Furnace transformer of HS9 HSZ9 HSP9 series

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Furnace transformer of HS9,HSZ9,HSP9 series


1.      Product introduction

The furnacetransformer is a transformer for power supply to furnace electricalsource. It is used to reducea voltage from a high voltage to an operational voltage needed by furnace.
Accordingto many different types of furnaces,there are many varieties of furnace transformer to fit it.At present, the furnace transformers produced by ourcompany are: electric arc furnace transformerused for steel-making furnace (including on-load and no-loadhigh voltage and enclosing reactortype); the furnace transformers (single - -phase, three-phase on-load and no-load voltage regulating) usedfor smelting various ferroalloy, silicon compounds, mineralssuch as calcium carbide, are all the low-loss energy-saving products.

2.    Technicalparameter

Furnacetransformer of HS9 HSZ9 HSP9 series

Rated Capacity

(kVA)

With series reactor

Without series reactor

Primary voltage

kV

Second voltage

kV

Rated second current

A

Voltage regulation mode

Label of connection

Impedance of short-circuit

%

Series reactor

No-load lossKw

load

lossKw

No-load current%

No-load loss

Kw

Load loss

Kw

No-load current%

Rated Capacity

(kVA)

Reactance voltage drop

%

630

800 1000

6

6.3

10

10.5

11

200 1700 116 98

1819

2309 2887

No load voltage regulating

Dd0

Dy11

8-9

120

150

190

19

2.4

2.7

3.1

8.6

11

14

3.0

2.9

2.9

2.2

2.7

3.1

11.0

13.5

16.0

3.0

2.9

2.8

1250 1600 2000

210 180 121 104

3437

4399 5499

200

260

320

16

3.6

4.1

4.6

17.5

22

27

2.6

2.5

2.4

3.7

4.6

5.6

18.5

24

28

2.6

2.5

2.4

2500 3150

220 190 127 110

6561

8267

280

350

11.2

5.2

6.0

32

39

2.3

2.2

6.7

8.0

34.5

41.5

2.3

2.2

4000 5000

240 210 139 121

9623 12028

340

360

8.5

7.6

8.4

46

54

2.1

2.0

6300 8000

260 240 210 139

13990 17765

7-8

430

460

5.7

11.8

15.0

63

74

1.9

1.8


HSZ9 series 35kV on-load voltage regulatingelectric-arc-furnace transformer technical parameter

Type

Primary voltage

kV

Secondary voltage

Secondary level voltage

V

Rated secondary current

A

Voltage regulating levels

Label of  connection

%

impedance of  short-circuit

%

Cooling

No-load loss

Kw

Load loss

Kw

No-load  current%

Constant power

Constant current

10000

35

38.5

280-240

240-100

10

24056

19 levels first 5  levels are constant power output and last 14 levels are constant current  output

Dd0

Yd11

YNd11

7-8

OFWF

or OFAF

20

130

1.4

12500

314-270

270-116

11

26729

23

150

1.3

16000

353-35

305-157

12

30287

28

180

1.1

20000

392-340

340-158

13

33962

6-7

32

210

1.0

25000

436-380

380-184

14

27984

39

240

0.9

31500

489-425

425-201

16

42792

45

290

0.8

40000

547-475

485-223

18

4819

52

350

0.7

50000

610-530

350-250

20

54467

61

410

0.7

63000

673-585

585-288

22

62176

68

480

0.6

80000

760-660

660-310

25

69982

80

580

0.6


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Q:How to test the transformer capacity
Generally, the same method as the short-circuit impedance test is used to test the temperature rise. The secondary side of the transformer is short-circuited and the voltage is applied once, and the secondary current reaches the rated value. Wait until the transformer temperature changes less than 1 degree / hour later, and then maintain 3 hours, this time the transformer temperature rise is the final test value.  
Q:630 kilovolt transformer secondary side of the current is how much
630kVA that his rated capacity, it is ideal to say that with the load of 630kW (but actually can not bring so much) Rated current calculation formula: rated current = capacity / rated voltage / root number 3 (for 3-phase transformer) 630 / 0.4 / 1.732 = 909.3 A Pure hand to play, hope to adopt ~
Q:630KVA transformer can use the maximum number of power appliances
600KVA box can generally increase the number of households with electricity
Q:Three-phase transformer how to calculate the current, ah, the formula is?
For example: 10KV / 0.4KV, 500KVA transformer Primary current: I1 = 500 / (1.732 * 10) = 28.9A Secondary current: I2 = 500 / (1.732 * 0.4 = 721.7A
Q:What is the critical load of the economic operation of the transformer?
In order to increase the amount of active loss caused by the active loss of the system in the power system, a conversion factor is introduced, that is, the reactive power economic equivalent. Reactive power economic equivalent, is that the power system to send 1kvar reactive power, the power system will increase the number of active power loss kw, the symbol kq, unit kw / kvar. This kq value and the power system capacity, structure and calculation of the specific location and other factors. For the factory substation, reactive power economic equivalent kq = 0.02 ~ 0.15; Kq = 0.05 ~ 0.08; for the three or more transformer factory, take kq = 0.1 ~ 0.15 ~ 0.04; for the two-stage transformer factory, take kq = The
Q:Transformer copper row selection
A general use of cable out, select 120mm2 line can be; second to use two 100 × 10mm copper, copper is to tin treatment. Copper specifications table recommended to find the national standard, see the copper discharge capacity and installation methods, after all, the transformer must be produced to meet the national standard.
Q:Will the installation of transformers, power supply bureau is how the charges?
Electricity Bureau to receive construction, survey, design, equipment, measurement fees. The plane fee for the capacity of the general 250KVA above the transformer to charge capacity fee 1KVA a month generally 25 yuan. Industrial electricity is divided into peak and valley electricity price difference, agricultural electricity regardless. Details can call me
Q:What is the rated current of the 80KVA transformer?
Rated current formula: I = P / 1.732 × U = 80000 / 1.732 × 380 ≈ 122 (A) (Current contains active and reactive current)
Q:How do I choose a transformer? The
Select the transformer, you can not choose too large, can not choose too small According to the following method to select the transformer capacity.    There is a very important data is not provided, is the load at the same time coefficient. Transformer with the actual load and equipment rated power ratio is called the transformer load factor. The concept of the load factor for the load is the probability that the load is used at the same time, also called the coefficient or the simultaneous coefficient, which is the probability of simultaneous use of the device. The total load is 300kw, but they are used at the same time there is a probability that the probability is the load factor. It is impossible to always use at the same time. Of course, you can according to the actual situation to calculate their own, power factor selected 0.80.   With this formula, s = p * kX / cosφ Transformer capacity s = device rated power p × transformer load rate kx / power factor cosφ = (300 × kX) × 0.80 This is the transformer capacity.    Do not know what the motor is used under the circumstances, so it can not be calculated, and only calculate the same time after the generation of the formula into the above formula. If the coefficient is 0.9, then the results of about 340 kVA, are used at the same time, that is, 375KVA, consider a certain margin and spare capacity, that can choose 400KVA. In addition to the above considerations, in particular, consider the motor starting current factor, select the transformer more reliable. As far as possible by the big do not rely on small. And the standard capacity of the transformer level 315,400,500,630KVA, 315KVA a little small, 500KVA big and no need. So choose 400kvA it
Q:100kVA above the transformer grounding resistance is how much
National specifications for electrical equipment grounding resistance requirements: 4.1.1 parallel operation of the transformer and other power equipment capacity of not more than 100 kVA, the grounding resistance should not exceed 10 Europe.

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