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+{
+"cells": [
+ {
+		   "cell_type": "markdown",
+	   "metadata": {},
+	   "source": [
+       "# Chapter 4: Steam and two phase systems"
+	   ]
+	},
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 4.10: mass_of_steam_and_water.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"clear;\n",
+"clc;\n",
+"disp('Example .10');\n",
+"\n",
+"// aim : To determine\n",
+"// (a) the mass of steam entering the heater\n",
+"// (b) the mass of water entering the heater\n",
+"\n",
+"//  Given values\n",
+"x = .95;//  Dryness fraction\n",
+"P = .7;//  pressure,[MN/m^2]\n",
+"d = 25;//  internal diameter of heater,[mm]\n",
+"C = 12; //  steam velocity in the pipe,[m/s]\n",
+"\n",
+"//  solution\n",
+"//  from steam table at .7 MN/m^2 pressure\n",
+"hf = 697.1;//  [kJ/kg]\n",
+"hfg = 2064.9;//  [kJ/kg]\n",
+"hg = 2762.0; //  [kJ/kg]\n",
+"vg = .273; //  [m^3/kg]\n",
+"\n",
+"//  (a)\n",
+"v = x*vg; //  [m^3/kg]\n",
+"ms_dot = %pi*(d*10^-3)^2*C*3600/(4*v);//  mass of steam entering, [kg/h]\n",
+"mprintf('\n (a) The mass of steam entering the heater is  =  %f kg/h \n',ms_dot);\n",
+"\n",
+"//  (b)\n",
+"h = hf+x*hfg;//  specific enthalpy of steam entering heater,[kJ/kg]\n",
+"//  again from steam tables\n",
+"hf1 = 376.8;//  [kJ/kg] at 90 C\n",
+"hf2 = 79.8;//  [kJ/kg] at 19 C\n",
+"\n",
+"//  using energy balance,mw_dot*(hf1-hf2)=ms_dot*(h-hf1)\n",
+"mw_dot = ms_dot*(h-hf1)/(hf1-hf2);//  mass of water entering to heater,[kg/h]\n",
+"\n",
+"mprintf('\n (b) The mass of water entering the heater is  =   %f  kg/h \n',mw_dot);\n",
+"\n",
+"//  End"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 4.11: change_of_internal_energy.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"clear;\n",
+"clc;\n",
+"disp('Example 4.11');\n",
+"\n",
+"// aim: To determine\n",
+"//  the change of internal energy\n",
+"\n",
+"//  Given values\n",
+"m = 1.5;//  mass of steam,[kg]\n",
+"P1 = 1;//  initial pressure, [MN/m^2]\n",
+"t = 225;//  temperature, [C]\n",
+"P2 = .28;//  final pressure, [MN/m^2]\n",
+"x = .9;//  dryness fraction of steam at P2\n",
+"\n",
+"//  solution\n",
+"\n",
+"// from steam table at P1\n",
+"h1 = 2886;//  [kJ/kg]\n",
+"v1 = .2198; //  [m^3/kg]\n",
+"//  hence\n",
+"u1 = h1-P1*v1*10^3;// internal energy [kJ/kg]\n",
+"\n",
+"// at P2\n",
+"hf2 = 551.4;// [kJ/kg]\n",
+"hfg2 = 2170.1;// [kJ/kg]\n",
+"vg2 = .646; //  [m^3/kg]\n",
+"//  so\n",
+"h2 = hf2+x*hfg2;//  [kj/kg]\n",
+"v2 = x*vg2;//   [m^3/kg]\n",
+"\n",
+"// now\n",
+"u2 = h2-P2*v2*10^3;//  [kJ/kg]\n",
+"\n",
+"//  hence change in specific internal energy is\n",
+"del_u = u2-u1;//  [kJ/kg]\n",
+"\n",
+"del_u = m*del_u;//  [kJ];\n",
+"mprintf('\n The change in internal energy is  =  %f  kJ \n',del_u);\n",
+"\n",
+"//  End"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 4.12: dryness_fraction.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"clear;\n",
+"clc;\n",
+"disp('Example 4.12');\n",
+"\n",
+"//  aim : To determine \n",
+"//  the dryness fraction of steam after throttling\n",
+"\n",
+"//  given values\n",
+"P1 = 1.4;//  pressure before throttling, [MN/m^2]\n",
+"x1 = .7;//  dryness fraction before throttling\n",
+"P2 = .11;//  pressure after throttling, [MN/m^2]\n",
+"\n",
+"//  solution\n",
