SZ9 ON-load Tap-changer Power Transformer
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- Shanghai
- Payment Terms:
- TT OR LC
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- Supply Capability:
- 1000sets set/month
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1.Product Feature The sz9 on-load tap-changer power transformer is widely applies to urban and rural electric network construetion. The iron core is in ladder shape and changes the traditional piling way of iron core as well as the inner distribution of magnetic way. It also decrease the no-load consurmption, no-load current consurmption and noise pollution all by 20%. The new product applies CAD tool and marked with advanced design, reasonable structure, high quality material. It accords with GB1094-1996 and GB/T64511999. 2.Technical Specification
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- 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????
- In principal the AC (varying) part of a varying DC signal will be transmitted through a transformer, but in practice this is not much use unless the transformer is appropriate. Sometimes transformers are used with a DC bias (so varying AC signal superimposed on a DC signal), but it is avoided where possible. Example? SIngle ended transformer coupled amplifiers, commonly used in early transistor amplifiers and valve amplifiers. Varying a rheostat like that would result in very low frequency variations, and the lower the frequency the more unlikely it is to pass through a typical off the shelf transformer. Remember, only the variations can pass through. The overall current has to be low enough that the transformer's core (the iron part) does not become saturated. The DC part is blocked by the transformer. Normally a DC signal like this can be transmitted through a transformer by chopping it up into small parts, so it becomes more like an AC signal, then is passed through the transformer, and then can be converted back to an equivalent of transformed DC using a rectifier and filters. This idea is used in isolation amplifiers, DC - DC converters, chopper amplifiers, which you can look up for more detail. Transformers must have a varying magnetic field to work. DC fields only bias the core, so are usually considered a problem because the core must be designed (larger) for this condition. So depending on your goal, this is not a very valid thing to be doing, despite the examples above. I have to mention there are also magnetic amplifiers, where the transformer is powered by AC, and a DC bias (usually separate DC bias and varying signal windings) varies the saturation of the core. Thus the varying DC signal controls the amount of AC allowed to pass, and the AC output can be rectified and filtered to get a replica of the varying control signal with greater power.
- Q: Actually both amplifies the input voltage. A step up transformer amplifies the input voltage and a transistor also amplifies. So what is the differece????
- A okorder
- Q: I am completing an essay on how to improve the performance of a transformer. Please could someone advise me on the problems incurred in transformers and ways that have been designed to reduce these problems. Thanks in advance
- To improve the performance of a transformer, you have to:- [1] Reduce the copper losses, and [2] reduce the iron losses. Copper losses can be reduced by using copper wire of the highest purity; and by using copper wire that has the thinnest possible layer of insulating material on it. These measures, taken together, reduce the length of the wire and therefore its resistance. Iron losses are reduced by using magnetically soft iron in the laminations. This is the type of iron that has a tall and narrow hysteresis loop. Using copper wire with thin insulation (as mentioned above), keeps the windings close to the iron and so reduces magnetic leakage. A third measure is to use very thin laminations and to insulate them with very thin material as this will reduce eddy currents. It is very expensive to build a 99% efficient 250 MVA transformer. The losses in a 94% efficient version, are often cheaper to manage than the extra cost of making the 99% version.
- Q: Branch circuit cables are rated for 75?C and feeder cables are rated for90°C. This will be a 3-phase, 575-volt system with four induction motors, specified asfollows:Motor 1: 60 hp 0.90 p.f. squirrel cage motorMotor 2: 60 hp 0.90 p.f. squirrel cage motorMotor 3: 40 hp 0.85 p.f. wound rotor motorMotor 4: 7-1/2 hp 0.80 p.f. wound rotor motorA) Assuming no line losses, find the capacity of the transformer required to supplypower to this system.B) Assume that it is desired to improve the overall power factor for this system to 0.95lagging. Determine, in kVAR, the required capacitance for this power factorcorrection.C) Assume these motors have their windings connected in a delta configuration. Whatwould be the line voltage if they were connected in wye?
