# Transformer Voltage Regulation and efficiency determination?

• nazia_f

#### nazia_f

A transformer is rated at 2300/230 V, 15 kVA and 50 Hz. Assume that the transformer is operating at 80% p.f., leading and at rated output the secondary terminal voltage is 230 V. The following parameters for the transformer are given:

N1 = 1500 turns
N2 = 150 turns
r1 = 2.7 ohms
r2 = 0.024 ohms
x1 = 9.1 ohms
x2 = 0.088 ohms
In = 0.15 A
Core loss = 92 W

Calculate voltage regulation and efficiency of the transformer.

Please be as detailed as possible. I solved this problem but the examiner didn't give me marks for it. No matter how much I try I can't figure out my mistake. Now I am thinking if I have understood this topic in the wrong way. So please help me to solve the problem and final results of V.R. and efficiency will be very much appreciated since I can compare my answers with the correct one. Thank you.

Show us what work you can do - let us help you see where the issues lie rather than just giving you the answer.
- my \$0.02

I am actually confused about voltage regulation. The way I solved it is below -

M=N1/N2 = 1500/150=10
Primary current, Ip=15000/2300= 6.52A
Secondary current, Is=15000/230= 65.22A
Equivalent resistance referred to the secondary side, Rs = {R1/(M^2)}+R2 = (2.7/100)+0.024 = 0.051 ohms
Equivalent reactance referred to the secondary side, Xs = {X1/(M^2)}+X2 = (9.1/100)+0.088 = 0.179 ohms
cos(Theta) = .8
Theta = 36.9
No-load voltage on the secondary side, Vp/M = (230<0) + (0.051*65.22<36.9) + {(0.179<90)*65.22<36.9)} = 225.65<2.88

Voltage Regulation = {(225.65 - 230)/230} * 100% = -1.89%

now have I made any mistake? I checked with a similar problem of Chapman's book and I can't seem to find my mistake. Please help me find it.

Vp/M = (230<0) + (0.051*65.22<36.9) + {(0.179<90)*65.22<36.9)} = 225.65<2.88

To find voltage regulation, you need to apply the same source voltage as in the loaded case. The source voltage is not 2300V. You need to calculate it in the loaded case and apply it back to the no-load case.

Will it be like this? -

Voltage drop across the load (referred to the primary), M*Vs= 10*230= 2300V
Equivalent resistance referred to the primary, Rp= (M^2)*Rs = 100*0.051= 5.1 ohms
Equivalent reactance referred to the primary, Xp= (M^2)*Xs = 100*0.179= 17.9 ohms
Current passing through the resistance, reactance and load = Is/M = 65.2/10 = 6.52 A
Source voltage = Voltage drop across load + voltage drop across resistance + voltage drop across reactance = 2300<0 + (5.1*6.52<36.9) + {(17.9<90)*(6.52<36.9)} = 2256.52<2.87

Now, source voltage referred to the secondary = 2256.52/10 = 225.652

V.R. = {(225.652-230)/230}*100% = -1.89%

I am really sorry if I can't understand it right. I hope you will be patient with me and help me do this problem correctly. And I know I am asking for too much but I'll really appreciate it if you could solve the problem for me. I think I can understand it better if I see the pattern of solving it. Thanks a lot for replying to my thread.

Unfortunately, I cannot solve the problem for you, or I will get another warning. However, here's a hint on the steps you should follow:

1) Don't calculate Rp and Xp, keep them separated since grouping them is an approximation.
2) Calculate Rmag and Xmag from usual formulas found in your book and put them on the 2300V side.
3) Calculate the load impedance at 230V.
4) Draw the circuit: you have R1+jX1 on the 2300V side, R2+jX2 on the 230V side and Rmag in parallel with jXmag on the 2300V side and an ideal transformer in the middle (the usual T-circuit model). Also add the load on the 230V side.
5) The voltage at the load is known when the transformer is loaded. It is 230V.

*The definition of voltage regulation is the difference of voltage between unloaded and loaded states on the load side if the source voltage is kept constant. So, the first unknown to find is the source voltage that would give you 230V<0deg at the load.

6) By using KCL and KVL you can go back to the source and find the voltage at the source.

7) Apply this same source voltage on the unloaded transformer (**this means that Iload=0=I2, so that I1=Imag). Calculate the load voltage in this case.

8) Calculate the voltage regulation, since the unloaded voltage was the only unknown (loaded voltage is 230V).

