Rearrange equations to calculate the Xpu impedance

[David J]

Homework Statement

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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

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

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

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

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

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: -

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

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

 

[David J]

Can anyone help with this please?

 

[David J]

I will close this as my time has run out now. Its marked un solved but not solved.