Download No-load current interruption

Background
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Motor Description: 13.2kv, 7000hp, 3600rpm, FLA=253, LRA = 1563A
Performed simulation by interrupting at current zero’s. A first, then B/C at their next
current zero
Phase A = Black, B=Red, C = Blue. Each slide shows line voltages (thick lines),
pre-interruption currents (thin lines) and post-interruption motor-regenerated
voltages (thin lines). You can distinguish between the latter two apart based on
where they occur in time (before or after interruption)
All voltages are referenced to power supply neutral (although there appears not
much difference between power supply neutral and motor neutral). Continuity
between motor and power supply after interruption for purposes of determining
voltages is maintained by switching contacts to very high resistance value (rather
than open circuit) which draws an insiginficant current. Capacitive effects are not
considered, but I suspect capacitive effects will keep steady state motor neutral
near earth ground (excluding possible high frequency ringing during interruption)
which should be close to power supply neutral voltage anyway
Voltages are normalized by dividing by peak line to neutral voltage =
13,200*sqrt(2/3). Currents are normalized by dividing by peak locked rotor current
= 1563*sqrt(2)
1
Interruption from no-load condition steady-state
Motor Terminal Voltages
LineVoltages
Currents
1
0.8
0.6
0.4
0.2
0
-0.25.99
-0.4
-0.6
-0.8
-1
Speed (hz)/60
Vap_pn/10778
Vbp_pn/10778
Vcp_pn/10778
Vam_pn/10778
Vbm_pn/10778
Vcm_pn/10778
6
6.01
6.02
6.03
6.04
I1as/2210
I1bs/2210
I1cs/2210
Interrupt
Phase A
Interrupt
time
B and C
2
Interruption from 80% load steady state
Motor Terminal Voltages
LineVoltages
Currents
1
0.8
0.6
0.4
0.2
0
-0.23.99
-0.4
-0.6
-0.8
Interrupt
-1 Phase A
Speed (hz)/60
Vap_pn/10778
Vbp_pn/10778
Vcp_pn/10778
Vam_pn/10778
Vbm_pn/10778
Vcm_pn/10778
4
4.01
Interrupt
B and
time C
4.02
4.03
4.04
I1as/2210
I1bs/2210
I1cs/2210
3
Interrupt from 90% speed during acceleration unloaded.
Motor Terminal Voltages
LineVoltages
Currents
1
0.8
0.6
0.4
0.2
0
4.43
-0.24.42
-0.4
-0.6
-0.8
Interrupt
-1 Phase A
Speed (hz)/60
Vap_pn/10778
Vbp_pn/10778
Vcp_pn/10778
Vam_pn/10778
Vbm_pn/10778
Vcm_pn/10778
4.44
time
Interrupt
B and C
4.45
4.46
4.47
I1as/2210
I1bs/2210
I1cs/2210
4
Interrupt from 60% speed during acceleration (load prop to
speed^2 and 60% at full speed)
LineVoltages
Currents
1
0.8
0.6
0.4
0.2
0
-0.23.99
-0.4
-0.6
-0.8
Interrupt
-1 A
Phase
Motor Terminal Voltages
Speed (hz)/60
Vap_pn/10778
Vbp_pn/10778
Vcp_pn/10778
Vam_pn/10778
Vbm_pn/10778
Vcm_pn/10778
4
4.01
Interrupt
B and C
time
4.02
4.03
4.04
I1as/2210
I1bs/2210
I1cs/2210
5
Proposed intuitive explanation
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See equivalent circuit next slide, and choose synrchonous ref frame
Currents are initially dc.
At moment of interruption the D and Q stator supplies are both open circuited.
For no-load condition, the “dc” current which had been flowing through stator and magnetizing
branch simply switches to the rotor branch (flows in same direction through magnetizing
branch… direction of current flow in rotor branch is opposite normal). Magnitude of
magnetizing branch current is unchanged and therefore magnitude of residual terminal
voltage is unchanged from pre-interruption condition
For loaded condition, similar phenomenon occurs, but the magnetizing branch must first
overcome the initial rotor current through rotor leakage inductance flowing in normal direction.
The higher inductance of magnetizing branch still wins, but the resulting magnetizing current
is reduced.
For starting condition, again similar, but now the magnetizing branch current must overcome a
large rotor current flowing through rotor leakage inductance. Product of current and
inductance is same order of magnitude (but opposite direction0, resulting in very small
remaining magnetizing component and small remaining residual voltage.
Note we have ignored the rotor induced speed voltage term (voltage source), which is
approximately constant under the assumption that there is no change in magnetizing branch
current… good assumption for no-load but not so good for interruption during start. When we
consider the rotor induced voltage term it brings about cross-coupling in the 2 circuits and
makes it too hard to solve intuitively… requires simulation
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Krause’s equivalent circuit for transient analysis of SCIM.
0
0
qs = Lls*Iqs+Lm*(Iqs+Iqr)
ds = Lls*Ids+Lm*(Ids+Idr)
qr = Llr*Iqr+Lm*(Iqs+Iqr)
dr = Llr*Idr+Lm*(Ids+Idr)
w = ref frame radian speed *
wr  (Poles/2)* 2*pi* Rotor Speed **
Torque=1.5*P/2*Lm*(Iqs*Idr-Ids*Iqr)
* Most convenient reference frame is synchronous
i.e w = we = 2*pi*LF. In that case Vqs and Vds are dc.
(With proper choice of supply phase, Vds = 0)
** note wr is NOT 2*pi*RotorSpeed, instead it is adjusted
by number of pole pairs so that it matches we when
rotor speed is zero
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Practical application
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We know breaker should be capable of interrupting motor at
any time (no-load, full load, locked rotor, acceleration). But we
are still trying to determine if some interrupting conditions
may be more severe in terms of voltage across recentlyopened contacts.
• Simplistic Conclusion:
1. For interruption near running speed, residual voltage is close
to line voltage. Since motor field is very near sync speed, it
doesn’t drift out of phase during time-frame of interest, no
significant voltage across contacts
2. For interruption during start, residual voltage drifts out of
phase much quicker, but is much smaller. In most cases
expect less than 1*Vln across contact. Note for phase-tophase fault we would see 1*Vln or 1*VLL across the open
contact.
3. Neglected capacitive effects… simplistic analysis
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