Transformers Under Load and No Load

[ErnieChicken]

Homework Statement

I have a step down transformer (6 : 1)
Copper Loss of 110W at full Load
Primary Coil: I = 3.33A R = 0.35 Ohms Power Loss = 4W
Secondary Coil: I = 20A R = 0.265 Power Loss = 106W
Power Drawn Over the Primary = 75W

9kVA Transformer

Homework Equations

The Equations i have are:

E1/E2 = N1/N2 = I2/I1

Power Loss = I^2 x R

The Attempt at a Solution

• No Load Current is 5% less than Full Load Current

• Primary Copper Loss is less than 1/400 of full load copper loss

Do i need to calculate the flux? Or Calculate the other Eddy Currents or Magnetic Flux?

Your help will be appreciated

Thanks

 

[Hesch]

Secondary Coil: I = 20A R = 0.265 Power Loss = 106W

That is at full load. At no load, I₂=0A, resulting in a copper loss = 0W

Primary Coil: I = 3.33A R = 0.35 Ohms Power Loss = 4W

Again at full load. I think, that your 5% comes from that the no load current is 5% of the full load current. This no load current keeps up the flux and delivers power to the iron-losses, which are the same at full or no load.

Power Drawn Over the Primary = 75W

That must be at no load and is the sum of primary copper losses and iron losses. You don't have to calculate the flux. Iron losses = 75W - ( primary copper losses ).

 

[ErnieChicken]

So calculating the Load and No Load Current.
My Full Load Current would be the 3.33A on the primary side since there is minimal current across the secondary side.

My Load current, Would this involve inserting a resistor for load or would i use the Coil Resistance for both Primary and Secondary side?

Thanks for your help!

 

[Hesch]

My Full Load Current would be the 3.33A on the primary side since there is minimal current across the secondary side.

At full load the currents are: I₁ = 3.33A, I₂ = 20A.

My (full) Load current, Would this involve inserting a resistor for load or would i use the Coil Resistance for both Primary and Secondary side?

You can just assume, that a resistor ( or other load) has been inserted. The value is not of interest in your calculations as for the transformer.

Primary and secondary coil resistances are not "loads".

I don't know what you are asked to calculate, but maybe you need to know cos φ and V₁, V₂.

Edit: Sorry, I've seen it now (V1, V2): 9 kVA transformer.

 

[ErnieChicken]

Hi Hesch,

I have been asked to calculate the Non Load Current and Load Current for the transformer.

Can i find this out using the formula's i have provided above since there is no voltage stated?

Thanks

 

[Hesch]

Well, I assumed that the load currents were given by the exercise (problem statement): 3.33A, 20A. ?

If not: Please attach the exercise.

 

[ErnieChicken]

Everything i have given to you i what i have been given and calculated!
I have already completed these steps

A 6/1 step down transformer has a full load secondary current of 20A and is rated at 9kVA. The copper loss at full load is 110W. The primary winding has a resistance of 0.35Ω.
a) Find the resistance of the secondary coil and the power loss in the secondary coil.
b) 75W of power is drawn by the primary when the secondary is open circuit. Calculate the efficiency at the rated load if the power factor is 0.75.

But i need to do this
Find if The no-load primary current is ≈ 5% less than the full-load primary current.
Find if The primary copper loss on no-load is less than ≈ 1/400 of that on full-load.

There is no mention of voltage but i have the Turns, Amps, Resistance on the windings (R1 = 0.35, R2 = 0.265) , Primary Power Drawn on Open Circuit(75w), Power Loss of Primary (4w) & Secondary (106w), Cooper Loss at Full Load (110w)

Thanks

 

[ErnieChicken]

So if my Full load current or Closed Circuit current is
I1 = 3.33A
I2 = 20A

Can i use the 75w that is drawn under a no load condition since it is open circuit, So If i was to use the 75w / 0.35 Ohms = 214A --> sqrt(214) = 14.127A this give me the 14.127A / 6 = 2.355A you can see within the simulation. This has given me the simulation value i have been given. Also within the simulation, i have used a current source since i have no voltage. Would this be correct?

 

[Hesch]

If the rated power = 9 kVA, and the full load (rated) current, I₂ = 20A, V₂ must be 450V.

You have solved a) correct. So now you know, that full load primary copper losses = 4W.

