エンジニアへの道: 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:

$R a t e_{I n} - R a t e_{O u t} + R a t e_{G e n e r a t e d} - R a t e_{C o n s u m e d} = R a t e_{A c c u m u l a t e d}$

　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:

$R a t e_{M a s s I n} - R a t e_{M a s s O u t} + R a t e_{M a s s G e n e r a t e d} - R a t e_{M a s s C o n s u m e d}$

　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:

$- Q_{O u t} = Q_{A c c u m u l a t e d}$

　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:

$\frac{1}{2} ρ {v_{x}}^{2} + ρ g z_{x} + p_{x} = \frac{1}{2} ρ {v_{y}}^{2} + ρ g z_{y} + p_{y}$

　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:

P r e s s u r e d i f f e r e n c e (∆ P) = (p_{a t m} + p_{h y d r o s t a t i c}) - (p_{a t m}) = ρ g h

　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:

F l o w r a t e (Q) = \frac{D r i v i n g F o r c e (D F)}{R e s i s \tan c e (R)}

　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{ρ g h}{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{π r^{4}}{8 η 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 η L}{π 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_{O u t} = \frac{π r^{4}}{8 η L} ρ g h

　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:

$- \frac{h ρ g}{R} = A \frac{d h}{d t}$

　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:

$- Q_{O u t} = Q_{A c c u m u l a t e d}$

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

$- h ρ g (\frac{π r^{4}}{8 η L}) = A \frac{d h}{d t}$

or, just simply as:

$- \frac{h ρ g}{R} = A \frac{d h}{d t}$

　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.

$\frac{h}{h_{0}} = e^{- (\frac{ρ g}{A R}) t}$
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.

Previous Article: What does Chemical engineers do??
Next Article: Water tanks "in series" and steady state

Water tank system and mass conservation

Road to a Chemical Engineer: #1

Recap

Water tank system and mass conservation

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??

How to quantify the rate of water flow?

What is the resistance here??

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

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: -QOut = QAccumulated I suppose you know a different way of expressing the flow rates in terms of height of water level.

-hρg(πr48ηL)=Adhdt or, just simply as:

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

　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

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

$- Q_{O u t} = Q_{A c c u m u l a t e d}$

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

$- h ρ g (\frac{π r^{4}}{8 η L}) = A \frac{d h}{d t}$

or, just simply as:

$- \frac{h ρ g}{R} = A \frac{d h}{d t}$

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