Maximum secondary winding resistance for 2% regulation

[Numbskull]

Homework Statement

A 415V to 11 kV transformer has a rating of 200 kVA. The winding resistance and leakage reactance when referred to the primary are 0.014 $\Omega$ and 0.057 $\Omega$ respectively.

(a) Determine the % regulation of the transformer at 0.8 power factor lagging.

(b) In designing a particular 415V to 11 kV, 200 kVA transformer, the primary winding resistance is to be 10 m$\Omega$. Find the maximum winding resistance of the secondary winding if the transformer is to have 2% regulation at unity power factor.

Homework Equations

% Regulation = $\frac {{VA}_{rating}} {V_1^2}$ x (R’p cos $\theta$ + X’p sin $\theta$) x 100%

The Attempt at a Solution

(a)
$\frac {200 \times 10^3}{(415)^2}$ x ((0.014 $\times$ 0.8) + (0.057 $\times$ $\sqrt{1 - 0.8^2}))$ x 100% = 5.272% regulation

(b)
Voltage regulation (%) $\approx \frac {200 \times 10^3}{(415)^2}$ x R'p x 100%

Rearranging to make R'p the subject, I get 17.2225 m$\Omega$ (as the primary reflected impedance), and this is only an approximation according to the formula provided in the textbook.

Trying to analyse this part in my head, I think that I need to get that primary impedance down to 10m$\Omega$. This means that I need to alter (calculate) the maximum current / voltage on the secondary side taking the 2% voltage drop, leakage reactance and 200kVA max power into consideration?

I'm keen to learn this as per the forum principles, but some gentle guidance would be welcome

 

[NascentOxygen]

A 415V to 11 kV transformer has a rating of 200 kVA.

This would be a 3-phase transformer, even though not stated explicitly. Does your textbook give their answer?

In part (b) there is no data given concerning reactance, so you'll regard reactance as zero.

 

[Numbskull]

This would be a 3-phase transformer, even though not stated explicitly. Does your textbook give their answer?

In part (b) there is no data given concerning reactance, so you'll regard reactance as zero.

I'm interpreting the question as a single phase transformer with the constraints in the first paragraph, but with the special considerations laid out in part (b). I do not think it is anything other than a single phase transformer. What would make you think it might be a 3-phase transformer, would it be because of the voltages involved are typical of distribution grid values?

 

[NascentOxygen]

What would make you think it might be a 3-phase transformer, would it be because of the voltages involved are typical of distribution grid values?

Those are the precise 3ɸ voltages seen on our 240V system here (in Australia). What country was your textbook printed for?

 

[Numbskull]

It's a UK print.

 

[NascentOxygen]

It's a UK print.

Then ditto.

 

[CWatters]

Just curious but where are 415V to 11KV step up transformers used most? Small wind farms?

 

[NascentOxygen]

Just curious but where are 415V to 11KV step up transformers used most? Small wind farms?

Maybe, I don't know.
But turned around they are used as 11kV:415V step-down on powerpoles. ☺

 

[magoo]

Numbscull-

I'm not sure I agree with your first equation for voltage regulation. Was it given to you?

If you examine the units, it doesn't make sense to me.

Also, I usually refer to the higher voltage winding as the primary. As a distribution transformer supplying load, I'd expect 11 kV would be the primary winding.

 

[Numbskull]

The question was exactly as phrased, and the formula taken from the text where there's a very similar example problem solved in the course material with slightly different values:

For (b), what's wrong with a step-up transformer? From the assignment:

 

[magoo]

Sorry to question your equation, but I am used to working in per unit.

In per unit, you select the phase to phase voltage in kV and the MVA base.

So, in your case,
kV = 0.415
MVA = 0.2
Zbase = kV^2/MVA = 0.861125

All impedances in ohms have to be divided by the base Z or Zbase to express them in per unit.
r(pu) = 0.014/.861125 = 0.016258
x(pu) = 0.057/.861125 = 0.066192

So my voltage drop equation becomes:

V Drop = I * (r cos th + x sin th)
I = 1 per unit for 100% load
cos th = 0.8
sin th = 0.6

V Drop = 1*(0.016258 * 0.8 + 0.066192 * 0.6)
= 0.052722
To express it in %, just multiply it by 100, so you get 5.2722% - which is exactly what you got by the equation you used.

I just wanted to show that your equation method is OK even though it is foreign to me.