Relative size and Arc Flash Hazards of 50 Hz versus 60 Hz transformers at same KVA

[Sahib]

Good lad! Well done!
You have finally taken note of what I posted in #6 on the first page of this thread.

To keep the flux density the same. Flux is dependent on the volts times time. At 50Hz the volt second area for any half cycle is 20% greater than at 60Hz. So, for the same flux density, you need 20% more core.

Your point is entirely different: it is about increasing the core size for use of 60hz transformer on 50hz; mine is about keeping the same core size for the same purpose.

But not a lot of good if you wanted what the transformer was designed to give - 480V output.
To exemplify how specious your point is, you could use the transformer with 115V into give 4V out.

Here the discussion is whether the use of 60hz transformer on 50hz is possible in principle.

 

[Besoeker]

mine is about keeping the same core size for the same purpose.

You simply can't.
You have already demonstrated that point:

So a transformer rated at 13800/480V, 60Hz may be used at 11500/400V, 50Hz with its KVA derated by 17%.

Different voltages and a different kVA rating. You cannot reasonably claim that an 83kVA transformer serves the same purpose. Nor that a 400V output serves the same purpose a 480V output.

Here the discussion is whether the use of 60hz transformer on 50hz is possible in principle.

That may be your discussion but, in any case, you again put that issue to bed yourself.

So the no-load current under 50hz operation is 0.05 x 81=4.05 times transformer rated current.

Not tenable.

 

[Sahib]

So the no-load current under 50hz operation is 0.05 x 81=4.05 times transformer rated current.
Not tenable.

Why?

 

[Besoeker]

Why?

In the infamous words of John McEnroe.."You can?t be serious, man. YOU CANNOT BE SERIOUS! ...."

 

[Sahib]

B:

Of late, what flaw you detected in my reasoning?

Perhaps it dawned on you that the source impedance and transformer resistance would prevent the current from reaching that high?

 

[Besoeker]

B:

Of late, what flaw you detected in my reasoning?

Perhaps it dawned on you that the source impedance and transformer resistance would prevent the current from reaching that high?

And perhaps the moon really is made of green cheese.
:roll:

 

[Sahib]

B:
Please explain in technical language.
I am not understanding you.

 

[Besoeker]

B:
Please explain in technical language.
I am not understanding you.

That I guessed.
You and I both know that the moon isn't actually made of green cheese.
And I know that that you know that the 50Hz option isn't viable.

Drop it.

 

[Sahib]

Okay.
Let us continue to arc flash hazard.
The 13800/480V, 60Hz transformer when used at 11500/400V, 50Hz has a higher arc hazard because its source impedance is lower.

 

[Besoeker]

Okay.
Let us continue to arc flash hazard.
The 13800/480V, 60Hz transformer when used at 11500/400V, 50Hz has a higher arc hazard because its source impedance is lower.

And the voltage is also lower.

 

[Sahib]

The voltage drop across an arc from a 13800/480V transformer or a 11500/400V transformer may be same.
If I1 and Z1 are the arc current and impedance of 11500/400V transformer and
I2 and Z2 are the arc current and impedance of 13800/480V transformer, then

I1=400/Z1 ...........(1)

I2=400/Z2

So for equality of voltage drops,

400 -I1*Z1=480-I2*Z2
I2*Z2-I1*Z1=80
I2*(Z2/Z1)-I1=80/Z1
I2*[0.2*(I1/I2)=80/Z1
0.2*I1=80/Z1
I1=400/Z1 same as (1) above.

 

[Besoeker]

The voltage drop across an arc from a 13800/480V transformer or a 11500/400V transformer may be same.
If I1 and Z1 are the arc current and impedance of 11500/400V transformer and
I2 and Z2 are the arc current and impedance of 13800/480V transformer, then

I1=400/Z1 ...........(1)

From which you would get:
I1*Z1 = 400.

Surely you don't mean that?

So for equality of voltage drops,

OK. You want equal volt drops?
Let's pick a number. Say 20V
And, very obviously
400 -20 is not equal to 480-20

 

[Sahib]

The calculation was done for the case of arc voltage drop assumed to be zero.
For non-zero arc voltage drop, the argument may be made along similar line.

 

[Besoeker]

The calculation was done for the case of arc voltage drop assumed to be zero.
For non-zero arc voltage drop, the argument may be made along similar line.

From which you would get:
I1*Z1 = 400.

Surely you don't mean that?

Or do you?

 

[Sahib]

From which you would get:
I1*Z1 = 400.

Surely you don't mean that?

Or do you?

If the arc voltage drop is assumed to be zero, then I1*Z1 = 400V. Surely I mean that, because I have no intention to challenge Kirchoff's voltage law.

 

[Besoeker]

If the arc voltage drop is assumed to be zero, then I1*Z1 = 400V. Surely I mean that, because I have no intention to challenge Kirchoff's voltage law.

If

I1*Z1 = 400V

then it follows that

400 -I1*Z1=0

and your equation:

400 - I1*Z1 = 480 - I2*Z2

boils down to

0 = 0

 

[Sahib]

Yes, but it does not contradict the initial assumption that arc voltage drop is zero.

 

[Besoeker]

Yes, but it does not contradict the initial assumption that arc voltage drop is zero.

Here's the contradiction:

So for equality of voltage drops,

400 -I1*Z1=480-I2*Z2

Either the voltage drops, the one most people would take to be the IZ term, are not equal or that equation is wrong.

 

[GoldDigger]

Here's the contradiction:



Either the voltage drops, the one most people would take to be the IZ term, are not equal or that equation is wrong.

Either I₁ is not equal to I₂ or Z₁ is not equal to Z₂ or both. No problem with that.

 

[Besoeker]

Either I₁ is not equal to I₂ or Z₁ is not equal to Z₂ or both. No problem with that.

But, for equal volt drops, the IZ products have to be equal whatever combination of I and Z you use.