Question

Ql(a) A triangular tubeſ with the dimension as in Figure Q1 is immersed into an unknown fluid. If the surface tension, contact angle and density of the fluid is given as os, and p respectively, form an equation that expresses the capillary rise h as a function of a, b, Os, and p. (4 marks) (b) Given that the surface tension, contact angle and density of the fluid is 0.073 Nm-?, 60° and 950 kg/m² respectively. The value of a is given as 0.5 cm. If the intended capillary rise is 20 cm, determine the value of b required. (6 marks) ! a a Fig. 21

please don't skip any step thanks.

Answer 1

a) Weight of liquid column = Vρg = (1/2)*a*b*2a*g*ρ

= ρgba²

Vertical component of surface tension = σ_s *2*(a²+b²)^1/2*cosθ

ρgba² =  σ_s *2*(a²+b²)^1/2*cosθ

b = 2σ_s (a²+b²)^1/2*cosθ/ρga²

b² = (2 σ_s cosθ/ρga²)²*(a²+b²)

opening the brackets and rearranging, the equation becomes

b² = (2 σ_s cosθ/ρg)²/(1-(2 σ_s cosθ/ρga²)² )

= 1/ (1/(2 σ_s cosθ/ρg)² - (1/a²)²)

b = [1/ (1/(2 σ_s cosθ/ρg)² - (1/a²)²)]² ---------------------------------------- (1)

b) σ_s = 0.073Nm^-1

θ = 60^o

ρ = 950 kg/m³

a = 0.5 cm = 0.005 m

substituting the following in the above equation (1)

b = 1/((1.6*10¹⁰)-(16*10⁸))^1/2

= 8.33*10^-6m = 8.33*10^-4cm