Faults on Power System - Free Electrical Engineering Board Exam Reviewer

LessonAbout 2 h

Faults Calculations

Fault is any abnormal connection or situation in an electrical power system that causes voltage and current imbalance among the three phases.

1. Symmetrical Components

Fortescue Theorem: any unbalanced 3-phase vector set breaks into 3 balanced vector sets
Sequence Sets: positive sequence, negative sequence, and zero sequence components
Symmetrical Analysis: decouples unbalanced system equations into independent sequence networks

2. Positive Sequence Components

Definition: 3 equal vectors displaced by \(120^\circ\) having original \(a-b-c\) phase sequence
Subscript Convention: denoted by subscript 1 (\(V_{a1}\), \(V_{b1}\), \(V_{c1}\))
Operator Relations: \(V_{b1}\) lags \(V_{a1}\) by \(120^\circ\), \(V_{c1}\) leads \(V_{a1}\) by \(120^\circ\)

Positive Sequence Vector Relations

\[ V_{b1} = a^2 V_{a1} \qquad\qquad V_{c1} = a V_{a1} \]

Where: \(V_{a1}, V_{b1}, V_{c1}\) = positive sequence phase voltages (V), \(a = 1\angle 120^\circ\) = complex vector operator, \(a^2 = 1\angle 240^\circ = 1\angle -120^\circ\)

3. Negative Sequence Components

Definition: 3 equal vectors displaced by \(120^\circ\) having reversed \(a-c-b\) phase sequence
Subscript Convention: denoted by subscript 2 (\(V_{a2}\), \(V_{b2}\), \(V_{c2}\))
Operator Relations: \(V_{b2}\) leads \(V_{a2}\) by \(120^\circ\), \(V_{c2}\) lags \(V_{a2}\) by \(120^\circ\)

Negative Sequence Vector Relations

\[ V_{b2} = a V_{a2} \qquad\qquad V_{c2} = a^2 V_{a2} \]

Where: \(V_{a2}, V_{b2}, V_{c2}\) = negative sequence phase voltages (V), \(a = 1\angle 120^\circ\) = complex vector operator, \(a^2 = 1\angle 240^\circ\)

EE Board Exam March 1998

Example. The three-phase unbalanced currents are \(I_a = 10\angle -30^\circ\ \text{A}\), \(I_b = 0\), \(I_c = 10\angle 150^\circ\ \text{A}\). Find the negative sequence current of phase a.

Solution.

\[ I_{a2} = \dfrac{1}{3}(I_a + a^2 I_b + a I_c) = \dfrac{1}{3}[10\angle -30^\circ + 0 + (1\angle 120^\circ)(10\angle 150^\circ)] = 5.77\angle -60^\circ\ \text{A} \]

Answer: \(5.77\angle -60^\circ\ \text{A}\)

4. Zero Sequence Components

Definition: 3 equal vectors equal in magnitude and in phase with each other
Subscript Convention: denoted by subscript 0 (\(V_{a0}\), \(V_{b0}\), \(V_{c0}\))
Ground Path: zero sequence current requires a neutral or ground return connection

Zero Sequence Vector Relations

\[ V_{a0} = V_{b0} = V_{c0} \]

Where: \(V_{a0}, V_{b0}, V_{c0}\) = zero sequence phase voltages (V)

EE Board Exam October 1998

Example. The three-phase unbalanced currents are \(I_a = 10\angle -30^\circ\ \text{A}\), \(I_b = 0\), \(I_c = 10\angle 150^\circ\ \text{A}\). Find the zero sequence current.

Solution.

\[ I_{a0} = \dfrac{1}{3}(I_a + I_b + I_c) = \dfrac{1}{3}(8.66 - j5 + 0 - 8.66 + j5) = 0\ \text{A} \]

Answer: \(0\ \text{A}\)

5. General Formulas on Symmetrical Components

Vector Operator \(a\): rotates a phasor counterclockwise by \(120^\circ\)
Zero Vector Sum: \(1 + a + a^2 = 0\)
Synthesis Equations: reconstruct original phase quantities from sequence components

Complex Operator \(a\) Identities

\[ a = 1\angle 120^\circ = -0.5 + j0.866 \qquad\qquad a^2 = 1\angle 240^\circ = -0.5 - j0.866 \qquad\qquad 1 + a + a^2 = 0 \]

