Batch mass transfer and mass conservation

Road to a Chemical Engineer: #3

Recap

　In the last article, we have learned about the steady state in a flow system and the state of equilibrium in a batch system. Moreover, we have seen how to interpret the value during the non-steady state and the fact that any constant continuous flow system will eventually reach a steady state.

　We have been using the water tank system as an example of mass conservation, specifically the volumetric conservation under same density. Today, we will be starting a new example for discussing mass conservation.

Conservation law for mass transfer

Idea and model of mass transfer

　You might be wondering what is mass transfer. It is quite simple. It is a system where small molecules move. Let's imagine a following situation.

　Two containers are sitting next to each other. One contains saturated salt solution, and the another contains pure water. Now, the wall between the containers are given to be hollowed wall such that it only allows the salt molecule to be passed through. What do you expect to happen??

　If this system is difficult to imagine, let's say the salt solution is now a tea bag. It contains tea molecules and the bag has a small holes that allows the tea molecules to transfer. What do you see??

　You will probably see the pure water gets stained by the tea color, slowly over time. That is what you will see for the above diagram as well. Which means that the salt molecules transfer from salt solution to the pure water. That is the mass transfer, namely it is 'diffusion'. The above hollowed wall has a name also, that is 'semipermeable membrane'.

　You may heard of diffusion and semi-permeable membrane before. For me, I heard it in Grade 9 Human Biology. I learned that diffusion of oxygen and carbon dioxide occurs between lung and blood vessel, where the blood vessel is a semipermeable membrane. Woah, I do remember things from Grade 9!!

　Anyway, that is just a slice of real use of mass transfer. It is used and exists in many places of our life, and reading this article will lead you to understand more concepts. So, let's get started!!

Constructing mass conservation equation for mass transfer

　Now, I suppose you have understood what kind of system mass transfer can be. However, you might not quite understand the connection to mass conservation. What mass is conserved here?

　If you recall the mass conservation for the water tank system, we were looking at the volume being conserved. Here, for mass transfer, what we are focusing is the number of salt molecules being conserved. Think about it, the mass of salt molecules within the system is conserved, and also salt molecules have constant molar mass. So, it is reasonable to say that number of salt molecules are conserved.

$number of salt molecules, n = mass/Molar mass$

$rate\left(n_{in}\right)-rate\left(n_{out}\right)+rate\left(n_{generated}\right)-rate\left(n_{consemed}\right)=rate\left(n_{accumulated}\right)$

　The unit of the equation is mol per second. Here, for our model, we are only concerned with the in and out of the salt molecules, so we can equate the rate of generation and consumption with 0. Before we do anything, let's reconstruct the model with simplified diagram and put named variables and constants.

Our new model for mass transfer

　Here, you see two compartments next to each other. One contains V₀ m³ of salt solution at concentration C₀, another contains V₀ m³ of salt solution at concentration C₀. Both are assumed to be well stirred and has uniform concentration. In addition, for right compartment, C₀ is initially 0 (at t = 0 s).

　Left compartment:    $-rate\left(n_{out}\right)=rate\left(n_{accumulated}\right)$

　Right compartment:   $rate\left(n_{in}\right)=rate\left(n_{accumulated}\right)$

Exploring rate of accumulation

　Now, we need to determine the rate at which the number of salt molecules change for above equation. However, we have a huge problem. How do we determine the rate of number of salt molecules. Can we see them? Can we count them? Then, what do we do?

　If you recall the water tank system, we did not actually use the volume of water. What we did was to use change in height, as volume = height * area. Which means, here, if we can equate the number of salt molecules to some measurable variable, that solves the problem.

　The answer was actually implied in the diagram above. Do you remember that I mentioned about the salt concentration in each compartment? Look at the equation below (you probably know / learned this equation before).

$Concentration, C = n / Volume$

　This equation indicates that instead of measuring unobservable quantity (number of salt molecules), we can measure the concentration of the solution over a known volume. Then, rate of number of salt molecules accumulated can be rewritten as follows:

$rate\left(n_{accumulated}\right)=\frac{dn}{dt}=V\frac{dC}{dt}$

　Let's recall the equation for rate accumulated for water tank system. There we had rate of volume of water accumulated, and dV/dt was rewritten as A*dh/dt. Aren't these so similar??

