If you are asking about an actual motor it is best to read the motor's rating plate.
If you are asking how to figure out the current in Amps drawn by a motor which produces a power output of 3 horsepower (hp) this question is way too vague to be answered accurately without knowing a lot more technical details.
The current drawn will depend on (a) the type of motor and (b) the Voltage that the motor is designed to run on. Regarding motor type, is it a Direct Current motor or, if it is an Alternating Current motor, is it a single-phase motor or a 3-phase motor. If it is a single-phase motor, then is it a squirrel-cage induction motor or a commutator motor of the type often used in vacuum cleaners or power tools? Another type of alternating current motor is the slip-ring synchronous motor but that type is not often used to supply only 3 horsepower.
A very rough way to calculate a possible current draw if you know the motor's operating voltage is to use the fact that 1 hp is approximately equal to 0.75 kiloWatts. So 3 hp = 3 x 0.75 = 2.25 kW.
Now, knowing that the actual efficiency of a motor depends entirely upon its design type, we shall have to assume an efficiency factor for the conversion of electrical power to mechanical power.
If we assume the motor is only 50% efficient then, to produce 3 hp, it will have to be supplied with electrical power of at least 2 x 2.25 = 4.5 kW.
Since Power P (Watts) = V (Volts) x I (Current in Amps), the current drawn I would therefore be given by I= P/V:
At 12 Volts DC: 4,500 / 12 = 375 Amps. (This would be a typical peak starting current drawn by, say, the starter motor of a small car engine in Summer. In Winter it might get to 500 Amps or more!)
At 120 Volts RMS AC: 4,500 / 120 = 37.5 Amps RMS single phase
At 240 Volts RMS AC: 4,500 / 240 = 18.75 Amps RMS single phase
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If you are asking about an actual motor it is best to read the motor's rating plate.
The current drawn will depend on (a) the type of motor and (b) the Voltage that the motor is designed to run on.
A very rough way to calculate a possible current draw if you know the motor's operating voltage is to use the fact that 1 hp is approximately equal to 0.75 kiloWatts. So 3 hp = 3 x 0.75 = 2.25 kW.
Now, knowing that the actual efficiency of a motor depends entirely upon its design type, we shall have to assume an efficiency factor for the conversion of electrical power to mechanical power.
If we assume the motor is only 50% efficient then, to produce 3 hp, it will have to be supplied with electrical power of at least 2 x 2.25 = 4.5 kW.
Since Power P (Watts) = V (Volts) x I (Current in Amps), the current drawn I would therefore be given by I= P/V:
At 240 Volts RMS AC: 4,500 / 240 = 18.75 Amps RMS which is a possible answer to this question. At 120 Volts RMS AC: 4,500 / 120 = 37.5 Amps RMS At 12 Volts DC: 4,500 / 12 = 375 Amps. (This would be a typical peak starting current drawn by, say, the starter motor of a small car engine in Summer. In Winter it might get to 500 Amps or more!)
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For a 240 v single-phase motor, allow 7 amps per HP, so the estimated current for a 3HP motor is 21 amps.
You don't use "kilowatts per hour". You use energy at the rate of some number of watts or kilowatts.
1 horsepower = 746 watts.
If the motor is delivering energy at the rate of 3 horsepower, then that can be expressed as
(3 x 746) = 2,238 watts.
You might think that this is exactly how much power you have to put into the motor. But unfortunately,
you always get less power out than what you put in.
(output power) divided by (input power) is called the motor's 'efficiency'. Every motor can
have a different efficiency. The only thing we're sure of for every motor is that it's always
less than 100%.
We know that this motor is outputting 2,238 watts.
(Input power) times (efficiency) = 2,238 watts
The Input Power used is [ (2,238) divided by (efficiency of the motor) ]
or (2.238 kilowatts) divided by (efficiency of the motor)
The electrical code book states that a 3HP 230 volts single phase motor is rated at 17 amps. Three phase 230 volts is not a common voltage any more but the book lists the amperage as 9.6, 460 volt motor is 4.8 and a 575 volt motor is 3.9 amps.
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The answer varies widely depending on whether the draw is measure at startup (substantially more) or once the motor has reached steady state. It further varies depending on the mechanical load placed on the motor - the greater the load, the greater the draw to maintain speed/momentum. Finally, the efficiency and design/architecture of the motor itself has quite a lot of bearing on it's current draw.
Nearly all motors of this size (3hp / 220v) will feature operating specs which should indicate base amperage draw under no-load - if not, consult the manufacturer.
If its 3ph then around 9.6 amps, 1ph then around 17.
1 hp means 746 watts, hence 5 hp consumes 3730 watts. (This is irrespective of any voltage) i.e 1 hp =746 watts
Officially a horse-power is 746 watts, so 3 HP is 2238 watts, but there will be some electrical losses in the motor so the power taken is about 2500 watts or 2.5 kW.
Question is incorrect. in a 240 Volt single phase circuit, how can you have A phase and B phase?
Amps are not directly convertible to horse power, which is a measure of power. Power is current times voltage. Therefore on a 240 v supply, 9.2 amps equals 2208 watts. One horse power is 746 watts.
Volt amps is the same way of saying watts. This is seen in the formula Watts = Volts x Amps. Using the formula I = W/E we transpose I = 20000/240 = 83.33 amps.
The formula you are looking for is I = W/E. Amps = Watts/Volts.
no you can not it would ruin your system for good. You have to get a heaters with the same watts