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diff --git a/Engineering_&_Chemical_Thermodynamics/ch4.ipynb b/Engineering_&_Chemical_Thermodynamics/ch4.ipynb new file mode 100755 index 00000000..dd2a21ec --- /dev/null +++ b/Engineering_&_Chemical_Thermodynamics/ch4.ipynb @@ -0,0 +1,434 @@ +{ + "metadata": { + "name": "" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter 4 : Equation of states and intermolecular forces" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 4.1 Page No : 220" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "import math \n", + "from numpy import *\n", + "\n", + "# Variables\n", + "H2O = 1. ;\n", + "NH3 = 2. ;\n", + "CH4 = 3. ;\n", + "CH3Cl = 4. ;\n", + "CCl4 = 5. ;\n", + "\n", + "M_11 = 1.85\n", + "alp_12 = 14.80\n", + "I_13 = 12.62 ;\n", + "M_12 = 1.47\n", + "alp_22 = 22.20\n", + "I_23 = 10.07 ;\n", + "M_31 = 0.00\n", + "alp_32 = 26.00\n", + "I_33 = 12.61 ;\n", + "M_41 = 1.87\n", + "alp_42 = 45.30\n", + "I_43 = 11.26 ; \n", + "M_51 = 0.00\n", + "alp_52 = 105.0\n", + "I_53 = 11.47 ;\n", + "\n", + "# Calculations\n", + "k = 1.38 * 10**-16 ; \t\t\t#[ J/K]\n", + "T = 298. \t\t\t#[K]\n", + "A =[[M_11 , alp_12 , I_13, 0,0,0,0],[M_12 , alp_22 , I_23,0,0,0,0 ],[M_31 , alp_32 , I_33,0,0,0,0],[M_41 , alp_42 , I_43,0,0,0,0],[M_51 , alp_52 , I_53,0,0,0,0 ]]\n", + "print A\n", + "\n", + "print (\" Molecule M alp*10**25 I C*10**60 Cd_d Cind Cdis\") ;\n", + "for i in range(5):\n", + " A[i][4] = ceil( 2./3 * A[i][0]**4 / (k * T) * 10**-12) ;\n", + " A[i][5] = ceil(2 * A[i][1] * A[i][0]**2 * 10**-1) ; \n", + " A[i][6] = ceil(3./4 * A[i][1]**2 * A[i][2] * 1.6 * 10**-2) ;\n", + " A[i][3] = ceil(A[i][4] + A[i][5] + A[i][6]) ; \t\t\t# ....E4.1D\n", + "\n", + "\n", + "print \" H2O %8.2f %5.1f %5.2f %7d %7d %5d %d \"%(A[0][0],A[0][1],A[0][2],A[0][3],A[0][4],A[0][5],A[0][6]) ;\n", + "print \" NH3 %8.2f %5.1f %5.2f %7d %7d %5d %d \"%(A[1][0],A[1][1],A[1][2],A[1][3],A[1][4],A[1][5],A[1][6]) ;\n", + "print \" CH4 %8.2f %5.1f %5.2f %7d %7d %5d %d \"%(A[2][0],A[2][1],A[2][2],A[2][3],A[2][4],A[2][5],A[2][6]) ;\n", + "print \" CH3Cl %8.2f %5.1f %5.2f %7d %7d %5d %d \"%(A[3][0],A[3][1],A[3][2],A[3][3],A[3][4],A[3][5],A[3][6]) ;\n", + "print \" CCl4 %8.2f %5.1f %5.2f %7d %7d %5d %d \"%(A[4][0],A[4][1],A[4][2],A[4][3],A[4][4],A[4][5],A[4][6]) ;\n", + "\n", + "print \"Even though it is non polar , CCl4 exhibit the largest intermolecular forces . It is due to the large polarizability accociated with the four Cl atom in CCl4 .