# 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. | |||||

BE
| susceptanceinduced 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] | Ss
| voltamperesslip 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 **V _{1}** applied across

**N**primary turns and secondary voltage

_{1}**V**appearing across

_{2}**N**secondary turns:

_{2}**V _{1} / V_{2} = N_{1} / N_{2}**

The primary current **I _{1}** and secondary current

**I**are related by:

_{2}**I _{1} / I_{2} = N_{2} / N_{1} = V_{2} / V_{1}**

For an ideal step-down auto-transformer with primary voltage **V _{1}** applied across

**(N**primary turns and secondary voltage

_{1}+ N_{2})**V**appearing across

_{2}**N**secondary turns:

_{2}**V _{1} / V_{2} = (N_{1} + N_{2}) / N_{2}**

The primary (input) current **I _{1}** and secondary (output) current

**I**are related by:

_{2}**I _{1} / I_{2} = N_{2} / (N_{1} + N_{2}) = V_{2} / V_{1}**

Note that the winding current is **I _{1}** through the

**N**section and

_{1}**(I**through the

_{2}– I_{1})**N**section.

_{2}For a single-phase transformer with rated primary voltage **V _{1}**, rated primary current

**I**, rated secondary voltage

_{1}**V**and rated secondary current

_{2}**I**, the voltampere rating

_{2}**S**is:

**S = V _{1}I_{1} = V_{2}I_{2}**

For a balanced **m**-phase transformer with rated primary phase voltage **V _{1}**, rated primary current

**I**, rated secondary phase voltage

_{1}**V**and rated secondary current

_{2}**I**, the voltampere rating

_{2}**S**is:

**S = mV _{1}I_{1} = mV_{2}I_{2}**

The primary circuit impedance **Z _{1}** referred to the secondary circuit for an ideal transformer with

**N**primary turns and

_{1}**N**secondary turns is:

_{2}**Z _{12} = Z_{1}(N_{2} / N_{1})^{2}**

The secondary circuit impedance **Z _{2}** referred to the primary circuit for an ideal transformer with

**N**primary turns and

_{1}**N**secondary turns is:

_{2}**Z _{21} = Z_{2}(N_{1} / N_{2})^{2}**

The voltage regulation **DV _{2}** 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

**E**unloaded and

_{2}**V**at rated load, the per-unit voltage regulation

_{2}**DV**is:

_{2pu}**DV _{2pu} = (E_{2} – V_{2}) / V_{2}**

Note that the per-unit base voltage is usually **V _{2}** and not

**E**.

_{2}*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**) of the transformer:

_{Fe}**P _{oc} = P_{Fe}**

The per-phase star values of the shunt magnetising admittance **Y _{m}**, conductance

**G**and susceptance

_{m}**B**of an

_{m}**m**-phase transformer are calculated from the open-circuit test results for the per-phase primary voltage

**V**, per-phase primary current

_{1oc}**I**and input power

_{1oc}**P**using:

_{oc}**Y _{m} = I_{1oc} / V_{1oc}**

**G _{m} = mV_{1oc}^{2} / P_{oc}**

**B _{m} = (Y_{m}^{2} – G_{m}^{2})^{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 **P _{sc}** represents the load loss (primary copper loss

**P**, secondary copper loss

_{1Cu}**P**and stray loss

_{2Cu}**P**) of the transformer:

_{stray}**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**and stray loss

_{2Cu}**P**:

_{stray}**P _{1Cu} = mI_{1sc}^{2}R_{1star}**

**P _{2Cu} = mI_{2sc}^{2}R_{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**and reactance

_{s1}**X**of an

_{s1}**m**-phase transformer are calculated from the short-circuit test results for the per-phase primary voltage

**V**, per-phase primary current

_{1sc}**I**and input power

_{1sc}**P**using:

_{sc}**Z _{s1} = V_{1sc} / I_{1sc} = Z_{1} + Z_{2}(N_{1}^{2} / N_{2}^{2})**

**R _{s1} = P_{sc} / mI_{1sc}^{2} = R_{1} + R_{2}(N_{1}^{2} / N_{2}^{2})**

**X _{s1} = (Z_{s1}^{2} – R_{s1}^{2})^{½} = X_{1} + X_{2}(N_{1}^{2} / N_{2}^{2})**

where **Z _{1}**,

**R**and

_{1}**X**are primary values and

_{1}**Z**,

_{2}**R**and

_{2}**X**are secondary values

_{2}*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**to flow, then the resistance

_{dc}**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**or

_{star}**R**are:

_{delta}**R _{star} = R_{test} / 2**

**R _{delta} = 3R_{test} / 2**

The per-phase star primary and secondary winding resistances **R _{1star}** and

**R**of an

_{2star}**m**-phase transformer may be used to calculate the separate primary and secondary copper losses

**P**and

_{1Cu}**P**:

_{2Cu}**P _{1Cu} = mI_{1}^{2}R_{1star}**

**P _{2Cu} = mI_{2}^{2}R_{2star}**

Note that if the primary and secondary copper losses are equal, then the primary and secondary resistances **R _{1star}** and

**R**are related by:

_{2star}**R _{1star} / R_{2star} = I_{2}^{2} / I_{1}^{2} = N_{1}^{2} / N_{2}^{2}**

The primary and secondary winding resistances **R _{1}** and

**R**may also be used to check the effect of stray loss on the total series resistance referred to the primary,

_{2}**R**, calculated from the short circuit test results:

_{s1}**R _{s1} = R_{1} + R_{2}(N_{1}^{2} / N_{2}^{2})**

#### Induction Machines

The synchronous rotational speed **n _{s}** and synchronous angular speed

**w**of a machine with

_{s}**p**pole pairs running on a supply of frequency

**f**are:

_{s}**n _{s} = 60f_{s} / p**

**w _{s} = 2pf_{s} / p = 2pn_{s} / 60**

The per-unit slip **s** of an induction machine of synchronous rotational speed **n _{s}** running at rotational speed

**n**is:

_{m}**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**is:

_{M}**T _{M} = P_{M} / w_{m} = 60P_{M} / 2pn_{m}**

For an induction machine with **N _{s}** stator turns and

**N**rotor turns running at slip

_{r}**s**on a supply of voltage

**E**and frequency

_{s}**f**, the rotor induced voltage and frequency

_{s}**E**and

_{r}**f**are:

_{r}**E _{r} = sE_{s}N_{r} / N_{s}**

**f _{r} = sf_{s}**

For a rotor current **I _{r}**, the equivalent stator current

**I**is:

_{rs}**I _{rs} = I_{r}N_{r} / N_{s}**

Note that the rotor / stator ratios are **N _{s} / N_{r}** for current,

**sN**for voltage and

_{r}/ N_{s}**s**for frequency.

For an induction machine with rotor resistance **R _{r}** and locked rotor leakage reactance

**X**, the rotor impedance

_{r}**Z**at slip

_{r}**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**and slip

_{m}**s**, the airgap transfer power

**P**, rotor copper loss

_{t}**P**and gross output power

_{r}**P**for a gross output torque

_{m}**T**are related by:

_{m}**P _{t} = w_{s}T_{m} = P_{r} / s = P_{m} / (1 – s)**

**P _{r} = sP_{t} = sP_{m} / (1 – s)**

**P _{m} = w_{m}T_{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.

#### Synchronous Machines

The synchronous rotational speed **n _{s}** and synchronous angular speed

**w**of a machine with

_{s}**p**pole pairs running on a supply of frequency

**f**are:

_{s}**n _{s} = 60f_{s} / p**

**w _{s} = 2pf_{s} / p**

The output power **P _{m}** for a load torque

**T**is:

_{m}**P _{m} = w_{s}T_{m}**

The rated load torque **T _{M}** for a rated output power

**P**is:

_{M}**T _{M} = P_{M} / w_{s} = P_{M}p / 2pf_{s} = 60P_{M} / 2pn_{s}**

*Synchronous Generator*

For a synchronous generator with stator induced voltage **E _{s}**, stator current

**I**and synchronous impedance

_{s}**Z**, the terminal voltage

_{s}**V**is:

**V = E _{s} – I_{s}Z_{s} = E_{s} – I_{s}(R_{s} + jX_{s})**

where **R _{s}** is the stator resistance and

**X**is the synchronous reactance

_{s}*Synchronous Motor*

For a synchronous motor with stator induced voltage **E _{s}**, stator current

**I**and synchronous impedance

_{s}**Z**, the terminal voltage

_{s}**V**is:

**V = E _{s} + I_{s}Z_{s} = E_{s} + I_{s}(R_{s} + jX_{s})**

where **R _{s}** is the stator resistance and

**X**is the synchronous reactance

_{s}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 **E _{a}**, armature current

**I**and armature resistance

_{a}**R**, the terminal voltage

_{a}**V**is:

