What is a kVa?

[CharlieH]

Sorry if this is really obvious, but I have been looking at hiring generators online (music-video we are thinking of shooting) and some of them have power outputs specified in watts, and some are specified in kVa (Kilovolt Amps I'm guessing). Obviously I am perfectly familiar with Watts, but I have never come across kVa before http://www.blue-room.org.uk/public/style_emoticons/default/huh.gif

 

I know that Power = Current * Voltage, so therefore surely a kVa is just another form of Watts (KIlovolts * Current). If this is the case then why bother with them at all? What is the difference between them?

 

I'm sure this is something really obvious, and I did do a quick google but the responses didn't quite make sense, hence why I am asking here. http://www.blue-room.org.uk/public/style_emoticons/default/wink.gif

 

Thanks for your help,

Charlie

[Brian]

Have a read up on Real and Apparent Power.

[Simon Lewis]

Also, it should be expressed as kVA, not kVa...

[CharlieH]

Also, it should be expressed as kVA, not kVa...

Ahh, thank you. http://www.blue-room.org.uk/public/style_emoticons/default/smile.gif

 

Have a read up on Real and Apparent Power.

 

I had a look at the Wikipedia page and it was way above my level I'm afraid - made no sense whatsoever! I did google Real and Apparent Power, and one page explained it very basically, which was good and I managed to follow it until it reached the 'Why Does it Matter' subtitle which again threw me.

 

I'm sorry for my lack of knowledge - GCSE Physics does very little on industrial electrical power (and absolutely nothing on 3-Phase) and nowhere does any courses on this sort of stuff until your 18 (understandably) http://www.blue-room.org.uk/public/style_emoticons/default/dry.gif. I've kind of got my head around 3-phase, but kVA has completely lost me! http://www.blue-room.org.uk/public/style_emoticons/default/wink.gif

[Smiffy]

As with everything it's a learning process Charlie. Wait till you start getting to think about Delta and Y (or is it Wye, or Why, or...?) Things that still confuse me.

 

If you are going to be an Electrician Electrician, as opposed to a Chief Electrician (not all of them are actually Electricians!) then it's important stuff. I would urge getting some electrical qualifications too. Makes you much more marketable as a freelance tech.

 

Regardless, a sound basic electrical understanding is a good thing. So good on you for asking the question. For most of the average LD's day to day stuff, you need to be able to calculate Current requirements, and possibly your kVA pull on a Genny. An ability to balance phases on paper is also handy, as it helps keep your dimmer patches tidy (and avoid bringing down the local power grid). Being able to meter out the sockets before plugging a desk / dimmer in are also handy skills to have. Moreso if you can understand what the meter is telling you hehehe.

 

Good question though...

 

Cheers

 

Smiffy

[CharlieH]

If you are going to be an Electrician Electrician, as opposed to a Chief Electrician (not all of them are actually Electricians!) then it's important stuff. I would urge getting some electrical qualifications too. Makes you much more marketable as a freelance tech.

I am hoping to get them in the future. I don't particularly want to be an electrician, however having the qualifications will (I imagine) make life much easier when things go wrong as I would be allowed to take apart dimmers etc to fix them, as well as looking good on a CV http://www.blue-room.org.uk/public/style_emoticons/default/wink.gif

 

 

Regardless, a sound basic electrical understanding is a good thing. So good on you for asking the question. For most of the average LD's day to day stuff, you need to be able to calculate Current requirements, and possibly your kVA pull on a Genny. An ability to balance phases on paper is also handy, as it helps keep your dimmer patches tidy (and avoid bringing down the local power grid). Being able to meter out the sockets before plugging a desk / dimmer in are also handy skills to have. Moreso if you can understand what the meter is telling you hehehe.

That's what I really am after. Current requirements are fine - I frequently do these at school when rigging extra dimmers in our drama studio to check what I have to work with (and full credit to the site electrician - he lets me do it all; including calculations, rewiring gear etc and just checks over it before I plug it in), but my 3-phase understanding is sketchy, which I want to sort out at some point, and pretty much everything else on your post I have no understanding of at all. Maybe I should do a course when I'm 18 http://www.blue-room.org.uk/public/style_emoticons/default/smile.gif

[Brian]

... until it reached the 'Why Does it Matter' subtitle which again threw me.

 

Slightly simplified example...

 

Let's take the example from that Wikipedia page...

