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chicago

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Why is the terminal current = 1.0/_0 ?

I think the zero degrees is because of unity power factor, correct?

But I_base = 150MVA / 13.8kV = 6.27kA

So, then the magnitude I_pu should be equal to I_actual / I_base, isn't it?

What values do the term "rated" MVA and "rated" voltage refer to in the question?

 
chicago said:
Why is the terminal current = 1.0/_0 ?
I think the zero degrees is because of unity power factor, correct?

But I_base = 150MVA / 13.8kV = 6.27kA

So, then the magnitude I_pu should be equal to I_actual / I_base, isn't it?

What values do the term "rated" MVA and "rated" voltage refer to in the question?
Click to expand...
Sheat!!!!...Sorry I cannot help right now but I will try. As far as I remember rated MVA and rated Voltage are referred to the MVA and the Voltage at the nameplate of the equipment.

 
chicago said:
Why is the terminal current = 1.0/_0 ?
I think the zero degrees is because of unity power factor, correct?
Click to expand...
Correct. The unity power factor results in no lead/lag for the current.

chicago said:
But I_base = 150MVA / 13.8kV = 6.27kA
So, then the magnitude I_pu should be equal to I_actual / I_base, isn't it?
Click to expand...
For this problem, you never have to find the base current. If you are at rated Voltage and MVA, you're pu current is 1.0 by definition.

Note: Looks like you included the sqrt 3 value in calculating the rated current but didn't type it here.

chicago said:
What values do the term "rated" MVA and "rated" voltage refer to in the question?
Click to expand...
They are basically telling you that the values for power, voltage, and current are all 1.0 pu.

 
I might be totally wrong here someone correct me if I am and if there is more to this. But I got the answer having never attempted this or seen it until this post.

Maybe this was just dumb luck or an example of knowing your resources in this case John J Grainger.

I started the same way..calculated IBase then reread the question and noticed the reactance was all 2.5 pu for 3 of 4 answers.

The question asks what values to use to perform a simple transient study (i dont have it here) if you have Grainger see page 136. In brief for steady state use synchronous reactance (Xd) for X"d is the Subtransient reactance. So use Xd' is the Transient reactance.

Solving for the Generator Internal Voltage E=Vt+jXd'I = 1/_0xj.25(1 as Jim said)=1.0307

Sorry if I just restated what was already said or in the solution.

John

 
jdd18vm said:
I might be totally wrong here someone correct me if I am and if there is more to this. But I got the answer having never attempted this or seen it until this post.
Maybe this was just dumb luck or an example of knowing your resources in this case John J Grainger.

I started the same way..calculated IBase then reread the question and noticed the reactance was all 2.5 pu for 3 of 4 answers.

The question asks what values to use to perform a simple transient study (i dont have it here) if you have Grainger see page 136. In brief for steady state use synchronous reactance (Xd) for X"d is the Subtransient reactance. So use Xd' is the Transient reactance.

Solving for the Generator Internal Voltage E=Vt+jXd'I = 1/_0xj.25(1 as Jim said)=1.0307

Sorry if I just restated what was already said or in the solution.

John
Click to expand...
John,

Looks like you have this one nailed to me. The key to the problem is that you want to use the transient reactance (X'd) rather than the subtransient reactance (X''d).

Jim

 

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