Triangulene: a highly reactive form of graphene

The original article is here. If the previous blog post is any indicator, graphene and its alternate forms are majorly popular in the field of materials science and engineering. Every research institution will invariably have a group doing work on some form of graphene. Now, what makes this particular form of graphene interesting?

Carbon likes to form 4 sp3 hybridized orbitals, or in the case of graphene, 3 sp2 hybrid orbitals plus one pi bond allowing it to form a planar molecule. The structure of triangulene is such that it cannot form a stabilizing pi bond which leaves two lone pairs of electrons floating around the middle carbon rings. This process had to be done in a vacuum because the resulting molecule is so reactive that any contaminates would instantly react with it.

Something to bear in mind with this research, which is said in the article, but I’ll reiterate here: this molecule was neither characterized nor was a mass production process developed. Meaning that none of the triangulene’s material properties were determined, the research was done as a proof of concept. Mass production will be an especially difficult problem to overcome because each molecule was made one at a time with each individual hydrogen bond cleaved under SEM. There’s not much else I can say about this, the research is mostly proof of concept. Next blog post will be a materials science related tutorial. Thanks for reading!

Back to the Basics: Differential Equations

Calculus and differential equations have immense utility in any field of engineering. Any quality engineer should take it upon themselves to occasionally review their academic work so that they maintain their knowledge base. I will try to skip over the mathematical theory and get right to solving certain types of differential equations since as engineers, we typically only need the concrete concepts.

Here is a practical example of a first order mixing solution problem. We are mixing a salt in a cylindrical tank. We have several variables: the concentration of salt going into and leaving the tank Cin and Cout respectively, the rate of water going into and leaving the tank, rin and rout respectively, the amount of water in the tank S(t), and the amount of salt in the water at time t, Q(t). Assuming that water leaving the tank is the same concentration as the water in the tank, Cout can be simplified by Q(t)/S(t). How do we find the amount of salt in the tank at time t?

Here are some initial conditions to help get us started. At time t=0 the amount of water in the tank is 500 liters, there are 100 grams of salt in the tank, the rate of water going into and out of the tank is 5 liters/min and the concentration going in is 2 grams/min. We know now that the level of water is not going to change since the flow rates are equal. Let’s look at our equation to determine salt levels over time.

Q'(t) = rin*Cin – rout*Q(t)/S(t)

Now we replace values that we already know and get:

Q'(t) = 5 * 2 – 5 * Q(t) / 500 or Q'(t) = 10 – Q(t) /100

Group like variables on opposite sides and you should get:

Q'(t) + Q(t) / 100 = 10

At this point we use a substitution method to simplify the equation. This step is especially important if you have length functions and derivatives. We simplify the function such that the derivative has no coefficient as is already done. This is the form we are aiming for:

μ(t)*Q'(t) +   μ(t)*p(t)*Q(t)/100 = g(t)

Here p(t) is 1/100 and g(t) is 10. We are going to solve for  μ(t). μ must satisfy the equation μ'(t)=p(t)*μ(t). The simplified form is below:

μ'(t)/ μ(t) = 10

Integrate both sides. Solve for  μ(t) by eliminating with the exponential function. You should get.

μ(t) = et/100

This allows us to integrate the right side to get μ(t)*Q(t) the left side is ∫g(t)*μ(t)dt. Note the dt in that equation. dt in this case is 100. This is what we are left with:

Q(t)=et/100 * ∫10*100*et/100

The result of this integral will produce a constant too, which will require the initial condition so it can be solved. Here is the solved form:

Q(t) = 1000+C*e-t/100

Since we know at time t = 0 Q is 100, we can then solve for C and then we can use this equation to determine salt levels over time. I’m going to leave that to you do to do for practice if you so choose. I realize my math is a bit sloppy, let me know if you have any improvements. Thanks for reading!

Rice Labs Develops Graphene Foam

In this article here you can read about Rice University’s graphene foam. Today I will go over the basics of graphene and its material properties. I will explain how its chemical structure creates such high strengths and speak briefly about the article.

Mechanical Properties of Graphene

Graphene itself is just a molecule of carbon like diamond or graphite but with a different crystal structure. It forms a 2-D hexagonal structure with 3 hybrid orbital bonds and 1 pi bond that extends orthogonally to the 2-D plane.  This is the source for graphene’s exceptional mechanical properties, at least in two directions. This image should explain why; any force applied orthogonally to the plane of graphene molecules will have a different stress/strain relationship than a force applied in parallel to the plane. Materials that exhibit this property are called anisotropic; their material properties vary by direction.

Overcoming Anisotropic Limitations

As you can see from the graphic in the abstract the multiwalled carbon nanotubes (MWCNTs) and sintered Nickel are used in conjunction to create a template for the graphene shell. Note the different configurations of the 2-D graphene layers; by layering different orientations of graphene planes researchers overcame the anisotropy of the graphene.



Materials Engineering Review Intro

Good day. I will be taking a variety of Materials Engineering or engineering related topics and explaining them in simple terms. I will also occasionally post tutorials on how to approach problems from a Materials standpoint.

If you have any suggestions for tutorials or topics for me to write about let me know. I tend to write concisely, if I leave crucial information out, let me know and I can write more about that topic.

Credit to image goes to: Scm83x at the English language Wikipedia, CC BY-SA 3.0, Link