Transformer voltage regulation and calculations

[Numbskull]

This particular question is for an assignment. However, I believe that I have the final (correct) answer, but want to check the understanding.

I am required to calculate the resistance of the secondary winding of a transformer. The transformer is a 415:11000 V step-up, rated at 200kVA but with a unity power factor, so I can conveniently treat this as 200kW.

The question asks for the resistance value of the secondary winding, when the output across the transformer has 2% voltage regulation. This means that the voltage across the terminals will be 10780 V.

As I have the power rating, I can divide by the voltage to determine the current. Then, when I have the current I can divide the voltage by the current to get the resistance.

The question is, do I determine the secondary winding resistance using the 11000 or 10780 voltage? Obviously these will both give a different answer for the resistance, but if the question was specific about the rating (kVA), then I would assume that it needs to be constant. Examples:

200,000 W / 11,000 V = 18.1818 A, then 11,000 V / 18.1818 A = 605 Ohms.
200,000 W / 10,780 V = 18.5528 A, then 10,780 V / 18.5528 A = 581.04 Ohms.

Both equate to 200 kW, but are different answers for the resistance. The question also requires me to factor in a reflected resistance and leakage reactance (from the primary side), but those two figure need to be the total of the secondary winding so that I can calculate (by removing the reflected component) the secondary winding resistance only.

Thanks for any assistance.

 

[Fernando Perfumo]

Calculate the current on the secundary. Suposing null the series inductance, the secondary resistance is the voltage drop divided by the intensity.
Lots of better problem like this in a book from Parker Smidth if I remember well.

 

[rude man]

Looks like you're computing load resistance rather than transformer resistance. So you're way off base.
How about drawing a picture with a resistance in series with a load resistance.
Remember the definition of kVA rating for a real load.

The above assumes zero leakage reactances i.e. the regulation factor is due to secondary resistance only. Including primary resistance and pri + sec leakage reactances complicates the calculations.

 

[magoo]

415 V and 11 kV sounds like its a three phase transformer. Your calculations are for a single phase transformer. Is the 200 kVA transformer single phase or three phase?

 

[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 Ω and 0.057 Ω 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 mΩ. Find the maximum winding resistance of the secondary winding if the transformer is to have 2% regulation at unity power factor.

Homework Equations

$R's = Rs + \frac{Rp}{n^2} \leftarrow$ this is to calculate the resistance reflected from the primary to the secondary winding.

$L's = Ls + \frac{Lp}{n^2} \leftarrow$ this is to calculate the reactance reflected from the primary to the secondary winding.

$n^{2} = \frac{415}{11000} = 0.0014233471 \leftarrow$ this is the value of $n^{2}$

The Attempt at a Solution

Firstly, with unity power factor, I believe that I can effectively treat this as 200kW because with a part reactive load we would have a non-unity power factor. Therefore calculations are based upon this assumption.

To calculate the primary (reflected to the secondary) impedance we use $R's = Rs + \frac{Rp}{n^2}$ which using $\frac{Rp}{n^2}$ and thus $\frac{10 mΩ}{0.00142...}$ gives us only the reflected portion of the impedance which is 40.06 (X) - we have not added the Rs

Similarly, the primary (reflected to the secondary) reactance is given by $L's = Ls + \frac{Lp}{n^2}$ which using $\frac{Lp}{n^2}$ and thus $\frac{0.057Ω}{0.00142...}$ gives us only the reflected portion of the reactance which is 7.0256 (R) - we have not added the Ls

As my previous method of calculation appears to be incorrect, I'm not sure what to do now. All I can see is that to achieve a power of 200kW at 11kV, would require a current of 18.181818 Amps. This gives a total impedance across the winding (i.e. the load) of 605 Ohms.

I have the feeling that I'm missing some key relationship between the reflected impedance and the only possible (maximum) value of the secondary winding. I don't know what the load value is, or the secondary winding resistance and reactance, so obviously the necessary data is in the given information.

There have been several topics on this question in this forum, all of which I have tried to learn from, but inevitably they become protracted or focus on a different part of the question. And like those other posters have said, the material supplied is often confusing, incomplete or simply inadequate.

