Biomedical Engineering Reference
In-Depth Information
drops have a slightly higher vapour pressure in comparison to a planar surface due
to their curvature. For this reason, they evaporate also in a saturated atmosphere.
This is quantified by the Kelvin Equation [41]
P
V
=
P
0
e
λ/R
,
(4)
where the vapour pressure of the liquid in the drop is
P
V
, and the parameter
λ
is a
function of the temperature and the nature of the liquid. Thus the vapour pressure
increases with decreasing drop size. As an example, a planar water surface has a
vapour pressure
P
0
=
31
.
69 mbar at NPT. If the surface is curved and the radius of
curvature is
R
=
1 µm, the vapour pressure is
P
V
=
31
.
72 mbar, and if
R
=
100 nm,
P
V
=
32
.
02 mbar. The difference between the planar and the curved surfaces is
small, but it is high enough for the drop to evaporate.
3. Deformable Surfaces
The profile of a deformable elastic surface close to the TPCL was first calculated
by Lester [42]. He shows that the Neumann's force triangle is valid to account for
the surface force distribution, and Young's equation is assumed valid only for small
deformations. Later Rusanov [43] developed expressions for the complete profile
of an elastic surface by the action of a sessile drop. He considered a stress vector
P
as a combination of the surface tension at the TPCL and the capillary pressure at
the liquid/solid interface. With this, he calculated the vector
z
(x,y)
of the vertical
displacement at all points of the surface by using the theory of elasticity [44]. In a
one-dimensional representation (see Fig. 2), Rusanov calculated the surface profile
z(r)
for three surface sections: (i) the deformation underneath the drop (
r
a
);
(ii) the deformation at the TPCL. The TPCL is assumed to be a thin line, but of finite
thickness
t(a
r
a
+
t)
; and (iii) the deformation of the surface not covered by
the drop (
r
a
+
t
):
PaE
r
a
aE
r
a
t)E
r
a
+
t
v
2
)
4
(
1
−
γ
L
sin
t
z(r)
=
+
−
(a
+
π
E
−
v)
γ
L
cos
,
+
π
(
1
−
2
v)
4
(
1
r
a,
(5a)
PrG
a
r
rG
a
r
t)E
r
a
−
v
2
)
π
4
(
1
γ
L
sin
t
z(r)
=
+
−
(a
+
E
+
t
r)
,
+
π
(
1
−
2
v)
v)
γ
L
cos
(a
+
t
−
4
t(
1
−
a
r
a
+
t,
(5b)
PrG
a
r
r
G
a
r
G
a
,
v
2
)
4
(
1
−
γ
L
sin
t
+
t
=
+
−
z(r)
π
E
r
r
a
+
t,
(5c)
Search WWH ::
Custom Search