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Topic Title: kw to amps, 3 phase calc
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Created On: 04 December 2007 05:58 PM
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 04 December 2007 05:58 PM
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steveth

Posts: 26
Joined: 17 September 2007

just a quick one guys, forgotten how to do the calc to convert 3 phase kw to amps. i.e if a machine stated 25kw 3 phase, what would the current per phase being drawn be?

cheers
 04 December 2007 06:18 PM
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rocknroll

Posts: 8824
Joined: 03 October 2005

(kW x 1000 / sqrt3 x UL x cos)

IL = IP

regards

-------------------------
"Take nothing but a picture,
leave nothing but footprints!"
-------------------------
"Oh! The drama of it all."
-------------------------
"You can throw all the philosophy you like at the problem, but at the end of the day it's just basic electrical theory!"
-------------------------

Edited: 05 December 2007 at 08:13 AM by rocknroll
 04 December 2007 08:44 PM
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Ricicle

Posts: 833
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Originally posted by: rocknroll

Star



(kW x 1000 / sqrt3 x UL x cos)



IL = IP



regards


Not this old chestnut again

If the machine is rated at 25kW the use the formula RnR posted, but this applies regardless of how the load is connected (star or delta). 25kW is 25kW

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Empty barrels make the most noise.
 04 December 2007 09:11 PM
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DALEC

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HI STEVETH , By my maths ( DOH !) I= 36 PER PHASE ..... regards dale

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 04 December 2007 09:23 PM
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Jaymack

Posts: 4582
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Originally posted by: Ricicle

Originally posted by: rocknroll
Star
(kW x 1000 / sqrt3 x UL x cos)
IL = IP


Not this old chestnut again

If the machine is rated at 25kW the use the formula RnR posted, but this applies regardless of how the load is connected (star or delta). 25kW is 25kW


And we still can't get it right.........assuming the "machine" is a motor: -

Power (Watts) = 1.73 × Line Volts × Line Amps × PF × Efficiency


Jaymack
 04 December 2007 09:32 PM
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Ricicle

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Yeah and don't forget efficiency if it's a motor

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Empty barrels make the most noise.
 04 December 2007 09:38 PM
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DALEC

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whoops.... 63 amp per phase .... ( no cos or efficiency )

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master of puppets is pulling your strings
--------------------------------------------
http://www.elliselectricalservices.com
 04 December 2007 10:16 PM
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BigEddie

Posts: 22
Joined: 20 April 2007

Dalec ,you were right first time at 36A .
Use line volts @ 400V
(If 415v = 34.8A)
All O.K if you want it exact with PF and efficiency.

As a rough guide you can't go far wrong with the old multiply by four and divide by three. It gives you an idea if your maths are not too good.
25000W x 4 /3 = 33.33A.
 04 December 2007 10:16 PM
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Ricicle

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Originally posted by: DALEC

whoops.... 63 amp per phase .... ( no cos or efficiency )



You were closer the first time (but still not right)

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Empty barrels make the most noise.
 04 December 2007 10:25 PM
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rocknroll

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Not this old chestnut again

Okay for the students lets elaborate on the OP idea.

A 25kW machine with a power factor of 0.9 amd am efficiency of 85%. What is the phase current?

regards

-------------------------
"Take nothing but a picture,
leave nothing but footprints!"
-------------------------
"Oh! The drama of it all."
-------------------------
"You can throw all the philosophy you like at the problem, but at the end of the day it's just basic electrical theory!"
-------------------------
 04 December 2007 10:30 PM
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extrasockets

Posts: 392
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HP = AMPS X VOLTS X EFF X 1.73 X PF/746
KW = AMPS X VOLTS X 1.73 X PF/1OOO

I think... to many tonight

EDIT
and after reading the op again.
KW X 1000 / 1.73 X VOLTS X PF

Edited: 04 December 2007 at 10:33 PM by extrasockets
 05 December 2007 09:49 PM
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Ricicle

Posts: 833
Joined: 23 October 2006

Sorry couldn't resist
47.36Amps (@ 400V)

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