Any help greatly appreciated!

 

[Fernando Perfumo]

Hi again.
In the case b), leakage inductances are not given, but they ask you for a maximun posible value the secondary winding can have, when the power factor is 1. It is not the same transformer that in the a case.
Voltage drop will be 2%.

Then referring all to secondary,
(Rsec + R'prim) * Isec_full_load <= 0.02 * Vsec_nominal,
or
Rsec <= (0.02 * Vsec_nominal / Isec_full_load) - R'prim
where Rsec is the secondary winding resistance, Vsec_nominal the secondary voltage when in open circuit; R'prim is the primary resistance reflected on the secondary.

Isec_full_load = kVA_load /( 0.98* 11)

A real transformer will have not null series inductance, but they ask for a 'maximun value of secondary resistance'.

 

[Numbskull]

OK so far.Where did this come from? Isec full load = 11000V/605Ω = 18.1818A which you have already calculated in post 5.

The quotes are from two different posters. Only the reference to post #5 is mine.

 

[Fernando Perfumo]

The quotes are from two different posters. Only the reference to post #5 is mine.

the voltage on the load is a 2% inferior to the nominal voltage. If we had 11000 volts unloaded, at full load we will have 11000 * ( 1- 0.02 )
The 0.02 is the 2% of regulation, other way of putting 2/100
Talking kilo-whatever wise :) we have
Intensity on the secondary at full load = 200000 / (11000 * 0.98) or kilo-wised : 200 kVA / (11*0.98) that is :
200/(0.98*11) = 18.55288 A
The voltage on the load is 11000 * 0.98 = 10780 V
The fall from no load to full load is 11000 - 10780 = 220 V, this voltage drop is only due to the total series resistance.

Of course, adding more turns to the secondary winding we could get the nominal voltage at full load, but then the transformation relation must be adjusted.

 

[rude man]

The quotes are from two different posters. Only the reference to post #5 is mine.

Sorry, got it.
So looks like two separate problems, one in post 1 and another in post 5.
Did you ever get done with post 1? I addressed that one in my post #3.
Then we can look at post 5 which is more involved since it adds reactances to the picture.

 

[Numbskull]

So looks like two separate problems, one in post 1 and another in post 5.
Did you ever get done with post 1? I addressed that one in my post #3.
Then we can look at post 5 which is more involved since it adds reactances to the picture.

Sorry for the confusion! It is one problem which I have detailed in post #5, and specifically I am looking for assistance with part (b). I was attempting to establish some fundamental values for the circuit, hence post #1 doesn't mention the reactances, but is an attempt at a general approach to the problem.

I've gained some obvious information from Fernando that I (very embarrassingly) didn't see, in that unloaded, we have a secondary voltage of 11kV, but loaded with 2% regulation we have 10.78kV, which of course tells me that the drop across the load is 220V.

Am I going in the right direction? i.e. would this information be essential to solve the problem?

 

[Fernando Perfumo]

Sorry for the confusion! It is one problem which I have detailed in post #5, and specifically I am looking for assistance with part (b). I was attempting to establish some fundamental values for the circuit, hence post #1 doesn't mention the reactances, but is an attempt at a general approach to the problem.

I've gained some obvious information from Fernando that I (very embarrassingly) didn't see, in that unloaded, we have a secondary voltage of 11kV, but loaded with 2% regulation we have 10.78kV, which of course tells me that the drop across the load is 220V.

Am I going in the right direction? i.e. would this information be essential to solve the problem?

I'm not sure you understand that the 220 volts drop is 'into the transformer box' while the voltage across the load is 10.78 kV, that is the voltage measured at the tranformer's output terminals.

 

[Numbskull]

I'm not sure you understand that the 220 volts drop is 'into the transformer box' while the voltage across the load is 10.78 kV, that is the voltage measured at the tranformer's output terminals.

Yes, I do understand, just got the figures the wrong way around.

 

[rude man]

Sorry for the confusion! It is one problem which I have detailed in post #5, and specifically I am looking for assistance with part (b). I was attempting to establish some fundamental values for the circuit, hence post #1 doesn't mention the reactances, but is an attempt at a general approach to the problem.

