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diff --git a/Thermodynamics:_From_concepts_to_applications/Chapter17_2.ipynb b/Thermodynamics:_From_concepts_to_applications/Chapter17_2.ipynb new file mode 100755 index 00000000..b7dd1d21 --- /dev/null +++ b/Thermodynamics:_From_concepts_to_applications/Chapter17_2.ipynb @@ -0,0 +1,312 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:33e389243f3edbfa798e02801dff08e24da71a0ad10959698b7564f561b867d4"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter17-Ideal solutions"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex1-pg 480"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#intilization variables\n",
+ "x=0.25\n",
+ "pa=40\n",
+ "pb=50\n",
+ "ya=0.25\n",
+ "alpha=1.25\n",
+ "#calculation\n",
+ "P=x *pa+(1-x)*pb\n",
+ "y=x*pa/P\n",
+ "yb=(1-y)\n",
+ "xa=alpha*y/(1+(alpha-1)*y)\n",
+ "xb=(1-x)\n",
+ "#results\n",
+ "print'%s %.2f %s'%('total pressure of an ideal solution',P,'kpa')\n",
+ "print'%s %.2f %s %.2f %s '%('composition of the gaseous phase',y,'' and ' ',yb,'')\n",
+ "print'%s %.2f %s %.2f %s '%('the composition of last drop',xa,' ' and ' ',xb,'') "
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "total pressure of an ideal solution 47.50 kpa\n",
+ "composition of the gaseous phase 0.21 0.79 \n",
+ "the composition of last drop 0.25 0.75 \n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example2-pg484"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "##initialisation of variables\n",
+ "T= 290 ##K\n",
+ "xa= 0.4\n",
+ "xb= 0.6\n",
+ "P= 600 ##kPa\n",
+ "V= 60 ##L\n",
+ "R= 8.314 ##J/mol K\n",
+ "Mp= 44 ##kg/kmol\n",
+ "Mb= 58.12 ##kg/kmol\n",
+ "vp= 0.00171 ##m**3/kg\n",
+ "vb= 0.00166 ##m**3/kg\n",
+ "na= 0.1 ##kmol\n",
+ "nb= 0.15 ##kmol\n",
+ "V1= 0.04000 ##m**3\n",
+ "xa= 0.4 \n",
+ "np= 2\n",
+ "Vc= 0.1 ##m**3\n",
+ "##CALCULATIONS\n",
+ "Pasat= math.e**(14.435-(2255/T))\n",
+ "Pbsat= math.e**(14.795-(2770/T))\n",
+ "P1= xa*Pasat+xb*Pbsat\n",
+ "Na1= P*V/(100*R*T)\n",
+ "Vp= vp*Mp\n",
+ "Vb= vb*Mb\n",
+ "V= na*Vp+nb*Vb\n",
+ "Vv= V1-V\n",
+ "nv= P1*Vv/(R*T)\n",
+ "ya= xa*Pasat/P\n",
+ "yb=1-ya\n",
+ "Na= na+ya*nv\n",
+ "Nb= nb+yb*nv\n",
+ "##RESULTS\n",
+ "print'%s %.2f %s'% (' initial pressure= ',P1,' kPa')\n",
+ "print'%s %.2f %s'% (' moles of propane= ',Na1,' kmol')\n",
+ "print'%s %.2f %s'% (' initial mole of propane= ',Na,' kmol')\n",
+ "print'%s %.2f %s'% (' initial mole of butane= ',Nb,' kmol')\n",
+ "print'%s %.2f %s'% (' numbar of phases= ',np,'')\n",
+ "print'%s %.2f %s'% (' volume in final state=',Vc,' m^3')"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ " initial pressure= 874.89 kPa\n",
+ " moles of propane= 0.15 kmol\n",
+ " initial mole of propane= 0.11 kmol\n",
+ " initial mole of butane= 0.15 kmol\n",
+ " numbar of phases= 2.00 \n",
+ " volume in final state= 0.10 m^3\n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example3-pg489"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\t\n",
