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diff --git a/Basic_Engineering_Thermodynamics_by_R_Joel/1-General_Introduction.ipynb b/Basic_Engineering_Thermodynamics_by_R_Joel/1-General_Introduction.ipynb
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+{
+"cells": [
+ {
+		   "cell_type": "markdown",
+	   "metadata": {},
+	   "source": [
+       "# Chapter 1: General Introduction"
+	   ]
+	},
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 1.10: Thermal_efficiency.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"clear;\n",
+"clc;\n",
+"disp('Example 1.10');\n",
+"//  Given values\n",
+"m_dot = 3.045; //  use of coal, [tonne/h]\n",
+"c = 28; // calorific value of the coal, [MJ/kg]\n",
+"P_out = 4.1; //  output of turbine, [MW]\n",
+"//  solution\n",
+"m_dot = m_dot*10^3/3600; // [kg/s]\n",
+"P_in = m_dot*c; //  power input by coal, [MW]\n",
+"n = P_out/P_in; //  thermal efficiency formula\n",
+"mprintf('\n Thermal efficiency of the plant is  =  %f \n',n);\n",
+"//End"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 1.11: Power_output.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"clear;\n",
+"clc;\n",
+"disp('Example 1.11');\n",
+"//  Given values\n",
+"v = 50; //  speed, [km/h]\n",
+"F = 900; //  Resistance to the motion of a car\n",
+"//  solution\n",
+"v = v*10^3/3600; //  [m/s]\n",
+" Power = F*v; //  Power formula, [W]\n",
+"mprintf('\n The power output of the engine is  =  %f  kW\n',Power*10^-3);\n",
+" \n",
+" //  End\n",
+" "
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 1.12: Power_output.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"clear;\n",
+"clc;\n",
+"disp('Example 1.12');\n",
+"//  Given values\n",
+"V = 230; //  volatage, [volts]\n",
+"I = 60; //  current, [amps]\n",
+"n_gen = .95; //  efficiency of generator\n",
+"n_eng = .92; //  efficiency of engine\n",
+"//  solution\n",
+"P_gen = V*I; //  Power delivered by generator, [W]\n",
+"P_gen=P_gen*10^-3; //  [kW]\n",
+"P_in_eng=P_gen/n_gen;//Power input from engine,[kW]\n",
+"P_out_eng=P_in_eng/n_eng;//Power output from engine,[kW]\n",
+"mprintf('\n The power output from the engine is  =  %f  kW\n',P_out_eng);\n",
+"//  End"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 1.13: Current.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"clear;\n",
+"clc;\n",
+"disp('Example 1.13');\n",
+"//  Given values\n",
+"V = 230; // Voltage, [volts]\n",
+"W = 4; //  Power of heater, [kW]\n",
+"//  solution\n",
+"//  using equation P=VI\n",
+"I = W/V; //  current, [K amps]\n",
+"mprintf('\n The current taken by heater is  =  %f  amps \n',I*10^3);\n",
+"//  End"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 1.14: Mass_of_coal_burnt.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"clear;\n",
+"clc;\n",
+"disp('Example 1.14');\n",
+"//  Given values\n",
+"P_out = 500; //  output of power station, [MW]\n",
+"c = 29.5; //  calorific value of coal, [MJ/kg]\n",
+"r=.28; \n",
+"//  solution\n",
+"//  since P represents only 28 percent of energy available from coal\n",
+"P_coal = P_out/r; //  [MW]\n",
+" \n",
+"m_coal = P_coal/c; //  Mass of coal used, [kg/s]\n",
+"m_coal = m_coal*3600; //  [kg/h]\n",
+"//After one hour\n",
+"m_coal = m_coal*1*10^-3; //  [tonne]\n",
+"mprintf('\n Mass of coal burnt by the power station in 1 hour is  =  %f  tonne \n',m_coal);\n",
+"//  End"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 1.1: Work_done.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"clear ;\n",
+"clc;\n",
+"disp('Example 1.1');\n",
+"// Given values\n",
+"P = 700;   //pressure,[kN/m^2]\n",
+"V1 = .28;   //initial volume,[m^3]\n",
+"V2 = 1.68;   //final volume,[m^3]\n",
+"//solution\n",
+"W = P*(V2-V1);// // Formula for work done at constant pressure is, [kJ]\n",
+"mprintf('\n The Work done is = %f MJ\n',W*10^-3);\n",
+"//End"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 1.2: Volume_of_the_gas.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"clear;\n",
+"clc;\n",
+"disp('Example 1.2');\n",
+"//Given values\n",
+"P1 = 138;  // initial pressure,[kN/m^2]\n",
+"V1 = .112;  //initial volume,[m^3]\n",
+"P2 = 690; //  final pressure,[kN/m^2]\n",
+"Gama=1.4; //  heat capacity ratio\n",
+"//  solution\n",
+"//  since gas is following, PV^1.4=constant,hence\n",
+"V2 =V1*(P1/P2)^(1/Gama); //   final volume, [m^3] \n",
+"mprintf('\n The new volume of the gas is = %f m^3\n',V2)\n",
+"//End"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 1.3: Work_done.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"clear;\n",
+"clc;\n",
+"disp('Example 1.3');\n",
+"//  Given values\n",
+"P1 = 2070;  //  initial pressure,  [kN/m^2]\n",
+"V1 = .014;  //  initial volume,  [m^3]\n",
+"P2 = 207;  //  final pressure, [kN/m^2]\n",
+"n=1.35;  //  polytropic index\n",
+"//  solution\n",
+"//  since gas is following PV^n=constant\n",
+"//  hence \n",
+"V2 = V1*(P1/P2)^(1/n);  // final volume,  [m^3]\n",
+"//  calculation of workdone\n",
+"W=(P1*V1-P2*V2)/(1.35-1);  // using work done formula for polytropic process, [kJ]\n",
+"mprintf('\n The Work done by gas during expansion is  =  %f kJ\n',W);\n",
+"//End"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 1.4: Final_Pressure_and_work_done.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"clear;\n",
+"clc;\n",
+"disp('Example 1.4');\n",
+"//  Given values\n",
+"P1 = 100;  //  initial pressure, [kN/m^2]\n",
+"V1 = .056;  //  initial volume,  [m^3]\n",
+"V2 = .007;  //  final volume,  [m^3]\n",
+"//  To know  P2\n",
+"//  since process is hyperbolic so, PV=constant\n",
+"//  hence\n",
+"P2 = P1*V1/V2;  //  final pressure, [kN/m^2]\n",
+"mprintf('\n  The final pressure is  =  %f  kN/m^2\n',P2);\n",
+"// calculation of workdone\n",
+"W = P1*V1*log(V2/V1); //  formula for work done in this process, [kJ]\n",
+"mprintf('\n  Work done on the gas is  =  %f  kJ\n',W);\n",
+"//End"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 1.5: Heat_transfer.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"clear;\n",
+"clc;\n",
+"disp('Example 1.5');\n",
+"//  Given values\n",
+"m = 5; //  mass,  [kg]\n",
+"t1 = 15; //  inital temperature, [C]\n",
+"t2 = 100; //  final temperature, [C]\n",
+"c = 450; //  specific heat capacity, [J/kg K]\n",
+"//  solution\n",
+"//  using heat transfer equation,[1]\n",
+"Q = m*c*(t2-t1); //  [J]\n",
+"mprintf('\n The heat required is  =  %f  kJ\n',Q*10^-3);\n",
+"//End"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 1.6: Heat_transfer.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"clear;\n",
+"clc;\n",
+"disp('Example 1.6');\n",
+"//  Given values\n",
+"m_cop = 2; //  mass of copper vessel, [kg]\n",
+"m_wat = 6; //  mass of water, [kg]\n",
+"c_wat = 4.19; //  specific heat capacity of water,  [kJ/kg K]\n",
+"t1 = 20; //  initial temperature, [C]\n",
+"t2 = 90; //  final temperature, [C]\n",
+"// From the table of average specific heat capacities\n",
+"c_cop = .390;  // specific heat capacity of copper,[kJ/kg k]\n",
+"// solution\n",
+"Q_cop = m_cop*c_cop*(t2-t1); //  heat required by copper vessel, [kJ]\n",
+"Q_wat = m_wat*c_wat*(t2-t1); //  heat required by water, [kJ]\n",
+"// since there is no heat loss,so total heat transfer is sum of both\n",
+"Q_total = Q_cop+Q_wat ; //  [kJ]\n",
+"mprintf(' \n Required heat transfer to accomplish the change  =  %f  kJ\n',Q_total);\n",
+"//End"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 1.7: Temperature.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"clear;\n",
+"clc;\n",
+"disp('Example 1.7');\n",
+"//  Given values\n",
+"m = 10; //  mass of iron casting, [kg]\n",
+"t1 = 200; //  initial temperature, [C]\n",
+"Q = -715.5; //  [kJ], since heat is lost in this process\n",
+"// From the table of average specific heat capacities\n",
+"c = .50; //  specific heat capacity of casting iron, [kJ/kg K]\n",
+"//  solution\n",
+"//  using heat equation\n",
+"//  Q = m*c*(t2-t1)\n",
+"t2 = t1+Q/(m*c); // [C]\n",
+"mprintf('\n The final temperature is  t2 =  %f C\n',t2);\n",
+"// End"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 1.8: Specific_heat_capacity.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"clear;\n",
+"clc;\n",
+"disp('Example 1.8');\n",
+" \n",
+"// Given values\n",
+"m = 4; //  mass of the liquid, [kg]\n",
+"t1 = 15; //  initial temperature, [C]\n",
+"t2 = 100; //  final temperature, [C]\n",
+"Q = 714; //  [kJ],required heat to accomplish this change\n",
+"//  solution\n",
+"//  using heat equation\n",
+"//  Q=m*c*(t2-t1)\n",
+"//  calculation of c\n",
+"c=Q/(m*(t2-t1)); //  heat capacity, [kJ/kg K] \n",
+"mprintf('\n The specific heat capacity of the liquid is  c =  %f  kJ/kg K\n',c);\n",
+"//End"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 1.9: Power_output_and_energy_rejected.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"clear;\n",
+"clc;\n",
+"disp('Example 1.9');\n",
+"//  Given values\n",
+"m_dot = 20.4; //  mass flowrate of petrol, [kg/h]\n",
+"c = 43; //  calorific  value of petrol, [MJ/kg]\n",
+"n = .2; //  Thermal efficiency of engine\n",
+"//  solution\n",
+"m_dot = 20.4/3600; // [kg/s]\n",
+"c = 43*10^6; //  [J/kg]\n",
+"//  power output\n",
+"P_out = n*m_dot*c; //  [W]\n",
+"mprintf('\n The power output of the engine is  =  %f  kJ\n',P_out*10^-3);\n",
+" \n",
+"//  power rejected\n",
+"P_rej = m_dot*c*(1-n); //  [W]\n",
+"P_rej = P_rej*60*10^-6; //  [MJ/min]\n",
+"mprintf('\n The energy rejected by the engine is  =   %f  MJ/min \n',P_rej);\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
+}
diff --git a/Basic_Engineering_Thermodynamics_by_R_Joel/10-Steam_plant.ipynb b/Basic_Engineering_Thermodynamics_by_R_Joel/10-Steam_plant.ipynb
new file mode 100644
index 0000000..b888739
--- /dev/null
+++ b/Basic_Engineering_Thermodynamics_by_R_Joel/10-Steam_plant.ipynb
@@ -0,0 +1,695 @@
+{
+"cells": [
+ {
+		   "cell_type": "markdown",
+	   "metadata": {},
+	   "source": [
+       "# Chapter 10: Steam plant"
+	   ]
+	},
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 10.10: mass_and_thermal_efficiency.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"clear;\n",
+"clc;\n",
+"disp('Example 10.10');\n",
+"\n",
+"// aim : To determine\n",
+"// (a) the mass of steam bled to each feed heater in kg/kg of supply steam\n",
+"// (b) the thermal efficiency of the arrangement\n",
+"\n",
+"// given values\n",
+"P1 = 7;// steam initial pressure, [MN/m^2]\n",
+"T1 = 273+500;// steam initil temperature, [K]\n",
+"P2 = 2;// pressure at stage 1, [MN/m^2]\n",
+"P3 = .5;// pressure at stage 2, [MN/m^2]\n",
+"P4 = .05;// condenser pressure,[MN/m^2]\n",
+"SE = .82;// stage efficiency of turbine\n",
+"\n",
+"// solution\n",
+"// from the enthalpy-entropy chart(Fig10.23) values of specific enthalpies are\n",
+"h1 = 3410;// [kJ/kg]\n",
+"h2_prim = 3045;// [kJ/kg]\n",
+"// h1-h2=SE*(h1-h2_prim), so\n",
+"h2 = h1-SE*(h1-h2_prim);// [kJ/kg]\n",
+"\n",
+"h3_prim = 2790;// [kJ/kg]\n",
+"// h2-h3=SE*(h2-h3_prim), so\n",
+"h3 = h2-SE*(h2-h3_prim);// [kJ/kg]\n",
+"\n",
+"h4_prim = 2450;// [kJ/kg]\n",
+"// h3-h4 = SE*(h3-h4_prim), so\n",
+"h4 = h3-SE*(h3-h4_prim);// [kJ/kg]\n",
+"\n",
+"// from steam table\n",
+"// @ 2 MN/m^2\n",
+"hf2 = 908.6;// [kJ/kg]\n",
+"// @ .5 MN/m^2\n",
+"hf3 = 640.1;// [kJ/kg] \n",
+"// @ .05 MN/m^2\n",
+"hf4 = 340.6;// [kJ/kg]\n",
+"\n",
+"// (a) \n",
+"// for feed heater1\n",
+"m1 = (hf2-hf3)/(h2-hf3);// mass of bled steam, [kg/kg supplied steam]\n",
+"// for feed heater2\n",
+"m2 = (1-m1)*(hf3-hf4)/(h3-hf4);// \n",
+"mprintf('\n (a) The mass of steam bled in feed heater 1 is  =  %f  kg/kg supply steam\n',m1);\n",
+"mprintf('\n      The mass of steam bled in feed heater 2 is  =  %f kg/kg supply steam\n',m2);\n",
+"\n",
+"// (b)\n",
+"W = (h1-h2)+(1-m1)*(h2-h3)+(1-m1-m2)*(h3-h4);// theoretical work done, [kJ/kg]\n",
+"Eb = h1-hf2;// energy input in the boiler, [kJ/kg]\n",
+"TE1 = W/Eb;// thermal efficiency\n",
+"mprintf('\n (b) The thermal efficiency of the arrangement is  =  %f percent\n',TE1*100);\n",
+"\n",
+"// If there is no feed heating\n",
+"hf5 = hf4;\n",
+"h5_prim = 2370;// [kJ/kg]\n",
+"// h1-h5 = SE*(h1-h5_prim), so\n",
+"h5 = h1-SE*(h1-h5_prim);// [kJ/kg]\n",
+"Ei = h1-hf5;//energy input, [kJ/kg]\n",
+"W = h1-h5;//  theoretical work, [kJ/kg]\n",
+"TE2 = W/Ei;// thermal efficiency\n",
+"mprintf('\n      The thermal efficiency if there is no feed heating is  =  %f percent\n',TE2*100);\n",
+"\n",
+"//  End "
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 10.1: equivalent_evaporation.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"clear;\n",
+"clc;\n",
+"disp('Example 10.1');\n",
+"\n",
+"//  aim : To determine\n",
+"//   the equivalent evaporation\n",
+"\n",
+"//  Given\n",
+"P = 1.4;//  [MN/m^2]\n",
+"m = 8;//  mass of water,[kg]\n",
+"T1 = 39;//  entering temperature,[C]\n",
+"T2 = 100;//  [C]\n",
+"x = .95;//dryness fraction \n",
+"\n",
+"//  solution\n",
+"hf = 830.1;//  [kJ/kg]\n",
+"hfg = 1957.7;//  [kJ/kg]\n",
+"//  steam is wet so specific enthalpy of steam is\n",
+"h = hf+x*hfg;//  [kJ/kg]\n",
+"\n",
+"//  at 39 C\n",
+"h1 = 163.4;//  [kJ/kg]\n",
+"//  hence\n",
+"q = h-h1;//  [kJ/kg]\n",
+"Q = m*q;//  [kJ]\n",
+"\n",
+"evap = Q/2256.9;//  equivalent evaporation[kg steam/(kg coal)]\n",
+"\n",
+"mprintf('\n The equivalent evaporation, from and at 100 C is  =  %f  kg steam/kg coal\n ',evap);\n",
+"\n",
+"// End"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 10.2: mass_fraction_of_enthalpy_drop_and_heat_transfer.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"clear;\n",
+"clc;\n",
+"disp('Example 10.2');\n",
+"\n",
+"// aim : To determine \n",
+"//  the mass of oil used per hour and the fraction of enthalpy drop through the turbine\n",
+"//  heat transfer available per kilogram of exhaust steam\n",
+"\n",
+"//  Given values\n",
+"ms_dot = 5000;// generation of steam, [kg/h]\n",
+"P1 = 1.8;// generated steam pressure, [MN/m^2]\n",
+"T1 = 273+325;// generated steam temperature, [K]\n",
+"Tf = 273+49.4;// feed temperature, [K]\n",
+"neta = .8;// efficiency of boiler plant \n",
+"c = 45500;// calorific value, [kJ/kg]\n",
+"P = 500;// turbine generated power, [kW]\n",
+"Pt = .18;// turbine exhaust pressure, [MN/m^2]\n",
+"x = .98;// dryness farction of steam\n",
+"\n",
+"//  solution\n",
+"//  using steam table at 1.8 MN/m^2\n",
+"hf1 = 3106;// [kJ/kg]\n",
+"hg1 = 3080;// [kJ/kg]\n",
+"//  so\n",
+"h1 = hf1-neta*(hf1-hg1);// [kJ/kg]\n",
+"//  again using steam table specific enthalpy of feed water is\n",
+"hwf = 206.9;// [kJ/kg]\n",
+"h_rais = ms_dot*(h1-hwf);// energy to raise steam, [kJ]\n",
+"\n",
+"h_fue = h_rais/neta;// energy from fuel per hour, [kJ]\n",
+"m_oil = h_fue/c;// mass of fuel per hour, [kg]\n",
+"\n",
+"//  from steam table at exhaust\n",
+"hf = 490.7;// [kJ/kg]\n",
+"hfg = 2210.8;//  [kJ/kg]\n",
+"//  hence\n",
+"h = hf+x*hfg;// [kJ/kg]\n",
+"//  now\n",
+"h_drop = (h1-h)*ms_dot/3600;// specific enthalpy drop in turbine [kJ]\n",
+"f = P/h_drop;// fraction ofenthalpy drop converted into work\n",
+"//  heat transfer available in exhaust is\n",
+"Q = h-hwf;// [kJ/kg]\n",
+"mprintf('\n The mass of oil used per hour is  =  %f  kg\n',m_oil);\n",
+"mprintf('\n The fraction of the enthalpy drop through the turbine that is converted into useful work is  =  %f\n',f);\n",
+"mprintf('\n The heat transfer available in exhaust steam above 49.4 C is  =  %f  kJ/kg\n',Q);\n",
+"\n",
+"//  End"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 10.3: efficiency_equivalent_evaporation_and_coal_consumption.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"clear;\n",
+"clc;\n",
+"disp('Example 10.3');\n",
+"\n",
+"//  aim : To determine\n",
+"//  (a) the thermal efficiency of the boiler\n",
+"//  (b) the equivalent evaporation of the boiler\n",
+"//  (c) the new coal consumption \n",
+"\n",
+"//  given values\n",
+"ms_dot = 5400;// steam feed rate, [kg/h]\n",
+"P = 750;// steam pressure, [kN/m^2]\n",
+"x = .98;// steam dryness fraction\n",
+"Tf1 = 41.5;// feed water temperature, [C]\n",
+"CV = 31000;// calorific value of coal used in the boiler, [kJ/kg]\n",
+"mc1 = 670;// rate of burning of coal/h, [kg]\n",
+"Tf2 = 100;// increased water temperature, [C]\n",
+"\n",
+"// solution\n",
+"//   (a)\n",
+"SRC = ms_dot/mc1;// steam raised/kg coal, [kg]\n",
+"hf = 709.3;// [kJ/kg]\n",
+"hfg = 2055.5;// [kJ/kg]\n",
+"h1 = hf+x*hfg;// specific enthalpy of steam raised, [kJ/kg]\n",
+"//  from steam table \n",
+"hfw = 173.9;// specific enthalpy of feed water, [kJ/kg]\n",
+"EOB = SRC*(h1-hfw)/CV;// efficiency of boiler\n",
+"mprintf('\n (a) The thermal efficiency of the boiler is  =  %f percent\n',EOB*100);\n",
+"\n",
+"// (b)\n",
+"he = 2256.9;// specific enthalpy of evaporation, [kJ/kg]\n",
+"Ee = SRC*(h1-hfw)/he;// equivalent evaporation[kg/kg coal]\n",
+"mprintf('\n (b) The equivalent evaporation of boiler is  =  %f  kg/kg coal\n',Ee);\n",
+"\n",
+"// (c)\n",
+"hw = 419.1;// specific enthalpy of feed water at 100 C, [kJ/kg]\n",
+"Eos = ms_dot*(h1-hw);// energy of steam under new condition, [kJ/h]\n",
+"neb = EOB+.05;// given condition new efficiency of boiler if 5%more than previous\n",
+"Ec = Eos/neb;// energy from coal, [kJ/h]\n",
+"mc2 = Ec/CV;// mass of coal used per hour in new condition, [kg]\n",
+"mprintf('\n (c) Mass of coal used in new condition is  =  %f kg\n',mc2);\n",
+"mprintf('\n      The saving in coal per hour is  =  %f  kg\n',mc1-mc2);\n",
+"\n",
+"//  End"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 10.4: heat_transfer_and_volume.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"clear;\n",
+"clc;\n",
+"disp('Example 10.4');\n",
+"\n",
+"//  aim : To determine the\n",
+"//  (a) Heat transfer in the boiler\n",
+"//   (b) Heat transfer in the superheater\n",
+"//  (c) Gas used\n",
+"\n",
+"//  given values\n",
+"P = 100;// boiler operating pressure, [bar]\n",
+"Tf = 256;// feed water temperature, [C]\n",
+"x = .9;// steam dryness fraction.\n",
+"Th = 450;// superheater exit temperature, [C]\n",
+"m = 1200;// steam generation/h, [tonne]\n",
+"TE = .92;// thermal efficiency\n",
+"CV = 38;// calorific value of fuel, [MJ/m^3]\n",
+"\n",
+"// solution\n",
+"// (a)\n",
+"// from steam table\n",
+"hw = 1115.4;// specific enthalpy of feed water, [kJ/kg]\n",
+"// for wet steam\n",
+"hf = 1408;// specific enthalpy, [kJ/kg]\n",
+"hg = 2727.7;// specific enthalpy, [kJ/kg]\n",
+"//  so\n",
+"h = hf+x*(hg-hf);// total specific enthalpy of wet steam, [kJ/kg]\n",
+"//  hence\n",
+"Qb = m*(h-hw);// heat transfer/h for wet steam, [MJ]\n",
+"mprintf('\n (a) The heat transfer/h in producing wet steam in the boiler is  =  %f  MJ\n',Qb);\n",
+"\n",
+"// (b)\n",
+"// again from steam table\n",
+"// specific enthalpy of superheated stem at given condition is,\n",
+"hs = 3244;// [kJ/kg]\n",
+"\n",
+"Qs = m*(hs-h);// heat transfer/h in superheater, [MJ]\n",
+"mprintf('\n (b) The heat transfer/h in superheater is  =  %f  MJ\n',Qs);\n",
+"\n",
+"// (c)\n",
+"V = (Qb+Qs)/(TE*CV);// volume of gs used/h, [m^3]\n",
+"mprintf('\n (c) The volume of gas used/h is  =  %f m^3\n',V);\n",
+"\n",
+"//  There is calculation mistake in the book so our answer is not matching\n",
+"\n",
+"//  End"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 10.5: flow_rate.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"clear;\n",
+"clc;\n",
+"disp('Example 10.5');\n",
+"\n",
+"//aim : To determine \n",
+"// the flow rate of cooling water\n",
+"\n",
+"//Given values\n",
+"P=24;//pressure, [kN/m^2]\n",
+"ms_dot=1.8;//steam condense rate,[tonne/h]\n",
+"x=.98;//dryness fraction\n",
+"T1=21;//entrance temperature of cooling water,[C]\n",
+"T2=57;//outlet temperature of cooling water,[C]\n",
+"\n",
+"//solution\n",
+"//at 24 kN/m^2, for steam\n",
+"hfg=2616.8;//[kJ/kg]\n",
+"hf1=268.2;//[kJ/kg]\n",
+"//hence\n",
+"h1=hf1+x*(hfg-hf1);//[kJ/kg]\n",
+"\n",
+"//for cooling water\n",
+"hf3=238.6;//[kJ/kg]\n",
+"hf2=88.1;//[kJ/kg]\n",
+"\n",
+"//using equation [3]\n",
+"//ms_dot*(hf3-hf2)=mw_dot*(h1-hf1),so\n",
+"mw_dot=ms_dot*(h1-hf1)/(hf3-hf2);//[tonne/h]\n",
+"disp('tonne/h',mw_dot,'The flow rate of the cooling water is =')\n",
+"\n",
+"//End"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 10.6: energy_supplied_dryness_fraction_and_Rankine_efficiency.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"clear;\n",
