{ "cells": [ { "cell_type": "markdown", "metadata": {}, "source": [ "# 5.7 – Reference States and State Properties \n", "\n", "---" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "## 5.7.0 – Learning Objectives\n", "\n", "By the end of this section you should be able to:\n", "\n", "1. Understand reference states and their importance.\n", "2. Understand the steam table chart and its reference point.\n", "\n", "---" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "## 5.7.1 – Introduction\n", "\n", "Often, we do not know the absolute value of internal energy or enthalpy ($\\hat{U}, \\hat{H}$). To give us a way of measuring these values, a __reference state__ is chosen to represent __0__ and the values are determined as a __difference to the reference state.__ \n", "\n", "This notebook will go over the concept of reference state and state properties.7\n", "\n", "---" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "## 5.7.2 – Internal energy of Water \n", "\n", "Let's look at water at 0 C$^\\circ$ and 1 atm. If asked to find the __specific__ energy $U$ of this water, you might think right away that you can use $C_p\\Delta T$ but this is wrong. We cannot use this formula because we need to consider the $\\Delta T$ which refers to a change in temperature. \n", "\n", "What we can do is use the 0 C$^\\circ$ 1 atm water as a __reference__ and compare it to changes in energy.\n", "\n", "We can take liquid water at 50 C$^\\circ$ 1 atm and say that __in reference to 0__ C$^\\circ$ __water__, 100 C$^\\circ$ water has $210 \\space \\frac{kJ}{kg}$ more energy than 0 C$^\\circ$ liquid water. \n", "\n", "Likewise, 100 C$^\\circ$ liquid water at 1 atm has $420 \\space \\frac{kJ}{kg}$ more energy than water at 0 C$^\\circ$. \n", "\n", "By setting water at 0 C$^\\circ$. We can use the referenced values as variables known as __state variables__ also known as $\\hat{U}$ and $\\hat{H}$. \n", "\n", "A state variable is a variable that its property only depends on its __initial and final state__.\n" ] }, { "cell_type": "code", "execution_count": 8, "metadata": {}, "outputs": [ { "data": { "text/html": [ "
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Temperature (K)Pressure (MPa)Volume (l/mol)Internal Energy (kJ/mol)Enthalpy (kJ/mol)Entropy (J/mol*K)
06000.0infinite51.41556.404500.13
16001.04.886151.12356.009130.11
26002.02.389150.81155.589123.82
36003.01.554850.47755.142119.88
46004.01.136150.11954.664116.88
56005.00.883449.73454.151114.36
66006.00.7135549.31853.599112.11
76007.00.5908748.86453.000110.03
86008.00.4974148.36652.345108.04
96009.00.4231647.81151.620106.06
1060010.00.3619547.18350.802104.05
\n", "
" ], "text/plain": [ " Temperature (K) Pressure (MPa) Volume (l/mol) Internal Energy (kJ/mol) \\\n", "0 600 0.0 infinite 51.415 \n", "1 600 1.0 4.8861 51.123 \n", "2 600 2.0 2.3891 50.811 \n", "3 600 3.0 1.5548 50.477 \n", "4 600 4.0 1.1361 50.119 \n", "5 600 5.0 0.8834 49.734 \n", "6 600 6.0 0.71355 49.318 \n", "7 600 7.0 0.59087 48.864 \n", "8 600 8.0 0.49741 48.366 \n", "9 600 9.0 0.42316 47.811 \n", "10 600 10.0 0.36195 47.183 \n", "\n", " Enthalpy (kJ/mol) Entropy (J/mol*K) \n", "0 56.404 500.13 \n", "1 56.009 130.11 \n", "2 55.589 123.82 \n", "3 