Electrical Machine Applications

Notation
The symbol font is used for some notation and formulae. If the Greek symbols for alpha beta delta do not appear here [ a b d ] the symbol font needs to be installed for correct display of notation and formulae.
B E f G I j k m N n P p R	susceptance induced voltage frequency conductance current j-operator coefficient number of phases number of turns rotational speed power pole pairs resistance	[siemens, S] [volts, V] [hertz, Hz] [siemens, S] [amps, A] [1Ð90°] [number] [number] [number] [revs/min] [watts, W] [number] [ohms, W]		S s T V X Y Z d F f h q w	voltamperes slip torque terminal voltage reactance admittance impedance loss angle magnetic flux phase angle efficiency temperature angular speed	[volt-amps, VA] [per-unit] [newton-metres, Nm] [volts, V] [ohms, W] [siemens, S] [ohms, W] [degrees, °] [webers, Wb] [degrees, °] [per-unit] [centigrade, °C] [radians/sec]

Transformers

For an ideal two-winding transformer with primary voltage V₁ applied across N₁ primary turns and secondary voltage V₂ appearing across N₂ secondary turns:
V₁ / V₂ = N₁ / N₂
The primary current I₁ and secondary current I₂ are related by:
I₁ / I₂ = N₂ / N₁ = V₂ / V₁

For an ideal step-down auto-transformer with primary voltage V₁ applied across (N₁ + N₂)V₂ appearing across N₂ secondary turns:
V₁ / V₂ = (N₁ + N₂) / N₂
The primary (input) current I₁ and secondary (output) current I₂ are related by:
I₁ / I₂ = N₂ / (N₁ + N₂) = V₂ / V₁
Note that the winding current is I₁ through the N₁ section and (I₂ - I₁) through the N₂ section. primary turns and secondary voltage

For a single-phase transformer with rated primary voltage V₁, rated primary current I₁, rated secondary voltage V₂ and rated secondary current I₂, the voltampere rating S is:
S = V₁I₁ = V₂I₂

For a balanced m-phase transformer with rated primary phase voltage V₁, rated primary current I₁, rated secondary phase voltage V₂ and rated secondary current I₂, the voltampere rating S is:
S = mV₁I₁ = mV₂I₂

The primary circuit impedance Z₁ referred to the secondary circuit for an ideal transformer with N₁ primary turns and N₂ secondary turns is:
Z₁₂ = Z₁(N₂ / N₁)²

The secondary circuit impedance Z₂ referred to the primary circuit for an ideal transformer with N₁ primary turns and N₂ secondary turns is:
Z₂₁ = Z₂(N₁ / N₂)²

The voltage regulation DV₂ of a transformer is the rise in secondary voltage which occurs when rated load is disconnected from the secondary with rated voltage applied to the primary. For a transformer with a secondary voltage E₂ unloaded and V₂ at rated load, the per-unit voltage regulation DV_2pu is:
DV_2pu = (E₂ - V₂) / V₂
Note that the per-unit base voltage is usually V₂ and not E₂.

Open Circuit Test
If a transformer with its secondary open-circuited is energised at rated primary voltage, then the input power P_oc represents the core loss (iron loss P_Fe) of the transformer:
P_oc = P_Fe

The per-phase star values of the shunt magnetising admittance Y_m, conductance G_m and susceptance B_m of an m-phase transformer are calculated from the open-circuit test results for the per-phase primary voltage V_1oc, per-phase primary current I_1oc and input power P_oc
Y_m = I_1oc / V_1oc
G_m = mV_1oc² / P_oc
B_m = (Y_m² - G_m²)^½ using:

Short Circuit Test
If a transformer with its secondary short-circuited is energised at a reduced primary voltage which causes rated secondary current to flow through the short-circuit, then the input power P_sc represents the load loss (primary copper loss P_1Cu, secondary copper loss P_2Cu and stray loss P_stray) of the transformer:
P_sc = P_1Cu + P_2Cu + P_stray
Note that the temperature rise should be allowed to stabilise because conductor resistance varies with temperature.

