I have an application where I need to seal an enclosure that is housing electronics. Given an initial temperature and humidity when sealed, is there a quick calculation to give the internal humidity of the enclosure when exposed to lower temperatures? Furthermore, how would the power output of the electronics affect the internal humidity?
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Humidity Calculations
- Thread starter Piewee
- Start date
Hi Piewee
the easist way for you to see whats going on is to get a psychrometric chart.
Enter your initial humidity and temp to get your start point, then trace along the constant absolute humidity line until you get to your lower temperature. Then read off the relative humidity.
The electronics will tend to raise the temperature in the box, which will tend to lower the relative humidity. The final temperature will be reached when the heat input from the electronics equals heat rejection rate from the enclosure. You will need to work this out.
Cheers
Steve
the easist way for you to see whats going on is to get a psychrometric chart.
Enter your initial humidity and temp to get your start point, then trace along the constant absolute humidity line until you get to your lower temperature. Then read off the relative humidity.
The electronics will tend to raise the temperature in the box, which will tend to lower the relative humidity. The final temperature will be reached when the heat input from the electronics equals heat rejection rate from the enclosure. You will need to work this out.
Cheers
Steve
Piewee,
One word of caution. "Sealed" electronic enclosures do "a bad thing" in high humidity environments where the temperature cycles up and down (think outdoors in Florida). Any small leaks will want to make the partial pressures of the gasses equalize and the effect is to precipitate liquid water inside the box. And the amount of liquid water and how short a time it takes to happen is astonishing.
One way to solve this is to fully pot the electronics in potting rubber. Alternately, let the box breathe and conformal coat the electronics.
One word of caution. "Sealed" electronic enclosures do "a bad thing" in high humidity environments where the temperature cycles up and down (think outdoors in Florida). Any small leaks will want to make the partial pressures of the gasses equalize and the effect is to precipitate liquid water inside the box. And the amount of liquid water and how short a time it takes to happen is astonishing.
One way to solve this is to fully pot the electronics in potting rubber. Alternately, let the box breathe and conformal coat the electronics.
- Thread starter
- #4
Thanks for the replies Steve & sreid.
Our enclosure will be plastic that is hot platen welded. A cable will protrude from the rear using a submersion proof connector. The operating environment is between 45 - 140 F and up to 100% humidity. We are trying to avoid potting and conformal coating.
Looking at the psychrometric chart. Assuming a temperature of 75F when sealed, the RH during sealing should be roughly 30% to avoid condensation at low temperatures, correct?
Any further insight would be appreciated.
Our enclosure will be plastic that is hot platen welded. A cable will protrude from the rear using a submersion proof connector. The operating environment is between 45 - 140 F and up to 100% humidity. We are trying to avoid potting and conformal coating.
Looking at the psychrometric chart. Assuming a temperature of 75F when sealed, the RH during sealing should be roughly 30% to avoid condensation at low temperatures, correct?
Any further insight would be appreciated.
sp humidity = 0.622 pv/(P-pv)
Lets say you start at 50 % RH and 75 F. Steam tables at 75F give a saturation pressure of 0.4298. Since we have 50 pct RH to start, use 50 pct of this pressure or 0.2149 in the above equation. Assuming P = 14.7, this gives 0.00925 lb water / lb dry air.
Now cool to 50 F. Steam tables at 50 F give a saturation pressure of 0.178. Plug this into the above equation and you get 0.00763 lb water / lb dry air.
Since this is less than the 0.00925 you started with, you are now at 100 pct RH and have condensed out (0.00925 - 0.00763 = 0.00162 lb water / lb dry air.
Lets say you start at 50 % RH and 75 F. Steam tables at 75F give a saturation pressure of 0.4298. Since we have 50 pct RH to start, use 50 pct of this pressure or 0.2149 in the above equation. Assuming P = 14.7, this gives 0.00925 lb water / lb dry air.
Now cool to 50 F. Steam tables at 50 F give a saturation pressure of 0.178. Plug this into the above equation and you get 0.00763 lb water / lb dry air.
Since this is less than the 0.00925 you started with, you are now at 100 pct RH and have condensed out (0.00925 - 0.00763 = 0.00162 lb water / lb dry air.
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