Introduction

Modern society and living conditions are made possible by advances in science, technology, engineering and mathematics (STEM). The subjects studied in these fields have become so relevant that any person not familiar with their basic principles is at a disadvantage. Consequently, the study of STEM fields must be encouraged throughout academic education. Even then, no one should stop there. Real things are dynamic, always changing, so one must engage in lifelong learning. In fact, this is a personal motivation to start this blog.

So what lies ahead?

During many years in higher education, I have learned that complex problems can be solved by breaking them down into simpler problems. In STEM fields, this means analyzing the problems from the fundamental concepts. In other cases, students might also need a reference in topics that are of great importance. As a result, the entries in this blog will present these concepts with thorough explanations that can help any student at different academic levels.

Statement of Purpose

Provide students of all ages and levels with articles and information that can help them understand a variety of concepts of STEM fields. I will do my best to organize and label entries with key words, so anyone can easily find what they look for.

Anyone reading the blog is welcomed to comment, make suggestions and point out errors. I hope this becomes helpful to many around the world.

Enjoy!

Stoichiometry in Combustion (Part 2 - Example with Excess Air)

Determine the air-fuel ratio on a mass basis for the complete combustion of octane, C8H18, with 150% theoretical air (50% excess air).

Solution:

The first part of the solution is to obtain the equation for the complete combustion of octane with the theoretical amount of air. Find the coefficients of the equation.

The second part of the solution is using the left hand side of the chemical equation to obtain the new coefficients of the products when 150% theoretical air is used. Now, the air is multiplied by 1.5, and there is excess air that appears as uncombined oxygen in the products.

The third part of the solution is actually calculating the air-fuel ratio which is what the problem asked for.

The ratio of the amount of air in this reaction to the amount of fuel is about 22.7.

Bibliography:

Michael J. Moran, Howard N. Shapiro. Fundamentals of Engineering Thermodynamics. John Wiley & Sons, Inc., 2008.

Stoichiometry in Combustion

Stoichiometry is the study of the relationship between quantities of reactants and products in chemical equations. It is based on the principle of conservation of mass. Therefore, it involves balancing both sides of the equation by assuring there are the same number of atoms for all elements involved. 

In the field of mechanical engineering, stoichiometry is mostly used in the field of thermodynamics when studying combustion reactions. In this type of reactions, a fuel reacts chemically with oxygen in order to release energy and combustion products. When a combustion is complete, all the carbon present in the fuel is burned to carbon dioxide, all the hydrogen is burned to water, all the sulfur is burned to sulfur dioxide, and all other combustible elements are fully oxidize. If these conditions are not met, the combustion is said to be incomplete. Now, we will observe a couple of examples.

Example 1:

Balance the equation below for the complete combustion of methane.

Solution:

By applying the principle of conservation of mass, we can create equations to balance the number of atoms of the same elements on both sides in terms of the coefficients of each compound in the reaction.

By solving the system of four equations and four unknowns, the balanced chemical equation is:

Example 2:

For the balanced equation in the previous example, obtain the air-fuel ratio.

Solution:

The ratio of the amount of air in this reaction to the amount of fuel is about 17.24.

Bibliography:

Michael J. Moran, Howard N. Shapiro. Fundamentals of Engineering Thermodynamics. John Wiley & Sons, Inc., 2008.

Basic Rules of Differentiation (Part 2)

In Part 1, we learned two basic rules that allow us to differentiate polynomial functions. Now we will learn to deal with functions that involve the multiplication or division of functions. We will learn about product rule and quotient rule.

Product Rule:

Suppose we have a function f(x) that is the result of the multiplication (known as a product) between two functions g(x) and h(x). Then, the derivative of f(x) is given by:

Many teachers and students play with words to memorize this rule. Anyone can make up his/her own way to remember it, but it basically says the derivative of f(x) is "the first times derivative of the last, plus the last times derivative of the first."

Quotient Rule:

Suppose we have a function f(x) that is the result of the division (known as a quotient) of the functions g(x) by h(x). Then, the derivative of f(x) is given by:

In this case, we can say the derivative of f(x) is "the bottom times derivative of the top, minus the top times derivative of the bottom. All divided by the bottom squared."

Examples:

These two rules are quite popular, so let's look at some examples. Notice that we will apply the rules learned in Part 1 when dealing with the terms of the functions.

Now, as seen in the examples, we can differentiate more complicated functions!!!