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# Electrical Machines • Notation
• Transformers
• Induction Machines
• Synchronous Machines
• Direct Current Machines
• Efficiency
• Temperature Rise
• Dielectric Dissipation Factor
 Notation The symbol font is used for some notation and formulae. If the Greek symbols for alpha beta delta do notappear here [ a b d ] the symbol font needs to be installed for correct display of notation and formulae. BEfGIjkmNnPpR susceptanceinduced voltagefrequencyconductancecurrentj-operatorcoefficientnumber of phasesnumber of turnsrotational speedpowerpole pairsresistance [siemens, S][volts, V][hertz, Hz][siemens, S][amps, A][1Ð90°][number][number][number][revs/min][watts, W][number][ohms, W] SsTVXYZdFfhqw voltamperessliptorqueterminal voltagereactanceadmittanceimpedanceloss anglemagnetic fluxphase angleefficiencytemperatureangular 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) primary turns and secondary voltage 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.

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 DV2pu is:

DV2pu = (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 Poc represents the core loss (iron loss PFe) of the transformer:

Poc = PFe

The per-phase star values of the shunt magnetising admittance Ym, conductance Gm and susceptance Bm of an m-phase transformer are calculated from the open-circuit test results for the per-phase primary voltage V1oc, per-phase primary current I1oc and input power Poc using:

Ym = I1oc / V1oc

Gm = mV1oc2 / Poc

Bm = (Ym2 – Gm2)½

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 Psc represents the load loss (primary copper loss P1Cu, secondary copper loss P2Cu and stray loss Pstray) of the transformer:

Psc = P1Cu + P2Cu + Pstray

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 P1Cu, secondary copper loss P2Cu and stray loss Pstray:

P1Cu = mI1sc2R1star

P2Cu = mI2sc2R2star

Pstray = Psc – P1Cu – P2Cu

If the stray loss is neglected, the per-phase star values referred to the primary of the total series impedance Zs1, resistance Rs1 and reactance Xs1 of an m-phase transformer are calculated from the short-circuit test results for the per-phase primary voltage V1sc, per-phase primary current I1sc and input power Psc using:

Zs1 = V1sc / I1sc = Z1 + Z2(N12 / N22)

Rs1 = Psc / mI1sc2 = R1 + R2(N12 / N22)

Xs1 = (Zs12 – Rs12)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 Vdc causes current Idc to flow, then the resistance R is:

R = Vdc / Idc

If the winding under test is a fully connected balanced star or delta and the resistance measured between any two phases is Rtest, then the equivalent winding resistances Rstar or Rdelta are:

Rstar = Rtest / 2

Rdelta = 3Rtest / 2

The per-phase star primary and secondary winding resistances R1star and R2star of an m-phase transformer may be used to calculate the separate primary and secondary copper losses P1Cu and P2Cu:

P1Cu = mI12R1star

P2Cu = mI22R2star

Note that if the primary and secondary copper losses are equal, then the primary and secondary resistances R1star and R2star are related by:

R1star / R2star = 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, Rs1, calculated from the short circuit test results:

Rs1 = R1 + R2(N12 / N22)

#### Induction Machines

The synchronous rotational speed ns and synchronous angular speed ws of a machine with p pole pairs running on a supply of frequency fs are:

ns = 60fs / p

ws = 2pfs / p = 2pns / 60

The per-unit slip s of an induction machine of synchronous rotational speed ns running at rotational speed nm is:

s = (ns – nm) / ns

Rearranging for rotational speed nm:

nm = (1 – s)ns

Using angular speed w instead of rotational speed n:

wm = (1 – s)ws

The rated load torque TM for a rated output power PM is:

TM = PM / wm = 60PM / 2pnm

For an induction machine with Ns stator turns and Nr rotor turns running at slip s on a supply of voltage Es and frequency fs, the rotor induced voltage and frequency Er and fr are:

Er = sEsNr / Ns

fr = sfs

For a rotor current Ir, the equivalent stator current Irs is:

Irs = IrNr / Ns

Note that the rotor / stator ratios are Ns / Nr for current, sNr / Ns for voltage and s for frequency.

For an induction machine with rotor resistance Rr and locked rotor leakage reactance Xr, the rotor impedance Zr at slip s is:

Zr = Rr + jsXr

The stator circuit equivalent impedance Zrf for a rotor / stator frequency ratio s is:

Zrf = Rrs / s + jXrs

For an induction motor with synchronous angular speed ws running at angular speed wm and slip s, the airgap transfer power Pt, rotor copper loss Pr and gross output power Pm for a gross output torque Tm are related by:

Pt = wsTm = Pr / s = Pm / (1 – s)

Pr = sPt = sPm / (1 – s)

Pm = wmTm = (1 – s)Pt

The power ratios are:

Pt : Pr : Pm = 1 : s : (1 – s)

The gross motor efficiency hm (neglecting stator and mechanical losses) is:

hm = Pm / Pt = 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.