I've gained some obvious information from Fernando that I (very embarrassingly) didn't see, in that unloaded, we have a secondary voltage of 11kV, but loaded with 2% regulation we have 10.78kV, which of course tells me that the drop across the load is 220V.

Am I going in the right direction? i.e. would this information be essential to solve the problem?

I'm still confused. In post 5, is the part (a) transformer the same as the part (b) transformer? If so, do the resistances and reactances of part (a) comprise both primary and secondary resistances & reactances? I still read part (b) to be a separate transformer, without reactances, just resistances.

 

[Fernando Perfumo]

I'm still confused. In post 5, is the part (a) transformer the same as the part (b) transformer? If so, do the resistances and reactances of part (a) comprise both primary and secondary resistances & reactances? I still read part (b) to be a separate transformer, without reactances, just resistances.

Yes, they are different transformers. Read it carefully, and yes, the reflected secondary series resistance and its leakage inductance added to the primary winding ones.

 

[rude man]

Yes, they are different transformers. Read it carefully, and yes, the reflected secondary series resistance and its leakage inductance added to the primary winding ones.

There is no mention made of leakage inductance in 5 (b). There are no numbers given for leakage inductance.
Forget leakage inductance and re-read my post #3 which is for a slightly different problem but the advice is still apposite. In both cases, certain resistances are given and you are asked for how much extra secondary resistance is tolerable for 2% regulation. Read and think.

 

[Numbskull]

There is no mention made of leakage inductance in 5 (b). There are no numbers given for leakage inductance.
Forget leakage inductance and re-read my post #3 which is for a slightly different problem but the advice is still apposite. In both cases, certain resistances are given and you are asked for how much extra secondary resistance is tolerable for 2% regulation. Read and think.

Just to clarify, Fernando is not the OP. I've edited the question into my interpretation of what they're asking. I did mention how the study material often falls short, you can see how the assignment questions are not much better

A 415V to 11 kV transformer has a rating of 200 kVA. The winding resistance and leakage reactance when referred to the primary are 0.010 Ω and 0.057 Ω respectively. Find the maximum winding resistance of the secondary winding if the transformer is to have 2% regulation at unity power factor.

That is the basis upon which I intend to answer; it appears to be a single transformer, with the second part of the question giving a special case.

 

[rude man]

Just to clarify, Fernando is not the OP. I've edited the question into my interpretation of what they're asking. I did mention how the study material often falls short, you can see how the assignment questions are not much better

A 415V to 11 kV transformer has a rating of 200 kVA. The winding resistance and leakage reactance when referred to the primary are 0.010 Ω and 0.057 Ω respectively. Find the maximum winding resistance of the secondary winding if the transformer is to have 2% regulation at unity power factor.

That is the basis upon which I intend to answer; it appears to be a single transformer, with the second part of the question giving a special case.

OK, that is a new statement far as I can make out. Stay tuned.

 

[rude man]

Primary winding & leakage impedance is 0.01 + j0.056 Ω. Multiplied by n² with n = 11000/415 gives you your secondary winding & leakage impedance.

So draw your secondary circuit with the secondary winding & leakage impedance in series with the load resistance and the extra allowable secondary winding resistance to give you 2% regulation at full load current = 200kV/11000V.

 

[Numbskull]

Primary winding & leakage impedance is 0.01 + j0.056 Ω. Multiplied by n² with n = 11000/415 gives you your secondary winding & leakage impedance.

Shouldn't that be 415/11000 as it's a step-up transformer?

 

[rude man]

You can define n either way. But obvously the higher-turn winding will reflect higher, not lower, impedances. So the .01ohm primary resistance is 0.01 x (11000/415)² = 7.03 ohms referred to the secondary.

I believe it is std practice to define n >= 1.

