| OCR Text |
Show 2. The Boundary Case Between a Melting and a Non- melting Surface Further discussion requires discrimination between a melting and a non-melting snow surface. In the case of a non- melting surface, ablation is possible only through evaporation from ice ( S= 0). V must be derived using the heat of vaporization of ice, rE . The surface temperature can have values r$- = 0° C. A melting surface on the other hand has <$•= 0° C, while S^ O and V must be derived from rw , the heat of vaporization of water. The next melting- non- melting boundary case to be considered here is the one for ice at T> = 0° C at the surface and S= 0. V must be derived with r£ . Substituting L according to Equation ( 2) and V according to Equation ( 3), with r- for r, the heat balance equation then reads Q + B - XL ( fro - » L) - °' 623 TE * L ( EQ - eL) = 0, ( 6) pcp where E0 = 4.58 torr, the vapor pressure at the ice surface for ^ = 0° C, In the special case where Q+ B = 0, 0,623 rE n 0,623 rE „ & L + - - « L = $ o - r - - E0, ( 7) pep pep that is, the equivalent temperature of the air derived with r^ is 10.6° C when p = 760 torr, valid for the given temperature ^ and vapor pressure EQ of the ice surface. From this value the following pairs of air humidity fL and air temperature i> r are found fL 100 80 60 40 20 0 & L 0,0 1,3 2,8 4,7 7,1 10.6 ° C, 0/ 0 |