- I would look in the NEC for the full load amperes of these motor sizes. These are: 7.5 HP9 amps 40 HP.41 amps 60 HP.62 amps The 7.5 HP motor KVA will be KVA 575 * 9 * 1.732 8.9631 The 7.5 HP motor KW will be KW 575 * 9 * 1.732 * 0.8 7.17048 The KVAR of this motor is Sqrt(8.9631^2 - 7.1705^2) 5.3778 The 40 HP motor KVA will be KVA 575 * 41 * 1.732 40.8319 The 40 HP motor KW will be KW 575 *41 * 1.732 * 0.8 5 34.707115 The KVAR for this motor is Sqrt(40.8319^2 - 34.7071^2) 21.5095 The 60 HP motor KVA will be KVA 575 * 62 * 1.732 61.7458 The 60 HP motor KW will be KW 575 *62 * 1.732 * 0.8 5 55.5712 The KVAR for this motor is Sqrt(61.7458^2 - 55.5712^2) 26.9141 The total KVA requirement for all motors running at once is 8.9631 + 40.8319 + 61.7458 + 61.7458, which is 173.2866 KVA (note this is the requirement from a transformer, not the size of the transformer The total KW of all the motors is 7.1701 + 34.7071 + 55.5712 + 55.5712 153.0196 KW The total KVAR of all the motors is 5.3778 + 21,5095 + 26,9141 + 26.9141 80.7145 KVAR The power factor for all the motor is KW / KVA 153.0196 / 173.2866 0.883 I took a short cut at this point and used an application i wrote to calculate the required capacitor KVAR. The results follows: At 0.95 power factor, the motors' KVA will be 161.065 The motors' KW will be the same. The motors' KVAR will be 50.2926 The required capacitor's KVAR will be 53.988 The reactance of the capacitor will be 10.65 ohms The capacitance will be 3.49059248 E -4 Farads You can calculate these values as I did above, if you want to. EDIT Forgot the last answer. The line voltage is the same for both delta and wye. The leg voltage for the wye motor will be 575 / 1.732 331.98 volts TexMav
- Q: i need a little help with my homework. if you bank three single phase dual voltage transformers in a delta connection for the primary and do the same for the secondary, do the kva of primary add together or do the kva stay the same because it is connected in delta.it's a 2:1 480v - 240v each transformer is 5 kvais the total kva when connected in delta 15 kva or 5 kva and do you use the formula Pa line current times line voltage times 1.73 to find the rated load current.thanks in advance
- With 3-5KVA transformers connected inbank, your total capacity is 15Kva.To find the line current: Kva 1.732[V](I)/1000. Line current, I KVA{1000]/1.732{V) I 15[1000]/480*1.732 I 18.042, volts at primary.
- Q: I have been reading about pulse transformers and i am curious if there exists a pulse transformer that would take 1 pulse/second or even less and convert to very low voltage high current. If not, would it be possible to make one? Any info is very appreciated! Thx!
- Pulse transformers can be wound to any turns ratio that is desired. That being said pulse transformers have small cores and therefore handle small amounts of power. You can find cores on those old mother boards that are no longer in service and junk computer power supplies also have a couple of cores. These toroids were normally used as filter inductors to remove spikes from DC or low frequency lines. You can easily rewind them for pulse transformers.
- Q: How does the core material affect a transformer? Such as if you use steel vs. air. Or wood vs. magnet.
- A transformer is a gadget that converts an alternating (A/C) modern of a definite voltage to an alternating modern of diverse voltage, without substitute of frequency, via electromagnetic induction. A 'step up' transformer gets a low voltage and converts right into a miles better voltage, and a 'step down' transformer does in basic terms the opposite.
- Q: 10 (6) /0.4kv three-phase transformer what does it mean
- Therefore, the decision to 60 and 35kV winding connection to be careful when the connection must meet the transmission system voltage phasor requirements. The connection between 60 and 35kV windings is determined according to the relative relationship of the voltage phasors. Otherwise, even if the capacity of the voltage ratio is also right, the transformer can not be used, connected wrong, the transformer can not be connected with the transmission system. 3). Domestic 10,6,3 and 0.4kV transmission and distribution system phase also has two phases. In the Shanghai area, there is a 10kV and 110kV transmission system voltage phase difference of 60 ° electrical angle, this time can be used 110/35 / 10kV voltage ratio and YN, yn0, y10 connection three-phase three-winding power transformer, but limited Three-phase three-core core. 4). But note that: single-phase transformer in the combination of three-phase group connection, can not use YNy0 connected three-phase group. Three-phase shell-type transformer can not use YNy0 connection.
- Q: i wish to construct a transformer with 220/120v,60hz,5watt,plz help me
- Please don't! By the time you buy the special core material, the 2 types of copper wire, a good house content and life insurance for yourself, you can buy a finished and safe one from Maplin (if you are in the UK) for Pds.15.99 and that is a 45 Watt type. And finally, you say you want 60 Hz ? Then you also need 60Hz at the input, because a transformer doesn't change from 50Hz to 60Hz. Still want to know how it works? Read hyperphysics.phy-astr.gsu.edu/hba
- Q: I hate feeling the need to justify myself, but I've seen a lot of answers telling askers to do their own homework. I'm trying to do so, but can't find any equations in the book relating power to voltage as well as turn number. Is it PviN? I've just seen Pvi, but that wouldn't apply to multiple loops. This is for ideal transformers by the way, so power for the secondary and primary coils is the same right? But are their vi's equal, or their viN's?The primary coil of a transformer has 200 turns and the secondary coil has 800 turns. The power supplied to the primary coil is 400 watts. What is the power generated in the secondary coil if it is terminated by a 20-ohm resistor?
- The turns ratio is 4:1 of the primary. The power supplied to the primary is 400W. So regardless of anything else, in a ideal transformer, 400W is available on the secondary. But voltage will be 4x higher, current 4x smaller, than the voltage and current in the primary. Disregarding all that, if there is 400W of power flowing in the primary, then 400W will be being dissipated in the 20 ohm resistor. Watts in watts out, but secondary load (resistance) controls the wattage. There is no way to tell the output voltage and current, without knowing the input voltage and current. But whatever they are, the product of voltage x current will be 400W, in this instance.
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SZ9 ON-load Tap-changer Power Transformer
- Ref Price:
-
- Loading Port:
- Shanghai
- Payment Terms:
- TT OR LC
- Min Order Qty:
- -
- Supply Capability:
- 1000sets set/month
OKorder Service Pledge
Quality Product, Order Online Tracking, Timely Delivery
OKorder Financial Service
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