Basics of Fluid Mechanics, 2014a

1.7. SURFACE TENSION 41
Example 1.19:
Calculate the diameter of a water droplet to attain pressure difference of 1000[N/m 2 ].
You can assume that temperature is 20 ◦ C.
Solution
The pressure inside the droplet is given by equation (1.47).
D =2R = 22σ
ΔP = 4 × 0.0728 ∼ 2.912 10 −4 [m]
1000
End Solution
Example 1.20:
Calculate the pressure difference between a droplet of water at 20 ◦ C when the droplet
has a diameter of 0.02 cm.
Solution
using equation
ΔP = 2 σ
r ∼ 2 × 0.0728 ∼ 728.0[N/m 2 ]
0.0002
End Solution
Example 1.21:
Calculate the maximum force necessary to lift a thin wire ring of 0.04[m] diameter from
a water surface at 20 ◦ C. Neglect the weight of the ring.
Solution
F = 2(2 πrσ) cos β
The actual force is unknown since the contact angle is unknown. However, the maximum
Force is obtained when β =0and thus cos β =1. Therefore,
F =4πrσ=4× π × 0.04 × 0.0728 ∼ .0366[N]
In this value the gravity is not accounted for.
End Solution
Example 1.22:
A small liquid drop is surrounded with the air and has a diameter of 0.001 [m]. The
pressure difference between the inside and outside droplet is 1[kPa]. Estimate the surface
tension?
Solution