Fault Calculation - Per Unit System 

By on March 27th, 2013

Symbol   Definition
Image(11)_thumb[1] - per unit method current base
Image(25)_thumb[1] - per unit method power base
Image(26)_thumb[1] - per unit method voltage base
Image(17)_thumb[1] - per unit method admittance base
Image(14)_thumb[1] - per unit method impedance base

image37 - percentage impedance

image34 - per unit impedance
image35 - actual impedance
image36 - per unit base impedance

Per unit fault calculations is a method whereby system impedances and quantities are normalised across different voltage levels to a common base.  By removing the impact of varying voltages, the necessary calculations are simplified.  

To use the per unit method, we normalise all the system impedances (and admittances) within the network under consideration to a common base.  These normalised impedances are know as per unit impedances.  Any per unit impedance will have the same value on both the primary and secondary of a transformer and is independent of voltage level.

A network of per unit impedances can then be solved using standard network analysis (see the example).  From this fault level can be readily determined. 

In applying the per unit method, the first step is to select an arbitrary voltage (Vbase) and power (Pbase) base. 

Tip: while the base power and voltage be any value, typically it would make sense to select values related to the system under construction (for example 11 kV and 20 MVA may be appropriate for a distribution type system)

Per Unit Method

Having selected a base power and voltage, the base per unit values of impedance, admittance and current can be calculated from:

Per Unit Single Phase Three Phase

Image(11)

Image(12)

Image(13)

Image(14)

Image(15)

Image(16)

Image(17)

Image(18)

Image(18)

 

Dividing a system element by it’s per-unit base value gives the per-unit value of the element, for example

image53

Some times per-unit values are available for a given base kV, but the problem being solved is using a different base.  In this instance it is possible to convert the unit:

{Z_{pu(new)}} = {Z_{pu(old)}}{\left( {\frac{{baseK{V_{old}}}}{{baseK{V_{new}}}}} \right)^2} \times \left( {\frac{{baseKV{A_{new}}}}{{basekV{A_{old}}}}} \right)

Fault calculation problems typically deal wit power sources, generators, transformers and system impedances.  Per-unit values for these elements can be quickly derived from:

 

Element Per-Unit Value
Source impedance image55
Generators image57
Transformers image58
Impedances image56, where V is in kV

 

Example - calculating per unit values

Consider a system of source impedance 4.48 Ω connected to a 20 MVA transformer (11/0.4 kV) at 6% impedance.  We want to find the fault level at the transformer secondary.

Selecting Pbase as 20 MVA and Vbase as 11 kV and using the above equations:

{I_{base}} = \frac{{{P_{base}}}}{{{V_{base}}\sqrt 3 }} = \frac{{20}}{{11 \times \sqrt 3 }} = 1.049\quad kA

and

 Image(20)

 

Image(21)

 

Image(22)

 

Image(23)

 

  the Line-Neutral voltage on the secondary of the transformer is 0.4/√3 = 0.230 kV, giving:

Image(24)

Three Phase Fault Example

PerUnitThreePhaseFault
Three Phase Fault Example

Per unit analysis can be used to calculate system three phase fault levels and the current distributions.  To gain a better understanding, it is worth running through the typical steps required to solve a fault calculation problem.  

Given the system single line diagram, construct and simplify the per unit impedance diagram.

The fault level at the point under consideration is given by:

image59

Where Zpu, is the total impedance between the source and the fault.

PerUnitThreePhaseFaultEx1

Fault flow through parallel branches is given by the ratio of impedances. As illustrated this can enable fault flows to be found through each branch.

PerUnitThreePhaseFaultEx2

Having calculated the fault flow in each branch, it is then relatively simple to find the current distribution using:

image60

where:

image61

Related Notes



Steven McFadyen's avatar Steven McFadyen

Steven has over twenty five years experience working on some of the largest construction projects. He has a deep technical understanding of electrical engineering and is keen to share this knowledge. About the author


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  1. FELIX JR.'s avatar FELIX JR. says:
    5/21/2013 7:52 AM

    greetings !, hello !

    this article which you have shared is a good explanation for short circuit current calculations using the P.U. Method based on a given kVA base, as this is the old school way (i.e. during my college days and my early prof. work with my employer).

    But how do you find and comment on the IEC-60909 way of using the ohmic method and also that of the MVA method, alternative ? So, which of the three (3) are more easy to use and accurate ?

    felix jr.

  2. gilly1's avatar gilly1 says:
    5/29/2013 6:41 PM

    Just wondering which method do you think the most accurate for calculating single phase-earth faults, the Ohmic Method or Symmetrical Component Method.
    The Symm. Comp. method is quite complex and while a lot of people use it to solve complex types of unbalanced faults it is not real in the sense that there aren't really 3 currents flowing at once : Pos, Neg and Zero sequence currents, whereas the Ohmic Method is based in reality with real impedances???

  3. Steven's avatar Steven says:
    5/30/2013 3:03 AM

    My opinion is that symmetrical components would be more accurate (especially for any unbalanced faults like phase to earth).

    If your looking for more practical guidance on using symmetrical components, then I would reefer to IEC 60909. This standard is also implemented in many software programs.


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