+"//  from steam table\n",
+"hf1 = 830.1;//  [kJ/kg]\n",
+"hfg1 = 1957.7;//  [kJ/kg]\n",
+"h1 = hf1 + x1*hfg1; //  [kJ/kg]\n",
+"\n",
+"hf2 = 428.8;//  [kJ/kg]\n",
+"hfg2 = 2250.8;//  [kJ/kg]\n",
+"\n",
+"//  now for throttling,\n",
+"//  hf1+x1*hfg1=hf2+x2*hfg2; where x2 is dryness fraction after throttling\n",
+"\n",
+"x2=(h1-hf2)/hfg2; // final dryness fraction\n",
+"\n",
+"mprintf('\n Dryness fraction of steam after throttling is  =  %f \n',x2);\n",
+"\n",
+"//  End"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 4.13: condition_of_steam_and_internal_diameter.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"clear;\n",
+"clc;\n",
+"disp('Example 4.13');\n",
+"\n",
+"//  aim : To determine \n",
+"//  the dryness fraction of steam \n",
+"//  and the internal diameter of the pipe\n",
+"\n",
+"//  Given values\n",
+"\n",
+"//  steam1\n",
+"P1 = 2;// pressure before throttling, [MN/m^2]\n",
+"t = 300;//  temperature,[C]\n",
+"ms1_dot = 2;// steam flow rate, [kg/s]\n",
+"P2 = 800;//  pressure after throttling, [kN/m^2]\n",
+"\n",
+"//  steam2\n",
+"P = 800;// pressure, [N/m^2]\n",
+"x2 = .9;//  dryness fraction\n",
+"ms2_dot = 5; //  [kg/s]\n",
+"\n",
+"//  solution\n",
+"//  (a)\n",
+"//  from steam table specific enthalpy of steam1 before throttling is\n",
+"hf1 = 3025;//  [kJ/kg]\n",
+"//  for throttling process specific enthalpy will same so final specific enthalpy of steam1 is\n",
+"hf2 = hf1;\n",
+"// hence\n",
+"h1 = ms1_dot*hf2;// [kJ/s]\n",
+"\n",
+"//  calculation of specific enthalpy of steam2\n",
+"hf2 = 720.9;//  [kJ/kg]\n",
+"hfg2 = 2046.5;//  [kJ/kg]\n",
+"//  hence\n",
+"h2 = hf2+x2*hfg2;//  specific enthalpy, [kJ/kg]\n",
+"h2 = ms2_dot*h2;//  total enthalpy, [kJ/s]\n",
+"\n",
+"//  after mixing\n",
+"m_dot = ms1_dot+ms2_dot;//  total mass of mixture,[kg/s]\n",
+"h = h1+h2;//  Total enthalpy of the mixture,[kJ/s]\n",
+"h = h/7;//  [kJ/kg]\n",
+"\n",
+"//  At pressure 800 N/m^2 \n",
+"hf = 720.9;//  [kJ/kg]\n",
+"hfg = 2046.5;//  [kJ/kg]\n",
+"//  so total enthalpy is,hf+x*hfg, where x is dryness fraction of mixture and which is equal to h\n",
+"//  hence\n",
+"x = (h-hf)/hfg;// dryness fraction after mixing\n",
+"mprintf('\n (a) The condition of the resulting mixture is dry with dryness fraction  =  %f \n',x);\n",
+"\n",
+"//  (b)\n",
+"// Given\n",
+"C = 15;//  velocity, [m/s]\n",
+"//  from steam table\n",
+"v = .1255;//  [m^/kg]\n",
+"A = ms1_dot*v/C;//  area, [m^2]\n",
+"//  using ms1_dot = A*C/v, where A is cross section area in m^2 and\n",
+"//  A = %pi*d^2/4, where d is diameter of the pipe \n",
+"\n",
+"//  calculation of d\n",
+"d = sqrt(4*A/%pi); // diameter, [m]\n",
+"\n",
+"mprintf('\n (b) The internal diameter of the pipe is  =  %f mm \n',d*1000);\n",
+"\n",
+"//  End"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 4.14: dryness_fraction.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"clear;\n",
+"clc;\n",
+"disp('Example 4.14');\n",
+"\n",
+"//  aim : To estimate \n",
+"//  the dryness fraction\n",
+"\n",
+"//  Given values\n",
+"M = 1.8;//  mass of condensate, [kg]\n",
+"m = .2;//  water collected, [kg]\n",
+"\n",
+"//  solution\n",
+"x = M/(M+m);//  formula for calculation of dryness fraction using seprating calorimeter\n",
+"\n",
