Furnace transformer of HS9 HSZ9 HSP9 series

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Furnace transformer of HS9 HSZ9 HSP9 series

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 loss（Kw）

load

loss（Kw）

No-load current（%）

No-load loss

（Kw）

Load loss

（Kw）

No-load current（%）

Rated Capacity

(kVA)

Reactance voltage drop

（%）

630

800 1000

6.3

10.5

200 1700 116 98

1819

2309 2887

No load voltage regulating

Dd0

Dy11

8-9

120

150

190

2.4

2.7

3.1

8.6

3.0

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

3.6

4.1

4.6

17.5

2.6

2.5

2.4

3.7

4.6

5.6

18.5

2.6

2.5

2.4

2500 3150

220 190 127 110

6561

8267

280

350

11.2

5.2

6.0

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

2.1

2.0

6300 8000

260 240 210 139

13990 17765

7-8

430

460

5.7

11.8

15.0

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（%）
Type	Primary voltage （kV）	Constant power	Constant current	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（%）
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

Q: actually we are trying to make a dc transformer,in which we want variable dc voltage at primary sidewe are doing this by using a rheostatwhich will change the value of current continously thus accordingly the value of voltage varyare we thinking right or wrong????: A dc transformer is impossible.Voltage transformation happens when there is an operating frequency and dc circuits has zero frequency. You may use voltage divider network for voltage reduction.

Q: I am learning about transformers and one of the items to calculate is the impedance. I need to calculate regular 60 Hz Core and Coil Shell Type Transformers Al wire in secundary, Cu Wire primary and use Epoxy paper for insulation.: It's pretty difficult to calculate the impedances (there's more than one) entirely from the transformer's design data. It's easier and more usual to decide on an equivalent circuit (there are plenty available depending on how well you want to model the transformer) and then to determine the parameters by a mix of calculation and measurement. A pretty basic equivalent circuit which refers all impedances to the primary side, has the primary terminals connected first by Xm and Rc in parallel (the magnetising reactance and core loss resistance) and then by a third parallel branch containg a series connection of leakage reactance Xl, winding resistance Rw and an ideal transformer of turns ratio Np/Ns. Xm is best obtained from an open circuit test but could be calculated as the inductance of the primary winding. For the latter you'd need to know core dimensions, number of winding turns and the magnetising characteristic of the core iron. If you have conductor sizes and conductivities you can calculate the winding resistance Rw, referring the secondary part to the primary by multiplying it by (Np/Ns)?. You can deduce Rc from the losses measured on open circuit at nominal voltage and Rw from dc resistance measurements on the windings. If you know the specific hysteresis and eddy current losses of the core material, you can also have a good stab at calculating Rc. The leakage reactance is quite difficult to calculate from first principles - even designers usually resort to some empirical factors. Basically it's determined from the short circiut test which is at nominal secondary current. The ratio of primary volts to secondary surrent (referred of course) on secondary short circuit will get you close to Xl - you can adjust for Rw which can be determined from the losses on short circuit (core losses are absent here!) or by the two methods indicated above. It's normally Xl that's referred to as the transformer reactance and together with Rc and Rw, the impedance.

Q: What is the reason why the dry transformer is loud: 500kva below the dry-type transformer noise 56 to 58 dB or so, 630kva above 62 to 65 dB or so! This is our factory data. The dry-type transformer sound is great. Big like cattle called the same, because the core is loose!

