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ElekKai
modified 1 year ago

Δ-Y 230-400V

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276
02:55:22
On a 3-phase induction motor, a type plate is (usually) found. An example: https://www.sersef.com/wp-content/uploads/2019/08/41204120M.webp On this type plate, 2 voltages (230/400V) and 2 currents (4,2/2,41A) are descripted with 1 mechanical power rating (1,1kW). The lowest descripted voltage (230V) is the maximum allowed coil voltage at 50Hz. When connecting the power supply, the line voltage (measured between phases) must be known. When the line voltage is 230V, the motor should be connected in a Δ circuit. In a Δ circuit, each coil is connected to 2 phases, which means the coil voltage (measured over a coil) is the same as the line voltage. When the line voltage is 400V, the motor should be connected in a Y circuit. In a Y circuit, 1 end of all 3 coils are connected together. Now, each coil is only connected to 1 phase, which means the coil voltage is now "line voltage÷√3" (400V÷√3)=230V. For the current it works vice versa. In a Δ circuit the line current (supplied current of each phase) of 4,2A will devide between the 2 connected coils of each phase to become coil current (current through a coil).. This will be "line current÷√3" (4,2A÷√3)=2,41A. In a Y circuit, the coil current will be the same as the line current (2,41) because the current will not devide over 2 coils. In either Δ or Y circuit, the coil voltage will be 230V and the coil current will be 2,41A. In the circuit are 2 voltmeters and 2 ammeters to show how the voltage or current will be devided. At start the Δ circuit is active with: Blue=line voltage Green=line current Orange=coil voltage Red=coil current Toggle the SPST switch to switchover to Y circuit and see how the voltages and currents will behave
published 1 year ago

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