+"mprintf(' \n The dryness fraction of the steam entering seprating calorimeter is  =  %f \n',x);\n",
+"\n",
+"//  End"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 4.15: dryness_fraction.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"clear;\n",
+"clc;\n",
+"disp('Example 4.15');\n",
+"\n",
+"//  aim : To determine\n",
+"//  the dryness fraction of the steam at 2.2 MN/m^2\n",
+"\n",
+"//  Given values\n",
+"P1 = 2.2;//  [MN/m^2]\n",
+"P2 = .13;//  [MN/m^2]\n",
+"t2 = 112;//  [C]\n",
+"tf2 = 150;//  temperature, [C]\n",
+"\n",
+"// solution\n",
+"// from steam table, at 2.2 MN/m^2\n",
+"//  saturated steam at 2 MN/m^2 Pressure\n",
+"hf1 = 931;//  [kJ/kg]\n",
+"hfg1 = 1870;//  [kJ/kg]\n",
+"hg1 = 2801;//  [kJ/kg]\n",
+"\n",
+"// for superheated steam\n",
+"//  at .1 MN/m^2\n",
+"hg2 = 2675;//  [kJ/kg]\n",
+"hg2_150 = 2777;// specific enthalpy at 150 C, [kJ/kg]\n",
+"tf2 = 99.6;//  saturation temperature, [C]\n",
+"\n",
+"// at .5 MN/m^2\n",
+"hg3 = 2693;//  [kJ/kg]\n",
+"hg3_150 = 2773;// specific enthalpy at 150 C, [kJ/kg]\n",
+"tf3 = 111.4;//  saturation temperature, [C]\n",
+"\n",
+"Table_P_h1 = [[.1,.5];[hg2,hg3]];// where, P in MN/m^2 and h in [kJ/kg]\n",
+"hg = interpln(Table_P_h1,.13);//  specific entahlpy at .13 MN/m^2, [kJ/kg]\n",
+"\n",
+"Table_P_h2 = [[.1,.5];[hg2_150,hg3_150]];//  where, P in MN/m^2 and h in [kJ/kg]\n",
+"hg_150 = interpln(Table_P_h2,.13);//  specific entahlpy at .13 MN/m^2 and 150 C, [kJ/kg]\n",
+"\n",
+"Table_P_tf = [[.1,.5];[tf2,tf3]];// where, P in MN/m^2 and h in [kJ/kg]\n",
+"tf = interpln(Table_P_tf,.13);//  saturation temperature, [C]\n",
+"\n",
+"//  hence\n",
+"h2 = hg+(hg_150-hg)/(t2-tf)/(tf2-tf);//  specific enthalpy at .13 MN/m^2 and 112 C, [kJ/kg]\n",
+"\n",
+"// now since process is throttling so h2=h1\n",
+"// and h1 = hf1+x1*hfg1, so\n",
+"x1 = (h2-hf1)/hfg1;// dryness fraction\n",
+"mprintf(' \n The dryness fraction of steam is  =  %f \n',x1);\n",
+"\n",
+"// There is a calculation mistake in book so answer is not matching\n",
+"\n",
+"//  End"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 4.16: minimum_dryness_fraction.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"clear;\n",
+"clc;\n",
+"disp('Example 4.16');\n",
+"\n",
+"//  aim : To determine \n",
+"//  the minimum dryness fraction of steam\n",
+"\n",
+"//  Given values\n",
+"P1 = 1.8;//  testing pressure,[MN/m^2]\n",
+"P2 = .11;//  pressure after throttling,[MN/m^2]\n",
+"\n",
+"//  solution\n",
+"//  from steam table\n",
+"//  at .11 MN/m^2 steam is completely dry and specific enthalpy is\n",
+"hg = 2680;//  [kJ/kg]\n",
+"\n",
+"//  before throttling steam is wet, so specific enthalpy is=hf+x*hfg, where x is dryness fraction\n",
+"//  from steam table\n",
+"hf = 885;//  [kJ/kg]\n",
+"hfg = 1912;//  [kJ/kg]\n",
+"\n",
+"//  now for throttling process,specific enthalpy will same before and after\n",
+"//  hence\n",
+"x = (hg-hf)/hfg;\n",
+"mprintf('\n The minimum dryness fraction of steam is x  =  %f \n',x);\n",
+"\n",
+"//  End\n",
+""
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 4.17: mass_dryness_fraction_and_heat_transfer.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"clear;\n",
+"clc;\n",
+"disp('Example 4.17');\n",
+"\n",
+"//  aim : To determine the\n",
+"//  (a) mass of steam in the vessel\n",