55.142 119.88 \n", "4 54.664 116.88 \n", "5 54.151 114.36 \n", "6 53.599 112.11 \n", "7 53.000 110.03 \n", "8 52.345 108.04 \n", "9 51.620 106.06 \n", "10 50.802 104.05 " ] }, "execution_count": 8, "metadata": {}, "output_type": "execute_result" } ], "source": [ "import pandas as pd\n", "import numpy as np\n", "import matplotlib.pyplot as plt\n", "\n", "df = pd.read_excel('../figures/Module-5/isoT600k.xlsx', sheet_name='Sheet1', index_col=0, na_values=['NA'])\n", "df\n", "df.truncate(before=0, after=10)\n" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "If you recall the steam tables\" from module 4: \n", "[NIST](http://webbook.nist.gov/cgi/fluid.cgi?Action=Load&ID=C7732185&Type=SatP&Digits=5&THigh=373.946&TLow=0.01&TInc=5&RefState=DEF&TUnit=C&PUnit=MPa&DUnit=kg%2Fm3&HUnit=kJ%2Fkg&WUnit=m%2Fs&VisUnit=uPa*s&STUnit=N%2Fm) steam tables from module 4, these values are all state variables referenced to **water at its triple point**.\n", "\n", "---" ] }, { "cell_type": "markdown", "metadata": { "collapsed": true }, "source": [ "\n", "## 5.73 – Example question\n", "\n", "Water is partially condensed out of a gas stream containing 66.9 mole% water vapor and the balance nitrogen.\n", "\n", "![Figure 1.](../figures/Module-5/module-5-example-5.7.1.svg)\n", "\n", "### Step 1 - Perform all required material balances\n", "\n", "- While none are needed, it might be helpful to fill in the missing mole fractions for nitrogen\n", "\n", "### Step 2 - Write the appropriate form of the energy balance (open or closed) and reduce to simplest form (get rid of any negligible terms)\n", "\n", "$$\\dot{Q} - \\dot{W} = \\Delta\\dot{H} + \\dot{E_k} + \\dot{E_p}$$\n", "\n", "$$\\dot{E_k} = 0$$\n", "\n", "$$\\dot{E_p} = 0$$\n", "\n", "$$\\dot{W} = 0$$\n", "\n", "$$\\dot{Q} = \\Delta\\dot{H}$$\n", "\n", "$$\\Delta\\dot{H} = \\sum_{out}\\dot{n_i}\\hat{H_i} - \\sum_{in}\\dot{n_i}\\hat{H_i}$$\n", "\n", "### Step 3 - Choose a reference state for the constituents\n", "\n", "\n", "**Water**\n", "\n", "$l$, 25 °C, $1 \\space atm$,\n", "\n", "**Nitrogen**\n", "\n", "$g$, 25 °C, $1 \\space atm$\n", "\n", "### Step 4 - For a closed constant- volume system, construct a table with columns for initial and final amounts of each species ($m_i$ or $n_i$) and specific internal energies relative to the chosen reference states. For an open system, construct a table with columns for inlet and outlet stream component flow rates ($\\dot{m_i}$ or $\\dot{n_i}$) and specific enthalpies relative to the chosen reference states.\n", "\n", "\n", "| Substance | $n_{in}$ | $\\hat{H_{in}}$ | $n_{out}$ | $\\hat{H_{out}}$ |\n", "|-----------|:--------:|--------------:|-----------|-----------------|\n", "| H2O(v) | 0.669 | $\\hat{H_{1}}$ | 3.35 | $\\hat{H_{3}}$ |\n", "| H2O(l) | | | 63.55 | |\n", "| N2(v) | 0.331 | $\\hat{H_{2}}$ | 33.1 | $\\hat{H_{4}}$ |\n", "\n", " \n", "### Step 5 - Calculate all required values of $\\hat{U}$ (or $\\hat{H}$ ) and insert the values in the appropriate places in the table.\n", "\n", "\n", "\n", "### Step 6 - Calculate the appropriate energy balances. Solve for any variable (kinetic, potential) that is not negligible." ] } ], "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 }