Transformers Under Load and No Load

[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