Electrical System Formulas

Contents

Notation

B C E f G h I j L P Q

susceptance capacitance voltage source frequency conductance h-operator current j-operator inductance active power reactive power

(Siemens) (Farads) (Volts) (Hertz) (Siemens) (1Ð120°) (Amps) (1Ð90°) (Henrys) (Watts) (VArs)

R S t V W X Z f h w

resistance apparent power time voltage drop energy reactance impedance phase angle efficiency angular frequency

(Ohms) (VA) (seconds) (Volts) (Joules) (Ohms) (Ohms) (degrees) (per-unit) (rad/sec)

Reactive Loads

When a voltage V (taken as reference) is applied across the reactive load Z, the resulting current I flowing through the load is:
I = VY = V(R / |Z|² - jX / |Z|²) = VR / |Z|² - jVX / |Z|² = I_P - jI_Q
The active current I_P and the reactive current I_Q are:
I_P = VR / |Z|² = |I|cosf
I_Q = VX / |Z|² = |I|sinf

The apparent power S, active power P and reactive power Q are:
S = V|I| = V² / |Z| = |I|²|Z|
P = VI_P = I_P²|Z|² / R = V²R / |Z|² = |I|²R
Q = VI_Q = I_Q²|Z|² / X = V²X / |Z|² = |I|²X

Resistance and Shunt Reactance
The impedance Z of a reactive load comprising resistance R and shunt reactance X is found from:
1 / Z = 1 / R + 1 / jX
Converting to the equivalent admittance Y:
Y = 1 / Z = 1 / R - j / X

When a voltage V (taken as reference) is applied across the reactive load Y, the resulting current I flowing through the load is:
I = VY = V(1 / R - j / X) = V / R - jV / X = I_P - jI_Q
The active current I_P and the reactive current I_Q are:
I_P = V / R = |I|cosf
I_Q = V / X = |I|sinf

The apparent power S, active power P and reactive power Q are:
S = V|I| = |I|²|Z| = V² / |Z|
P = VI_P = I_P²R = |I|²|Z|² / R = V² / R
Q = VI_Q = I_Q²X = |I|²|Z|² / X = V² / X

Complex Power

For an inductive load:
Z = R + jX_L
I = I_P - jI_Q
cosf = R / |Z| (lagging)
I* = I_P + jI_Q
S = P + jQ
An inductive load is a sink of lagging VArs (a source of leading VArs).

For a capacitive load:
Z = R - jX_C
I = I_P + jI_Q
cosf = R / |Z| (leading)
I* = I_P - jI_Q
S = P - jQ
A capacitive load is a source of lagging VArs (a sink of leading VArs).

Three Phase Power

Star connected load:
V_star = V_L / Ö3
I_star = I_L
Z_star = V_L / Ö3I_L

Delta connected load:
V_delta = V_L
I_delta = I_L / Ö3
Z_delta = Ö3V_L / I_L

Apparent power S, active power P, reactive power Q and power factor cosf:
S = Ö3V_LI_L = 3V_starI_star = 3V_deltaI_delta
P = Ö3V_LI_Lcosf = Scosf
Q = Ö3V_LI_Lsinf = Ssinf
cosf = P / S
S² = P² + Q²

The Per-unit System

Select rated system values as base values, usually power in MVA and voltage in kV. It is convenient to operate on 'per phase' (equivalent star) quantities, so set the base voltage to the rated phase voltage:
S_base = rated MVA
E_base = rated line kV / Ö3 = rated phase kV

The base values for line current in kA and impedance in Ohms/phase are then:
I_base = S_base / 3E_base (kA)
Z_base = 3E_base² / S_base (Ohms/phase)

The base values for voltage, current and impedance satisfy Ohm's Law:
E_base = I_baseZ_base

Per-unit value is relative to 1 and percentage value is relative to 100, therefore per-unit value is equal to percentage value divided by 100:
Z_pu = Z_% / 100

Transformers
The primary and secondary MVA ratings of a transformer are the same, but the primary and secondary voltages (and currents) are usually different. Using subscript ₁ for the primary and subscript ₂ for the secondary:
Ö3E_1LI_1L = S = Ö3E_2LI_2L
Converting to base values:
3E_1baseI_1base = S_base = 3E_2baseI_2base
E_1base / E_2base = I_2base / I_1base
Z_1base / Z_2base = (E_1base / E_2base)²

The impedance Z_21pu referred to the primary side, equivalent to an impedance Z_2pu on the secondary side, is:
Z_21pu = Z_2pu(E_1base / E_2base)²

Symmetrical Components

The positive, negative and zero sequence currents are calculated from the line currents using:
I_a1 = (I_a + hI_b + h²I_c) / 3
I_a2 = (I_a + h²I_b + hI_c) / 3
I_a0 = (I_a + I_b + I_c) / 3

