Hey all, We have a Trane Climate 1

[imok2]

Hey all,
We have a Trane Climate Changer air handler,chilled water system.The plumbers ran the condensate lines off the unit,due to limited space the traps turn down 5 inches over 4 inches and back up 4 inches then slope to floor drain.The unit has three 1 1/2" nipples all are trapped then tied together to floor drain. Problem is they won't drain properly,unit holds a ton of water and overflows.Trane has a bulletin that shows how to make a proper trap.The unit in question runs at 10 inches of static.So using their formula the trap should go down 18 inches over 4 inches and back up 5.5 inches.The unit is in the penthouse of a hospital so it is possible to drill through the floor to make the trap bigger, but not the easiest thing to do in a hospital.The boss asked why can't we just make the trap wider instead of deeper?Trane says they don't know why it won't work, but they say it can't be done that way.Does anybody have a thought or opinion?

 

[cjw81]

Chasbean1- Of course I take it the right way, that's why I like this site so much- I am in the operations end of engineering ans also happen to work currently in a hospital. I am a particular fan of your posts and have followed many threads in the site. Many systems currently in use are poorly functional due do poor engineering, installation, poor or misunderstood design intent, or poor operation. I am at a point in my career where I need to seek sound engineering solutions to everyday problems. My best tools are observation and knowledge. I apply what I learn here to practical solutions. You guys are teriffic in taking the time out of your busy day to enlighten the rest of us. Many thanks.

PS. How about explaining your formula, you'll help me and save me a lot of time. Thanks

 

[ChasBean1]

It's just pressure at a submersion point, h units below surface.

P = rho * g * h

P is the pressure
rho is the density (62.4 lbm/ft3 for water)
g is gravity (32.2 ft/s2)

If something is 1 foot below water level, the pressure at that point is:

rho g h
P = (62.4 lbm/ft3)(32.2 ft/s2)(1 ft)(lbf s2/32.2 ft lbm)(ft2/144 in2)

So pressure is 0.43 psig for something submersed 1 foot deep. The last two terms are the pain-in-the-but gc (english units gravitational constant - notice that the 32.2's cancel but the units become corrected) and conversion from lbf/ft2 to lbf/in2, or psi.

Going back to inches water:

(0.43 psi)(29.92 in Hg/14.696 in Hg)(13.6 in w.c./1 in Hg)
= 12 inches (hey, back to 1 foot!)

Anything 12 inches below water level will experience a pressure of 12 in. w.c. regardless if it's 12 inches deep in a lake or 12 inches down a straw (width & trap horizontals don't matter).

This sounds kind of basic (and it is) but it helps with going from inches to psi, etc. Knowing unit coversions goes a long way.

Thanks for the compliment & best regards, -CB