S (G)B 10H grade insulating dry-type power transformer
G -----------------------------non-encapsulation dry type
B----------------------------- LV foil coil
10 ----------------------------- performance code
□----------------------------- rated capacity (KVA)
□----------------------------- rated voltage（HVKV）
SG10 type-this series dry-type distributing transformer has strong withstanding of thermal shock, overload capability. Beside, it has many good points such as flame-retardant, low-loss, and low partial discharge capacity, low noise, no harmful gas and no pollution. It is insensitive of humidity and dust, compacted, no crack and so it is easy to be maintained. Therefore, it is suitable for the using in the adverse circumstances of high requirement for fire protection, high degree of load fluctuation and dirty, moist, such as airport , power plant, metallurgy, hospital, skyscraper, shopping center, and residential areas and other special environments like oil, chemical industry, nuclear plant, nuke, etc.
1. Technical parameters
- Q:What is the transformer capacity ratio? Capacity than 100%.
- The ratio of the maximum capacity to the minimum capacity. The ratio of capacity to 100% is the ratio of maximum capacity to minimum capacity of 1: 1.
- Q:Average transformer voltage ? rated power rating?
- A transformer rating is specified as a number of volt-amps (VA) where A is the RMS current. In cases where there are rectifiers, the RMS current may be far greater than the average current. This means that the power that can be delivered will be far less than the VA rating.
- Q:Transformer size problem (kVA)?
- There is no actual formula that will size the x-former for a building but I can give you some help::: Add up all the loads lets say you have(main IDEA of the question) 20000 watts at 120 v 10000 watts at 240v for a total of 30000 watts calculating the secondary amps you use ohms law for single phase PI x E solving for I 20000/120167 amps solving for I 10000/240 41 amps You have to go to the NEC code book to look up the wire size but its about 4/0 (you MUST have a local expert do this for you never trust anyone on the net) Now you know what size wire feeds your panel from the x-former lets work this backwards The 480/277 Y is a 3 phase system Im sure you will select the higher votage of 480 to supply the transformer this lowers the wire size proportionally. The formula for OHM's 3 phase is P I x E x Sqrt of three or 1.732 the output kva is 30kva so you need 30 kva going in minimum solving 30000 I x 480 x1.732 I 30000/831.3636 amps now lets go bac to the code book to size the wire 10 is 30 amps #8 is 40+ so lets feed the primary side with #8. Sizing the overcurrent device is done the same way Since we did all that fluff.to answer your question Let me say this, Always oversize the x-former 10-20% . and always check your addition for the totals on the loads.There are a bunch of appliances that can be devaluated and in winter your /ac doesn't run and all that. so before you jump right into this try to get a real good handle on what's going on.Its not complicated at all you just need to be accurateWell Have a good day.From the E
- Q:Switching Transformer, Alternative to MPPT Charge Controller? (wind+solar)?
- Transformers don't work on DC. You need a DC to DC converter.
- Q:Power is generated at genaration station with 33kv,then by using step up transformer the voltage raises to?
- Gawd I don't know where people go to school to make up rubbish about the power being too high for use in homes. Just get a few facts and make the rest up. Ok When electricity is generated in a central power station the electricity needs to be transmitted many miles. The most efficient way of doing this is to use as high a voltage as possible. So at the power station a transformer is used to convert the generated voltage into a high voltage for transmission. When the electricity arrives at the destination it is transformed back down to whatever voltage the end user requires. The reason for using the high transmission voltage is to reduce resistive losses in the cables. I'll give an example Suppose we need to go 100 miles and have a resistive load of 1 ohm/mile We need to deliver 1MW down to the end of the line and the end voltage required is 100V We'll ignore the voltage drop for the example, but you can do homework to take it into account. If we transmit at 100 volts then applying W VI we need I 10^6/10^2 10^4 amps We have 10 ohms in the transmission path so resistive losses are I^2 . R (10^4)^2 . 10 10^9 Watts That is one heck of a loss! Now, if we transmit as 100kV 10^5V then our current is now 10^6/10^5 10 amps Applying I^2 . R for our resistive losses we now get 10^2 . 10 10^3 watts You can see the VAST difference that stepping up the voltage for transmission does now for the transmission loss.
- Q:why transformer draws lagging current in summers?
- it does not have a clue about seasons. transformers are inherently inductive devices and thisn means current lags any day of the year
- Q:Power Transformer help!?
- If you assume no losses, the power remains constant. If the voltage is stepped up, the current is stepped down in the same ratio. The input current will depend on the output current drawn by the load. With a transformer stepping up 1:4, 25 volts becomes 100 volts and the current is determined by the resistance of the load. If you assume it to be 100 ohms, for example, the current will be 1 amp. This is reflected as a current of 4 amps at the input.
- Q:Mall Cop Vs Transformer 2?
- Transformers 2 was better. The action was good and so was the story. Plus transformers has Megan Fox. Mall Cop wasn't that funny to be honest.
- Q:Proteus transformer how to adjust the parameters, the 220V into 14V output.
- Transformer parameter setting method is:
1, the calculation: According to the ideal transformer primary turns ratio ratio n1 / n2 = V1 / V2 = 220/14 ≈ 15.7,
Let the primary winding sense L1 = 1H, then the secondary winding inductance L2 = 1 / (15.7 ^ 2) ≈ 0.00406H,
- Q:can ferro alloy smelting furnace transformer can be converted to higher voltage to handle ferronickel smelting?
- You need to provide more information about the original use of the transformer. Nickel smelting is a very specialized application that calls for a 'submerged arc'. Basically, the arc is fairly short and will essentially be submerged in a layer of slag on top of the molten material. It is a semi-continuous process in that periodically, some of molten nickel is poured off. The stream of molten nickel is blasted with water that causes the nickel to solidify into small pellets. Steel melting furnaces, by contrast, involve a much longer arc. The slag layer in the furnace is much thinner, and much of the arc is above the slag. It is a batch process in that once the 'campaign' has been completed, the entire melt is 'tapped' - poured into a tundish either to be cast into ingots or to feed a continuous casting process to make billets, slabs or even plates. There are other smelting processes for other materials - for example, silicon is produced via a smelting process. I would guess that if the voltage requirements of the nickel smelting process are different from the requirements of the original utilization, then it would be necessary that the transformer be redesigned and rewound - something that I would not expect to be very practical.
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