+"//  (b) final dryness of the steam\n",
+"//  (c) amount of heat transferrred during the cooling process\n",
+"\n",
+"//  Given values\n",
+"V1 = .8;//  [m^3]\n",
+"P1 = 360;//  [kN/m^2]\n",
+"P2 = 200;//  [kN/m^2]\n",
+"\n",
+"//  solution\n",
+"\n",
+"//  (a)\n",
+"// at 360 kN/m^2\n",
+"vg1 = .510;//  [m^3]\n",
+"m = V1/vg1;//  mass of steam,[kg]\n",
+"mprintf('\n (a) The mass of steam in the vessel is =  %f kg\n',m);\n",
+"\n",
+"//  (b)\n",
+"//  at 200 kN/m^2\n",
+"vg2 = .885;// [m^3/kg]\n",
+"//  the volume remains constant so\n",
+"x = vg1/vg2;// final dryness fraction\n",
+"mprintf('\n (b) The final dryness fraction of the steam is  =  %f \n',x);\n",
+"\n",
+"// (c)\n",
+"//  at 360 kN/m^2\n",
+"h1 = 2732.9;// [kJ/kg]\n",
+"//  hence\n",
+"u1 = h1-P1*vg1;//  [kJ/kg]\n",
+"\n",
+"//  at 200 kN/m^2\n",
+"hf = 504.7;// [kJ/kg]\n",
+"hfg=2201.6;//[kJ/kg]\n",
+"//  hence\n",
+"h2 = hf+x*hfg;// [kJ/kg]\n",
+"//  now\n",
+"u2 = h2-P2*vg1;//  [kJ/kg]\n",
+"//  so\n",
+"del_u = u2-u1;//  [kJ/kg]\n",
+"//  from the first law of thermodynamics del_U+W=Q, \n",
+"W = 0;//  because volume is constant\n",
+"del_U = m*del_u;//  [kJ]\n",
+"//  hence\n",
+"Q = del_U;//  [kJ]\n",
+"mprintf('\n (c) The amount of heat transferred during cooling process is  =  %f kJ \n',Q);\n",
+"\n",
+"//  End"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 4.18: specific_heat.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"clear;\n",
+"clc;\n",
+"disp('Example 4.18');\n",
+"\n",
+"//  aim : To determine\n",
+"//  the heat received by the steam per kilogram\n",
+"\n",
+"// Given values\n",
+"// initial\n",
+"P1 = 4;// pressure, [MN/m^2]\n",
+"x1 = .95; //  dryness fraction\n",
+"\n",
+"//  final\n",
+"t2 = 350;//  temperature,[C]\n",
+"\n",
+"//  solution\n",
+"\n",
+"// from steam table, at 4 MN/m^2 and x1=.95\n",
+"hf = 1087.4;//  [kJ/kg]\n",
+"hfg = 1712.9;//  [kJ/kg]\n",
+"//  hence\n",
+"h1 = hf+x1*hfg;//  [kJ/kg]\n",
+"\n",
+"//  since pressure is kept constant ant temperature is raised so at this condition\n",
+"h2 = 3095;//  [kJ/kg]\n",
+"\n",
+"//  so by energy balance\n",
+"Q = h2-h1;//  Heat received,[kJ/kg]\n",
+"mprintf('\n The heat received by the steam is  =  %f  kJ/kg \n',Q);\n",
+"\n",
+"//  End"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 4.19: condition_of_steam.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"clear;\n",
+"clc;\n",
+"disp('Example 4.19');\n",
+"\n",
+"//  aim : To determine the condition of the steam after \n",
+"//  (a) isothermal compression to half its initial volume,heat rejected\n",
+"//  (b) hyperbolic compression to half its initial volume\n",
+"\n",
+"//  Given values\n",
+"V1 = .3951;//  initial volume,[m^3]\n",
+"P1 = 1.5;//  initial pressure,[MN/m^2]\n",
+"\n",
+"//  solution\n",
+"\n",
+"//  (a)\n",
+"//  from steam table, at 1.5 MN/m^2 \n",
+"hf1 = 844.7;//  [kJ/kg]\n",
+"hfg1 = 1945.2;//  [kJ/kg]\n",
+"hg1 = 2789.9;// [kJ/kg]\n",
+"vg1 = .1317;//  [m^3/kg]\n",
+"\n",
+"//  calculation\n",
+"m = V1/vg1;//  mass of steam,[kg]\n",
+"vg2b = vg1/2;// given,[m^3/kg](vg2b is actual specific volume before compression)\n",
+"x1 = vg2b/vg1;//  dryness fraction\n",
+"h1 = m*(hf1+x1*hfg1);//  [kJ]\n",
+"Q = m*x1*hfg1;//  heat loss,[kJ]\n",
