Calculating Per Unit Fault Impedance: Finding Errors in Given Relationships

In summary, the given relationships show that Xpu can be calculated using the formula Xpu = (MVAbase / MVArated) * (X% / 100) and that MVAfault can be calculated using the formula MVAfault = sqrt(3) * V * Ifault. Additionally, the formula Xohms = [V / sqrt(3)] / Ifault can be used to calculate Xpu, which is equivalent to the formula Xpu = [MVAbase / V^2] * Xohms. However, there is a typo in the given equation for Xohms, which should be Xohms = [V / sqrt(3)]
  • #1
acw260
14
0

Homework Statement


Given the relationships:

Xpu = MVAbase / MVAfault

Xpu = ( MVAbase / MVArated ) * ( X% / 100 )

MVAfault = sqrt(3) * V * Ifault

Xohms = [ V / sqrt(3) ] * Ifault

Homework Equations



Show that:

Xpu = [ MVAbase / V^2 ] * Xohms

The Attempt at a Solution



No matter how I approach this problem I keep finding that one of the given relationships must be incorrect. Could someone with better algebra skills then me please take a look and confirm whether (a) there is an error in the set problem or (b) I'm just being dense?

Thanks
 
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  • #2
I have no idea of what you are trying to ask - but "Xohms = [ V / sqrt(3) ] * Ifault" seems very wrong. Resistance (ohms) is always Voltage divided by current.
 
  • #3
Please see the attachment for the problem in its original format, which should make things clearer.

I have solved the derivation of equation 5(a).

I can solve the derivation of equation 5(b) no problem using "V=IX" (ignoring the resistance of the cables for the purposes of the fault calculation, as instructed in the problem).

What I can't see is how the given equation "Xohms = [ V / sqrt(3) ] * Ifault" fits into the derivation of equation 5(b). This equation seems wrong.
 

Attachments

  • per unit impedance problem.pdf
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  • #4
acw260 said:
What I can't see is how the given equation "Xohms = [ V / sqrt(3) ] * Ifault" fits into the derivation of equation 5(b). This equation seems wrong.
A typo. It should be "Xohms = V /[sqrt(3) * Ifault]".
 
  • #5
acw260 said:
Xohms = [ V / sqrt(3) ] * Ifault

Svein said:
A typo. It should be "Xohms = V /[sqrt(3) * Ifault]".

Or, equivalently, Xohms = [ V / sqrt(3) ] / Ifault
 

1. What is per unit fault impedance?

Per unit fault impedance refers to the normalized value of fault impedance in a power system. It is used to compare the impedance of different components in the system, regardless of their actual values.

2. Why is per unit fault impedance important in power systems?

Per unit fault impedance is important because it allows for easy comparison of fault levels in different parts of a power system. It also helps in determining the fault current and voltage levels in the system, which is crucial for protection and coordination of equipment.

3. How is per unit fault impedance calculated?

Per unit fault impedance is calculated by dividing the actual impedance of a component by the base impedance of the system. The base impedance is usually determined based on the rating of the power system and can be expressed in ohms or per unit values.

4. What factors can affect per unit fault impedance?

The main factors that can affect per unit fault impedance include the type and length of the transmission lines, transformer ratings, and the type of fault (e.g. three-phase, single-phase, etc.). The impedance also varies with the frequency of the system.

5. How is per unit fault impedance used in fault analysis?

Per unit fault impedance is used in fault analysis to determine the fault current and voltage levels at different points in the power system. This information is crucial for designing protection schemes and ensuring the safe and efficient operation of the system.

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