Question

A capillary tube is immersed vertically in a water container. Knowing that water starts to evaporate when the pressure drops below 4 kPa, determine the maximum capillary rise and tube diameter for this maximum-rise case. Take the contact angle at the inner wall of the tube to be 6° and the surface tension to be 1.00 N/m.

Answer 1

Answer:

$d=0.414\times 10^{-4}\ m$

Explanation:

Given that

P = 4 KPa

Contact angle = 6°

Surface tension = 1 N/m

Lets assume that atmospheric pressure = 100 KPa

Lets take that density of water =$1000\ kg/m^3$

So the capillarity rise h

$h=\dfrac{\Delta P}{\rho g}$

$h=\dfrac{100\times 1000-4\times 1000}{1000\times 10}$

h= 9.61 m

We know that for capillarity rise h

$h=\dfrac{2\sigma cos\theta }{r\rho g}$

$r=\dfrac{2\sigma cos\theta }{h\rho g}$

$r=\dfrac{2\times 1 cos4^{\circ} }{9.61\times 1000\times 10}$

$r=0.207\times 10^{-4}\ m$

$d=0.414\times 10^{-4}\ m$