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N

non-sparky

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Hello

I have a report I need to do, I have added up the amount of Watts that pile of electrical devices (computer equipment) i have and it adds up to 6.2KW. If I want to convert this to KVA is it a straight swap, i.e. 6.2 KVA?

Also, i want to calculate how much it will cost to run, do I need KVA or KW for this?

Thanks!
 
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N

non-sparky

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  • #3
um...

ok, so watts is no good to me maybe?

Thanks
 
S

Shakey

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  • #4
Hello

I have a report I need to do, I have added up the amount of Watts that pile of electrical devices (computer equipment) i have and it adds up to 6.2KW. If I want to convert this to KVA is it a straight swap, i.e. 6.2 KVA?

Also, i want to calculate how much it will cost to run, do I need KVA or KW for this?

Thanks!
basically VA is apparant power this is volts supplied times amps used


W is true power. this is amps used times voltage developed acrross the purely resistive element of the load

VAR (volts amps reactive) is amps used times V across purely reactive element of the
load

The power factor is W/VA or VAR/W

now in practice, you will malways have resistance and reactance in a circuit (unless it is something like a pure resistor (heater))

so the PF will always be less than 1.

Around 0.8 is normal.

If you were using a lot of inductive loads such as motors (in a factory) then your PF would decrease and you would be wasting a lot of the supplied VA (which is what you are paying for!), in which case you can improve the PF by adding capacitance to the circuit

What this means is, even for computers, unless they are the ONLY thing on the supply, the VA (and the PF) of the total load will be influenced by EVERYTHING on the circuit, this side of the electricity meter.

So lets say you have a PF of 0.8 , this gives you a ratio of 3:4:5 of VAR:W:VA

so for every 4W you have 5VA

so your 6.2KW divided by 4 = 1550

times 5 for VA = 7750 VA = 7.75KVA @ 0.8 PF

And, the important thing is, your electric bill is worked out on the VA supplied, not the Watts used

Or of course, i could just be making this up from stuff i dreamt..........its been known...:p:p


hope this helps;)
 
R

rumrunner

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  • #5
What a player you are shakey ,that answer explains it better than any text book ive ever read ,well done ,i was thinking along the same lines but couldnt have put it into an understandable format,top marks star performer:)
 
N

non-sparky

  • Thread Starter Thread Starter
  • #6
Wow Shakey - thanks, I think...hehe

So in the manual for a computer, say it says it runs at an average of 300W, what use is this figure - what does it actually tell me??

So pf for that computer is 300/(220Vx1.5A) = 300/330 =0.9 right?

But VA is the useful figure so 330VA = 0.3 KVA ?

Going back to my figures of 6.2KW then

V = 220 to 240, lets use 230
A = 33.1 total
W= 6.2KW = 6200

PF= (W = 6200) / (V = 230 x A= 31.1)

6200/ 7153 = 0.87

PF = 0.87
VA = 7.153 KVA

So my equipment is using 7.2 KVA right? Close to your figure but I never really got the ratio part...I'm happy with 7.2KVA for now :)

SOOOoooo..... how do I get this KVA figure into the almighty £££?

Thanks ever so much!
 
S

Shakey

  • Thread Starter Thread Starter
  • #7
Wow Shakey - thanks, I think...hehe

So in the manual for a computer, say it says it runs at an average of 300W, what use is this figure - what does it actually tell me??

So pf for that computer is 300/(220Vx1.5A) = 300/330 =0.9 right?

But VA is the useful figure so 330VA = 0.3 KVA ?

Going back to my figures of 6.2KW then

V = 220 to 240, lets use 230
A = 33.1 total
W= 6.2KW = 6200

PF= (W = 6200) / (V = 230 x A= 31.1)

6200/ 7153 = 0.87

PF = 0.87
VA = 7.153 KVA

So my equipment is using 7.2 KVA right? Close to your figure but I never really got the ratio part...I'm happy with 7.2KVA for now :)

SOOOoooo..... how do I get this KVA figure into the almighty £££?

Thanks ever so much!

erm... not sure where you've got your amps figures from.

the book says the computer runs at an average of 300W - this relates to the power consumed by the computer (watts) it bears no relation to the power supplied to the circuit (VA) - some of which is lost 'running' the circuit i.e. dur to the natrually inductive (and therefore reactive) nature of much of what you have plugged in

it really only comes into play with large factories, where the electricity supplier will penalise you for having a poor power factor i.e where you are wasting a lot of the supplied VA, for your circuit it would make little difference, and as i say, you could only truly calculate it if you had a supply connected to the computers AND NOTHING ELSE!!

the reality is that your circuit VA is changing evertime you introduce or remove a load (which is happening all the time)

HOWEVER; your electricity supplier charges you in KWh kilowatts per hour - they are really charging you KVA per hour, must most people understand Watts better!