+"clc;\n",
+"disp('Example 10.6');\n",
+"\n",
+"//  aim : To determine\n",
+"//  (a) the energy supplied in the boiler\n",
+"//  (b) the dryness fraction of the steam entering the condenser\n",
+"//  (c) the rankine efficiency\n",
+"\n",
+"//  given values\n",
+"P1 = 3.5;// steam entering pressure, [MN/m^2]\n",
+"T1 = 273+350;// entering temperature, [K]\n",
+"P2 = 10;//steam exhaust pressure, [kN/m^2]\n",
+"\n",
+"// solution\n",
+"//  (a)\n",
+"//  from steam table, at P1 is,\n",
+"hf1 = 3139;// [kJ/kg]\n",
+"hg1 = 3095;// [kJ/kg]\n",
+"h1 = hf1-1.5/2*(hf1-hg1);\n",
+"// at Point 3\n",
+"h3 = 191.8;// [kJ/kg]\n",
+"Es = h1-h3;// energy supplied, [kJ/kg]\n",
+"mprintf('\n (a) The energy supplied in boiler/kg steam is  =  %f kJ/kg\n',Es);\n",
+"\n",
+"// (b)\n",
+"// at P1\n",
+"sf1 = 6.960;// [kJ/kg K]\n",
+"sg1 = 6.587;// [kJ/kg K]\n",
+"s1 = sf1-1.5/2*(sf1-sg1);// [kJ/kg K]\n",
+"// at P2\n",
+"sf2 = .649;// [kJ/kg K] \n",
+" sg2 = 8.151;// [kJ/kg K]\n",
+" // s2=sf2+x2(sg2-sf2)\n",
+" // theoretically expansion through turbine is isentropic so s1=s2\n",
+" // hence\n",
+" s2 = s1;\n",
+" x2 = (s2-sf2)/(sg2-sf2);// dryness fraction\n",
+" mprintf('\n (b) The dryness fraction of steam entering the condenser is =  %f \n',x2);\n",
+" \n",
+" // (c)\n",
+" // at point 2\n",
+" hf2 = 191.8;// [kJ/kg]\n",
+" hfg2 = 2392.9;// [kJ/kg]\n",
+" h2 = hf2+x2*hfg2;// [kJ/kg]\n",
+" Re = (h1-h2)/(h1-h3);// rankine efficiency\n",
+" mprintf('\n (c) The Rankine efficiency is  =  %f percent\n',Re*100);\n",
+" \n",
+" //  End"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 10.7: Rankine_efficiency_and_specific_work_done.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"clear;\n",
+"clc;\n",
+"disp('Example 10.7');\n",
+"\n",
+"// aim : To determine\n",
+"//  the specific work done and compare this with that obtained when determining the rankine effficiency\n",
+"\n",
+"// given values\n",
+"P1 = 1000;// steam entering pressure, [kN/m^2]\n",
+"x1 = .97;// steam entering dryness fraction\n",
+"P2 = 15;//steam exhaust pressure, [kN/m^2]\n",
+"n = 1.135;// polytropic index\n",
+"\n",
+"// solution\n",
+"// (a)\n",
+"// from steam table, at P1 is\n",
+"hf1 = 762.6;// [kJ/kg]\n",
+"hfg1 = 2013.6;// [kJ/kg]\n",
+"h1 = hf1+hfg1; // [kJ/kg]\n",
+"\n",
+"sf1 = 2.138;// [kJ/kg K]\n",
+"sg1 = 6.583;// [kJ/kg K]\n",
+"s1 = sf1+x1*(sg1-sf1);// [kJ/kg K]\n",
+"\n",
+"// at P2\n",
+"sf2 = .755;// [kJ/kg K] \n",
+" sg2 = 8.009;// [kJ/kg K]\n",
+"// s2 = sf2+x2(sg2-sf2)\n",
+"// since expansion through turbine is isentropic so s1=s2\n",
+" // hence\n",
+" s2 = s1;\n",
+" x2 = (s2-sf2)/(sg2-sf2);// dryness fraction\n",
+" \n",
+"  // at point 2\n",
+" hf2 = 226.0;// [kJ/kg]\n",
+" hfg2 = 2373.2;// [kJ/kg]\n",
+" h2 = hf2+x2*hfg2;// [kJ/kg]\n",
+" \n",
+"// at Point 3\n",
+"h3 = 226.0;// [kJ/kg]\n",
+"\n",
+"// (a)\n",
+" Re = (h1-h2)/(h1-h3);// rankine efficiency\n",
+" mprintf('\n (a) The Rankine efficiency is  =  %f percent\n',Re*100);\n",
+" \n",
+"// (b)\n",
+"vg1 = .1943;// specific volume at P1, [m^3/kg]\n",
+"vg2 = 10.02;// specific volume at P2, [m^3/kg]\n",
+"V1 = x1*vg1;// [m^3/kg]\n",
+"V2 = x2*vg2;// [m^3/kg]\n",
+"\n",
+"W1 = n/(n-1)*(P1*V1-P2*V2);// specific work done, [kJ/kg]\n",
+"\n",
+"//  from rankine cycle\n",
+"W2 = h1-h2;// [kJ/kg]\n",
+"mprintf('\n (b) The specific work done is  =  %f kJ/kg\n',W1);\n",
+"mprintf('\n     The specific work done (from rankine) is  =  %f kJ/kg\n',W2);\n",
+"\n",
+"// there is calculation mistake in the book so our answer is not matching\n",
+"\n",
+"//  End"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 10.8: Rankine_efficiency_steam_consumption_and_Carnot_efficiency.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"clear;\n",
+"clc;\n",
+"disp('Example 10.8');\n",
+"\n",
+"//  aim : To determine\n",
+"//  (a) the rankine fficiency\n",
+"//  (b) the specific steam consumption\n",
+"//  (c) the carnot efficiency of the cycle\n",
+"\n",
+"// given values\n",
+"P1 = 1100;// steam entering pressure, [kN/m^2]\n",
+"T1 = 273+250;// steam entering temperature, [K]\n",
+"P2 = 280;// pressure at point 2, [kN/m^2]\n",
+"P3 = 35;// pressure at point 3, [kN/m^2]\n",
+"\n",
+"// solution\n",
+"// (a)\n",
+"// from steam table, at P1  and T1 is\n",
+"hf1 = 2943;// [kJ/kg]\n",
+"hg1 = 2902;// [kJ/kg]\n",
+"h1 = hf1-.1*(hf1-hg1); // [kJ/kg]\n",
+"\n",
+"sf1 = 6.926;// [kJ/kg K]\n",
+"sg1 = 6.545;// [kJ/kg K]\n",
+"s1 = sf1-.1*(sf1-sg1);// [kJ/kg K]\n",
+"\n",
+"// at P2\n",
+"sf2 = 1.647;// [kJ/kg K] \n",
+" sg2 = 7.014;// [kJ/kg K]\n",
+"// s2=sf2+x2(sg2-sf2)\n",
+"// since expansion through turbine is isentropic so s1=s2\n",
+" // hence\n",
+" s2  = s1;\n",
+" x2 = (s2-sf2)/(sg2-sf2);// dryness fraction\n",
+" \n",
+" // at point 2\n",
+" hf2 = 551.4;// [kJ/kg]\n",
+" hfg2 = 2170.1;// [kJ/kg]\n",
+" h2 = hf2+x2*hfg2;// [kJ/kg]\n",
+" vg2 = .646;// [m^3/kg]\n",
+" v2 = x2*vg2;// [m^3/kg]\n",
+" \n",
+" // by Fig10.20.\n",
+" A6125 = h1-h2;// area of 6125, [kJ/kg]\n",
+" A5234 = v2*(P2-P3);// area 5234, [kJ/kg]\n",
+" W = A6125+A5234;// work done \n",
+" hf = 304.3;// specific enthalpy of water at condenser pressuer, [kJ/kg]\n",
+" ER = h1-hf;// energy received, [kJ/kg]\n",
+" Re = W/ER;// rankine efficiency\n",
+" mprintf('\n (a) The rankine efficiency is  =  %f percent\n',Re*100);\n",
+" \n",
+" // (b)\n",
+" kWh = 3600;// [kJ]\n",
+" SSC = kWh/W;// specific steam consumption, [kJ/kWh]\n",
+" mprintf('\n (b) The specific steam consumption is = %f kJ/kWh\n',SSC);\n",
+" \n",
+" // (c)\n",
+" // from steam table \n",
+"T3 = 273+72.7;// temperature at point 3\n",
+"CE = (T1-T3)/T1;// carnot efficiency\n",
+"mprintf('\n (c) The carnot efficiency of the cycle is  =  %f percent\n',CE*100);\n",
+"\n",
+"//  End"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 10.9: power_and_thermal_efficiency.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"clear;\n",
+"clc;\n",
+"disp('Example 10.9');\n",
+"\n",
+"// aim : To determine\n",
+"// (a) the theoretical power of steam passing through the turbine\n",
+"// (b) the thermal efficiency of the cycle\n",
+"// (c) the thermal efficiency of the cycle assuming there is no reheat\n",
+"\n",
+"// given values\n",
+"P1 = 6;// initial pressure, [MN/m^2]\n",
+"T1 = 450;// initial temperature, [C]\n",
+"P2 = 1;// pressure at stage 1, [MN/m^2]\n",
+"P3 = 1;// pressure at stage 2, [MN/m^2]\n",
+"T3 = 370;// temperature, [C]\n",
+"P4 = .02;// pressure at stage 3, [MN/m^2]\n",
+"P5 = .02;// pressure at stage 4, [MN/m^2]\n",
+"T5 = 320;// temperature, [C]\n",
+"P6 = .02;// pressure at stage 5, [MN/m^2]\n",
+"P7 = .02;// final pressure , [MN/m^2]\n",
+"\n",
+"// solution\n",
+"// (a) \n",
+"// using Fig 10.21\n",
+"h1 = 3305;// specific enthalpy, [kJ/kg]\n",
+"h2 = 2850;// specific enthalpy, [kJ/kg]\n",
+"h3 = 3202;// specific enthalpy, [kJ/kg]\n",
+"h4 = 2810;// specific enthalpy, [kJ/kg]\n",
+"h5 = 3115;// specific enthalpy, [kJ/kg]\n",
+"h6 = 2630;// specific enthalpy, [kJ/kg]\n",
+"h7 = 2215;// specific enthalpy, [kJ/kg]\n",
+"W = (h1-h2)+(h3-h4)+(h5-h6);// specific work through the turbine, [kJ/kg]\n",
+"mprintf('\n (a) The theoretical power/kg steam/s is  =  %f kW\n',W);\n",
+"\n",
+"// (b)\n",
+"// from steam table\n",
+"hf6 = 251.5;// [kJ/kg]\n",
+"\n",
+"TE1 = ((h1-h2)+(h3-h4)+(h5-h6))/((h1-hf6)+(h3-h2)+(h5-h4));// thermal efficiency\n",
+"mprintf('\n (b) The thermal efficiency of the cycle is  =  %f percent\n',TE1*100);\n",
+"\n",
+"// (c)\n",
+"// if there is no heat\n",
+"hf7 = hf6;\n",
+"TE2 = (h1-h7)/(h1-hf7);// thermal efficiency\n",
+"mprintf('\n (c) The thermal efficiency of the cycle if there is no heat is  =  %f percent\n',TE2*100);\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
+}
diff --git a/Basic_Engineering_Thermodynamics_by_R_Joel/11-The_steam_engine.ipynb b/Basic_Engineering_Thermodynamics_by_R_Joel/11-The_steam_engine.ipynb
new file mode 100644
index 0000000..703f063
--- /dev/null
+++ b/Basic_Engineering_Thermodynamics_by_R_Joel/11-The_steam_engine.ipynb
@@ -0,0 +1,586 @@
+{
+"cells": [
+ {
+		   "cell_type": "markdown",
+	   "metadata": {},
+	   "source": [
+       "# Chapter 11: The steam engine"
+	   ]
+	},
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 11.1: bore_stroke_and_speed.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"clear;\n",
+"clc;\n",
+"disp('Example 11.1')\n",
+"\n",
+"//  aim : To determine the \n",
+"// (a) bore of the cylinder\n",
+"// (b) piston stroke\n",
+"// (c) speed of the engine\n",
+"\n",
+"//  Given values\n",
+"P_req = 60;// power required to develop, [kW]\n",
+"P = 1.25;// boiler pressure, [MN/m^2]\n",
+"Pb = .13;// back pressure, [MN/m^2]\n",
+"cut_off = .3;// [stroke]\n",
+"k = .82;// diagram factor\n",
+"n = .78;// mechanical efficiency\n",
+"LN = 3;// mean piston speed, [m/s]\n",
+"\n",
+"// solution\n",
+"// (a)\n",
+"r = 1/cut_off;// expansion ratio\n",
+"Pm = P/r*(1+log(r))-Pb;// mean effective pressure, [MN/m^2]\n",
+"P_ind = P_req/n;// Actual indicated power developed, [kW]\n",
+"P_the = P_ind/k;// Theoretical indicated power developed, [kW]\n",
+"\n",
+"//  using indicated_power=Pm*LN*A\n",
+"//  Hence\n",
+"A = P_the/(Pm*LN)*10^-3;// piston area,[m^2]\n",
+"d = sqrt(4*A/%pi)*10^3;// bore ,[mm]\n",
+"mprintf('\n (a) The bore of the cylinder is  =  %f mm\n',d);\n",
+"\n",
+"// (b)\n",
+"// given that stroke is 1.25 times bore\n",
+"L = 1.25*d;// [mm]\n",
+"mprintf('\n (b) The piston stroke is  =  %f mm\n',L);\n",
+"\n",
+"// (c)\n",
+"// LN=mean piston speed, where L is stroke in meter and N is 2*rev/s,(since engine is double_acting)\n",
+"//  hence\n",
+"rev_per_sec = LN/(2*L*10^-3);// [rev/s]\n",
+"\n",
+"rev_per_min = rev_per_sec*60;// [rev/min]\n",
+"mprintf('\n (c) The speed of the engine is  =  %f rev/min\n',rev_per_min);\n",
+"\n",
+"//  End"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 11.2: diameter_and_stroke.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"clear;\n",
+"clc;\n",
+"disp('Example 11.2')\n",
+"\n",
+"//  aim : To determine the \n",
+"//  (a) the diameter of the cylinder\n",
+"//  (b) piston stroke\n",
+"//  (c) actual steam consumption and indicated thermal efficiency\n",
+"\n",
+"//  Given values\n",
+"P = 900;// inlet pressure, [kN/m^2]\n",
+"Pb = 140;//  exhaust pressure, [kN/m^2]\n",
+"cut_off =.4;// [stroke]\n",
+"k = .8;// diagram factor\n",
+"rs = 1.2;// stroke to bore ratio\n",
+"N = 4;// engine speed, [rev/s]\n",
+"ip = 22.5;// power output from the engine, [kW]\n",
+"\n",
+"// solution\n",
+"// (a)\n",
+"r = 1/cut_off;// expansion ratio\n",
+"Pm = P/r*(1+log(r))-Pb;// mean effective pressure, [kN/m^2]\n",
+"Pm = Pm*k;// actual mean effective pressure, [kN/m^2]\n",
+"\n",
+"// using ip=Pm*L*A*N\n",
+"// and L=r*d; where L is stroke and d is bore\n",
+"d = (ip/(Pm*rs*%pi/4*2*N))^(1/3);// diameter of the cylinder, [m]\n",
+"\n",
+"mprintf('\n (a) The diameter of the cylinder is  =  %f  mm\n',d*1000);\n",
+"\n",
+"// (b)\n",
+"L = rs*d;// stroke, [m]\n",
+"mprintf('\n (b) The piston stroke is  =  %f  mm\n',L*1000);\n",
+"\n",
+"// (c)\n",
+"SV = %pi/4*d^2*L;// stroke volume, [m^3]\n",
+"V = SV*cut_off*2*240*60;// volume of steam consumed per  hour, [m^3]\n",
+"v = .2148;// specific volume at 900 kN/m^2, [m^3/kg]\n",
+"SC = V/v;// steam consumed/h, [kg]\n",
+"ASC = 1.5*SC;// actual steam consumption/h, [kg]\n",
+"mprintf('\n (c) The actual steam consumption/h is  =  %f kg\n',ASC);\n",
+"\n",
+"m_dot = ASC/3600;// steam consumption,[kg/s] \n",
+"// from steam table\n",
+"hg = 2772.1;// specific enthalpy of inlet steam, [kJ/kg]\n",
+"hfe = 458.4;// specific liquid enthalpy at exhaust pressure, [kJ/kg]\n",
+"\n",
+"ITE = ip/(m_dot*(hg-hfe));// indicated thermal efficiency\n",
+"mprintf('\n     The indicated thermal efficiency is  =  %f  percent\n',ITE*100);\n",
+"\n",
+"//  End"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 11.3: diagram_factor_and_indicated_thermal_efficiency.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"clear;\n",
+"clc;\n",
+"disp('Example 11.3');\n",
+"\n",
+"//  aim : To determine\n",
+"//  (a) the diagram factor\n",
+"//  (b) the indicated thermal efficiency of the engine\n",
+"\n",
+"// given values\n",
+"d = 250*10^-3;// cylinder diameter, [m]\n",
+"L = 375*10^-3;// length of stroke, [m]\n",
+"P = 1000;// steam pressure , [kPa]\n",
+"x = .96;// dryness fraction of steam\n",
+"Pb = 55;// exhaust pressure, [kPa]\n",
+"r = 6;// expansion ratio\n",
+"ip = 45;// indicated power developed, [kW]\n",
+"N = 3.5;// speed of engine, [rev/s]\n",
+"m = 460;// steam consumption, [kg/h]\n",
+"\n",
+"// solution\n",
+"// (a)\n",
+"Pm = P/r*(1+log(r))-Pb;// [kN/m^3]\n",
+"A = %pi*(d)^2/4;// area, [m^2]\n",
+"tip = Pm*L*A*N*2;// theoretical indicated power, [kW]\n",
+"k = ip/tip;// diagram factor\n",
+"mprintf('\n (a) The diagram factor is  =  %f\n',k);\n",
+"\n",
+"// (b)\n",
+"// from steam table at 1 MN/m^2\n",
+"hf = 762.6;// [kJ/kg]\n",
+"hfg = 2013.6;// [kJ/kg]\n",
+"// so \n",
+"h1 = hf+x*hfg;// specific enthalpy of steam at 1MN/m^2, [kJ/kg]\n",
+"// minimum specific enthalpy in engine at 55 kPa \n",
+"hf = 350.6;// [kJ/kg]\n",
+"// maximum energy available in engine is\n",
+"h = h1-hf;// [kJ/kg]\n",
+"ITE = ip/(m*h/3600)*100;// indicated thermal efficiency\n",
+"mprintf('\n (b) The indicated thermal efficiency is  =  %f  percent\n ',ITE);\n",
+"\n",
+"//  End"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 11.4: steam_consumption.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"clear;\n",
+"clc;\n",
+"disp('Example 11.4');\n",
+"\n",
+"// aim : To determine\n",
+"// steam consumption\n",
+"\n",
+"// given values\n",
+"P1 = 11;// power, [kW]\n",
+"m1 = 276;// steam use/h when developing power P1,[kW]\n",
+"ip = 8;// indicated power output, [kW]\n",
+"B = 45;// steam used/h at no load, [kg]\n",
+"\n",
+"// solution\n",
+"// using graph of Fig.11.9 \n",
+"A = (m1-B)/P1;// slop of line, [kg/kWh]\n",
+"W = A*ip+B;// output, [kg/h]\n",
+"mprintf('\n The steam consumption is  =  %f  kg/h\n ',W);\n",
+"\n",
+"//  End"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 11.5: pressure_power_output_and_steam_consumption.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"clear;\n",
+"clc;\n",
+"disp('Example 11.5');\n",
+"\n",
+"//  aim : To determine\n",
+"// (a) the intermediate pressure\n",
+"// (b) the indicated power output\n",
+"// (c) the steam consumption of the engine\n",
+"\n",
+"// given values\n",
+"P1 = 1400;// initial pressure, [kN/m^2]\n",
+"x = .9;// dryness fraction\n",
+"P5 = 35;// exhaust pressure\n",
+"k = .8;// diagram factor of low-pressure cylindaer\n",
+"L = 350*10^-3;// stroke of both the cylinder, [m]\n",
+"dhp = 200*10^-3;// diameter of high pressure cylinder, [m]\n",
+"dlp = 300*10^-3;// diameter of low-pressure cylinder, [m]\n",
+"N = 300;// engine speed, [rev/min]\n",
+"\n",
+"// solution\n",
+"// taking reference Fig.11.13\n",
+"Ahp = %pi/4*dhp^2;// area of high-pressure cylinder, [m^2]\n",
+"Alp = %pi/4*dlp^2;// area of low-pressure cylinder, [m^2]\n",
+"// for equal initial piston loads\n",
+"// (P1-P7)Ahp=(P7-P5)Alp\n",
+"deff('[x]=f(P7)','x=(P1-P7)*Ahp-(P7-P5)*Alp');\n",
+"P7 = fsolve(0,f);// intermediate pressure, [kN/m^2]\n",
+"mprintf('\n (a) The intermediate pressure is  =  %f kN/m^2\n ',P7);\n",
+"\n",
+"// (b)\n",
+"V6 = Ahp*L;// volume of high-pressure cylinder, [m^3]\n",
+"P2 = P1;\n",
+"P6 = P7;\n",
+"// using P2*V2=P6*V6\n",
+"V2 = P6*V6/P2; // [m^3]\n",
+"V1 = Alp*L;// volume of low-pressure cylinder, [m^3]\n",
+"R = V1/V2;// expansion ratio\n",
+"Pm = P1/R*(1+log(R))-P5;// effective pressure of low-pressure cylinder, [kn/m^2]\n",
+"Pm = k*Pm;// actual effective pressure, [kN/m^2]\n",
+"ip = Pm*L*Alp*N*2/60;// indicated power, [kW]\n",
+"mprintf('\n (b) The indicated power is  =  %f kW\n',ip);\n",
+"\n",
+"// (c) \n",
+"COV = V1/ R;// cut-off  volume in high-pressure cylinder, [m^3]\n",
+"V = COV*N*2*60;// volume of steam admitted/h\n",
+"// from steam table\n",
+"vg = .1407;// [m^3/kg]\n",
+"AV = x*vg;// specific volume of admission steam, [m^3/kg]\n",
+"m = V/AV;// steam consumption, [kg/h]\n",
+"mprintf('\n (c) The steam consumption of the engine is  =  %f  kg/h\n',m);\n",
+"\n",
+"//  End "
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 11.6: power_output_diameter_and_intermediate_pressure.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"clear;\n",
+"clc;\n",
+"disp('Example 11.6');\n",
+"\n",
+"// aim : To determine\n",
+"// (a) the indicated power output\n",
+"// (b) the diameter of high-pressure cylinder of the engine\n",
+"// (c) the intermediate pressure\n",
+"\n",
+"// given values\n",
+"P = 1100;// initial pressure, [kN/m^2]\n",
+"Pb = 28;// exhaust pressure\n",
+"k = .82;// diagram factor of low-pressure cylindaer\n",
+"L = 600*10^-3;// stroke of both the cylinder, [m]\n",
+"dlp = 600*10^-3;// diameter of low-pressure cylinder, [m]\n",
+"N = 4;// engine speed, [rev/s]\n",
+"R = 8;// expansion ratio\n",
+"\n",
+"// solution\n",
+"// taking reference Fig.11.13\n",
+"// (a)\n",
+"Pm = P/R*(1+log(R))-Pb;// effective pressure of low-pressure cylinder, [kn/m^2]\n",
+"Pm = k*Pm;// actual effective pressure, [kN/m^2]\n",
+"Alp = %pi/4*dlp^2;// area of low-pressure cylinder, [m^2]\n",
+"ip = Pm*L*Alp*N*2;// indicated power, [kW]\n",
+"mprintf('\n (a) The indicated power is  =  %f kW\n',ip);\n",
+"\n",
+"// (b)\n",
+"// work done by both cylinder is same as area of diagram\n",
+"w = Pm*Alp*L;// [kJ]\n",
+"W = w/2;// work done/cylinder, [kJ]\n",
+"V2 = Alp*L/8;// volume, [m63]\n",
+"P2 = P;// [kN/m^2]\n",
+"//  using area A1267=P2*V2*log(V6/V2)=W\n",
+"V6 = V2*exp(W/(P2*V2));// intermediate volume, [m^3]\n",
+"// using Ahp*L=%pi/4*dhp^2*L=V6\n",
+"dhp = sqrt(V6*4/L/%pi);// diameter of high-pressure cylinder, [m]\n",
+"mprintf('\n (b) The diameter of high-pressure cylinder is  =  %f  mm\n',dhp*1000);\n",
+"\n",
+"// (c)\n",
+"// using P2*V2=P6*V6\n",
+"P6 = P2*V2/V6; // intermediate pressure, [kN/m^2]\n",
+"mprintf('\n (c) The intermediate opressure is =   %f  kN/m^2\n',P6);\n",
+"\n",
+"//  End "
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 11.7: speed_and_diameter.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"clear;\n",
+"clc;\n",
+"disp('Example 11.7');\n",
+"\n",
+"// aim : To determine\n",
+"// (a) The speed of the engine\n",
+"// (b) the diameter of the high pressure cylinder\n",
+"\n",
+"// given values\n",
+"ip = 230;// indicated power, [kW]\n",
+"P = 1400;// admission pressure, [kN/m^2]\n",
+"Pb = 35;// exhaust pressure, [kN/m^2]\n",
+"R = 12.5;// expansion ratio\n",
+"d1 = 400*10^-3;// diameter of low pressure cylinder, [m]\n",
+"L = 500*10^-3;// stroke of both the cylinder, [m]\n",
+"k = .78;// diagram factor\n",
+"rv = 2.5;// expansion ratio of high pressure cylinder\n",
+"\n",
+"// solution\n",
+"// (a)\n",
+"Pm = P/R*(1+log(R))-Pb;// mean effective pressure in low pressure cylinder, [kN/m^2]\n",
+"ipt = ip/k;// theoretical indicated power, [kw]\n",
+"// using ip=Pm*L*A*N\n",
+"A = %pi/4*d1^2;// area , [m^2]\n",
+"N = ipt/(Pm*L*A*2);// speed, [rev/s]\n",
+"mprintf('\n (a) The engine speed is  =  %f  rev/s\n',N);\n",
+"\n",
+"// (b)\n",
+"Vl = A*L;// volume of low pressure cylinder, [m^3]\n",
+"COV = Vl/R;// cutt off volume of hp cylinder, [m^3]\n",
+"V = COV*rv;// total volume, [m^3]\n",
+"\n",
+"//  V = %pi/4*d^2*L, so\n",
+"d = sqrt(4*V/%pi/L);// diameter of high pressure cylinder, [m]\n",
+"mprintf('\n (b) The diameter of the high pressure cylinder is  =  %f  mm\n',d*1000);\n",
+"\n",
+"//  End"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 11.8: mean_effective_pressures_diagram_factor_and_indicated_power.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"clear;\n",
+"clc;\n",
+"disp('Example 11.8');\n",
+"\n",
+"// aim : To determine\n",
+"// (a) the actual and hypothetical mean effective pressures referred to the low-pressure cylinder\n",
+"// (b) the overall diagram factor\n",
+"// (c) the indicated power \n",
+"\n",
+"// given values\n",
+"P = 1100;// steam supply pressure, [kN/m^2]\n",
+"Pb = 32;// back pressure, [kN/m^2]\n",
+"d1 = 300*10^-3;// cylinder1 diameter, [m]\n",
+"d2 = 600*10^-3;// cylinder2 diameter, [m]\n",
