Amps to HP

[lclark2074]

Amps to HP can anyone tell me where I can find an amp to hp chart. I am looking at 3.9 amps at 120 volt. Thanks.

[Zork]

Is this to find the HP of a compressor?? or what?

[lclark2074]

I found the formula it is 1 hp = 745.699872 Watts
I am sorry about my spelling.
3.9A x 115V = 448.5 watts / 745 = 0.60 HP
-----------------------------------------
But when I bought it and took it apart I saw 1HP on label?

[craig588]

thats fraud right there.

semi OT: I just remember that 1HP=~6.5 amps on a 115V line, thats an incredibly accurate number of watts you found though. :D

[lclark2074]

I agree 745 / 115 = 6.47 amps
The listing a the dehumidifier seas 3.9 amps

The rating on compressor seas 1HP 60hz 115 LRA 17
Model number is QA050CBA
SER NO is 21E00263-A-3ML21-801311

I have not looked at any tec sheets yet

[matttheniceguy]

This is a real shot in the dark, but is it possable that the compresor is on a voltage regulator and not getting the full 115?

also, is it possable that the power rating (3.9 amps) is for the average operation of the unit, where the compressor may only be on half of the time?

[Mustanley]

Look at the rating on my chiller sporting a Danfoss compressor.

http://virginstudent.com/fileserver/upload/151/149/331/cooler4.jpg

According to that formula this is nearly 3/4 HP, but yet the label lists 1/4 HP :confused:
Go figure

[greenman100]

Look at the rating on my chiller sporting a Danfoss compressor.

http://virginstudent.com/fileserver/upload/151/149/331/cooler4.jpg

According to that formula this is nearly 3/4 HP, but yet the label lists 1/4 HP :confused:
Go figure


Yes, because the voltage and current are out of phase. Go Google if you care to learn more. Those specifications are correct.

[Mustanley]

I'm not questioning the specifications, just the application of that formula to those specifications. From what little I understood about measuring power when voltage and current are out of phase, one would have to use a true watt (or watt/hour) meter.

[matttheniceguy]

If you know the phase angle you can calculate everything you want. The problem is finding the angle. If you can't find any other information on the electricl properties of the device, then just measuring the power with some sort of meter is probably the easiest way.

[lclark2074]

The formula was taken from google.com I put in (watts to hp converters) and it gave me a websight that gave me that formula?
And my amps listed on case on dehumidifier and HP is listed on compressor
Yours is more realistic remember there a fan and control circuitTo remember.

[lclark2074]

Look at the rating on my chiller sporting a Danfoss compressor.

http://virginstudent.com/fileserver/upload/151/149/331/cooler4.jpg

According to that formula this is nearly 3/4 HP, but yet the label lists 1/4 HP :confused:
Go figure

I agree but yours maybe talking about starting amps not running amps ?

[lclark2074]

No I did not see a voltage regulator just cap start unit

[Sean Lindstrom]

Wherever I see horsepower rating directed at the consumer, I ignore it. It's unregulated and often grossly overstated. It can mean "developed horsepower" for example, an engineer's trick, which might be a heavy flywheel spun by the motor in question, then the force needed to brake the flywheel is the horsepower rating boasted on the box.

Stick with amps, and remember larger motors are more efficient due to better windings and clearances.

[squeakygeek]

thats fraud right there.

semi OT: I just remember that 1HP=~6.5 amps on a 115V line, thats an incredibly accurate number of watts you found though. :D

Precise? Yes. Accurate? Who knows?

[lclark2074]

I got amp probe so when I get home in couple of days I will check the true amps out. Maybe this will get to bottom of this

[lclark2074]

I checked it with amp probe and draws 3.8 amps

[BioTuned]

I found the formula it is 1 hp = 745.699872 Watts
I am sorry about my spelling.
3.9A x 115V = 448.5 watts / 745 = 0.60 HP
-----------------------------------------
But when I bought it and took it apart I saw 1HP on label?

isn't this the physics formula. in physics doesnt watts mean something else. and 745.699872 watts = 1hp is in i believed used in U.S american system. i might be wrong. their are few kinds of systems for HP. such as Boiler HP, continental HP, Electical HP, Water HP, Metric HP, as well as U.S HP.

[lclark2074]

isn't this the physics formula. in physics doesnt watts mean something else. and 745.699872 watts = 1hp is in i believed used in U.S american system. i might be wrong. their are few kinds of systems for HP. such as Boiler HP, continental HP, Electical HP, Water HP, Metric HP, as well as U.S HP.
Probably your right but :thup:
I would think thy should be still close not about 1/3 off.