+"mprintf('\n (a) The Quantity of steam present is  =  %f kg \n',m);\n",
+"mprintf('\n      Dryness fraction is  =  %f \n',x1);\n",
+"mprintf('\n      The enthalpy is  =  %f kJ \n',h1);\n",
+"mprintf('\n      The heat loss is  =  %f kJ \n',Q);\n",
+"\n",
+"//  (b)\n",
+"V2 = V1/2;\n",
+"//  Given compression is according to the law PV=Constant,so\n",
+"P2 = P1*V1/V2;//  [MN/m^2]\n",
+"//  from steam table at P2\n",
+"hf2 = 1008.4;// [kJ/kg]\n",
+"hfg2 = 1793.9;//  [kJ/kg]\n",
+"hg2 = 2802.3;//  [kJ/kg]\n",
+"vg2 = .0666;//  [m^3/kg]\n",
+"\n",
+"//  calculation\n",
+"x2 = vg2b/vg2;//  dryness fraction\n",
+"h2 = m*(hf2+x2*hfg2);//  [kJ]\n",
+"\n",
+"mprintf('\n (b) The dryness fraction is  =  %f \n',x2);\n",
+"mprintf('\n     The enthalpy is  =  %f kJ\n',h2);\n",
+"\n",
+"//  End"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 4.1: specific_enthalpies.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"clear;\n",
+"clc;\n",
+"disp('Example 4.1');\n",
+"\n",
+"//  aim : To determine\n",
+"//  the enthalpy\n",
+"\n",
+"//  Given values\n",
+"P = .50;// Pressure, [MN/m^2]\n",
+"\n",
+"//  solution\n",
+"\n",
+"//  From steam tables, at given pressure\n",
+"hf = 640.1;//  specific liquid enthalpy ,[kJ/kg]\n",
+"hfg = 2107.4;//  specific enthalpy of evaporation ,[kJ/kg]\n",
+"hg = 2747.5; //  specific enthalpy of dry saturated steam ,[kJ/kg]\n",
+"tf = 151.8; //  saturation temperature,[C]\n",
+"\n",
+"mprintf('\n The specific liquid enthalpy is  =  %f kJ/kg \n',hf);\n",
+"mprintf('\n The specific enthalpy of evaporation is  =  %f  kJ/kg \n',hfg);\n",
+"mprintf('\n The specific enthalpy of dry saturated steam is  =  %f kJ/kg \n',hg);\n",
+"\n",
+"//  End"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 4.20: mass_work_change_in_internal_energy_and_heat_transfer.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"clear;\n",
+"clc;\n",
+"disp('Example 4.20');\n",
+"\n",
+"//  aim : To determine the\n",
+"//  (a) mass of steam \n",
+"//  (b) work transfer\n",
+"//  (c) change of internal energy\n",
+"//  (d) heat exchange b/w the steam and surroundings\n",
+"\n",
+"//  Given values\n",
+"P1 = 2.1;//  initial pressure,[MN/m^2]\n",
+"x1 = .9;//  dryness fraction\n",
+"V1 = .427;//  initial volume,[m^3]\n",
+"P2 = .7;//  final pressure,[MN/m^2]\n",
+"//  Given process is polytropic with\n",
+"n = 1.25; // polytropic index\n",
+"\n",
+"//  solution\n",
+"//  from steam table\n",
+"\n",
+"//  at 2.1 MN/m^2\n",
+"hf1 = 920.0;//  [kJ/kg]\n",
+"hfg1=1878.2;//  [kJ/kg]\n",
+"hg1=2798.2;//  [kJ/kg]\n",
+"vg1 = .0949;//  [m^3/kg]\n",
+"\n",
+"//  and at .7 MN/m^2\n",
+"hf2 = 697.1;//  [kJ/kg]\n",
+"hfg2 = 2064.9;//  [kJ/kg]\n",
+"hg2 = 2762.0;// [kJ/kg]\n",
+"vg2 = .273;//  [m^3/kg]\n",
+"\n",
+"//  (a)\n",
+"v1 = x1*vg1;//  [m^3/kg]\n",
+"m = V1/v1;//  [kg]\n",
+"mprintf('\n (a) The mass of steam present is  =  %f kg\n',m);\n",
+"\n",
+"//  (b)\n",
+"//  for polytropic process\n",
+"v2 = v1*(P1/P2)^(1/n);//  [m^3/kg]\n",
+"\n",
+"x2 = v2/vg2;//  final dryness fraction\n",
+"//  work transfer\n",
+"W = m*(P1*v1-P2*v2)*10^3/(n-1);//  [kJ]\n",
+"mprintf('\n (b) The work transfer is  =  %f  kJ\n',W);\n",
+"\n",
+"//  (c)\n",
+"//  initial\n",
+"h1 = hf1+x1*hfg1;//  [kJ/kg]\n",
+"u1 = h1-P1*v1*10^3;//  [kJ/kg]\n",
+"\n",
+"//  final\n",