If the resistance of each winding is determined by winding resistance tests immediately after the short circuit test, then the load loss of an m-phase transformer may be split into primary copper loss P_1Cu, secondary copper loss P_2Cu and stray loss P_stray:
P_1Cu = mI_1sc²R_1star
P_2Cu = mI_2sc²R_2star
P_stray = P_sc - P_1Cu - P_2Cu

If the stray loss is neglected, the per-phase star values referred to the primary of the total series impedance Z_s1, resistance R_s1 and reactance X_s1 of an m-phase transformer are calculated from the short-circuit test results for the per-phase primary voltage V_1sc, per-phase primary current I_1sc and input power P_sc using:
Z_s1 = V_1sc / I_1sc = Z₁ + Z₂(N₁² / N₂²)
R_s1 = P_sc / mI_1sc² = R₁ + R₂(N₁² / N₂²)
X_s1 = (Z_s1² - R_s1²)^½ = X₁ + X₂(N₁² / N₂²)
where Z₁, R₁ and X₁ are primary values and Z₂, R₂ and X₂ are secondary values

Winding Resistance Test
The resistance of each winding is measured using a small direct current to avoid thermal and inductive effects. If a voltage V_dc causes current I_dc to flow, then the resistance R is:
R = V_dc / I_dc

If the winding under test is a fully connected balanced star or delta and the resistance measured between any two phases is R_test, then the equivalent winding resistances R_star or R_delta are:
R_star = R_test / 2
R_delta = 3R_test / 2

The per-phase star primary and secondary winding resistances R_1star and R_2star of an m-phase transformer may be used to calculate the separate primary and secondary copper losses P_1Cu and P_2Cu:
P_1Cu = mI₁²R_1star
P_2Cu = mI₂²R_2star
Note that if the primary and secondary copper losses are equal, then the primary and secondary resistances R_1star and R_2star are related by:
R_1star / R_2star = I₂² / I₁² = N₁² / N₂²

The primary and secondary winding resistances R₁ and R₂ may also be used to check the effect of stray loss on the total series resistance referred to the primary, R_s1, calculated from the short circuit test results:
R_s1 = R₁ + R₂(N₁² / N₂²)

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Induction Machines

The synchronous rotational speed n_s and synchronous angular speed w_s of a machine with pf_s are:
n_s = 60f_s / p
w_s = 2pf_s / p = 2pn_s / 60 pole pairs running on a supply of frequency

The per-unit slip s of an induction machine of synchronous rotational speed n_s running at rotational speed n_m is:
s = (n_s - n_m) / n_s
Rearranging for rotational speed n_m:
n_m = (1 - s)n_s
Using angular speed w instead of rotational speed n:
w_m = (1 - s)w_s

The rated load torque T_M for a rated output power P_M is:
T_M = P_M / w_m = 60P_M / 2pn_m

For an induction machine with N_s stator turns and N_r rotor turns running at slip s on a supply of voltage E_s and frequency f_s, the rotor induced voltage and frequency E_r and f_r are:
E_r = sE_sN_r / N_s
f_r = sf_s
For a rotor current I_r, the equivalent stator current I_rs is:
I_rs = I_rN_r / N_s
Note that the rotor / stator ratios are N_s / N_r for current, sN_r / N_s for voltage and s for frequency.

For an induction machine with rotor resistance R_r and locked rotor leakage reactance X_r, the rotor impedance Z_r at slip s is:
Z_r = R_r + jsX_r
The stator circuit equivalent impedance Z_rf for a rotor / stator frequency ratio s is:
Z_rf = R_rs / s + jX_rs

For an induction motor with synchronous angular speed w_s running at angular speed w_m and slip s, the airgap transfer power P_t, rotor copper loss P_r and gross output power P_m for a gross output torque T_m are related by:
P_t = w_sT_m = P_r / s = P_m / (1 - s)
P_r = sP_t = sP_m / (1 - s)
P_m = w_mT_m = (1 - s)P_t
The power ratios are:
P_t : P_r : P_m = 1 : s : (1 - s)
The gross motor efficiency h_m (neglecting stator and mechanical losses) is:
h_m = P_m / P_t = 1 - s

An induction machine can be operated as a generator, a motor or a brake:
- for negative slip (speed above synchronous) the machine is a generator,
- for positive slip between 0 and 1 (speed below synchronous) the machine is a motor,
- for positive slip greater than 1 (speed negative) the machine is a brake,
In all cases the magnetizing current (at lagging power factor) is provided by the supply system.