Pcore = 92 W
S = 0.15*2300 = 345 VA
Q = sqrt{(345)^2 - (92)^2} = 332.5 var

Rmag = (2300)^2/92 = 57500 ohms
Xmag = (2300)^2/332.5 = 15909.77 ohms

Impedance of the load = (230<0°)/(65.2<36.9°) = 3.53<-36.9° ohms

I hope I have done right upto this but I can't understand after that. How and where do I apply KVL and KCL? Do I have to bring the secondary to the primary side and then calculate source voltage? If that is so, then can I eliminate excitation branch while calculating the source voltage? What about the current through r1+jx1 and r2+jx2? Is primary current 6.52A flowing through the whole 'referred to primary equivalent circuit' now?

There are also some things I am confused about. '2300/230' refers to rated primary and secondary voltage and 'Ip/Is' refers to rated primary and secondary current, right? While calculating voltage regulation we are taking primary voltage 2300 V as constant and secondary voltage 230 V as full-load voltage. Then why are we calculating source voltage again? And while going through some problems from the book I noticed that in some problems rated secondary voltage is taken as the no-load voltage and they have calculated full-load voltage using load impedance, also the load impedance they calculated doesn't give the same result as I get using the formula 'Z = Vs/Is'. Then have I calculated load impedance in our problem wrong? How do I calculate it? (In the book '2000 KW resistive load' was given and they calculated load impedance by 'Z = (Vs)^2/2000 KW'. But in our problem no such information is given)
What is the difference between rated secondary current and full-load current. Isn't secondary current 65.2A passing through the load?
Different books have different method to solve voltage regulation and I am getting too confused. I can't understand this voltage regulation properly. Now I feel like I need more clear conception of transformer.
Thank you for helping me. I hope you can help me clear my confusions too.

Yes, you have done it correctly.

You can bring back impedances on the 2300V side or not, it doesn't matter. You don't need to eliminate the excitation branch, since you know 1 voltage in the circuit. The key here is that you need to use the equations of the ideal transformer (N1/N2=V1/V2=I2/I1). Hint: you already calculated I2 and from I2, you can also find V2. Then work your way back to I1,V1. Remember that 230V is the voltage at the load, which is different from V2.

About the impedance of the load, in your problem, it is assumed that the transformer is fully loaded (even though it is not stated). The rating of the transformer is 15 kVA with leading PF of 0.8. So the power is 15000<-acos(0.8) and the load is Z=V^2/(S*)=3.53<-36.9° ohms, just like you calculated.

Applying KVL in the secondary loop,
V2 = (0.024+j0.088)*(65.2<36.9°) + 230<0° = 227.88<1.39° V

N1/N2 = V1/V2
so, V1 = (N1/N2)*V2 = 2278.8<1.39° V
Similarly,
I1 = (N2/N1)*I2 = 6.52<36.9° A

Now,
Vp' = (2.7+j9.1)*(6.52<36.9°) + 2278.8<1.39° = 2256.6<2.87° V

At no-load condition, I1 = Imag = 0.15<36.9° A
V1' = Vp' - (2.7+j9.1)*(0.15<36.9°) = 2256.6<2.87° - 1.42<110.37° = 2254.26<2.84° V

V1'/V2' = N1/N2
so, V2' = (N2/N1)*V1' = 225.43<2.84° V

Voltage regulation = {(225.43-230)/230}*100% = -1.98%

Have I done correct this time?

Not exactly. You forgot to remove Imag before here:
Vp' = (2.7+j9.1)*(6.52<36.9°) + 2278.8<1.39° = 2256.6<2.87° V

The current in R1+jX1 is not (6.52<36.9°). Apply V1 to Rmag and Xmag.

Also, since the load is capacitive, I think it is normal that you get an higher voltage when loaded.

Ir-mag = (2278.8<1.39°)/(57500<0°) = 0.04<1.39° A
Ix-mag = (2278.8<1.39°)/(15909.43<90°) = 0.14<-88.61° A
Imag = Ir-mag + Ix-mag = 0.15<-72.66° A
Ip' = I1+ Imag = 6.52<36.9° + 0.15<-72.66° = 6.47<35.6° A [Since the excitation branch comes after r1+jx1 is connected in series]
Now,
Vp' = (2.7+j9.1)*(6.47<35.6°) + 2278.8<1.39° = 2258.04<2.87° V

At no-load condition, Ip' = Imag = 0.15<-72.66° A
V1' = Vp' - (2.7+j9.1)*(0.15<-72.66° A) = 2258.04<2.87° - 1.42<0.84° = 2253.8<2.87° V

V1'/V2' = N1/N2
so, V2' = (N2/N1)*V1' = 225.38<2.87° V

Voltage regulation = {(225.38-230)/230}*100% = -2.01%

Is it correct now?