+"L = 400*10^-3;// common stroke of both cylinder, [m]\n",
+"\n",
+"A1 = 12.5;//  average area of indicated diagram for HP, [cm^2]\n",
+"A2 = 11.4;//  average area of indicated diagram for LP, [cm^2]\n",
+"\n",
+"P1 = 270;// indicator calibration, [kN/m^2/ cm]\n",
+"P2 = 80;// spring calibration, [kN/m^2/ cm]\n",
+"N = 2.7;// engine speed, [rev/s]\n",
+"l = .75;// length of both diagram, [m]\n",
+"\n",
+"// solution\n",
+"// (a)\n",
+"// for HP cylinder\n",
+"Pmh = P1*A1/7.5;// [kN/m^2]\n",
+"F = Pmh*%pi/4*d1^2;// force on HP, [kN]\n",
+"PmH = Pmh*(d1/d2)^2;// pressure referred to LP cylinder, [kN/m^2]\n",
+"PmL = P2*A2/7.5;// pressure for LP cylinder, [kN/m^2]\n",
+"PmA = PmH+PmL;// actual effective pressure referred to LP cylinder, [kN/m^2]\n",
+"\n",
+"Ah = %pi/4*d1^2;// area of HP cylinder, [m^2]\n",
+"Vh = Ah*L;// volume of HP cylinder, [m^3]\n",
+"CVh = Vh/3;// cut-off volume of HP cylinder, [m^3]\n",
+"Al = %pi/4*d2^2;// area of LP cylinder, [m^2]\n",
+"Vl = Al*L;// volume of LP cylinder, [m^3]\n",
+"\n",
+"R = Vl/CVh;// expansion ratio\n",
+"Pm = P/R*(1+log(R))-Pb;// hypothetical mean effective pressure referred to LP cylinder, [kN/m^2]\n",
+"\n",
+"mprintf('\n (a) The actual mean effective pressure referred to LP cylinder is  =  %f kN/m^2\n',PmA);\n",
+"mprintf('\n     The hypothetical mean effective pressure referred to LP cylinder is  =  %f kN/m^2\n',Pm);\n",
+"\n",
+"// (a)\n",
+"ko = PmA/Pm;// overall diagram factor\n",
+"mprintf('\n (b) The overall diagram factor is  =  %f\n',ko);\n",
+"\n",
+"// (c) \n",
+"ip = PmA*L*Al*N*2;// indicated power, [kW]\n",
+"mprintf('\n (c) The indicated power is  =  %f kW\n',ip);\n",
+"\n",
+"//   End"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 11.9: EX11_9.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"clear;\n",
+"clc;\n",
+"disp('Example 11.9');\n",
+"\n",
+"// aim : To determine\n",
+"// (a) the actual and hypothetical mean effective pressures referred to the low-pressure cylinder\n",
+"// (b) the overall diagram factor\n",
+"// (c) the pecentage of the total indicated power developed in each cylinder\n",
+"\n",
+"// given values\n",
+"P = 1400;// steam supply pressure, [kN/m^2]\n",
+"Pb = 20;// back pressure, [kN/m^2]\n",
+"Chp = .6;// cut-off in HP cylinder, [stroke]\n",
+"dh = 300*10^-3;// HP diameter, [m]\n",
+"di = 500*10^-3;// IP diameter, [m]\n",
+"dl = 900*10^-3;// LP diameter, [m]\n",
+"\n",
+"Pm1 = 590;// actual pressure of HP cylinder, [kN/m^2]\n",
+"Pm2 = 214;// actual pressure of IP cylinder, [kN/m^2]\n",
+"Pm3 = 88;// actual pressure of LP cylinder, [kN/m^2]\n",
+"\n",
+"// solution\n",
+"// (a)\n",
+"// for HP cylinder\n",
+"PmH = Pm1*(dh/dl)^2;// PmH referred to LP cylinder, [kN/m^2]\n",
+"// for IP cylinder\n",
+"PmI = Pm2*(di/dl)^2;// PmI referred to LP cylinder, [kN/m^2]\n",
+"PmA = PmH+PmI+Pm3;// actual mean effective pressure referred to LP cylinder, [kN/m^2]\n",
+"\n",
+"R = dl^2/(dh^2*Chp);// expansion ratio\n",
+"Pm = P/R*(1+log(R))-Pb;// hypothetical mean effective pressure referred to LP cylinder, [kN/m^2]\n",
+"\n",
+"mprintf('\n (a) The actual mean effective pressure referred to LP cylinder is  =  %f  kN/m^2\n',PmA);\n",
+"mprintf('\n      The hypothetical mean effective pressure referred to LP cylinder is  =  %f  kN/m^2\n',Pm);\n",
+"\n",
+"// (b)\n",
+"ko = PmA/Pm;// overall diagram factor\n",
+"mprintf('\n (b) The overall diagram factor is  =  %f\n',ko);\n",
+"\n",
+"// (c)\n",
+"HP = PmH/PmA*100;// %age of indicated power developed in HP\n",
+"IP = PmI/PmA*100; // %age of indicated power developed in IP\n",
+"LP = Pm3/PmA*100; // %age of indicated power developed in LP\n",
+"mprintf('\n (c) The pecentage of the total indicated power developed in HP cylinder is  =  %f percent\n',HP);\n",
+"mprintf('\n     The pecentage of the total indicated power developed in IP cylinder is  =  %f percent\n',IP);\n",
+"mprintf('\n     The pecentage of the total indicated power developed in LP cylinder is  =  %f  percent\n',LP);\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
+}
diff --git a/Basic_Engineering_Thermodynamics_by_R_Joel/12-Nozzle.ipynb b/Basic_Engineering_Thermodynamics_by_R_Joel/12-Nozzle.ipynb
new file mode 100644
index 0000000..094eaa5
--- /dev/null
+++ b/Basic_Engineering_Thermodynamics_by_R_Joel/12-Nozzle.ipynb
@@ -0,0 +1,274 @@
+{
+"cells": [
+ {
+		   "cell_type": "markdown",
+	   "metadata": {},
+	   "source": [
+       "# Chapter 12: Nozzle"
+	   ]
+	},
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 12.1: area_and_Mach_number.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"clear;\n",
+"clc;\n",
+"disp('Example 12.1');\n",
+"\n",
+"//  aim : To determine the\n",
+"//  (a) throat area\n",
+"//   (b) exit area\n",
+"//   (c) Mach number at exit\n",
+"\n",
+"// Given values\n",
+"P1 = 3.5;// inlet pressure of air, [MN/m^2]\n",
+"T1 = 273+500;// inlet temperature of air, [MN/m^2]\n",
+"P2 = .7;// exit pressure, [MN/m^2]\n",
+"m_dot = 1.3;// flow rate of air, [kg/s]\n",
+"Gamma = 1.4;// heat capacity ratio\n",
+"R = .287;// [kJ/kg K]\n",
+"\n",
+"// solution\n",
+"// given expansion may be considered to be adiabatic and to follow the law PV^Gamma=constant\n",
+"// using ideal gas law\n",
+"v1 = R*T1/P1*10^-3;// [m^3/kg]\n",
+"Pt = P1*(2/(Gamma+1))^(Gamma/(Gamma-1));// critical pressure, [MN/m^2]\n",
+"\n",
+"//  velocity at throat is\n",
+"Ct = sqrt(2*Gamma/(Gamma-1)*P1*10^6*v1*(1-(Pt/P1)^(((Gamma-1)/Gamma))));// [m/s]\n",
+"vt = v1*(P1/Pt)^(1/Gamma);// [m^3/kg]\n",
+"// using m_dot/At=Ct/vt\n",
+"At = m_dot*vt/Ct*10^6;// throat area, [mm^2]\n",
+"mprintf('\n (a) The throat area is  =  %f mm^2\n',At);\n",
+"\n",
+"// (b)\n",
+"//  at exit\n",
+"C2 = sqrt(2*Gamma/(Gamma-1)*P1*10^6*v1*(1-(P2/P1)^(((Gamma-1)/Gamma))));// [m/s]\n",
+"v2 = v1*(P1/P2)^(1/Gamma);// [m^3/kg]\n",
+"A2 = m_dot*v2/C2*10^6;// exit area, [mm^2]\n",
+"\n",
+"mprintf('\n (b) The exit area is  =  %f  mm^2\n',A2);\n",
+"\n",
+"//  (c)\n",
+"M = C2/Ct;\n",
+"mprintf('\n (c) The Mach number at exit is  =  %f\n',M);\n",
+"\n",
+"//  End"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 12.2: increases_in_pressure_temperature_and_internal_energy.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"clear;\n",
+"clc;\n",
+"disp('Example 12.2');\n",
+"\n",
+"//  aim : To determine the increases in pressure, temperature and internal energy per kg of air\n",
+"\n",
+"//  Given values\n",
+"T1 = 273;// [K]\n",
+"P1 = 140;// [kN/m^2]\n",
+"C1 = 900;// [m/s]\n",
+"C2 = 300;// [m/s]\n",
+"cp = 1.006;// [kJ/kg K]\n",
+"cv =.717;// [kJ/kg K]\n",
+"\n",
+"// solution\n",
+"R = cp-cv;// [kJ/kg K]\n",
+"Gamma = cp/cv;// heat capacity ratio\n",
+"//  for frictionless adiabatic flow, (C2^2-C1^2)/2=Gamma/(Gamma-1)*R*(T1-T2)\n",
+"\n",
+"T2 =T1-((C2^2-C1^2)*(Gamma-1)/(2*Gamma*R))*10^-3; //  [K]\n",
+"T_inc = T2-T1;// increase in temperature [K]\n",
+"\n",
+"P2 = P1*(T2/T1)^(Gamma/(Gamma-1));// [MN/m^2]\n",
+"P_inc = (P2-P1)*10^-3;// increase in pressure,[MN/m^2]\n",
+"\n",
+"U_inc = cv*(T2-T1);//  Increase in internal energy per kg,[kJ/kg]\n",
+"mprintf('\n The increase in pressure is  =  %f  MN/m^2\n',P_inc);\n",
+"mprintf('\n Increase in temperature is  =  %f  K\n',T_inc);\n",
+"mprintf('\n Increase in internal energy is  =   %f  kJ/kg\n',U_inc);\n",
+"\n",
+"// there is minor variation in result\n",
+"\n",
+"//  End"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 12.3: throat_area_and_degree_of_undercooling.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"clear;\n",
+"clc;\n",
+"disp('Example 12.3');\n",
+"\n",
+"// aim : To determine the \n",
+"//  (a) throat and exit areas\n",
+"//   (b) degree of undercooling at exit\n",
+"// Given values\n",
+"P1 = 2;// inlet pressure of air, [MN/m^2]\n",
+"T1 = 273+325;// inlet temperature of air, [MN/m^2]\n",
+"P2 = .36;// exit pressure, [MN/m^2]\n",
+"m_dot = 7.5;// flow rate of air, [kg/s]\n",
+"n = 1.3;// polytropic index\n",
+"\n",
+"// solution\n",
+"// (a)\n",
+"//  using steam table\n",
+"v1 = .132;// [m^3/kg]\n",
+"// given expansion following law PV^n=constant\n",
+"\n",
+"Pt = P1*(2/(n+1))^(n/(n-1));// critical pressure, [MN/m^2]\n",
+"\n",
+"//velocity at throat is\n",
+"Ct = sqrt(2*n/(n-1)*P1*10^6*v1*(1-(Pt/P1)^(((n-1)/n))));// [m/s]\n",
+"vt = v1*(P1/Pt)^(1/n);// [m^3/kg]\n",
+"// using m_dot/At=Ct/vt\n",
+"At = m_dot*vt/Ct*10^6;// throat area, [mm^2]\n",
+"mprintf('\n (a) The throat area is =  %f  mm^2\n',At);\n",
+"\n",
+"// at exit\n",
+"C2 = sqrt(2*n/(n-1)*P1*10^6*v1*(1-(P2/P1)^(((n-1)/n))));// [m/s]\n",
+"v2 = v1*(P1/P2)^(1/n);// [m^3/kg]\n",
+"A2 = m_dot*v2/C2*10^6;// exit area, [mm^2]\n",
+"\n",
+"mprintf('\n      The exit area is  =  %f  mm^2\n',A2);\n",
+"\n",
+"//  (b)\n",
+"T2 = T1*(P2/P1)^((n-1)/n);//outlet temperature, [K]\n",
+"t2 = T2-273;//[C]\n",
+"//  at exit pressure saturation temperature is\n",
+"ts = 139.9;// saturation temperature,[C]\n",
+"Doc = ts-t2;// Degree of undercooling,[C]\n",
+"mprintf('\n (b) The Degree of undercooling at exit is  =  %f C\n',Doc);\n",
+"\n",
+"// There is some calculation mistake in the book so answer is not matching\n",
+"\n",
+"//  End"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 12.4: velocities_and_areas.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"clear;\n",
+"clc;\n",
+"disp('Example 12.4');\n",
+"\n",
+"// aim : To determine the \n",
+"//  (a) throat and exit velocities\n",
+"//  (b) throat and exit areas\n",
+"\n",
+"// Given values\n",
+"P1 = 2.2;// inlet pressure, [MN/m^2]\n",
+"T1 = 273+260;// inlet temperature, [K]\n",
+"P2 = .4;// exit pressure,[MN/m^2]\n",
+"eff = .85;// efficiency of the nozzle after throat\n",
+"m_dot = 11;// steam flow rate in the nozzle, [kg/s]\n",
+"\n",
+"// solution\n",
+"// (a)\n",
+"// assuming steam is following same law as previous question 12.3\n",
+"Pt = .546*P1;// critical pressure,[MN/m^2]\n",
+"// from Fig. 12.6\n",
+"h1 = 2940;// [kJ/kg]\n",
+"ht = 2790;// [kJ/kg]\n",
+"\n",
+"Ct = sqrt(2*(h1-ht)*10^3);// [m/s]\n",
+"\n",
+"//  again from Fig. 12.6\n",
+"h2_prime = 2590;// [kJ/kg]\n",
+"// using eff = (ht-h2)/(ht-h2_prime)\n",
+"\n",
+"h2 = ht-eff*(ht-h2_prime); // [kJ/kg]\n",
+"\n",
+"C2 = sqrt(2*(h1-h2)*10^3);// [m/s]\n",
+"\n",
+"// (b)\n",
+"// from chart\n",
+"vt = .16;// [m^3/kg]\n",
+"v2 = .44;// [m^3/kg]\n",
+"//  using m_dot*v=A*C\n",
+"At = m_dot*vt/Ct*10^6;// throat area, [mm^2]\n",
+"\n",
+"A2 = m_dot*v2/C2*10^6;// throat area, [mm^2]\n",
+"\n",
+"mprintf('\n (a) The throat velocity is  =  %f  m/s\n',Ct);\n",
+"mprintf('\n      The exit velocity is  =  %f  m/s\n',C2);\n",
+"mprintf('\n (b) The throat area is  =  %f  mm^2\n',At);\n",
+"mprintf('\n      The throat area is  =  %f  mm^2\n',A2);\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
+}
diff --git a/Basic_Engineering_Thermodynamics_by_R_Joel/13-Steam_turbines.ipynb b/Basic_Engineering_Thermodynamics_by_R_Joel/13-Steam_turbines.ipynb
new file mode 100644
index 0000000..f74dd9d
--- /dev/null
+++ b/Basic_Engineering_Thermodynamics_by_R_Joel/13-Steam_turbines.ipynb
@@ -0,0 +1,284 @@
+{
+"cells": [
+ {
+		   "cell_type": "markdown",
+	   "metadata": {},
+	   "source": [
+       "# Chapter 13: Steam turbines"
+	   ]
+	},
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 13.1: power_developed_and_kinetic_energy.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"clear;\n",
+"clc;\n",
+"disp('Example 13.1');\n",
+"\n",
+"//  aim : To determine \n",
+"//  the power developed for a steam flow of 1 kg/s at the blades and the kinetic energy of the steam finally leaving the wheel\n",
+"\n",
+"//  Given values\n",
+"alfa = 20;//  blade angle, [degree]\n",
+"Cai = 375;// steam exit velocity in the nozzle,[m/s]\n",
+"U = 165;// blade speed, [m/s]\n",
+"loss = .15;//  loss of velocity due to friction\n",
+"\n",
+"//  solution\n",
+"//  using Fig13.12,\n",
+"Cvw = 320;// change in velocity of whirl, [m/s]\n",
+"cae = 132.5;// absolute velocity at exit, [m/s]\n",
+"Pds = U*Cvw*10^-3;// Power developed for steam flow of 1 kg/s, [kW]\n",
+"Kes = cae^2/2*10^-3;// Kinetic energy change of steam, [kW/kg] \n",
+"\n",
+"mprintf('\n The power developed for a steam flow of 1 kg/s is  =  %f kW\n',Pds)\n",
+"mprintf('\n The energy of steam finally leaving the wheel is  =  %f  kW/kg\n',Kes);\n",
+"\n",
+"//  End"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 13.2: angle_of_blade_work_done_diagram_efficiency_and_end_thrust.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"clear;\n",
+"clc;\n",
+"disp('Example 13.2');\n",
+"\n",
+"// aim : To determine\n",
+"// (a) the entry angle of  the blades\n",
+"// (b) the work done per kilogram of steam per second\n",
+"// (c) the diagram efficiency\n",
+"// (d) the end-thrust per kilogram of steam per second\n",
+"\n",
+"// given values\n",
+"Cai = 600;// steam velocity, [m/s]\n",
+"sia = 25;// steam inlet angle with blade, [degree]\n",
+"U = 255;// mean blade speed, [m/s]\n",
+"sea = 30;// steam exit angle with blade,[degree] \n",
+"\n",
+"// solution\n",
+"// (a)\n",
+"// using Fig.13.13(diagram for example 13.2)\n",
+"eab = 41.5;// entry angle of blades, [degree]\n",
+"mprintf('\n (a) The angle of blades is  =  %f  degree\n',eab);\n",
+"\n",
+"// (b)\n",
+"Cwi_plus_Cwe = 590;// velocity of whirl, [m/s]\n",
+"W = U*(Cwi_plus_Cwe);// work done on the blade,[W/kg]\n",
+"mprintf('\n (b) The work done on the blade is  =  %f kW/kg\n',W*10^-3);\n",
+"\n",
+"// (c)\n",
+"De = 2*U*(Cwi_plus_Cwe)/Cai^2;// diagram efficiency \n",
+"mprintf('\n (c) The diagram efficiency is  =  %f percent\n',De*100);\n",
+"\n",
+"// (d)\n",
+"// again from the diagram\n",
+"Cfe_minus_Cfi = -90;// change invelocity of flow, [m/s]\n",
+"Eth = Cfe_minus_Cfi;// end-thrust, [N/kg s]\n",
+"mprintf('\n (d) The End-thrust is  =  %f  N/kg',Eth);\n",
+"\n",
+"//  End"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 13.3: power_output_and_diagram_efficiency.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"clear;\n",
+"clc;\n",
+"disp('Example 13.3');\n",
+"\n",
+"// aim : To determine\n",
+"// (a) the power output of the turbine\n",
+"// (b) the diagram efficiency\n",
+"\n",
+"// given values\n",
+"U = 150;// mean blade speed, [m/s]\n",
+"Cai1 = 675;// nozzle speed, [m/s]\n",
+"na = 20;// nozzle angle, [degree]\n",
+"m_dot = 4.5;// steam flow rate, [kg/s]\n",
+"\n",
+"// solution\n",
+"// from Fig. 13.15(diagram 13.3)\n",
+"Cw1 = 915;// [m/s]\n",
+"Cw2 = 280;// [m/s]\n",
+"\n",
+"// (a)\n",
+"P = m_dot*U*(Cw1+Cw2);// power of turbine,[W]\n",
+"mprintf('\n (a) The power of turbine is  =  %f kW\n',P*10^-3);\n",
+"\n",
+"// (b)\n",
+"De = 2*U*(Cw1+Cw2)/Cai1^2;// diagram efficiency\n",
+"mprintf('\n (b) The diagram efficiency is  =  %f percent\n',De*100);\n",
+"\n",
+"//  End"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 13.4: power_output_specific_enthalpy_drop_and_increase_in_relative_velocity.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"clear;\n",
+"clc;\n",
+"disp('Example 13.4');\n",
+"\n",
+"// aim : To determine\n",
+"// (a) the power output of the stage\n",
+"// (b) the specific enthalpy drop in the stage\n",
+"// (c) the percentage increase in relative velocity in the moving blades due to expansion in the bladse\n",
+"\n",
+"// given values\n",
+"N = 50;// speed, [m/s]\n",
+"d = 1;// blade ring diameter, [m]\n",
+"nai = 50;// nozzle inlet angle, [degree]\n",
+"nae = 30;// nozzle exit angle, [degree]\n",
+"m_dot = 600000;// steam flow rate, [kg/h]\n",
+"se = .85;// stage efficiency\n",
+"\n",
+"// solution\n",
+"// (a)\n",
+"U = %pi*d*N;// mean blade speed, [m/s]\n",
+"// from Fig. 13.17(diagram 13.4)\n",
+"Cwi_plus_Cwe = 444;// change in whirl speed, [m/s]\n",
+"P = m_dot*U*Cwi_plus_Cwe/3600;// power output of the stage, [W]\n",
+"mprintf('\n (a) The power output of the stage is  =  %f MW\n',P*10^-6);\n",
+"\n",
+"// (b)\n",
+"h = U*Cwi_plus_Cwe/se;// specific enthalpy,[J/kg]\n",
+"mprintf('\n (b) The specific enthalpy drop in the stage is  =  %f  kJ/kg\n ',h*10^-3);\n",
+"\n",
+"// (c)\n",
+"// again from diagram\n",
+"Cri = 224;// [m/s]\n",
+"Cre = 341;// [m/s]\n",
+"Iir = (Cre-Cri)/Cri;// increase in relative velocity\n",
+"mprintf('\n (c) The increase in relative velocity is  =  %f  percent\n',Iir*100);\n",
+"\n",
+"//  End"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 13.5: blade_height_power_developed_and_specific_enthalpy_drop.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"clear;\n",
+"clc;\n",
+"disp('Example 13.5');\n",
+"\n",
+"// aim : To determine\n",
+"// (a) the blade height of the stage\n",
+"// (b) the power developed in the stage\n",
+"// (c) the specific enthalpy drop at the stage\n",
+"\n",
+"// given values\n",
+"U = 60;// mean blade speed, [m/s]\n",
+"P = 350;// steam pressure, [kN/m^2]\n",
+"T = 175;// steam temperature, [C]\n",
+"nai = 30;// stage inlet angle, [degree]\n",
+"nae = 20;// stage exit angle, [degree] \n",
+"\n",
+"// solution\n",
+"// (a)\n",
+"m_dot = 13.5;// steam flow rate, [kg/s]\n",
+"// at given T and P\n",
+"v = .589;// specific volume, [m^3/kg]\n",
+"// given H=d/10, so\n",
+"H = sqrt(m_dot*v/(%pi*10*60));// blade height, [m]\n",
+"mprintf('\n (a) The blade height at this stage is  =  %f  mm\n',H*10^3);\n",
+"\n",
+"// (b)\n",
+"Cwi_plus_Cwe = 270;// change in whirl speed, [m/s]\n",
+"P = m_dot*U*(Cwi_plus_Cwe);// power developed, [W]\n",
+"mprintf('\n (b) The power developed is  =  %f  kW\n',P*10^-3);\n",
+"\n",
+"// (c)\n",
+"s = .85;// stage efficiency\n",
+"h = U*Cwi_plus_Cwe/s;// specific enthalpy,[J/kg]\n",
+"mprintf('\n (a) The specific enthalpy drop in the stage is  =  %f  kJ/kg',h*10^-3);\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
+}
diff --git a/Basic_Engineering_Thermodynamics_by_R_Joel/14-Air_and_gas_compressors.ipynb b/Basic_Engineering_Thermodynamics_by_R_Joel/14-Air_and_gas_compressors.ipynb
new file mode 100644
index 0000000..0282db3
--- /dev/null
+++ b/Basic_Engineering_Thermodynamics_by_R_Joel/14-Air_and_gas_compressors.ipynb
@@ -0,0 +1,472 @@
+{
+"cells": [
+ {
+		   "cell_type": "markdown",
+	   "metadata": {},
+	   "source": [
+       "# Chapter 14: Air and gas compressors"
+	   ]
+	},
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 14.1: EX14_1.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"clear;\n",
+"clc;\n",
+"disp(' Example 14.1');\n",
+"\n",
+"// aim : To determine \n",
+"// (a) the free air delivered\n",
+"// (b) the volumetric efficiency\n",
+"// (c) the air delivery temperature\n",
+"// (d) the cycle power\n",
+"// (e) the isothermal efficiency\n",
+"\n",
+"// given values\n",
+"d = 200*10^-3;// bore, [m]\n",
+"L = 300*10^-3;// stroke, [m]\n",
+"N = 500;// speed, [rev/min]\n",
+"n = 1.3;// polytropic index\n",
+"P1 = 97;// intake pressure, [kN/m^2]\n",
+"T1 = 273+20;// intake temperature, [K]\n",
+"P3 = 550;// compression pressure, [kN/m^2]\n",
+"\n",
+"// solution\n",
+"// (a)\n",
+"P4 = P1;\n",
+"P2 = P3;\n",
+"Pf = 101.325;// free air pressure, [kN/m^2]\n",
+"Tf = 273+15;// free air temperature, [K]\n",
+"SV = %pi/4*d^2*L;// swept volume, [m^3]\n",
+"V3 = .05*SV;// [m^3]\n",
+"V1 = SV+V3;// [m^3]\n",
+"V4 = V3*(P3/P4)^(1/n);// [m^3]\n",
+"ESV = (V1-V4)*N;// effective swept volume/min, [m^3]\n",
+"// using PV/T=constant\n",
+"Vf = P1*ESV*Tf/(Pf*T1);// free air delivered, [m^3/min]\n",
+"mprintf('\n (a) The free air delivered is  =  %f  m^3/min\n',Vf);\n",
+"\n",
+"// (b)\n",
+"VE = Vf/(N*(V1-V3));// volumetric efficiency\n",
+"mprintf('\n (b) The volumetric efficiency is  =  %f  percent\n',VE*100);\n",
+"\n",
+"// (c)\n",
+"T2 = T1*(P2/P1)^((n-1)/n);//  free air temperature, [K]\n",
+"mprintf('\n (c) The air delivery temperature is  =  %f  C\n',T2-273);\n",
+"\n",
+"//  (d)\n",
+"CP = n/(n-1)*P1*(V1-V4)*((P2/P1)^((n-1)/n)-1)*N/60;// cycle power, [kW]\n",
+" mprintf('\n (d) The cycle power is  =  %f  kW\n',CP);\n",
+"\n",
+"// (e)\n",
+"// neglecting clearence\n",
+"W = n/(n-1)*P1*V1*((P2/P1)^((n-1)/n)-1)\n",
+"Wi = P1*V1*log(P2/P1);// isothermal efficiency\n",
+"IE = Wi/W;// isothermal efficiency\n",
+"mprintf('\n (e) The isothermal efficiency neglecting clearence  is  =  %f percent\n',IE*100);\n",
+"\n",
+"// End"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 14.2: intermediate_pressure_volume_and_cycle_power.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"clear;\n",