\"\n", + "\n", + "# Note : Answer are slightly different because of rounding off error." + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "[[1.85, 14.8, 12.62, 0, 0, 0, 0], [1.47, 22.2, 10.07, 0, 0, 0, 0], [0.0, 26.0, 12.61, 0, 0, 0, 0], [1.87, 45.3, 11.26, 0, 0, 0, 0], [0.0, 105.0, 11.47, 0, 0, 0, 0]]\n", + " Molecule M alp*10**25 I C*10**60 Cd_d Cind Cdis\n", + " H2O 1.85 14.8 12.62 235 190 11 34 \n", + " NH3 1.47 22.2 10.07 146 76 10 60 \n", + " CH4 0.00 26.0 12.61 103 0 0 103 \n", + " CH3Cl 1.87 45.3 11.26 509 199 32 278 \n", + " CCl4 0.00 105.0 11.47 1518 0 0 1518 \n", + "Even though it is non polar , CCl4 exhibit the largest intermolecular forces . It is due to the large polarizability accociated with the four Cl atom in CCl4 .\n" + ] + } + ], + "prompt_number": 41 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 4.2 Page No : 222" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Variables\n", + "import math\n", + "C6_Ar_HCl_tab = 76 * 10**-60 ;\t\t\t#From table E4.2\n", + "C6_Ar_Ar_tab = 52 * 10**-60 ;\t\t\t#From table E4.2\n", + "C6_HCl_HCl_tab = 134 * 10**-60 ;\t\t\t#From table E4.2\n", + "\n", + "# Calculations\n", + "C6_Ar_HCl_gmean = math.sqrt(C6_Ar_Ar_tab * C6_HCl_HCl_tab) ; \t\t\t#[erg/cm**6]\n", + "x = (C6_Ar_HCl_gmean - C6_Ar_HCl_tab) / C6_Ar_HCl_tab * 100 ;\n", + "\n", + "# Results\n", + "print \"The geometric mean is different from that in table E4.2 by %d %%\"%(x)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The geometric mean is different from that in table E4.2 by 9 %\n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 4.4 Page No : 230" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Variables\n", + "import math\n", + "Psat_wat_25 = 3.169 * 10**3 ;\t\t\t# From steam table\n", + "Psat_wat_50 = 1.235 * 10**4 ;\t\t\t# From steam table\n", + "Psat_wat_100 = 1.014 * 10**5 ;\t\t\t# From steam table\n", + "A =11.9673 ;\n", + "B = 3626.55 ;\n", + "C = -34.29 ;\n", + "T1 = 25 ; \t\t\t#[*C]\n", + "T2 = 50 ; \t\t\t#[*C]\n", + "T3 = 100 ; \t\t\t#[*C]\n", + "\n", + "# Calculations\n", + "M = [[T1, Psat_wat_25 , 0],[T2 , Psat_wat_50, 0],[T3 , Psat_wat_100, 0]]\n", + "#M = array(M)\n", + "\n", + "print (\" T(*C) Water(Pa) Methanol(Pa)\")\n", + "for i in range(3):\n", + " M[i][2] = math.exp(A - B / (M[i][0] + 273 + C)) * 10**5 ;\n", + " print \"%5d %7.3e %7.2e\"%(M[i][0],M[i][1],M[i][2])\n", + " #t.append(math.exp(A - B / (M[i][0] + 273 + C)) * 10**5)\n", + "#M.append(t)\n", + "\n", + "# Results\n", + "\n", + "print \"1) Water can form two hydrogen bonds . While CH4Oh can form only one . Thus at\\\n", + " a given temperature, water has stronger attractive forces in the liquid and a lower vapour pressure .\"\n", + "print \"2) Since the Maxwell-Boltzmann distribution depends exponentially on temperature,\\\n", + " Psat also increses exponentially with temperature .\"\n", + "\n", + "# Note: Answers may vary because of rounding off error." + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " T(*C) Water(Pa) Methanol(Pa)\n", + " 25 3.169e+03 1.68e+04\n", + " 50 1.235e+04 5.52e+04\n", + " 100 1.014e+05 3.53e+05\n", + "1) Water can form two hydrogen bonds . While CH4Oh can form only one . Thus at a given temperature, water has stronger attractive forces in the liquid and a lower vapour pressure .