0
$\begingroup$

I try not to be a complete moron but am not an engineer so please bear with me. I have this LP record ultrasonic cleaner system that uses a solution stored in a "tank" which is a simple water container and can't stand much pressure (atmospheric tank). The ultrasonic cleaner is open to the air during operation; a pump moves the solution in and out of the tank through a filter; the sequence of operation is controlled by microprocessor. The solution is distilled water/Methanol/Isopropyl alcool plus a few drops of wetting agent that is heated to 43°C for cleaning. Methanol and especially 99% Isopropyl are not cheap so I would like to have an intervention-free system to reduce evaporation of the solution while not having to open/close a lid during solution transfers and cool down after a cleaning session.

In a previous post I learned about storage tank "conservation vents" used in the industry which control vacuum or pressure build-up. I figured I could rig something similar with both pressure/vacuum proportional reliefs but agree this all seems overkill as was mentionned in replies. As was suggested and as I read more on storage tank operating principles, turns out an adequately sized vent might fit the bill. I have started reading API 2000 standard but I don't have a copy of the Crane TP-410 reference used in the industry.

From what I understand, in my situation the vent has a compressible flow and the pressure drop associated with a vent in this mode is proportional to the flow pressure, the Darcy friction factor (which takes into account the inner surfaces roughness of the vent piping), the pipe length, and inversely proportional to the piping inner diameter and the operating pressure. There is also a minor loss (K) factor added that accounts for vent entry, valves, bends etc... So a gooseneck vent with two right angle bends has a higher pressure drop than a straight pipe vent (eh... dah!)

Do I understand correctly the mecanism and the following conclusions:

  1. If the pressure drop through a vent is equal to the vapor pressure of the solution at storage temperature then there would be no evaporation at that temp? E.g. the solution reaches an equilibrium with the outside air pressure at some point.

  2. Conversely, when the solution is back into the tank after cleaning a record, there will be evaporation since the solution temperature after the process is much higher (about 37°C) and so is it's vapor pressure, until the solution cools down and the vapor pressure equalizes with the vent's pressure drop. I calculated the solution's vapor pressure (Raoult's law) at that temp to be 7.1 kPa, slightly higher than water only (6.2).

  3. During transfer operations, the pump will force more air flow through the vent, which will raise the pressure drop; at some point a dynamic equilibrium will be reached during transfer where the output of the pump will be reduced by the work needed to overcome the vent pressure drop.

  4. Many small vent holes would allow more total air flow during transfer operations while reducing evaporation since each would have a higher pressure loss than a single vent of a cross-sectional area equivalent to the sum of the small holes'. Small holes are short in length (thickness of the cap material), but minor losses play a bigger role, each hole adding it's own entry point loss.

Or I am completely lost? Any insights appreciated!

At the moment I hesitate between a gooseneck vent made from of about 30 cm of 4mm I/D PVC tubing, or 7-10 1mm diameter holes through a 3mm cap thickness. I used the "empirical" formula from this site to evaluate pressure drop: https://www.engineeringtoolbox.com/pressure-drop-compressed-air-pipes-d_852.html but I'm not sure it applies; I'm still searching for a good calculator online, suggestions are more than welcome.

Many thanks. -Joe

$\endgroup$
1
  • $\begingroup$ you could add an electric vent valve ... or a trap filled with distilled water $\endgroup$
    – jsotola
    Sep 28, 2022 at 0:20

1 Answer 1

0
$\begingroup$

Pressure drop is nonzero only in case of some flow (for really small flow rates, it is proportional to the flow rate), so the equilibrium in your case would always involve some evaporation. What you need is a relief valve with a set pressure. This causes the valve to be closed until the difference between the inside and outside fluid pressures is greater than the set pressure.

For a proper choice of relief valve, you need 2 parameters:

  • the set pressure - when should the valve open (in case of storage tanks this can be chosen based on allowable pressures that the tank can safely withstand)
  • relieving capacity (how fast can the pressure difference be brought below the set pressure - this is also important, because in some cases, pressure buildup can be fast and a valve with small capacity would fail to relief it even when open)

The most basic relief valve designs are one way, but I think you would require two of these or a "two way relief valve", maybe one of these or something similar.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.