 

Example: The real power is 700 W and the phase angle between voltage and current is 45.6°. The power factor is cos(45.6°) = 0.700. The apparent power is then: 700 W / cos(45.6°) = 1000 VA.

 

So the real power, which is the power that actually does work, is 700W which means a current of around 2.9A but because of the power factor we have an apparent power of 1000W which is a current of 4A.

 

We get nothing extra from the 'lost' 300W but we still have to find the extra current from our generator. And our cabling needs to be bigger along with our connectors.

[CharlieH]

Slightly simplified example...

 

Let's take the example from that Wikipedia page...

 

Example: The real power is 700 W and the phase angle between voltage and current is 45.6°. The power factor is cos(45.6°) = 0.700. The apparent power is then: 700 W / cos(45.6°) = 1000 VA.

 

So the real power, which is the power that actually does work, is 700W which means a current of around 2.9A but because of the power factor we have an apparent power of 1000W which is a current of 4A.

 

We get nothing extra from the 'lost' 300W but we still have to find the extra current from our generator. And our cabling needs to be bigger along with our connectors.

Great, thank you! That all makes sense except 'Power Factor' which I assume is something I should know, but I'm afraid I don't. http://www.blue-room.org.uk/public/style_emoticons/default/huh.gif What is it, and how do you calculate it?

[Brian]

...except 'Power Factor'...

In a resistive load ie a resistor or in our case a filament lamp the current is in phase with the voltage: when the voltage is at its maximum value so is the current. This is a power factor of 1.

 

However, in inductive (transformer/magnetic ballast/motor) loads or capacitive (switch mode supply) loads the current and voltage are not in phase. The 'reactance' of the load can mean the voltage is already on the downward bit of the sinewave whilst the current is still rising. Or the other way around. You can measure the angle between the two waveforms and the power factor is the cosine of this.

 

 

If you want to read up on electrical fundamentals I'd recommend a secondhand copy of 'Hughes Electrical Technology'. You can get a copy of the 7th edition for about £5 including delivery on amazon.

 

http://images-eu.amazon.com/images/P/0582226961.01.MZZZZZZZ.jpg

[Matthew Robinson]

Power factor is the ratio of real power to imaginary power. When I did GCSE, we stayed well clear of AC, so I assume that you've done the same. Basically, when you have a capacitive of (more likely) inductive load, because of the way that AC interacts with these two properties, they store energy which is then pushed back into the system at different times within a cycle. This is imaginary power. Capacitors and inductors can be considered opposite, so the way to correct power factor is to put inductors in parallel with a capacitive load or, as I said before, capacitors in parallel with an inductive load.

 

Edit: Beaten to it by Brian.

[CharlieH]

...except 'Power Factor'...

In a resistive load ie a resistor or in our case a filament lamp the current is in phase with the voltage: when the voltage is at its maximum value so is the current. This is a power factor of 1.

 

However, in inductive (transformer/magnetic ballast/motor) loads or capacitive (switch mode supply) loads the current and voltage are not in phase. The 'reactance' of the load can mean the voltage is already on the downward bit of the sinewave whilst the current is still rising. Or the other way around. You can measure the angle between the two waveforms and the power factor is the cosine of this.

 

 

If you want to read up on electrical fundamentals I'd recommend a secondhand copy of 'Hughes Electrical Technology'. You can get a copy of the 7th edition for about £5 including delivery on amazon.

Ahh, It makes sense now! So its all down to what's plugged in, and whether it is resistive or inductive.

So if I wanted to power a dimmer from a gennie, that would be inductive (?) and therefore it is irrelevant and I just take the kVA as kW?

I will be buying that book to go next Schiller's on my book shelf when I next have some money http://www.blue-room.org.uk/public/style_emoticons/default/wink.gif

[Brian]

...I just take the kVA as kW?

 

Dimmers pose a whole different set of issues. In the extreme you might come across a situation where a gennie is only supplying stage lighting and it's all conventional stuff fed from dimmers. The 'bad thing to do' is to bump from a nothing-on state to a loads-of-lamps-on state. Gennies don't really like that as they have to work really hard to maintain their speed when the load changes. And often don't manage it.

 

I will be buying that book to go next Schiller's on my book shelf when I next have some money.

I've been doing serious electrics/electronics for years and I still sit underneath a large shelf of 'classic' books...

 

[jmaudio]

Power factor refers to the ratio between the angles of the current and voltage waveform.

