1 In a vertical cylindrical wetted-wall column for gas absorption, a solvent (which is a Newtonian liquid) flows down over the inside surface of the column of radius R. The flow is steady, laminar and the thickness of the liquid film is constant.
(a) Draw the flow geometry and the control volume at an appropriate location in the liquid film, and show clearly all the forces acting on it.
(b) Derive the equation for the velocity distribution across the liquid film in terms or the radial distance (r) measured from the centreline of the column, the film thickness (a), gravitational acceleration (g) and appropriate physical properties of the solvent. State clearly all the assumptions made with justifications in deriving the equation.
(c) Sketch the velocity and shear stress profiles across the liquid film.
2 (a) A liquid of viscosity of 0.03 N s m 2 is flowing under laminar conditions through a convergent, tapered tube for which the tube diameter changes linearly with length. The tube length is 0.50 m and its diameter varies from 0.01 m at the entrance to 0.005 m at the exit. Determine the pressure drop which is required to maintain the flow at a rate of 10 m3 s 1. The effect of kinetic energy due to the change in velocity in the tube may be neglected. Show the derivation of the final equation used in solving this problem.
(b) Water flows from an open cylindrical tank of 0.5 m in diameter to a vessel which is maintained at atmospheric pressure through a horizontal pipe of smooth inside surface attached to the tank wall near its bottom. The pipe is 0.025 m in diameter and 2 m long. The initial liquid level in the tank is 2 m above the pipe entrance. Estimate the time required to discharge 0.157 m3 of water.
The flow in the pipe is turbulent and a quasi steady-state head loss (hi) of 0.67 m of water in the pipe can be used in order to determine friction factor using the pipe friction factor chart (Figure Q2) provided on page 3. The head losses at the pipe entrance and exit can be neglected. The viscosity of water is 1.1 x 10 3 kg m s ‘ and density is 1000 kg m-3. Gravitational acceleration (g) = 9.81 m s-2.

Sample Solution