 

[Babadag]

I agree with rude man, since it is a step-up transformer the primary voltage it is 415 V. The voltage regulation is:
sqrt(3)*Irat*[Rtrf*cos(fi)+Xtrf*sin(fi)] -it is an approximate formula but the error is negligible.
I think the job it is to find a supplementary resistance to be inserted in the secondary [11 kV] in order to achieve 2% regulation.
We cannot separate the resistance of the secondary from primary-this 0.01 ohm it is the equivalent resistance across the transformer:
Rtrf=Rpr+R'sec R'sec=Rsec*0.415^2/11^2
If the voltage regulation is 2% that means 2%*415=8.3 V [line-to-line]
The new resistance viewed from primary side[415 V] has to be:
New Rtrf=DV/sqrt(3)/Irat-Xtrf*sin(fi) fi=0 [cos(fi)=1]
Irat=200/0.415/sqrt(3)=278.24 A
New Rtrf=8.3/sqrt(3)/278.24=0.017223 ohm
The new inserted resistance will be 0.017223-0.01=0.007223 ohm
Viewed from 11 kV the -actual- resistance will be 11^2/0.415^2*0.007223 =5.0747 ohm.

 

[Numbskull]

I'm a little confused; if $\frac{415}{11000^2}$ is 0.00142 and $\frac{11000}{415^2}$ is 702.56 then they will give completely different answers. The formula I have in the textbook for calculating the impedance reflected from the primary winding to the secondary winding is $R's = Rs + \frac{Rp}{n^2}$ so it appears to me that $n^2$ can only be one value to give the correct answer.

 

[rude man]

I'm a little confused; if $\frac{415}{11000^2}$ is 0.00142 and $\frac{11000}{415^2}$ is 702.56 then they will give completely different answers. The formula I have in the textbook for calculating the impedance reflected from the primary winding to the secondary winding is $R's = Rs + \frac{Rp}{n^2}$ so it appears to me that $n^2$ can only be one value to give the correct answer.

Well, you're right, and value is n² = (N2/N1)² = 702.56.
Perhaps in your textbook the transformer was a step-down, not step-up. Or, secondary impedances were reflected into the primary. In any case the primary impedances reflect higher into the secondary if the transformer is a step-up.

Look at it this way: assume primary current = ip.
Then primary resistance power loss is Pp = ip² Rp
But if we reflect into the secondary the power loss must be the same: Ps = is² Rs.
But is/ip = N1/N2 so Ps = (N1/N2)² ip² Rs = ip² Rp.
So Rp = (N1/N2)² Rs
or Rs = (N2)/N1)² Rp. For your step-up transformer, (N2/N1) = 702.56
so Rs = 702.56 Rp.

I advise not using n ever again. Use N1 aand N2 and you won't be confused.

 

[Numbskull]

I advise not using n ever again. Use N1 aand N2 and you won't be confused.

And simply using $\frac{N1}{N2}$ or $\frac{N2}{N1}$ is correct for step-up and step-down respectively?

 

[rude man]

N1 is the number of turns in one winding. N2 is the number of turns in the other winding. Conventionally, N1 is the primary and N2 the secondary, but you can put the input voltage on either winding and the load on the other, making either a step-up or step-down transformer.

If N2 > N1 it's called a step-up transformer. If N2 < N1 it's called a step-down transformer. In either case, V2/V1 = N2/N1 = I1/I2 for an ideal transformer. Don't use n = N2/N1 or N1/N2, youi'll confuse yourself.

 

[Numbskull]

So far then I have primary impedance: $Z = \sqrt{0.01^2 + 0.057^2}$ which equals $0.05787\Omega$

As $R's = Rs + \frac{Rp}{n^2}$ then $R's = Rs + \frac{0.05787}{702.569}$ is $R's = Rs + 82.369\mu\Omega$

Am I on the right track? That primary reflected result ($82.369\mu\Omega$) looks awfully small?

 

[rude man]

So far then I have primary impedance: $Z = \sqrt{0.01^2 + 0.057^2}$ which equals $0.05787\Omega$
As $R's = Rs + \frac{Rp}{n^2}$ then $R's = Rs + \frac{0.05787}{702.569}$ is $R's = Rs + 82.369\mu\Omega$
Am I on the right track? That primary reflected result ($82.369\mu\Omega$) looks awfully small?

No. You're not paying attention. Look again at my post 23. In in I said "Rs = 702.56 Rp".