Sequence Component Transformation Formulas

\[ V_{a0} = \dfrac{1}{3}(V_a + V_b + V_c) \]

\[ V_{a1} = \dfrac{1}{3}(V_a + a V_b + a^2 V_c) \]

\[ V_{a2} = \dfrac{1}{3}(V_a + a^2 V_b + a V_c) \]

Where: \(V_{a0}, V_{a1}, V_{a2}\) = zero, positive, and negative sequence components of phase a (V), \(V_a, V_b, V_c\) = unbalanced phase voltages (V)

Phase Synthesis Formulas

\[ V_a = V_{a0} + V_{a1} + V_{a2} \]

\[ V_b = V_{b0} + V_{b1} + V_{b2} = V_{a0} + a^2 V_{a1} + a V_{a2} \]

\[ V_c = V_{c0} + V_{c1} + V_{c2} = V_{a0} + a V_{a1} + a^2 V_{a2} \]

Where: \(V_a, V_b, V_c\) = synthesized phase voltages (V), \(V_{a0}, V_{a1}, V_{a2}\) = sequence components of phase a (V)

Example. Given three unbalanced three-phase voltages \(V_a = 150 + j0\ \text{V}\), \(V_b = -90 - j120\ \text{V}\), \(V_c = -120 + j90\ \text{V}\). Determine \(V_{a1}\).

Solution.

\[ V_{a1} = \dfrac{1}{3}(V_a + a V_b + a^2 V_c) = \dfrac{1}{3}[150\angle 0^\circ + (1\angle 120^\circ)(150\angle -126.87^\circ) + (1\angle 240^\circ)(150\angle 143.13^\circ)] = 145.62 + j13.66\ \text{V} \]

Answer: \(145.62 + j13.66\ \text{V}\)

6. Per Unit Impedance Using a Different Power and Voltage Base

Per Unit Definition: ratio of actual quantity to chosen base quantity
Base Change: converts per-unit impedance to a new MVA or kV base

Per Unit Impedance Base Conversion

\[ Z_{\text{pu2}} = Z_{\text{pu1}} \left(\dfrac{V_{\text{base1}}}{V_{\text{base2}}}\right)^2 \left(\dfrac{S_{\text{base2}}}{S_{\text{base1}}}\right) \]

\[ Z_{\text{pu}} = \dfrac{Z_{\Omega} S_{\text{base}}}{(V_{\text{base}})^2} \]

Where: \(Z_{\text{pu1}}, Z_{\text{pu2}}\) = original and new per-unit impedances, \(V_{\text{base1}}, V_{\text{base2}}\) = original and new voltage bases (kV), \(S_{\text{base1}}, S_{\text{base2}}\) = original and new apparent power bases (kVA or MVA), \(Z_{\Omega}\) = actual impedance (\(\Omega\))

Example. A per-unit impedance \(Z_{\text{pu1}} = 0.03\ \text{pu}\) based on \(34.5\ \text{kV}\) and \(50\ \text{MVA}\). Convert to a new base of \(33\ \text{kV}\) and \(100\ \text{MVA}\).

Solution.

\[ Z_{\text{pu2}} = 0.03 \left(\dfrac{34.5}{33}\right)^2 \left(\dfrac{100}{50}\right) = 0.0549\ \text{pu} = 5.49\% \]

Answer: \(5.49\%\)

7. Sequence Networks: Zero, Positive, Negative

Sequence Network: per-phase equivalent circuit corresponding to one sequence set
Voltage Source Location: internal generated emfs (\(E_a\)) exist in positive sequence network only
Passive Networks: negative and zero sequence networks contain no internal source voltages

Sequence Network Voltage Equations

\[ V_{a1} = E_a - I_{a1} Z_1 \qquad\qquad V_{a2} = -I_{a2} Z_2 \qquad\qquad V_{a0} = -I_{a0} Z_0 \]

Where: \(V_{a1}, V_{a2}, V_{a0}\) = sequence terminal voltages of phase a (pu or V), \(E_a\) = pre-fault open-circuit phase a voltage (pu or V), \(I_{a1}, I_{a2}, I_{a0}\) = sequence currents of phase a (pu or A), \(Z_1, Z_2, Z_0\) = positive, negative, and zero sequence Thevenin impedances (pu or \(\Omega\))

8. Single Line to Ground Fault

Series Connection: positive, negative, and zero sequence networks connected in series
Equal Sequence Currents: \(I_{a1} = I_{a2} = I_{a0}\)
Total Fault Current: sum of three sequence currents (\(I_f = 3 I_{a0}\))