No Load Test
If an induction machine with its rotor unloaded is energised at rated voltage, then the input power represents the sum of the iron loss and mechanical loss of the machine.

Locked Rotor Test
If an induction machine with its rotor locked is energised at a reduced voltage which causes rated current input, then the input power represents the sum of the full load copper loss and stray loss of the machine.

Stator Resistance Test
The resistance of the stator winding is measured using a small direct current.

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Synchronous Machines

The synchronous rotational speed n_s and synchronous angular speed w_s of a machine with pf_s are:
n_s = 60f_s / p
w_s = 2pf_s / p pole pairs running on a supply of frequency

The output power P_m for a load torque T_m is:
P_m = w_sT_m

The rated load torque T_M for a rated output power P_M is:
T_M = P_M / w_s = P_Mp / 2pf_s = 60P_M / 2pn_s

Synchronous Generator
For a synchronous generator with stator induced voltage E_s, stator current I_s and synchronous impedance Z_s, the terminal voltage V is:
V = E_s - I_sZ_s = E_s - I_s(R_s + jX_s)
where R_s is the stator resistance and X_s is the synchronous reactance

Synchronous Motor
For a synchronous motor with stator induced voltage E_s, stator current I_s and synchronous impedance Z_s, the terminal voltage V is:
V = E_s + I_sZ_s = E_s + I_s(R_s + jX_s)
where R_s is the stator resistance and X_s is the synchronous reactance

Note that the field excitation of a parallelled synchronous machine determines its power factor:
- an under-excited machine operates with a leading power factor,
- an over-excited machine operates with a lagging power factor.
The field excitation of an isolated synchronous generator determines its output voltage.

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Direct Current Machines

Shunt Generator
For a shunt generator with armature induced voltage E_a, armature current I_a and armature resistance R_a, the terminal voltage V is:
V = E_a - I_aR_a
The field current I _f for a field resistance R _f is:
I _f = V / R _f

The armature induced voltage E_a and torque T with magnetic flux F at angular speed w are:
E_a = k _fFw = k_mw
T = k _fFI_a = k_mI_a
where k _f and k_m are design coefficients of the machine.

Note that for a shunt generator:
- induced voltage is proportional to speed,
- torque is proportional to armature current.

The airgap power P_e for a shunt generator is:
P_e = wT = E_aI_a = k_mw I_a

Shunt Motor
For a shunt motor with armature induced voltage E_a, armature current I_a and armature resistance R_a, the terminal voltage V is:
V = E_a + I_aR_a
The field current I _f for a field resistance R _f is:
I _f = V / R _f

The armature induced voltage E_a and torque T with magnetic flux F at angular speed w are:
E_a = k _fFw = k_mw
T = k _fFI_a = k_mI_a
where k _f and k_m are design coefficients of the machine.

Note that for a shunt motor:
- induced voltage is proportional to speed,
- torque is proportional to armature current.

The airgap power P_e for a shunt motor is:
P_e = wT = E_aI_a = k_mw I_a

Series Motor
For a series motor with armature induced voltage E_a, armature current I_a, armature resistance R_a and field resistance R _f, the terminal voltage V is:
V = E_a + I_aR_a + I_aR _f = E_a + I_a(R_a + R _f)
The field current is equal to the armature current.

The armature induced voltage E_a and torque T with magnetic flux F at angular speed w are:
E_a = k _fFw I_a = k_mw I_a
T = k _fFI_a² = k_mI_a²
where k _f and k_m are design coefficients of the machine.

Note that for a series motor:
- induced voltage is proportional to both speed and armature current,
- torque is proportional to the square of armature current,
- armature current is inversely proportional to speed for a constant induced voltage.

The airgap power P_e for a series motor is:
P_e = wT = E_aI_a = k_mw I_a²

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Efficiency

The per-unit efficiency h of an electrical machine with input power P_in, output power P_out and power loss P_loss is:
h = P_out / P_in = P_out / (P_out + P_loss) = (P_in - P_loss) / P_in

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

For an electrical machine with output power P_out (proportional to current) and power loss P_lossP_fix (independent of current) plus a variable loss P_var (proportional to square of current) the efficiency is a maximum when P_var is equal to P_fix. comprising a fixed loss

For a transformer, P_fix is the iron loss and P_var is the copper loss plus the stray loss.

