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{
"metadata": {
"name": "",
"signature": "sha256:0bd1584504a5182c99f46d576a77cfaa07f83a047faf3b55eaafd0cea74518f2"
},
"nbformat": 3,
"nbformat_minor": 0,
"worksheets": [
{
"cells": [
{
"cell_type": "heading",
"level": 1,
"metadata": {},
"source": [
"Chapter17-Ideal solutions"
]
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example1-pg367"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math\n",
"#calculate Total pressure and composition of vapour phases and composition of last drop liquids\n",
"##initialisation of variables\n",
"Pa= 40. ##kPa\n",
"Pb= 50. ##kPa\n",
"na= 2. ##moles\n",
"nb= 6. ##moles\n",
"##CALCULATIONS\n",
"a= Pb/Pa\n",
"xa= na/(na+nb)\n",
"xb= 1.-xa\n",
"p= xa*Pa+xb*Pb\n",
"y= (xa*Pa)/p\n",
"ya= 1.-y\n",
"Xa= a*xa/(1+(a-1)*xa)\n",
"Xb= 1.-Xa\n",
"##RESULTS\n",
"print'%s %.1f %s'%('Total pressure=',p,'kPa')\n",
"print'%s %.4f %s'%('composition of vapour phase=',y,'')\n",
"print'%s %.4f %s'%('composition of vapour phase=',ya,'')\n",
"print'%s %.4f %s'%('composition of last drop of liquid=',Xa,'')\n",
"print'%s %.4f %s'%('composition of last drop of liquid=',Xb,'')\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Total pressure= 47.5 kPa\n",
"composition of vapour phase= 0.2105 \n",
"composition of vapour phase= 0.7895 \n",
"composition of last drop of liquid= 0.2941 \n",
"composition of last drop of liquid= 0.7059 \n"
]
}
],
"prompt_number": 1
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example2-pg371"
]
},
{
"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": [
"Example3-pg373"
]
},
{
"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": [
"Example4-pg376"
]
},
{
"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": [
"Example5-pg377"
]
},
{
"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)))\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= 2160.4 kW\n",
"irreversibility= 2160.4 kW\n"
]
}
],
"prompt_number": 7
}
],
"metadata": {}
}
]
}
|
