Transcript ht-7 asipp

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HT-7
The Design and Analysis of Poloidal Field Grid
Power Supply System of HT-7 Superconducting
Tokamak
L.W.Xu et al.,
Institute of Plasma Physics, Chinese Academy of
Sciences, P.R.China
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Outline
• Introduction
• The parameter selection of PF grid
power supply system
• The design of reactive compensation
and harmonic suppress
• Experiment results analysis
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Introduction
In past experiments, the electric power which the PF coils in HT-7
tokamak were supplied by a 120MW ac flywheel generator. Because
the ac flywheel generator was in idling at most times, so the ratio of
electric utility is very lower. In the meantime, we found that it has the
disadvantage of large noise, complex operation, difficult maintenance
and high cost (about 6000 RMB a day) .Based on the actual running
parameters of HT-7 PF and the capacity of our transformer
substation , we puts forward a new project – PF grid power supply
system to replace ac flywheel generator power supply system .
Through two experiment campaigns of 2001-2002 d 2002-2003, the
running results Show that this project satisfied the requirements of
HT-7 experiment and criterions
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The parameter selection of PF grid
power supply system
1. Parameter selection of three-winding transformer
The HT-7 PF includes two parts: OH heating field (HF) and EF vertical field
(VF), The HF is supplied by two rectifiers in series, and the VF is supplied by one
rectifier. Because the ac flywheel generator outputs 6 phases that their phasic
difference is 300, so the converter transformer is adopted reverse-three-windingtransformer. In the experiments of HT-7, the current of HF and VF are IP/80+0.5
and 0.031×Ip (KA) respectively, where Ip (KA) is the discharge current of plasma.
Because Ip is less than 200KA in most experiments, so we take 200KA as the
maximum current of plasma. When the output voltage of ac flywheel generator is
1700V, the maximum dc output voltage of HF rectifier and VF rectifier are 668V
and 438V. In actual experiment, when IP ≤ 200KA, the trigger angle αof HF
rectifier and
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VF rectifier are more than 500 and 400 respectively. So the required dc
output voltage of HF rectifier and VF rectifier are
UdHF=668×cos500=430V.
UdVF＝438×cos400=336V.
¦ ¤/¦ ¤-Y-12-11
10KV/1.4KV
¦ ¤-Y/¦ ¤-12-11
2.2KV/0.32KV
¦¤
Y
¦¤
Y
¦¤
¦¤
¦¤
Y
¦ ¤ Y-¦ ¤/¦ ¤-11-12
2.2KV/0.32KV
OH Heating Field Coils
¦¤
Y
¦¤
Y-¦ ¤/¦ ¤-11-12
2.2KV/0.42KV
EF Vertical Field Coils
Fig .1 Connection diagram of PF power supply system
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We take α=150 when dc output voltage of HF rectifier and VF rectifier are
430V and 336V , so the maximum dc output voltage (α=00) of them are
445V and 348V. and the secondary line voltage of rectifier transformer can
be obtain:
U2HF＝165 V
U2VF＝258 V
Through above analysis and based on the turn ratio of rectifier transformer,
the primary line voltage is:
U1HF＝165×6.875＝1134 V
U1VF＝258×5.24＝1351V
In order to satisfy the requirement of HF and VF simultaneously,
So the parameter of three-winding transformer is selected as
U2=1400 V
S=5 MVA
Uk=5 %
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2. Capacity calculation of HT-7 PF power supply system
The maximum dc output voltage (α=00) of HF and VF are 550V
and 360V respectively. According to the design parameters of PF, the
resistance of HF coil and VF coil is 35mΩ. Based on this data, the
active power, reactive power and total capacity of PF can be
calculated. Fig .1 is the curves of PF power versus IP variation, when
IP=200KA, the active power, reactive power and total capacity of PF
are 1.66MW, 3.4MVAR,and 3.78MVA respectively. QC is the reactive
compensation capacity when the power factor is compensated to 0.8.
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4.0
PQS
3.5
S
3.0
Q
2.5
2.0
QC
1.5
1.0
P
0.5
0.0
0
50
100
150
200
IP(KA)
Fig .2 the curve of PF power versus IP
P－active power (MW) Q－reactive power (MVAR)
S－total capacity (MVA)
QC－reactive compensation capacity (0.8) (MVAR)
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The design of reactive compensation
and harmonic suppress
1.Harmonic current calculation of PF power supply system
Because the OH HF and EF VF coils are large inductance loads, the
i
current waveform
of rectifier transformer secondary coil is square wave.
