# 4.3: Species Velocity

In our representation of the axioms for the mass of multicomponent systems, we have introduced the concept of a species velocity indicating that individual molecular species move at their own velocities designated by $$\mathbf{v}_{A}$$ where $$A=1, 2, .. N$$. In order to begin thinking about the species velocity, we consider a lump of sugar (species $$A$$) placed in the bottom of a tea cup which is very carefully filled with water (species $$B$$). If we wait long enough, the solid sugar illustrated in Figure $$\PageIndex{1}$$ will dissolve and become uniformly distributed throughout the cup. This is a clear indication that the velocity of the sugar molecules is different from the velocity of the water molecules, i.e.,

$\mathbf{v}_{Sugar} \neq \mathbf{v}_{Water} \label{40}$

If the solution in the cup is not stirred, the velocity of the sugar molecules will be very small and the time required for the sugar to become uniformly distributed throughout the cup will be very long. We generally refer to this process as diffusion and diffusion velocities are generally very small. If we stir the liquid in the teacup, the sugar molecules will be transported away from the sugar cube by convection as we have illustrated in Figure $$\PageIndex{2}$$. In this case, the sugar will become uniformly distributed throughout the cup in a relatively short time and we generally refer to this process as mixing. Mechanical mixing can accelerate the process by which the sugar becomes uniformly distributed throughout the teacup; however, a true mixture of sugar and water could never be achieved unless the velocities of the two species were different. The difference between species velocities is crucial. It is responsible for mixing, for separation and purification, and it is necessary for chemical reactions to occur. If all species velocities were equal, life on earth would cease immediately.

In addition to mixing the sugar and water as indicated in Figures $$\PageIndex{1}$$ and $$\PageIndex{2}$$, we can also separate the sugar and water by allowing the water to evaporate. In that case all the water in the tea cup would appear in the surrounding air and the sugar would remain in the bottom of the cup. This separation would not be possible unless the velocity of the water were different than the velocity of the sugar. While the difference between species velocities is of crucial interest to chemical engineers, there is a class of problems for which we can ignore this difference and still obtain useful results. In the following paragraphs we want to identify this class of problems.

To provide another example of the difference between species velocities and the effect of diffusion, we consider the process of absorption of $$\ce{SO2}$$ in a falling film of water as illustrated in Figure $$\PageIndex{3}$$. The gas mixture entering the column consists of air, which is essentially insoluble in water, and $$\ce{SO2}$$, which is soluble in water.

Because of the absorption of $$\ce{SO2}$$ in the water, the exit gas is less detrimental to the local environment. It should be intuitively appealing that the species velocities in the axial direction are constrained by

$\mathbf{v}_{\ce{SO2} } \cdot \mathbf{k}\approx \mathbf{v}_{air} \cdot \mathbf{k} \label{41}$

in which $$\mathbf{k}$$ is the unit vector pointing in the $$z$$-direction. The situation for the radial components of $$\mathbf{v}_{\ce{SO2} }$$ and $$\mathbf{v}_{air}$$ is quite different because the radial components are normal to the gas-liquid interface. Since the air is insoluble in water, the component of $$\mathbf{v}_{air}$$ in the radial direction must be zero at the gas-liquid interface, i.e.,

$\mathbf{v}_{air} \cdot \mathbf{n}=0 , \quad \text{ at the gas-liquid interface} \label{42}$

On the other hand, the sulfur dioxide is crossing the interface as it leaves the gas stream and enters the liquid stream. The radial component of $$\mathbf{v}_{\ce{SO2}}$$ must therefore be positive and we express this idea as

$\mathbf{v}_{\ce{SO2}} \cdot \mathbf{n} >0 , \quad \text{ at the gas-liquid interface} \label{43}$

It should be clear that $$\mathbf{v}_{\ce{SO2}} \cdot \mathbf{n}$$ is a velocity associated with a diffusion process while $$\mathbf{v}_{\ce{SO2}} \cdot \mathbf{k}$$ is a velocity associated with a convection process and that the latter is generally much, much larger than the former, i.e.,

$\underbrace{\mathbf{v}_{\ce{SO2}} \cdot \mathbf{k}}_{convection}>>\underbrace{\mathbf{v}_{\ce{SO2}} \cdot \mathbf{n}}_{diffusion} \label{44}$

The motion of a chemical species can result from a force applied to the fluid, i.e., a fan might be used to move the gas mixture through the tube illustrated in Figure $$\PageIndex{3}$$. The motion of a chemical species can also result from a concentration gradient such as the gradient that causes the sugar to diffuse throughout the teacup illustrated in Figure $$\PageIndex{1}$$. Because the motion of chemical species can be caused by both applied forces and concentration gradients, it is reasonable to decompose the species velocity into two parts: the mass average velocity and the mass diffusion velocity. We represent this decomposition as

$\mathbf{v}_{\ce{SO2}} =\underbrace{ \mathbf{v} }_{\text{mass average velocity}} +\underbrace{ \mathbf{u}_{\ce{SO2}} }_{\text{mass diffusion velocity}} \label{45}$

At entrances and exits, such as those illustrated in Figure $$\PageIndex{3}$$, the diffusion velocity in the $$z$$-direction is usually small compared to the mass average velocity in the $$z$$-direction and Equation can be approximated by

$\mathbf{v}_{\ce{SO2}} \cdot \mathbf{k}=\mathbf{v} \cdot \mathbf{k} , \quad \text{negligible diffusion velocity} \label{46}$

In this text we will repeatedly make use of this simplification in order to solve a variety of problems without the need to predict the diffusion velocity. However, in subsequent courses diffusion at fluid-fluid interfaces will be studied in detail, and those studies will require a complete understanding of the mass diffusion velocity.