[matttheniceguy]

It's also possable the the power rating on the comressor is talking about it's cooling power, and not it's power consumption. It's quite common for a phase change system to have a COP above 1. The compressor may be rated at 1 hp, and have that much cooling capacity, but only draw 1/2 hp of electrical power.

[rashal]

HP is basically moving wights over time, which is rather flexible and can be misleading. Take this for example:

You have a truck and some sport car. They both generate a peak HP output of... Lets say 700HP.

So what's the difference? Trouqe. While the truck can move very heavy loads very slowly, the car can move very light loads very fast. You wont be able to tow 2 tons of metal with the sports car much like you wont be able to get to 100mph in 5 secounds with the truck. Different results, same HP.

Amps to HP conversion especially impractical because we have 'pure' energy on the one hand and 'mechanical' energy on the other.

[Damian]

HP is basically moving wights over time, which is rather flexible and can be misleading. Take this for example:

You have a truck and some sport car. They both generate a peak HP output of... Lets say 700HP.

So what's the difference? Trouqe. While the truck can move very heavy loads very slowly, the car can move very light loads very fast. You wont be able to tow 2 tons of metal with the sports car much like you wont be able to get to 100mph in 5 secounds with the truck. Different results, same HP.

Amps to HP conversion especially impractical because we have 'pure' energy on the one hand and 'mechanical' energy on the other.
Actually, that's not quite a good comparison either. Provided that the car and truck weigh the same, with the right gearing both will accelerate just as quickly since gearing can increase or decrease torque.

[lclark2074]

Hears a link about HP (http://auto.howstuffworks.com/horsepower.htm) as you can see its weight over time moved = power
ps goto page 2

[squeakygeek]

Actually, that's not quite a good comparison either. Provided that the car and truck weigh the same, with the right gearing both will accelerate just as quickly since gearing can increase or decrease torque.

But they have different gearing. That's his point.

[kayl]

the deeper the vacuum the compressor runs the less power it will consume.
You cant go by AMPs x voltage to work out what Hp a compressor.
if different conditions a compressor can draw more or less power than what its rated to.

best way to know is to check the spec sheets for size of a compressor.

[Prandtl]

the formula to go from watt to hp is accurate, what is not, is the way you calculate your electrical power.
P = I * V is only good for a purely resistive load, in the case of an inductive load, the formula is P = I * V / cos(phase angle), ie in the case of a resistive load, the "phase angle" is 0, then cos(phase angle)=1 and hence, P = I*V

[GigaForce310]

Actually, that formula is wrong. If it were correct, as the phase angle increases the power would increase too. In reality, when the phase angle increases, the power decreases.

That decrease isn't because the energy is being converted to heat or mechanical energy but stored as magnetic flux or capacitive potential energy.

The correct formulas are:

P(VA) = V*I, where P is imaginary power in Volt Amps (VA)

P(real) = V*I*cosθ, where P is real power in some unit (watts, HP, ect), θ is the phase angle between voltage and current and is subject to -90° < θ < 90°

Therefore P(VA) = P(real) only when θ = 0° (resistive, or like resistive circuit)

Motors are tricky beasts to solve for outside of physical testing. There are many factors that effect the results such as: copper resistive losses, inductive reactance, Flux effect and gap calculations, core eddy current losses, induced currents in the rotor and resistive losses, and the easiest is bearing losses. The end result after heaps of headaches is the output mechanical power. All of the above can be found with a no load and a locked rotor test.

Of course compressors freeze so they need to have a low locked rotor current. A good motor won't burn out if you turn the compressor on while locked.

[lclark2074]

Actually, that formula is wrong. If it were correct, as the phase angle increases the power would increase too. In reality, when the phase angle increases, the power decreases.

That decrease isn't because the energy is being converted to heat or mechanical energy but stored as magnetic flux or capacitive potential energy.

The correct formulas are:

P(VA) = V*I, where P is imaginary power in Volt Amps (VA)

P(real) = V*I*cosθ, where P is real power in some unit (watts, HP, ect), θ is the phase angle between voltage and current and is subject to -90° < θ < 90°

Therefore P(VA) = P(real) only when θ = 0° (resistive, or like resistive circuit)

Motors are tricky beasts to solve for outside of physical testing. There are many factors that effect the results such as: copper resistive losses, inductive reactance, Flux effect and gap calculations, core eddy current losses, induced currents in the rotor and resistive losses, and the easiest is bearing losses. The end result after heaps of headaches is the output mechanical power. All of the above can be found with a no load and a locked rotor test.

Of course compressors freeze so they need to have a low locked rotor current. A good motor won't burn out if you turn the compressor on while locked.
Do you mean frees as in cold or stop? If you mean cold then the compressor is one of the hottest parts of ref loop. If you mean stop that is another problem.