CM3201 Problem Set 3 (2015/2016)

Question 8

(i) Calculate the equivalent diameter for a rectangular slit with width W and height H where H << W

(ii) Show that the equivalent diameter for the annular ring between two concentric tubes with diameters D and d is equal to Deff = D - d. (In the derivation, assume that the tube wall thickness t << d, D and can be neglected).

Question 9

Calculate the pressure drop in a HPLC column under the following operating conditions:Flow rate 1 cm3/min; particle size 5 μm; column diameter (inner diameter, ID) 4.6 mm; column length 250 mm. Packing void fraction: 0.35. The viscosity of the solvent is 0.3 mPa·s.

Question 10

Design a heat exchanger (condenser) for a distillation column that has to condense 10,000 bbl/d (barrel per day) of gasoline (octane). The gasoline has been stripped with steam, and the amount of water in the condensate is equal to 10 vol% of the gasoline (at room temperature). The temperature of the condensate is 110 oC (adjusted by using an appropriate overpressure). Decide what medium to use as coolant, and find the relevant property data (ΔHV, ρ, etc).

Use heuristics of engineering rules of thumb for the design, e.g., for fluid velocity in a tube-and-shell condenser:

Liquid - Tube side: 3 – 7 ft/s; maximum is 13 ft/s if need to reduce fouling; Water typically 5 – 8 ft/s. Shell side: 1 – 3 ft/s.

Vapour – Vacuum: 164 – 230 ft/s; Atmospheric: 33 – 98 ft/s; High pressure:16 – 33 ft/s.

You may have to change your assumptions and iterate the design.

Question 11

Calculate the pressure drop for the vapour phase in a laboratory distillation column. In distillation, the mixture to be separated is evaporated in the reboiler; the vapour rises to the top where it is condensed. A part of the condensed vapor is returned to the column as “reflux”; the rest is taken off as distillate. Reflux is essential for separation because the countercurrent contact between the rising vapour and the down coming liquid establishes the vapour/liquid equilibrium. This has to take place multiple times along the height of the column to achieve the separation in fractional distillation. Each such equilibration stage is

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CM3201 Problem Set 3 (2015/2016)

one theoretical plate (or mass transfer stage); in tray columns, the trays will correspond to the plates, but will generally have an efficiency less than 1.

Assume that the “reflux ratio” (ratio of reflux to product flow) is 2. The amount of product is desired to be 1 ml/min. The molecular weight of the compound is 100, the liquid density 0.8 g/cm3, and the boiling point 100 oC. The column is a cylindrical tube with 2 cm inner diameter and 50 cm length, filled with 3 mm spherical beads. The beads are there to ensure good contact between the rising vapour and the down flowing liquid phase. Assume that the initial void volume is 0.4. In operation, every bead is covered by a thin liquid film of 0.1 mm thickness.

(i) what is the void fraction of the column in operation, and what is the “liquid hold-up”? Hint: calculate how many spheres are in the column, then add the layer of liquid – ignore that there are contact points between the spheres.

(ii) what is the pressure drop in this column based on this final void fraction, when the viscosity of the vapour is 1·10-5 Pa·s. Hint: you need the superficial velocity of the vapor, which is the volume flow rate divided by the empty column cross section.

Kinetic theory of gases leads to this formula for the viscosity (check that it is dimensionally correct!).

where

is the kinetic cross-section of the molecule (d is the hard sphere diameter of the molecule)

You can draw some quite amazing conclusions:

(a) the viscosity increases as T increases (common experience with liquids is that viscosity decreases on heating).

(b) The viscosity should increase with mass as m1/2, but then there is the σ in the denominator. Sigma is the collision cross section, and a bigger molecule also has a bigger cross section. If the molecule is modeled as a sphere, the mass is proportional to the volume (=R3), and the cross section is proportional to R2. We then have m1/2 in the numerator and m2/3 in the denominator, so that the viscosity is predicted to change as m(1/2-2/3) = m-1/6. That is, the viscosity of gases is almost independent of the mass, but if anything, will decrease for bigger molecules!

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CM3201 Problem Set 3 (2015/2016)

(c) Also, density and pressure do not appear in this formula, only mass and temperature. The viscosity of an ideal gas is independent of pressure! Of course, that has to break down if you go to pressures so high that the density of the gas is comparable to liquid densities – for a molecule with molecular mass of 100, that will be about 200 bar.

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