Single Line to Ground (SLG) Fault Equations

\[ I_{a1} = I_{a2} = I_{a0} = \dfrac{E_a}{Z_1 + Z_2 + Z_0 + 3Z_f} \]

\[ I_f = I_a = 3 I_{a0} \qquad\qquad I_b = 0 \qquad\qquad I_c = 0 \]

Where: \(I_{a1}, I_{a2}, I_{a0}\) = sequence currents of phase a (pu or A), \(E_a\) = pre-fault phase voltage (pu or V), \(Z_1, Z_2, Z_0\) = sequence impedances (pu or \(\Omega\)), \(Z_f\) = fault impedance (pu or \(\Omega\)), \(I_f\) = total line-to-ground fault current (pu or A)

EE Board Exam October 1998

Example. At a certain point of the system network the positive, negative and zero sequence impedances are \(0.25\ \text{pu}\), \(0.25\ \text{pu}\) and \(0.30\ \text{pu}\) respectively. The base MVA is \(100\). The voltage level at that point is \(34.5\ \text{kV}\). Determine the zero sequence current for a single line to ground fault.

Solution.

\[ I_{a0} = \dfrac{E_{\text{pu}}}{Z_1 + Z_2 + Z_0} = \dfrac{1.0}{0.25 + 0.25 + 0.30} = 1.25\ \text{pu} \implies I_{a0} = 1.25 \times \dfrac{100 \times 10^6}{\sqrt{3}(34.5 \times 10^3)} = 2091.81\ \text{A} \]

Answer: \(2091.81\ \text{A}\)

9. Line to Line Fault

Parallel Connection: positive and negative sequence networks connected in parallel
Zero Sequence Current: zero sequence current is identically zero (\(I_{a0} = 0\))
Fault Current Relation: fault line current equals \(\sqrt{3} I_{a1}\)

Line to Line (LL) Fault Equations

\[ I_{a1} = \dfrac{E_a}{Z_1 + Z_2 + Z_f} \qquad\qquad I_{a2} = -I_{a1} \qquad\qquad I_{a0} = 0 \]

\[ I_a = 0 \qquad\qquad I_b = -I_c = (a^2 - a) I_{a1} = -j\sqrt{3} I_{a1} \]

Where: \(I_{a1}, I_{a2}, I_{a0}\) = sequence currents of phase a (pu or A), \(E_a\) = pre-fault phase voltage (pu or V), \(Z_1, Z_2\) = positive and negative sequence impedances (pu or \(\Omega\)), \(Z_f\) = fault impedance (pu or \(\Omega\)), \(I_b, I_c\) = line fault currents in phases b and c (pu or A)

Example. A \(20\ \text{MVA}\), \(13.8\ \text{kV}\), \(60\ \text{Hz}\), three-phase synchronous turbo alternator has positive, negative and zero sequence reactances of \(0.25\), \(0.35\) and \(0.15\) respectively. If lines b and c are short-circuited, determine the subtransient fault current in line b.

Solution.

\[ I_{\text{base}} = \dfrac{20 \times 10^6}{\sqrt{3}(13.8 \times 10^3)} = 836.7\ \text{A} \qquad\qquad I_{a1} = \dfrac{1.0}{j0.25 + j0.35} = -j1.667\ \text{pu} \]

\[ I_b = \sqrt{3} I_{a1} I_{\text{base}} = \sqrt{3}(1.667)(836.7) = 2414.7\ \text{A} \approx 2415\ \text{A} \]

Answer: \(2415\ \text{A}\)

10. Double Line to Ground Fault

Parallel Branching: positive, negative, and zero sequence networks connected in parallel
Ground Return Current: neutral/ground current equals \(I_n = I_b + I_c = 3 I_{a0}\)

Double Line to Ground (2LG) Fault Equations

\[ I_{a1} = \dfrac{E_a}{Z_1 + Z_f + \frac{(Z_2 + Z_f)(Z_0 + Z_f + 3Z_g)}{Z_2 + Z_0 + 2Z_f + 3Z_g}} \]

\[ I_{a2} = -\dfrac{V_{a1} - I_{a1} Z_f}{Z_2 + Z_f} \qquad\qquad I_{a0} = -\dfrac{V_{a1} - I_{a1} Z_f}{Z_0 + Z_f + 3Z_g} \]

\[ I_n = I_b + I_c = 3 I_{a0} \]

Where: \(I_{a1}, I_{a2}, I_{a0}\) = sequence currents (pu or A), \(E_a\) = pre-fault voltage (pu or V), \(Z_1, Z_2, Z_0\) = sequence impedances (pu or \(\Omega\)), \(Z_f\) = fault impedance (pu or \(\Omega\)), \(Z_g\) = neutral-to-ground impedance (pu or \(\Omega\)), \(I_n\) = total current to ground (pu or A)

Example. In a short circuit analysis, the positive, negative and zero sequence impedances are \(0.20\ \text{pu}\), \(0.20\ \text{pu}\) and \(0.25\ \text{pu}\) respectively, using a base MVA of \(50\). Estimate the fault current on the faulted lines if a double line to ground fault at the \(138\ \text{kV}\) level occurs.