 

For example a light bulb, a very simple electrical appliance, is a purely resistive load. This means that its current and voltage demands are in phase. ie, when the incoming voltage is at a maximum, (every 1/25th of a second in the UK) the current demand from the light bulb is also at a maximum. We can deduce from this information that the power factor of a lightbulb = 1.

 

This is because the angle between the curves is zero. And cos0=1.

 

For other appliances however, their power factors are less than one ( 0 ≤ power factor ≥ 1)

 

Things like motors, switch mode power supplies, florescent/discharge lamp ballasts, all have a power factor of less than 1. So the current demand from the appliance lags behind the incoming voltage. If we know the power factor of an appliance, we can find how large this lag is in terms of an angle. so if a unit has a PF of 0.94. then we know

cosx=0.94

rearrange to give

x=arccos0.94

x=19.9 degrees.

 

With this information we can start to see the relevance of power factor. If we know the power rating of the above appliance (in watts), we can find the total power which is needed to be generated and transmitted for this appliance.

 

So, say the real output power of the appliance is 1500 watts.

We calculate:

1500/cos19.9 = 1595VA. or ≈ 1.6kVA.

 

This is the basics of it. And about the limit of my knowledge, but explaining it to makes it clearer in my head. Equally if anyone disputes my maths/physics, then feel free to correct me!

 

HTH.

[CharlieH]

Dimmers pose a whole different set of issues. In the extreme you might come across a situation where a gennie is only supplying stage lighting and it's all conventional stuff fed from dimmers. The 'bad thing to do' is to bump from a nothing-on state to a loads-of-lamps-on state. Gennies don't really like that as they have to work really hard to maintain their speed when the load changes. And often don't manage it.

Okay, and I suppose then you start to see the advantage of discharge fixtures, which as long as you stagger them when lamping on you can snap from a blackout to a whiteout without any extra strain on your gennie (assuming there is no generic fixtures in the grid).

 

I've been doing serious electrics/electronics for years and I still sit underneath a large shelf of 'classic' books...

Wow......enough bedtime reading to keep you going for a while http://www.blue-room.org.uk/public/style_emoticons/default/wink.gif

 

 

Power factor refers to the ratio between the angles of the current and voltage waveform.

 

For example a light bulb, a very simple electrical appliance, is a purely resistive load. This means that its current and voltage demands are in phase. ie, when the incoming voltage is at a maximum, (every 1/25th of a second in the UK) the current demand from the light bulb is also at a maximum. We can deduce from this information that the power factor of a lightbulb = 1.

 

This is because the angle between the curves is zero. And cos0=1.

 

For other appliances however, their power factors are less than one ( 0 ≤ power factor ≥ 1)

 

Things like motors, switch mode power supplies, florescent/discharge lamp ballasts, all have a power factor of less than 1. So the current demand from the appliance lags behind the incoming voltage. If we know the power factor of an appliance, we can find how large this lag is in terms of an angle. so if a unit has a PF of 0.94. then we know

cosx=0.94

rearrange to give

x=arccos0.94

x=19.9 degrees.

 

With this information we can start to see the relevance of power factor. If we know the power rating of the above appliance (in watts), we can find the total power which is needed to be generated and transmitted for this appliance.

 

So, say the real output power of the appliance is 1500 watts.

We calculate:

1500/cos19.9 = 1595VA. or ≈ 1.6kVA.

 

This is the basics of it. And about the limit of my knowledge, but explaining it to makes it clearer in my head. Equally if anyone disputes my maths/physics, then feel free to correct me!

 

HTH.

That makes sense. So to calculate the kVA of something, you need to know its Power Factor, or the difference between the spikes of its current and voltage. How would you know this, is it given to you on the device? I guess you could calculate the difference with a voltage/time graph and a current/time graph, which you could get from an oscilloscope http://www.blue-room.org.uk/public/style_emoticons/default/huh.gif Please say if I'm wrong!

[jmaudio]

 

That makes sense. So to calculate the kVA of something, you need to know its Power Factor, or the difference between the spikes of its current and voltage. How would you know this, is it given to you on the device? I guess you could calculate the difference with a voltage/time graph and a current/time graph, which you could get from an oscilloscope http://www.blue-room.org.uk/public/style_emoticons/default/huh.gif Please say if I'm wrong!

 

 

Yeah, for professional items, either on the kit itself, or on the datasheet, the power factor should be stated. Either as a the ratio (the power factor) or as an angle such as cosx, where x is the angle between the current and voltage curves.