+ "#calculate pressure of the phase of pure A\n",
+ "##initialisation of variables\n",
+ "p0= 10. ##Mpa\n",
+ "R= 8.314 ##J/mol K\n",
+ "T= 30. ##C\n",
+ "va= 0.02 ##m^3/kmol\n",
+ "xa= 0.98\n",
+ "##CALCULATIONS\n",
+ "p= p0+(R*(273.15+T)*math.log(xa)/(va*1000.))\n",
+ "##RESULTS\n",
+ "print'%s %.2f %s'%('Pressure of the phase of pure A=',p,'Mpa')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Pressure of the phase of pure A= 7.45 Mpa\n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example4-pg491"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#calculate the boiling point elevation\n",
+ "##initialisation of variables\n",
+ "hfg= 2257.0 ##kJ/kg\n",
+ "Tb= 100 ##C\n",
+ "R= 8.314 ##J/mol K\n",
+ "m2= 10 ##gms\n",
+ "M2= 58.5 ##gms\n",
+ "m1= 90. ##gms\n",
+ "M1= 18. ##gms\n",
+ "##CALCULATIONS\n",
+ "x2= (m2/M2)/((m2/M2)+(m1/M1))\n",
+ "dT= R*math.pow(273.15+Tb,2)*x2/(M1*hfg)\n",
+ "##RESULTS\n",
+ "print'%s %.3f %s'%(' Boiling point elevation=',dT,'C')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ " Boiling point elevation= 0.942 C\n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example5-pg494"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#calculate Osomatic pressures\n",
+ "##initialisation of variables\n",
+ "M1= 18.02 ##gms\n",
+ "m1= 0.965 ##gms\n",
+ "m2= 0.035 ##gms\n",
+ "M2= 58.5 ##gms\n",
+ "R= 8.314 ##J/mol K\n",
+ "M= 18.02 ##kg\n",
+ "T= 20. ##C\n",
+ "vf= 0.001002 ##m^3\n",
+ "x21= 0.021856 ##m^3\n",
+ "##CALCULATIONS\n",
+ "n1= m1/M1\n",
+ "n2= m2/M2\n",
+ "x1= n1/(n1+n2)\n",
+ "x2= n2/(n2+n1)\n",
+ "P= R*(273.15+T)*x2/(M*vf)\n",
+ "P1= R*(273.15+T)*x21/(M*vf)\n",
+ "##RESULTS\n",
+ "print'%s %.1f %s'%('Osmotic pressure=',P,'kpa')\n",
+ "print'%s %.1f %s'%('Osmotic pressure=',P1,'kpa')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Osmotic pressure= 1491.4 kpa\n",
+ "Osmotic pressure= 2950.2 kpa\n"
+ ]
+ }
+ ],
+ "prompt_number": 5
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example6-pg495"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#what is useful work in the process and heat interaction and maximum work and irreversibility\n",
+ "##initialisation of variables\n",
+ "W= 0.\n",
+ "Q= 0.\n",
+ "R= 8.314 ##J/mol K\n",
+ "T0= 300. ##K\n",
+ "x= 5./13.\n",
+ "n1= 0.5 ##kmol/s\n",
+ "n2= 0.8 ##kmol/s\n",
+ "##CALCULATIONS\n",
+ "W1= (n1+n2)*R*T0*(x*math.log(1/x)+(1-x)*math.log(1./(1.-x)))+470\n",
+ "I= W1\n",
+ "##RESULTS\n",
+ "print'%s %.f %s'%('useful work of the process=',W,'kW') \n",
+ "print'%s %.f %s'%('heat interaction=',Q,'kW') \n",
+ "print'%s %.1f %s'%('maximum work=',W1,'kW') \n",
+ "print'%s %.1f %s'%('irreversibility=',I,'kW')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "useful work of the process= 0 kW\n",
+ "heat interaction= 0 kW\n",
+ "maximum work= 2630.4 kW\n",
+ "irreversibility= 2630.4 kW\n"
+ ]
+ }
+ ],
+ "prompt_number": 8
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
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