Q: I understand that a 70-volt transformer is used in a commercial environment, to connect several 70-volt speakers to a 70-volt amplifier. But why does a 70-volt transformer have all of those different colored wires coming from it? What do they connect to, and what are they for? How would I connect a 70-volt transformer to an amplifier, and where I connect the cable that goes to the speakers? Please explain this to me in an easy to understand way, because I was curious about this for a long time.: The okorder

Q: I use transformers in my circuitry all the time but when a transformer is say a secondary 12v 1a transformer than what does it mean by 1 ampis it that the secondary winding wire will pop after 1 amp doo to the simple thinness of the wire? or that the ohms of the secondary coil, when using iv/r, 112/12so does this tell you that the coil is 12 ohms?please don't give me other information about the transformer just answer the question please.: NO, IT DOES NOT MEAN IT HAS 12 OHMS IMPEDANCE. IF YOU SHORT THE 12V TERMINALS, IT COULD PRODUCE MORE THAN 1A BEFORE IT TURNS INTO SMOKE. 1A IS THE MAXIMUM SAFETY CURRENT LIMITATION TO BE OBTAINED FROM 12V TERMINALS TO MAINTAIN 12 VOLTS WITHOUT DROPPING.

Q: An AC adapter for a telephone answering machine uses a transformer to reduce the line voltage of 120 V to a voltage of 8.00 V. The RMS current delivered to the answering machine is 570 mA. If the primary (input) coil of the transformer has 600 turns, then how many turns are there on the secondary (output) coil?What is the power drawn from the electric outlet, if the transformer is assumed to be ideal?What is the power drawn by the transformer, if 13.5 percent of the input power is dissipated as heat in the coils and in the iron core of the transformer?: Ns/Np Vs/Vp Ns (8/120) * 600 40 turns P Pp Ps Vs * Is (an ideal transformer) 8 * 0.570 4.56 watt 13.5 percent of the input power is dissipated, then the power from output coil (100 - 13.5) % 86.5% of input power. ---Po 86.5% Pi 4.56 0.86 Pi ---Pi 4.56/0.86 5.30 watt

Q: A 120-V, 60-H air conductor is to be operated in a remote area where the voltage drop in the long transmission line results in a utilization voltage of 102-V. Determine:A) the required step-up voltage ratio for the satisfactory performance.B) The voltage ratio of a standard buck-boost transformer that most closely meets the requirements of the loadC) The voltage at the load with the buck-boost transformer installedD) Sketch the appropriate connection diagram and show the NEWM standard terminal markings: required step up voltage ratio load voltage/ primary(or) input voltage k 120/1021.176. a standard buck-boost transformer uses a 120 X 240V primary and a 12X24V or 16X32V secondary. available buck-boost transformer ratio: as the input supply is 102V(around 120V), we'll consider 120V on primary side of auto transformer. a(120+12)/1201.100 a(120+24)/1201.200 a(120+16)/1201.133 a(120+32)/1201.266 hence, the voltage ratio that closely meets the requirement of load is a'1.200 hence, a buck-boost t/f with 24V secondary is required the voltage at load with B-B transformer a' X input supply 1.2 X 102122.4V

Q: Exactly what is the storyline of the Transformers. As far back as I know, there were two brother, Unicron and Primus who fought each other because Unicron wanted to suck up Universes. But where did these two come from? Does it even say?: I am a fan but I haven't been able to find that out myself, if you find out let me know.

Q: I have a Whirlwind microphone splitter box with a direct out and a transformer isolated output. What's the reason for the transformer isolated output?: This Site Might Help You. RE: What does an isolated transformer output on a microphone splitter do? I have a Whirlwind microphone splitter box with a direct out and a transformer isolated output. What's the reason for the transformer isolated output?

Q: How does a transformer effect amplitude modulation in an am transmitter. i would really love this to be explained from the basics as i am not so advanced in electronics. but i would really like to comprehend how this happens form the basics. thank you: The collector of the output transistor or plate of the output vacuum tube in a high power AM transmitter has DC voltage and an audio signal applied to it. The audio signal causes the the transistor/ tube to create AM. The AM radio frequency is shunted off to a tank circuit and then coupled to the antenna. A transformer is used to couple the audio signal to the transistor's collector/ tube's plate. A high power audio signal is applied to the primary side of a transformer and the secondary of the transformer is connected in series to the DC power source to the output device. This creates a varying DC voltage to the output device which creates Amplitude Modulation.

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