SatCon Technology Corporation
Technology Center
Cambridge, MA 02142
Real and Reactive Power Balance in the Slip Ring Machine
c
2001
SatCon Technology Corporation
1
January 17, 2004
Introduction
In preparation for using a doubly fed (slip ring) machine for an uninterruptible power supply, it
is necessary to understand the real and reactive power flows in the machine. The purpose of this
note is to derive the relationships between energy conversion (power converted from mechanical
to electrical form) and power flows through the electrical terminals of the machine. As it turns
out, there are fairly simple relationships between these power flows: generally they are scaled by
frequency. One could think of the machine as a transformer with one little twist: the frequencies
seen by the different windings are scaled by rotation of the machine. So while currents are as one
would expect in a transformer– the same on both sides of the machine when turns ratio is taken
into account, voltages are proportional to electrical frequency in each winding. And of course the
frequencies of rotor and stator differ by the rotational speed of the machine. Thus if the machine
is operating at a frequency less than synchronous and is called for to generate (to take energy
out of rotational kinetic form and turn it into electrical), it is necessary to inject power into the
rotor terminals in proportion to the output power multiplied by the per-unit slip. Similarly, if
the machine is supersynchronous, it is necessary to sink power from the rotor terminals. Reactive
power follows similar rules.
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Machine description
The slip ring motor is a physical embodiment of the ’tranformer model’ of the induction machine
that is used in a first subject to describe induction motors. It has a conventional stator winding
and a nearly identical winding on the rotor. Both windings are generally polyphase and balanced.
The rotor winding is connected to a set of (generally 3) slip rings which are engaged by brushes
and so brought out to the stationary frame.
When one does the Park’s Transformation on the machine, it can be shown in a straightforward
but tedious fashion that the voltages in the (virtual) d- and q- axis windings of the machine are
described by:
Stator:
vds =
vqs =
dλds
− ωe λq + ra ids
dt
dλqs
+ ωe λd + ra iqs
dt
Rotor:
vdr =
dλdr
− ωr λq + rr idr
dt
1
dλqr
+ ωr λd + rr iqr
dt
vqr =
It is well known that mechanical torque in an electric machine is:
Te =
3
p (λds iqs − λqs ids )
2
where p is the number of pole pairs of the machine. And the flux-current relationships for the two
axes of the machine are similar:
3
"
λds
λdr
#
"
λqs
λqr
#
=
"
Ls M
M Lr
#"
ids
idr
#
=
"
Ls M
M Lr
#"
iqs
iqr
#
Real Power
Assume steady state operation and ignore resistance. We may define an equivalent mechanical
frequency by using the number of pole pairs p and mechanical rotational speed Ω:
ωm = pΩ
then mechanical power is:
3
p (λds iqs − λqs ids )
2
Now look at real power into the stator terminals of the machine:
Pm = ΩTe =
Ps =
=
=
3
vds ids + vqs iqs
2
3
ωe (λds iqs − λqs ids )
2
ωe
ωe
Pm =
Te
ωm
p
That is, the electrical frequency into the stator terminals is the same as the mechanical power,
scaled by the ratio of electrical frequency to mechanical ’frequency’. This is familiar to induction
motor people. Now, it is also well known that the rotor frequency ω r , stator frequency ωe and
mechanical speed ωm are related by:
ωe = ω m + ω r
and when dealing with induction motors it is usual to define a ’per-unit slip’ (generally just called
’slip’) as
ωr
s=
ωe
so that
ωm = ωe (1 − s)
2
then we have the well-known relationship:
Pm = Ps (1 − s)
Now look at rotor input power:
Pr =
=
3
(vdr idr + vqr iqr )
2
3
ωr (λdr iqr − λqr idr )
2
this result looks intuitively right as it is just like the same result for the stator and the rotor
and stator are physically similar. But to relate those results we must invoke the flux/current
relationships:
λdr = M ids + Lr idr
λqr = M iqs + Lr iqr
λds = Ls ids + M idr
λqs = Ls iqs + M iqr
Now if we use these expressions in place of the fluxes in the expressions for power above, we
obtain for rotor electrical power:
Pr =
=
3
ωr ((M ids + Lr idr )iqr − (M iqs + Lr iqr )idr )
2
3
ωr (M (ids iqr − iqs idr ))
2
Note that two terms (±Lr idr iqr ) cancel.
The same calculation done for stator input power is:
Ps =
=
3
ωe ((Ls ids + M idr )iqs − (Ls iqs + M iqr )ids )
2
3
ωe M (idr iqs − iqr ids )
2
Now, comparing these two expressions, we can see that rotor input power is simply related to stator
input power (and mechanical output power) by:
Pr = −
ωr
ωr
ωr
Ps = −
Pm = − Te
ωe
ωm
p
This stands to reason: in an ordinary induction motor, in which the rotor is turning a bit
less than synchronous speed so all of the frequencies here are positive, with the motor producing
positive mechanical power and consuming positive electrical (stator) power, the rotor terminals are
producing slip power (which is of course being dissipated in the rotor resistance). This is the basis
for slip energy recovery circuits which attempt to salvage this rotor power. On the other hand,
if the machine is generating, meaning it is producing electrical power from the stator terminals
and absorbing mechanical power, either the slip is negative (ω m > ωe ) or the rotor input power is
positive, meaning power must be injected.
