The Design and Analysis of Poloidal Field Grid Power Supply System of HT-7 Superconducting Tokama

k

HT-7 ASIPP

L.W.Xu et al.,

Institute of Plasma Physics, Chinese Academy of Sciences, P.R.China

Outline

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• Introduction• The parameter selection of PF grid power supply system• The design of reactive compensation and harmonic suppress • Experiment results analysis

Introduction

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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

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-winding-transformer. 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. Becau

se 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

The parameter selection of PF grid power supply system

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2.2KV/0.42KVY-¦¤/¦¤-11-12

2.2KV/0.32KV

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10KV/1.4KV¦¤/¦¤-Y-12-11 ¦¤

EF Vertical Field CoilsOH Heating Field Coils

¦¤ Y

¦¤¦¤Y

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Fig .1 Connection diagram of PF power supply system

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.

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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 %

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

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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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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)

The design of reactive compensation and harmonic suppress

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1.Harmonic current calculation of PF power supply system

Because the OH HF and EF VF coils are large inductance loads, the current waveform of rectifier transformer secondary coil is square wave. According to connection diagram of Fig .1, the HF current IHF that injects

to three-winding transformer is a two-step-ladder wave . Through the Fourier series analysis, the PF current wave can be expressed as:

PFi

tttt 11sin858.287sin10.265sin544.36sin44.317PFi

t13sin418.24

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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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L5 R5 L5 R5 L7 R7 L7 R7 L5 R7 L7 R7

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L5 R5 L5 R5 L7 R7 L7 R7 L5 R7 L7 R7

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Fig .3 Connection diagram of filter

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3. The device of PF grid power supply system

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Experiment results analysis

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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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1. The analysis of harmonic current and voltage in 10KV

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.

0 200 400 600 800 10000

200

400

600

800

1000

Frequency (Hz)

Am

plit

ud

e

-1000-500

0500

1000

0 200 400 600 800 1000Frequency (Hz)

An

gle

(de

g)

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

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300MVA

5000KVA35/10KV

Uk=7.29%

35KV

55.82MVA

3150KVA35/10KV

Uk=7.12%

38.555MVA

3MVA10000/1400V

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Uk=(2 5)%

23.47 30.67MVA

Fig. 5 PF grid power supply distribution diagram

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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

0.20 0.21 0.22 0.23 0.24 0.25 0.26-15

-10

-5

0

5

10

15

I(*46.5A)

U(*1388)

0.00201(36.40)

U10

I 10

t(s)

Fig. 6 The voltage and current curve when IP=199KA

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100 120 140 160 180 200 220 2400

1

2

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5

COS¦Õ

Q

P

S

S,

P,

Q,C

OS

¦Õ

IP (KA)

Fig. 7 The curve of capacity, active power, reactive power

and power factor versus IP

Thank You!