Chapter 4, Problem 4.6P

Interpretation Introduction

Interpretation: The thickness of the layer of water on a vertical cylinder reactor is to be calculated.

Concept introduction: In one form of layer flow, when the liquid layer has a free surface and flows under the gravity of an inclined or vertical surface, the flow is in steady state and the thickness of the layer is constant.

The thickness here is given as,

  $\text{δ}\text{=}{\left(\frac{\text{3<mover accent="true"> <mo>m </mo><mo>˙</mo></mover> μ}}{{\text{ρ}}^{\text{2}}\text{gbcos}\varphi }\right)}^{\frac{\text{1}}{\text{3}}}\mathrm{.......}\text{(1)}$

In the above formulae, the notations used are,

  $\begin{array}{l}\text{δ}\text{=}\text{Thickness}\text{of}\text{the}\text{layer}\\ \text{<mover accent="true"><mo> m </mo><mo>˙</mo></mover>}\text{=}\text{Mass}\text{flow}\text{rate}\text{of}\text{fluid}\\ \text{μ}\text{=}\text{Viscosity}\text{of}\text{fluid}\\ \text{ρ}\text{=}\text{Density}\text{of}\text{fluid}\\ \text{g}\text{=}\text{Acceleration}\text{due}\text{to}\text{gravity}\text{=}\text{9}\text{.8}{\text{m/s}}^{\text{2}}\\ \text{b}\text{=}\text{width}\text{of}\text{cylindrical}\text{reactor}\text{=}\text{πD}\\ \text{D}\text{=}\text{Diameter}\text{of}\text{cylindrical}\text{reactor}\\ \varphi \text{=}\text{Angle}\text{with}\text{vertical}\text{surface}\end{array}$