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3.1
Power Balance
Another way of looking at this is to note that our ordinary notion of power conservation holds:
Pm = P s + P r = T e
ωs − ω r
p
Now consider running this machine with positive power into both the stator and rotor terminals,
so that stator electrical frequency is less than synchronous speed. Suppose we note the fraction as
α, so that
ωe = αωm
This would suggest that rotor frequency is
ωr = (α − 1) ωm
So that stator and rotor power inputs are
Ps = αPm
Pr = (1 − α) Pm
This would suggest that, for example, a 2-pole machine operating at a speed of 10,800 RPM
and a stator frequency of 60 Hz (equivalent to 1/3 of the mechanical speed) would require a rotor
input frequency of 120 Hz. Stator input power would be 1/3 of mechanical power and rotor input
power would be 2/3 of mechanical power.
Conversely, operating as a generator, the rotor would have to provide reactive power (excitation)
and both the rotor and stator would have negative (motor sense) power meaning that real power
would come out of the machine from both sets of terminals, again in the ration of 1:2, stator:rotor.
4
Reactive Power
Reactive power is the product of those components of voltage and current that are out of phase by
90 degrees. Positive reactive power is defined so that current lags voltage.
In the d-q coordinate system with positive frequency, reactive power at some terminal set is
defined by:
3
Q = (vq id − vd iq )
2
This is consisent with the notion that reactive power is positive for currents which lag voltages.For
negative frequency, as would occur at the rotor terminals for supersynchronous rotational speeds,
the sign of Q is reversed (or the roles of d- and q- axes).
Now, reactive power at the stator terminals is:
3
Qs = ωe (λds ids + λqs iqs )
2
Now if we use the flux-current relationship to substitute for fluxes in these expressions, we find,
Qs =
=
3
ωe ((Ls ids + M idr )ids + (Ls iqs + M iqr )iqs )
2
3 ωe Ls (i2ds + i2qs ) + M (idr ids + iqr iqs )
2
4
L1
L2
i
i
ds
i
φd
dr
M
Figure 1: Transformer Model
Now, with reference to Figure 1, note we can relate stator and rotor currents to magnetizing current, and note that the total stator and rotor inductances are the sum of a magnetizing inductance
and a leakage inductance:
Ls = M + L 1
Lr = M + L 2
iφd = ids + idr
iφq = iqs + iqr
It is usual to think of these expressions as applying to a model in which everything has been referred
through the turns ratio so that there would be an ideal transformer somewhere. This is not, as it
turns out, necessary, but if there is a turns ratio some of the leakage inductances turn out to have
negative value and that makes intuition difficult. Stator reactive power becomes:
Qs =
=
=
3 ωe (L1 + M )(i2ds + i2qs ) + M (ids (iφd − ids ) + iqs (iφq − iqs ))
2
3 ωe L1 (i2ds + i2qs ) + M (ids iφd + iqs iφq )
2
3 ωe L1 (i2ds + i2qs ) + M (i2φd + i2φq ) − M (iφd idr + iφq iqr )
2
This may be interpreted as three different physical quantities: First, there is reactive power consumed by stator leakage:
3
Q1 = ωe L1 (i2ds + i2qs )
2
Then there is reactive power required to excite the air-gap of the machine:
3
Qm = ωe M (i2φd + i2φq )
2
the third term is a reactive power contribution from the rotor side. We will call it:
Qg =
3
ωe M (iφd idr + iφq iqr )
2
So that the reactive power at the stator terminals is:
Qs = Q 1 + Q m − Q g
5
Now consider reactive power at the rotor terminals:
Qr =
=
=
=
3
|ωr | (λdr idr + λqr iqr )
2
3
|ωr | (L2 + M )(i2dr + i2qr ) + M (idr ids + iqr iqs )
2
3
|ωr | L2 (i2dr + i2qr ) + M (i2dr + i2qr ) + M (idr (iφd − idr ) + iqr (iφq − iqr )
2
3
|ωr | L2 (i2dr + i2qr ) + M (idr iφd + iqr iφq )
2
Note that the last term is just a scaled version of Q g , so that if we note reactive power consumed
by rotor leakage as:
3
Q2 = |ωr |L2 (i2dr + i2qr )
2
Then rotor reactive power is:
|ωr |
Qg
Qr = Q 2 +
ωe
Now, note that if the machine is called on to supply some reactive power at its load terminals,
so that
Qs = −QL
The reactive power that must be supplied to the rotor terminals to supply the load reactive power
and excite the machine is:
|ωr |
(QL + Q1 + Qm )
Qr = Q 2 +
ωe
and of course the real power relationship is the same:
Pr =
ωr
PL
ωe
Note that these expressions will hold true even if the rotor frequency is negative! In that case,
the rotor must sink real power, in proportion to the load side real power. The same is not true of
reactive power, however, as when rotor frequency is negative the phase relationship between voltage
and current reverses.
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