Electrical amp-power factor-wattage question

slicksilver

Golden Member
Mar 14, 2000
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A machine in our factory has the following power ratings :

Phase : 3
Voltage : 400
Cycles : 50hz
Wattage : 4.5 KW
Amp :18.5

Power formula is P = V*I*Power factor

In the case of machines with motors power factor is 0.8.

So using the above numbers P = 400*18.5*0.8 = 9.964KW

My question is why does the machine label say 4.5KW then?

Thanks
 

brainhulk

Diamond Member
Sep 14, 2007
9,376
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anything over 9000! W is attenuated by 50% if it doen't have a dragon sphere resistor.

i'm just saiyan
 

Colt45

Lifer
Apr 18, 2001
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I don't really remember how three phase math works, but you need a sqrt(3) in there somewhere.
 

ShawnD1

Lifer
May 24, 2003
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A machine in our factory has the following power ratings :

Phase : 3
Voltage : 400
Cycles : 50hz
Wattage : 4.5 KW
Amp :18.5

In the case of machines with motors power factor is 0.8.
4500W / 0.8 = 5625VA
5625VA / 18.5A / 1.73 = 176V line to line
176V / 0.92 voltage drop = 191V from wall with no load
What the hell country do you live in? The voltage is fucked up and the power is only 50Hz. Sounds like North Korea or something.

400V * 18.5A * 1.73 = 12.802kVA. That puts your power factor around 1/3. Worst motor EVER.
 
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trungma

Senior member
Jul 1, 2001
466
36
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4500W / 0.8 = 5625VA
5625VA / 18.5A / 1.73 = 176V line to line
176V / 0.92 voltage drop = 191V from wall with no load
What the hell country do you live in? The voltage is fucked up and the power is only 50Hz. Sounds like North Korea or something.

400V * 18.5A * 1.73 = 12.802kVA. That puts your power factor around 1/3. Worst motor EVER.

Or it could be that the efficiency of the motor is really bad. ~0.44%
 

slicksilver

Golden Member
Mar 14, 2000
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4500W / 0.8 = 5625VA
5625VA / 18.5A / 1.73 = 176V line to line
176V / 0.92 voltage drop = 191V from wall with no load
What the hell country do you live in? The voltage is fucked up and the power is only 50Hz. Sounds like North Korea or something.

400V * 18.5A * 1.73 = 12.802kVA. That puts your power factor around 1/3. Worst motor EVER.

I'm in India. What the the downsides of having a really low power factor?
 

PowerEngineer

Diamond Member
Oct 22, 2001
3,606
786
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I'm in India. What the the downsides of having a really low power factor?

Ratings are one thing; actual quantities are another. I'd measure the actual voltage and current before getting overly excited about the resulting 12.8 KVA value. It may be that the motor has higher amp ratings so that it doesn't burn up when voltages are substantially lower than the rated voltage.

The biggest downside to poor power factor is that it takes more current to deliver the same amount of real power, and therefore you must deal with higher resistive losses and bigger voltage drops.
 

ShawnD1

Lifer
May 24, 2003
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I'm in India. What the the downsides of having a really low power factor?

Bad power factor means a lot of the current through the wires is useless. As a result, the wires need to be a lot bigger. Suppose the power factor is 0.5 and you're pulling 100A. You'll need a big 100A conductor, but only 50A of it is useful. How much of a cost difference do you think there is between 50A and 100A cable? It's a lot.
 

Howard

Lifer
Oct 14, 1999
47,982
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The amperage is FLA - full load amps.

EDIT: Are there multiple motors inside, some of lower voltage? There might be a step-down transformer, and single-phase motors have worse PF.
 
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slicksilver

Golden Member
Mar 14, 2000
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I went to the specific and yes there are multiple motors. Electric heating elements and turbo still should be ignored as the machine is not fitted with this option. Please check the image below :



Uploaded with ImageShack.us
 

PowerEngineer

Diamond Member
Oct 22, 2001
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786
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My question is why does the machine label say 4.5KW then?

