10000 = 415 * i
i = 10000/415 = 24 a
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The formula you are looking for when KVA is known is I = KVA x 1000/1.73 x E = 10000/1.73 x 415 = 10000/240 = 41.6 amps.
to calculate the motor full load current one should know motor power factor, supply voltage and phase of supply. Assuming your motor is with 0.85 pf and 3 phase, 415V source then P = (Sqrt3)*V*I*pf 18500 = 1.732*415*0.85*I hence I = 30.2A
15 amps X 415volts = 6225watts or 6.225Kw
In short No,The UK and most of Europe uses 230V ac single phase for domestic and light commercial installations and 400V ac three phase for industrial and heavy commercial requirements.The confusion comes the fact that a lot of cables are rated for use up to 600 or 650V ac and have markings on the insulation showing this.The wiring regulations allow for systems up to 1000V ac but these are rare. Distribution circuits ie the underground or over head supply cables between power stations and substations use 11,000 33,000, 125,000 and even 400,000 Volts.The UK previously used 240 / 415V and a lot of Europe used 220 / 380V but this was harmonised in 1988 with effect from 1995 (these changes make no practical difference for the user) but a lot of equipment is still labeled 240V or 415V.
Turns ratio test is very important in order to find out that the transformer has the right ratio corresponding on its rated voltage in primary and secondary. For example, three phase transformer is subjected to turns ratio test when each phase has equal turns ratio then the transformer is balance.The test is performed to ensure that overheating or overcurrent conditions have not shorted turns in the transformer windings, which would distort the desired output or input voltage. This test is particularly important for instrument transformers, i.e. CTs and PTs which are connected to protective relays. These relays are extremely sensitive, and operate on very small fluctuations in secondary voltage or current. Shorted turns in these transformers can lead to big problems if they cause the relay to operate abnormally, or not at all.
for 415V phase to earth min clearance is 25mm pallavCommentYou cannot have a voltage between a 'phase' and earth. You should be asking the clearance between a 'line' and earth.
415V 3 phase is the line to line voltage. The line to neutral of this supply is 230V single phase. Therefore you use one of the phases and the neutral.
In India its 415V, 50Hz.
Current shall increase while you apply 415V for operating it and will decrease the rpm to a much low level.
Yes, it will run drawing little more extra current in similar conditions. The motor is expected to be running in lesser rpm as well.
The single phase voltage in India is 230v when we check with the phase and the nutral single line
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.
apply 3 phase voltage(415V) to the winding whose magnetising current is to be found and open circuit another winding. now measure current using tong tester or connecting an ammeter in series between supply and winding.
5 hp moto 1500 rpm 3 phase winding calculation
0.073 assuming 0.85 power factor
The voltage of three phase is 415v and the colours are brown black and greyAnswerThe nominal line voltage is 400 V, and the nominal phase voltage is 230 V.
Divide 2.2 kW by (an assumed, because you did not state it) power factor of 0.85 to get 2.6 kVA. Divide that by 3 to get 0.863 KVA per winding. Divide that by 415 volts to get 2.08 amperes per winding. If you are running star, then that is the phase current. If you are running delta, then multiply by the square root of 3 (1.732) to get 3.6 amperes per phase.
If it has one load, it isn't balanced.AnswerThe voltages are determined by the supply, not by the load -so they would remain at 415 V (line voltage) and 240 V (phase voltage) regardless of your load. But, assuming your load specification is per phase, then you have a balanced load and the phase and line currents are easily calculated.The first step is to determine the impedance per phase, which is the vector sum of the resistance and reactance -the reactance can be calculated from the inductance value (63 ohms) -this works out at 118 ohms.Next, we find the phase current, which is the phase voltage (240 V) divided by the phase impedance, 240/118 = 2 A. Finally, we calculate the line current, which is 1.732 times the phase current = 3.5 A.