$rate\left(V_{accumulated}\right)=\frac{dV}{dt}=A\frac{dh}{dt}$

　Anyway, now we know how to write the rate of accumulation in terms of concentration, our observing variable. So, now we can rewrite the conservation equation again for each compartment.

　Left compartment:    $-rate\left(n_{out}\right)=V\frac{d{C}_0}{dt}$

　Right compartment:   $rate\left(n_{in}\right)=V\frac{d{C}_1}{dt}$

Exploring rate of diffusion

　In above conservation equation, rate of salt molecules in and out of the compartments are equal to the rate of diffusion as follows:

$rate of diffusion=rate\left(n_{in}\right)=rate\left(n_{out}\right)$

　Here, you might wonder that what is the rate of diffusion. To understand this, you have to first know what causes diffusion. Rate of diffusion is usually written in terms of flux which is a flow per unit area. The equation is written as following:

$Flux, J=\frac{Driving Force, DF}{Resis\tan ce, R}$

　Diffusion is caused by some driving force, and that is concentration gradient. The reason is because salt molecules are constantly moving and bouncing off each other. Sometimes, those molecules move from left to right compartment since semipermeable membrane allows them to.

　Of course, once molecules entered left compartment, some of those may go back. However, the rate of moving to the right is so great such that you observe the salt concentration decreases in left compartment and increases in right compartment.

　Now, we can rewrite the rate of diffusion as follows:

$rate of diffusion=JA=\frac{C_0-C_1}{R_m}A$

　Let's combine this to the conservation equation:

　Left compartment:    $-\frac{C_0-C_1}{R_m}A=V\frac{d{C}_0}{dt}$

　Right compartment:   $\frac{C_0-C_1}{R_m}A=V\frac{d{C}_1}{dt}$

Determine the change in concentration over time

　To solve for C₀ and C₁ for the above equation, as we discussed in previous article, it is above the mathematical level for now. So, let's simplify the model a bit.

　For the same diagram, now we say that C₀ is kept constant. You can put over saturated solution in left compartment to make sure that is true. Then, let's see what will happen to the conservation equation we derived above:

　Left compartment:    $\frac{d{C}_0}{dt}=0$

　Right compartment:   $\frac{C_0-C_1}{R_m}A=V\frac{d{C}_1}{dt}$

　It looks like nothing much has changed, but you have to realize that now we can integrate the equation for right compartment to find C₁(t). Just recall that C₁(0) = 0 for eliminating the constant. (calculation is omitted)

$C_1\left(t\right)=C_0(1-e^{-\frac{A}{V{R}_m}t})$

　If you plot this graph, you will see something similar to the following graph:

　From the graph, you can see that when we assume the C₀ to be kept constant, pure water will reach C₀ eventually, i.e. equilibrium

Determining the membrane resistance, R_m

　Lastly, the membrane resistance has not covered yet. Think about what could bring a resistance to diffusion. One obvious factor is the membrane width (∆x_w) because thicker the membrane is, harder for the salt molecules to transfer. So, membrane width and membrane resistance are proportional. Are there anything else??

　The another factor is diffusibility (D). This is unique to the membrane and the type of salt, and it gives how easy the salt molecules can transfer across the membrane.. Greater the diffusibility is, more easier the diffusion occurs. Thus, the diffusibility is inversely proportional to the membrane resistance. If we combine these two factors, we get membrane resistance as following:

$membrane resis\tan ce, R_m=\frac{membrane width, ∆x_w}{Diffusibility, D}$

　Using the above equation we derived, this membrane resistance can be calculated. How ever will the membrane resistance equals to what we calculate with the equation for membrane resistance?? This will be discussed in next article, so you can start think and get a head around the resistance that is other than membrane resistance.

Endnote

　Today, we looked into the mass transfer and mass conservation. We saw many similarities with water tank system and applying the previous knowledge to explore a different type of model / system. It was somewhat manageable, right?

　This article is named as 'batch mass transfer' for a reason. So, the next article will be in flow system where given concentration of solution flows in and out within the subsystems. They will be more complex but more fun at the same time. Please wait for the upcoming article!! I hope you enjoyed reading, see you next time!!


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