Solution.

\[ I_{a1} = \dfrac{1.0}{j0.20 + j\frac{(0.20)(0.25)}{0.20 + 0.25}} = -j3.214\ \text{pu} \implies I_b = 4.831\ \text{pu} \]

\[ I_f = 4.831 \times \dfrac{50 \times 10^6}{\sqrt{3}(138 \times 10^3)} = 1011\ \text{A} \]

Answer: \(1011\ \text{A}\)

11. Three-Phase Fault

Symmetrical Fault: all 3 phases shorted together with equal fault currents
Positive Sequence Only: \(I_{a2} = 0\) and \(I_{a0} = 0\)
Most Severe Fault: generally yields maximum short-circuit MVA and current

Three-Phase (\(3\phi\)) Symmetrical Fault Equations

\[ I_{a1} = \dfrac{E_a}{Z_1 + Z_f} \qquad\qquad I_a = I_{a1} \qquad\qquad I_b = a^2 I_{a1} \qquad\qquad I_c = a I_{a1} \]

Where: \(I_{a1}\) = positive sequence fault current (pu or A), \(E_a\) = pre-fault phase voltage (pu or V), \(Z_1\) = positive sequence impedance (pu or \(\Omega\)), \(Z_f\) = fault impedance (pu or \(\Omega\))

EE Board Exam March 1998

Example. In a short circuit study, the positive, negative and zero sequence impedances are \(0.25\ \text{pu}\), \(0.25\ \text{pu}\) and \(0.3\ \text{pu}\) respectively. The base MVA is \(100\). Determine the fault current for a three-phase fault at the \(115\ \text{kV}\) level.

Solution.

\[ I_f = \dfrac{E_{\text{pu}}}{X_1} I_{\text{base}} = \dfrac{1.0}{0.25} \times \dfrac{100 \times 10^6}{\sqrt{3}(115 \times 10^3)} = 4 \times 502 = 2008\ \text{A} \approx 2000\ \text{A} \]

Answer: \(2000\ \text{A}\)

11.1 kV-Base Method

Ohmic Calculations: convert all network impedances to ohms on a common kV base
Ohm’s Law Application: fault current calculated directly using \(I_f = V_f / Z_{\text{total}}\)

kV-Base Method Equations

\[ Z_c = Z_a \left(\dfrac{V_c}{V_a}\right)^2 \qquad\qquad I_f = \dfrac{V_f}{Z_{\text{total}}} \qquad\qquad I_a = I_f \left(\dfrac{V_c}{V_a}\right) \]

Where: \(Z_c\) = impedance on common kV base (\(\Omega\)), \(Z_a\) = actual impedance (\(\Omega\)), \(V_c\) = common kV base (kV), \(V_a\) = original voltage base (kV), \(Z_{\text{total}}\) = equivalent network impedance (\(\Omega\)), \(V_f\) = phase voltage at fault point (V), \(I_f\) = fault current (A)

EE Board Exam March 1998

Example. A \(5\ \text{MVA}\), \(13.8\ \text{kV}/480\ \text{V}\), \(5\%\) impedance transformer is tapped at \(13.8\ \text{kV}\) line where the Thevenin’s equivalent impedance is \(1/2\ \Omega\). Determine the fault current at the primary for a three-phase fault at the secondary.