The positive, negative and zero sequence currents are combined to give the line currents using:
I_a = I_a1 + I_a2 + I_a0
I_b = I_b1 + I_b2 + I_b0 = h²I_a1 + hI_a2 + I_a0
I_c = I_c1 + I_c2 + I_c0 = hI_a1 + h²I_a2 + I_a0

The neutral current I_n is equal to the total zero sequence current:
I_n = I_a0 + I_b0 + I_c0 = 3I_a0 = I_a + I_b + I_c

The h-operator
The h-operator is the complex cube root of unity. Note that the j-operator is the complex fourth root of unity because it is the square root of negative unity. Some useful properties of the h-operator are:
h = - 1 / 2 + jÖ3 / 2 = 1Ð120° = 1Ð-240°
h² = - 1 / 2 - jÖ3 / 2 = 1Ð240° = 1Ð-120°
1 + h + h² = 0
h + h² = - 1 = 1Ð180°
h - h² = jÖ3 = Ö3Ð90°
h² - h = - jÖ3 = Ö3Ð-90°

Fault Calculations

For each type of short-circuit fault occurring on an unloaded system:
- the first column states the phase voltage and line current conditions at the fault,
- the second column states the phase 'a' sequence current and voltage conditions at the fault,
- the third column provides formulas for the phase 'a' sequence currents at the fault,
- the fourth column provides formulas for the fault current and the resulting line currents.

I _f = fault current
I_e = earth fault current
E_a = normal phase voltage at the fault location
Z₁ = positive phase sequence network impedance to the fault
Z₂ = negative phase sequence network impedance to the fault
Z₀ = zero phase sequence network impedance to the fault

By convention, the faulted phases are selected for fault symmetry with respect to reference phase 'a'.

Single phase to earth - fault from phase 'a' to earth:

Va = 0 Ib = Ic = 0 I f = Ia = Ie

Ia1 = Ia2 = Ia0 = Ia / 3 Va1 + Va2 + Va0 = 0

Ia1 = Ea / (Z1 + Z2 + Z0) Ia2 = Ia1 Ia0 = Ia1

I f = 3Ia0 = 3Ea / (Z1 + Z2 + Z0) = Ie Ia = 3Ea / (Z1 + Z2 + Z0)

Double phase - fault from phase 'b' to phase 'c':

Vb = Vc Ia = 0 I f = Ib = - Ic

Ia1 + Ia2 = 0 Ia0 = 0 Va1 = Va2

Ia1 = Ea / (Z1 + Z2) Ia2 = - Ia1 Ia0 = 0

I f = h2Ia1 + hIa2 + Ia0 = - jÖ3Ia1 Ib = I f = - jÖ3Ea / (Z1 + Z2) Ic = - I f = jÖ3Ea / (Z1 + Z2)

Double phase to earth - fault from phase 'b' to phase 'c' to earth: (S_zz = Z₁Z₂ + Z₂Z₀ + Z₀Z₁)

Vb = Vc = 0 Ia = 0 I f = Ib + Ic = Ie

Ia1 + Ia2 + Ia0 = 0 Va1 = Va2 = Va0

Ia1 = Ea(Z2 + Z0) / Szz Ia2 = - EaZ0 / Szz Ia0 = - EaZ2 / Szz

I f = 3Ia0 = - 3EaZ2 / Szz = Ie Ib = I f / 2 - jÖ3Ea(Z2 / 2 + Z0) / Szz Ic = I f / 2 + jÖ3Ea(Z2 / 2 + Z0) / Szz

Three phase (and three phase to earth) - fault from phase 'a' to phase 'b' to phase 'c' (and to earth):

Va = Vb = Vc (= 0) Ia + Ib + Ic = 0 (= Ie) I f = Ia = hIb = h2Ic

Va0 = Va (= 0) Va1 = Va2 = 0

Ia1 = Ea / Z1 Ia2 = 0 Ia0 = 0

I f = Ia1 = Ea / Z1 = Ia Ib = Eb / Z1 Ic = Ec / Z1

Power Factor Correction

To find tanf from cosf:
tanf = (1 / cos²f - 1)^½

The leading (capacitive) reactive power demand Q which must be connected across the load is:
Q = Q_u - Q_c = P (tanf_u - tanf_c)

Comparing uncorrected and corrected load currents and apparent power demands for the same active power demand P:
I_c / I_u = S_c / S_u = cosf_u / cosf_c

If the load is required to have a corrected power factor of unity:
Q = Q_u = P tanf_u
I_c / I_u = S_c / S_u = cosf_u

Efficiency

Rearranging the efficiency equations:
P_in = P_out + P_loss = P_out / h = P_loss / (1 - h)
P_out = P_in - P_loss = hP_in = hP_loss / (1 - h)
P_loss = P_in - P_out = (1 - h)P_in = (1 - h)P_out / h