acu compressor motor window type aircon 45 uf capacitor for 2.5 hp compressor 50 uf for 3 ton acu 60 uf for 1.5 hp compressor 35 uf
You can generate electricity with a 27KVA alternator, a big wheel, and a 2HP motor, but you will not be able to even come close to generating 27KVA. The best you could hope for is about 1KVA or so, because a 2HP motor can only do about 1.5KW of work, and efficiency and power factor losses will make the output be even less.
Apart from anything else, it's necessary to know the rated votage of the pump!
Use this formula W = A x V . Assuming that the saw is using 120 volts. 15 x 120 = 1800. Start up current on a motor will draw up to 300 % instantaneously so a 2000 watt generator might just do it. It would lug at the start but then catch up and run the saw. On motor from generator uses higher wattage is better.
Full load amps is found on the motor's nameplate, and is unique to that particular brand and model number. A Magnetec 2hp motor will draw different amps than a Baldor 2hp, etc. There are high efficiency motors, and not so high as well. Consult the nameplate. If you want a "rule of thumb" to estimate the RLA (Running Load Amps - this is the term most often found on the nameplate - same as full load amps), then use this: Figure about 1400 watts per horsepower. Divide Watts by volts to get Amps, for example: * 3hp motor, 120V * 3hp X 1400 W = 4200 W * 4200 W / 120 V = 35 A This formula is only for single-phase motors! I looked up a typical 3hp 120 V Baldor motor in the Grainger's catalog, and the RLA for it was 32.0, so this will get you in the ballpark. I have seen 3hp 120 V motors as low as 25 A and as high as 40, so DO NOT use the rule of thumb to size branch circuit conductors or overcurrent protection! For 3-phase motors: Figure about 1100 watts per horsepower (3-phase is a little more efficient). Divide Watts by Volts, then divide the answer by 1.73 to get Amps, for example: * 3hp motor, 208V 3Ph * 3hp X 1100 W = 3300 W * 3300 W / 208 V = 15.86 * 15.86 / 1.73 = 9.16 A This formula is only for 3-phase motors! My Grainger's catalog motor with the same Voltage and HP had a nameplate rating of 8.34 A.
12/2 with ground.
enough oil is what you need
acu compressor motor window type aircon 45 uf capacitor for 2.5 hp compressor 50 uf for 3 ton acu 60 uf for 1.5 hp compressor 35 uf
50/1
You can generate electricity with a 27KVA alternator, a big wheel, and a 2HP motor, but you will not be able to even come close to generating 27KVA. The best you could hope for is about 1KVA or so, because a 2HP motor can only do about 1.5KW of work, and efficiency and power factor losses will make the output be even less.
The ratio is 50:1.
It would depend on the efficiency and design of the motor, as well as the operating conditions such as load distribution and friction. In general, a 2HP 4-pole motor may not have enough torque to lift 300kg vertically, but it could potentially move the weight horizontally on wheels or a conveyor. It's important to consider the motor's continuous duty rating and consult with a mechanical engineer for a more accurate assessment.
Find the amperage that your pump motor operates on (should be on the motor nameplate) and multiply it by the voltage that the motor is connected to (120 or 240 volts). This will give you the wattage of the motor. It is usage of watts that the utility company bills you on. Find from your utility bill what you are charged per kilowatt hour. Take your pump wattage, times the amount of time that the pump runs, times the kilowatt rate you are charged by the utility company and the result will be how much it costs you to run the pump motor.
i have 7.5kw dynmo and 2hp motor the motor attach with home electicity and dynmo show 225 watt but not working dynmo to motor i dont no tell me plz
well the motor will like it, it will run cool but the power bill will be high. it will not have the flow of a 2hp impeller
The motor is a 1947 2hp. Great little motors. I have two of my own.
Since (in North America, at least), the horsepower of a motor describes its rated output power, you need to know its efficiency in order to determine its input power, then its rated voltage, to determine its full-load current.