Solution.

\[ Z_L = \dfrac{(0.05)(13.8 \times 10^3)^2}{5 \times 10^6} = 1.9044\ \Omega \implies Z_t = 0.5 + 1.9044 = 2.4044\ \Omega \]

\[ I_f = \dfrac{13800 / \sqrt{3}}{2.4044} = 3313\ \text{A} \approx 3300\ \text{A} \]

Answer: \(3300\ \text{A}\)

11.2 kVA-Base Method

Per Unit Conversion: convert all per-unit impedances to a common kVA/MVA base
Short Circuit MVA: calculate fault power \(\text{kVA}_f\) and fault current \(I_f\)

kVA-Base Method Equations

\[ Z_{c\text{pu}} = Z_{a\text{pu}} \left(\dfrac{\text{kVA}_c}{\text{kVA}_a}\right) \qquad\qquad \text{kVA}_f = \dfrac{\text{kVA}_c}{Z_{\text{total,pu}}} \qquad\qquad I_f = \dfrac{\text{kVA}_f}{\sqrt{3} V_f} \]

Where: \(Z_{c\text{pu}}\) = per-unit impedance on common kVA base, \(Z_{a\text{pu}}\) = original per-unit impedance, \(\text{kVA}_c\) = chosen common kVA base, \(\text{kVA}_a\) = original kVA rating, \(\text{kVA}_f\) = short-circuit fault MVA/kVA, \(V_f\) = line-to-line voltage at fault point (kV), \(I_f\) = total fault current (A)

EE Board Exam March 1998

Example. At a certain location in an electric system, the available fault MVA is \(400\ \text{MVA}\). A \(15\ \text{MVA}\), \(34.5\ \text{kV}/6.24\ \text{kV}\), \(2.5\%\) impedance, wye-wye grounded transformer is installed at that location. Determine the short circuit MVA at the secondary side of the transformer.

Solution.

\[ Z_{\text{t,pu}} = \dfrac{\text{MVA}_c}{\text{MVA}_f} + Z_{\text{trans,pu}} = \dfrac{15}{400} + 0.025 = 0.0625\ \text{pu} \implies \text{MVA}_{\text{new}} = \dfrac{15}{0.0625} = 240\ \text{MVA} \]

Answer: \(240\ \text{MVA}\)

12. Equivalence of Line to Line Fault and Three-Phase Fault on Unloaded Generator

Unloaded Generator Property: positive sequence reactance equals negative sequence reactance (\(Z_1 = Z_2\))
Current Ratio Factor: line-to-line fault current equals \(\sqrt{3}/2\) (or \(86.6\%\)) of 3-phase fault current

Line-to-Line and Three-Phase Fault Equivalence

\[ I_{LL} = \dfrac{\sqrt{3}}{2} I_{3\phi} = 0.866 I_{3\phi} \]

Where: \(I_{LL}\) = line current during line-to-line fault (A), \(I_{3\phi}\) = line current during 3-phase fault (A)

EE Board Exam October 1982

Example. The reactance of a three-phase alternator is \(8\%\). If the alternator is rated \(25\ \text{MVA}\) at \(13.25\ \text{kV}\) output voltage, line to phase, solve for the magnitude of the fault current generated when a short circuit occurs between 2 phases at the terminals.

Solution.

\[ I_{3\phi} = \dfrac{25000 / \sqrt{3}(\sqrt{3} \times 13.25)}{0.08} = 7861.6\ \text{A} \implies I_{LL} = \dfrac{\sqrt{3}}{2}(7861.6) = 6808\ \text{A} \]

Answer: \(6808\ \text{A}\)

