Presumably you mean 'rated full-load current'? The thing to remember is that motors, whether rated in horsepower (North America) or watts (practically everywhere else), are rated according to their output power. Because motors are not 100% efficient, the input power is always somewhat higherthan the output power. 0.5 horsepower is equivalent to about 373 W output power. If we assume that the motor runs at, say, 90% efficiency at its rated output, this means its input power will be around 414 W. To determine its input current you must then divide its input power by the rated supply voltage.
If it was 440 v single-phase, use half the number of turns in the windings, but use thicker wire of double the cross-section. Or to use the same wire again, fold the wire once into half its length and use that.
Single phase induction motor is not self starting because, when the main winding is excited from a single phase supply, it produces an alternating magnetic field instead of rotating magnetic field in two phase and 3 phase induction motor. According to double field revolving theory any alternating vector can be resolved into two vectors rotating in opposite directions, each having magnitude equal to one half the magnitude of the actual vector. The vectors will be rotating in such a way that their resultant will be zero at every instant. Thus a single phase induction motor fails to produce a net starting torque, and is not self starting.It can be made self starting by any of the methods below1) By using an aux winding in series with a very high resistance2) using two capacitors,in series with the aux wdg,one with intermittent duty and the other one with continuous duty3) Using shaded poles
Single-phase ac has current that changes direction 100 times a second (if it's a 50 Hz supply). It only needs two wires to distribute a single-phase supply. Imagine it as a battery that is continually being reversed.
Yes. 230V line to line is considered single phase power. The 115v half part of that is called split phase. It should not be called two phase, as it is one phase that has been center tapped and grounded at the neutral point - it is still one phase power.Answer230 V is the standard nominal voltage for residences in Europe. 230 V is the line-to-neutral voltage (phase voltage). Low-voltage distribution is by three-phase, four-wire, system with a nominal line voltage of 400 V and a nominal phase voltage of 230 V.
It depends on whether or not it is a half wave or full wave rectifier. For a single phase 60 Hz rectifier, a half wave rectifier will be 60 Hz while a full wave rectifier will be 120 Hz. A three phase full wave rectifier will be 360 Hz.
One And a Half Horse Power Obviously.
to calculate: 1hp=745W 3hp=745*3=2.2Kw assuming a 3 phase motor power is V*I*pf*the square root of 3 I would gues a PF of 0.8 if its a resonable motor which gives about 7.2 amps per phase. if the PF is bad say .65 it could pull up to 9 amps in short, too little information to give you the actual answer.
Two phase motors are a thing of the past and are not used today if any still exist. Two phase legs are classified as a single phase system. If the question refers to a dual voltage motor then the answer depends upon what supply voltages are available to operate the motor. Operating a motor at the higher of the two voltages means that smaller control equipment and wire feeders can be used to operate the motor. If the nameplate is checked, in the amperage box there will be two current ratings, Amps/Amps. One will be half of the other one. Look at the supply voltage box and you will see the corresponding Voltage/Voltage. The first number in the voltage box corresponds to the first number in the amperage box. The higher the voltage the less line loss will occur over a distance. If possible always try and use the highest voltage to operate motors.
If it was 440 v single-phase, use half the number of turns in the windings, but use thicker wire of double the cross-section. Or to use the same wire again, fold the wire once into half its length and use that.
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.
A pool pump motor which is drawing half the amps listed on its nameplate can indicate a problem with the windings or a lack of incoming current. Pumps will only draw as many amps as are required to operate under the current load.
Single phase induction motor is not self starting because, when the main winding is excited from a single phase supply, it produces an alternating magnetic field instead of rotating magnetic field in two phase and 3 phase induction motor. According to double field revolving theory any alternating vector can be resolved into two vectors rotating in opposite directions, each having magnitude equal to one half the magnitude of the actual vector. The vectors will be rotating in such a way that their resultant will be zero at every instant. Thus a single phase induction motor fails to produce a net starting torque, and is not self starting.It can be made self starting by any of the methods below1) By using an aux winding in series with a very high resistance2) using two capacitors,in series with the aux wdg,one with intermittent duty and the other one with continuous duty3) Using shaded poles
Depending on the type (three phase, single phase), there are generic power factor and efficiency values you can estimate if you haven't chosen a specific motor, but need to size wiring ahead of time. You'll need to double check after the motor is specified, though. Cart before the horse. You can find these values by simply comparing a half dozen motors similar to what you're thinking (compare phase, horsepower/kw rating, duty factor, etc.).
It all depends upon the type of motor you are using. Not all motors are the same. Just because it says "single phase one half horsepower electric motor" in the description, does not mean that all these motors are the same. They are not the same. There are differences from one make and model compared to another. One country might use 60-HZ, and another country might use 50HZ. The power source must be matched to the motor. There should be a Tag on the Motor itself, telling you exactly what the power requirements are. If you are in the United States, and if the motor is rated at 120-VAC, 60-HZ, all you have to do is hook-up a circuit breaker in series with one of the power leads, and attach a ground lead to the case of the motor, and plug it into any standard wall outlet. Make sure the power source is matched to the information on the Motor, otherwise the motor could be damaged, and some of the electrical hookup wiring could be burned, and your safety might be at risk. Don't take chances. Make sure the tag on the motor matches the power source you are using.
If you make the connection, the breaker on the generator will trip. The motor will not get up to running RPM. Check and see if the motor is dual voltage. If it is, run it on the higher voltage. This will reduce your run current to about half of what it is on the lower voltage.
A half wave rectifier does not make a stable voltage. A single phase half wave creates a "bumpy road" where voltage modulates between sine wave maximum and zero. A three phase half wave will create a more stable, but ultimately "unclean", voltage.
Power in an electric, AC circuit is the product of Volts, Amps, and the Cosine of the angle that separates them. When the Amps lag behind the Volts by 60 degrees, the product of Volts, Amps, and the Cosine of the angle between them provides half the power that would otherwise be available without the 60 degree angle. At 60 degrees, the cosine is 0.5 and at 90 degrees it is zero. So the product of Volts and Amps whenever they are 90 degrees out of phase will result in zero power.