+"clc;\n",
+"disp(' Example 14.2');\n",
+"\n",
+"// aim : To determine \n",
+"// (a) the intermediate pressure\n",
+"// (b) the total volume of each cylinder\n",
+"// (c) the cycle power\n",
+"\n",
+"// given values\n",
+"v1 = .2;// air intake, [m^3/s]\n",
+"P1 = .1;// intake pressure, [MN/m^2]\n",
+"T1 = 273+16;// intake temperature, [K]\n",
+"P3 = .7;// final pressure, [MN/m^2]\n",
+"n = 1.25;// compression index\n",
+"N = 10;// speed, [rev/s]\n",
+"\n",
+"// solution\n",
+"// (a)\n",
+"P2 = sqrt(P1*P3);// intermediate pressure, [MN/m^2]\n",
+"mprintf('\n (a) The intermediate pressure is  =  %f  MN/m^2\n',P2);\n",
+"\n",
+"// (b)\n",
+"V1  = v1/N;// total volume,[m^3]\n",
+"// since intercooling is perfect so 2 lie on the isothermal through1, P1*V1=P2*V2\n",
+"V2 = P1*V1/P2;// volume, [m^3]\n",
+"mprintf('\n (b) The total volume of the HP cylinder is  =  %f  litres\n',V2*10^3);\n",
+"\n",
+" // (c)\n",
+" CP = 2*n/(n-1)*P1*v1*((P2/P1)^((n-1)/n)-1);// cycle power, [MW]\n",
+"  mprintf('\n (c) The cycle power is  =  %f  MW\n',CP*10^3);\n",
+" \n",
+" // there is calculation mistake in the book so answer is not matching\n",
+" \n",
+" //  End"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 14.3: intermediate_pressures_effective_swept_volume_temperature_and_work_done.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"clear;\n",
+"clc;\n",
+"disp(' Example 14.3');\n",
+"\n",
+"// aim : To determine \n",
+"// (a) the intermediate pressures\n",
+"// (b) the effective swept volume of the LP cylinder\n",
+"// (c) the temperature and the volume of air delivered per stroke at 15 bar\n",
+"// (d) the work done per kilogram of air\n",
+"\n",
+"// given values\n",
+"d = 450*10^-3;//  bore , [m]\n",
+"L = 300*10^-3;//  stroke, [m]\n",
+"cl = .05;//  clearence\n",
+"P1 = 1; // intake pressure, [bar]\n",
+"T1 = 273+18;// intake temperature, [K]\n",
+"P4 = 15;// final delivery pressure, [bar]\n",
+"n = 1.3;//  compression and expansion index\n",
+"R = .29;// gas constant, [kJ/kg K]\n",
+"\n",
+"// solution\n",
+"// (a)\n",
+"k=(P4/P1)^(1/3); \n",
+"// hence\n",
+"P2 = k*P1;//  intermediare pressure, [bar]\n",
+"P3 = k*P2;//  intermediate pressure, [bar]\n",
+"\n",
+"mprintf('\n (a) The intermediate pressure is P2 =  %f  bar\n',P2);\n",
+"mprintf('\n     The intermediate pressure is  P3=  %f  bar\n',P3);\n",
+"\n",
+"// (b)\n",
+"SV = %pi*d^2/4*L;//  swept volume of LP cylinder, [m^3]\n",
+"// hence\n",
+"V7 = cl*SV;// volume, [m^3]\n",
+"V1 = SV+V7;// volume, [m^3]\n",
+"// also\n",
+"P7 = P2;\n",
+"P8 = P1;\n",
+"V8 = V7*(P7/P8)^(1/n);//  volume, [m^3]\n",
+"ESV = V1-V8;// effective swept volume of LP cylinder, [m^3]\n",
+"\n",
+"mprintf('\n (b) The effective swept volume of the LP cylinder is  =  %f litres\n',ESV*10^3);\n",
+"\n",
+"// (c)\n",
+"T9 = T1;\n",
+"P9 = P3;\n",
+"T4 = T9*(P4/P9)^((n-1)/n);// delivery temperature, [K]\n",
+"// now using P4*(V4-V5)/T4=P1*(V1-V8)/T1\n",
+"V4_minus_V5 = P1*T4*(V1-V8)/(P4*T1);// delivery volume, [m^3]\n",
+" \n",
+"mprintf('\n (c) The delivery temperature is =  %f  C\n',T4-273);\n",
+"mprintf('\n      The delivery volume is =  %f  litres\n',V4_minus_V5*10^3);\n",
+"\n",
+"//  (d)\n",
+"\n",
+"W = 3*n*R*T1*((P2/P1)^((n-1)/n)-1)/(n-1);//  work done/kg ,[kJ]\n",
+"mprintf('\n (d) The work done per kilogram of air is =  %f  kJ\n',W);\n",
+" \n",
+"// End"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 14.4: pressure_temperature_and_energy.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"clear;\n",
+"clc;\n",
+"disp(' Example 14.4');\n",
+"\n",
+"// aim : To determine \n",
+"// (a) the final pressure and temperature\n",
+"// (b) the energy required to drive the compressor\n",
+"\n",
+"// given values\n",
+"rv = 5;// pressure compression ratio\n",
+"m_dot = 10;// air flow rate, [kg/s]\n",
+"P1 = 100;// initial pressure, [kN/m^2]\n",
+"T1 = 273+20;// initial temperature, [K]\n",
+"n_com = .85;// isentropic efficiency of compressor\n",
+"Gama = 1.4;// heat capacity ratio\n",
+"cp = 1.005;// specific heat capacity, [kJ/kg K]\n",
+"\n",
+"// solution\n",
+"// (a)\n",
+"T2_prim = T1*(rv)^((Gama-1)/Gama);// temperature after compression, [K]\n",
+"// using isentropic efficiency=(T2_prim-T1)/(T2-T1)\n",
+"T2 = T1+(T2_prim-T1)/n_com;//  final temperature, [K]\n",
+"P2 = rv*P1;// final pressure, [kN/m^2]\n",
+"mprintf('\n (a) The final temperature is  =  %f  C\n',T2-273);\n",
+"mprintf('\n (b) The final pressure is  =  %f  kN/m^2\n',P2);\n",
+"\n",
+"// (b)\n",
+"E = m_dot*cp*(T1-T2);// energy required, [kW]\n",
+"mprintf('\n (b) The energy required to drive the compressor is  =  %f  kW',E);\n",
+"if(E<0)\n",
+"    disp('The negative sign indicates energy input');\n",
+"else\n",
+"    disp('The positive sign indicates energy output');\n",
+"end\n",
+"\n",
+" //  End\n",
+"\n",
+""
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 14.5: power_developed.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"clear;\n",
+"clc;\n",
+"disp(' Example 14.5');\n",
+"\n",
+"// aim : To determine \n",
+"//  the power absorbed in driving the compressor\n",
+"\n",
+"// given values\n",
+"FC = .68;// fuel consumption rate, [kg/min]\n",
+"P1 = 93;// initial pressure, [kN/m^2]\n",
+"P2 = 200;// final pressure, [kN/m^2]\n",
+"T1 = 273+15;// initial temperature, [K]\n",
+"d = 1.3;// density of mixture, [kg/m^3]\n",
+"n_com = .82;// isentropic efficiency of compressor\n",
+"Gama = 1.38;// heat capacity ratio\n",
+"\n",
+"// solution\n",
+"R = P1/(d*T1);// gas constant, [kJ/kg K]\n",
+"// for mixture\n",
+"cp = Gama*R/(Gama-1);// heat capacity, [kJ/kg K]\n",
+"T2_prim = T1*(P2/P1)^((Gama-1)/Gama);// temperature after compression, [K]\n",
+"// using isentropic efficiency=(T2_prim-T1)/(T2-T1)\n",
+"T2 = T1+(T2_prim-T1)/n_com;//  final temperature, [K]\n",
+"m_dot = FC*15/60;// given condition, [kg/s]\n",
+"P = m_dot*cp*(T2-T1);// power absorbed by compressor, [kW]\n",
+"mprintf('\n The power absorbed by compressor is  =  %f kW\n',P);\n",
+"\n",
+"// End"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 14.6: power.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"clear;\n",
+"clc;\n",
+"disp(' Example 14.6');\n",
+"\n",
+"// aim : To determine \n",
+"//  the power required to drive the blower\n",
+"\n",
+"// given values\n",
+"m_dot = 1;// air capacity, [kg/s]\n",
+"rp = 2;// pressure ratio\n",
+"P1 = 1*10^5;// intake pressure, [N/m^2]\n",
+"T1 = 273+70;// intake temperature, [K]\n",
+"R = .29;// gas constant, [kJ/kg k]\n",
+"\n",
+"// solution\n",
+"V1_dot = m_dot*R*T1/P1*10^3;// [m^3/s]\n",
+"P2 = rp*P1;// final pressure, [n/m^2]\n",
+"P = V1_dot*(P2-P1);// power required, [W]\n",
+"mprintf('\n The power required to drive the blower is  =  %f kW\n',P*10^-3);\n",
+"\n",
+"// End"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 14.7: power.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"clear;\n",
+"clc;\n",
+"disp(' Example 14.7');\n",
+"\n",
+"// aim : To determine \n",
+"//  the power required to drive the vane pump\n",
+"\n",
+"// given values\n",
+"m_dot = 1;// air capacity, [kg/s]\n",
+"rp = 2;// pressure ratio\n",
+"P1 = 1*10^5;// intake pressure, [N/m^2]\n",
+"T1 = 273+70;// intake temperature, [K]\n",
+"Gama = 1.4;// heat capacity ratio\n",
+"rv = .7;// volume ratio\n",
+"\n",
+"// solution\n",
+"V1 = .995;// intake pressure(as given previous question),[m^3/s] \n",
+"// using P1*V1^Gama=P2*V2^Gama, so\n",
+"P2 = P1*(1/rv)^Gama;// pressure, [N/m^2]\n",
+"V2 = rv*V1;// volume,[m^3/s]\n",
+"P3 = rp*P1;// final pressure, [N/m^2]\n",
+"P = Gama/(Gama-1)*P1*V1*((P2/P1)^((Gama-1)/Gama)-1)+V2*(P3-P2);// power required,[W]\n",
+"mprintf('\n The power required to drive the vane pump is  =  %f  kW\n',P*10^-3);\n",
+"\n",
+"//  End"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 14.8: power_temperature_and_pressure.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"clear;\n",
+"clc;\n",
+"disp(' Example 14.8');\n",
+"\n",
+"// aim : To determine \n",
+"//  the total temperature and pressure of the mixture\n",
+"\n",
+"// given values\n",
+"rp = 2.5;// static pressure ratio\n",
+"FC = .04;// fuel consumption rate, [kg/min]\n",
+"P1 = 60;// inilet pressure, [kN/m^2]\n",
+"T1 = 273+5;// inilet temperature, [K]\n",
+"n_com = .84;// isentropic efficiency of compressor\n",
+"Gama = 1.39;// heat capacity ratio\n",
+"C2 = 120;//exit velocity from compressor, [m/s]\n",
+"rm = 13;// air-fuel ratio\n",
+"cp = 1.005;// heat capacity ratio\n",
+"\n",
+"// solution\n",
+"P2 = rp*P1;// given condition, [kN/m^2]\n",
+"T2_prim = T1*(P2/P1)^((Gama-1)/Gama);// temperature after compression, [K]\n",
+"// using isentropic efficiency=(T2_prim-T1)/(T2-T1)\n",
+"T2 = T1+(T2_prim-T1)/n_com;//  final temperature, [K]\n",
+"m_dot = FC*(rm+1);// mass of air-fuel mixture, [kg/s]\n",
+"P = m_dot*cp*(T2-T1);// power to drive compressor, [kW]\n",
+"mprintf('\n The power required to drive compressor is  =  %f  kW\n',P);\n",
+"\n",
+"Tt2 = T2+C2^2/(2*cp*10^3);// total temperature,[K]\n",
+"Pt2 = P2*(Tt2/T2)^(Gama/(Gama-1));// total pressure, [kN/m^2]\n",
+"mprintf('\n The temperature in the engine is  =  %f  C\n',Tt2-273);\n",
+"mprintf('\n The pressure in the engine cylinder is  =  %f  kN/m^2\n',Pt2);\n",
+"\n",
+"//  There is calculation mistake in the book\n",
+"\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
+}
diff --git a/Basic_Engineering_Thermodynamics_by_R_Joel/15-Ideal_gas_power_cycles.ipynb b/Basic_Engineering_Thermodynamics_by_R_Joel/15-Ideal_gas_power_cycles.ipynb
new file mode 100644
index 0000000..511bac3
--- /dev/null
+++ b/Basic_Engineering_Thermodynamics_by_R_Joel/15-Ideal_gas_power_cycles.ipynb
@@ -0,0 +1,1218 @@
+{
+"cells": [
+ {
+		   "cell_type": "markdown",
+	   "metadata": {},
+	   "source": [
+       "# Chapter 15: Ideal gas power cycles"
+	   ]
+	},
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 15.10: maximum_temperature_and_thermal_efficiency.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 maximum temperature attained during the cycle\n",
+"// (b) the thermal efficiency of the cycle\n",
+"\n",
+"// given value\n",
+"rva  =7.5;// volume ratio of adiabatic expansion\n",
+"rvc  =15;// volume ratio of compression\n",
+"P1 = 98;// initial pressure, [kn/m^2]\n",
+"T1 = 273+44;// initial temperature, [K]\n",
+"P4 = 258;// pressure at the end of the adiabatic expansion, [kN/m^2]\n",
+"Gama = 1.4;//  heat capacity  ratio\n",
+"\n",
+"// solution\n",
+"// by seeing diagram\n",
+"// for process 4-1, P4/T4=P1/T1\n",
+"T4 = T1*(P4/P1);// [K]\n",
+"// for process 3-4\n",
+"T3 = T4*(rva)^(Gama-1);\n",
+"mprintf('\n (a) The maximum temperature during the cycle is  =  %f  C\n',T3-273);\n",
+"\n",
+"// (b)\n",
+"\n",
+"// for process 1-2,\n",
+"T2 = T1*(rvc)^(Gama-1);// [K]\n",
+"n_the = 1-(T4-T1)/((Gama)*(T3-T2));// thermal efficiency\n",
+"mprintf('\n (b) The thermal efficiency of the cycle is  =  %f percent\n',n_the*100);\n",
+"\n",
+"//  End"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 15.11: thermal_efficiency_and_indicated_power.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"clear;\n",
+"clc;\n",
+"disp('Example 15.11');\n",
+"\n",
+"// aim : To determine\n",
+"// (a) the thermal efficiency of the cycle\n",
+"// (b) the indicared power of the cycle\n",
+"\n",
+"// given values\n",
+"// taking basis one second\n",
+"rv = 11;// volume ratio\n",
+"P1 = 96;// initial pressure , [kN/m^2]\n",
+"T1 = 273+18;// initial temperature, [K]\n",
+"Gama = 1.4;// heat capacity ratio\n",
+"\n",
+"// solution\n",
+"// taking reference  Fig. 15.24\n",
+"// (a)\n",
+"Beta = 2;// ratio of V3 and V2\n",
+"TE = 1-(Beta^(Gama)-1)/((rv^(Gama-1))*Gama*(Beta-1));// thermal efficiency\n",
+"mprintf('\n (a) the thermal efficiency of the cycle is  =  %f percent\n ',TE*100);\n",
+"\n",
+"// (b) \n",
+"// let V1-V2=.05, so\n",
+"V2 = .05*.1;// [m^3]\n",
+"// from this\n",
+"V1 = rv*V2;// [m^3]\n",
+"V3 = Beta*V2;// [m^3]\n",
+"V4 = V1;// [m^3]\n",
+"P2 = P1*(V1/V2)^(Gama);// [kN/m^2]\n",
+"P3 = P2;// [kn/m^2]\n",
+"P4=P3*(V3/V4)^(Gama);// [kN/m^2]\n",
+"// indicated power\n",
+"W = P2*(V3-V2)+((P3*V3-P4*V4)-(P2*V2-P1*V1))/(Gama-1);// indicated power, [kW]\n",
+"mprintf('\n (c) The indicated power of the cycle is  =  %f kW\n',W);\n",
+"\n",
+"//  End"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 15.12: pressures_and_temperatures.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"clear;\n",
+"clc;\n",
+"disp('Example 15.12');\n",
+"\n",
+"// aim : To determine\n",
+"// (a) the pressure and temperature at the end of compression\n",
+"// (b) the pressure and temperature at the end of the constant volume process\n",
+"// (c) the temperature at the end of constant pressure process\n",
+"\n",
+"// given values\n",
+"P1 = 103;// initial pressure, [kN/m^2]\n",
+"T1 = 273+22;// initial temperature, [K]\n",
+"rv = 16;// volume ratio of the compression\n",
+"Q = 244;//heat added, [kJ/kg]\n",
+"Gama = 1.4;// heat capacity ratio\n",
+"cv = .717;// heat capacity, [kJ/kg k]\n",
+"\n",
+"// solution\n",
+"//  taking reference as Fig.15.26\n",
+"// (a)\n",
+"// for compression\n",
+"// rv = V1/V2\n",
+"P2 = P1*(rv)^Gama;// pressure at end of compression, [kN/m^2]\n",
+"T2 = T1*(rv)^(Gama-1);// temperature at end of compression, [K]\n",
+"mprintf('\n (a) The pressure at the end of compression is  =  %f  MN/m^2\n',P2*10^-3);\n",
+"mprintf('\n      The temperature at the end of compression is  =  %f C\n',T2-273);\n",
+"\n",
+"// (b)\n",
+"// for constant volume process, \n",
+"// Q = cv*(T3-T2), so\n",
+"T3 = T2+Q/cv;// temperature at the end of constant volume, [K]\n",
+"\n",
+"// so for constant volume, P/T=constant, hence\n",
+"P3 = P2*(T3/T2);// pressure at the end of constant volume process, [kN/m^2]\n",
+"mprintf('\n (b) The pressure at the end of constant volume process is  =  %f  MN/m^2\n ',P3*10^-3);\n",
+"mprintf('\n     The temperature at the end of constant volume process is  =  %f C\n',T3-273);\n",
+"\n",
+"// (c)\n",
+"S = rv-1;// stroke\n",
+"// assuming \n",
+"V3 = 1;// [volume]\n",
+"//so\n",
+"V4 = V3+S*.03;// [volume]\n",
+"// also for constant process V/T=constant, hence\n",
+"T4 = T3*(V4/V3);// temperature at the end of constant presure process, [k] \n",
+"mprintf('\n (c) The temperature at the end of constant pressure process is  =  %f C\n',T4-273);\n",
+"\n",
+"//  End"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 15.13: EX15_13.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"clear;\n",
+"clc;\n",
+"disp('Example 15.13');\n",
+"\n",
+"// aim : To determine\n",
+"// (a) the pressure, volume and temperature at cycle process change points\n",
+"// (b) the net work done \n",
+"// (c)  the thermal efficiency\n",
+"// (d) the heat received\n",
+"// (e)  the work ratio\n",
+"// (f)  the mean effective pressure\n",
+"// (g)  the carnot efficiency\n",
+"\n",
+"\n",
+"// given values\n",
+"rv = 15;// volume ratio\n",
+"P1 = 97*10^-3;// initial pressure , [MN/m^2]\n",
+"V1 = .084;// initial volume, [m^3]\n",
+"T1 = 273+28;// initial temperature, [K]\n",
+"T4 = 273+1320;// maximum temperature, [K]\n",
+"P3 = 6.2;// maximum pressure, [MN/m^2]\n",
+"cp = 1.005;// specific heat capacity at constant pressure, [kJ/kg K]\n",
+"cv = .717;// specific heat capacity at constant volume, [kJ/kg K]\n",
+"\n",
+"// solution\n",
+"// taking reference  Fig. 15.27\n",
+"// (a)\n",
+"R = cp-cv;// gas constant, [kJ/kg K]\n",
+"Gama = cp/cv;// heat capacity ratio\n",
+"// for process 1-2\n",
+"V2 = V1/rv;// volume at stage2, [m^3] \n",
+"// using PV^(Gama)=constant for process 1-2\n",
+"P2 = P1*(V1/V2)^(Gama);// pressure at stage2,. [MN/m^2]\n",
+"T2 = T1*(V1/V2)^(Gama-1);// temperature at stage 2, [K]\n",
+"\n",
+"// for process 2-3\n",
+"// since volumee is constant in process 2-3 , so using P/T=constant, so\n",
+"T3 = T2*(P3/P2);// volume at stage 3, [K]\n",
+"V3 = V2;// volume at stage 3, [MN/m^2]\n",
+"\n",
+"// for process 3-4\n",
+"P4 = P3;// pressure at stage 4, [m^3]\n",
+"// since in stage 3-4 P is constant, so V/T=constant, \n",
+"V4 = V3*(T4/T3);// temperature at stage  4,[K]\n",
+"\n",
+"// for process 4-5\n",
+"V5 = V1;// volume at stage 5, [m^3]\n",
+"P5 = P4*(V4/V5)^(Gama);// pressure at stage5,. [MN/m^2]\n",
+"T5 = T4*(V4/V5)^(Gama-1);// temperature at stage 5, [K]\n",
+"\n",
+"mprintf('\n (a)  P1  =  %f kN/m^2,    V1  =  %f m^3,       t1  =  %f C,\n      P2  =  %f MN/m^2,    V2  =  %f m^3,        t2  =  %f C,\n      P3  =  %f MN/m^2,    V3  =  %f m^3,         t3  =  %f C,\n      P4  =  %f MN/m^2,      V4  =  %f m^3,       t4  =  %f C,\n      P5  =   %fkN/m^2,    V5  =  %fm^3,        t5  =  %fC\n',P1*10^3,V1,T1-273,P2,V2,T2-273,P3,V3,T3-273,P4,V4,T4-273,P5*10^3,V5,T5-273);\n",
+"\n",
+"\n",
+"// (b)\n",
+"W = (P3*(V4-V3)+((P4*V4-P5*V5)-(P2*V2-P1*V1))/(Gama-1))*10^3;// work done, [kJ]\n",
+"mprintf('\n (b) The net work done is  =  %f kJ\n',W);\n",
+"\n",
+"// (c) \n",
+"TE = 1-(T5-T1)/((T3-T2)+Gama*(T4-T3));// thermal efficiency\n",
+"mprintf('\n (c) The thermal efficiency is  =  %f percent\n',TE*100);\n",
+"\n",
+"// (d)\n",
+"Q = W/TE;// heat received, [kJ]\n",
+"mprintf('\n (d) The heat received is  =  %f kJ\n',Q);\n",
+"\n",
+"// (e)\n",
+"PW = P3*(V4-V3)+(P4*V4-P5*V5)/(Gama-1)\n",
+"WR = W*10^-3/PW;//  work ratio\n",
+"mprintf('\n (f) The work ratio is  =  %f\n',WR);\n",
+"\n",
+"// (e)\n",
+"Pm = W/(V1-V2);// mean effective pressure, [kN/m^2]\n",
+"mprintf('\n (e) The mean effective pressure is  =  %f kN/m^2\n',Pm);\n",
+"\n",
+"// (f)\n",
+"CE = (T4-T1)/T4;// carnot efficiency\n",
+"mprintf('\n (f) The carnot efficiency is  =  %f percent\n',CE*100);\n",
+"\n",
+"//  End"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 15.14: thermal_efficiency_heat_work_done_work_ratio_and_mean_effective_pressure.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"clear;\n",
+"clc;\n",
+"disp('Example 15.14');\n",
+"\n",
+"// aim : To determine\n",
+"// (a)  the thermal efficiency\n",
+"// (b)  the heat received\n",
+"// (c) the heat rejected\n",
+"// (d) the net work \n",
+"// (e)  the work ratio\n",
+"// (f)  the mean effective pressure\n",
+"// (g)  the carnot efficiency\n",
+"\n",
+"\n",
+"// given values\n",
+"P1 = 101;// initial pressure , [kN/m^2]\n",
+"V1 = 14*10^-3;// initial volume, [m^3]\n",
+"T1 = 273+15;// initial temperature, [K]\n",
+"P3 = 1850;// maximum pressure, [kN/m^2]\n",
+"V2 = 2.8*10^-3;// compressed volume, [m^3]\n",
+"Gama = 1.4;// heat capacity\n",
+"R = .29;// gas constant, [kJ/kg k]\n",
+"\n",
+"// solution\n",
+"// taking reference  Fig. 15.29\n",
+"// (a)\n",
+"// for process 1-2\n",
+"// using PV^(Gama)=constant for process 1-2\n",
+"P2 = P1*(V1/V2)^(Gama);// pressure at stage2,. [MN/m^2]\n",
+"T2 = T1*(V1/V2)^(Gama-1);// temperature at stage 2, [K]\n",
+"\n",
+"// for process 2-3\n",
+"// since volumee is constant in process 2-3 , so using P/T=constant, so\n",
+"T3 = T2*(P3/P2);// volume at stage 3, [K]\n",
+"\n",
+"// for process 3-4\n",
+"P4 = P1;\n",
+"T4 = T3*(P4/P3)^((Gama-1)/Gama);// temperature\n",
+"\n",
+"TE = 1-Gama*(T4-T1)/(T3-T2);// thermal efficiency\n",
+"mprintf('\n (a) The thermal efficiency is  =  %f percent\n',TE*100);\n",
+"\n",
+"// (b)\n",
+"cv = R/(Gama-1);// heat capacity at copnstant volume, [kJ/kg k]\n",
+"m = P1*V1/(R*T1);// mass of gas, [kg]\n",
+"Q1 = m*cv*(T3-T2);// heat received, [kJ/cycle]\n",
+"mprintf('\n (b) The heat received is  =  %f kJ/cycle\n',Q1);\n",
+"\n",
+"// (c)\n",
+"cp = Gama*cv;// heat capacity at constant at constant pressure, [kJ/kg K]\n",
+"Q2 = m*cp*(T4-T1);// heat rejected, [kJ/cycle]\n",
+"mprintf('\n (c) The heat rejected is  =  %f kJ/cycle\n',Q2);\n",
+"\n",
+"// (d)\n",
+"W = Q1-Q2;// net work , [kJ/cycle]\n",
+"mprintf('\n (d) The net work is  =  %f  kJ/cycle\n',W);\n",
+"\n",
+"// (e)\n",
+"// pressure is constant for process 1-4, so V/T=constant\n",
+"V4 = V1*(T4/T1);// volume, [m^3]\n",
+"V3 = V2;// for process 2-3\n",
+"P4 = P1;// for process 1-4\n",
+"PW = (P3*V3-P1*V1)/(Gama-1);// positive work done, [kJ/cycle]\n",
+"WR = W/PW;//  work ratio\n",
+"mprintf('\n (e) The work ratio is  =  %f\n',WR);\n",
+"\n",
+"// (f)\n",
+"Pm = W/(V4-V2);// mean effective pressure, [kN/m^2]\n",
