Per-Unit Method Short Circuit Calculation

I. Introduction

In a previous post, the Impedance Method of Short-Circuit Calculation has been presented. To provide a comparative analysis on the accuracy of each calculation, the example provided in the impedance method will be used in this article.

Per unit value of any quantity is defined as the ratio of actual value to the chosen base value in the same unit.

Per Unit Impedance

$Z_{pu}~=~{Z}/{Z_b}~=~Z~*~{{MVA_b}/{{kV_b}^2}}~~~~~{right}~eq.1$

where:

$Z_{pu}~=~per~unit~impedance$

Change of Base

The impedances are usually specified on the rating of the equipment. These impedances need to be changed into PU values from the base of the equipment rating (old value) to that of the chosen system base (new value).

$Z_{pu}_{new}~=~Z_{pu}_{old}~*~{{MVA_b_new}/{{kV_b}^2}_new}~*~{{{{kV_b}^2}_old}/{MVA_b_old}}~~~~~{right}~eq.2$

$X_{pu}_{new}~=~X_{pu}_{old}~*~{{MVA_b_new}/{{kV_b}^2}_new}~*~{{{{kV_b}^2}_old}/{MVA_b_old}}~~~~~{right}~eq.3$

II. Example Calculation

Figure 1 Impedance Method Example — Figure 1 Per-Unit Method Example

From Figure 1, given the parameters below, calculate the 3-phase fault and X/R ratio at Fault 1, Fault 2 and Fault 3:

Utility
Available Fault = 436 MVA
X/R = 15
Cable 1
Length = 1300 m
Unit Resistance = 0.39 ohms/km
Unit Reactance = 0.039 ohms/km
Number of parallel conductors / phase = 1
Transformer
Primary Voltage = 13.8 kV
Secondary Voltage = 0.48 kV
Capacity = 2 MVA
%Z = 5.75%
X/R = 5.662
Cable 2
Length = 500 m
Unit Resistance = 0.048 ohms/km
Unit Reactance = 0.029 ohms/km
Number of parallel conductors / phase = 2

Conversion of Actual Values to Per Unit

Choosing some arbitrary base values

At 13.8kV (Transformer Primary Side):
MVA_b = 2MVA
kV_b = 13.8kV
At 0.48kV (Transformer Secondary Side):
MVA_b = 2MVA
kV_b = 0.48kV

From the above selected base values, the PU equivalent of actual values will be:

Utility
Available Fault = 436 MVA
X/R = 15

PU values

Zpu1 = 0.004587156 pu

Rpu1 = 0.000305133 pu

Xpu1 = 0.004576996 pu
Cable 1
Length = 1300 m
Unit Resistance = 0.39 ohms/km
Unit Reactance = 0.039 ohms/km
Number of parallel conductors / phase = 1

PU values

Zpu2 = 0.005351068 pu

Rpu2 = 0.005324512 pu

Xpu2 = 0.000532451 pu
Transformer
Primary Voltage = 13.8 kV
Secondary Voltage = 0.48 kV
Capacity = 2 MVA
%Z = 5.75%
X/R = 5.662

PU values

Zpu3 = 0.0575 pu

Rpu3 = 0.010000644 pu

Xpu3 = 0.056623645 pu
Cable 2
Length = 500 m
Unit Resistance = 0.048 ohms/km
Unit Reactance = 0.029 ohms/km
Number of parallel conductors / phase = 2

PU values

Zpu4 = 0.121702039 pu

Rpu4 = 0.104166667 pu

Xpu4 = 0.062934028 pu

Please note that the base kV for this cable is 0.48kV

III. Fault Calculations

From the above calculated PU values, it is now possible to calculate the short-circuit currents at the different fault locations.

Fault 1 (13.8kV)

$R_{pu}~=~R1_{pu}~+~R2_{pu}$
$R_{pu}~=~0.000305133~+~0.005324512$
$R_{pu}~=~0.005629645$

$X_{pu}~=~X1_{pu}~+~X2_{pu}$
$X_{pu}~=~0.004576996~+~0.000532451$
$X_{pu}~=~0.005109447$

$Z_{pu}~=~R_pu~+~jX_pu$
$Z_{pu}~=~0.005629645~+~j0.005109447$
$Z_{pu}~=~0.007602588{lt}{42.2}^{o}$

From eq. 1
${Z}~=~0.007602588~*~{{13.8^2}/{2}}$

Calculating the current at Fault1 @ 13.8kV:
$I_{fault1}~=~{1000*kV}/{sqrt{3}*Z}$
$I_{fault1}~=~{1000*13.8}/{sqrt{3}*0.723918475}$

$I_{fault1}~=~11,005.98202~A$

${X/R}~=~tan(42.2^o)~=~0.91$

Fault 2 (0.48kV)

$R_{pu}~=~R1_{pu}~+~R2_{pu}~+~R3_{pu}$
$R_{pu}~=~0.000305133~+~0.005324512~+~0.010000644$
$R_{pu}~=~0.015630288$

$X_{pu}~=~X1_{pu}~+~X2_{pu}~+~X3_{pu}$
$X_{pu}~=~0.004576996~+~0.000532451~+~0.056623645$
$X_{pu}~=~0.063681085$

$Z_{pu}~=~R_pu~+~jX_pu$
$Z_{pu}~=~0.015630288~+~j0.063681085$
$Z_{pu}~=~0.063681085{lt}{75.79}^{o}$

From eq. 1
${Z}~=~0.063681085~*~{{0.48^2}/{2}}$

Calculating the current at Fault2 @ 0.48kV:
$I_{fault2}~=~{1000*kV}/{sqrt{3}*Z}$
$I_{fault2}~=~{1000*0.48}/{sqrt{3}*0.007336061}$

$I_{fault2}~=~ 37,776.15~A$

${X/R}~=~tan(75.79^o)~=~3.95$

Fault 3 (0.48kV)

$R_{pu}~=~R1_{pu}~+~R2_{pu}~+~R3_{pu}~+~R4_{pu}$
$R_{pu}~=~0.000305133~+~0.005324512~+~0.010000644~+~0.104166667$
$R_{pu}~=~0.119796955$

$X_{pu}~=~X1_{pu}~+~X2_{pu}~+~X3_{pu}~+~X4_{pu}$
$X_{pu}~=~0.004576996~+~0.000532451~+~0.056623645~+~0.062934028$
$X_{pu}~=~0.12466712$

$Z_{pu}~=~R_pu~+~jX_pu$
$Z_{pu}~=~0.119796955~+~j0.12466712$
$Z_{pu}~=~0.172896504{lt}{46.14}^{o}$

From eq. 1
${Z}~=~0.172896504~*~{{0.48^2}/{2}}$

Calculating the current at Fault2 @ 0.48kV:
$I_{fault3}~=~{1000*kV}/{sqrt{3}*Z}$
$I_{fault3}~=~{1000*0.48}/{sqrt{3}*0.019917677}$

$I_{fault3}~=~ 13,913.68 ~A$

${X/R}~=~tan(46.14^o)~=~1.04$

IV. Impedance Method vs. Per Unit Method

The results of the Impedance Method and Per Unit Method are the same hence the accuracy of Per Unit Method is comparable to the Impedance Method.