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!!


Previous Article: Water tanks "in series" and steady state
Next Article: Coming Soon

Water tanks "in series" and steady state

Road to a Chemical Engineer: #2

Recap

　In the last article, we have learned a different way of looking at mass conservation, using water tank system as an example. We thought of an equation of water outflowing tank as:

$Q_{Accumulated}=-Q_{Out}$

　Then, we managed to figure out the same equation can be written in terms of change in height over measure time period. The equation that we derived is:

$A\frac{dh}{dt}=-\frac{\rho gh}{R}$

　Today, this article will consider about the tanks that are in align. In the previous article, we only focused on the tank that water outflows. However, how about the mass conservation is the tank that receives the water, or what if there is more tanks connected and so on. 

　This will be the last section for using the water tank example. So, let's cover this mass conservation up with this article!!

Water tanks in series

　Let's imagine a situation like above. Here, you can formulate multiple equations for the tanks' mass conservation.

For Tank A:

$A_1\frac{d{h}_1}{dt}=Q_0-\frac{\rho g{h}_1}{R_1}$

For Tank B:

$A_2\frac{d{h}_2}{dt}=\frac{\rho g{h}_1}{R_1}-\frac{\rho g(h_2-h_3)}{R_2}$

For Tank C:

$A_3\frac{d{h}_3}{dt}=\frac{\rho g(h_2-h_3)}{R_2}-Q_3$

　If you do not understand how these equations are brought up, please look back to the previous articles. This is just an application of ideas expanded to multiple water tanks. It is not easy at first but will get there eventually.

　So, if you imagine such situation, what can you say about this system? Can anything be drawn from this equation??

　One thing that can be thought is the situation where this system reaches steady state. Before we dig into the steady state and their equations, let's see what is steady state.

Steady state and Equilibrium

　People usually have some misunderstanding between equilibrium and steady state. This is because both indicate the state at which the system has no change in variable, height. However, their difference is quite simple. It is whether there is a flow in the system or not. In other words, whether Driving Force (DF) = 0 or not.

　Let's imagine a water tank where the water level is just at the outlet of hole. Will there be any water flow?? No, right. So, DF = 0, and we call such situation, 'equilibrium'. And, this occurs ing batch process, i.e. do not have continuous flow.

　This means 'steady state' is when there is no change in height with DF ≠ 0. Is such situation possible?? And, it is. Let's say there is water flowing into a tank, and water is also flowing out of the tank. What happens when the flow rates of inflows and outflows are the same. 

$A_1\frac{dh}{dt}=Q_0-Q_1 \{where Q_0=Q_1\}=0$

　You see that the rate of change in height is 0, right? Yet, there is a flow ongoing. And, that is the situation we call, 'steady state'.

Steady state and water tanks in series

　Now, we are back to the idea of water tanks in series. So, as we discussed now, since there is a continuous flow in this model, only steady state can be achieved. That is when all the rate of change of height in all tanks go to 0. Thus, for each tank, following should be achieved.

For Tank A:

$$Q0=ρgh1R1

For Tank B:

$\frac{}{}$ρgh1R1=ρg(h2-h3)R2

For Tank C:

$\frac{{\displaystyle \rho g(h_2-h_3)}}{{\displaystyle R_2}}=Q_3$

　Do you see anything surprising? Or, is it too obvious??

At steady state:

$Q_0=\frac{\rho g{h}_1}{R_1}=\frac{{\displaystyle \rho g(h_2-h_3)}}{{\displaystyle R_2}}=Q_3$

　This is what you can draw from this water tanks in series, at steady state. So, at steady state, it might be obvious but all the flow rates must be constant, and that can be formulated using equations!!


　Moreover, If you want to know the height at which tank A achieves steady state, you can simply, reforming and using the equation above, such that:

$h_1=\frac{Q_0{R}_1}{\rho g}$

　However, for other tanks, it is possible to find the difference between (h₂ - h₃) but not individually with my current knowledge. If you know how please share it with me, or otherwise, let's come back sometimes in the future with more knowledge on chemical engineering and mathematics.

Tank A at non-steady state

　So, we just saw what would happen at the steady state. What about at non-steady state? Is is possible to formulate something about it??

　It is actually possible with integration of equations that we formulated above for each tank. However, here, we will only focus on Tank A as it only consists h₁ as a variable. On the other hands, here again, for Tank B and Tank C, both consists h₂ and h₃, so it will require higher level of integration skills, which I will not cover for this article.

　Let's see what happens if we integration Tank A equation:

$A_1\frac{d{h}_1}{dt}=Q_0-\frac{\rho g{h}_1}{R_1}$

　In the integration of this equation, we assume that, at t = 0, h₁ = h₀, and at t = t, h₁ = h. However, I will not cover the method of integration here, so please try it out by yourself.

　It will be similar to the integration in last article but this time, we have Q₀ to consider. You will eventually get equation similar to the below:

$h=\frac{Q_0{R}_1}{\rho g}(1-e^{-\frac{\rho g}{R_1{A}_1}t})+h_0 e^{-\frac{{\displaystyle \rho g}}{{\displaystyle R_1{A}_1}}t}$

　"But... what is so special about this equation... this is just even more complex..."