+"mprintf('\n (f) The mean effefctive pressure is  =  %f  kN/m^2\n',Pm);\n",
+"\n",
+"// (g)\n",
+"CE = (T3-T1)/T3;// carnot efficiency\n",
+"mprintf('\n (g) The carnot efficiency is  =  %f percent\n',CE*100);\n",
+"\n",
+"// there is minor variation in answer reported in the book\n",
+"\n",
+"//  End"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 15.15: work_done_and_thermal_efficiency.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"clear;\n",
+"clc;\n",
+"disp('Example 15.15');\n",
+"\n",
+"// aim : To determine\n",
+"// (a) the net work done\n",
+"// (b) the ideal thermal efficiency\n",
+"// (c) the thermal efficiency if the process of generation is not included\n",
+"\n",
+"// given values\n",
+"P1 = 110;// initial pressure, [kN/m^2)\n",
+"T1 = 273+30;// initial temperature, [K]\n",
+"V1 = .05;// initial volume, [m^3]\n",
+"V2 = .005;// volume, [m^3]\n",
+"T3 = 273+700;// temperature, [m^3]\n",
+"R = .289;// gas constant, [kJ/kg K]\n",
+"cv = .718;// heat capacity, [kJ/kg K]\n",
+"\n",
+"// solution\n",
+"// (a)\n",
+"m = P1*V1/(R*T1);// mass , [kg]\n",
+"W = m*R*(T3-T1)*log(V1/V2);// work done, [kJ]\n",
+"mprintf('\n (a) The net work done is  =  %f kJ\n',W);\n",
+"\n",
+"// (b)\n",
+"n_the = (T3-T1)/T3;// ideal thermal efficiency\n",
+"mprintf('\n (b) The ideal thermal efficiency is  =  %f percent\n',n_the*100);\n",
+"\n",
+"// (c)\n",
+"V4 = V1;\n",
+"V3 = V2;\n",
+"T4 = T3;\n",
+"T2 = T1;\n",
+"\n",
+"Q_rej = m*cv*(T4-T1)+m*R*T1*log(V1/V2);// heat rejected\n",
+"Q_rec = m*cv*(T3-T2)+m*R*T3*log(V4/V3);// heat received\n",
+"\n",
+"n_th = (1-Q_rej/Q_rec);// thermal efficiency\n",
+"mprintf('\n (c) the thermal efficiency if the process of regeneration is not included is  =  %f percent\n',n_th*100);\n",
+"\n",
+"//  End"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 15.16: maximum_temperature_work_done_and_thermal_efficiency.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"clear;\n",
+"clc;\n",
+"disp('Example 15.16');\n",
+"\n",
+"// aim : To determine\n",
+"// (a) the maximum temperature\n",
+"// (b) the net work done\n",
+"// (c) the ideal thermal efficiency\n",
+"// (d) the thermal efficiency if the process of regeneration is not included\n",
+"\n",
+"// given values\n",
+"P1 = 100;// initial pressure, [kN/m^2)\n",
+"T1 = 273+20;// initial temperature, [K]\n",
+"V1 = .08;// initial volume, [m^3]\n",
+"rv = 5;// volume ratio\n",
+"R = .287;// gas constant, [kJ/kg K]\n",
+"cp = 1.006;// heat capacity, [kJ/kg K]\n",
+"V3_by_V2 = 2;\n",
+"\n",
+"// solution\n",
+"// (a)\n",
+"// using Fig.15.33\n",
+"// process 1-2 is isothermal\n",
+"T2 = T1;\n",
+"// since process 2-3 isisobaric, so V/T=constant\n",
+"T3 = T2*(V3_by_V2);// maximumtemperature, [K]\n",
+"mprintf('\n (a) The maximum temperature is  =  %f C\n',T3-273);\n",
+"\n",
+"// (b)\n",
+"m = P1*V1/(R*T1);// mass , [kg]\n",
+"W = m*R*(T3-T1)*log(rv);// work done, [kJ]\n",
+"mprintf('\n (b) The net work done is  =  %f  kJ\n',W);\n",
+"\n",
+"// (c)\n",
+"TE = (T3-T1)/T3;// ideal thermal efficiency\n",
+"mprintf('\n (c) The ideal thermal efficiency is  =  %f  percent\n',TE*100);\n",
+"\n",
+"// (d)\n",
+"T4 = T3;\n",
+"T2 = T1;\n",
+"\n",
+"Q_rej = m*cp*(T4-T1)+m*R*T1*log(rv);// heat rejected\n",
+"Q_rec = m*cp*(T3-T2)+m*R*T3*log(rv);// heat received\n",
+"\n",
+"n_th = (1-Q_rej/Q_rec);// thermal efficiency\n",
+"mprintf('\n (d) the thermal efficiency if the process of regeneration is not included is  =  %f  percent\n',n_th*100);\n",
+"\n",
+"//  End"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 15.17: work_done_and_thermal_efficiency.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"clear;\n",
+"clc;\n",
+"disp('Example 15.17');\n",
+"\n",
+"// aim : To determine \n",
+"// (a) the net work done\n",
+"// (b) thethermal efficiency\n",
+"\n",
+"// given values\n",
+"m = 1;// mass of air, [kg]\n",
+"T1 = 273+230;// initial temperature, [K]\n",
+"P1 = 3450;// initial pressure, [kN/m^2]\n",
+"P2 = 2000;// pressure, [kN/m^2]\n",
+"P3 = 140;// pressure, [kN/m^2]\n",
+"P4 = P3;\n",
+"Gama = 1.4; // heat capacity ratio\n",
+"cp = 1.006;// heat capacity, [kJ/kg k]\n",
+"\n",
+"// solution\n",
+"T2 =T1;// isothermal process 1-2\n",
+"// process 2-3 and 1-4 are adiabatic so\n",
+"T3 = T2*(P3/P2)^((Gama-1)/Gama);// temperature, [K] \n",
+"T4 = T1*(P4/P1)^((Gama-1)/Gama);// [K]\n",
+"R = cp*(Gama-1)/Gama;// gas constant, [kJ/kg K]\n",
+"Q1 = m*R*T1*log(P1/P2);// heat received, [kJ]\n",
+"Q2 = m*cp*(T3-T4);// heat rejected\n",
+"\n",
+"//hence\n",
+"W = Q1-Q2;// work done\n",
+"mprintf('\n (a) The net work done is  =  %f kJ\n',W);\n",
+"\n",
+"// (b)\n",
+"TE = 1-Q2/Q1;// thermal efficiency\n",
+"mprintf('\n (b) The thermal efficiency is  =  %f  percent\n',TE*100);\n",
+"\n",
+"//  End"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 15.18: thermal_efficiency_and_Carnot_efficiency.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"clear;\n",
+"clc;\n",
+"disp('Example 15.18');\n",
+"\n",
+"// aim : To determine \n",
+"// thermal eficiency\n",
+"// carnot efficiency\n",
+"\n",
+"// given values\n",
+"rv = 5;// volume ratio\n",
+"Gama = 1.4;// heat capacity ratio\n",
+"\n",
+"// solution\n",
+"// under given condition\n",
+"\n",
+"TE = 1-(1/Gama*(2-1/rv^(Gama-1)))/(1+2*((Gama-1)/Gama)*log(rv/2));// thermal efficiency\n",
+"mprintf('\n The thermal efficiency is  =  %f percent\n',TE*100);\n",
+"\n",
+"CE = 1-1/(2*rv^(Gama-1));// carnot efficiency\n",
+"mprintf('\n The carnot efficiency is  =  %f \n',CE*100);\n",
+"\n",
+"//  End"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 15.1: thermal_efficiency.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"clear;\n",
+"clc;\n",
+"disp('Example 15.1');\n",
+"\n",
+"// aim : To determine \n",
+"// the thermal efficiency of the cycle\n",
+"\n",
+"// given values\n",
+"T1 = 273+400;//  temperature limit, [K]\n",
+"T3 = 273+70;// temperature limit, [K]\n",
+"\n",
+"//  solution\n",
+"//  using equation [15] of section 15.3\n",
+"n_the = (T1-T3)/T1*100;// thermal efficiency \n",
+"mprintf('\n The thermal efficiency of the cycle is  =  %f percent\n',n_the);\n",
+"\n",
+"//  End"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 15.2: volume_ratios_and_thermal_efficiency.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"clear;\n",
+"clc;\n",
+"disp('Example 15.2');\n",
+"\n",
+"// aim : To determine\n",
+"// (a) the volume ratios of the isothermal and adiabatic processes\n",
+"// (b) the thermal efficiency of the cycle\n",
+"\n",
+"// given values\n",
+"T1 = 273+260;// temperature, [K]\n",
+"T3 = 273+21;// temperature, [K]\n",
+"er = 15;// expansion ratio\n",
+"Gama = 1.4;// heat capacity ratio\n",
+"\n",
+"// solution\n",
+"// (a)\n",
+"T2 = T1;\n",
+"T4 = T3;\n",
+"// for adiabatic process\n",
+"rva = (T1/T4)^(1/(Gama-1));// volume ratio of adiabatic\n",
+"rvi = er/rva;// volume ratio of isothermal\n",
+"mprintf('\n (a) The volume ratio of the adiabatic process is  =  %f\n',rva);\n",
+"mprintf('\n      The volume ratio of the isothermal process is  =  %f\n',rvi);\n",
+"\n",
+"// (b)\n",
+"n_the = (T1-T4)/T1*100;// thermal efficiency\n",
+"mprintf('\n (b) The thermal efficiency of the cycle is  =  %f percent\n',n_the);\n",
+"\n",
+"//  End"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 15.3: pressure_volume_temperature_thermal_efficiency_work_done_and_work_ratio.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"clear;\n",
+"clc;\n",
+"disp('Example 15.3');\n",
+"\n",
+"// aim : To determine\n",
+"// (a) the pressure, volume and temperature at each corner of the cycle\n",
+"// (b) the thermal efficiency of the cycle\n",
+"// (c) the work done per cycle\n",
+"// (d) the work ratio\n",
+"\n",
+"// given values\n",
+"m = 1;// mass of air, [kg]\n",
+"P1 = 1730;// initial pressure of carnot engine, [kN/m^2]\n",
+"T1 = 273+300;// initial temperature, [K]\n",
+"R = .29;// [kJ/kg K]\n",
+"Gama = 1.4;// heat capacity ratio\n",
+"\n",
+"// solution\n",
+"// taking reference  Fig. 15.15\n",
+"// (a)\n",
+"// for the isothermal process 1-2\n",
+"// using ideal gas law\n",
+"V1 = m*R*T1/P1;// initial volume, [m^3]\n",
+"T2 = T1;\n",
+"V2 = 3*V1;// given condition\n",
+"// for isothermal process, P1*V1=P2*V2, so\n",
+"P2 = P1*(V1/V2);// [MN/m^2]\n",
+"// for the adiabatic process 2-3\n",
+"V3 = 6*V1;// given condition\n",
+"T3 = T2*(V2/V3)^(Gama-1);\n",
+"// also for adiabatic process, P2*V2^Gama=P3*V3^Gama, so\n",
+"P3 = P2*(V2/V3)^Gama;\n",
+"// for the isothermal process 3-4\n",
+"T4 = T3;\n",
+"// for both adiabatic processes, the temperataure ratio is same, \n",
+"// T1/T4 = T2/T3=(V4/V1)^(Gama-1)=(V3/V2)^(Gama-1), so\n",
+"V4 = 2*V1;\n",
+"// for isothermal process, 3-4, P3*V3=P4*V4, so\n",
+"P4 = P3*(V3/V4);\n",
+"disp('(a) At line 1');\n",
+"mprintf('\n V1  =  %f  m^3, t1  =  %f  C,  P1  =  %f  kN/m^2\n',V1,T1-273,P1);\n",
+"\n",
+"disp('At line 2');\n",
+"mprintf('\n V2  =  %f  m^3, t2  =  %f  C,  P2  =  %f  kN/m^2\n',V2,T2-273,P2);\n",
+"\n",
+"disp('At line 3');\n",
+"mprintf('\n V3  =  %f  m^3, t3  =  %f  C,  P3  =  %f  kN/m^2\n',V3,T3-273,P3);\n",
+"\n",
+"\n",
+"disp('At line 4');\n",
+"mprintf('\n V4  =  %f  m^3, t4  =  %f  C,  P4  =  %f  kN/m^2\n',V4,T4-273,P4);\n",
+"\n",
+"\n",
+"// (b)\n",
+"n_the = (T1-T3)/T1;// thermal efficiency\n",
+"mprintf('\n (b) The thermal efficiency of the cycle is  =  %f percent\n',n_the*100);\n",
+"\n",
+"// (c)\n",
+"W = m*R*T1*log(V2/V1)*n_the;//  work done, [J]\n",
+"mprintf('\n (c) The work done per cycle is  =  %f  kJ\n',W);\n",
+"\n",
+"//  (d)\n",
+"wr = (T1-T3)*log(V2/V1)/(T1*log(V2/V1)+(T1-T3)/(Gama-1));// work ratio\n",
+"mprintf('\n (d) The work ratio is  =  %f\n',wr);\n",
+"\n",
+"// there is calculation mistake in the book so answer is not matching\n",
+"\n",
+"//   End\n",
+""
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 15.4: EX15_4.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"clear;\n",
+"clc;\n",
+"disp('Example 15.4');\n",
+"\n",
+"// aim : To determine\n",
+"// (a) the pressure, volume and temperature at cycle state points\n",
+"// (b) the heat received\n",
+"// (c) the work done \n",
+"// (d) the thermal efficiency\n",
+"// (e) the carnot efficiency\n",
+"// (f) the work ration\n",
+"// (g) the mean effective pressure\n",
+"\n",
+"// given values\n",
+"ro = 8;// overall volume ratio;\n",
+"rv = 6;// volume ratio of adiabatic compression\n",
+"P1 = 100;// initial pressure , [kN/m^2]\n",
+"V1 = .084;// initial volume, [m^3]\n",
+"T1 = 273+28;// initial temperature, [K]\n",
+"Gama = 1.4;// heat capacity ratio\n",
+"cp = 1.006;// specific heat capacity, [kJ/kg K]\n",
+"\n",
+"// solution\n",
+"// taking reference  Fig. 15.18\n",
+"// (a)\n",
+"V2 = V1/rv;// volume at stage2, [m^3] \n",
+"V4 = ro*V2;// volume at stage 4;[m^3]\n",
+"// using PV^(Gama)=constant for process 1-2\n",
+"P2 = P1*(V1/V2)^(Gama);// pressure at stage2,. [kN/m^2]\n",
+"T2 = T1*(V1/V2)^(Gama-1);// [K]\n",
+"\n",
+"P3 = P2;// pressure at stage 3, [kN/m^2]\n",
+"V3 = V4/rv;// volume at stage 3, [m^3]\n",
+"// since pressure is constant in process 2-3 , so using V/T=constant, so\n",
+"T3 = T2*(V3/V2);// temperature at stage 3, [K]\n",
+"\n",
+"// for process 1-4\n",
+"T4 = T1*(V4/V1);// temperature at stage4, [K\n",
+"P4 = P1;// pressure at stage4, [kN/m^2]\n",
+"\n",
+"mprintf('\n (a)  P1  =  %f kN/m^2,    V1  =  %f m^3,       t1  =  %f C,\n      P2  =    %f kN/m^2,    V2  =  %f m^3,      t2  =  %f C,\n      P3  =  %f kN/m^2,    V3  =  %f m^3,       t3  =  %f C,\n      P4  =  %f kN/m^2,      V4  =  %f m^3,       t4  =  %f C\n',P1,V1,T1-273,P2,V2,T2-273,P3,V3,T3-273,P4,V4,T4-273);\n",
+"\n",
+"// (b)\n",
+"R = cp*(Gama-1)/Gama;// gas constant, [kJ/kg K]\n",
+"m = P1*V1/(R*T1);// mass of gas, [kg]\n",
+"Q = m*cp*(T3-T2);// heat received, [kJ]\n",
+"mprintf('\n (b) The heat received is  =  %f kJ\n',Q);\n",
+"\n",
+"// (c) \n",
+"W = P2*(V3-V2)-P1*(V4-V1)+((P3*V3-P4*V4)-(P2*V2-P1*V1))/(Gama-1);// work done, [kJ]\n",
+"mprintf('\n (c) The work done is  =  %f  kJ\n',W);\n",
+"\n",
+"//  (d)\n",
+"TE = 1-T1/T2;// thermal efficiency\n",
+"mprintf('\n (d) The thermal efficiency is  =  %f  percent\n',TE*100);\n",
+"\n",
+"// (e)\n",
+"CE = (T3-T1)/T3;// carnot efficiency\n",
+"mprintf('\n (e) The carnot efficiency is  =  %f  percent\n',CE*100);\n",
+"\n",
+"// (f)\n",
+"PW = P2*(V3-V2)+(P3*V3-P4*V4)/(Gama-1);// positive work done, [kj]\n",
+"WR = W/PW;//  work ratio\n",
+"mprintf('\n (f) The work ratio is  =  %f\n',WR);\n",
+"\n",
+"// (g)\n",
+"Pm = W/(V4-V2);// mean effective pressure, [kN/m^2]\n",
+"mprintf('\n (g) The mean effective pressure is  =  %f  kN/m^2\n',Pm);\n",
+"\n",
+"//  there is minor variation in answer reported in the book\n",
+"\n",
+"//  End"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 15.5: thermal_efficiency_and_specific_fuel_consumption.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"clear;\n",
+"clc;\n",
+"disp('Example 15.5');\n",
+"\n",
+"// aim : To determine\n",
+"// (a) the actual thermal efficiency of the turbine\n",
+"// (b) the specific fuel consumption of the turbine in kg/kWh\n",
+"\n",
+"// given values\n",
+"P2_by_P1 = 8;\n",
+"n_tur = .6;// ideal turbine thermal efficiency\n",
+"c = 43*10^3;// calorific value of fuel, [kJ/kg]\n",
+"Gama = 1.4;// heat capacity ratio\n",
+"\n",
+"// solution\n",
+"// (a)\n",
+"rv = P2_by_P1;\n",
+"n_tur_ide = 1-1/(P2_by_P1)^((Gama-1)/Gama);// ideal thermal efficiency\n",
+"ate = n_tur_ide*n_tur;// actual thermal efficiency\n",
+"mprintf('\n (a) The actual thermal efficiency of the turbine is  =  %f percent\n',ate*100);\n",
+"\n",
+"// (b)\n",
+"ewf = c*ate;// energy to work fuel, [kJ/kg]\n",
+"kWh = 3600;// energy equivalent ,[kJ]\n",
+"sfc = kWh/ewf;// specific fuel consumption, [kg/kWh]\n",
+"mprintf('\n (b) The specific fuel consumption of the turbine is  =  %f  kg/kWh',sfc);\n",
+"\n",
+"//  End"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 15.6: relative_efficiency.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"clear;\n",
+"clc;\n",
+"disp('Example 15.6');\n",
+"\n",
+"// aim : To determine\n",
+"// the relative efficiency of the engine\n",
+"\n",
+"// given values\n",
+"d = 80;// bore, [mm]\n",
+"l = 85;// stroke, [mm]\n",
+"V1 = .06*10^6;// clearence volume, [mm^3]\n",
+"ate = .22;// actual thermal efficiency of the engine\n",
+"Gama = 1.4;// heat capacity ratio\n",
+"\n",
+"// solution\n",
+"sv = %pi*d^2/4*l;// stroke volume, [mm^3]\n",
+"V2 = sv+V1;// [mm^3]\n",
+"rv = V2/V1;\n",
+"ite = 1-(1/rv)^(Gama-1);// ideal thermal efficiency\n",
+"re = ate/ite;// relative thermal efficiency\n",
+"mprintf('\n The relative efficiency of the engine is  =  %f percent\n',re*100);\n",
+"\n",
+"//  End"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 15.7: EX15_7.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"clear;\n",
+"clc;\n",
+"disp('Example 15.7');\n",
+"\n",
+"// aim : To determine\n",
+"// (a) the pressure, volume and temperature at each cycle process change points\n",
+"// (b) the heat transferred to air\n",
+"// (c) the heat rejected by the air\n",
+"// (d) the ideal thermal efficiency\n",
+"// (e) the work done \n",
+"// (f) the mean effective pressure\n",
+"\n",
+"// given values\n",
+"m = 1;// mass of air, [kg]\n",
+"rv = 6;// volume ratio of adiabatic compression\n",
+"P1 = 103;// initial pressure , [kN/m^2]\n",
+"T1 = 273+100;// initial temperature, [K]\n",
+"P3 = 3450;// maximum pressure, [kN/m^2]\n",
+"Gama = 1.4;// heat capacity ratio\n",
+"R = .287;// gas constant, [kJ/kg K]\n",
+"\n",
+"// solution\n",
+"// taking reference  Fig. 15.20\n",
+"// (a)\n",
+"// for point 1\n",
+"V1 = m*R*T1/P1;// initial volume, [m^3]\n",
+"\n",
+"// for point 2\n",
+"V2 = V1/rv;// volume at point 2, [m^3] \n",
+"// using PV^(Gama)=constant for process 1-2\n",
+"P2 = P1*(V1/V2)^(Gama);// pressure at point 2,. [kN/m^2]\n",
+"T2 = T1*(V1/V2)^(Gama-1);// temperature at point 2,[K]\n",
+"\n",
+"// for point 3\n",
+"V3 = V2;// volume at point 3, [m^3]\n",
+"// since volume is constant in process 2-3 , so using P/T=constant, so\n",
+"T3 = T2*(P3/P2);// temperature at stage 3, [K]\n",
+"\n",
+"// for point 4\n",
+"V4 = V1;//  volume at point 4, [m^3]\n",
+"P4 = P3*(V3/V4)^Gama;// pressure at point 4, [kN/m^2] \n",
+"// again  since volume is constant in process 4-1 , so using P/T=constant, so\n",
+"T4 = T1*(P4/P1);// temperature at point 4, [K]\n",
+"\n",
+"mprintf('\n (a)  P1  =  %f kN/m^2,    V1  =  %f m^3,       t1  =  %f C,\n      P2  =  %f kN/m^2,    V2  =  %f m^3,      t2  =  %f C,\n      P3  =  %f kN/m^2,    V3  =  %f m^3,       t3  =  %f C,\n      P4  =  %f kN/m^2,      V4  =  %f m^3,       t4  =  %f C\n',P1,V1,T1-273,P2,V2,T2-273,P3,V3,T3-273,P4,V4,T4-273);\n",
+"\n",
+"//  (b)\n",
+"cv = R/(Gama-1);//  specific heat capacity, [kJ/kg K]\n",
+"Q23 = m*cv*(T3-T2);// heat transferred, [kJ]\n",
+"mprintf('\n (b) The heat transferred to the air is  =  %f  kJ\n',Q23);\n",
+"\n",
+"// (c) \n",
+"Q34 = m*cv*(T4-T1);// heat rejected by air, [kJ]\n",
+"mprintf('\n (c) The heat rejected by the air is  =  %f  kJ\n',Q34);\n",
+"\n",
+"// (d)\n",
+"TE = 1-Q34/Q23;// ideal thermal efficiency\n",
+"mprintf('\n (d) The ideal thermal efficiency is  =  %f percent\n',TE*100);\n",
+"\n",
+"// (e)\n",
+"W = Q23-Q34;// work done ,[kJ]\n",
+"mprintf('\n (e) The work done  is  =  %f  kJ\n',W);\n",
+"\n",
+"// (f)\n",
+"Pm = W/(V1-V2);// mean effective pressure, [kN/m^2]\n",
+"mprintf('\n (f)  The mean effefctive pressure is  =  %f kN/m^2\n',Pm);\n",
+"\n",
+"//  End"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 15.8: EX15_8.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"clear;\n",
+"clc;\n",
+"disp('Example 15.8');\n",
+"\n",
+"// aim : To determine\n",
+"// (a) the pressure, volume and temperature at cycle state points\n",
+"// (b) the thermal efficiency\n",
+"// (c) the theoretical output\n",
+"// (d) the mean effective pressure\n",
+"// (e) the carnot efficiency\n",
+"\n",
+"// given values\n",
+"rv = 9;// volume ratio\n",
+"P1 = 101;// initial pressure , [kN/m^2]\n",
+"V1 = .003;// initial volume, [m^3]\n",
+"T1 = 273+18;// initial temperature, [K]\n",
+"P3 = 4500;// maximum pressure, [kN/m^2]\n",
+"N = 3000;\n",
+"cp = 1.006;// specific heat capacity at constant pressure, [kJ/kg K]\n",
+"cv = .716;// specific heat capacity at constant volume, [kJ/kg K]\n",
+"\n",
+"// solution\n",
+"// taking reference  Fig. 15.20\n",
+"// (a)\n",
+"// for process 1-2\n",
+"Gama = cp/cv;// heat capacity ratio\n",
+"R = cp-cv;// gas constant, [kJ/kg K]\n",
+"V2 = V1/rv;// volume at stage2, [m^3] \n",
+"// using PV^(Gama)=constant for process 1-2\n",
+"P2 = P1*(V1/V2)^(Gama);// pressure at stage2,. [kN/m^2]\n",
+"T2 = T1*(V1/V2)^(Gama-1);// [K]\n",
+"\n",
+"// for process 2-3\n",
+"V3 = V2;// volume at stage 3, [m^3]\n",
+"// since volume is constant in process 2-3 , so using P/T=constant, so\n",
+"T3 = T2*(P3/P2);// temperature at stage 3, [K]\n",
+"\n",
+"// for process 3-4\n",
+"V4 = V1;// volume at stage 4\n",
+"// using PV^(Gama)=constant for process 3-4\n",
+"P4 = P3*(V3/V4)^(Gama);// pressure at stage2,. [kN/m^2]\n",
+"T4 = T3*(V3/V4)^(Gama-1);// temperature at stage  4,[K]\n",
+"\n",
+"mprintf('\n (a)  P1  =  %f kN/m^2,    V1  =  %f m^3,       t1  =  %f C,\n      P2  =  %f kN/m^2,    V2  =  %f m^3,      t2  =  %f C,\n      P3  =  %f kN/m^2,    V3  =  %f m^3,       t3  =  %f C,\n      P4 =  %f kN/m^2,      V4  =  %f m^3,       t4  =  %f C\n',P1,V1,T1-273,P2,V2,T2-273,P3,V3,T3-273,P4,V4,T4-273);\n",
+"\n",
+"// (b)\n",
+"TE = 1-(T4-T1)/(T3-T2);// thermal efficiency\n",
+"mprintf('\n (b) The thermal efficiency is  =  %f percent\n',TE*100);\n",
+"\n",
+"// (c)\n",
+"m = P1*V1/(R*T1);// mass os gas, [kg] \n",
+"W = m*cv*((T3-T2)-(T4-T1));// work done, [kJ]\n",
+"Wt = W*N/60;// workdone per minute, [kW]\n",
+"mprintf('\n (c) The theoretical output  is  =  %f  kW\n',Wt);\n",
+"\n",
+"// (d)\n",
+"Pm = W/(V1-V2);// mean effective pressure, [kN/m^2]\n",
+"mprintf('\n (g) The mean effefctive pressure is  =  %f  kN/m^2\n',Pm);\n",
+"\n",
+"// (e)\n",
+"CE = (T3-T1)/T3;// carnot efficiency\n",
+"mprintf('\n (e) The carnot efficiency is  =  %f percent\n',CE*100);\n",
+"\n",
+"\n",