R₁ ≈ 4W / I₁² = 0,36Ω

75W of power is drawn by the primary when the secondary is open circuit.

Now there is no load. There are no copper losses in the secondary as I₂ = 0. The 75W loss is due to primary copper losses + iron losses.

Iron losses = 75W - 4W = 71W.

R_iron ≈ (6*V₂)² / 71W = 103kΩ

Calculate the efficiency at the rated load if the power factor is 0.75.

There it was: Power factor = cos φ.

But there is a problem: We don't know if the phase ( 41.4° ) is due to the load or the reactances in the transformer? So I really don't know what to do, concerning the efficiency. The efficiency must be calculated as active outputpower/inputpower, and copper losses are calculated by absolute currents. That's why a short-circuit test of a transformer is done: Then you know about reactance in the transformer.

 

[ErnieChicken]

Hi Hersch,

Thanks for that, The efficiency is not an issue as i do not need to calculate that for this example but maybe required if he decides they want to know more.

Ok So i got the voltage part since i had reverted the KvA formula. My V1 = 2700V, V2 = 450V

Now the Iron Losses have been worked out, I have the voltages. Using Ohms law i should be able to use that Resistance to calculate the current, Its gives me a far lower value if i am looking for a 5% value of Non Loaded - Loaded Current.

I really do appreciate your help Hersch. Helping me make more sense of it than i have been taught.

Thanks

 

[Hesch]

Using Ohms law i should be able to use that Resistance to calculate the current,

No, you cannot do this, using only resistors. A model of a transformer also includes inductance, and you don't know the values:

 

[ErnieChicken]

Thanks for the reply,

So the Unloaded current cannot be calculated this way. If we have no circuit on the secondary side, we are only considering the Primary Side Resistance from the windings (0.35 ohms). I am sorry if i am missing the obvious.

Thanks again!

 

[Hesch]

In the model, I_p = I_s + I_e. But you don't know I_e as you don't know the value of X_m. Thus, not knowing I_p, you don't know the primary copper losses.

 

[NascentOxygen]

Now there is no load. There are no copper losses in the secondary as I2 = 0. The 75W loss is due to primary copper losses + iron losses.

Iron losses = 75W - 4W = 71W.

At no load the primary copper losses will drop to a tiny fraction of their full-load value, so almost zero here.

I think it's reasonable to assume the quoted power factor refers to the pf of the secondary's load.

The question involves a low-power power transformer, so transformer reactances are probably neglected for these calculations.

 

[ErnieChicken]

Hi Nascent, Thanks for the reply.

So i have my primary Closed Circuit Current which is the 3.33A i had calculated earlier since i was able to calculate the secondary current. Is that correct?
Via the simulation, to get the calculated values i had, the current source was placed in at 4.71 which gave me the values of the current i needed.

The No Load is Current is 5% less than the full load current. From what Hersch was explaining earlier i need to have an Ie and Xm but from the data i have given above this is all i have.

Thanks

 

[NascentOxygen]

There are 75w of iron losses. These don't change [much] from no load to full load. Have you used this to calculate R_e to account for these iron losses?

How did you come up with a primary current of 4.71A for your simulation?

The No Load is Current is 5% less than the full load current.

Is this stated in the question, or is it something you interjected?

 

[ErnieChicken]

Hi Nascent, Thanks for the reply,

This was stated earlier on in the question since the Load Current is 5% higher than the non load current.

Thanks

 

[ErnieChicken]

From my earlier calculations,
I can confirm the 110W copper loss with I1 = 3.33A R1 = 0.35ohms I2 = 20A R2 = 0.265 Ohms but this is at full load. Which to me says that those currents are for full load.
I have a 75W loss at open circuit on the primary side so can i not rework this to find the current on the primary side?
As for the Full Load Copper Load, i can assume that this is the 110W since it has been calculated at the full load values.

 

[NascentOxygen]

This was stated earlier on in the question since the Load Current is 5% higher than the non load current.

But as stated it doesn't make sense. So we have to try to work out what was intended, and it's not proving easy.

First, have you translated this from another language?

 

[NascentOxygen]

From my earlier calculations,
I can confirm the 110W copper loss with I1 = 3.33A R1 = 0.35ohms I2 = 20A R2 = 0.265 Ohms but this is at full load. Which to me says that those currents are for full load.

Yes, those all seem right.

I have a 75W loss at open circuit on the primary side so can i not rework this to find the current on the primary side?