+"h2 = hf2+x2*hfg2;//  [kJ/kg]\n",
+"u2 = h2-P2*v2*10^3;//  [kJ/kg]\n",
+"\n",
+"del_U = m*(u2-u1);//  [kJ]\n",
+"mprintf('\n (c) The change in internal energy is  =  %f kJ',del_U);\n",
+"if(del_U<0)\n",
+"    disp('since del_U<0,so this is loss of internal energy')\n",
+"else\n",
+"    disp('since del_U>0,so this is gain in internal energy')\n",
+"end\n",
+"\n",
+"//  (d)\n",
+"Q = del_U+W;//  [kJ]\n",
+"mprintf('\n (d) The heat exchange between the steam and surrounding is  =  %f  kJ',Q);\n",
+"if(Q<0)\n",
+"    disp('since Q<0,so this is loss of heat energy to surrounding')\n",
+"else\n",
+"    disp('since Q>0,so this is gain in heat energy to the steam')\n",
+"end\n",
+"\n",
+"// there are minor vairations in the values reported in the book\n",
+"\n",
+"//  End"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 4.21: volume_dryness_fraction_and_change_of_internal_energy.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"clear;\n",
+"clc;\n",
+"disp('Example 4.21');\n",
+"\n",
+"//  aim : To determine the \n",
+"//  (a) volume occupied by steam\n",
+"//  (b)(1) final dryness fraction of steam\n",
+"//       (2) Change of internal energy during expansion\n",
+"\n",
+"//  (a)\n",
+"//  Given values\n",
+"P1 = .85;//  [mN/m^2]\n",
+"x1 = .97;\n",
+"\n",
+"//  solution\n",
+"//  from steam table, at .85 MN/m^2,\n",
+"vg1 = .2268;//  [m^3/kg]\n",
+"//  hence\n",
+"v1 = x1*vg1;//  [m^3/kg]\n",
+"mprintf('\n (a) The volume occupied by 1 kg of steam is  =  %f m^3/kg\n',v1);\n",
+"\n",
+"// (b)(1)\n",
+"P2 = .17;//  [MN/m^2]\n",
+"// since process is polytropic process with\n",
+"n = 1.13; //  polytropic index\n",
+"// hence\n",
+"v2 = v1*(P1/P2)^(1/n);// [m^3/kg]\n",
+"\n",
+"// from steam table at .17 MN/m^2\n",
+"vg2 = 1.031;// [m^3/kg]\n",
+"// steam is wet so\n",
+"x2 = v2/vg2;//  final dryness fraction\n",
+"mprintf('\n (b)(1) The final dryness fraction of the steam is  =  %f \n',x2);\n",
+"\n",
+"//  (2)\n",
+"W = (P1*v1-P2*v2)*10^3/(n-1);// [kJ/kg]\n",
+"//  since process is adiabatic, so\n",
+"del_u = -W;// [kJ/kg]\n",
+"mprintf('\n     (2) The change in internal energy of the steam during expansion is  =  %f kJ/kg  (This is a loss of internal energy)\n',del_u);\n",
+"// There are minor variation in the answer\n",
+"\n",
+"//  End"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 4.2: Saturation_temperature_and_specific_enthalpies.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"clear;\n",
+"clc;\n",
+"disp('Example 4.2');\n",
+"\n",
+"//  aim : To determine \n",
+"//  saturation temperature and enthalpy\n",
+"\n",
+"//  Given values\n",
+"P = 2.04;// pressure, [MN/m^2]\n",
+"\n",
+"//  solution\n",
+"//  since in the steam table values of enthalpy and saturation temperature at 2 and 2.1 MN?m^2 are   given, so for knowing required values at given pressure,there is need to do interpolation\n",
+"\n",
+"//  calculation of saturation temperature\n",
+"//  from steam table\n",
+"Table_P_tf = [[2.1,2.0];[214.9,212.4]]; //  P in [MN/m^2] and tf in [C]\n",
+"// using interpolation\n",
+"tf = interpln(Table_P_tf,2.04);//  saturation temperature at given condition\n",
+"mprintf('\n The Saturation temperature is  =  %f C \n',tf);\n",
+"\n",
+"// calculation of specific liquid enthalpy\n",
+"//  from steam table\n",
