Product Description:
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 steelmaking furnace (including onload and noloadhigh voltage and enclosing reactortype); the furnace transformers (single  phase, threephase onload and noload voltage regulating) usedfor smelting various ferroalloy, silicon compounds, mineralssuch as calcium carbide, are all the lowloss energysaving 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 shortcircuit （%）  Series reactor  Noload loss（Kw）  load loss（Kw）  Noload current（%）  Noload loss （Kw）  Load loss （Kw）  Noload 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  89  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  78  430 460  5.7  11.8 15.0  63 74  1.9 1.8    
HSZ9 series 35kV onload voltage regulatingelectricarcfurnace 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 shortcircuit （%）  Cooling  Noload loss （Kw）  Load loss （Kw）  Noload current（%） 
Constant power  Constant current 
10000  35 38.5  280240  240100  10  24056  19 levels， first 5 levels are constant power output and last 14 levels are constant current output  Dd0 Yd11 YNd11  78  OFWF or OFAF  20  130  1.4 
12500  314270  270116  11  26729  23  150  1.3 
16000  35335  305157  12  30287  28  180  1.1 
20000  392340  340158  13  33962  67  32  210  1.0 
25000  436380  380184  14  27984  39  240  0.9 
31500  489425  425201  16  42792  45  290  0.8 
40000  547475  485223  18  4819  52  350  0.7 
50000  610530  350250  20  54467  61  410  0.7 
63000  673585  585288  22  62176  68  480  0.6 
80000  760660  660310  25  69982  80  580  0.6 
 Q:How to test the transformer capacity
 Generally, the same method as the shortcircuit impedance test is used to test the temperature rise. The secondary side of the transformer is shortcircuited 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 3phase 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:Threephase 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 twostage 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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