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We currently add hot water to a kettle at ~170 degrees F. The vapors from this "hot" water rise up and collect onto an overhead auger where dry material drops into the kettle, causing a "build-up" due to caking and the presence of water at that junction. The question is: If we add cooler water, 170, 160, 140, will the amount of vapor rising be significantly less? Logic would say yes but if it is not boiling we question how much less vapor we would actually get. We cant go below 140. I guess Im looking for a graph or equation?
- Rich (age 31)
Manufacturing Plant, MN
Anyway, although there are sevral complications in getting the rate of
water evaporation, probably the most imporatnt factor here is the
equilibrium vapor pressure of the water. You can look that up in
tables. Here's some numbers for how that vapor pressure depends on
100 F: 51 (mm Hg units)
140 F: 150
150 F: 191
160 F: 244
170 F: 297.
There's an equation, called Clausius-Clapeyron, which captures this
behavior well, but for your purposes these numbers should suffice. You
can do better by about a factor of two by dropping to 140 F.
(republished on 07/24/06)
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