13. Formula Summary

Formula Name	Formula	Variables
Positive Sequence Vector \(V_{b1}\)	\(V_{b1} = a^2 V_{a1}\)	\(V_{b1}, V_{a1}\) = positive sequence voltages (V) \(a = 1\angle 120^\circ\)
Negative Sequence Vector \(V_{b2}\)	\(V_{b2} = a V_{a2}\)	\(V_{b2}, V_{a2}\) = negative sequence voltages (V) \(a = 1\angle 120^\circ\)
Zero Sequence Identity	\(V_{a0} = V_{b0} = V_{c0}\)	\(V_{a0}, V_{b0}, V_{c0}\) = zero sequence voltages (V)
Zero Sequence Voltage Component	\(V_{a0} = \dfrac{1}{3}(V_a + V_b + V_c)\)	\(V_{a0}\) = zero sequence voltage (V) \(V_a, V_b, V_c\) = phase voltages (V)
Positive Sequence Voltage Component	\(V_{a1} = \dfrac{1}{3}(V_a + a V_b + a^2 V_c)\)	\(V_{a1}\) = positive sequence voltage (V) \(V_a, V_b, V_c\) = phase voltages (V)
Negative Sequence Voltage Component	\(V_{a2} = \dfrac{1}{3}(V_a + a^2 V_b + a V_c)\)	\(V_{a2}\) = negative sequence voltage (V) \(V_a, V_b, V_c\) = phase voltages (V)
Phase Voltage Synthesis \(V_a\)	\(V_a = V_{a0} + V_{a1} + V_{a2}\)	\(V_a\) = phase a voltage (V) \(V_{a0}, V_{a1}, V_{a2}\) = sequence components (V)
Per Unit Base Conversion	\(Z_{\text{pu2}} = Z_{\text{pu1}} \left(\dfrac{V_{\text{base1}}}{V_{\text{base2}}}\right)^2 \left(\dfrac{S_{\text{base2}}}{S_{\text{base1}}}\right)\)	\(Z_{\text{pu1}}, Z_{\text{pu2}}\) = per-unit impedances \(V_{\text{base}}\) = base voltage (kV) \(S_{\text{base}}\) = base power (MVA)
Positive Sequence Network Voltage	\(V_{a1} = E_a - I_{a1} Z_1\)	\(V_{a1}\) = sequence terminal voltage (pu) \(E_a\) = generated emf (pu) \(I_{a1}\) = current (pu) \(Z_1\) = impedance (pu)
SLG Fault Sequence Current	\(I_{a1} = I_{a2} = I_{a0} = \dfrac{E_a}{Z_1 + Z_2 + Z_0 + 3Z_f}\)	\(I_{a1}, I_{a2}, I_{a0}\) = sequence currents (pu) \(E_a\) = pre-fault voltage (pu) \(Z_1, Z_2, Z_0\) = sequence impedances (pu) \(Z_f\) = fault impedance (pu)
SLG Fault Line Current	\(I_f = 3 I_{a0}\)	\(I_f\) = total SLG fault current (A or pu) \(I_{a0}\) = zero sequence current (A or pu)
LL Fault Sequence Current	\(I_{a1} = \dfrac{E_a}{Z_1 + Z_2 + Z_f}\)	\(I_{a1}\) = positive sequence current (pu) \(E_a\) = pre-fault voltage (pu) \(Z_1, Z_2\) = impedances (pu) \(Z_f\) = fault impedance (pu)
LL Fault Line Current	\(I_b = -j\sqrt{3} I_{a1}\)	\(I_b\) = fault line current in phase b (pu) \(I_{a1}\) = positive sequence current (pu)
2LG Fault Sequence Current \(I_{a1}\)	\(I_{a1} = \dfrac{E_a}{Z_1 + Z_f + \dfrac{(Z_2 + Z_f)(Z_0 + Z_f + 3Z_g)}{Z_2 + Z_0 + 2Z_f + 3Z_g}}\)	\(I_{a1}\) = positive sequence current (pu) \(E_a\) = pre-fault voltage (pu) \(Z_1, Z_2, Z_0\) = sequence impedances (pu) \(Z_f, Z_g\) = fault/ground impedances (pu)
2LG Fault Ground Return Current	\(I_n = 3 I_{a0}\)	\(I_n\) = neutral/ground fault current (pu or A) \(I_{a0}\) = zero sequence current (pu or A)
Three-Phase Fault Sequence Current	\(I_{a1} = \dfrac{E_a}{Z_1 + Z_f}\)	\(I_{a1}\) = fault current (pu) \(E_a\) = pre-fault voltage (pu) \(Z_1\) = positive sequence impedance (pu) \(Z_f\) = fault impedance (pu)
kV-Base Impedance Conversion	\(Z_c = Z_a \left(\dfrac{V_c}{V_a}\right)^2\)	\(Z_c\) = converted impedance (\(\Omega\)) \(Z_a\) = actual impedance (\(\Omega\)) \(V_c, V_a\) = common and original kV bases
kVA-Base Impedance Conversion	\(Z_{c\text{pu}} = Z_{a\text{pu}} \left(\dfrac{\text{kVA}_c}{\text{kVA}_a}\right)\)	\(Z_{c\text{pu}}\) = converted per-unit impedance \(Z_{a\text{pu}}\) = original per-unit impedance \(\text{kVA}_c, \text{kVA}_a\) = kVA bases
Short Circuit Fault MVA	\(\text{kVA}_f = \dfrac{\text{kVA}_c}{Z_{\text{total,pu}}}\)	\(\text{kVA}_f\) = fault MVA/kVA \(\text{kVA}_c\) = base kVA \(Z_{\text{total,pu}}\) = total per-unit impedance
Line-to-Line Fault Equivalence Ratio	\(I_{LL} = \dfrac{\sqrt{3}}{2} I_{3\phi}\)	\(I_{LL}\) = line-to-line fault current (A) \(I_{3\phi}\) = 3-phase fault current (A)