motor hp. calculation

[bpk]

I was recently working with a hydraulics tech. on 4 250hp. pumps. He wanted to amp draw them as they go through there pump cycle,they all would draw approx. 240 amps at 480 volts. The technician asked me what approx. horsepower the motor is useing at its highest amp draw and I came up with 154 horsepower (1hp.=746 watts). The FLA. on the motor nameplate is only 260 amps at 480 volts, so useing this equation the motor running at FLA would only produce about 167hp. and to get its rated 250hp. it would need to run at 404 amps. What am I missing?

 

[bpk]

I alredy realized 1 thing I did wrong when I calculated the watts I didnt use 1.732, that gets me a little closer to the rated hp.

 

[Jraef]

Well, you can't accurately correlate amps directly to HP anyway. While it's true that HP/746 = Watts, you can't directly determine Watts from Amps alone; you also need the power factor. You could assume a .8pf when fully loaded, but if the motor is running at less than full load the power factor will be lower than that, or even whatever is on the nameplate (if it is at all), because nameplate data is always based on full load. So from that standpoint, knowing Amps only is kind of a "hand grenade" estimate of HP.

Now the challenge will be, trying to explain power factor to your hydraulics tech...:wink:

 

[GeorgeB]

Well, you can't accurately correlate amps directly to HP anyway. While it's true that HP/746 = Watts, you can't directly determine Watts from Amps alone; you also need the power factor. You could assume a .8pf when fully loaded, but if the motor is running at less than full load the power factor will be lower than that, or even whatever is on the nameplate (if it is at all), because nameplate data is always based on full load. So from that standpoint, knowing Amps only is kind of a "hand grenade" estimate of HP.

Now the challenge will be, trying to explain power factor to your hydraulics tech...:wink:

We Electro Hydraulics guys find it hard to explain that motor load (amps) is not linear with shaft load to many plant electricians. I have a really hard time with PF, even to experienced MECHANICAL engineers and most electricians. I bet that fewer than 10% understand PF correction capacitors or the line inductors we use in servo and VF systems.

The Baldor website gives, for most motors, amps at %Load ...

Taking their M4408TS-4, 447TS frame (a good choice for a hydraulic system), we find DARN IT, no performance data on the website!

I selected a 200HP a none of the 250's have performance data ... M4416-T does.
0HP, 52A
50HP, 77.3A (35%)
100HP, 119A (54%)
150HP, 168A (76%)
200HP, 220A (100%)
230HP, 252A (S.F)
250HP, 274A
300HP, 331A

Large motors are pretty linear from half load up ... interesting thing here is that the no load is 52A, but the power consumed is probably in the 2 to 3 HP range, virtually all in iron and copper losses.

 

[kingpb]

To accurately determine what you are trying to find, you must use the performance curves.

 

[mayanees]

HP Approximation.....

HP Approximation.....

bpk,

I think your answer to the tech regarding HP for a 240 amp reading on 260 amp fla/250 HP motor is logically 240/260 * 250 HP = 230 HP.

If you're attempting to calculate the hP based on amperage readings you need to take into account pf, as mentioned in previous posts, AND also efficiency.

Realize that HP is a mechanical term relating to shaft output. Motor current will vary for 250 HP motors based on the power factor and efficiency of the motor.

JM

 

[gar]

080603-1245 EST

bpk:

Instead of using an ammeter use a wattmeter. Then you are only concerned with motor efficiency.

I have an electronic, meaning it uses Hall devices for current sensing and multiplication with voltage, three phase wattmeter that I have use to evaluate nut torque on a machine that did not have a torque transdiucer. Worked quite well on a standard motor, but not on the leads from a vector drive to the motor (the frequency components were too high).

Instantaneous power is e*i. If this is averaged you get a measure of the rate of doing work over the averaging period. When you are working with sine waves and you multiple two in phase waves together the result is a DC component and a double frequency component. Averaging this extracts the DC component. Multiple a sin and cos together and ther is no DC component and only a double frequency component. Average that and you get zero.

Thus, a multiplier of instantaneous voltage and current averaged will be a measure of power. The Hall device is inherently a multiplier.

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