How many amps are in horse power


Old 02-21-05, 10:07 AM
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How many amps are in horse power

how many amps are in 1/2horse power.Sorry if i didnt word this right.i am as green as grass.Thanks.
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Old 02-21-05, 10:13 AM
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Around 9.8 amps @ 120 VAC, 4.9 @ 240 VAC, assuming single phase AC motor
Old 02-21-05, 10:14 AM
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The answer depends on what the voltage is.

1 horsepower = 745.699872 Watts (Google horsepower to obtain this number)

The formula is P=VI, or Power = Voltage x current. This becomes P / V = A, or Power divided by voltage equals current.

At 120 volts, 745.699872 Watts / 120 volts equals 6.214 amps.

At 240 volts, 745.699872 Watts / 240 volts = 3.107 amps.

Now do understand that with motors, there is also an efficiency rating that comes into play...
Old 02-21-05, 10:21 AM
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The NEC has a table that lists horsepower at various voltages and phases. This table lists 1/2 HP as 9.8 amps @ 120v/1-phase, and 4.9 amps @ 240v/1-phase, confirming what Jedi9 said.

Old 02-21-05, 06:13 PM
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Yeah,the 746 watts per hp is an ambiguous formula. Reality is no where near it.
I'm not sure of it's origin, but we were all taught it at one time.
Old 02-21-05, 09:30 PM
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I'm pretty sure the "746 watts per horsepower" was developed when they were discussing steam engines which didn't have the problems of inductive reactance and power factor ratings.
We always ballpark figure it at one kva per Hp. This is assuming fully loaded and, admittedly, is a little overkill.

Hey Gary....What's the application? Everyone here (including me) is assuming an AC motor. You're not doing some sort of DC Frankenstein project, are you?
Old 02-22-05, 09:41 AM
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The horsepower formula takes into account power factor and efficiency of the motor. If you do not have this information you sort of have to ballpark those figures to complete the calculation.

I usually use a power factor of .85 and an efficiency of .9.

Most people use the NEC tables. However, if available the NEC says the manufacturer's nameplate values for full load amps (FLA) and recommended breaker or fuse size (overcurrent protective devise, or OCPD) rules.

An interesting note: In Europe and other parts of the world I have heard that they do not rate motors in horsepower, but in kilowatts. In my work I often have to convert HP to kW so I can add up all my motor loads along with resistive loads. So I created a conversion table, and a pattern started to form. For "quick and dirty" ballparking of motor loads you can generally equate 1 HP to 1 kW. This is only if precision is not required, because the relationship is not linear. For instance, at 480v/3-phase, a 1 HP motor is 1.5 kW. At 10 HP it reaches near parity, at 9.9 kW. Above that, kW decreases from the 1:1 ratio, for instance, a 20 HP motor is 19.1 kW, a 30 HP is 28.3, a 100 HP is 87.6.

Anyway, it always amazed me that in the 21st century we still talk about such things in terms of horsepower and, believe it or not, we still describe lighting levels in terms of candlepower. But then again, here in the U.S. we are still using inches, which were standardized by King Edward II of England (1324) who decreed the inch to be equal to three barleycorns end to end.

Old 02-22-05, 12:44 PM
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Horsepower and watts are both units of power, just from different systems of measurement. One horsepower (interestingly the horsepower was defined by James Watt) is 33000 lbf*ft/min (lbf = pound force). One watt (of course named after James Watt) is defined as 1 N*m/s (N=newton). To convert simply convert each underlying unit:

33000 lbf*ft/min * (1 min / 60 s) * (4.448 N / 1 lbf) * (1 m / 3.28 ft) = 745.85 N*m/s = 745.85 W

It doesn't matter what type of power we are talking about, the conversion between HP and W is fixed.

The difficulty when dealing with electric motors is (1) motors are rated by the number of HP they deliver (under very specific conditions) and (2) we are generally concerned with the current which they draw. The 745.85 W/HP conversion factor would only apply for a perfect (100% efficient) motor which presents a purely resistive load. A real motor will have less than 100% efficiency (consuming more power than is delivered to the load) and will present a reactive load (meaning that the current draw is higher than the number of watts consumed would indicate). thus the difference between racraft's theoretical calculation and the NEC tables.

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