p= V*I*COS FIEE SAY COS PYE IS UNITY
10*1000=230*I*1
I= 10000/230
I=43.4 A
Comment10 kW is the output power of the machine. You will need to know the machine's efficiency in order to determine its input power before applying the above equation.
It depends on the voltage and power factor.
If you were using 240 single phase, you could divide 10KW by 240V and get 42A. Then you divide by power factor, lets assume a typical value of 0.85, and you get 49A.
Now, 49A on 240 single phase is a bit unrealistic, though doable, so lets look at 480 three phase running delta. You divide 49A by 6 (3 for the 3 phase, and 2 for the 480 step up) getting 8A per winding current, then you multiply by square root of 3 to get the per phase current of 14A - call it 15A - and size the circuit for 20A.
The 'line' current (not the 'phase' current -they are two different things!) depends on the input power of the machine and the voltage. Since you give no information on the efficiency of the machine (so, we cannot calculate its input power) or its voltage, it is impossible to answer your question.
To answer this question two values are needed, what voltage the motor operates on and whether the motor is single or three phase. Starting current is usually three times the full load amps but depends on the type and duty of the motor.
Need the voltage of the motor to answer this question.
A single phase 10 HP motor will draw aproximately 50 amps. A three phase 10 HP motor will draw aproximately 28 amps.
What is the continuous current of 10HP star delta motor
you need the current of motor or the KW/HP rating
5 hp equals 3730 watts, and on a 3 phase 480 v system the line voltage is 277, so the current times 277 times 3 equals 3730. The answer in theory is 4.5 amps. But you have to allow for 10% power-loss in the motor, and also the power factor, which could be 0.8. Therefore the current is probably 6 to 6½ amps. Maybe 10 amps on starting up. <<>> The formula you are looking for when amperage is desired when horsepower is shown is - I = HP x 746 / 1.73 x volts x % efficiency x power factor. A standard motor's efficiency between 5 to 100 HP is .84 to .91. A standard motor's power factor between 10 to 100 HP is .86 to .92. Amps = 5 x 746 = 3730/1.73 x 480 = 3730/830 x .84 x .84 = 3730/586 = 6.37 amps. Starting current will be 300% of the motors run current.
The first thing you have to do is find the full load amps of the motor. The wire size feeding the motor has to be 125% of the full load current. The breaker is usually 250% of the full load current. If the voltage and amperage had been added to the question the exact breaker size could have been calculated.
A single phase 10 HP motor will draw aproximately 50 amps. A three phase 10 HP motor will draw aproximately 28 amps.
YES
it uses 45A at 415V in pakistan and starting current might be near 60A. <<>> The equation you are looking for is, Amperage when horsepower is shown; HP x 746/1.73 x V x %eff x pf. As you can see that an answer can not be given without the voltage of the motor being stated. Once you have the voltage use the formula. Use .89.5 for the % eff of a standard 25 HP motor and .89 for the power factor of a standard 25 HP motor. Using the above answer's voltage, the motors amperage equals 32 amps. A motors starting current can reach 300% on start up. The starting current can reach 96 amps instantaneous until the motor amperage drops to its run amperage.
What is the continuous current of 10HP star delta motor
HP = (Current x Voltage)/746, or HP = (IE)/746 (disregarding %Efficiency) So, you have to solve for I, current: I = (746 x HP)/E
no
you need the current of motor or the KW/HP rating
5 hp equals 3730 watts, and on a 3 phase 480 v system the line voltage is 277, so the current times 277 times 3 equals 3730. The answer in theory is 4.5 amps. But you have to allow for 10% power-loss in the motor, and also the power factor, which could be 0.8. Therefore the current is probably 6 to 6½ amps. Maybe 10 amps on starting up. <<>> The formula you are looking for when amperage is desired when horsepower is shown is - I = HP x 746 / 1.73 x volts x % efficiency x power factor. A standard motor's efficiency between 5 to 100 HP is .84 to .91. A standard motor's power factor between 10 to 100 HP is .86 to .92. Amps = 5 x 746 = 3730/1.73 x 480 = 3730/830 x .84 x .84 = 3730/586 = 6.37 amps. Starting current will be 300% of the motors run current.
5-10 hp gas motor should be more than adequate. An electric motor would alos work well, especially on small lakes or sloughs without significant current.
For a single-phase induction motor, allow 7 amps on a 240 v for a 1-HP motor. Therefore the formula is: current = 7 X HP x 240 / voltage
10 hp and above motor power rating....
current = power/ voltage current = 3700/ 240 = 15.4 amps