1 HP electric motor = 1.5 HP hydraulic motor so for a 3 HP electric motor you would need a 4.5 HP hydraulic motor. Also for your information a 1 HP hydraulic motor = 1 2/3 HP gasoline engine.
1 HP = 2446 btu/hr and 1 kWh=3413 btu/hr so 1kWh = 1.413 HP, thus 45kWh x 1.413 = 63.585 HP.
1 HP = 746 watts. 16000 watts / 746 = 21.4 HP. Nominal HP is probably 20 HP.
According to Wiki s.com to convert horsepower to kilowatts you multiple the horsepower by a factor of 0.746. How do you convert horsepower to kilowatts However, you are saying that your 1 HP motor uses 1.84KW instead of 0.746KW which doesn't make sense to me. You will be shocked by how much your electric bill will rise by the use of the pool pump. To figure out how much, find out the Kilowatt (KW) rating of your main pool pump and your pool sweep pump and multiply this by the number of hours each runs. Then multiply this by your electric bill Kilowatt-hour (KWh) rate. For my 16,000 gallon pool which uses a 1 HP motor at 1.84 KW running 8-10 hours per day plus a 0.75 HP motor at 1.495 KW running 3 hours per day for the pool sweep, the cost is $75-$90 per month (at 13 cents per KWh). I have a 15000 gallon pool with a 1 HP motor running 10-12 hours per day. The electric bill is $100-120 per month.
No.
if your talking about a generator its not the hp of the engine its the size, rpm, and gearing of the generator motor
1hp = 746 watts, so 2hp would be 1492 watts, or almost 1.5 kw kWh does not convert to hp, as kWh has a time component.
A single phase 10 HP motor will draw aproximately 50 amps. A three phase 10 HP motor will draw aproximately 28 amps.
5 HP MOTOR WOULD CONSUME ENERGY OF 1342800 JOULES IN AN HOUR.EXPLANATION :-1 HP = 746 WATTSTHEREFORE 5 HP = 3730 WATTS.= 373O Joules/sec (since 1WATT= 1 Joules/sec)=3730 x 3600= 13428000 Joules/ hour.
no
1.34 hp (electric)
AWG #10 copper on a 30 amp breaker.
1 hp = 746 watts2 hp = 1,492 watts2 hp x 1 hour = 1,492 watt-hours = 1.492 kWh(Note: The "2 hp" rating on the pump refers to the useful output, that is,the water that the pump moves. The '1.492 kWh' calculated is simply theequivalent of 2 hp-hour, expressed in a different unit. The consumptionfrom the electric utility will be more than that, because no machine is100% efficient, that is, a machine's useful output work/energy is alwaysless than the input to it.To calculate the input energy required to operate this pump at full loadfor an hour, divide 1.492 kWh by the efficiency of the pump.)
10 hp and above motor power rating....
10 hp is its output power. To determine its inputpower, you need to know its efficiency, then use the following equation:efficiency = (output power)/(input power)You will first have to convert 10 hp into watts. There are 746 W in one horsepower.
1 HP electric motor = 1.5 HP hydraulic motor so for a 3 HP electric motor you would need a 4.5 HP hydraulic motor. Also for your information a 1 HP hydraulic motor = 1 2/3 HP gasoline engine.
A kilowatt is a unit of power, and energy is measured in kilowatt-hours, in other words 1 kilowatt flowing for 1 hour is one kWh of energy. For an ordinary heat engine it is normal to use a round-number estimate of consumption as half a pound of fuel for one horse-power-hour, which is 0.746 kWh.