A 1-HP motor is reckoned to draw 7 amps at 240 v single-phase. The same power of motor would draw 3.5 amps at 480 v single-phase, but a 480 v supply could most likely be a three-phase suppy, and the current in that case would be reckoned as 2 amps.
To answer this question the pump's voltage is needed.
At peak power it should draw 1.36 amp at power factor 1 or more realistically 1.7 amp at power factor 1.7.
To answer this question accurately the amperage of the motor must be given. Using 746 watt equals one horsepower will not be as accurate but will be close to the correct wire size. The equation for amps when HP is known is I = HP x 746/E x %eff x pf. Amps = 746 x 3 = 2238/240 x .84 x .86 = 2238/173 = 13 amps. 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. A #4 copper conductor will limit the voltage drop to 3% or less when supplying 13 amps for 700 feet on a 240 volt system. The electrical code book states that a 3 HP motor on 230 volts draws 17 amps. If this is closer to what you pump draws then the wire size will be; A #4 copper conductor will limit the voltage drop to 3% or less when supplying 17 amps for 700 feet on a 240 volt system.
It would depend on how deep your well is, and how efficient the pump.
24.87amps 1 hp=746 watts P=IxV ... (746x8)/240
106 amps
A 1-HP motor is reckoned to draw 7 amps at 240 v single-phase. The same power of motor would draw 3.5 amps at 480 v single-phase, but a 480 v supply could most likely be a three-phase suppy, and the current in that case would be reckoned as 2 amps.
To answer this question the pump's voltage is needed.
A single phase 10 HP motor will draw aproximately 50 amps. A three phase 10 HP motor will draw aproximately 28 amps.
To answer this question the voltage of the motor must be stated.
It depends on the voltage-- I think at 110v it's 4 amps per hp
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
AWG #10 copper on a 30 amp breaker.
HP = (Current x Voltage)/746, or HP = (IE)/746 (disregarding %Efficiency) So, you have to solve for I, current: I = (746 x HP)/E
current = power/ voltage current = 3700/ 240 = 15.4 amps
Look at the motor nameplate and it shoud have the amp draw on it. If the nameplate is missing, then the amp draw depends on what type of motor it is. The basic calculation to get you in the ball park would be as follows: 1 HP = 0.75 KW 7.5 HP = 5.63 KW Assume the efficiency of the motor is 80%, then the power supplied will need to be 5.63/0.8 = 7.04 KW amp draw = 7040/220 = 32 amps <<>> For calculation purposes the electrical code book states that a 7.5 HP motor draws 40 amps.