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.

#### Synchronous Machines

The synchronous rotational speed ns and synchronous angular speed ws of a machine with p pole pairs running on a supply of frequency fs are:

ns = 60fs / p

ws = 2pfs / p

The output power Pm for a load torque Tm is:

Pm = wsTm

The rated load torque TM for a rated output power PM is:

TM = PM / ws = PMp / 2pfs = 60PM / 2pns

Synchronous Generator

For a synchronous generator with stator induced voltage Es, stator current Is and synchronous impedance Zs, the terminal voltage V is:

V = Es – IsZs = Es – Is(Rs + jXs)

where Rs is the stator resistance and Xs is the synchronous reactance

Synchronous Motor

For a synchronous motor with stator induced voltage Es, stator current Is and synchronous impedance Zs, the terminal voltage V is:

V = Es + IsZs = Es + Is(Rs + jXs)

where Rs is the stator resistance and Xs 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.

#### Direct Current Machines

Shunt Generator

For a shunt generator with armature induced voltage Ea, armature current Ia and armature resistance Ra, the terminal voltage V is:

V = Ea – IaRa

The field current I f for a field resistance R f is:

I f = V / R f

The armature induced voltage Ea and torque T with magnetic flux F at angular speed w are:

Ea = k fFw = kmw

T = k fFIa = kmIa

where k f and km 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 Pe for a shunt generator is:

Pe = wT = EaIa = kmw Ia

Shunt Motor

For a shunt motor with armature induced voltage Ea, armature current Ia and armature resistance Ra, the terminal voltage V is:

V = Ea + IaRa

The field current I f for a field resistance R f is:

I f = V / R f

The armature induced voltage Ea and torque T with magnetic flux F at angular speed w are:

Ea = k fFw = kmw

T = k fFIa = kmIa

where k f and km 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 Pe for a shunt motor is:

Pe = wT = EaIa = kmw Ia

Series Motor

For a series motor with armature induced voltage Ea, armature current Ia, armature resistance Ra and field resistance R f, the terminal voltage V is:

V = Ea + IaRa + IaR f = Ea + Ia(Ra + R f)

The field current is equal to the armature current.

The armature induced voltage Ea and torque T with magnetic flux F at angular speed w are:

Ea = k fFw Ia = kmw Ia

T = k fFIa2 = kmIa2

where k f and km 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 Pe for a series motor is:

Pe = wT = EaIa = kmw Ia2

#### Efficiency

The per-unit efficiency h of an electrical machine with input power Pin, output power Pout and power loss Ploss is:

h = Pout / Pin = Pout / (Pout + Ploss) = (Pin – Ploss) / Pin

Rearranging the efficiency equations:

Pin = Pout + Ploss = Pout / h = Ploss / (1 – h)

Pout = Pin – Ploss = hPin = hPloss / (1 – h)

Ploss = Pin – Pout = (1 – h)Pin = (1 – h)Pout / h

For an electrical machine with output power Pout (proportional to current) and power loss Ploss comprising a fixed loss Pfix (independent of current) plus a variable loss Pvar (proportional to square of current) the efficiency is a maximum when Pvar is equal to Pfix.

For a transformer, Pfix is the iron loss and Pvar is the copper loss plus the stray loss.

For an induction machine, Pfix is the iron loss plus the mechanical loss and Pvar 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.

#### 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 / (q1q0) = R2 / (q2q0) = (R2 – R1) / (q2q1)

where q0 is the extrapolated temperature for zero resistance.

The ratio of resistances R2 and R1 is:

R2 / R1 = (q2q0) / (q1q0)

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 = q2q1 = (q1q0) (R2 – R1) / R1

Rearranging for per-unit change in resistance DRpu relative to R1:

DRpu = (R2 – R1) / R1 = (q2q1) / (q1q0) = Dq / (q1q0)

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 = q2q1 = (q1 + 234.5) (R2 – R1) / R1

If q1 is 20 °C and Dq is 1 degC:

DRpu = (R2 – R1) / R1 = Dq / (q1q0) = 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 = q2q1 = (q1 + 228) (R2 – R1) / R1

If q1 is 20 °C and Dq is 1 degC:

DRpu = (R2 – R1) / R1 = Dq / (q1q0) = 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.

#### 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 RS, then the voltage VR across RS and the voltage VC across C due to the resulting current I are:

VR = IRS

VC = IXC

V = (VR2 + VC2)½

The dielectric dissipation factor of the insulation system is the tangent of the dielectric loss angle d between VC and V:

tand = VR / VC = RS / XC = 2pfCRS

RS = XCtand = tand / 2pfC

Note that an increase in the dielectric losses of a insulation system (from an increase in the series loss resistance RS) 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 QC by:

P = I2RS = I2XCtand = QCtand

The power factor of the insulation system is the cosine of the phase angle f between VR and V:

cosf = VR / 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 RS is not related to the shunt leakage resistance of the insulation system (which is measured using direct current).