Chilling Pressure Drop

Please register or login

Welcome to ScubaBoard, the world's largest scuba diving community. Registration is not required to read the forums, but we encourage you to join. Joining has its benefits and enables you to participate in the discussions.

Benefits of registering include

  • Ability to post and comment on topics and discussions.
  • A Free photo gallery to share your dive photos with the world.
  • You can make this box go away

Joining is quick and easy. Log in or Register now!

spectrum

Dive Bum Wannabe
ScubaBoard Supporter
Messages
11,395
Reaction score
827
Location
The Atlantic Northeast (Maine)
# of dives
500 - 999
How long would folks estimate it takes for the pressure drop from immersion cooling of a cylinder to occur? It's probably a range since air and water tems can vary widely with the seasons making the delta T anything from zero 40+degrees F.

I ask since I have read that "wet filling" is an ineffective tank cooling. I recognize that during a dive the tank is fully immersed and the ocean (take your pick) is one big big heat sink.Yet a recent post mentioned watching for this phenomenon as soon as 30 seconds into the dive.

I'm sure someone has done tests and it probably varies by tank. Assume the tank was fully cooled to ambient, not hot off the whip.

Pete
 
the drop stabilizes inside of a minute in 70 F water

i would imagine it will take longer for the full effect in colder water?
 
I notice the effect immediately upon entering the water. I don't know how long it takes to stabilize as I'm usually under water by that point anyway.

Wet filling and refrigerating the air in the buffers does seem to help.

R..
 
P1/T1=P2/T2
So the change in pressure is proportional to the change in temperature.
TYPICAL:
-------------------------------------
3000psi = 204.1atm
Tank temp on boat 80F (300 degrees kelvin)
Water temp 70F(294 deg K)
p1/p2=t1/t2

p2=p1(t2/t1)

p2=204.1*(294/300)=200Atm

p2=200*14.7=2940psi

So if you drop about 10 degrees F it will drop the tank about 60psi.
------------------------------------------------

Lastly do a 1 degree analysis.

p2=p1(t2/t1)
P2=204(299/300)
p2=2988psi.

so 1 degree F change is about 12psi.


Edit: Just saw you wanted "time" not pressure.
 

Back
Top Bottom