Conductors for Chiller

[mike n]

I have recently been given the task to size out a chiller that will be replacing the 30 year old existing.I am really not confident in the process.
The ratings are as follows;
450 ton 460v/3ph/60hz
JOB FLA 381
MOTOR FLA 371
LRA 2598
INRUSH 381
MIN CIRC AMPS 476
MAX FUSE 800

I will use the MCA-476a to size the conductors,I want to run parallel conductors.
350kcmil are good for 240a continuous @75 degrees?
What am I missing. Thanks for your help.

 

[barbeer]

Being a continuous load, the conductors ae to be sized 125% larger than MCA- @ 90 deg C 350's may be.(no code book right here)

Other than that I see no gross errors!:smile:

 

[jatrottpe]

Assuming the JOB FLA is for the entire chiller package (all motors, controls, heaters, etc)

JOB FLA x 125% = 476A

If the power source circuit breaker or fuse is relatively new, they should have 75 degree terminations, so you could use the 75 deg C rating in the table and use 2 sets of 250kcmil.

However if you do not know the vintage of the source circuit breaker or fuse to confirm termination rating you should be conservative and utilize the 60 deg C rating and use 2 sets of 300 kcmil.

If the JOB FLA is only for the motors and the MCA of 476A includes all heaters, motors, controls, etc on the package, than the 476A needs to be multiplied by 125% which would give you 350kcmil at 75 deg or 500kcmil at 60 deg.

 

[spsnyder]

Noce post Jatrottpe. I would only add derating for the 6 CCC if in the same conduit. 80% of 250MCM if it's a 90 degree cable is 464 amps. Would have to increase to 300 MCM or 512amps. Not concidering temperature derating or voltage drop.

Regards,

 

[buck33k]

What is JOB?

 

[jatrottpe]

My previous post assumed that there would be 3 phase, 1 gnd in each of two conduits due to the wire size and difficulty of pulling 7 wires through conduit. If all 7 were to be pulled in one conduit then the ampacity would indeed need to be derated by the 80%.

I also assumed the JOB acronym was what the particular manufacturer referred to as the entire chiller skid load (including lube oil pumps, heaters, control panel). The assumption was based on the motor FLA being 371 and JOB FLA being 381. Coupled with the fact that new chillers run about (.6 -.65) kw per ton which equates to about 324-351 running amps.

I made a lot of assumptions due to the limited (yet pertinent) information made available in the original question

 

[infinity]

You should just use the MCA to size your conductors and forget about the other information. 25% has already been added to find the MCA so adding an additional 25% for a continuous load is not required. 2 sets of 250's in separate raceways are all that you need.