+"//  End"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 15.9: EX15_9.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"clear;\n",
+"clc;\n",
+"disp('Example 15.9');\n",
+"\n",
+"// aim : To determine\n",
+"// (a) the pressure and temperature at cycle process change points\n",
+"// (b) the work done \n",
+"// (c)  the thermal efficiency\n",
+"// (d)  the work ratio\n",
+"// (e)  the mean effective pressure\n",
+"// (f)  the carnot efficiency\n",
+"\n",
+"\n",
+"// given values\n",
+"rv = 16;// volume ratio of compression\n",
+"P1 = 90;// initial pressure , [kN/m^2]\n",
+"T1 = 273+40;// initial temperature, [K]\n",
+"T3 = 273+1400;// maximum temperature, [K]\n",
+"cp = 1.004;// specific heat capacity at constant pressure, [kJ/kg K]\n",
+"Gama = 1.4;// heat capacoty ratio\n",
+"\n",
+"// solution\n",
+"cv = cp/Gama;// specific heat capacity at constant volume, [kJ/kg K]\n",
+"R = cp-cv;// gas constant, [kJ/kg K]\n",
+"// for one kg of gas\n",
+"V1 = R*T1/P1;// initial volume, [m^3]\n",
+"// taking reference  Fig. 15.22\n",
+"// (a)\n",
+"// for process 1-2\n",
+"// using PV^(Gama)=constant for process 1-2\n",
+"// also rv = V1/V2\n",
+"P2 = P1*(rv)^(Gama);// pressure at stage2,. [kN/m^2]\n",
+"T2 = T1*(rv)^(Gama-1);// temperature at stage 2, [K]\n",
+"\n",
+"// for process 2-3\n",
+"P3 = P2;// pressure at stage 3, [kN/m^2]\n",
+"V2 = V1/rv;//[m^3]\n",
+"// since pressure is constant in process 2-3 , so using V/T=constant, so\n",
+"V3 = V2*(T3/T2);// volume at stage 3, [m^3]\n",
+"\n",
+"// for process 1-4\n",
+"V4 = V1;// [m^3]\n",
+"P4 = P3*(V3/V4)^(Gama)\n",
+"// since in stage 1-4 volume is constant, so P/T=constant, \n",
+"T4 = T1*(P4/P1);// temperature at stage  4,[K]\n",
+"\n",
+"mprintf('\n (a)  P1  =  %f kN/m^2,          t1  =  %f C,\n      P2  =  %f kN/m^2,        t2  =  %f C,\n      P3  =  %f kN/m^2,         t3  =  %f C,\n      P4  =  %f kN/m^2,          t4  =  %f C\n',P1,T1-273,P2,T2-273,P3,T3-273,P4,T4-273);\n",
+"\n",
+"// (b)\n",
+"W = cp*(T3-T2)-cv*(T4-T1);// work done, [kJ]\n",
+"mprintf('\n (b) The work done is  =  %f kJ\n',W);\n",
+"\n",
+"// (c) \n",
+"TE = 1-(T4-T1)/((T3-T2)*Gama);// thermal efficiency\n",
+"mprintf('\n (c) The thermal efficiency is  =  %f percent\n',TE*100);\n",
+"\n",
+"// (d)\n",
+"PW = cp*(T3-T2)+R*(T3-T4)/(Gama-1);// positive work done\n",
+"WR = W/PW;//  work ratio\n",
+"mprintf('\n (d) The work ratio is  =  %f\n',WR);\n",
+"\n",
+"// (e)\n",
+"Pm = W/(V1-V2);// mean effective pressure, [kN/m^2]\n",
+"mprintf('\n (e) The mean effefctive pressure is  =  %f kN/m^2\n',Pm);\n",
+"\n",
+"// (f)\n",
+"CE = (T3-T1)/T3;// carnot efficiency\n",
+"mprintf('\n (f) The carnot efficiency is  =  %f percent\n',CE*100);\n",
+"\n",
+"// value of t2 printed in the book is incorrect\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
+}
diff --git a/Basic_Engineering_Thermodynamics_by_R_Joel/16-Internal_combustion_engines.ipynb b/Basic_Engineering_Thermodynamics_by_R_Joel/16-Internal_combustion_engines.ipynb
new file mode 100644
index 0000000..e6ce289
--- /dev/null
+++ b/Basic_Engineering_Thermodynamics_by_R_Joel/16-Internal_combustion_engines.ipynb
@@ -0,0 +1,477 @@
+{
+"cells": [
+ {
+		   "cell_type": "markdown",
+	   "metadata": {},
+	   "source": [
+       "# Chapter 16: Internal combustion engines"
+	   ]
+	},
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 16.1: power_output_thermal_efficiency_and_work_ratio.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"clear;\n",
+"clc;\n",
+"disp('Example 16.1');\n",
+"\n",
+"// aim : To determine \n",
+"// (a) the net power output of the turbine plant if the turbine is coupled to the compresser\n",
+"// (b) the thermal efficiency of the plant\n",
+"// (c) the work ratio\n",
+"\n",
+"// Given values\n",
+"P1 = 100;// inlet pressure of compressor, [kN/m^2]\n",
+"T1 = 273+18;// inlet temperature, [K]\n",
+"P2 = 8*P1;// outlet pressure of compressor, [kN/m^2]\n",
+"n_com = .85;// isentropic efficiency of compressor\n",
+"T3 = 273+1000;//inlet temperature of turbine, [K]\n",
+"P3 = P2;// inlet pressure of turbine, [kN/m^2]\n",
+"P4 = 100;// outlet pressure of turbine, [kN/m^2]\n",
+"n_tur = .88;// isentropic efficiency of turbine\n",
+"m_dot = 4.5;// air mass flow rate, [kg/s]\n",
+"cp = 1.006;// [kJ/kg K]\n",
+"Gamma = 1.4;// heat capacity ratio\n",
+"\n",
+"// (a)\n",
+"// For the compressor\n",
+"T2_prime = T1*(P2/P1)^((Gamma-1)/Gamma);// [K]\n",
+"T2 = T1+(T2_prime-T1)/n_com;//  exit pressure of compressor, [K]\n",
+"\n",
+"// for turbine\n",
+"T4_prime = T3*(P4/P3)^((Gamma-1)/Gamma);// [K]\n",
+"T4 = T3-(T3-T4_prime)*n_tur;// exit temperature of turbine, [K]\n",
+"\n",
+"P_output = m_dot*cp*((T3-T4)-(T2-T1));// [kW]\n",
+"mprintf('\n (a) The net power output is  =  %f  kW\n',P_output);\n",
+"\n",
+"// (b)\n",
+"n_the = ((T3-T4)-(T2-T1))/(T3-T2)*100;// thermal efficiency\n",
+"mprintf('\n (b) The thermal efficiency of the plant is  =  %f percent\n',n_the);\n",
+"\n",
+"// (c)\n",
+"P_pos = m_dot*cp*(T3-T4);// Positive cycle work, [kW]\n",
+"\n",
+"W_ratio = P_output/P_pos;// work ratio\n",
+"mprintf('\n (c) The work ratio is  =  %f\n',W_ratio)\n",
+"\n",
+"//  End"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 16.2: pressure_ratio_work_output_thermal_efficiency_work_ratio_and_Carnot_efficiency.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"clear;\n",
+"clc;\n",
+"disp('Example 16.2');\n",
+"\n",
+"// aim : To determine\n",
+"// (a) the pressure ratiowhich will give the maximum net work output\n",
+"// (b) the maximum net specific work output\n",
+"// (c) the thermal efficiency at maximum work output\n",
+"// (d) the work ratio at maximum work output\n",
+"// (e) the carnot efficiency within the cycle temperature limits\n",
+"\n",
+"// Given values\n",
+"// taking the refrence as Fig.16.35\n",
+"T3 = 273+1080;// [K]\n",
+"T1 = 273+10;// [K]\n",
+"cp = 1.007;// [kJ/kg K]\n",
+"Gamma = 1.41;//  heat capacity ratio\n",
+"\n",
+"// (a)\n",
+"r_pmax = (T3/T1)^((Gamma)/(Gamma-1));// maximum pressure ratio\n",
+"// for maximum net work output\n",
+"r_p = sqrt(r_pmax);\n",
+"mprintf('\n (a) The pressure ratio which give the maximum network output is  =  %f\n',r_p);\n",
+"\n",
+"// (b)\n",
+"T2 = T1*(r_p)^((Gamma-1)/Gamma);// [K]\n",
+"// From equation [23]\n",
+"T4 = T2;\n",
+"W_max = cp*((T3-T4)-(T2-T1));// Maximum net specific work output, [kJ/kg]\n",
+"\n",
+"mprintf('\n (b) The maximum net specific work output is  =  %f kJ/kg\n',W_max);\n",
+"\n",
+"// (c)\n",
+"W = cp*(T3-T2);\n",
+"n_the = W_max/W;// thermal efficiency\n",
+"mprintf('\n (c) The thermal efficiency at maximum work output is  = %f percent\n ',n_the*100);\n",
+"\n",
+"// (d)\n",
+"// From the equation [26]\n",
+"W_ratio = n_the;// Work ratio\n",
+"mprintf('\n (d) The work ratio at maximum work output is  =  %f\n',W_ratio);\n",
+"\n",
+"// (e)\n",
+"n_carnot = (T3-T1)/T3*100;// carnot efficiency\n",
+"mprintf('\n (e) The carnot efficiency within the cycle temperature limits is  =  %f  percent\n',n_carnot);\n",
+"\n",
+"// End"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 16.3: power_output_temperature_thermal_efficiency_and_work_ratio.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"clc;\n",
+"disp('Example 16.3');\n",
+"\n",
+"// aim : To determine\n",
+"// (a) the net power output of the plant\n",
+"// (b) the exhaust temperature from the heat exchanger\n",
+"// (c) the thermal efficiency of the plant\n",
+"// (d) the thermal efficiency of the plant if there were no heat exchanger\n",
+"// (e) the work ratio\n",
+"\n",
+"// Given values\n",
+"T1 = 273+15;// temperature, [K]\n",
+"P1 = 101;// pressure, [kN/m^2]\n",
+"P2 = 6*P1; // [kN/m^2]\n",
+"eff = .65;// effectiveness of the heat exchanger, \n",
+"T3 = 273+870;// temperature, [K]\n",
+"P4 = 101;// [kN/m^2]\n",
+"n_com = .85;// efficiency of compressor, \n",
+"n_tur = .80;// efficiency of turbine\n",
+"m_dot = 4;// mass flow rate, [kg/s]\n",
+"Gama = 1.4;// heat capacity ratio\n",
+"cp = 1.005;// [kJ/kg K]\n",
+"\n",
+"// solution\n",
+"// (a)\n",
+"// For compressor\n",
+"T2_prim = T1*(P2/P1)^((Gama-1)/Gama);// [K]\n",
+"\n",
+"// using n_com = (T2_prim-T1)/(T2-T1)')\n",
+"\n",
+"T2 = T1+(T2_prim-T1)/n_com\n",
+"// For turbine\n",
+"P3 = P2;\n",
+"T4_prim = T3*(P4/P3)^((Gama-1)/Gama);// [K]\n",
+"\n",
+"T4=T3-n_tur*(T3-T4_prim); //  [K]\n",
+"P_out = m_dot*cp*((T3-T4)-(T2-T1));//  net power output, [kW]\n",
+"mprintf('\n (a) The net power output of the plant is  =  %f kW\n',P_out);\n",
+"\n",
+"// (b)\n",
+"mtd = T4-T2;// maximum temperature drop for heat transfer, [K]\n",
+"atd = eff*mtd;// actual temperature, [K]\n",
+"et = T4-atd;// Exhaust temperature from heat exchanger, [K]\n",
+"t6 = et-273;// [C]\n",
+"mprintf('\n (b) The exhaust temperature from the heat exchanger is  =  %f C\n',t6);\n",
+"\n",
+"// (c)\n",
+"T5 = T2+atd;// [K]\n",
+"n_the = ((T3-T4)-(T2-T1))/(T3-T5)*100;// thermal effficiency \n",
+"mprintf('\n (c) The thermal efficiency of the plant is  =  %f percent\n',n_the);\n",
+"\n",
+"// (d)\n",
+"// with no heat exchanger\n",
+"n_the = ((T3-T4)-(T2-T1))/(T3-T2)*100;// thermal efficiency without heat exchanger\n",
+"mprintf('\n (d) The thermal efficiency of the plant if there wereno heat exchanger is  =  %f  percent\n',n_the);\n",
+"\n",
+"// (e)\n",
+"P_pos = m_dot*cp*(T3-T4);// positive cycle work;// [kW]\n",
+"w_rat = P_out/P_pos;// work ratio\n",
+"mprintf('\n (e) The work ratio is  =  %f\n',w_rat)\n",
+"\n",
+"//  End"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 16.4: EX16_4.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"clear;\n",
+"clc;\n",
+"disp('Example 16.4');\n",
+"\n",
+"// aim : To determine\n",
+"// (a) the pressure and temperature as the air leaves the compressor turbine\n",
+"// (b) the power output from the free power turbine\n",
+"// (c) the thermal efficiency of the plant\n",
+"// (d) the work ratio\n",
+"// (e) the carnot efficiency within the cycle temperature limits\n",
+"\n",
+"// Given values\n",
+"T1 = 273+19;// temperature, [K]\n",
+"P1 = 100;// pressure, [kN/m^2]\n",
+"P2 = 8*P1; // [kN/m^2]\n",
+"P3 = P2;// [kN/m^2]\n",
+"T3 = 273+980;// temperature, [K]\n",
+"n_com = .85;// efficiency of rotary compressor\n",
+"P5 = 100;// [kN/m^2]\n",
+"n_cum = .88;// isentropic efficiency of combustion chamber compressor, \n",
+"n_tur = .86;// isentropic efficiency of turbine\n",
+"m_dot = 7;// mass flow rate of air, [kg/s]\n",
+"Gama = 1.4;// heat capacity ratio\n",
+"cp = 1.006;// [kJ/kg K]\n",
+"\n",
+"// solution\n",
+"// (a)\n",
+"// For compressor\n",
+"T2_prim = T1*(P2/P1)^((Gama-1)/Gama);// [K]\n",
+"\n",
+"T2 = T1+(T2_prim-T1)/n_com;// temperature, [K]\n",
+"\n",
+"// for compressor turbine\n",
+"// T3-T4 = T2-T1,because compressor turbine power=compressor power so\n",
+"T4 = T3-(T2-T1);//turbine exit temperature, [K]\n",
+"T4_prim = T3-(T3-T4)/n_cum;// [K]\n",
+"\n",
+"// For turbine\n",
+"// T4_prim = T3*(P4/P3)^((Gama-1)/Gama)\n",
+"P4 = P3*(T4_prim/T3)^(Gama/(Gama-1));// exit air pressure of air, [kN/m^2]\n",
+"\n",
+"mprintf('\n (a) The temperature as the air leaves the compressor turbine is  =  %f  C\n',T4-273);\n",
+"mprintf('\n      The pressure as the air leaves the compressor turbine is  =  %f  kN/m^2\n',P4);\n",
+"\n",
+"// (b)\n",
+"T5_prim = T4*(P5/P4)^((Gama-1)/Gama);// [K]\n",
+"\n",
+"\n",
+"T5 = T4-n_tur*(T4-T5_prim);// temperature, [K]\n",
+"\n",
+"PO = m_dot*cp*(T4-T5);// power output\n",
+"mprintf('\n (b) The power output from the free power turbine is  =  %f kW\n',PO);\n",
+"\n",
+"// (c)\n",
+"\n",
+"n_the = (T4-T5)/(T3-T2)*100;// thermal effficiency \n",
+"mprintf('\n (c) The thermal efficiency of the plant is  =  %f percent\n',n_the);\n",
+"\n",
+"// (d)\n",
+"\n",
+"WR = (T4-T5)/(T3-T5);// work ratio\n",
+"mprintf('\n (d) The work ratio is  =  %f\n',WR);\n",
+"\n",
+"// (e)\n",
+"CE = (T3-T1)/T3;// carnot efficiency\n",
+"mprintf('\n (e) The carnot efficiency is  =  %f percent\n',CE*100);\n",
+"\n",
+"//  End"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 16.5: pressure_temperature_and_power_developed.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"clear;\n",
+"clc;\n",
+"disp('Example 16.5');\n",
+"\n",
+"// aim : To determine\n",
+"// (a) the pressure and temperature of the air compression \n",
+"// (b) the power developed by the gas turbine\n",
+"// (c) the temperature and pressure of the airentering the exhaust jet as it leaves the gas turbine \n",
+"\n",
+"// Given values\n",
+"T1 = 273-22.4;// temperature, [K]\n",
+"P1 = 470;// pressure, [bar]\n",
+"P2 = 30*P1; // [kN/m^2]\n",
+"P3 = P2;// [kN/m^2]\n",
+"T3 = 273+960;// temperature, [K]\n",
+"r = 1.25;// ratio of turbine power to compressor power\n",
+"n_tur = .86;// isentropic efficiency of turbine\n",
+"m_dot = 80;// mass flow rate of air, [kg/s]\n",
+"Gama = 1.41;// heat capacity ratio\n",
+"cp = 1.05;// [kJ/kg K]\n",
+"\n",
+"// solution\n",
+"// (a)\n",
+"// For compressor\n",
+"T2_prim = T1*(P2/P1)^((Gama-1)/Gama);// [K]\n",
+"// using n_tur=(T2_prim-T1)/(T2-T1)\n",
+"T2 = T1+(T2_prim-T1)/n_tur;// temperature, [K]\n",
+"\n",
+"mprintf('\n (a) The pressure of the air after compression is  =  %f  bar\n',P2);\n",
+"\n",
+"mprintf('\n      The temperature of the air after compression is  =  %f  C\n',T2-273);\n",
+"\n",
+"// (b)\n",
+"Td  = r*(T2-T1);// temperature drop in turbine, [K]\n",
+"PO = m_dot*cp*Td;// power output, [kW]\n",
+"mprintf('\n (b) The power developed by the gas turbine is  =  %f  MW\n',PO*10^-3);\n",
+"\n",
+"// (c)\n",
+"t3 = T3-273;// [C]\n",
+"t4 = t3-Td;// temeprerature of air leaving turbine,[K]\n",
+"Tdi = Td/n_tur;// isentropic temperature drop, [K]\n",
+"T4_prim = t3-Tdi+273;// temperature, [K]\n",
+"// using T4_prim=T3*(P4/P3)^((Gama-1)/Gama)\n",
+"P4 = P3*(T4_prim/T3)^(Gama/(Gama-1));// exit air pressure of air, [kN/m^2]\n",
+"\n",
+"mprintf('\n (c) The air pressure as it leaves the gas turbine is  =  %f  bar\n',P4);\n",
+"\n",
+"// Result in the book is not matching because they have taken pressure in mbar  but in in question it is given in bar \n",
+"\n",
+"//   End\n",
+""
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 16.6: mass_theoretical_output_and_thermal_efficiency.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"clc;\n",
+"disp('Example 16.6');\n",
+"\n",
+"// aim : To determine\n",
+"// (a) the mass of fuel oil used by the gas turbine\n",
+"// (b) the mass flow of steam from the boiler \n",
+"// (c) the theoretical output from the steam turbine\n",
+"// (d) the overall theoretical thermal efficiency of the plant\n",
+"\n",
+"// given values\n",
+"Po = 150;// generating plant output, [MW]\n",
+"n_the1 = .35;// thermal efficiency\n",
+"CV = 43;// calorific value of fuel, [MJ]\n",
+"me = 400;// flow rate of exhaust gas, [kg/s]\n",
+"T = 90;// boiler exit temperature, [C]\n",
+"T1 = 550;// exhaust gas temperature, [C]\n",
+"P2 = 10;// steam generation pressure, [MN/m^2]\n",
+"T2 = 450;// boiler exit temperature, [C]\n",
+"Tf = 140;// feed water temperature, [C]\n",
+"n_tur = .86;// turbine efficiency\n",
+"P3 = .5;// exhaust temperature, [MN/m^2]\n",
+"n_boi = .92;// boiler thermal efficiency\n",
+"cp = 1.1;// heat capacity, [kJ/kg]\n",
+"\n",
+"\n",
+"//  solution\n",
+"// (a)\n",
+"ER = Po*3600/n_the1;// energy requirement from the fuel, [MJ/h]\n",
+"mf = ER/CV*10^-3;// fuel required, [tonne/h]\n",
+"mprintf('\n (a) The mass of fuel oil used by the gas is  =  %f  tonne/h\n',mf);\n",
+"\n",
+"// (b) \n",
+"\n",
+"ET = me*cp*(T1-T)*3600*n_boi;// energy transferred to steam,[kJ/h]\n",
+"// from steam table\n",
+"h1 = 3244;// specific enthalpy, [kJ/kg]\n",
+"hf = 588.5;// specific enthalpy, [kJ/kg]\n",
+"ERR = h1-hf;// energy required to raise steam, [kJ/kg]\n",
+"ms = ET/ERR*10^-3;// mass flow of steam, [tonne/h]\n",
+"mprintf('\n (b) The mass flow  rate of steam from the boiler is  =  %f  tonne/h\n',ms);\n",
+"\n",
+"// again from steam table\n",
+"s1 = 6.424;// specific entropy, [kJ/kg K]\n",
+"sf2 = 1.86;// specific entropy, [kJ/kg K\n",
+"sg2 = 6.819;// specific entropy, [kJ/kg K]\n",
+"\n",
+"hf2 = 640.1;// specific enthalpy,[kJ/kg]\n",
+"hg2 = 2747.5;// specific enthalpy, [kJ/kg]\n",
+"// for ths process s1=s2=sf2+x2*(sg2-sf2)\n",
+"s2 = s1;\n",
+"// hence\n",
+"x2 = (s2-sf2)/(sg2-sf2);// dryness fraction\n",
+"\n",
+"h2_prim = hf2+x2*(hg2-hf2);// specific enthalpy of steam, [kJ/kg]\n",
+"\n",
+"TO = n_tur*(h1-h2_prim);//theoretical steam turbine output, [kJ/kg]\n",
+"TOt = TO*ms/3600;// total theoretical steam turbine output, [MW]\n",
+"\n",
+"mprintf('\n (c) The theoretical output from the steam turbine is  =  %f  MW\n',TOt);\n",
+"\n",
+"// (d)\n",
+"n_tho = (Po+TOt)*n_the1/Po;// overall theoretical thermal efficiency\n",
+"mprintf('\n (d) The overall thermal efficiency is  =  %f  percent\n',n_tho*100);\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
+}
diff --git a/Basic_Engineering_Thermodynamics_by_R_Joel/17-Engine_and_plant_trials.ipynb b/Basic_Engineering_Thermodynamics_by_R_Joel/17-Engine_and_plant_trials.ipynb
new file mode 100644
index 0000000..7087ec2
--- /dev/null
+++ b/Basic_Engineering_Thermodynamics_by_R_Joel/17-Engine_and_plant_trials.ipynb
@@ -0,0 +1,455 @@
+{
+"cells": [
+ {
+		   "cell_type": "markdown",
+	   "metadata": {},
+	   "source": [
+       "# Chapter 17: Engine and plant trials"
+	   ]
+	},
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 17.1: indicated_and_brake_output_mechanical_efficiency_and_energy_balance.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"clear;\n",
+"clc;\n",
+"disp('Example 17.1');\n",
+"\n",
+"// aim : To determine\n",
+"// the indicated and brake output and the mechanicl efficiency\n",
+"// draw up an overall energy balance and as % age\n",
+"\n",
+"// given values\n",
+"h = 21;// height of indicator diagram, [mm]\n",
+"ic = 27;// indicator calibration, [kN/m^2 per mm]\n",
+"sv = 14*10^-3;// swept volume of the cylinder;,[m^3]\n",
+"N = 6.6;// speed of engine, [rev/s]\n",
+"ebl = 77;// effective brake load, [kg]\n",
+"ebr = .7;// effective brake radious, [m]\n",
+"fc = .002;// fuel consumption, [kg/s]\n",
+"CV = 44000;// calorific value of fuel, [kJ/kg]\n",
+"cwc = .15;// cooling water circulation, [kg/s]\n",
+"Ti = 38;// cooling water inlet temperature, [C]\n",
+"To = 71;// cooling water outlet temperature, [C]\n",
+"c = 4.18;// specific heat capacity of water, [kJ/kg]\n",
+"eeg = 33.6;// energy to exhaust gases, [kJ/s]\n",
+"g = 9.81;// gravitational acceleration, [m/s^2]\n",
+"\n",
+"// solution\n",
+"PM = ic*h;// mean effective pressure, [kN/m^2]\n",
+"LA = sv;// swept volume of the cylinder, [m^3]\n",
+"ip = PM*LA*N/2;// indicated power,[kW]\n",
+"T = ebl*g*ebr;// torque, [N*m]\n",
+"bp = 2*%pi*N*T;// brake power, [W]\n",
+"n_mech = bp/ip*10^-3;// mechanical efficiency\n",
+"mprintf('\n The Indicated power is  =  %f kW\n',ip);\n",
+"mprintf('\n The Brake power is  =  %f  kW\n',bp*10^-3);\n",
+"mprintf('\n The mechanical efficiency is  =  %f  percent\n',n_mech);\n",
+"\n",
+"ef = CV*fc;// energy from fuel, [kJ/s]\n",
+"eb = bp*10^-3;// energy to brake power,[kJ/s]\n",
+"ec = cwc*c*(To-Ti);// energy to coolant,[kJ/s]\n",
+"es = ef-(eb+ec+eeg);// energy to surrounding,[kJ/s]\n",
+"\n",
+"disp('Energy can be tabulated as :-');\n",
+"disp('----------------------------------------------------------------------------------------------------');\n",
+"disp('                                                  kJ/s                           Percentage   ')\n",
+"disp('----------------------------------------------------------------------------------------------------');\n",
+"mprintf('\n Energy from fuel                         %f                   %f\n Energy to brake power                %f                    %f\n Energy to coolant                       %f                    %f\n Energy to exhaust                      %f                    %f\n Energy to suroundings,etc.         %f                    %f\n',ef,ef/ef*100,eb,eb/ef*100,ec,ec/ef*100,eeg,eeg/ef*100,es,es/ef*100);\n",
+"\n",
+"//  End"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 17.2: EX17_2.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"clear;\n",
+"clc;\n",
+"disp('Example 17.2');\n",
+"\n",
+"// aim : To determine\n",
+"// (a) bp\n",
