{ "cells": [ { "cell_type": "markdown", "metadata": {}, "source": [ "# 5.1 – The First Law of Thermodynamics\n", "\n", "---" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "## 5.1.0 – Learning Objectives\n", "\n", "By the end of this section you should be able to:\n", "\n", "1. Understand the relationship between the first law and energy conservation.\n", "2. Understand the breakdown of the energy term\n", "3. Learn about a closed system and internal energy.\n", "\n", "---" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "## 5.1.1 – Introduction \n", "\n", "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.\n", "\n", "---" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "## 5.1.2 – The First Law\n", "\n", "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: \n", "\n", "$$E_{sys} = Q - W$$ \n", "\n", "and \n", "\n", "$$\\Delta E_{sys} = \\Delta Q-\\Delta W$$\n", "\n", "__Important! The system must be closed. Otherwise internal energy cannot be used alone__\n", "\n", "---" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "## 5.1.3 – Units of the first law\n", "\n", "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.\n", "\n", "Let's break down these terms down one-by-one.\n", "\n", "---" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "## 5.1.4 – Energy\n", "\n", "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.\n", "\n", "$$E_{tot} = E_{sys} = E_k + E_p + U$$ \n", "\n", "---" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "## 5.1.5 – Fermi question!\n", "\n", "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__" ] }, { "cell_type": "markdown", "metadata": { "collapsed": true }, "source": [ "### A solution to Fermi. \n", "\n", "I know that the Mount Everest is approximately 30000ft. The average person weighs 65kg. therefore, using the __potential energy__ formula and rough unit conversions: \n", "\n", "$$E_p = mgh$$ \n", "\n", "we obtain:\n", "\n", "$$ E_p = 65 \\space kg \\cdot \\bigg(30,000 \\space ft \\times 0.3 \\space \\frac{m}{ft} \\bigg) \\cdot 9.81 \\space \\frac{m}{s^2} = 5,738,850 \\space J$$\n", "\n", "Let's guess that an Oreo has 100 food calories (Cal) \n", "\n", "$$ \\therefore \\space 5,738,850 \\space J \\times \\frac{1 \\space Cal}{4,000 \\space J} \\times \\frac{1 \\space \\text{Oreo cookies}}{100 \\space Cal} = 14.3 \\space \\text{Oreo cookies}. $$\n", "\n", "A car weighs approximately 1000kg. Using the __kinetic energy__ formula:\n", "\n", "$$ E_k = \\frac{1}{2} \\cdot m *v^2 $$\n", "\n", "$$ 5,738,850 \\space J = \\frac{1}{2} \\cdot 1,000 \\space kg \\cdot v^2 $$ \n", "\n", "Solving for v results in $v = 107.13 \\space m/s$ which is just over the speed of the bullet train in Japan.\n", "\n", "Click here for more fermi questions [more fermi!](http://www.fermiquestions.com/play)" ] }, { "cell_type": "markdown", "metadata": { "collapsed": true }, "source": [ "## 5.1.6 – Problem Statement\n", "\n", "### Question\n", "\n", "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.\n", "\n", "![Figure 1.](../figures/Module-5/5.1-example-question1.svg)\n", "\n", "### Answer\n", "\n", "**Kinetic energy** - We will use m/s just to keep consistent with SI units. \n", "\n", "$$u = 2.00 \\space m^3/h \\cdot \\frac{1}{0.01^2\\pi} \\cdot \\frac{1hr}{3600s} = 1.77 \\space m/s$$\n", "\n", "$$\\dot{m} = 2.00 \\space m^3/h \\cdot \\frac{1000 \\space kg}{1 \\space m^3} \\cdot \\frac{1hr}{3600s} = 0.556 \\space kg/s$$\n", "\n", "$$\\dot{E_k} = \\frac{1}{2} \\cdot 0.556 \\space kg/s \\cdot 1.77 \\space m/s = 0.870 \\space J/s$$\n", "\n", "**Potential energy**\n", "\n", "$$m = 2.00 \\space m^3/h \\cdot \\frac{1000 \\space kg}{1 \\space m^3} \\cdot 2 hrs = 4000 \\space kg$$\n", "\n", "$$\\dot{E_p} = m \\cdot g \\cdot h$$\n", "\n", "$$\\dot{E_p} = 4000 \\space kg \\cdot 9.81 \\cdot 2m = 78484 \\space J$$" ] }, { "cell_type": "code", "execution_count": null, "metadata": { "collapsed": true }, "outputs": [], "source": [] } ], "metadata": { "anaconda-cloud": {}, "kernelspec": { "display_name": "Python 3", "language": "python", "name": "python3" }, "language_info": { "codemirror_mode": { "name": "ipython", "version": 3 }, "file_extension": ".py", "mimetype": "text/x-python", "name": "python", "nbconvert_exporter": "python", "pygments_lexer": "ipython3", "version": "3.6.1" } }, "nbformat": 4, "nbformat_minor": 2 }