In order to answer this question, you need to know (a) its efficiency, and (b) its power factor at full load. 11 kW (not 'Kw') is its output power, so you need to know its efficiency in order to determine its input power. Then, because, for a three-phase system, power is equal to 1.732 times the product of the line voltage, line current, and power factor, you also need to know its power factor.
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The line current would be the same if the motor were connected in delta. The current can be based on the rule of thumb which says 7 amps must be allowed for a 1-HP single-phase motor on 240 v. A 2.2 kW motor is three times as powerful, and on a three-phase supply of the same voltage (240/415) it would draw 7 amps.
The electrical code states that a 30 HP induction motor at 460 volts three phase will draw 40 amps. <<>> I = 33.34 AMPS IF EFF.= 95% AND P.F.= 85%
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Watts = Volts times Amps. Therefore, if the voltage was 220 volts, the motor would draw 500 amps. If the voltage was 4,000 volts, the motor would draw 27.5 amps. The voltages for large powerful motors tend to be relatively high, for example in the 380 Volts to 11,500 Volts range.
T430.247 of the NEC shows that a 1 hp motor operating at full load on 115v will draw 16 amps, called Full Load Current (FLC). Conductors supplying this motor are required to be 125% of FLC which is 20 amps. Motor circuits are complicated things and do not follow the rules of other circuits. This motor, while drawing a maximum of 16 amps at full load and supplied with #12 AWG copper conductors can be protected by a breaker of 40 amps.