According to connection diagram of Fig .1, the HF current IHF that injects
PF
to three-winding transformer is a two-step-ladder wave . Through the
Fourier series analysis, the PF current wave can be expressed as:
i PF  317.44 sin  t  36.544 sin 5 t  26.10 sin 7 t  28.858 sin 11 t 
24.418 sin 13 t
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2. Parameter design of reactive and harmonic compensation
From above analysis, we know that PF power supply system inject large
reactive power and harmonic current, so the measurements must be taken
to suppress these reactive and harmonic. In the design of this device, we
adopted fixed LC filter to compensate reactive power and suppress
harmonic current. Because the impedance and frequency of grid may be
fluctuated in the actual running, and it perhaps get rise to parallel resonance,
therefore, a resistance must be added to the filter branch to suppress the
harmonic over-voltage.According to Fig. 2, the required compensation
capacity is about 1MVar if the power coefficient is compensated to 0.8, so
we take 1.2Mvar as the compensation capacity.
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If the resistance is increased, the harmonic enlarge times can be limited in a
lower value, and the resonance is suppressed, on the other hand , this can result
in a worse filter characteristic, so we must consider two factors to reach a
satisfied effect.
The parameter of filter are
Total compensation capacity :
1.2MVar
Phase capacity: 200KVar ( each branch of 5th and 7th 100KVar
5th branch parameter: C5＝393.175μF L5＝1.032 mH R5＝32Ω (Q=20)
7th branch parameter: C7＝393.175μF L7＝0.526 mH R5＝23Ω (Q=20)
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2
3
4
A
A
5
C
C
B
D
B
L5
C
R5
L5
R5
L7
R7
L7
R7
L5
R7
N2
N1
L7
R7
L5
R5
L5
R5
L7
L7
Fig .3 Connection diagram of filter
R7
L5
R7
N2
N1
DH
B
R7
DH
L7
R7
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3. The device of PF grid power supply system
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Experiment results analysis
Through 2001-2002 and 2002-2003 two experiment campaigns, the PF grid
power supply system satisfied the requirement of HT-7 experiment
completely. From the analysis of acquired waves, it obtain the anticipated effect.
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Frequency (Hz)
Angle(deg)
0
100
200
300
400
500
600
700
800
1000
500
0
-500
-1000
120
1. The analysis of harmonic current and voltage in 10KV
100
Amplitude
80
64.26
60
40
42.28
28.53
40
20
0
0
100
200
300
400
500
600
700
800
Frequency (Hz)
Fig. 3 PF harmonic current analysis diagram when I P =143KA
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Table1 harmonic current value (10KV)
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When harmonic current is injected 10KV grid, it can produce
harmonic voltage because of grid impedance. This harmonic voltage may
bring harm to other user. So the harmonic voltage must be limited.
Angle(deg)
Frequency (Hz)
0
200
400
600
800
1000
0
200
400
600
800
1000
1000
500
0
-500
-1000
1000
Amplitude
800
600
400
200
0
Frequency (Hz)
Fig. 4 Harmonic voltage analysis diagram (10KV)when I P =199KA
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Table 2. Harmonic voltage coefficient ( 10KV)
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2.The analysis of voltage drop ratio
3
4
300MVA
A 35KV
5000KVA
35/10KV
Y
3150KVA
35/10KV
Y
Uk=7.29%
Uk=7.12%
B
55.82MVA
D
38.555MVA
Y
3MVA
10000/1400V
Y
C
Uk=(25)%
23.4730.67MVA
Fig. 5 PF grid power supply distribution diagram
5
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Table 3 . The voltage drop ratio (10KV)
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3. The analysis of capacity, active power, reactive
power and power factor
U(*1388)
15
I(*46.5A)
10
U10 I10
5
0
-5
-10
0
0.00201(36.4 )
-15
0.20
0.21
0.22
0.23
0.24
0.25
0.26
t(s)
Fig. 6 The voltage and current curve when IP=199KA
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5
S
4
S, P, Q,COS¦Õ
P
3
Q
2
1
COS¦Õ
0
100
120
140
160
180
200
220
240
IP (KA)
Fig. 7 The curve of capacity, active power, reactive power
and power factor versus IP
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