Contents |
Notation | ||||||
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is used for some notation and formulae. If the
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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] |
For an ideal step-down auto-transformer with primary voltage V_{1} applied across
(N_{1} + N_{2}) primary turns and secondary voltage V_{2}
appearing across N_{2} secondary turns:
V_{1} / V_{2} = (N_{1} + N_{2}) / N_{2}
The primary (input) current I_{1} and secondary (output) current I_{2} are
related by:
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_{1} section and
(I_{2} - I_{1}) through the N_{2} section.
For a single-phase transformer with rated primary voltage V_{1}, rated primary current
I_{1}, rated secondary voltage V_{2} and rated secondary current
I_{2}, the voltampere rating 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_{1}, rated secondary phase voltage V_{2} and rated
secondary current I_{2}, the voltampere rating 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_{1} primary turns and N_{2} secondary turns is:
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_{1} primary turns and N_{2} secondary turns is:
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_{2} unloaded and
V_{2} at rated load, the per-unit voltage regulation
DV_{2pu} is:
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_{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} using:
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_{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_{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_{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}
= 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_{1} and X_{1} are primary values and
Z_{2}, R_{2} and X_{2} 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} = 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_{2star} are related by:
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_{2} 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} = R_{1} + R_{2}(N_{1}^{2}
/ N_{2}^{2})
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_{s}N_{r} / N_{s}
f_{r} = sf_{s}
For a rotor current I_{r}, the equivalent stator current I_{rs} is:
I_{rs} = I_{r}N_{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_{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.
The output power P_{m} for a load torque T_{m} is:
P_{m} = w_{s}T_{m}
The rated load torque T_{M} for a rated output power P_{M} is:
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_{s} and synchronous impedance Z_{s}, the terminal voltage 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_{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_{s}Z_{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.
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_{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_{a}I_{a}
= k_{m}w 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_{a}R_{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_{ f}Fw
= k_{m}w
T = k_{ f}FI_{a} = k_{m}I_{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_{a}I_{a}
= k_{m}w 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_{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_{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_{a}I_{a}
= k_{m}w I_{a}^{2}
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_{loss} comprising a fixed loss P_{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}.
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.
The ratio of resistances R_{2} and R_{1} is:
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_{1} (usually ambient temperature) and the hot winding resistance
R_{2} at temperature q_{2}, using:
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.
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).
Updated 09 June 2008 Copyright ©1999-2008 BOWest Pty Ltd |
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