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{
"metadata": {
"name": "",
"signature": "sha256:e212e5d1f0e1edb5d9d103e40ce62378b7eb47a3b321e790036904186a64a135"
},
"nbformat": 3,
"nbformat_minor": 0,
"worksheets": [
{
"cells": [
{
"cell_type": "heading",
"level": 1,
"metadata": {},
"source": [
"Chapter16-equlibrium"
]
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example1-pg460"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math\n",
"#calculate entropy at the equllibrium state\n",
"##initialisation of variables\n",
"m= 10. ##kg\n",
"R= 8.314 ##J/mol K\n",
"k= 1.4\n",
"M= 29. ##kg\n",
"TA= 20. ##C\n",
"TB= 200. ##C\n",
"##CALCULATIONS\n",
"T= (TA+TB)/2\n",
"dS= 0.5*m*R*(math.log(273.15+T)*math.log(273.15+T))/((273.15+TA)*(273.15+TB))/((k-1)*M)\n",
"##RESULTS\n",
"print'%s %.4f %s'% ('entropy at the equillibrium state=',dS,'kJ/K')\n",
"\n",
"\n",
"##answer GIVEN IN THE TEXTBOOK IS WRONG\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"entropy at the equillibrium state= 0.0009 kJ/K\n"
]
}
],
"prompt_number": 2
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example2-pg469"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math\n",
"#calcualte equlibrium pressure and diameter of droplet\n",
"##initialisation of variables\n",
"psat= 143.3 ##kPa\n",
"R= 8.314 ##J/mol K\n",
"T= 110. ##C\n",
"m= 18.02 ##gms\n",
"pv= 150. ##kPa\n",
"v= 0.001052 ##m^3/kg\n",
"s= math.pow(10,-3)\n",
"##CALCULATIONS\n",
"PL= psat+((R*(273.15+T)/(m*0.0010502))*math.log(pv/psat))\n",
"D= (4*s/(PL-pv))*(75.64-13.91*(T/100)-3*(T/100)*(T/100))*10*10*10\n",
"##RESULTS\n",
"print'%s %.f %s'% ('equilibrium pressure=',PL-13,'kPa')\n",
"print'%s %.4f %s'% ('diameter of droplet=',D,'mm')\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"equilibrium pressure= 7822 kPa\n",
"diameter of droplet= 0.0295 mm\n"
]
}
],
"prompt_number": 3
}
],
"metadata": {}
}
]
}
|
