X/R Ratio and 3 phase fault

[katag]

I came across this question in a practice test and don't understand the concept, nor do I have the actual solution, just the answer.

What is the equivalent system impedance for a 3-phase fault at the high voltage bus of a 15/7.2kV transformer. The following information about the fault is given:

200MVA

X/R = 3.8 @ 15kV

A. .45 + j1.54 ohm

B. .29 + j1.09 ohm

C . 52 + j1.41ohm

D. .71 + j2.5 ohm

Thanks in advance.

 

[shilmag]

This is how I would solve it:

we know that Z= sqrt( R2 + X2),

we also know that Z = V2/S = (15,000KV)2/200MVA = 1.126

Using the X/R = 3.8 and the Z value in the first equation you can solve for R, and X = 3.8R

So answer is B. 0.29 + j1.09

 

[elminses]

Here is the extra math, I had to do it in order for it all to make sense to me.

z=sqrt( R^2 + (3.8R)^2)

1.126 = sqrt( R^2 + 14.46R^2)

1.268= 15.46R^2

.082=R^2

.286=R

1.126=sqrt(.29^2 + X^2)

1.1839= X^2

1.089=X

.29 + j1.09

 

[katag]

Thanks for the help. Makes sense now!

 

[ecbahr]

Another option:

Z=V2/S

Ɵ=tan-1(X/R)

X=ZsinƟ

R=ZcosƟ

Therefore:

Z=(15,000KV)2/200MVA = 1.126

Ɵ=tan-1(3.8)=75.25

X=1.126sin(75.25)=1.09

R=1.126cos(75.25)=0.29

 

[pbo064]

A practice problem like this is Complex Imaginary test 2 question 42. He uses ecbahr's method in the solutions.

 

[Guest]

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