

[msteiner]

Sorry, yes, the imaginary component is what I was referring to.