What is a kVa?

[CharlieH]

Brilliant, thankyou! It sort of makes sense now http://www.blue-room.org.uk/public/style_emoticons/default/biggrin.gif

[themadhippy]

How would you know this

unless you know different the guesstimate is normally 0.8

[maxjones2000]

wow I didnt know any of this!! :D

 

Thats quite useful!! :)

[Bobbsy]

Just to throw in a minor extra "why it matters", when you have a large commercial mains supply to a building (the last TV facility I built had a supply not far short of a megawatt) your contract specifies the power factor you need to achieve--and the rate you pay per kilowatt hour goes up quite a lot if you don't keep to this. It makes sense--the worse your power factor the more kVA they have to provide to achieve your KW.

 

Over a year, the numbers could make a big difference in cost.

 

Bob

[CharlieH]

Just to throw in a minor extra "why it matters", when you have a large commercial mains supply to a building (the last TV facility I built had a supply not far short of a megawatt) your contract specifies the power factor you need to achieve--and the rate you pay per kilowatt hour goes up quite a lot if you don't keep to this. It makes sense--the worse your power factor the more kVA they have to provide to achieve your KW.

 

Over a year, the numbers could make a big difference in cost.

 

Bob

So what actually happens to the extra power? If I had a device drawing 500w actual power, and 560w apparent power, which if I understand correctly is the same as 560kVA, (numbers made up off the top of my head), then where does the extra 60w go?

[Matthew Robinson]

Just to throw in a minor extra "why it matters", when you have a large commercial mains supply to a building (the last TV facility I built had a supply not far short of a megawatt) your contract specifies the power factor you need to achieve--and the rate you pay per kilowatt hour goes up quite a lot if you don't keep to this. It makes sense--the worse your power factor the more kVA they have to provide to achieve your KW.

 

Over a year, the numbers could make a big difference in cost.

 

Bob

 

Which is where power factor correction comes in, as I mentioned above. A building is likely to be an inductive load, so in many large buildings you'll find banks of capacitors in parallel with the load to correct power factor.

[blip2]

So what actually happens to the extra power? If I had a device drawing 500w actual power, and 560w apparent power, which if I understand correctly is the same as 560kVA, (numbers made up off the top of my head), then where does the extra 60w go?

 

 

That's the trap that lots of politicians fall into when talking about electricity. Real/Apparent power isn't a ratio rather a triangle (hence the cos in the calculations) illustrated by this diagram where S is apparent power, P is real power, Q is complex power (often called imaginary power) and the angle is the lag angle as discussed before.

 

Apparent Power = sqrt((Real Power)^2 + (Complex Power)^2)

Power Factor = cos(lag angle)

 

The 60W difference you speak of doesn't exist in physical terms.

[jmaudio]

So what actually happens to the extra power? If I had a device drawing 500w actual power, and 560w apparent power, which if I understand correctly is the same as 560kVA, (numbers made up off the top of my head), then where does the extra 60w go?

 

 

The 60W difference you speak of doesn't exist in physical terms.

 

It does exist. As in, it still has to be generated - you don't just see it do any useful work.

Most of the 60w you speak of will be lost as heat in the cable.

[blip2]

It does exist. As in, it still has to be generated - you don't just see it do any useful work.

Most of the 60w you speak of will be lost as heat in the cable.

 

Let me clarify, you cannot compare Apparent power and Real power, they are in different units (VA and W respectively).

 

560 VA - 500 W = 252.2 Var

 

This has nothing (or at least very little) to do with the conductive properties of the cables assuming you aren't talking transmission lines.

 

 

 

[Bobbsy]

 

Which is where power factor correction comes in, as I mentioned above. A building is likely to be an inductive load, so in many large buildings you'll find banks of capacitors in parallel with the load to correct power factor.

 

Been there, done that, bought the T-shirt. The hardest part is trying to explain to accountants the difference between real and apparent power and why spending money on Power Factor Correction will save money in the longer term.

[AdrianW]

It does exist. As in, it still has to be generated - you don't just see it do any useful work.

Most of the 60w you speak of will be lost as heat in the cable.

 

560 VA - 500 W = 252.2 Var

 

This has nothing (or at least very little) to do with the conductive properties of the cables assuming you aren't talking transmission lines.

 

Well it does, sorta. This 252VAr of reactive power still has to 'flow' in the circuit. If we assume that we're taking about a 250V system (lets make the maths easy), then just over 1A of reactive current must flow to 'supply' these VArs. This 1A is in addition to the 2A of current to supply the 500W of real power. However, these currents are not in phase, so the apparent (ie. the thing we could measure) current in the cable would be sqrt(1*1 + 2*2) = 2.24A. So an additional 240mA must flow to support this load. This additional current would result in additional losses (as heat) in the cables & transformers etc that supply this load (since the loss in the cable is I*I*R - where I is the apparent current & R is the resistance of the cable , not the load)..... so yes, reactive power does have a cost associated with it.

 

Adrian

[blip2]

Well it does, sorta. This 252VAr of reactive power still has to 'flow' in the circuit. If we assume that we're taking about a 250V system (lets make the maths easy), then just over 1A of reactive current must flow to 'supply' these VArs. This 1A is in addition to the 2A of current to supply the 500W of real power. However, these currents are not in phase, so the apparent (ie. the thing we could measure) current in the cable would be sqrt(1*1 + 2*2) = 2.24A. So an additional 240mA must flow to support this load. This additional current would result in additional losses (as heat) in the cables & transformers etc that supply this load (since the loss in the cable is I*I*R - where I is the apparent current & R is the resistance of the cable , not the load)..... so yes, reactive power does have a cost associated with it.

 

I would call 0.0053% power loss from a 10m length of TRS negligible, certainly when dealing with half a kilowatt.

[AdrianW]

I would call 0.0053% power loss from a 10m length of TRS negligible, certainly when dealing with half a kilowatt.

 

 

So would I, no dispute there :D ..... what I was getting at is that the flow of reactive power does cost the system - we can't/shouldn't forget about it. Looking at the losses another way... 1.5% of the cable losses for this particular load are due to the flow of VArs. This could be eliminated.

[CharlieH]

...due to the flow of VArs.

I take it a VAr is the extra current required to allow for the difference between apparent power and actual power?

[blip2]

VAr is the measure of complex power (proportional to the current though).