+"Table_P_hf = [[2.1,2.0];[920.0,908.6]];//  P in [MN/m^2] and hf in [kJ/kg]\n",
+"// using interpolation\n",
+"hf = interpln(Table_P_hf,2.04); //  enthalpy at given condition, [kJ/kg]\n",
+"mprintf('\n The Specific liquid enthalpy is  =  %f kJ/kg \n',hf);\n",
+"\n",
+"// calculation of specific enthalpy of evaporation\n",
+"//  from steam table\n",
+"Table_P_hfg = [[2.1,2.0];[1878.2,1888.6]];//  P in [MN/m^2] and hfg in [kJ/kg]\n",
+"// using interpolation\n",
+"hfg = interpln(Table_P_hfg,2.04); //  enthalpy at given condition, [kJ/kg]\n",
+"mprintf('\n The Specific enthalpy of evaporation is  =  %f kJ/kg \n',hfg);\n",
+"\n",
+"//  calculation of specific enthalpy of dry saturated steam\n",
+"//  from steam table\n",
+"Table_P_hg = [[2.1,2.0];[2798.2,2797.2]];//P in [MN/m^2] and hg in [kJ/kg]\n",
+"// using interpolation\n",
+"hg = interpln(Table_P_hg,2.04); //  enthalpy at given condition, [kJ/kg]\n",
+"mprintf('\n The Specific enthalpy of dry saturated steam is  =  %f kJ/kg \n',hg);\n",
+"\n",
+"//  End"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 4.3: specific_enthalpy.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"clear;\n",
+"clc;\n",
+"disp('Example 4.3');\n",
+"\n",
+"//  aim : To determine\n",
+"//  the specific enthalpy\n",
+"\n",
+"//  given values\n",
+"P = 2; //  pressure ,[MN/m^2]\n",
+"t = 250; //  Temperature, [C]\n",
+"cp = 2.0934; //  average value of specific heat capacity, [kJ/kg K]\n",
+"\n",
+"//  solution\n",
+"\n",
+"//  looking up steam table it shows that at given pressure saturation temperature is 212.4 C,so\n",
+"tf = 212.4; //  [C]\n",
+"//  hence,\n",
+"Degree_of_superheat = t-tf;//  [C]\n",
+"//  from table at given temperature 250 C\n",
+"h = 2902; //  specific enthalpy of steam at 250 C ,[kJ/kg]\n",
+"mprintf('\nThe specific enthalpy of steam at 2 MN/m^2 with temperature 250 C is  =  %f  kJ/kg \n',h);\n",
+"\n",
+"//  Also from steam table enthalpy at saturation temperature is\n",
+"hf = 2797.2 ;//  [kJ/kg]\n",
+"//  so enthalpy at given temperature is\n",
+"h = hf+cp*(t-tf);//  [kJ/kg]\n",
+"\n",
+"mprintf('\n The specific enthalpy at given T and P by alternative path is  =  %f kJ/kg \n',h);\n",
+"\n",
+"//  End"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 4.4: specific_enthalpy.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"clear;\n",
+"clc;\n",
+"disp('Example 4.4');\n",
+"\n",
+"//  aim : To determine\n",
+"//  the specific enthalpy of steam\n",
+"\n",
+"//  Given values\n",
+"P = 2.5;//  pressure, [MN/m^2]\n",
+"t = 320; //  temperature, [C]\n",
+"\n",
+"//  solution\n",
+"//  from steam table at given condition the saturation temperature of steam is 223.9 C, therefore steam is superheated\n",
+"tf = 223.9;//  [C]\n",
+"\n",
+"//  first let's calculate estimated enthalpy\n",
+"//  again from steam table \n",
+"\n",
+"hg = 2800.9;//  enthalpy at saturation temp, [kJ/kg]\n",
+"cp =2.0934;//  specific heat capacity of steam,[kJ/kg K]\n",
+"\n",
+"//  so enthalpy at given condition is\n",
+"h = hg+cp*(t-tf);//  [kJ/kg]\n",
+"mprintf('\n The estimated specific enthalpy is  =  %f kJ/kg \n',h);\n",
+"\n",
+"//  calculation of accurate specific enthalpy\n",
+"//  we need double interpolation for this\n",
+"\n",
+"//  first interpolation w.r.t. to temperature\n",
