The full load current of a 3 HP motor depends on its voltage and efficiency. For example, a typical 3 HP motor operating at 230 volts usually has a full load current of about 15-16 amps, while one running at 460 volts may have a current around 8-9 amps. To calculate the exact full load current, you can use the formula: Current (I) = Power (P) / (Voltage (V) × Efficiency × √3) for three-phase motors. Always refer to the motor's nameplate for the most accurate value.
It would probably be 20-30 amps. <<>> The electrical code book states a value of 18.7amps for a 3HP 208 volt motor. If the nameplate data is available it is always best to use it when calculating wire size and overload protection for the motor.
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.
A 3hp motor refers to an electric motor with a power output of three horsepower (hp). Horsepower is a unit of measurement that quantifies the power a motor can produce, with one horsepower roughly equivalent to 746 watts. Therefore, a 3hp motor can deliver about 2,238 watts of power, making it suitable for various applications requiring significant energy, such as pumps, compressors, and industrial machinery.
The power consumed can be calculated directly from the current draw, voltage, and power factor: (3A) x (400v line to line) x (sqrt(3)) x (.9pf) = 1.87kW. This would be a reasonable value for a 3hp motor that is not running under full load (the motor may be slightly over-sized for this application). The size of the motor is irrelevant to this calculation, although it shows that it is running at ~85% of its capability.Another AnswerSince a motor is a balanced load, we use the following equation to find its input power:power = 1.732 x line voltage x line current x power factor.So, in your example,power = 1.732 x 400 x 3 x 0.9 = 1870 WObviously, it cannot be delivering 2200 W when its input is just 1870 W. However, to determine its output power you will need to determine its efficiency at that input.
The largest 120V motor generally available is 3hp. A 3hp, 120V motor draws about 30 amperes. All true 5hp motors I have seen are 230V or higher. Note that some motors, such as compressors and vacuum cleaners, use artificially high hp numbers to entice you to buy their unit (they will use terms like peak-developed hp, etc.). These ratings are not true hp, and you just have to go by the nameplate rating on the unit. For instance, I have a Sanborn Air Compressor that claims 4 PEAK horsepower, yet only draws 17 amperes at 120V. Its really closer to a 2hp motor.
It would probably be 20-30 amps. <<>> The electrical code book states a value of 18.7amps for a 3HP 208 volt motor. If the nameplate data is available it is always best to use it when calculating wire size and overload protection for the motor.
i gota three 3hp seahorse johnson boat motor i need to get to head how does everthing come apart
How do you change the oil in a 1967 3Hp Evin-rude outboard motor?
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.
David Underhay The decal on my 1968 3hp Johnson says use either of: Champion J4J, AC M42K, Autolite A21X sparkplugs. You may have to find the current equivalents
A 3hp motor refers to an electric motor with a power output of three horsepower (hp). Horsepower is a unit of measurement that quantifies the power a motor can produce, with one horsepower roughly equivalent to 746 watts. Therefore, a 3hp motor can deliver about 2,238 watts of power, making it suitable for various applications requiring significant energy, such as pumps, compressors, and industrial machinery.
The recommended plug gap for the 1993 model, 3hp Evinrude outboard is .030 in.
1955 3hp
it depends on the load and size of the motor if the motor has to drive heavy loads from initial state it require more torque, for this we use star-delta starters where load on motor in permissible then we use a DOL starters DOL starters are direct supply feeders to the motor so there is no rule to use a starter. we can use anything to swith the motor. for protection we need a overload relay and you ask for a 1ph motor, i.e maximum 3hp motors are available for less than 2 hp motor you can use a 6amp switch used in household or a 6amp mcb (includes overload). for 3hp a 16 amp switch or a 10amp mcb is more than enough
A 3HP motor can reach speeds up to at least 30 mph depending on what you are using it for.
The power consumed can be calculated directly from the current draw, voltage, and power factor: (3A) x (400v line to line) x (sqrt(3)) x (.9pf) = 1.87kW. This would be a reasonable value for a 3hp motor that is not running under full load (the motor may be slightly over-sized for this application). The size of the motor is irrelevant to this calculation, although it shows that it is running at ~85% of its capability.Another AnswerSince a motor is a balanced load, we use the following equation to find its input power:power = 1.732 x line voltage x line current x power factor.So, in your example,power = 1.732 x 400 x 3 x 0.9 = 1870 WObviously, it cannot be delivering 2200 W when its input is just 1870 W. However, to determine its output power you will need to determine its efficiency at that input.
To run a 3HP (horsepower) 120V chainsaw, you would need an inverter that can handle the starting and running wattage. A 3HP motor typically requires around 2,200 to 3,000 watts for starting and about 2,000 watts for continuous operation. Therefore, you should look for an inverter with at least a 3,000-watt output capacity to ensure it can handle both the starting surge and the continuous load. Additionally, it's wise to choose an inverter with a bit of extra capacity for safety and efficiency.