+"// (b) ip\n",
+"// (c) mechanical efficiency\n",
+"// (d) indicated thermal efficiency\n",
+"// (e) brake specific steam consumption\n",
+"// (f) draw up complete energy account for the test one-minute basis taking 0 C as datum\n",
+"\n",
+"// given values\n",
+"d = 200*10^-3;// cylinder diameter, [mm]\n",
+"L = 250*10^-3;// stroke, [mm]\n",
+"N = 5;// speed, [rev/s]\n",
+"r = .75/2;// effective radious of brake wheel, [m]\n",
+"Ps = 800;// stop valve pressure, [kN/m^2]\n",
+"x = .97;// dryness fraction of steam\n",
+"BL = 136;// brake load, [kg]\n",
+"SL = 90;// spring balance load, [N]\n",
+"PM = 232;// mean effective pressure, [kN/m^2]\n",
+"Pc = 10;// condenser pressure, [kN/m^2]\n",
+"m_dot = 3.36;// steam consumption, [kg/min]\n",
+"CC = 113;// condenser cooling water, [kg/min]\n",
+"Tr = 11;// temperature rise of condenser cooling water, [K]\n",
+"Tc = 38;// condensate temperature, [C]\n",
+"C = 4.18;// heat capacity of water, [kJ/kg K]\n",
+"g = 9.81;// gravitational acceleration, [m/s^2]\n",
+"\n",
+"// solution\n",
+"// from steam table\n",
+"// at 800 kN/m^2\n",
+"tf1 = 170.4;// saturation temperature, [C]\n",
+"hf1 = 720.9;// [kJ/kg]\n",
+"hfg1 = 2046.5;// [kJ/kg]\n",
+"hg1 = 2767.5;// [kJ/kg]\n",
+"vg1 = .2403;// [m^3/kg]\n",
+"\n",
+"// at 10 kN/m^2\n",
+"tf2 = 45.8;// saturation temperature, [C]\n",
+"hf2 = 191.8;// [kJ/kg]\n",
+"hfg2 = 2392.9;// [kJ/kg]\n",
+"hg2 = 2584.8;// [kJ/kg]\n",
+"vg2 = 14.67;// [m^3/kg]\n",
+"\n",
+"// (a)\n",
+"T = (BL*g-SL)*r;// torque, [Nm]\n",
+"bp = 2*%pi*N*T*10^-3;// brake power,[W]\n",
+"mprintf('\n (a) The brake power is  =   %f  kW\n',bp);\n",
+"\n",
+"// (b)\n",
+"A = %pi*d^2/4;// area, [m^2]\n",
+"ip = PM*L*A*N*2;// double-acting so*2, [kW]\n",
+"mprintf('\n (b) The indicated power is  =  %f kW\n',ip);\n",
+"\n",
+"// (c)\n",
+"n_mec = bp/ip;// mechanical efficiency\n",
+"mprintf('\n (c) The mechanical efficiency is  =  %f  percent\n',n_mec*100);\n",
+"\n",
+"// (d)\n",
+"h = hf1+x*hfg1;// [kJ/kg]\n",
+"hf = hf2;\n",
+"ITE = ip/((m_dot/60)*(h-hf));// indicated thermal efficiency\n",
+"mprintf('\n (d) The indicated thermal efficiency is  =  %f percent\n',ITE*100);\n",
+"// (e)\n",
+"Bsc=m_dot*60/bp;// brake specific steam consumption, [kg/kWh]\n",
+"mprintf('\n (e) The brake steam consumption is  =  %f  kg/kWh\n',Bsc);\n",
+"\n",
+"// (f)\n",
+"// energy balanvce reckoned from 0 C\n",
+"Es = m_dot*h;// energy supplied, [kJ]\n",
+"Eb = bp*60;// energy to bp, [kJ]\n",
+"Ecc = CC*C*Tr;// energy to condensate cooling water, [kJ]\n",
+"Ec = m_dot*C*Tc;// energy to condensate, [kJ]\n",
+"Ese = Es-Eb-Ecc-Ec;// energy to surrounding,etc, [kJ]\n",
+"\n",
+"mprintf('\n (f) Energy supplied/min is  =  %f  kJ\n',Es);\n",
+"\n",
+"mprintf('\n    Energy to bp/min is  =  %f  kJ\n',Eb);\n",
+"mprintf('\n    Energy to condenser cooling water/min is  =  %f kJ\n',Ecc);\n",
+"mprintf('\n    Energy to condensate/min is  =  %f kJ\n',Ec);\n",
+"mprintf('\n    Energy to surrounding, etc/min is  =  %f  kJ\n',Ese);\n",
+"\n",
+"// answer in the book is misprinted\n",
+"\n",
+"//  End"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 17.3: brake_power_fuel_consumption_thermal_efficiency_and_energy_balance.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"clear;\n",
+"clc;\n",
+"disp('Example 17.3');\n",
+"\n",
+"// aim : To determine\n",
+"// (a) the brake power\n",
+"// (b) the brake specific fuel consumption\n",
+"// (c) the indicated thermal efficiency\n",
+"// (d) the energy balance, expressing the various items\n",
+"\n",
+"// given values\n",
+"t = 30;// duration of trial, [min]\n",
+"N = 1750;// speed of engine, [rev/min]\n",
+"T = 330;// brake torque, [Nm]\n",
+"mf = 9.35;// fuel consumption, [kg]\n",
+"CV = 42300;// calorific value of fuel, [kJ/kg]\n",
+"cwc = 483;// jacket cooling water circulation, [kg]\n",
+"Ti = 17;// inlet temperature, [C]\n",
+"To = 77;// outlet  temperature, [C]\n",
+"ma = 182;// air consumption, [kg]\n",
+"Te = 486;// exhaust temperature, [C]\n",
+"Ta = 17;// atmospheric temperature, [C]\n",
+"n_mec = .83;// mechanical efficiency\n",
+"c = 1.25;// mean specific heat capacity of exhaust gas, [kJ/kg K]\n",
+"C = 4.18;// specific heat capacity, [kJ/kg K]\n",
+"\n",
+"// solution\n",
+"// (a)\n",
+"bp = 2*%pi*N*T/60*10^-3;// brake power, [kW]\n",
+"mprintf('\n (a) The Brake power is  =  %f  kW\n',bp);\n",
+"\n",
+"// (b)\n",
+"bsf = mf*2/bp;//brake specific fuel consumption, [kg/kWh]\n",
+"mprintf('\n (b) The brake specific fuel consumption is  =  %f kg/kWh\n',bsf);\n",
+"\n",
+"// (c)\n",
+"ip = bp/n_mec;// indicated power, [kW]\n",
+"ITE = ip/(2*mf*CV/3600);// indicated thermal efficiency\n",
+"mprintf('\n (c) The indicated thermal efficiency is  =  %f percent\n',ITE*100);\n",
+"\n",
+"// (d)\n",
+"// taking  basis one minute \n",
+"ef = CV*mf/30;// energy from fuel, [kJ]\n",
+"eb = bp*60;// energy to brake power,[kJ]\n",
+"ec = cwc/30*C*(To-Ti);// energy to cooling water,[kJ]\n",
+"ee = (ma+mf)/30*c*(Te-Ta);// energy to exhaust, [kJ]\n",
+"es = ef-(eb+ec+ee);// energy to surrounding,etc,[kJ]\n",
+"\n",
+"mprintf('\n (d) Energy from fuel is  =  %f kJ\n',ef);\n",
+"mprintf('\n      Energy to brake power is  =  %f kJ\n',eb);\n",
+"mprintf('\n      Energy to cooling water is  =  %f  kJ\n',ec);\n",
+"mprintf('\n      Energy to exhaust is  =  %f  kJ\n',ee);\n",
+"mprintf('\n      Energy to surrounding, etc is  =  %f kJ\n',es);\n",
+" \n",
+"//  End"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 17.4: indicated_power_and_mechanical_efficiency.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"clear;\n",
+"clc;\n",
+"disp('Example 17.4');\n",
+"\n",
+"// aim : To determine\n",
+"// (a) the indicated power of the engine\n",
+"// (b) the mechanical efficiency of the engine\n",
+"\n",
+"// given values\n",
+"bp = 52;// brake power output, [kW]\n",
+"bp1 = 40.5;// brake power of cylinder cut1, [kW]\n",
+"bp2 = 40.2;// brake power of cylinder cut2, [kW]\n",
+"bp3 = 40.1;// brake power of cylinder cut3, [kW]\n",
+"bp4 = 40.6;// brake power of cylinder cut4, [kW]\n",
+"bp5 = 40.7;// brake power of cylinder cut5, [kW]\n",
+"bp6 = 40.0;// brake power of cylinder cut6, [kW]\n",
+"\n",
+"// sollution\n",
+"ip1 = bp-bp1;// indicated power of cylinder cut1, [kW]\n",
+"ip2 = bp-bp2;// indicated power of cylinder cut2, [kW]\n",
+"ip3 = bp-bp3;// indicated power of cylinder cut3, [kW]\n",
+"ip4 = bp-bp4;// indicated power of cylinder cut4, [kW]\n",
+"ip5 = bp-bp5;// indicated power of cylinder cut5, [kW]\n",
+"ip6 = bp-bp6;// indicated power of cylinder cut6, [kW]\n",
+"\n",
+"ip = ip1+ip2+ip3+ip4+ip5+ip6;// indicated power of engine,[kW]\n",
+"mprintf('\n (a) The indicated power of the engine is  =  %f  kW\n',ip);\n",
+"\n",
+"// (b)\n",
+"n_mec = bp/ip;// mechanical efficiency\n",
+"mprintf('\n (b) The mechanical  efficiency of the engine is  =  %f percent\n',n_mec*100);\n",
+"\n",
+"// End"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 17.5: brake_power_indicated_power_mechanical_efficiency_and_energy_balance.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"clear;\n",
+"clc;\n",
+"disp('Example 17.5');\n",
+"\n",
+"// aim : To determine\n",
+"// the brake power,indicated power and mechanicl efficiency\n",
+"// draw up an energy balance and as % age of the energy supplied\n",
+"\n",
+"// given values\n",
+"N = 50;// speed, [rev/s]\n",
+"BL = 267;// break load.,[N]\n",
+"BL1 = 178;// break load of cylinder cut1, [N]\n",
+"BL2 = 187;// break load of cylinder cut2, [N]\n",
+"BL3 = 182;// break load of cylinder cut3, [N]\n",
+"BL4 = 182;// break load of cylinder cut4, [N]\n",
+"\n",
+"FC = .568/130;// fuel consumption, [L/s]\n",
+"s = .72;// specific gravity of fuel\n",
+"CV = 43000;// calorific value of fuel, [kJ/kg]\n",
+"\n",
+"Te = 760;// exhaust temperature, [C]\n",
+"c = 1.015;// specific heat capacity of exhaust gas, [kJ/kg K]\n",
+"Ti = 18;// cooling water inlet temperature, [C]\n",
+"To = 56;// cooling water outlet temperature, [C]\n",
+"mw = .28;// cooling water flow rate, [kg/s]\n",
+"Ta = 21;// ambient tempearture, [C]\n",
+"C = 4.18;// specific heat capacity of cooling water, [kJ/kg K]\n",
+"\n",
+"// solution\n",
+"bp = BL*N/455;// brake power of engine, [kW]\n",
+"bp1 = BL1*N/455;// brake power of cylinder cut1, [kW]\n",
+"i1 = bp-bp1;// indicated power of cylinder cut1, [kW]\n",
+"bp2 = BL2*N/455;// brake power of cylinder cut2, [kW]\n",
+"i2 = bp-bp2;// indicated power of cylinder cut2, [kW]\n",
+"bp3 = BL3*N/455;// brake power of cylinder cut3, [kW]\n",
+"i3 = bp-bp3;// indicated power of cylinder cut3, [kW]\n",
+"bp4 = BL4*N/455;// brake power of cylinder cut4, [kW]\n",
+"i4 = bp-bp4;// indicated power of cylinder cut4, [kW]\n",
+"\n",
+"ip = i1+i2+i3+i4;// indicated power of engine, [kW]\n",
+"n_mec = bp/ip;// mechanical efficiency\n",
+"\n",
+"mprintf('\n The Brake power is  =  %f kW\n',bp);\n",
+"mprintf('\n The Indicated power is  =  %f kW\n',ip);\n",
+"mprintf('\n The mechanical efficiency is  =  %f percent\n',n_mec*100);\n",
+"\n",
+"mf = FC*s;// mass of fuel/s, [kg]\n",
+"ef = CV*mf;// energy from fuel/s, [kJ]\n",
+"me = 15*mf;// mass of exhaust/s,[kg],(given in condition)\n",
+"ee = me*c*(Te-Ta);// energy to exhaust/s,[kJ]\n",
+"ec = mw*C*(To-Ti);// energy to cooling water/s,[kJ]\n",
+"es = ef-(ee+ec+bp);// energy to surrounding,etc/s,[kJ]\n",
+"\n",
+"disp('Energy can be tabulated as :-');\n",
+"disp('----------------------------------------------------------------------------------------------------');\n",
+"disp('                                                  kJ/s                           Percentage   ')\n",
+"disp('----------------------------------------------------------------------------------------------------');\n",
+"mprintf('\n Energy from fuel                         %f                 %f\n Energy to brake power                 %f                   %f\n Energy to exhaust                       %f                   %f\n Energy to coolant                        %f                   %f\n Energy to suroundings,etc.           %f                  %f\n',ef,ef/ef*100,bp,bp/ef*100,ee,ee/ef*100,ec,ec/ef*100,es,es/ef*100);\n",
+"\n",
+"// there is minor variation in the result reported in the book\n",
+"//  End"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 17.6: brake_power_fuel_consumption_and_thermal_efficiency.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"clear;\n",
+"clc;\n",
+"disp('Example 17.6');\n",
+"\n",
+"// aim : To determine \n",
+"// (a) the break power of engine\n",
+"// (b) the fuel consumption of the engine\n",
+"// (c) the brake thermal efficiency of the engine\n",
+"\n",
+"// given values\n",
+"d = 850*10^-3;// bore , [m]\n",
+"L = 2200*10^-3;// stroke, [m]\n",
+"PMb = 15;// BMEP of cylinder, [bar]\n",
+"N = 95/60;// speed of engine, [rev/s]\n",
+"sfc = .2;// specific fuel oil consumption, [kg/kWh]\n",
+"CV = 43000;// calorific value of the fuel oil, [kJ/kg]\n",
+"\n",
+"// solution\n",
+"// (a)\n",
+"A = %pi*d^2/4;// area, [m^2]\n",
+"bp = PMb*L*A*N*8/10;// brake power,[MW]\n",
+"mprintf('\n (a) The brake power is  =  %f  MW\n',bp);\n",
+"\n",
+" // (b)\n",
+" FC = bp*sfc;// fuel consumption, [kg/h]\n",
+" mprintf('\n (b) The fuel consumption is  =  %f  tonne/h\n',FC);\n",
+" \n",
+" // (c)\n",
+" mf = FC/3600;// fuel used, [kg/s]\n",
+" n_the = bp/(mf*CV);// brake thermal efficiency\n",
+" mprintf('\n (c) The brake thermal efficiency is  =  %f  percent\n',n_the*100);\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
+}
diff --git a/Basic_Engineering_Thermodynamics_by_R_Joel/18-Refrigeration.ipynb b/Basic_Engineering_Thermodynamics_by_R_Joel/18-Refrigeration.ipynb
new file mode 100644
index 0000000..689c0f3
--- /dev/null
+++ b/Basic_Engineering_Thermodynamics_by_R_Joel/18-Refrigeration.ipynb
@@ -0,0 +1,182 @@
+{
+"cells": [
+ {
+		   "cell_type": "markdown",
+	   "metadata": {},
+	   "source": [
+       "# Chapter 18: Refrigeration"
+	   ]
+	},
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 18.1: coefficient_of_performance_mass_flow_and_cooling_water_requirement.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"clear;\n",
+"clc;\n",
+"disp('Example 18.1');\n",
+"\n",
+"// aim : To determine\n",
+"// (a) the coefficient of performance\n",
+"// (b) the mass flow of the refrigerant\n",
+"// (c) the cooling water required by the condenser\n",
+"\n",
+"// given values\n",
+"P1 = 462.47;// pressure limit, [kN/m^2]\n",
+"P3 = 1785.90;// pressure limit, [kN/m^2]\n",
+"T2 = 273+59;// entering saturation temperature, [K]\n",
+"T5 = 273+32;// exit temperature of condenser, [K]\n",
+"d = 75*10^-3;// bore, [m]\n",
+"L = d;// stroke, [m]\n",
+"N = 8;// engine speed, [rev/s]\n",
+"VE = .8;// olumetric efficiency\n",
+"cpL = 1.32;// heat capacity of liquid, [kJ/kg K]\n",
+"c = 4.187;// heat capacity of water, [kj/kg K]\n",
+"\n",
+"// solution\n",
+"// from given table\n",
+"// at P1\n",
+"h1 = 231.4;// specific enthalpy, [kJ/kg]\n",
+"s1 = .8614;// specific entropy,[ kJ/kg K\n",
+"v1 = .04573;// specific volume, [m^3/kg]\n",
+"\n",
+"// at P3\n",
+"h3 = 246.4;// specific enthalpy, [kJ/kg]\n",
+"s3 = .8093;// specific entropy,[ kJ/kg K\n",
+"v3 = .04573;// specific volume, [m^3/kg]\n",
+"T3= 273+40;// saturation temperature, [K]\n",
+"h4 = 99.27;// specific enthalpy, [kJ/kg]\n",
+"// (a)\n",
+"s2 = s1;// specific entropy, [kJ/kg k]\n",
+"// using s2=s3+cpv*log(T2/T3)\n",
+"cpv = (s2-s3)/log(T2/T3);// heat capacity, [kj/kg k]\n",
+"\n",
+"// from Fig.18.8\n",
+"T4 = T3;\n",
+"h2 = h3+cpv*(T2-T3);// specific enthalpy, [kJ/kg]\n",
+"h5 = h4-cpL*(T4-T5);// specific enthalpy, [kJ/kg]\n",
+"h6 = h5;\n",
+"COP = (h1-h6)/(h2-h1);// coefficient of performance\n",
+"mprintf('\n (a) The coefficient of performance of the refrigerator is  =  %f\n',COP);\n",
+"\n",
+"// (b)\n",
+"SV = %pi/4*d^2*L;// swept volume of compressor/rev, [m^3]\n",
+"ESV = SV*VE*N*3600;// effective swept volume/h, [m^3]\n",
+"m = ESV/v1;// mass flow of refrigerant/h,[kg]\n",
+"mprintf('\n (b) The mass flow of refrigerant/h is  =  %f  kg\n',m);\n",
+"\n",
+"// (c)\n",
+"dT = 12;// temperature limit, [C]\n",
+"Q = m*(h2-h5);// heat transfer in condenser/h, [kJ]\n",
+"// using Q=m_dot*c*dT, so\n",
+"m_dot = Q/(c*dT);// mass flow of water required, [kg/h]\n",
+"mprintf('\n (c) The mass flow of water required is  =  %f kg/h\n',m_dot);\n",
+"\n",
+"//  End"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 18.2: mass_flow_dryness_fraction_power_and_ratio_of_heat_transfer.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"clear;\n",
+"clc;\n",
+"disp('Example 18.2');\n",
+"\n",
+"// aim : To determine\n",
+"// (a) the mass flow of R401\n",
+"// (b) the dryness fraction of  R401 at the entry to the evaporator\n",
+"// (c) the power of driving motor\n",
+"// (d) the ratio of heat transferred from condenser to the power required to the motor\n",
+"\n",
+"// given values\n",
+"P1 = 411.2;// pressure limit, [kN/m^2]\n",
+"P3 = 1118.9;// pressure limit, [kN/m^2]\n",
+"Q = 100*10^3;// heat transfer from the condenser,[kJ/h]\n",
+"T2 = 273+60;// entering saturation temperature, [K]\n",
+"\n",
+"// given\n",
+"// from given table\n",
+"// at P1\n",
+"h1 = 409.3;// specific enthalpy, [kJ/kg]\n",
+"s1 = 1.7431;// specific entropy,[ kJ/kg K\n",
+"\n",
+"// at P3\n",
+"h3 = 426.4;// specific enthalpy, [kJ/kg]\n",
+"s3 = 1.7192;// specific entropy,[ kJ/kg K\n",
+"T3 = 273+50;// saturation temperature, [K]\n",
+"h4 = 265.5;// specific enthalpy, [kJ/kg]\n",
+"// (a)\n",
+"s2 = s1;// specific entropy, [kJ/kg k]\n",
+"// using s2=s3+cpv*log(T2/T3)\n",
+"cpv = (s2-s3)/log(T2/T3);// heat capacity, [kj/kg k]\n",
+"\n",
+"// from Fig.18.8\n",
+"h2 = h3+cpv*(T2-T3);// specific enthalpy, [kJ/kg]\n",
+"Qc = h2-h4;// heat transfer from condenser, [kJ/kg]\n",
+"mR401 = Q/Qc;// mass flow of R401, [kg]\n",
+" mprintf('\n (a) The mass flow of R401 is  =  %f kg/h\n',mR401);\n",
+"\n",
+"// (b)\n",
+"hf1 = 219;// specific enthalpy, [kJ/kg]\n",
+"h5 = h4;\n",
+"// using h5=hf1+s5*(h1-hf1),so\n",
+"x5 = (h5-hf1)/(h1-hf1);// dryness fraction\n",
+"mprintf('\n (b) The dryness fraction of R401 at the entry to the evaporator is  =  %f\n',x5);\n",
+"\n",
+"// (c)\n",
+"P = mR401*(h2-h1)/3600/.7;// power to driving motor, [kW]\n",
+" mprintf('\n (c) The power to driving motor is  =  %f kW\n',P);\n",
+"\n",
+"// (d)\n",
+"r = Q/3600/P;// ratio\n",
+"mprintf('\n (d) The ratio of heat transferred from condenser to the power required to the motor is  =  %f : 1\n',r);\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
+}
diff --git a/Basic_Engineering_Thermodynamics_by_R_Joel/19-Psychrometry.ipynb b/Basic_Engineering_Thermodynamics_by_R_Joel/19-Psychrometry.ipynb
new file mode 100644
index 0000000..3e042c9
--- /dev/null
+++ b/Basic_Engineering_Thermodynamics_by_R_Joel/19-Psychrometry.ipynb
@@ -0,0 +1,308 @@
+{
+"cells": [
+ {
+		   "cell_type": "markdown",
+	   "metadata": {},
+	   "source": [
+       "# Chapter 19: Psychrometry"
+	   ]
+	},
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 19.1: moisture_content.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"clear;\n",
+"clc;\n",
+"disp('Example 19.1');\n",
+"\n",
+"// aim : To compare the moisture content and the true specific volumes of  atmosphere air \n",
+"// (a) temperature is 12 C and the air is saturaded\n",
+"// (b) temperature is 31 C and air is .75 saturated\n",
+"\n",
+"// Given values\n",
+"P_atm = 101.4;// atmospheric pressure, [kN/m^2]\n",
+"R = .287;// [kJ/kg K]\n",
+"\n",
+"// solution\n",
+"// (a)\n",
+"T = 273+12;// air temperature, [K]\n",
+"// From steam table at 12 C\n",
+"p = 1.4;// [kN/m^2]\n",
+"vg = 93.9;// [m^3/kg]\n",
+"pa = P_atm-p;// partial pressure of the dry air, [kN/m^2]\n",
+"va = R*T/pa;// [m^3/kg]\n",
+"\n",
+"mw = va/vg;// mass of water vapor in the air,[kg]\n",
+"v = va/(1+mw);// specific volume of humid air, [m^3/kg]\n",
+"\n",
+"mprintf('\n (a) The mass of water vapor in the humid air is  =  %f  kg\n',mw);\n",
+"mprintf('\n      The specific volume of humid air is  =  %f m^3/kg\n',v);\n",
+"\n",
+"// (b)\n",
+"x = .75;// dryness fraction\n",
+"T = 273+31;// air temperature, [K]\n",
+"// From steam table\n",
+"p = 4.5;// [kN/m^2]\n",
+"vg = 31.1;// [m^3/kg]\n",
+"pa = P_atm-p;// [kN/m^2]\n",
+"va = R*T/pa;// [m^3/kg]\n",
+"mw1= va/vg;// mass of water vapor in the air, [kg]\n",
+"mw_actual = mw1*x;// actual mass of vapor, [kg]\n",
+"v = va/(1+mw_actual);// true specific volume of humid air,[m^3/kg] \n",
+"\n",
+"mprintf('\n (b) The mass of water vapor in the humid air is  =  %f  kg\n',mw1);\n",
+"mprintf('\n     The specific volume of humid air is  =  %f  m^3/kg\n',v);\n",
+"\n",
+"ewv = mw_actual/mw ;\n",
+"mprintf('\n On the warm day the air conteains %f  times the mass of water vapor as on the cool day \n',ewv);\n",
+"\n",
+"//  End"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 19.2: partial_pressures_specific_humidity_and_composition.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"clear;\n",
+"clc;\n",
+"disp('Example 19.2');\n",
+"\n",
+"// aim : To determine\n",
+"// (a) the partial pressures of the vapor and the dry air\n",
+"// (b) the specific humidity of the mixture\n",
+"// (c) the composition of the mixture\n",
+"\n",
+"//  Given values\n",
+"phi = .65;// Relative humidity\n",
+"T = 2733+20;// temperature, [K]\n",
+"p = 100;// barometric pressure, [kN/m^2]\n",
+"\n",
+"// solution\n",
+"// (a)\n",
+"//  From the steam table at 20 C\n",
+"pg = 2.34;// [kN/m^2]\n",
+"ps = phi*pg;// partial pressure of vapor, [kN/m^2]\n",
+"pa = p-ps;// partial pressure of dry air, [kN/m^2]\n",
+"mprintf('\n (a) The partial pressure of vapor is  =  %f  kN/m^2\n',ps);\n",
+"mprintf('\n     The partial pressure of dry air is  =  %f kN/m^2\n',pa);\n",
+"\n",
+"// (b)\n",
+"// from equation [15]\n",
+"omega = .622*ps/(p-ps);// specific humidity of the mixture\n",
+"mprintf('\n (b) The specific humidity of the mixture is  =  %f kg/kg dry air\n',omega);\n",
+"\n",
+"// (c)\n",
+"// using eqn [1] from section 19.2\n",
+"y = 1/(1+omega);// composition of the mixture\n",
+"mprintf('\n (c) The composition of the mixture is  =  %f\n',y);\n",
+"\n",