That's what I asked you! Have you done this calculation?

 

[ErnieChicken]

If i reworked this calculation. Since P = I^2 * R --> I = Sqrt(P / R) --> sqrt(75 / 0.35) = 14.63 but this does not seem right.

 

[ErnieChicken]

Or sqrt(75 x (R1||R2) = 3.36A

 

[NascentOxygen]

Guessing is never the advisable course. Perhaps you should refer back to the diagram that Hesch kindly provided and first identify exactly which resistance value you are trying to calculate?

 

[ErnieChicken]

Since i have only got a Resistance for both Windings i am not sure what i need to calculate.

 

[NascentOxygen]

You have already calculated the copper resistances, so you won't be recalculating either of them.

https://www.physicsforums.com/threads/transformers-under-load-and-no-load.811768/page-2#post-5096562

 

[ErnieChicken]

Ok i have the copper resistance. 4W for Primary, 106w for Secondary with the currents i had posted. If i am correct, this is for the circuit on load since there is a completed circuit of both sides but i need to find the unloaded values for both current and copper resistance.

 

[NascentOxygen]

There is no such thing as unloaded values for copper resistance. The model uses fixed resistances.

Iron losses are represented by r_c in this model: https://www.physicsforums.com/threads/transformers-under-load-and-no-load.811768/#post-5095883

All that is left for you to determine now is the value for r_c using that 75w data value. You can ignore the inductive reactances in this problem, it seems.

 

[ErnieChicken]

There is no such thing as unloaded values for copper resistance. The model uses fixed resistances.

Quoting From Hersch,
"Iron losses = 75W - 4W = 71W.
Riron ≈ (6*V2)2 / 71W = 103kΩ"

This is what Hersch Used, would that be the correct application ?

 

[ErnieChicken]

I am sorry to have wasted your time.

I am going through a rough time right now and my brain has turned to mush which isn't an excuse.

I would like to thank you guys for your help

 

[NascentOxygen]

Quoting From Hersch,
"Iron losses = 75W - 4W = 71W.
Riron ≈ (6*V2)2 / 71W = 103kΩ"

This is what Hersch Used, would that be the correct application ?

I pointed out that it is an error to subtract the 4W, otherwise, yes, that is how to calculate the iron loss resistor in the transformer model.

Don't forget, in the equivalent circuit model when there is no load there is no current in the secondary and therefore there is no current in the primary because at the heart of the equivalent circuit is an ideal transformer. So once you erase the ideal transformer from the circuit because it's taking zero current, all that's left is the path for magnetizing losses, here denoted r _c.

 

[NascentOxygen]

I am sorry to have wasted your time.

You haven't wasted anyone's time!

You have learned probably more than you realize, and besides just you there are a lot of others following this thread. Let me see ... 431 views at last count.

Persistence always pays off in the end.

 

[ErnieChicken]

Thanks man I appreciate it. It must seem like I am a pain as I am struggling to get my head around it.

So for the iron loss (6x450^2) /75w = 92.2kOhms --> 2700 / 97.2k = 0.02777A
Subtract this from my current I calculated earlier 3.33A - 0.2777A = 3.3A which is within the 5%
My question would be since my 3.33a was calculated under condition where the circuit was closed to calculate the secondary winding will it mean it is loaded or would I need to put a resistor inside to determine this?

Thanks for being so patient

 

[NascentOxygen]

So for the iron loss (6x450^2) /75w = 92.2kOhms --> 2700 / 97.2k = 0.02777A

Always write down in words/algebra what you are about to do, before substituting in numbers. This step will help you, and it will help the examiner. If you get the numbers wrong, then at least you may be given some marks for having the right idea. Here you incorrectly show only the 450 being squared. The answer comes to 97.2 kΩ

Subtract this from my current I calculated earlier 3.33A - 0.2777A = 3.3A which is within the 5%

No subtraction. The 3.33A is going through the primary of the ideal transformer in the equivalent circuit. The iron loss current is parallel to this.

My question would be since my 3.33a was calculated under condition where the circuit was closed to calculate the secondary winding will it mean it is loaded or would I need to put a resistor inside to determine this?

The 3.33A primary current is the full load current. (I don't know what you are asking, but I'd say you should never consider "putting a resistor" inside anything that is supplied with 2.7kV!)