　Don't worry.This is much more clear and meaningful when we plot on the graph. Let's see this.




　What do you see?? One thing that is clear is that, height of Tank A will eventually reach to the height for steady state (which we determined in above). In addition, from the values of Q₀ , R₁ , and h₀, it is also possible to know whether the height will start to decrease or increase, based on the comparison.

Endnote

　This article has covered what we call water tanks in series, and we determined that steady state is achieved when all the rates of flow are equal. Furthermore, we graphed what will the graph of Tank A be over time and checked that, it will eventually reach the steady state.

　However, at this moment, there is a lot that is yet to be detailed such as the similar graph for Tank B and Tank C. With those knowledge, it is possible to model a better model and have a better understanding of the mass conservation.

　Are you surprised how much we can dig into with just mass conservation?? However, this is just a scoop of what mass conservation is, and not even a lip of chemical engineering. Are you know more excited about chemical engineering??

　Please give me some comments and thank you for reading. Next article will be on still on mass conservation, but not water (Volumetric) anymore!! Molecular mass conservation for the next, so wait for more fun and more chemical engineering.


Previous Article: Water tank system and mass conservation
Next Article: Batch mass transfer and mass conservation

Water tank system and mass conservation

Road to a Chemical Engineer: #1

Recap

　In last article, we have covered two things. First, what is Chemical engineering about, and second, the conservation law. If you remember the conservation law for instantaneous time, correctly, it is the followings:

$Rat{e}_{In}-Rat{e}_{Out}+Rat{e}_{Generated}-Rat{e}_{Consumed}=Rat{e}_{Accumulated}$

　However, this article is not just about any conservation but "mass conservation". It is, literally, the conservation law applied upon mass. So, in mass conservation, the conservation law can be simply rewritten as follows:

$$\begin{gathered} Rat{e}_{Mass In}-Rat{e}_{Mass Out}+Rat{e}_{Mass Generated}-Rat{e}_{Mass Consumed}\text{ \\ =RateMass Accumulated} \end{gathered}$$

　Now, we have the equation for mass conservation. Let's check this out with a realistic experiment.

Water tank system and mass conservation

　Imagine two water tanks. From one of the tank, water flows into another tank. Here is a mass conservation going on, right??

　Before we proceeding any further, we need to note that since only water is being used, mass conservation is equivalent to volume being conserved in this system. Let's say we use a common symbol of Q as a rate of volume flow.

　Without any calculation, you probably intuitively know that the volume of water went out from Tank A equals to the volume of water decreased from Tank A. However, using equation we derived above, makes this idea more clear. 

　So, in such situation for Tank A, there is no volume coming into the Tank A, so Q _In = 0. That goes same for Q _Generated = 0 and Q _Consumed = 0 as there are nothing neither generated nor consumed here. Then, we are left out as following for Tank A:

$-Q_{Out} = Q_{Accumulated}$

　Isn't this something that we expected to happen??

　If you are to do this experiment to check the mass conservation for the water tank system, what would you do??

　One of the straightforward method is to measure the volume of water within the tank and measure the volume of water going out / into the tank. It is clear but it only tells you what it is. In other words, this does not expand into any other ideas or any other factors that may affect the result.

　So that, what if the initial water level in Tank A changes? What if now we use different pipes to allow the water flow?

　To answer these questions, the water tank system is now needed to be modeled with named variables for the conservation.

Modeling the water tank system

1. How does the rate of water flow determined??

• What causes water to flow??

　Well, before anything, let's me ask you a question. What causes the water flow??

　You can actually answer this differently but the core concept is due to the pressure difference. Particularly, the pressure difference at the outlet in and out side.

　Why is this?? Just to note that, from here, it is something that engineers do not necessary have to know but since I researched from curiosity, I will share it here.

　Do you know the fluid flows from high to low pressure? This is actually something to do with what I learned in my secondary school (IB diploma Physics HL Option B). So, here, Bernoulli equation comes in. The Bernoulli equation states that:

$\frac{1}{2}\rho {v_x}^2+\rho g{z}_x+p_x= \frac{{\displaystyle 1}}{{\displaystyle 2}}\rho {v_y}^2+\rho g{z}_y+p_y$

　We are not qualitatively analyze this equation. Yet, this still explains the water flow. Here, (ρgz + p) is basically the pressure, which means that having the pressure difference means this value differ at x and y. Then, what should be different between x and y to equate as in Bernoulli equation?

　The only variable there is the water speed, and if you think about the outlet, speed at the inside is 0, so there should be some speed of water at y. Thus, water flows out!!! I hope this make sense to you.

• How to quantify the rate of water flow?

　To find the pressure difference, we need to find the pressure at each point of outlet. First, at the surface of the water of pipe is really straightforward, and that is just atmospheric pressure (p_atm). 

　How about at just inside of the outlet. To consider this, just to let you know that at the same level, the pressure is same. And, that is basically because there is same mass of water per area is upon such level. The pressure, here, can be given to be hydrostatic pressure (p_hydrostatic) + p_atm.

　p_hydrostatic can be given thought out simply using the high school physics equations. Mass at gravitational acceleration is falling upon a level per area, i.e. p_{hydrostatic =} ρgh.

