**Air is introduced through a nozzle into a tank of water to form a stream of bubbles. If the bubbles are intended to have a diameter of 2 mm,
calculate how much the pressure of the air at the tip of the nozzle must exceed that of the surrounding water. Assume that the value of surface tension
between air and water as 72.7 x 10 ^{-3} N/m.**

**Data:**

Surface tension (s) = 72.7 x 10^{-3} N/m

Radius of bubble (r) = 1

**Formula:**

Dp = 2s/r

**Calculations:**

Dp = 2 x 72.7 x 10^{-3} / 1 = 145.4 N/m^{2}

That is, the pressure of the air at the tip of nozzle must exceed the pressure of surrounding water by **145.4 N/m ^{2}**

**A soap bubble 50 mm in diameter contains a pressure (in excess of atmospheric) of 2 bar. Find the surface tension in the soap film.
**

**Data:**

Radius of soap bubble (r) = 25 mm = 0.025 m

Dp = 2 Bar = 2 x 10^{5} N/m^{2}

**Formula:**

Pressure inside a soap bubble and surface tension (s) are related by,

Dp = 4s/r

**Calculations:**

s = Dpr/4 = 2 x 10^{5} x 0.025/4 = **1250 N/m**

**Water has a surface tension of 0.4 N/m. In a 3 mm diameter vertical tube if the liquid rises 6 mm above the liquid outside the tube, calculate the contact angle. **

**Data:**

Surface tension (s) = 0.4 N/m

Dia of tube (d) = 3 mm = 0.003 m

Capillary rise (h) = 6 mm = 0.006 m

**Formula:**

Capillary rise due to surface tension is given by

h = 4scos(q)/(rgd), where q is the contact angle.

**Calculations:**

cos(q) = hrgd/(4s) = 0.006 x 1000 x 9.812 x 0.003 / (4 x 0.4) = 0.11

Therfore, contact angle q = **83.7 ^{o}**

Last Modified on: 12-Sep-2014

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