Almost there!

You forgot that the magnetizing current is not the same when loaded, so you can't use:
Ip' = Imag = 0.15<-72.66° A

Instead, when the transformer is not loaded, you simply have a potential divider between Z1=R1+jX1 and Z2=Rmag//jXmag (shunt impedances). The voltage V1' will be V1'=Z2/(Z1+Z2)*Vp'. The rest is good.

Z2 = 1/{(1/57500<0°) + (1/15909.43<90°)} = 15332.7<74.54° Ω
Z1 = 9.492<73.47° Ω

V1' = {Z2/(Z1+Z2)}*Vp' = 2256.64<2.87° V

V1'/V2' = N1/N2
so, V2' = (N2/N1)*V1' = 225.664<2.87° V

Voltage regulation = {(225.664-230)/230}*100% = -1.89%

That seems correct (did not calculate, but the steps are good).

Thank you very much for helping me solve the problem and also for keeping patience with me. If you don't mind I'd like to ask some things that confused me while solving the math.

I didn't really get the concept of no-load current (In or Imag) = 0.15 A that was given in the problem. Is this the current that flows through excitation branch when the secondary is loaded or is it the current that flows only through the primary (r1, x1 and excitation branch) when secondary is not loaded?
I determined Rmag and Xmag using this "Imag = 0.15 A" assuming that when the secondary is not loaded this current is flowing through the primary. Then again, I measured Ip'= I1 + Imag when the secondary is loaded. I used this Imag both in no-load and loaded cases. So I can't really understand what you meant by this -
"You forgot that the magnetizing current is not the same when loaded, so you can't use:
Ip' = Imag = 0.15<-72.66° A"

I didn't really get the concept of no-load current (In or Imag) = 0.15 A that was given in the problem. Is this the current that flows through excitation branch when the secondary is loaded or is it the current that flows only through the primary (r1, x1 and excitation branch) when secondary is not loaded?

As the name implies, the no-load current is the current through the primary winding when the secondary winding is not loaded.

Then again, I measured Ip'= I1 + Imag when the secondary is loaded. I used this Imag both in no-load and loaded cases. So I can't really understand what you meant by this -
"You forgot that the magnetizing current is not the same when loaded, so you can't use:

Look at it this way. Draw your model with a 2300V source and a variable resistor load. The parameters of your transformer model are fixed. If the load is R=infinity, you get a no-load condition. By looking at your model, you will see that I2=0, so I1=0 and Ip'=Imag. Then, by diminishing your load resistance towards zero, the load current I2 will start to increase. In turn, this means that there will be a bigger voltage drop across R1+jX1 (because Ip'= I1 + Imag) that will diminish the voltage on Rmag and Xmag. This means that the current Imag will be reduced as the load current goes up, because Imag=V2/Zmag. At the limit, if R=0, you have a short-circuit and the impedance R2+jX2 will be in parallel with Rmag and Xmag. Because R2+jX2 is much smaller, Imag will be almost zero.

Physically, we can explain the phenomenon like this. In no-load, the transformer still draws some current to magnetize the core (from the connected winding), which has a finite permeability (and some losses). However, when you start to load the secondary winding, the current in this winding will create an "opposite magnetic flux" (check with the right-hand rule) to the one produced by the primary winding. Hence, there will be less flux in the core, thus less magnetizing current is needed. At the limit, if the secondary winding is shorted, the opposite flux generated by the short-circuit current in the secondary winding will be equal to the flux generated by the primary and there is no flux in the core (almost).

So given no-load current is the current that flows only through the primary when secondary is not loaded. Then it means -
Imag = Ir-mag + Ix-mag = 0.15<-72.66° A is the Imag when secondary is loaded and though we have no-load current (0.15 A) given, we don't know its phase angle which means we actually don't know proper no-load current which is why we can't use it here as Ip' = Imag = 0.15<-72.66° A.
I see...I get it now. :D thank you so much for clearing my confusions out. Thanks a lot again.