Now that you have determined all the unknowns in the transformer equivalent circuit, you are ready to proceed to part (b).

 

[ErnieChicken]

Thanks Nascant,

That has finally clicked in and i was wondering where i was going wrong. When it stated "The Non Load Current must be 5% of the Full Load Current" i was thinking of it being 5% less of the 3.33A but clearly i had misread it. Thanks for that clarification. So i have the 0.0277A (Missed a Decimal place) which is just less than 1% (0.033A) would this be good enough do you think as rounding up or down will affect that value? The Formula (Np x V2)^2 / (Ns * Power Drawn), This is done through P = V^2 / R then multiplied using the turns because of the losses right?

Secondly, The Full Copper loss is 110W across the whole transformer and isolated to the primary it is 4W out of the 110W. The Number i should be looking for is approximately 0.01w under non load since using the 1/400 --> 4 / 400 = 0.01w. Can i use the same formula tweaked?

Thanks for you help. It is much appreciated, It is slowly making sense in my mind.

 

[NascentOxygen]

When it stated "The Non Load Current must be 5% of the Full Load Current"

I don't know where that comes from. It sounds like someone's rule of thumb for a good transformer, no-load current should not exceed 5% of the full-load current, otherwise it is uneconomically lossy?

Secondly, The Full Copper loss is 110W across the whole transformer and isolated to the primary it is 4W out of the 110W. The Number i should be looking for is approximately 0.01w under non load since using the 1/400 --> 4 / 400 = 0.01w. Can i use the same formula tweaked?

No need to guess! Refer to the equivalent circuit. https://www.physicsforums.com/threads/transformers-under-load-and-no-load.811768/#post-5095883

Does it show iron loss current passing through the primary resistance? It does, though it is a simplified model.

Power engineers are refreshingly practical people. As a rule of thumb, they are happy with better than 10% inaccuracy. Ignoring the copper loss due to the magnetizing current introduces way less than 1% error, so ignore it. Consider iron losses as being fixed, regardless of load; copper losses as being due to load current (zero load current, zero copper losses).*

* but you'll find they do become significant when it comes to really big power transformers

 

[ErnieChicken]

I don't know where that comes from. It sounds like someone's rule of thumb for a good transformer, no-load current should not exceed 5% of the full-load current, otherwise it is uneconomically lossy?

This has been specified as a guideline to work to i believe. It just states that it needs to be "no-load primary current is ≈ 5% less than the full-load primary current." But if this is between 0% - 5% i guess this is ok since the high voltage and the low losses.

Does it show iron loss current passing through the primary resistance? It does, though it is a simplified model.

Yes, i understand this. my train of though with this is -> The Iron Loss is effectively in series with the Coil Loss of the 97.2K Ohms and 0.35Ohms which gives me my resistance, I know my Current at non load which is 0.0277A at the primary side. Would that be correct?

Thanks Again!

 

[NascentOxygen]

The Iron Loss is effectively in series with the Coil Loss of the 97.2K Ohms and 0.35Ohms which gives me my resistance, I know my Current at non load which is 0.0277A at the primary side. Would that be correct?

Yes, those are the figures that you've calculated.

 

[ErnieChicken]

Thanks,

Now i know that those figures have been correctly worked out. I have my 4W which is what the loss is at the primary side under full load. At non load, I can work out the resistance using the same method as before to give me a resistance then by using the current i can give myself a power figure for the non load?

 

[NascentOxygen]

I have my 4W which is what the loss is at the primary side under full load. At non load, I can work out the resistance using the same method as before to give me a resistance then by using the current i can give myself a power figure for the non load?

You have already worked these out. You're finished! The no-load loss is the 75W. It is represented by the 97.2kΩ. The copper loss resistance doesn't change.

There is nothing more to find out. Go onto part (b)...

 

[ErnieChicken]

Sorry for the confusion,

This is what i am trying to work out. Part B)

 

[NascentOxygen]

Calculate the efficiency at the rated load if the power factor is 0.75.

Use the resistance values for the transformer model already calculated.

 

[ErnieChicken]

So this would be what i am looking --> Rc (Iron Loss) + Resistance Primary( Np/Ns)^2 which is on the secondary side.

I am truly sorry for being a pain.

