Problems mixed mass-volume

In problems involving a gas, it is easier to measure the volume of the gas than the mass. Mixed mass-volume problems include both mass-volume problems, where you are given a mass and asked for a volume, and volume-mass problems, where you are given a volume and asked for a mass. You will need to do the molar conversions involving the molar volume of a gas, so review Lesson 7-2 if necessary. The basic steps for solving this type of problem remain the same ... 

This type of calculation does not have to be carried out for a plate column because the two phases are well mixed on each plate. This means that on each individual plate a state of equilibrium can be presumed. Therefore a volume element is identical to an equilibrium stage, and the height of the column can be obtained from the number of equilibrium stages required for a particular separation. This is a thermodynamic rather then mass transfer problem. This explains why a mass transfer device, such as a distillation column can be sized without any knowledge of the laws of mass transfer. 

Eq. (9.32) predicts that the the mass-transfer coefficient for the oxygen dissolution in water 25°C in a mixing vessel is 4.58 x lO m/s, regardless of the power consumption and gas-flow rate as illustrated in the previous example problem. Lopes De Figueiredo and Calderbank (1978) reported later that the value of kL varies from 7.3 x 10"4 to 3.4 x 1CT3 m/s, depending on the power dissipation by impeller per unit volume (Pm/v) as... 

Most biopolymers are produced as extracellular metabolites by fermentation in bioreactors leading to special technical problems caused by the very viscous solutions that make mass transfer and mixing in the fermentation fluids difficult. Large volumes of water and solvents are needed for dilution and extraction, respectively. 

From the viewpoint of both a mass balance or a mole balance for elements themselves, such as C, H, or O, the generation and consumption terms are not involved in a material balance. Finally, Eq. (2.1) should not be applied to a balance on a volume of material unless ideal mixing occurs (see Sec. 3.1) and the densities of the streams are the same. In this chapter, information about the generation and consumption terms for a chemical compound will be given a priori or can be inferred from the stoichiometric equations involved in the problem. Texts treating chemical reaction engineering describe how to calculate from basic principles gains and losses of chemical compounds. 

Apparently the choice of either method should not affect the course of kinetic curves, but each one has potential pitfalls that are probably responsible for the discussed above discrepancies. In the first method a series of dispersions with the same solid to liquid ratio can be easily prepared by mixing certain mass of the adsorbent with known mass or volume of the solution, but these dispersions can differ in particle size distribution, thus, the adsorption kinetics in particular dispersions can be different. Use of one dispersion eliminates this problem, but, e.g. stopping the shaker to collect one sample affects the further kinetics of adsorption. Moreover, unless the collected samples are very small, the solid to liquid ratio and the amount of adsorbate are not constant in course of the kinetic experiment, and this is in conflict with assumptions made in derivation of the... 

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