**V = E _{a} – I_{a}R_{a}**

The field current **I _{ f}** for a field resistance

**R**is:

_{ f}**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_{ f}Fw = k_{m}w**

**T = k _{ f}FI_{a} = k_{m}I_{a}**

where **k _{ f}** and

**k**are design coefficients of the machine.

_{m}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_{a}I_{a} = k_{m}w I_{a}**

*Shunt Motor*

For a shunt motor with armature induced voltage **E _{a}**, armature current

**I**and armature resistance

_{a}**R**, the terminal voltage

_{a}**V**is:

**V = E _{a} + I_{a}R_{a}**

The field current **I _{ f}** for a field resistance

**R**is:

_{ f}**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_{ f}Fw = k_{m}w**

**T = k _{ f}FI_{a} = k_{m}I_{a}**

where **k _{ f}** and

**k**are design coefficients of the machine.

_{m}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_{a}I_{a} = k_{m}w I_{a}**

*Series Motor*

For a series motor with armature induced voltage **E _{a}**, armature current

**I**, armature resistance

_{a}**R**and field resistance

_{a}**R**, the terminal voltage

_{ f}**V**is:

**V = E _{a} + I_{a}R_{a} + I_{a}R_{ 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_{ f}Fw I_{a} = k_{m}w I_{a}**

**T = k _{ f}FI_{a}^{2} = k_{m}I_{a}^{2}**

where **k _{ f}** and

**k**are design coefficients of the machine.

_{m}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_{a}I_{a} = k_{m}w I_{a}^{2}**

#### Efficiency

The per-unit efficiency **h** of an electrical machine with input power **P _{in}**, output power

**P**and power loss

_{out}**P**is:

_{loss}**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**comprising a fixed loss

_{loss}**P**(independent of current) plus a variable loss

_{fix}**P**(proportional to square of current) the efficiency is a maximum when

_{var}**P**is equal to

_{var}**P**.

_{fix}For a transformer, **P _{fix}** is the iron loss and

**P**is the copper loss plus the stray loss.

_{var}For an induction machine, **P _{fix}** is the iron loss plus the mechanical loss and

**P**is the copper loss plus the stray loss.

_{var}*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 **R _{1}** at temperature

**q**and

_{1}**R**at temperature

_{2}**q**, then:

_{2}**R _{1} / (q_{1} – q_{0}) = R_{2} / (q_{2} – q_{0}) = (R_{2} – R_{1}) / (q_{2} – q_{1})**

where **q _{0}** is the extrapolated temperature for zero resistance.

The ratio of resistances **R _{2}** and

**R**is:

_{1}**R _{2} / R_{1} = (q_{2} – q_{0}) / (q_{1} – q_{0})**

The average temperature rise **Dq** of a winding under load may be estimated from measured values of the cold winding resistance **R _{1}** at temperature

**q**(usually ambient temperature) and the hot winding resistance

_{1}**R**at temperature

_{2}**q**, using:

_{2}**Dq = q _{2} – q_{1} = (q_{1} – q_{0}) (R_{2} – R_{1}) / R_{1}**

Rearranging for per-unit change in resistance **DR _{pu}** relative to

**R**:

_{1}**DR _{pu} = (R_{2} – R_{1}) / R_{1} = (q_{2} – q_{1}) / (q_{1} – q_{0}) = Dq / (q_{1} – q_{0})**

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

*Copper Windings*

The value of **q _{0}** for copper is

**- 234.5 °C**, so that:

**Dq = q _{2} – q_{1} = (q_{1} + 234.5) (R_{2} – R_{1}) / R_{1}**

If **q _{1}** is

**20 °C**and

**Dq**is

**1 degC**:

**DR _{pu} = (R_{2} – R_{1}) / R_{1} = Dq / (q_{1} – q_{0}) = 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 **q _{0}** for aluminium is

**- 228 °C**, so that:

**Dq = q _{2} – q_{1} = (q_{1} + 228) (R_{2} – R_{1}) / R_{1}**

If **q _{1}** is

**20 °C**and

**Dq**is

**1 degC**:

**DR _{pu} = (R_{2} – R_{1}) / R_{1} = Dq / (q_{1} – q_{0}) = 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 **R _{S}**, then the voltage

**V**across

_{R}**R**and the voltage

_{S}**V**across

_{C}**C**due to the resulting current

**I**are:

**V _{R} = IR_{S}**

**V _{C} = IX_{C}**

**V = (V _{R}^{2} + V_{C}^{2})^{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_{C}tand = 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 = I ^{2}R_{S} = I^{2}X_{C}tand = Q_{C}tand**

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 – cos ^{2}f)^{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).