- #1

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## 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

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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##