Smart questions
Smart answers
Smart people
INTELLIGENT WORK FORUMS
FOR ENGINEERING PROFESSIONALS

Member Login




Remember Me
Forgot Password?
Join Us!

Come Join Us!

Are you an
Engineering professional?
Join Eng-Tips now!
  • Talk With Other Members
  • Be Notified Of Responses
    To Your Posts
  • Keyword Search
  • One-Click Access To Your
    Favorite Forums
  • Automated Signatures
    On Your Posts
  • Best Of All, It's Free!

Join Eng-Tips
*Eng-Tips's functionality depends on members receiving e-mail. By joining you are opting in to receive e-mail.

Donate Today!

Do you enjoy these
technical forums?
Donate Today! Click Here

Posting Guidelines

Promoting, selling, recruiting, coursework and thesis posting is forbidden.
Jobs from Indeed

Link To This Forum!

Partner Button
Add Stickiness To Your Site By Linking To This Professionally Managed Technical Forum.
Just copy and paste the
code below into your site.

gerdsen (Civil/Environmental) (OP)
21 Oct 03 4:54
Hi,
Can anyone tell me the formula for converting KvA to amps?
I often need to make these calculations for my job as an event organiser. For examble you have a 500 KvA generator so how many amps 3phase do you have?
I believe to find usable KW you times by 0.8 (as you can only always use 80% of the power for constant supply), but I still can't work out the amps. for examble if you have a 2.5 KvA fridge (2.5 x 0.8 = 2KW) can that be powered safely by a 13 amp plug?
I'm based in England if that makes any difference.
Thanks for your help

Camilla
RobWard (Industrial)
21 Oct 03 6:32
Well, power (Watts) divided by voltage gives you the current (amperes)

So if your generator gives you 230V you can run 2990W (2.99kW) on a 13A plugtop. Most people would say that's near as damn it 3kW.

However, check the voltage, as this could actually get down to around 220V (or even less from a generator) and that, on a 3kW load gives you 13.6A.

Now, without wanting to sound contradictory,in reality though most people I know (and myself) use a rule of thumb of 4A per kW, which tends to work out alright, if you aren't pushing the limits.

I hope that helps.

"I love deadlines. I love the whooshing noise they make as they go past." Douglas Adams

kokonut (Electrical)
21 Oct 03 7:20
For a three phase generator, the kVA is Current times Voltage times Sqrt(3), which makes the current that can be delivered equal to the Power divided by the product of Sqrt(3) and the voltage.
For a 500kVA, 3phase generator, the line voltage should be about 415V, which should give a deliverable Amps of about 695Amps, if you work it out.
For single phases voltages (of between 220 and 230V) the Sqrt(3) vanishes from the formula.
As you already know, your useful power is your VA times 0.8.
Cheers.
gerdsen (Civil/Environmental) (OP)
21 Oct 03 7:57
Hi Kokonut,

Thanks for your answer, but I still don't get it...sorry

So if you have a 500 kVA 3ph genny with 415 volts, to get the amps available you do the following
500 kVA divided by 3 (for the individual amps per phase?)divided by 415 volts = 0.4016
That's not right, is it...
Thank you very much for your patience
Helpful Member!  pablo02 (Electrical)
21 Oct 03 8:55
gerdsen -- remember to keep track of your units (k=1000):

for 3ø,  Amps = (kVA*1000)/Volts/1.732 (SqRoot of 3)

also, if you divide the kVA by 3 to get kVA per phase (assuming the load is balanced across all three phases), you don't use the 415v (line to line voltage), you use the line to neutral voltage (as kokonut was pointing out):

for instance, for 500 kVA at 480v 3ø, the formula is 500,000/480/1.732=601 Amps

if you assume the load is balanced and use 1/3 of 500 kVA, you must use the single phase voltage of 277v (=480/1.732) in the following formula:

(kVA*1000)/Volts = Amps:  (500,000/3)/277 = 601 Amps

either way, the result is the same... hope this helps with your calculations..
kokonut (Electrical)
21 Oct 03 9:00
No problem, gerdsen.
First you'll agree that squareroot of 3 is about 1.732.
So, for a 500kVA, 3ph genny delivering 415 line voltage, the current capacity is (500 x 1000) / (415 x 1.732) = 695A.
The 1000 is to convert from kilovolts to volts.
Once again, its VA / (V x sqrt (3)).
Hope this helps.
gerdsen (Civil/Environmental) (OP)
21 Oct 03 9:06
Thanks very much to you all.
It makes good sense now.

Have a lovely day

buzzp (Electrical)
21 Oct 03 12:06
Be careful with those calculations when your running an inductive load such as a motor. The links below give a good set of formulas.
http://www.reliance.com/mtr/flaclcmn.htm
http://www.iprocessmart.com/techsmart/formulas.htm
Helpful Member!  sparksski (Electrical)
21 Oct 03 14:25
gerdsen

As a very simple rule of thumb for a 3 phase generator the line current per phase can be found by Genny KVA X 1.4

So a 500KVA genny = 500 X 1.4 = 700A per phase.

if you want to know the KVA rating of the genny from the line current of 700 Amps.

Then KVA rating = 700/1.4 = 500KVA

Very similar to the above calcs but easier to work out if you dont understand the theory. This approximation works out for all gennys and transformers and gives a very comparable current when compared to the theory answer.

Regards

 
Lakey (Electrical)
23 Oct 03 5:25
Hi,

Please forgive my ignorance, but where does the square root of three come from?  Why square root of three?  As used in the above formulas.

Thanks & regards,

Always trying to learn new things!
pablo02 (Electrical)
23 Oct 03 9:20
From geometry/trigonometry:  3 phase power is 120° difference from phase to phase -- draw a Y with 120° between each of the three legs, if you draw a line X between any two legs, that length (X) represents the magnitude of the voltage phase to phase; each leg (Z) represents magnitude of the phase to neutral (center point of the Y) -- if you apply the 30/60/90 triangle functions, you'll find that X (phase to phase) equals 1.732 (SqR 3)* Z (phase to neutral)

This is the simple explanation...
Lakey (Electrical)
23 Oct 03 9:57
Thx mate.
A2a2a2a2 (Electrical)
28 Oct 03 6:23
Amp is power/vlt
as in ur example 2kw=<240 or 220> *I
the volt will depend on the generator supply.
but to be on the save side  15amps will be the best, to prevent overheating.
motorman (Electrical)
18 Nov 03 12:53
Helpful Member!  amptramp (Electrical)
18 Nov 03 14:51
pablo02:

Excellent explanation!

May I illustrate? Using the Law of Cosines:

c2 = a2 + b2 - 2(a)(b)cos(C)

where:

c = phase to phase vector magnitude
a,b = phase vector magnitude
C = angle between a and b

Let a and b be of unit magnitude, i.e., 1, and C = 120 then

c2 = 12 + 12 - 2(1)(1)cos(120)

c2 = 1 + 1 -2(-.5)

c = sqr(3)

Reply To This Thread

Posting in the Eng-Tips forums is a member-only feature.

Click Here to join Eng-Tips and talk with other members!

Close Box

Join Eng-Tips® Today!

Join your peers on the Internet's largest technical engineering professional community.
It's easy to join and it's free.

Here's Why Members Love Eng-Tips Forums:

Register now while it's still free!

Already a member? Close this window and log in.

Join Us             Close