Class 10KV S9-M series full-sealed transformer

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ted capacity (KVA)Voltageconnection group tabLossunload current(%)resistant voltageweightMeasure(MM)Distance of Din rail
high-voltage(kv)extend connectionlow-voltage(kv)unloadloadempty weightoil weighttotal weightlength(L)width(W)height(H)cross(M)×length



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Q:Transformer Uk = 4.5% What does it mean?
Transformer short-circuit impedance, that is, his own impedance. The impedance is large, the transformer needs to resist the short circuit current multiplier is small, relatively speaking, anti-short circuit capability, but the transformer's external characteristics (volt-ampere characteristics) soft (with the transformer output current increases, the output voltage - transformer Voltage drop is very powerful). On the contrary, short-circuit impedance is small, the transformer needs to resist the short-circuit current multiplier, requiring a strong anti-short circuit transformer capacity. Of course, the external characteristics of the transformer better. Another short-circuit impedance will also affect the transformer manufacturing costs. He is not the bigger the better, nor the smaller the better. To be considered, so there are strict rules in the national standard.
Q:Can the steel of a ferromagnetic transformer instead be fashioned as a set of windings, acting as a secondary?
An AC transformer connects AmpereTurns in the Primary to AmpereTurns in the Secondary using the time varying magnetic field in the core. The maths for the size of the core is well established. If you double back on turns they do not count as part of your ampereturns - unless you shield the doubled bit. Such methods are used in shaded pole motors. The only way to lighten the Tx is to use higher frequencies in the core. This 'dumps' the magnetic energy into the secondary more often so you can use a smaller core (bucket of energy!) The magnetic field is always perpendicular to the current and hence so are the core loops (in a Vector Notation sense - not necessarily physically) If you introduce Iron windings in the windings space you will get bigger Ohmic Losses. I hope this helps - you obviously have a plan to make hyper-efficient Tx but they are already near 0.999 efficiency.
Q:What is the meaning of the high voltage side and the low side of the transformer?
In general, the "high voltage side" refers to the "side of the high voltage side" (high pressure side) "low side" from the power supply to the primary side of the transformer (low voltage outlet) The From the wiring is also basically able to distinguish: high side (into the end) is generally "△" connection, low side (outlet) is generally "Y" connection.
Q:Transformers, Which is the first ever season???????
no, okorder
Q:Will the iPad 3 surpass the Transformer Prime?
I think you mean the iPad 2 HD. The rumour is that it will just be a refresh first, then iPad 3 at the end of next year. But who knows iPhone 5 got released, right? ;)
Q:Bridge rectifier, transformer vibrates and gets hot?
The transformer may be vibrating because the winding inside have loosened up and vibrating due to the varying magnetic field. But it is quite possible it may have a shorted winding this will definitely make it heat up and vibrate. Disconnect the filter capacitors and plug it back in if the transformer stops vibrating and/ or the transformer doesn't overheat you have a leaky filter capacitor.
Q:find transformer size?
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:transformers or simpsons?
Simpsons for sure:]
Q:Will I need a transformer?
If the charger/adapter says 100~240V it means you can plug it directly to the wall. Otherwise, you will have to buy a step-down transformer (240V to 120V). While you could buy a transformer at Walmart, Radioshack or BestBuy, I'd recommend buying it at any hardware store in Peru. They tend to be cheaper in Peru by some reason (less than $10). I would also recommend buying one of those plug adapters, since power outlets in Peru don't have a hole for the ground pin (they use 2-pin connectors).
Q:single phase transformer?
I2 I1[V1/V2] 9[2200/V2] I29[2200/20] 9[10] I2 90 amperes. V2 220 volts

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