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diff --git a/Industrial_Instrumentation/ch7.ipynb b/Industrial_Instrumentation/ch7.ipynb new file mode 100644 index 00000000..f35bb080 --- /dev/null +++ b/Industrial_Instrumentation/ch7.ipynb @@ -0,0 +1,210 @@ +{ + "metadata": { + "name": "" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter 7 : Thermodynamic Relations" + ] + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": [ + "Example 7.17 Page no : 370" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "'''\n", + "Determine the following :\n", + "(i) Work done on the copper during the process,\n", + "(ii) Change in entropy,\n", + "(iii) The heat transfer,\n", + "(iv) Change in internal energy, and\n", + "(v) (c p \u2013 c v ) for this change of state.\n", + "'''\n", + "\n", + "# Variables\n", + "B = 5.*10**(-5); \t\t\t# /K\n", + "K = 8.6*10**(-12); \t\t\t# m**2/N\n", + "v = 0.114*10**(-3); \t\t\t#m**3/kg\n", + "p2 = 800.*10**5; \t\t\t#Pa\n", + "p1 = 20.*10**5; \t\t\t#Pa\n", + "T = 288.; \t\t\t#K\n", + "\n", + "# Calculations and Results\n", + "W = -v*K/2*(p2**2-p1**2);\n", + "print (\"(i) Work done on the copper = %.3f\")%(W),(\"J/kg\")\n", + "\n", + "ds = -v*B*(p2-p1);\n", + "print (\"(ii) Change in entropy = %.3f\")% (ds), (\"J/kg K\")\n", + "\n", + "Q = T*ds;\n", + "print (\"(iii) The heat transfer = %.3f\")%(Q), (\"J/kg\")\n", + "\n", + "du = Q-W;\n", + "print (\"(iv) Change in internal energy = %.3f\")%(du),(\"J/kg\")\n", + "\n", + "R = B**2*T*v/K;\n", + "print (\"(v) cp \u2013 cv = %.3f\")%(R),(\"J/kg K\")\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "(i) Work done on the copper = -3.135 J/kg\n", + "(ii) Change in entropy = -0.445 J/kg K\n", + "(iii) The heat transfer = -128.045 J/kg\n", + "(iv) Change in internal energy = -124.909 J/kg\n", + "(v) cp \u2013 cv = 9.544 J/kg K\n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": [ + "Example 7.18 Page no : 371" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "'''\n", + "Using Clausius-Claperyon\u2019s equation, find enthalpy of vapourisation.\n", + "'''\n", + "\n", + "# Variables\n", + "vg = 0.1274; \t\t\t#m**3/kg\n", + "vf = 0.001157; \t\t\t#m**3/kg\n", + "# dp/dT = 32; \t\t\t#kPa/K\n", + "T3 = 473; \t\t\t#K\n", + "\n", + "# Calculations\n", + "h_fg = 32*10**3*T3*(vg-vf)/10**3;\n", + "\n", + "# Results\n", + "print (\" enthalpy of vapourisation = %.3f\")%(h_fg),(\"kJ/kg\")\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " enthalpy of vapourisation = 1910.814 kJ/kg\n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": [ + "Example 7.19 Page no : 372" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "'''\n", + "Determine the pressure an ice skate blade must exert to allow smooth ice skate at \u2013 10\u00b0C.\n", + "'''\n", + "\n", + "import math\n", + "\n", + "# Variables\n", + "h_fg = 334.; \t\t\t#kJ/kg\n", + "v_liq = 1.; \t\t\t#m**3/kg\n", + "v_ice = 1.01; \t\t\t#m**3/kg\n", + "T1 = 273.; \t\t\t#K\n", + "T2 = 263.; \t\t\t#K\n", + "p1 = 1.013*10**5; \t\t\t#Pa\n", + "\n", + "# Calculations\n", + "p2 = (p1+h_fg*10**3/(v_ice-v_liq)*math.log(T1/T2))/10**5;\n", + "\n", + "# Results\n", + "print (\"pressure = %.3f\")%(p2),(\"bar\")\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "pressure = 13.477 bar\n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": [ + "Example 7.20 Page no : 372" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "'''\n", + "Calculate the specific volume v g of saturation mercury vapour at 0.1 bar\n", + "'''\n", + "\n", + "# Variables\n", + "h_fg = 294.54; \t\t\t#kJ/kg\n", + "p = 0.1; \t\t\t#bar\n", + "T = 523; \t\t\t#K\n", + "\n", + "# Calculations\n", + "vg = h_fg*10**3/T/(2.302*3276.6*p*10**5/T**2 - 0.652*p*10**5/T);\n", + "\n", + "# Results\n", + "print (\"specific volume = %.3f\")%(vg),(\"m**3/kg\")\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "specific volume = 2.139 m**3/kg\n" + ] + } + ], + "prompt_number": 3 + } + ], + "metadata": {} + } + ] +}
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