+"//  At 2 MN/m^2\n",
+"Table_t_h = [[325,300];[3083,3025]];// where, t in [C] and h in [kJ/kg]\n",
+"h1 = interpln(Table_t_h,320); //  [kJ/kg]\n",
+"\n",
+"//  at 4 MN/m^2\n",
+"Table_t_h = [[325,300];[3031,2962]]; //  t in [C] and h in [kJ/kg]\n",
+"h2 = interpln(Table_t_h,320); //  [kJ/kg]\n",
+"\n",
+"//  now interpolation w.r.t. pressure\n",
+"Table_P_h = [[4,2];[h2,h1]]; //  where P in NM/m^2 and h1,h2 in kJ/kg\n",
+"h = interpln(Table_P_h,2.5);//  [kJ/kg]\n",
+"mprintf('\n The accurate specific enthalpy of steam at pressure of 2.5 MN/m^2 and with a temperature 320 C is  =   %f kJ/kg \n',h);\n",
+"\n",
+"//  End"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 4.5: specific_enthalpy.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"clear;\n",
+"clc;\n",
+"disp('Example 4.5');\n",
+"\n",
+"// aim : To determine \n",
+"//  the specific enthalpy \n",
+"\n",
+"//  Given values\n",
+"P = 70; //  pressure, [kn/m^2]\n",
+"x = .85; //  Dryness fraction\n",
+"\n",
+"// solution\n",
+"\n",
+"//  from steam table, at given pressure \n",
+"hf = 376.8;//  [kJ/kg]\n",
+"hfg = 2283.3;//  [kJ/kg]\n",
+"\n",
+"// now using equation [2]\n",
+"h = hf+x*hfg;//  specific enthalpy of wet steam,[kJ/kg]\n",
+"\n",
+"mprintf('\n The specific enthalpy of wet steam is  =  %f kJ/kg \n',h);\n",
+"\n",
+"// There is minor variation in the book's answer\n",
+"\n",
+"//  End"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 4.8: specific_volume.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"clear;\n",
+"clc;\n",
+"disp('Example 4.8');\n",
+"\n",
+"//  aim : To determine \n",
+"//  the specific volume of wet steam\n",
+"\n",
+"//  Given values\n",
+"P = 1.25; //  pressure, [MN/m^2]\n",
+"x = .9; //  dry fraction\n",
+"\n",
+"//  solution\n",
+"//  from steam table at given pressure\n",
+"vg = .1569;//  [m^3/kg]\n",
+"//  hence\n",
+"v = x*vg; //  [m^3/kg]\n",
+"\n",
+"mprintf('\nThe specific volume of wet steam is  =  %f  m^3/kg \n',v);\n",
+"\n",
+"//  End"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 4.9: specific_volume.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"clear;\n",
+"clc;\n",
+"disp('Example 4.9');\n",
+"\n",
+"//  aim : To determine\n",
+"//  the specific volume \n",
+"\n",
+"//  Given values\n",
+"t = 325; //  temperature, [C]\n",
+"P = 2; //  pressure, [MN/m^2]\n",
+"\n",
+"//  solution\n",
+"// from steam table at given t and P\n",
+"vf = .1321; //  [m^3/kg]\n",
+"tf = 212.4; //  saturation temperature, [C]\n",
+"\n",
+"mprintf('\n The specific volume of steam at pressure of 2 MN/m^2 and with temperature 325 C is  =  %f  m^3/kg \n',vf);\n",
+"doh= t-tf; //  degree of superheat, [C]\n",
+"mprintf('\n The degree of superheat is  =  %f C\n',doh);\n",
+"\n",
+"// End "
+   ]
+   }
+],
+"metadata": {
+		  "kernelspec": {
+		   "display_name": "Scilab",
+		   "language": "scilab",
+		   "name": "scilab"
+		  },
+		  "language_info": {
+		   "file_extension": ".sce",
+		   "help_links": [
+			{
+			 "text": "MetaKernel Magics",
+			 "url": "https://github.com/calysto/metakernel/blob/master/metakernel/magics/README.md"
+			}
+		   ],
+		   "mimetype": "text/x-octave",
+		   "name": "scilab",
+		   "version": "0.7.1"
+		  }
+		 },
+		 "nbformat": 4,
+		 "nbformat_minor": 0
+}