MIT Department of Mechanical Engineering
2.25 Advanced Fluid Mechanics
Problem 2.07
This problem is from “Advanced Fluid Mechanics Problems” by A.H. Shapiro and A.A. Sonin
F
air
air
plate
plate
h
liquid
R
A drop of liquid of volume V is squeezed between two parallel smooth plates until the liquid thickness h
is very small compared with the liquid’s radial extent R. The liquid/plate/air contact angle α, and the
liquid/air surface tension is σ. Gravitational effects are negligible.
(a) Derive an expression for the downward force F required to hold the plates in position. Express F in
terms V , α, σ, and R.
(b) If α = π radians (a perfectly nonwetting situation) and T = 0.07 N/m, say (representing a clean airwater interface), what downward force is required to press a 3 mm3 drop of liquid into a thin disc or
radius R = 2 cm?
2.25 Advanced Fluid Mechanics
1
c 2010, MIT
Copyright @
Surface Tension
A.H. Shapiro and A.A. Sonin 2.07
Solution:
Given:
liquid volume, V
liquid/plate/air contact angle, α
surface tension, σ
Unknown: downward force, F
Find Po − Pi by considering equilibrium across an interface:
1
1
Po − Pi = σ
+
R2
R1
−R1
h
2
α
R2
−R1
α
In this specific problem, (Assuming that the Bo number is small and therefore we can neglect gravity effects
on the shape of the free surface)
⎫
R2 = R
⎬
1
2 cos α
(since R » h)
⇒
P
−
P
=
σ
−
+
o
i
h
R
h
−R1 =
(Bo
1)⎭
2 cos α
2σ cos α
⇒ Po − Pi = ΔP = −
h
Now apply force balance on upper plate 1:
Fy = 0
⇒
F = −FΔP
= ΔP (πR2 )
=−
2σπR2 cos α
h
where h represents the gap between the plate and the solid surface. The above equation says: the smaller
the gap, the greater the force. This effect is known as capillary stiction.
(a) Since h = V /πR2 ,
F =−
1 Notice
2σπ 2 R4 cos α
V
(2.07a)
that the direction of the applied force was taken as going downwards
2.25 Advanced Fluid Mechanics
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c 2010, MIT
Copyright @
Surface Tension
A.H. Shapiro and A.A. Sonin 2.07
F < 0: (upward) when 0 < α <
→ Liquid is wetting
π
2
F > 0: (downward) when π2 < α < π → Liquid is nonwetting
(b) Plug in given values:
α = π,
σ = T = 0.07 N/m,
F =−
V = 3 mm3 = 3 × 10−9 m3 ,
R = 2 cm = 0.02 m
][ ][
]
2(0.07)π 2 (0.02)4 cos π [
Nm−1 m4 m−3
−9
3 × 10
= 74 N
F
h
α=π
perfectly nonwetting
D
Problem Solution by Sungyon Lee, MC(Updated), Fall 2008
2.25 Advanced Fluid Mechanics
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c 2010, MIT
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2.25 Advanced Fluid Mechanics
Fall 2013
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