5.1 – The First Law of Thermodynamics¶

5.1.0 – Learning Objectives¶

By the end of this section you should be able to:

Understand the relationship between the first law and energy conservation.
Understand the breakdown of the energy term
Learn about a closed system and internal energy.

5.1.1 – Introduction¶

As Chemical engineers, the knowledge of energy flows is integral to the industry. How will equipment affect the energy of a system? How does a reactor withstand an exothermic/endothermic reaction? Thermodynamics is the stepping stone in the study of energy flows. This notebook will give you an overview of the first law of thermodynamics.

5.1.2 – The First Law¶

The first law is the concept that energy is conserved. ** The first law states that the energy of a **closed system \(E_{sys}\), is equal to the sum of heat added to the system \(Q\) and the work done to or upon the system \(W\). This is written mathematically as:

\[E_{sys} = Q - W\]

and

\[\Delta E_{sys} = \Delta Q-\Delta W\]

Important! The system must be closed. Otherwise internal energy cannot be used alone

5.1.3 – Units of the first law¶

Energy in chemical engineering is often described as a \(\frac{Kilojoule}{Kilogram}\) or \(\frac{Joule}{mol}\). These units are used because of the massive amount of material that chemical engineers work with.

Let’s break down these terms down one-by-one.

5.1.4 – Energy¶

Energy comes in many different forms. Sitting up on a chair reading this, you have potential energy. When you run in class to hand in your homework, you are demonstrating kinetic energy. Your mere existence and the vibrations of the atoms inside of you, is a form of the internal energy. There are many other energies that can be classified but this course will be mainly focusing on these three.

\[E_{tot} = E_{sys} = E_k + E_p + U\]

5.1.5 – Fermi question!¶

Using only facts, you know yourself... How fast would a car have to drive to have the same energy to climb mount Everest? How many Oreo cookies would that be? An answer will be provided at the bottom

A solution to Fermi.¶

I know that the Mount Everest is approximately 30000ft. The average person weighs 65kg. therefore, using the potential energy formula and rough unit conversions:

\[E_p = mgh\]

we obtain:

\[E_p = 65 \space kg \cdot \bigg(30,000 \space ft \times 0.3 \space \frac{m}{ft} \bigg) \cdot 9.81 \space \frac{m}{s^2} = 5,738,850 \space J\]

Let’s guess that an Oreo has 100 food calories (Cal)

\[\therefore \space 5,738,850 \space J \times \frac{1 \space Cal}{4,000 \space J} \times \frac{1 \space \text{Oreo cookies}}{100 \space Cal} = 14.3 \space \text{Oreo cookies}.\]

A car weighs approximately 1000kg. Using the kinetic energy formula:

\[E_k = \frac{1}{2} \cdot m *v^2\]

\[5,738,850 \space J = \frac{1}{2} \cdot 1,000 \space kg \cdot v^2\]

Solving for v results in \(v = 107.13 \space m/s\) which is just over the speed of the bullet train in Japan.

Click here for more fermi questions more fermi!

5.1.6 – Problem Statement¶

Question¶

Water flows into a process unit through a 2-cm ID (internal diameter), 4m long pipe at a rate of 2.00 \(m^3/h\), at a height of 2m from the ground. Assuming 2 hours have passed and, calculate \(\dot{E_k}\), \(\dot{E_p}\) and the total energy for the system.

Figure 1.

Answer¶

Kinetic energy - We will use m/s just to keep consistent with SI units.

\[u = 2.00 \space m^3/h \cdot \frac{1}{0.01^2\pi} \cdot \frac{1hr}{3600s} = 1.77 \space m/s\]

\[\dot{m} = 2.00 \space m^3/h \cdot \frac{1000 \space kg}{1 \space m^3} \cdot \frac{1hr}{3600s} = 0.556 \space kg/s\]

\[\dot{E_k} = \frac{1}{2} \cdot 0.556 \space kg/s \cdot 1.77 \space m/s = 0.870 \space J/s\]

Potential energy

\[m = 2.00 \space m^3/h \cdot \frac{1000 \space kg}{1 \space m^3} \cdot 2 hrs = 4000 \space kg\]

\[\dot{E_p} = m \cdot g \cdot h\]

\[\dot{E_p} = 4000 \space kg \cdot 9.81 \cdot 2m = 78484 \space J\]

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