Power factor, UPS, computer

[Uncle Peter]

I was wondering why this 2nd hand UPS I bought says 1500VA, 980W. Are they assuming you don't have decent PFC on your computers? When did they start putting decent PFC in computers?

I tested (with one of those plug in energy meters) a computer running a good graphics card and CPU flat out with a cheap power supply (CIT) with "passive power factor correction" (along with a stereo playing loud and two old LCD screens), as this is where the UPS will be used. The meter indicated a PF of 0.71. I then tried a computer with a Corsair power supply with "active power factor correction", running a dual chip graphics card flat out, and got a PF of 0.98.

--
My friend drowned in a bowl of muesli - a strong currant pulled him in.

 

[Uncle Peter]

On Fri, 03 Jan 2014 12:52:45 -0000, Uncle Peter <no@spam.com> wrote:

I was wondering why this 2nd hand UPS I bought says 1500VA, 980W. Are they assuming you don't have decent PFC on your computers? When did they start putting decent PFC in computers?

I tested (with one of those plug in energy meters) a computer running a good graphics card and CPU flat out with a cheap power supply (CIT) with "passive power factor correction" (along with a stereo playing loud and two old LCD screens), as this is where the UPS will be used. The meter indicated a PF of 0.71. I then tried a computer with a Corsair power supply with "active power factor correction", running a dual chip graphics card flat out, and got a PF of 0.98.

This answer from Yahoo answers seems correct:

"This is of the specification written by UPS manufacturers is widely misunderstood by users.
The actual load that can be supported can be up to 1500 VA OR 980W. These figures MUST NOT be exceeded. Power factor is irrelevant.
Thus with a PF of 0.98 the load can be 980W/1000VA (both figures are within the limits above, limited by W)
With a PF of 0.5 the load can be 1500VA/750W (limited by VA)
Battery standby time is determined by the W (watts) of the load. Thus if it will give 5 mins at 980W you can approximate that it will give 10 mins at 490W."

Presumably the invertor has a wattage limit and the transformer has a VA limit.

--
Illegal is a big sick bird.

 

[Artem Bondarenko]

On Sunday, January 5, 2014 1:01:21 PM UTC+2, Dani...@teranews.com wrote:

connected to a resistive-capacitive-inductive load. So the phase shift,
reflected back from the secondary winding into the primary winding will,

almost certainly, not be 1, i.e. purely resistive.

The phase shift (cos(fi)) in not THD (total harmonic distortion). The computer have a truly active load. But this active load in not linear. This do non - sinusoidal current. Non-sinusoidal current = Harmonic distortion.

 

[Daniel47@teranews.com]

On 04/01/14 00:21, Uncle Peter wrote:

On Fri, 03 Jan 2014 12:52:45 -0000, Uncle Peter <no@spam.com> wrote:

I was wondering why this 2nd hand UPS I bought says 1500VA, 980W. Are
they assuming you don't have decent PFC on your computers? When did
they start putting decent PFC in computers?

I tested (with one of those plug in energy meters) a computer running
a good graphics card and CPU flat out with a cheap power supply (CIT)
with "passive power factor correction" (along with a stereo playing
loud and two old LCD screens), as this is where the UPS will be used.
The meter indicated a PF of 0.71. I then tried a computer with a
Corsair power supply with "active power factor correction", running a
dual chip graphics card flat out, and got a PF of 0.98.

This answer from Yahoo answers seems correct:

"This is of the specification written by UPS manufacturers is widely
misunderstood by users.
The actual load that can be supported can be up to 1500 VA OR 980W.
These figures MUST NOT be exceeded. Power factor is irrelevant.
Thus with a PF of 0.98 the load can be 980W/1000VA (both figures are
within the limits above, limited by W)
With a PF of 0.5 the load can be 1500VA/750W (limited by VA)
Battery standby time is determined by the W (watts) of the load. Thus
if it will give 5 mins at 980W you can approximate that it will give 10
mins at 490W."

Presumably the invertor has a wattage limit and the transformer has a VA
limit.

The transformer consists of two, inter-wound, inductors, one of which is
connected to a resistive-capacitive-inductive load. So the phase shift,
reflected back from the secondary winding into the primary winding will,
almost certainly, not be 1, i.e. purely resistive.

HTH

Daniel

 

[Daniel47@teranews.com]

On 06/01/14 00:38, Artem Bondarenko wrote:

On Sunday, January 5, 2014 1:01:21 PM UTC+2, Dani...@teranews.com wrote:
connected to a resistive-capacitive-inductive load. So the phase shift,
reflected back from the secondary winding into the primary winding will,

almost certainly, not be 1, i.e. purely resistive.

The phase shift (cos(fi)) in not THD (total harmonic distortion). The computer have a truly active load. But this active load in not linear. This do non - sinusoidal current. Non-sinusoidal current = Harmonic distortion.

Artem, I think English is not your first language, as I'm having trouble
understanding what you are trying to say here!!

Daniel