\n", + "2) Since the Maxwell-Boltzmann distribution depends exponentially on temperature, Psat also increses exponentially with temperature .\n" + ] + } + ], + "prompt_number": 27 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 4.6 Page No : 236" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Variables\n", + "Pc_B = 49.1 ; \t\t\t# [bar] , From table\n", + "Pc_T = 42.0 ; \t\t\t# [bar] , From table\n", + "Pc_C = 40.4 ; \t\t\t# [bar] , From table\n", + "Tc_B = 562 ; \t\t\t# [K] , From table\n", + "Tc_T = 594 ; \t\t\t# [K] , From table \n", + "Tc_C = 553 ; \t\t\t# [K] , From table\n", + "R = 8.314 ;\n", + "\n", + "# Calculations\n", + "A = [[Pc_B , Tc_B, 0,0],[Pc_T , Tc_T,0,0],[Pc_C , Tc_C,0,0]]\n", + "\n", + "# Results\n", + "print \" P_c T_c a b \"\n", + "for i in range(3):\n", + " A[i][2] = 27./64 * (R * A[i][1])**2 /( A[i][0] * 10**5) ;\n", + " A[i][3] = R * A[i][1] / (8 * A[i][0] * 10**5) ;\n", + " print \" %5.1f %5d %7.2f %7.2e\"%(A[i][0],A[i][1],A[i][2],A[i][3])\n", + "\n", + "print \"The attractive interactions of all three compounds are dominated by\\\n", + " print ersion interactions ( parameter a) , while size affects parameter b .\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " P_c T_c a b \n", + " 49.1 562 1.88 1.19e-04\n", + " 42.0 594 2.45 1.47e-04\n", + " 40.4 553 2.21 1.42e-04\n", + "The attractive interactions of all three compounds are dominated by print ersion interactions ( parameter a) , while size affects parameter b .\n" + ] + } + ], + "prompt_number": 32 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 4.11 Page No : 246" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Variables\n", + "import math\n", + "B = 0.0486 * 10**-3 ;\n", + "T1 = 20 + 273 ; \t\t\t#[K]\n", + "T2 = 500 + 273 ; \t\t\t#[K]\n", + "v1 = 7.11 ; \t\t\t# [cm**3/mol]\n", + "\n", + "# Calculations\n", + "v2 = v1 * math.exp( B * (T2 - T1)) ;\n", + "\n", + "# Results\n", + "print \" Molar volume of solid state 2 = %.2f cm**3/mol\"%( v2);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " Molar volume of solid state 2 = 7.28 cm**3/mol\n" + ] + } + ], + "prompt_number": 5 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 4.12 Page No : 248" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Variables\n", + "P_c = 37.9 * 10**5 ;\t\t\t#[N/m**2] , From compressibility chart\n", + "T_c = 425.2 \t\t\t# [K , From compressibility chart\n", + "P = 50. * 10**5 ; \t \t\t#N/m**2]\n", + "T = 333.2 ;\t\t\t #[K]\n", + "R = 8.314 ;\n", + "z_0 = 0.2148 ; \t\t\t\n", + "z_1 = -0.0855 ; \t\t\n", + "w = 0.199 ;\n", + "m = 10. ;\n", + "MW = 0.05812 ;\n", + "\n", + "# Calculations\n", + "a = (0.42748 * R**2 * T_c**2.5) / P_c ;\n", + "b = 0.08664 * R * T_c / P_c ;\n", + "A = P * T**(1./2) ;\n", + "B = -R * T**(3./2) ;\n", + "C = (a - P * T**(1./2) * b**2 - R * T**(3./2)*b) ;\n", + "D = - a * b;\n", + "\n", + "#mycoeff = [ D , C , B , A] ;\n", + "mycoeff = [ A,B,C,D]\n", + "#p = poly1d(mycoeff , \"v\" , \"coeff\" ); \n", + "M = roots(mycoeff);\n", + "\n", + "# Results\n", + "for i in range(3):\n", + " ans = sign(M[i])\n", + " if ans == 1:\n", + " V = m / MW *(M[i]) ;\n", + " #print \"Using Redlich Kwong equation the volume is = %.3f m**3\"%(V)\n", + "\n", + "z = z_0 + w * z_1 ; \n", + "v = z * R * T / P ;\n", + "V = m / MW * v ;\n", + "print \"Using compressibility chart the volume is = %.3f m**3\"%(V)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Using compressibility chart the volume is = 0.019 m**3\n" + ] + } + ], + "prompt_number": 8 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 4.13 Page No : 253" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Variables\n", + "import math\n", + "T = 100. + 273 ; \t\t\t#[K]\n", + "P = 70. * 10**5 ; \t\t\t#[N/m**2]\n", + "P_c = 42.2 * 10 ** 5 ;\n", + "T_c = 370. ; \t\t\t#[K]\n", + "w = 0.153 \t \t\t# Interpolating from table C.1 and C.2\n", + "z_0 = 0.2822 ;\t\t \t# Interpolating from table C.1 and C.2\n", + "z_1 = - 0.0670 ;\t\t\t# Interpolating from table C.1 and C.2\n", + "m = 20. * 10**3 ;\t\t\t#[g]\n", + "MW = 44. ; \t\t \t#[g/mol]\n", + "R = 8.314 ;\n", + "\n", + "# Calculations and Results\n", + "P_r = P / P_c ;\n", + "T_r = T / T_c ;\n", + "z = z_0 + w * z_1 ;\n", + "V = m / MW *z * R * T / P ;\n", + "\n", + "print \"1) Volume = %.4f m**3 \"%( V )\n", + "\n", + "\n", + "T = 295. ;\t\t\t#[K]\n", + "n = 50. ; \t\t\t# [mol]\n", + "a = 0.42748 * R**2 * T_c**2.5 / P_c ;\n", + "b = 0.08664 * R * T_c / P_c ;\n", + "v = 0.1 ;\n", + "P = R * T / (v - b) - a / (T**0.5 * v * (v + b)) ;\n", + "x = P * n * 10**-6 ;\n", + "\n", + "print \"2) Pressure = %.1f MPa \"%( x )\n", + "\n", + "y1 = 0.4 ;\n", + "y2 = 1 - y1 ;\n", + "n = 50. ;\n", + "P_c = 48.7 * 10**5 ;\t\t\t#[N/m**2]\n", + "T_c = 305.5 ; \t\t\t#[K]\n", + "a1 = a ;\n", + "b1 = b ;\n", + "a2 = 0.42748 * R**2 * T_c**2.5 / P_c ;\n", + "b2 = 0.08664 * R * T_c / P_c ;\n", + "\n", + "a_mix = y1**2 * a1 + 2 * y1 * y2 * math.sqrt(a1 * a2) + y2**2 * a2 ;\n", + "b_mix = y1 * b1 + y2 * b2 ;\n", + "P = R * T / (v - b_mix) - a_mix /(T**0.5 * v * (v + b_mix));\n", + "x = P * n * 10**-6 ;\n", + "\n", + "print \"3) Pressure = %.2f MPa \"%( x )\n", + "\n", + "# Note : Answers are slightly different because of rounding off error." + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "1) Volume = 0.0548 m**3 \n", + "2) Pressure = 1.2 MPa \n", + "3) Pressure = 1.22 MPa \n" + ] + } + ], + "prompt_number": 4 + } + ], + "metadata": {} + } + ] +}
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