# From the Source to the Machine



## mbednarik (Oct 10, 2011)

you have 32 meters and 40 feet. Which is it meters or feet? A meter i just over 3 feet.


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## wildleg (Apr 12, 2009)

#2 will work fine


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## gnuuser (Jan 13, 2013)

is this a test question or an actual installation?:laughing:

when calculating conductor size you need to take all conditions into account including environment conditions.
higher ambient air temps must also be considered.
if you are asking us to solve a problem while we are not there to see the actual equipment and environment conditions. then you need to provide all the relevant information including conduit sizes.


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## amantelnar (Apr 21, 2013)

*Thanks for your Responses*

Mr. Embednarik: Actual length or distance is 32.8 feet, but I can't see 32.8 feet in the link I've shown you http://www.engineeringtoolbox.com/am...uge-d_730.html That's why I used 40 feet instead.

Mr. Wildgeg: Thank You for giving your direct answer.

Mr. Gnuuser: That was given to me by my future employer in Dammam Saudi Arabia...Not knowing yet the higher ambient air temps or conduit size since I have not seen yet the place. Perhaps, this is a test for me that if I've got it correctly they will send me a Visa to work with them. 

Guys, thanks for your great help. I'm still looking forward for more precise computation.

God Bless!


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## gnuuser (Jan 13, 2013)

ok since you know the region 
this link will help
http://www.saudi-arabia.climatemps.com/

conduit may need to be over sized a little bit to allow more air inside.
but this will give you some info that will help

and this chart may help also
http://assets.bluesea.com/files/resources/reference/21731.jpg


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## amantelnar (Apr 21, 2013)

*Are you Familiar with these Electrical Machines?*

In relation to my previous post, my employer sent me 5 images of these plastic machines. Are you familiar with all these machines?

http://amantechksa.wix.com/machines

Can you name each one?


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## SteveBayshore (Apr 7, 2013)

Load calculation is not correct if machine is three phase.


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## gnuuser (Jan 13, 2013)

heres an example of 3 phase load calculation

In a three phase system, power P = U * I * sqrt(3) * cos(phi), where U is the voltage (rms), I the amperage (rms), sqrt(3) is the square root of three, and cos(phi) is cosine of phi, where phi is the angle between U and I. For pure resistive load, phi is zero (i.e. cos(phi) = 1), making the calculation pretty easy in this case. For inductive or capacitive loads such as motors or flourescent lights, phi moves to different values and must be estimated, measured, or calculated. 
Find the Amperage of an 7200w inductive (i.e. motors, pumps, fluorescents etc.) load on a 415 Volt, 
3 phase 
branch circuit. 
I = 7200 / (415 x 1.732) = 10.01 Amps per phase. 


In practical we have to consider P.F (i.e cos(phi) = 0.95) for inductive loads. 


I = 7200 / (415 x 1.732*0.95) = 13.83 Amps per phase.


hope this helps


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## JohnR (Apr 12, 2010)

amantelnar said:


> In relation to my previous post, my employer sent me 5 images of these plastic machines. Are you familiar with all these machines?
> 
> http://amantechksa.wix.com/machines
> 
> Can you name each one?


Can't name EVEN one.:laughing:

Is he asking you if you are familiar with all of these machines? Most of us have wired machines that we did not know the name of. We may find out later, Some machines are one of a kind.


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