Rearrange equations to calculate the Xpu impedance

  • Thread starter David J
  • Start date
  • #1
David J
Gold Member
138
15

Homework Statement


[/B]
Given the relationships:

##X pu =\frac{\text{MVA base}}{\text{MVA fault}}##

##=\frac{\text{MVA base}}{\text{MVA rated}}\frac{\text{X%}}{100}##

##\text{fault MVA}=\sqrt{3}\text{VIfault}##


##X\Omega=\frac{V}{\sqrt{3}}I fault##

Show that: -

(a) ##\text{fault MVA}=\frac{\text{MVA rated}}{V^2}##


(b) ##X pu=\frac{\text{MVA base}}{V^2}*X\Omega##

and hence calculate the pu impedance of 100 m of 3 single core 415 V cables in trefoil. Take the cable impedance to be 90 microhms per meter. Use a 1 MVA base. Neglect the resistance of the cables.

Homework Equations


[/B]
##X pu =\frac{\text{MVA base}}{\text{MVA fault}}##

##=\frac{\text{MVA base}}{\text{MVA rated}}\frac{\text{X%}}{100}##

##\text{fault MVA}=\sqrt{3}\text{VIfault}##


##X\Omega=\frac{V}{\sqrt{3}}I fault##

The Attempt at a Solution



My problem is that I do not fully understand these "show that" questions. I am given 4 equations to begin with and I believe I need to show how: -

##\text{fault MVA}=\frac{\text{MVA rated}}{V^2}##

and

##X pu=\frac{\text{MVA base}}{V^2}*X\Omega##

are derived from: -

##X pu =\frac{\text{MVA base}}{\text{MVA fault}}##

##=\frac{\text{MVA base}}{\text{MVA rated}}\frac{\text{X%}}{100}##

##\text{fault MVA}=\sqrt{3}\text{VIfault}##

##X\Omega=\frac{V}{\sqrt{3}}I fault##


Could someone please give me a clue as to how to go about solving this with perhaps a small clue as to where I should be concentrating first
[/B]
I have calculated the pu impedance of the cable to be: -

##X=100*90*10^{-6}=9*10^{-3}##

##X pu=9*10^{-3}*\frac{1}{(0.415)^{2}}=0.052257 pu##


 

Answers and Replies

  • #2
David J
Gold Member
138
15
Can anyone help with this please?

 
  • #3
David J
Gold Member
138
15
I will close this as my time has run out now. Its marked un solved but not solved.
 

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