[TheForge] winter storm and equipment

[Ron Childers]

I Googled the formula but it doesn't seem to vary in direct proportion. 
I thought there was a conversion factor like converting FPS to MPH. Alas....Have to look at the tables.

-----Original Message-----
From: theforge-bounces at mailman.qth.net [mailto:theforge-bounces at mailman.qth.net] On Behalf Of Bruce .
Sent: Monday, January 06, 2014 9:44 AM
To: Blacksmithing List Sponsored by ABANA
Subject: Re: [TheForge] winter storm and equipment

Re: wind chill.

Any object warmer than ambient will be cooled by wind chill till it is no longer above ambient.  (An object cooler than ambient can likewise be
heated.)

Any object wet with water or other volatile liquid will be cooled until either the liquid is gone or the "dew point" of the liquid in the air has been reached.  (This is the same phenomenon in use in A/C systems and
refrigerators.)  In principle, there is no limit to such cooling, aside from the dew point, but in practice the volatility of the liquid decreases as the temperature of the object (or the air) decreases, so there is an effective limit, approached asymptotically.

In the case of water, the dew point is an obvious parameter, related to relative humidity.  In the case of other liquids, it is unlikely that the air will reach the dew point w/Re: these liquids, so the calculation can assume all of the liquid will evaporate.

I'm not an engineer, so I won't submit a formula.  However the temperature drop relates to the heat capacity of the object being cooled, the heat capacity of the liquid, and the mass of each.

Bruce
NJ