 

[Numbskull]

But is/ip = N1/N2 so Ps = (N1/N2)2 ip2 Rs = ip2 Rp

Was trying to follow, but got a bit lost. I get that $\frac{Is}{Ip}=\frac{N_1}{N_2}$, but couldn't see how $Ps = \left[\frac{N_1}{N_2}\right]^2$

I'm sure it is just an algebra manipulation, but with some intermediate steps left out.

 

[Babadag]

The actual transformer diagram it is as in Fig.1
Let’s put E's=Es*N1/N2=Ep and V's=Vs*N1/N2
N1/N2=Vprat/Vsrat at no-load.
Zk=Rtot+Xtot*i i=sqrt(-1) and it is measured when secondary terminals are short-circuited and the current stays rated transformer current.
Since the power Sxfm transferred through magnetic field remains the same:
E's*I's=Es*Is I's=Is*Es/E's=Is*N2/N1
Voltage drop at secondary windings will be:
DV=Es-Vs then D's=E's-V's
Now if D'Vs=DVs*N1/N2 then Z's=D'V/I's=DVs/Is*(N1/N2)^2
R's=Rs*(N1/N2)^2 and X's=Xs*(N1/N2)^2
And now we can draw the diagram Fig.2
Since Io is negligible you may draw the Fig.3
So the total resistance viewed from primary side it is:
Rtotold=Rp+R’s=Rp+Rs*(N1/N2)^2
or if we shall add a supplementary resistance :
Rtotnew=Rp+ Rs*(N1/N2)^2+Rnew*(N1/N2)^2
Rnew=(Rtotnew-Rtotold)*(N1/N2)^2
The regulation is the voltage drop for a certain cosfi[p/f.] and current.
DVtot=sqrt(3)*Iload*[Rtot*cos(fi)+Xtot*sin(fi)] and since cos(fi)=1 sin(fi)=0
DVtot=sqrt(3)*Iload*Rtotnew

 

[Numbskull]

So I have $R_p = \sqrt {0.01^2 + 0.057^2}$ which is $0.05787 \Omega$

Multiplied by $n^2 (702.569)$ is $40.658\Omega$

With a secondary output of 11kV and a 200kVA rating, the (max) current is 18.1818 Amps.

With a voltage regulation of 2%, I have 11kV * 0.98 which is 10.78kV (across the load).

Therefore, we have 220v (11kV - 10.78kV) dropped internally. As $Z = \frac{V}{I}$ then $Z = 12.1\Omega$

With a max current of 18.1818 Amps, the total circuit impedance is $Z = \frac{11,000}{18.1818}$ then $Z = 605\Omega$

Then we have a load impedance of $Z = \frac{10,780}{18.1818}$ then $Z = 592.9\Omega$

I can see this doesn't work as the primary reflected impedance already appears to eclipse the allowance I've calculated for the non-load impedances.

More guidance please?

 

[rude man]

What you have done is essentially correct. You have approx. 40Ω secondary reactance which, even though essentially inductive rather than resistive, will still drop about 18A x 40Ω = 720V which well exceeds the alloted 220V.

So, there are but two possibilities:

1. you're confusing transformers. This transformer is to have only the 0.01Ω primary resistance, with no leakage impedance. In that case the secondary internal resistance is about 7Ω, putting the internal secondary drop to only 7Ω x 18A = 126V, well within the allotted 220V drop.

2. You have been given an impossible assignment.

 

[Numbskull]

Thank you rude man.

Following your reasoning, I have a total secondary impedance of ($605\Omega$), minus the impedance of the load ($592.9\Omega$), minus the reflected primary impedance ($7.0257\Omega$).

As the question asked for the maximum value of the secondary winding resistance, I am left with $5.0743\Omega$

That look good?

 

[rude man]

Thank you rude man.

Following your reasoning, I have a total secondary impedance of ($605\Omega$), minus the impedance of the load ($592.9\Omega$), minus the reflected primary impedance ($7.0257\Omega$).

As the question asked for the maximum value of the secondary winding resistance, I am left with $5.0743\Omega$

That look good?

That looks fine.

 

[shaundicko]

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 Ω and 0.057 Ω 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 mΩ. Find the maximum winding resistance of the secondary winding if the transformer is to have 2% regulation at unity power factor.