Thanks

 

[ErnieChicken]

From Reusing the KvA formula using P = 1000 x 3kVA x 0.75 = 6750W. And approximately 30kW is where it needs to be at full load. (75 x 400 = 30kW)

 

[NascentOxygen]

At non load, I can work out the resistance using the same method as before to give me a resistance then by using the current i can give myself a power figure for the non load?

I looked back to see what you are being asked. You are up to calculating the ratio (at no load) of copper losses : iron losses
You don't even need to involve a particular value of current, because the ratio is determined by the ratio of resistances you have already calculated. You see the current in the model goes though each resistance, there being no other path.

It's "no load" not "non load".

 

[NascentOxygen]

From Reusing the KvA formula using P = 1000 x 3kVA x 0.75 = 6750W. And approximately 30kW is where it needs to be at full load. (75 x 400 = 30kW)

This is something to do with full load when the load pf is 0.75? Can you explain these numbers you are using.

I'm puzzled by the appearance of a figure of 30kW, in light of the transformer rating being just 9kVA.

A 6/1 step down transformer has a full load secondary current of 20A and is rated at 9kVA. The copper loss at full load is 110W. The primary winding has a resistance of 0.35Ω.
a) Find the resistance of the secondary coil and the power loss in the secondary coil.

75W of power is drawn by the primary when the secondary is open circuit.
b) Calculate the efficiency at the rated load if the power factor is 0.75.

 

[ErnieChicken]

Hi Nascant, I can see the confusion. Let me take it back to the beginning and explain the question much better.

A 6/1 step down transformer has a full load secondary current of 20A and is rated at 9kVA. The copper loss at full load is 110W. The primary winding has a resistance of 0.35Ω.

a) Find the resistance of the secondary coil and the power loss in the secondary coil.

• Primary Current = Ns x Is / Np --> 1 x 20A / 6 = 3.33A
• Primary Power Loss = I1^2 x R1 --> 3.33a^2 x 0.35 = 4W
  
• Secondary Coil Resistance: Total Power = I1^2 x R1 + I2^2 x R2 --> 3.33a^2 x 0.35 + 20a^2 x R2 = 110w --> (110w - 3.33a^2 x 0.35) / 20a^2 = 0.265 Ohms
• Secondary Power Loss = I2^2 x R2 --> 20a^2 x 0.265 = 106W

b) 75W of power is drawn by the primary when the secondary is open circuit. Calculate the efficiency at the rated load if the power factor is 0.75

• Efficiency = (Rating x Power Factor) / (Rating x Power Factor) + (Coil Loss + Copper Loss) --> (9000 x 0.75) / (9000 x 0.75) + (75w + 110w) = 97%

The Above was my calculated values for the transformer. Which is where i got my figures from.

1) The no-load primary current is 5% less than the full-load primary current. (You need to establish full load impedance by way of calculations, Which was what i was given today!)

2) The primary copper loss on no-load is less than (1/400) of that on full-load.

Hope this makes more sense. I should have started off with this in the first place. I apologise for that.

 

[NascentOxygen]

That looks right. There may have been no need to calculate the actual value for r_c because it is usually good enough to consider that iron losses stay approximately constant. Though it's still unclear whether you were required to show that the fraction of < 1/400 applies here, or whether it was good enough to assume it does, without proving it.

So it's all finished now?

 

[ErnieChicken]

From what he was saying. "No Load current is less than 5% of the full current load". Whether or not the interpretation here is that it is the 0.028A calculated earlier or it is actually more than than because if so 5% of I1 at full load --> 3.33a = 3.165a

Have i actually calculated an impedance here then?

The 1/400 is the fraction to say that. The Load copper loss is Copper Loss / 400 at no load.

Thanks

 

[ErnieChicken]

1) The no-load primary current is 5% less than the full-load primary current. (You need to establish full load impedance by way of calculations, Which was what i was given today!)

• Primary Full Load Current = 3.33A
• Impedance Primary = 2700V / 3.33A = 810.8 Ohms
• Impedance Secondary = 450v / 20A = 22.5 Ohms
• Total Load Impedance = Ip + Is = 810.8 Ohms + 22.5 Ohms = 833.3 Ohms
• No Load Current = Voltage / R = 2700V / 833.3 Ohms = 3.24A

Would this be correct?

Then for my No load Primary Loss, that would be my 75W which is used above as it is the Open Circuit value for the Efficiency calculation
The Total Full Load Power = Power Factor x V2 x I2 --> 0.75 x 450V x 20A = 6750W