So: @ 300W, I=P/V = 300/230 = 1.3A

but the amps is irrelevent because they are charging Kw per hour

so lets say you use a 300W computer for 30 minutes

300W = 0.3Kw
30 mins = 0.5 hours

= 0.15 KWH

now lets say you pay 7p per KWH for your electricty. (for example)

thats 0.15 x 7 = 1.05p to run that computer for amount of time


basically take watts divided by 1000 (to give answer in Kw) times time in minuites divided by 60 (to give time in hours)

this gives KWH used, multiply this by KWH unit cost of electricity

Job done

Another product from theShakey school of 'made up' lecktricks:p:p:p
 
N

non-sparky

  • Thread Starter Thread Starter
  • #8
....Hold on a minute!! :)

My Amps was the total Amps added together for all 21 devices (31.1 Amps) , the calculation still seemed to work, but anyway...

So all i need is Watts?

300Watts = 300 WH = 0.3KWH

so for one hour a 300 Watt PC = 0.3KWH x 7p =2.1p = £0.02

so basically my 6.2KW from the start of the post is:

6.2 KW or actually 6.2 KWH and at 7p per unit, that is 43.4p per hour for the 21 devices on average, or £0.43/H

Soooo, £0.434 x 24 = £10 per day to run the devices?

man I wish I listened in Physics!!! :p are these ramblings correct?

Thanks Shakey!
 
S

Shakey

  • Thread Starter Thread Starter
  • #9
....Hold on a minute!! :)

My Amps was the total Amps added together for all 21 devices (31.1 Amps) , the calculation still seemed to work, but anyway...

So all i need is Watts?

300Watts = 300 WH = 0.3KWH

so for one hour a 300 Watt PC = 0.3KWH x 7p =2.1p = £0.02

so basically my 6.2KW from the start of the post is:

6.2 KW or actually 6.2 KWH and at 7p per unit, that is 43.4p per hour for the 21 devices on average, or £0.43/H

Soooo, £0.434 x 24 = £10 per day to run the devices?

man I wish I listened in Physics!!! :p are these ramblings correct?

Thanks Shakey!
yeah you got it!!!

now just change my imagined 7p per hour for what you are ACTUALLY paying and your there!:D:D
 
N

non-sparky

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  • #10
6.2 KWH x 12.44p = £18.51 per day, wow! £6,756.15 nice...

Ok so one more thing then if you would be so kind?

So if a server hosting rack (a metal cabinet full of PCs) has 2.4KVA electricity supplied with it, then why don't they just use Watts?

Do I then have to apply the power factor to it average 0.87 so really they supply 2.1KWH?

Thats it after that, honestly!! Thanks very much for your help, its much appreciated.
 
S

Shakey

  • Thread Starter Thread Starter
  • #11
6.2 KWH x 12.44p = £18.51 per day, wow! £6,756.15 nice...

Ok so one more thing then if you would be so kind?

So if a server hosting rack (a metal cabinet full of PCs) has 2.4KVA electricity supplied with it, then why don't they just use Watts?

Do I then have to apply the power factor to it average 0.87 so really they supply 2.1KWH?

Thats it after that, honestly!! Thanks very much for your help, its much appreciated.
the answer is in the 'supplied with it'

whenever we talk about supplies we talk about VA because this is quite literally V supplied by amps available. Transformers and generators are normally rated in VA, (or realistically KVA.)

Of course, depending on what type of load you apply will depend how much of this VA (apparant power) you use as Watts (true power) and how much you just waste through inductive reactance

so if your load had a pf of 0.8, then for your 2.4KVa supply, then you could use 1.9Kw of true power

and to be honest, at this level its hardly worth worrying about.

if you were doing the same thing for a factory running at 500KVa, and by introducing over excited motors as synchronous capacitors to improve PF correction from say, 0.75 to 0.8, then you would be talking serious bucks !!!;)
 
A

alex

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  • #12
:Dgo on shakey.............. your unstoppable.


i think i might start making stuff up to try and wind him up and catch him out.:p
 
S

Shakey

  • Thread Starter Thread Starter
  • #13
:Dgo on shakey.............. your unstoppable.


i think i might start making stuff up to try and wind him up and catch him out.:p
bring it on......

and remember you dont need to know the answer to the question

you just need to evaluate how much the person asking the question knows about the answer to the question;)

its the art of the woffler

just ramble on till their eyes glaze over and it all goes away:p:p:p:p
 
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