For an induction machine, P_fix is the iron loss plus the mechanical loss and P_var is the copper loss plus the stray loss.

Energy Conversion
Comparing megawatt-hours and gigajoules, 1 MWh is equivalent to 3.6 GJ. For an energy conversion process with a per-unit efficiency h, 1 MWh of energy output is obtained from (3.6 / h) GJ of energy input.

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Temperature Rise

The resistance of copper and aluminium windings increases with temperature, and the relationship is quite linear over the normal range of operating temperatures. For a linear relationship, if the winding resistance is R₁ at temperature q₁ and R₂ at temperature q₂, then:
R₁ / (q₁ - q₀) = R₂ / (q₂ - q₀) = (R₂ - R₁) / (q₂ - q₁)
where q₀ is the extrapolated temperature for zero resistance.

The ratio of resistances R₂ and R₁ is:
R₂ / R₁ = (q₂ - q₀) / (q₁ - q₀)

The average temperature rise Dq of a winding under load may be estimated from measured values of the cold winding resistance R₁ at temperature q₁ (usually ambient temperature) and the hot winding resistance R₂ at temperature q₂, using:
Dq = q₂ - q₁ = (q₁ - q₀) (R₂ - R₁) / R₁

Rearranging for per-unit change in resistance DR_pu relative to R₁:
DR_pu = (R₂ - R₁) / R₁ = (q₂ - q₁) / (q₁ - q₀) = Dq / (q₁ - q₀)

Note that the resistance values are measured using a small direct current to avoid thermal and inductive effects.

Copper Windings
The value of q₀ for copper is - 234.5 °C, so that:
Dq = q₂ - q₁ = (q₁ + 234.5) (R₂ - R₁) / R₁

If q₁ is 20 °C and Dq is 1 degC:
DR_pu = (R₂ - R₁) / R₁ = Dq / (q₁ - q₀) = 1 / 254.5 = 0.00393
The temperature coefficient of resistance of copper at 20 °C is 0.00393 per degC.

Aluminium Windings
The value of q₀ for aluminium is - 228 °C, so that:
Dq = q₂ - q₁ = (q₁ + 228) (R₂ - R₁) / R₁

If q₁ is 20 °C and Dq is 1 degC:
DR_pu = (R₂ - R₁) / R₁ = Dq / (q₁ - q₀) = 1 / 248 = 0.00403
The temperature coefficient of resistance of aluminium at 20 °C is 0.00403 per degC.

Note that aluminium has 61% of the conductivity and 30% of the density of copper, therefore for the same conductance (and same resistance) an aluminium conductor has 164% of the cross-sectional area, 128% of the diameter and 49% of the mass of a copper conductor.

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Dielectric Dissipation Factor

If an alternating voltage V of frequency f is applied across an insulation system comprising capacitance C and equivalent series loss resistance R_S, then the voltage V_R across R_S and the voltage V_C across C due to the resulting current I are:
V_R = IR_S
V_C = IX_C
V = (V_R² + V_C²)^½

The dielectric dissipation factor of the insulation system is the tangent of the dielectric loss angle d between V_C and V:
tand = V_R / V_C = R_S / X_C = 2pfCR_S
R_S = X_Ctand = tand / 2pfC
Note that an increase in the dielectric losses of a insulation system (from an increase in the series loss resistance R_S) results in an increase in tand. Note also that tand increases with frequency.

The dielectric power loss P is related to the capacitive reactive power Q_C by:
P = I²R_S = I²X_Ctand = Q_Ctand

The power factor of the insulation system is the cosine of the phase angle f between V_R and V:
cosf = V_R / V
so that d and f are related by:
d + f = 90°

tand and cosf are related by:
tand = 1 / tanf = cosf / sinf = cosf / (1 - cos²f)^½
so that when cosf is close to zero, tand » cosf

Note that the series loss resistance R_S is not related to the shunt leakage resistance of the insulation system (which is measured using direct current).