🚀 go-pugleaf

In article <Bvpu7q.1t4@unx.sas.com>, sasrbk@banana.unx.sas.com (Bart Killam) writes:

|> [...]
|>  Young's Equation 
|>    cos(theta)= (sigma_s_a - sigma_s_w)/sigma_a_w
|> 
|> where  
|> 
|>   theta = the connecting angle
|> 
|>   sigma_s_a = interfacial energy between the solid-air
|>               (the energy required to separate a unit
|>                area of air-solid interface)
|> 
|>   sigma_s_w = interfacial energy between the solid-water
|> 
|>   sigma_a_w = interfacial energy between the air-water
|> 
|> Suppose (using J/m**2 or any convenient units)
|> 
|>   sigma_s_a = 1, sigma_s_w = 2 , sigma_a_w = 1
|> 
|> Since it take twice the amount of energy to peel
|> a unit area of solid-water apart compared to
|> solid-air, or air-solid, it would seem that the
|> system would seek to minimize its energy by
|> spreading the water over the entire surface
|> (theta-0). However, these values give theta=180,
No I don't agree with you. If sigma_s_w equal 2,
this mean that the solid-water interface is a high
energy interface (at least higher energy than the
two others) and if the system want to minimize his
energy, it want to have a solid-water interface
as short as possible. So the contact angle will
be 180 degree (0 lenght water-solid interface) 
|> where the water is only touching at a single
|> point. This seems a much higher energy state.
|> 
|> What is wrong with my reasoning. Thank you
|> for any help on this.
|> 
|> --- Bart Killam,  SAS Institute Inc. 

Hope it will help

-- 
Umberto D'Ortona (umberto@aomc.jussieu.fr)
Labo. AOMC   Universite  P.  et  M.  Curie 
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