Homework Equations

$R's = Rs + \frac{Rp}{n^2} \leftarrow$ this is to calculate the resistance reflected from the primary to the secondary winding.

$L's = Ls + \frac{Lp}{n^2} \leftarrow$ this is to calculate the reactance reflected from the primary to the secondary winding.

$n^{2} = \frac{415}{11000} = 0.0014233471 \leftarrow$ this is the value of $n^{2}$

The Attempt at a Solution

Firstly, with unity power factor, I believe that I can effectively treat this as 200kW because with a part reactive load we would have a non-unity power factor. Therefore calculations are based upon this assumption.

To calculate the primary (reflected to the secondary) impedance we use $R's = Rs + \frac{Rp}{n^2}$ which using $\frac{Rp}{n^2}$ and thus $\frac{10 mΩ}{0.00142...}$ gives us only the reflected portion of the impedance which is 40.06 (X) - we have not added the Rs

Similarly, the primary (reflected to the secondary) reactance is given by $L's = Ls + \frac{Lp}{n^2}$ which using $\frac{Lp}{n^2}$ and thus $\frac{0.057Ω}{0.00142...}$ gives us only the reflected portion of the reactance which is 7.0256 (R) - we have not added the Ls

As my previous method of calculation appears to be incorrect, I'm not sure what to do now. All I can see is that to achieve a power of 200kW at 11kV, would require a current of 18.181818 Amps. This gives a total impedance across the winding (i.e. the load) of 605 Ohms.

I have the feeling that I'm missing some key relationship between the reflected impedance and the only possible (maximum) value of the secondary winding. I don't know what the load value is, or the secondary winding resistance and reactance, so obviously the necessary data is in the given information.

There have been several topics on this question in this forum, all of which I have tried to learn from, but inevitably they become protracted or focus on a different part of the question. And like those other posters have said, the material supplied is often confusing, incomplete or simply inadequate.

Any help greatly appreciated!

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 Ω and 0.057 Ω 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 mΩ. Find the maximum winding resistance of the secondary winding if the transformer is to have 2% regulation at unity power factor.

Homework Equations

$R's = Rs + \frac{Rp}{n^2} \leftarrow$ this is to calculate the resistance reflected from the primary to the secondary winding.

$L's = Ls + \frac{Lp}{n^2} \leftarrow$ this is to calculate the reactance reflected from the primary to the secondary winding.

$n^{2} = \frac{415}{11000} = 0.0014233471 \leftarrow$ this is the value of $n^{2}$

The Attempt at a Solution

Firstly, with unity power factor, I believe that I can effectively treat this as 200kW because with a part reactive load we would have a non-unity power factor. Therefore calculations are based upon this assumption.

To calculate the primary (reflected to the secondary) impedance we use $R's = Rs + \frac{Rp}{n^2}$ which using $\frac{Rp}{n^2}$ and thus $\frac{10 mΩ}{0.00142...}$ gives us only the reflected portion of the impedance which is 40.06 (X) - we have not added the Rs

Similarly, the primary (reflected to the secondary) reactance is given by $L's = Ls + \frac{Lp}{n^2}$ which using $\frac{Lp}{n^2}$ and thus $\frac{0.057Ω}{0.00142...}$ gives us only the reflected portion of the reactance which is 7.0256 (R) - we have not added the Ls

As my previous method of calculation appears to be incorrect, I'm not sure what to do now. All I can see is that to achieve a power of 200kW at 11kV, would require a current of 18.181818 Amps. This gives a total impedance across the winding (i.e. the load) of 605 Ohms.

I have the feeling that I'm missing some key relationship between the reflected impedance and the only possible (maximum) value of the secondary winding. I don't know what the load value is, or the secondary winding resistance and reactance, so obviously the necessary data is in the given information.

There have been several topics on this question in this forum, all of which I have tried to learn from, but inevitably they become protracted or focus on a different part of the question. And like those other posters have said, the material supplied is often confusing, incomplete or simply inadequate.

Any help greatly appreciated!

I'm actually on the same question now. What was your working out for question (a)?