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	   "source": [
       "# Chapter 2: Properties of Pure Substances"
	   ]
	},
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 2.1: saturated_water_is_vaporized.sce"
		   ]
		  },
  {
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"clc\n",
"//solution\n",
"//    initialization of variables\n",
"m=10;  // mass of saturated water in kg\n",
" // All the necessary values are taken from table C.2\n",
" // part (a)\n",
" \n",
"P=0.001; // Pressure in MPa\n",
"vf=0.001; // specific volume of saturated liquid at 0.001 Mpa in Kg/m^3\n",
"vg=129.2;//  specific volume of saturated vapour at 0.001 Mpa in Kg/m^3\n",
"deltaV=m*(vg-vf)//properties of pure substance \n",
"printf('The Volume change at pressure '+string(P)+' MPa is %.0f m^3 \n',deltaV)\n",
"\n",
"// part (b) \n",
"P=0.26;  // Pressure in MPa\n",
"vf=0.0011; //   specific volume of saturated liquid at 0.26 MPa( it is same from at 0.2 and 0.3 MPa upto 4 decimals)\n",
"vg=(P-0.2)*(0.6058-0.8857)/(0.3-0.2)+0.8857;  // specific volume of saturated vapour by interpolation of Values at 0.2 MPa and 0.3 MPa\n",
"deltaV=m*(vg-vf)\n",
"printf(' The Volume change at pressure '+string(P)+' MPa is %.2f m^3 \n',deltaV)\n",
"\n",
"// part (c) \n",
"P=10;  // Pressure in MPa\n",
"vf=0.00145;  // specific volume of saturated liquid at 10 MPa\n",
"vg=0.01803; //specific volume of saturated vapour at 10 MPa\n",
"deltaV=m*(vg-vf)\n",
"printf(' The Volume change at pressure '+string(P)+' MPa is %.4f m^3',deltaV)"
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 2.2: volume_of_vapour.sce"
		   ]
		  },
  {
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	   "execution_count": null,
	   "metadata": {
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"clc\n",
"//solution\n",
"//    initialization of variables\n",
"m=4// mass of water in kg\n",
"V=1 // volume in m^3\n",
"T=150 // temperature of water in degree centigrade\n",
"\n",
"// TABLE C.1 is used for values in wet region\n",
"// Part (a)\n",
"P=475.8// pressure in KPa in wet region at temperature of 150 *C\n",
"printf('The pressure is %.1f kPa \n',P)\n",
"\n",
"// Part (b)\n",
"// first we determine the dryness fraction\n",
"v=V/m// specific volume of water\n",
"vg=0.3928 //  specific volume of saturated vapour @150 degree celsius\n",
"vf=0.00109 // specific volume of saturated liquid @150 degree celsius\n",
"x=(v-vf)/(vg-vf); //dryness fraction\n",
"mg=m*x; // mass of vapour\n",
"printf(' The mass of vapour present is %.3f kg \n',mg)\n",
"\n",
"// Part(c) \n",
"Vg=mg*vg;// volume of vapour\n",
"printf(' The volume of vapour is %.3f m^3',Vg)"
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 2.3: the_final_volume_of_mixture.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
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"source": [
"clc\n",
"//solution\n",
"//    initialization of variables\n",
"m=2 // mass of water in kg\n",
"P=220 // pressure in KPa\n",
"x=0.8 // quality of steam\n",
"// Table C.2 is used for values\n",
"vg=(P-200)*(0.6058-0.8857)/(300-200)+0.8857 // specific volume of saturated vapour @ given pressure by interpolating\n",
"vf=0.0011 // specific volume of saturated liquid @ 220 KPa\n",
"v=vf+x*(vg-vf)// property of pure substance\n",
"V=m*v // total volume\n",
"printf('The Total volume of the mixture is '+string(V)+' m^3')"
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 2.4: constant_pressure_cylinder.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
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"source": [
"clc\n",
"//solution\n",
"//    initialization of variables\n",
"m=2 // mass of water in kg\n",
"P=2.2 // pressure in Mpa\n",
"T=800 // temperature in degree centigrade\n",
" // Table C.3 is used for values\n",
"v=0.2467+(P-2)*(0.1972-0.2467)/(2.5-2)// specific volue by interpolatin between 2 and 2.5 MPa\n",
"V=m*v // final volume\n",
"printf('The Final Volume is %.3f m^3',V)\n",
"\n",
""
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 2.5: mass_of_air_in_the_tire.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"clc\n",
"//solution\n",
"//    initialization of variables\n",
"V=0.6 // volume of tyre in m^3\n",
"Pgauge=200 // gauge pressure in KPa\n",
"T=20+273 // temperature converted to kelvin\n",
"Patm=100 // atmospheric pressure in KPa\n",
"R=287 // gas constant in Nm/kg.K\n",
"Pabs=(Pgauge+Patm)*1000 // calculating absolute pressue in Pa \n",
"\n",
"m=Pabs*V/(R*T)// mass from ideal gas equation\n",
"printf('The Mass of air is %.2f Kg',m)"
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 2.6: the_van_der_Waals_equation.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
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	   "outputs": [],
"source": [
"clc\n",
"//solution\n",
"//    initialization of variables\n",
"T=500+273 // temperature of steam in kelvin\n",
"rho=24 // density in Kg/m^3\n",
"R=0.462 // gas constant from Table B.2\n",
"v=1/rho // specific volume and density relation\n",
"// PART (a)\n",
"P=rho*R*T // from Ideal gas equation\n",
"printf('PART (a) The Pressure is '+string(P)+' KPa \n')\n",
"// answer is approximated in textbook\n",
"\n",
"// PART (b)\n",
"a=1.703 //  van der Waal's constant a value from Table B.7\n",
"b=0.00169 // van der Waal's constant b value from Table B.7\n",
"P=(R*T/(v-b))-(a/v**2) // Pressure from van der Waal's equation\n",
"printf(' PART (b) The Pressure is '+string(P)+' KPa \n')\n",
"// answer is approximated in textbook\n",
"\n",
"// PART (c)\n",
"a=43.9 //  van der Waal's constant a value from Table B.7\n",
"b=0.00117 // van der Waal's constant b value from Table B.7\n",
"\n",
"P=(R*T/(v-b))-(a/(v*(v+b)*sqrt(T)))// Redlich-Kwong equation\n",
"printf(' PART (c) The Pressure is '+string(P)+' KPa \n')\n",
"// answer is approximated in textbook\n",
"\n",
"// PART (d)\n",
"Tcr=947.4 // compressibilty temperature from table B.3\n",
"Pcr=22100 // compressibility pressure from table B.3\n",
"\n",
"TR=T/Tcr // reduced temperature\n",
"PR=P/Pcr // reduced pressure\n",
"Z=0.93 // from compressiblility chart\n",
"P=Z*R*T/v // Pressure in KPa\n",
"printf(' PART (d) The Pressure is '+string(P)+' KPa \n')\n",
"// answer is approximated in textbook\n",
"\n",
"// PART (e)\n",
"P=8000 // pressure from steam table @ 500*c and v= 0.0417 m^3\n",
"printf(' PART (e) The Pressure is '+string(P)+' KPa \n')\n",
"// answer is approximated in textbook"
   ]
   }
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