　Then, the pressure difference at the outlet can be written as:

$Pressure difference (∆P) = (p_{atm}+p_{hydrostatic})-\left(p_{atm}\right)=\rho gh$

　Since this pressure difference is driving the water flow, we call this "Driving Force (DF)". Now how do we get the flow rate of water from this value??

　To get the flow rate, we need an another equation: 

$Flow rate \left(Q\right) =\frac{Driving Force \left(DF\right)}{Resis\tan ce \left(R\right)}$

　This equation may not be familiar but I am sure that you know it from your Physics class. Do you remember I = V / R, where I is the current, or in other words, flow rate of charges, and V for potential difference and R for resistance. The same concepts lies here.

　So, flow rate of water can be given as follows:

$Q=\frac{∆P}{R}=\frac{\rho gh}{R}$

• What is the resistance here??

　Now, we have managed to integrate the rate of water flow into variables. Let's go father by looking at what Resistance could be.

　To find it out, to simply use Hagen-Poiseuille equation:

$Q=\frac{\pi r^4}{8\eta L}∆P$

where r is the radius of the pipe, η is the viscosity of fluid (water),

 and L is the length of the pipe

You may have realized that this is somewhat similar to the equation we got in the previous section. The only difference is 1 / R and  (πr⁴) / (8ηL), and this is why we can equate them as to give resistance to be:

$R=\frac{8\eta L}{\pi r^4}$

　Let's think about this intuitively. What happens if the length of the pipe increases? You can imagine more surface friction to be against the water flows, and therefore, the resistance increases. That agrees with the equation above!

　What about if the radius of the pipe increases? There will be more area of the pipe for the water to flow, so the resistance should decreases, right? And, this also agrees with the equation we have just got!!

• Lastly... what is the rate of flow of water??

　Did you realize that we can now determine the Q _Out differently without measure the volume of the water??

　This is just a combination of equations we have derived (partially):

$Q_{Out}=\frac{\pi r^4}{8\eta L}\rho gh$

　This equation itself is not too important as it is just a mix of concepts. However, what this implies is quite significant. That is the fact which height (h) being the only variable to determine the water flow rate. All symbols in (ρgπr⁴) / (8ηL) are constant in one water tank system.

　Hence, without weighting the water tank, with just the height, we now know that we can determine the water flow rate!!!

2. How else can accumulated water flow rate be written??

　Now we manage to write the rate of water flow out from Tank 1 in terms of more general way. We also know that it can be determined using height.

　Well, in that case, it would be great if we can also define Q _Accumulated in terms of height. This is actually not too hard. Look at the followings:

$-\frac{h\rho g}{R}=A\frac{dh}{dt}$

　Does this make sense to you?? Area of water tank does not change, so rate of volume of water change can also be thought as area times the rate of water level change.

3. How can the conservation law be expressed for water tank system??

　Now, let's look back to the conservation equation derived above for Tank A:

$-Q_{Out} = Q_{Accumulated}$

　I suppose you know a different way of expressing the flow rates in terms of height of water level.

$-h\rho g\left(\frac{\pi r^4}{8\eta L}\right)=A\frac{dh}{dt}$

or, just simply as:

$-\frac{h\rho g}{R}=A\frac{dh}{dt}$

　However, what is so good about being able to express this way?? Previous equation looked much simpler.

　Yet, I think you will understand if you think about doing such water tank system experiment. Which will be easier and more accurate: measuring the volume or measuring the height??

　I guess it is clear that it is much easier to find the change in height over time. One method of taking could be just filming the experiment and analyze the video later. However, volume cannot be measured in such way. You need to read out the volume, which is likely to be ambiguous, every few seconds.

　Thus, this equation is useful in experimental analysis. 

　Moreover, this new equation is useful as it consists different variables. So, it is possible to determine how the each variable can affect the rate of change in height, etc. 


Analyze the model obtained

　This is not it. Did you thought about integrating the equation we have obtained?? Let's see what we will get. 

$\frac{h}{h_0}=e^{-\left(\frac{\rho g}{AR}\right)t}$
where h = h₀ at t = 0

　This result is quite useful because the usefulness and the beauty of this equation can be illustrated by drawing this graph.



　This indicates that over time, the h will eventually reach a point where no more flow occurs. And, that is indeed true because as I mentioned previously, the pressure difference reaches 0 when the pressure difference at the outlet inside and outside equates (i.e. water level reaches the outlet level).

End note

　I do understand that this article was neither well written nor well explained but I hope you got a grasp something about chemical engineering. In addition, first read might not make sense to you but over time over days, reading the same thing can create a meaning to you and sometimes that works for me.

　Here, you might have realized there are tons of equations coming up. Yet, I have to say that this is the chemical engineering. Despite the sound of "chem", in this basic state we have not even came across any chemical equation. It certainly will but not get misunderstood. Yet, I hope you have enjoyed chemical engineering and will enjoy this with me further.


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