:confused:

Even without the still and heater elements, the sum of pump and motor loads is 8.36 kW.

Exactly where are you getting the equipment "power ratings" listed in your original post?
 

slicksilver

Golden Member
Mar 14, 2000
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In the summary of the techinical specs. Also the machine has a plate on the back which has the same numbers except for voltage which is 380V. My machine is the steam version.



Uploaded with ImageShack.us
 
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ShawnD1

Lifer
May 24, 2003
15,987
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In the summary of the techinical specs. Also the machine has a plate on the back which has the same numbers except for voltage which is 380V. My machine is the steam version.



Uploaded with ImageShack.us
I'm curious what you mean by saying it's a steam version. Do you mean like it has heating elements to create steam and the steam drives a motor? If so, that would explain why it's horrendously inefficient. Gasoline and diesel engines are heat engines like that and their efficiency is never higher than 50%. Even 40% efficiency from a heat engine is considered good.


The "electric version" numbers look a little better
400V * 1.73 * 36 = 24.9kVA and 16kW usable power.
If you assumed the power factor and efficiency are equally responsible for the difference, that would make it roughly 80% efficient and 80% power factor. That's still a pretty low budget motor, but it sounds realistic.
 

jagec

Lifer
Apr 30, 2004
24,442
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I'm curious what you mean by saying it's a steam version. Do you mean like it has heating elements to create steam and the steam drives a motor? If so, that would explain why it's horrendously inefficient. Gasoline and diesel engines are heat engines like that and their efficiency is never higher than 50%. Even 40% efficiency from a heat engine is considered good.


The "electric version" numbers look a little better
400V * 1.73 * 36 = 24.9kVA and 16kW usable power.
If you assumed the power factor and efficiency are equally responsible for the difference, that would make it roughly 80% efficient and 80% power factor. That's still a pretty low budget motor, but it sounds realistic.
It probably means that heat is supplied externally via a steam line for his version, versus being supplied by the equipment with a resistance heater in the electric version. That's why the electric version uses that much more power. Also, the unity power factor of the heaters means that the .8 PF for the overall system still includes some abysmally inefficient motors.

As Howard mentioned, the amp rating may not represent what the motor actually consumes when the equipment is running. If it's FLA (sum of Full Load Amps for every motor in the system), it's unlikely that every motor will be running at 100% rated power at all times, whereas if it's LRA (locked rotor amps, often the figure listed for refrigeration equipment), that current will only be drawn on startup.

You have to size your connectors for max amps, but the kW rating will tell you how much power you'll actually be paying for. Unless you're in an industrial setting where you have to pay for power factor, which complicates matters again.

If you DO have to pay for power factor, you might want to have a power engineer come over and size some capacitors to correct the awful power factors on those motors.
 

ShawnD1

Lifer
May 24, 2003
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As Howard mentioned, the amp rating may not represent what the motor actually consumes when the equipment is running. If it's FLA (sum of Full Load Amps for every motor in the system), it's unlikely that every motor will be running at 100% rated power at all times
The current and power on the name plate are both based on maximum power. Example: I have a blender that is rated for 1HP and 10A. It doesn't always use 10A and it doesn't always do 1HP of work. If I put a bunch of ice in it and it's running at full power (1HP), it will consume 10A. The numbers should still add up correctly.

whereas if it's LRA (locked rotor amps, often the figure listed for refrigeration equipment), that current will only be drawn on startup.
I don't think I've ever seen a motor that lists the starting current as a number. What I usually see is a code letter that indicates a multiple of the full load current.
http://www.engineeringtoolbox.com/locked-rotor-code-d_917.html
If my 10A blender says it's a code G motor, it would be 10A * 5.9 = 59A starting current.
 

slicksilver

Golden Member
Mar 14, 2000
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Lot of knowledge power here. Thank you all. I'm basically trying to figure out how many units the machine consumes per hour realistically.