The efficiency of a simple transformer is limited by resistive loss in the wiring, and by hysteresis (magnet related) losses in the transformer core. You may limit the resistance loss by using superconductors at very low temperatures. But not practical for most situations. (Yet!)
It depends on the design of the transformer but 1 MVA is a common size for an 11 kV / 415 v three-phase transformer.
the efficiency of transformer is more than alternator, because 1- the alternator have friction and windage loss but in transformer this not occure. 2- in the alternator the flux leak is more than the transformer, because in case of alternator the flux move through the air which is dimagnet and in transformer flux move through silicon steel. 3- in the other hand alternator have dc compound generator for exciation (depend on other electric course 220volt) but in t/f is not need.
Perhaps an electrical AC transformer can have 99 percent efficiency. A loudspeaker can have only 1 percent efficiency.
The efficiency of the Otto cycle is given by the formula: Efficiency 1 - (1 / compression ratio)(-1), where is the specific heat ratio. To optimize the efficiency of the Otto cycle for maximum efficiency, you can increase the compression ratio, improve combustion efficiency, reduce heat losses, and use higher octane fuel.
Power factor ranges from zero to a maximum of 1. At 1 the current and voltage waveforms are in phase and operate at maximum efficiency.
The maximum efficiency of a heat engine, such as a steam engine, operating between two temperatures can be determined using the Carnot efficiency formula: ( \eta = 1 - \frac{T_C}{T_H} ), where ( T_C ) is the cold reservoir temperature and ( T_H ) is the hot reservoir temperature, both in Kelvin. For a steam engine operating between 300 K (cold) and 1500 K (hot), the maximum efficiency is: [ \eta = 1 - \frac{300}{1500} = 1 - 0.2 = 0.8 \text{ or } 80%. ] Thus, the maximum efficiency of the steam engine is 80%.
The kVA represents the power-handling capability of the transformer.So, if you were using a 1 kVA transformer at 110 volts, you could roughly estimate the maximum output to be 9 amps.
CONDITIONS REQUIRED FOR TRANSFORMER PARALLEL operation are.. 1. Voltage turn ratio. 2. phase sequence 3. Impedence should be same 4. polarity Hitesh
Short Answer: Dew point measurement is necessary to ensure the moisture content of the cellulose insulation is low enough to safely energize the transformer (after filling with oil). Maximum acceptable moisture content varies with unit size/age, but less than 1% moisture content is generally acceptable.
because the copper is saved in the auto transformer that is the ratio of weights of copper in auto transformer to normal transformer is (1-(1/a)) where a is transformer ratio
You will have a 1:1 ratio isolation transformer.
The maximum efficiency of a heat engine is determined by the Carnot efficiency, which is defined by the temperatures of the heat reservoir and the cold sink. It is given by the formula ( \eta = 1 - \frac{T_C}{T_H} ), where ( T_H ) is the absolute temperature of the hot reservoir and ( T_C ) is the absolute temperature of the cold sink, both measured in Kelvin. This theoretical maximum efficiency can never be achieved in practice due to irreversibilities and other losses in real engines.