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? 1 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 that the direction of the applied force was taken as going downwards 2 (2.07a) 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 3