+"// End"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 19.3: specific_humidity_dew_point_degree_of_superheat_mass_of_condensate.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"clear;\n",
+"clc;\n",
+"disp('Example 19.3');\n",
+"\n",
+"// aim : To determine\n",
+"// (a) the specific humidity\n",
+"// (b) the dew point\n",
+"// (c) the degree of superheat of the superheated vapor\n",
+"// (d) the mass of condensate formed per kg of dry air if the moist air is cooled to 12 C\n",
+"\n",
+"// Given values\n",
+"t = 25;// C\n",
+"T = 273+25;// moist air temperature, [K]\n",
+"phi = .6;// relative humidity\n",
+"p = 101.3;// barometric pressure, [kN/m^2]\n",
+"R = .287;// [kJ/kg K]\n",
+"\n",
+"// solution\n",
+"// (a)\n",
+"// From steam table at 25 C\n",
+"pg = 3.17;// [kN/m^2]\n",
+"ps = phi*pg;// partial pressure of the vapor, [kN/m^2]\n",
+"omega = .622*ps/(p-ps);// the specific humidity of air\n",
+"\n",
+"mprintf('\n (a) The specific humidity is  =  %f  kg/kg air\n',omega);\n",
+"\n",
+"// (b)\n",
+"// Dew point is saturated temperature at ps is,\n",
+"t_dew = 16+2*(1.092-1.817)/(2.062-1.817);// [C]\n",
+"mprintf('\n (b) The dew point is  =  %f C\n',t_dew);\n",
+"\n",
+"// (c)\n",
+"Dos = t-t_dew;// degree of superheat, [C]\n",
+"mprintf('\n (c) The degree of superheat is  =  %f C\n',Dos);\n",
+"\n",
+"// (d)\n",
+"// at 25 C\n",
+"pa = p-ps;// [kN/m^2]\n",
+"va = R*T/pa;// [m^3/kg]\n",
+"// at 16.69 C\n",
+"vg = 73.4-(73.4-65.1)*.69/2;// [m^3/kg]\n",
+"ms1= va/vg; \n",
+"// at 12 C\n",
+"vg = 93.8;// [m^3/kg]\n",
+"ms2 = va/vg;\n",
+"\n",
+"m = ms1-ms2;// mas of condensate\n",
+"mprintf('\n (d) The mass of condensate is  =  %f  kg/kg dry air\n',m);\n",
+"\n",
+"//  there is calculation mistake in the book so answer is no matching\n",
+"\n",
+"//  End"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 19.4: volume_mass_and_heat_transfer.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"clear;\n",
+"clc;\n",
+"disp(' Example 19.4');\n",
+"\n",
+"// aim : To determine\n",
+"// (a) the volume of external saturated air\n",
+"// (b) the mass of air\n",
+"// (c) the heat transfer\n",
+"// (d) the heat transfer required by the combind water vapour\n",
+"\n",
+"// given values\n",
+"Vb = 56000;// volume of building, [m^3]\n",
+"T2 = 273+20;// temperature of air in thebuilding, [K]\n",
+"phi = .6;// relative humidity\n",
+"T1 = 8+273;// external air saturated temperature, [K]\n",
+"p0 = 101.3;// atmospheric pressure, [kN/m^2]\n",
+"cp = 2.093;// heat capacity of saturated steam, [kJ/kg K]\n",
+"R = .287;// gas constant, [kJ/kg K]\n",
+"\n",
+"// solution\n",
+"// from steam table at 20 C saturation pressure of steam is,\n",
+"pg = 2.34;// [kN/m^2]\n",
+"\n",
+"// (a)\n",
+"pvap = phi*pg;// partial pressure of vapor, [kN/m^2] \n",
+"P = p0-pvap;// partial pressure of air, [kN/m^2]\n",
+"V = 2*Vb;// air required, [m^3]\n",
+"// at 8 C saturation pressure ia\n",
+"pvap = 1.072;// [kN/m^2]\n",
+"P2 = p0-pvap;// partial pressure of entry at 8 C, [kN/m^2]\n",
+"\n",
+"// using P1*V1/T1=P2*V2/T2;\n",
+"V2 = P*V*T1/(T2*P2);// air required at 8 C, [m^3/h]\n",
+"mprintf('\n (a) The volume of air required is  =  %f  m^3/h\n',V2);\n",
+"\n",
+"// (b)\n",
+"//  assuming\n",
+"pg = 1.401;// pressure, [kN/m^2]\n",
+"Tg = 273+12;// [K]\n",
+"vg = 93.8;// [m^3/kg]\n",
+"// at constant pressure\n",
+"v = vg*T2/Tg;// volume[m^3/kg]\n",
+"mv = V/v;// mass of vapor in building at 20 C, [kg/h]\n",
+"// from steam table at 8 C\n",
+"vg2 = 121;// [m^3/kg]\n",
+"mve = V2/vg2;// mass of vapor supplied with saturated entry air, [kg/h]\n",
+"mw = mv-mve;// mass of water added, [kg/h]\n",
+"mprintf('\n (b) The mass of water added is  =  %f  kg/h\n ',mw);\n",
+"\n",
+"// (c)\n",
+"// for perfect gas\n",
+"m = P2*V2/(R*T1);// [kg/h]\n",
+"Cp = .287;// heat capacity, [kJ/kg K]\n",
+"Q = m*Cp*(T2-T1);// heat transfer by dry air,[kJ/h]\n",
+"mprintf('\n (c) The heat transfer required by dry air is  =  %f MJ/h\n',Q*10^-3);\n",
+"\n",
+"// (d)\n",
+"// from steam table\n",
+"h1 = 2516.2;// specific enthalpy of saturated vapor at 8 C,[kJ/kg]\n",
+"hs = 2523.6;// specific enthalpy of saturated vapor at 20 C, [kJ/kg]\n",
+"h2 = hs+cp*(T2-T1);// specific enthalpy of vapor at 20 c, [kJ/kg]\n",
+"Q1 = mve*(h2-h1);// heat transfer required for vapor, [kJ]\n",
+"\n",
+"// again from steam table\n",
+"hf1 = 33.6;// [kJ/kg]\n",
+"hg3 = 2538.2;// [kJ/kg]\n",
+"Q2 = mw*(hg3-hf1);// heat transfer required for water, [kJ/h]\n",
+"Qt = Q1+Q2;// total heat transfer, [kJ/h]\n",
+"mprintf('\n (d) The heat transferred required for vapor+supply water is  =  %f  MJ/h\n',Qt*10^-3);\n",
+"\n",
+"//  there is minor variation in the answer reported in the book\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
+}
diff --git a/Basic_Engineering_Thermodynamics_by_R_Joel/2-Systems.ipynb b/Basic_Engineering_Thermodynamics_by_R_Joel/2-Systems.ipynb
new file mode 100644
index 0000000..13741d6
--- /dev/null
+++ b/Basic_Engineering_Thermodynamics_by_R_Joel/2-Systems.ipynb
@@ -0,0 +1,266 @@
+{
+"cells": [
+ {
+		   "cell_type": "markdown",
+	   "metadata": {},
+	   "source": [
+       "# Chapter 2: Systems"
+	   ]
+	},
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 2.1: Change_in_total_energy.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"clear;\n",
+"clc;\n",
+"disp('Example 2.1');\n",
+"//  Given values\n",
+"Q = 2500; // Heat transferred into the system, [kJ]\n",
+"W = 1400; //  Work transferred from the system,  [kJ]\n",
+" \n",
+"//   solution\n",
+"//  since process carried out on a closed system, so using equation [4]\n",
+"del_E = Q-W; //  Change in total energy, [kJ]\n",
+"mprintf('\n The Change in total energy is, del_E =  %f  kJ\n',del_E);\n",
+"if(del_E&gt;0)\n",
+"    disp('Since del_E is positive, so there is an increase in total enery')\n",
+"else\n",
+"    disp('Since del_E is negative, so there is an decrease in total enery')\n",
+"end\n",
+"// There is mistake in the book's results unit\n",
+"//  End"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 2.2: Heat_transferred.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"clear;\n",
+"clc;\n",
+"disp('Example 2.2');\n",
+"//  Given values\n",
+"del_E = 3500; // Increase in total energy of the system, [kJ]\n",
+"W = -4200; // Work transfer into the system, [kJ]\n",
+"//  solution\n",
+"//  since process carried out on a closed system, so using equation [3]\n",
+"Q = del_E+W;// [kJ]\n",
+"mprintf('\n The Heat transfer is, Q =  %f kJ \n',Q);\n",
+"if(Q&gt;0)\n",
+"    disp('Since Q&gt;0, so heat is transferred into the system')\n",
+"else\n",
+"    disp('Since Q&lt;0, so heat is transferred from the system')\n",
+"end\n",
+"//  End"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 2.3: Work_done.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"clear;\n",
+"clc;\n",
+"disp('Example 2.3');\n",
+"//  Given values\n",
+"Q = -150; // Heat transferred out of the system, [kJ/kg]\n",
+"del_u = -400; // Internal energy decreased ,[kJ/kg]\n",
+"//  solution\n",
+"//  using equation [3],the non flow energy equation\n",
+"//  Q=del_u+W\n",
+"W = Q-del_u; //  [kJ/kg]\n",
+"mprintf('\n The Work done is, W =  %f kJ/kg \n',W);\n",
+"if(W&gt;0)\n",
+"      disp('Since W&gt;0, so Work done by the engine per kilogram of working substance')\n",
+"else\n",
+"   disp('Since &lt;0, so Work done on the engine per kilogram of working substance')\n",
+"end\n",
+"//  End"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 2.4: Power_of_the_system.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"clear;\n",
+"clc;\n",
+"disp('Example 2.4');\n",
+"//  Given values\n",
+"m_dot = 4; //  fluid flow rate, [kg/s]\n",
+"Q = -40;  //  Heat loss to the surrounding, [kJ/kg]\n",
+"//  At inlet \n",
+"P1 = 600;  // pressure  ,[kn/m^2]\n",
+"C1 = 220;  // velocity  ,[m/s]\n",
+"u1 = 2200;  // internal energy,  [kJ/kg]\n",
+"v1 = .42; // specific volume, [m^3/kg]\n",
+"//  At outlet\n",
+"P2 = 150; // pressure, [kN/m^2]\n",
+"C2 = 145;  // velocity,  [m/s]\n",
+"u2 = 1650;  // internal energy,  [kJ/kg]\n",
+"v2 = 1.5;  // specific volume,  [m^3/kg]\n",
+"//  solution\n",
+"//  for steady flow energy equation for the open system is given by\n",
+"//  u1+P1*v1+C1^2/2+Q=u2+P2*v2+C2^2/2+W\n",
+"//  hence\n",
+"W = (u1-u2)+(P1*v1-P2*v2)+(C1^2/2-C2^2/2)*10^-3+Q; // [kJ/kg]\n",
+"mprintf('\n workdone is, W =  %f kJ/kg ',W);\n",
+"if(W&gt;0)\n",
+"      disp('Since W&gt;0, so Power is output from the system')\n",
+"else\n",
+"   disp('Since &lt;0, so Power is input to the system')\n",
+"end\n",
+" \n",
+"//  Hence\n",
+"P_out = W*m_dot; // power out put from the system, [kW]\n",
+"mprintf('\n The power output from the system is  =  %f  kW \n',P_out);\n",
+"//  End"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 2.5: Temperature_rise.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"clear;\n",
+"clc;\n",
+"disp('Example 2.5');\n",
+"//  Given values\n",
+"del_P = 154.45; // pressure difference across the die, [MN/m^2]\n",
+"rho = 11360; // Density of the lead, [kg/m^3]\n",
+"c = 130; // specific heat capacity of the lead, [J/kg*K]\n",
+"//  solution\n",
+"//  since there is no cooling and no externel work is done, so energy balane becomes\n",
+"//  P1*V1+U1=P2*V2+U2 ,so\n",
+"//  del_U=U2-U1=P1*V1-P2*V2\n",
+"//  also, for temperature rise, del_U=m*c*t, where, m is mass; c is specific heat capacity; and t is temperature rise\n",
+"//  Also given that lead is incompressible, so V1=V2=V and assuming one m^3 of lead\n",
+"//  using above equations\n",
+"t = del_P/(rho*c)*10^6 ;// temperature rise [C]\n",
+"mprintf('\n The temperature rise of the lead is  =  %f  C\n',t);\n",
+"//  End"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 2.6: Area_velocity_and_power.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"clear;\n",
+"clc;\n",
+"disp('Example 2.6');\n",
+"//  Given values\n",
+"m_dot = 4.5; // mass flow rate of air, [kg/s]\n",
+"Q = -40; // Heat transfer loss, [kJ/kg]\n",
+"del_h = -200; // specific enthalpy reduce, [kJ/kg]\n",
+"C1 = 90; // inlet velocity, [m/s]\n",
+"v1 = .85; // inlet specific volume, [m^3/kg]\n",
+"v2 = 1.45; //  exit specific volume, [m^3/kg]\n",
+"A2 = .038;  //  exit area of turbine, [m^2]\n",
+"//  solution\n",
+"//  part (a)\n",
+"//  At inlet, by equation[4], m_dot=A1*C1/v1\n",
+"A1 = m_dot*v1/C1;//inlet area, [m^2]\n",
+"mprintf('\n (a) The inlet area is, A1 =  %f  m^2 \n',A1);\n",
+"//  part (b), \n",
+"//  At outlet, since mass flow rate is same, so m_dot=A2*C2/v2, hence\n",
+"C2 = m_dot*v2/A2; // Exit velocity,[m/s]\n",
+"mprintf('\n (b) The exit velocity is, C2 =  %f  m/s \n',C2);\n",
+"//  part (c)\n",
+"//  using steady flow equation, h1+C1^2/2+Q=h2+C2^2/2+W\n",
+"W = -del_h+(C1^2/2-C2^2/2)*10^-3+Q; //  [kJ/kg]\n",
+"//  Hence power developed is\n",
+"P = W*m_dot;//  [kW]\n",
+"mprintf('\n (c) The power developed by the turbine system is  =  %f kW \n',P);\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
+}
diff --git a/Basic_Engineering_Thermodynamics_by_R_Joel/4-Steam_and_two_phase_systems.ipynb b/Basic_Engineering_Thermodynamics_by_R_Joel/4-Steam_and_two_phase_systems.ipynb
new file mode 100644
index 0000000..aa80cfc
--- /dev/null
+++ b/Basic_Engineering_Thermodynamics_by_R_Joel/4-Steam_and_two_phase_systems.ipynb
@@ -0,0 +1,1096 @@
+{
+"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
+}
diff --git a/Basic_Engineering_Thermodynamics_by_R_Joel/5-Gases_and_single_phase_systems.ipynb b/Basic_Engineering_Thermodynamics_by_R_Joel/5-Gases_and_single_phase_systems.ipynb
new file mode 100644
index 0000000..49e1cef
--- /dev/null
+++ b/Basic_Engineering_Thermodynamics_by_R_Joel/5-Gases_and_single_phase_systems.ipynb
@@ -0,0 +1,1365 @@
+{
+"cells": [
+ {
+		   "cell_type": "markdown",
+	   "metadata": {},
+	   "source": [
+       "# Chapter 5: Gases and single phase systems"
+	   ]
+	},
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 5.10: temperature.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"clear;\n",
+"clc;\n",
+"disp('Example 5.10');\n",
+"\n",
+"//  aim : To determine\n",
+"//  the new temperature of the gas\n",
+"\n",
+"//  Given values\n",
+"V1 = .015;//  original volume,[m^3]\n",
+"T1 = 273+285;//  original temperature,[K]\n",
+"V2 = .09;//  final volume,[m^3]\n",
+"\n",
+"//  solution \n",
+"//  Given gas is following the law,P*V^1.35=constant\n",
+"//  so process is polytropic with\n",
+"n = 1.35; // polytropic index\n",
+"\n",
+"// hence\n",
+"T2 = T1*(V1/V2)^(n-1);//  final temperature, [K]\n",
+"\n",
+"t2 = T2-273;//  [C]\n",
+"\n",
+"mprintf('\n The new temperature of the gas is  =  %f C \n',t2);\n",
+"\n",
+"// there is minor error in book's answer\n",
+"\n",
+"//  End"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 5.11: volume_pressure_and_temperature.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"clear;\n",
+"clc;\n",
+"disp('Example 5.11');\n",
+"\n",
+"//  aim : To determine the\n",
+"//  (a) original and final volume of the gas\n",
+"//  (b) final pressure of the gas\n",
+"//  (c) final temperature of the gas\n",
+"\n",
+"//  Given values\n",
+"m = .675;//  mass of the gas,[kg]\n",
+"P1 = 1.4;//  original pressure,[MN/m^2]\n",
+"T1 = 273+280;//  original temperature,[K]\n",
+"R = .287;//gas constant,[kJ/kg K]\n",
+"\n",
+"//  solution\n",
+"\n",
+"//  (a)\n",
+"//  using characteristic equation, P1*V1=m*R*T1\n",
+"V1 = m*R*T1*10^-3/P1;//  [m^3]\n",
+"//  also Given \n",
+"V2 =  4*V1;// [m^3]\n",
+"mprintf('\n (a) The original volume of the gas is  =  %f  m^3\n',V1);\n",
+"mprintf('\n      and The final volume of the gas is  =  %f  m^3\n',V2);\n",
+"\n",
+"//  (b)\n",
+"//  Given that gas is following the law P*V^1.3=constant\n",
+"//  hence process is polytropic with \n",
+"n = 1.3; //  polytropic index\n",
+"P2 = P1*(V1/V2)^n;//  formula for polytropic process,[MN/m^2]\n",
+"mprintf('\n (b) The final pressure of the gas is  =  %f  kN/m^2\n',P2*10^3);\n",
+"\n",
+"//  (c)\n",
+"//  since mass is constant so,using P*V/T=constant\n",
+"//  hence\n",
+"T2 = P2*V2*T1/(P1*V1);//  [K]\n",
+"t2 = T2-273;//  [C]\n",
+"mprintf('\n (c) The final temperature of the gas is  =  %f C\n',t2);\n",
+"\n",
+"//  End"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 5.12: change_of_internal_energy_work_done_and_heat_transfer.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"clear;\n",
+"clc;\n",
+"disp('Example 5.12');\n",
+"\n",
+"//  aim : T0 determine \n",
+"//  (a) change in internal nergy of the air\n",
+"//  (b) work done\n",
+"//  (c) heat transfer\n",
+"\n",
+"//  Given values\n",
+"m = .25;//  mass, [kg]\n",
+"P1 = 140;//  initial pressure, [kN/m^2]\n",
+"V1 = .15;//  initial volume, [m^3]\n",
+"P2 = 1400;//  final volume, [m^3]\n",
+"cp = 1.005;//  [kJ/kg K]\n",
+"cv = .718;//  [kJ/kg K]\n",
+"\n",
+"//  solution\n",
+"\n",
+"//  (a)\n",
+"//  assuming ideal gas\n",
+"R = cp-cv;//  [kJ/kg K]\n",
+"//  also, P1*V1=m*R*T1,hence\n",
+"T1 = P1*V1/(m*R);//  [K]\n",
+"\n",
+"//  given that process is polytropic with \n",
+"n = 1.25; //  polytropic index\n",
+"T2 = T1*(P2/P1)^((n-1)/n);//  [K]\n",
+"\n",
+"//  Hence, change in internal energy is,\n",
+"del_U = m*cv*(T2-T1);//  [kJ]\n",
+"mprintf('\n (a) The change in internal energy of the air is del_U  =  %f  kJ',del_U);\n",
+"if(del_U>0)\n",
+"    disp('since del_U>0, so it is gain of internal energy to the air')\n",
+"else\n",
+"    disp('since del_U<0, so it is gain of internal energy to the surrounding')\n",
+"end\n",
+"\n",
+"//  (b)\n",
+"W = m*R*(T1-T2)/(n-1);//  formula of work done for polytropic process,[kJ]\n",
+"mprintf('\n (b) The work done is W =  %f  kJ',W);\n",
+"if(W>0)\n",
+"    disp('since W>0, so the work is done by  the air')\n",
+"else\n",
+"    disp('since W<0, so the work is done on the air')\n",
+"end\n",
+"\n",
+"//  (c)\n",
+"Q = del_U+W;//  using 1st law of thermodynamics,[kJ]\n",
+"mprintf('\n (c) The heat transfer is Q  =  %f  kJ',Q);\n",
+"if(Q>0)\n",
+"    disp('since Q>0, so the heat is received by  the air')\n",
+"else\n",
+"    disp('since Q<0, so the heat is rejected by the air')\n",
+"end\n",
+"\n",
+"//  End"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 5.13: volume_work_done_and_change_of_internal_energy.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"clear;\n",
+"clc;\n",
+"disp('Example 5.13');\n",
+"\n",
+"//  aim : To determine the\n",
+"//  final volume, work done and the change in internal energy\n",
+"\n",
+"//  Given values\n",
+"P1 = 700;//  initial pressure,[kN/m^2]\n",
+"V1 = .015;// initial volume, [m^3]\n",
+"P2 = 140;//  final pressure, [kN/m^2]\n",
+"cp = 1.046;//  [kJ/kg K]\n",
+"cv = .752; //  [kJ/kg K]\n",
+"\n",
+"//  solution\n",
+"\n",
+"Gamma = cp/cv;\n",
+"//  for adiabatic expansion, P*V^gamma=constant, so\n",
+"V2 = V1*(P1/P2)^(1/Gamma);//  final volume, [m^3]\n",
+"mprintf('\n The final volume of the gas is V2  =  %f  m^3\n',V2);\n",
+"\n",
+"//  work done\n",
+"W = (P1*V1-P2*V2)/(Gamma-1);//  [kJ]\n",
+"mprintf('\n The work done by the gas is  =  %f  kJ\n',W);\n",
+"\n",
+"//  for adiabatic process\n",
+"del_U = -W;//  [kJ]\n",
+"mprintf('\n The change of internal energy is  =  %f kJ',del_U);\n",
+"if(del_U>0)\n",
+"    disp('since del_U>0, so the the gain in internal energy of the gas ')\n",
+"else\n",
+"    disp('since del_U<0, so this is a loss of internal energy from the gas')\n",
+"end\n",
+"\n",
+"//  End"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 5.14: heat_transfer_change_of_internal_energy_and_mass.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"clear;\n",
+"clc;\n",
+"disp('Example 5.14');\n",
+"\n",
+"//  aim : To determine the\n",
+"//  (a)heat transfer\n",
+"//  (b)change of internal energy\n",
+"//  (c)mass of gas\n",
+"\n",
+"//   Given values\n",
+"V1 = .4;//  initial volume, [m^3]\n",
+"P1 = 100;//  initial pressure, [kN/m^2]\n",
+"T1 = 273+20;//  temperature, [K]\n",
+"P2 = 450;//  final pressure,[kN/m^2]\n",
+"cp = 1.0;//  [kJ/kg K]\n",
+"Gamma = 1.4; //  heat capacity ratio\n",
+"\n",
+"//  solution\n",
+"\n",
+"//  (a)\n",
+"//  for the isothermal compression,P*V=constant,so\n",
+"V2 = V1*P1/P2;//  [m^3]\n",
+"W = P1*V1*log(P1/P2);//  formula of workdone for isothermal process,[kJ]\n",
+"\n",
+"//  for isothermal process, del_U=0;so\n",
+"Q = W;\n",
+"mprintf('\n (a) The heat transferred during compression is Q  =  %f kJ\n',Q);\n",
+"\n",
+"\n",
+"//  (b)\n",
+"V3 = V1;\n",
+"//  for adiabatic expansion\n",
+"//  also\n",
+"\n",
+"P3 = P2*(V2/V3)^Gamma;//  [kN/m^2]\n",
+"W = -(P3*V3-P2*V2)/(Gamma-1);//  work done formula for adiabatic process,[kJ]\n",
+"//  also, Q=0,so using Q=del_U+W\n",
+"del_U = -W;//  [kJ]\n",
+"mprintf('\n (b) The change of the internal energy during the expansion is,del_U  =  %f kJ\n',del_U);\n",
+"\n",
+"//  (c)\n",
+"//  for ideal gas\n",
+"//  cp-cv=R, and cp/cv=gamma, hence\n",
+"R = cp*(1-1/Gamma);//  [kj/kg K]\n",
+"\n",
+"//  now using ideal gas equation\n",
+"m = P1*V1/(R*T1);//  mass of the gas,[kg]\n",
+"mprintf('\n (c) The mass of the gas is,m  =  %f  kg\n',m);\n",
+"\n",
+"// There is calculation mistake in the book\n",
+"\n",
+"\n",
+"//  End\n",
+""
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 5.15: heat_transfer_and_polytropic_heat_capacity.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"clear;\n",
+"clc;\n",
+"disp('Example 5.15');\n",
+"\n",
+"//  aim : To determine \n",
+"//  the heat transferred and polytropic specific heat capacity\n",
+"\n",
+"//  Given values\n",
+"P1 = 1;//  initial pressure, [MN/m^2]\n",
+"V1 = .003;//  initial volume, [m^3]\n",
+"P2 = .1;//  final pressure,[MN/m^2]\n",
+"cv = .718;//  [kJ/kg*K]\n",
+"Gamma=1.4;//  heat capacity ratio\n",
+"\n",
+"//  solution\n",
+"//  Given process is polytropic with\n",
+"n = 1.3;//  polytropic index\n",
+"//  hence\n",
+"V2 = V1*(P1/P2)^(1/n);//  final volume,[m^3]\n",
+"W = (P1*V1-P2*V2)*10^3/(n-1);//  work done,[kJ]\n",
+"//  so\n",
+"Q = (Gamma-n)*W/(Gamma-1);//  heat transferred,[kJ]\n",
+"\n",
+"mprintf('\n The heat received or rejected by the gas during this process is Q  =  %f kJ',Q);\n",
+"if(Q>0)\n",
+"    disp('since Q>0, so heat is received by the gas')\n",
+"else\n",
+"    disp('since Q<0, so heat is rejected by the gas')\n",
+"end\n",
+"\n",
+"//  now\n",
+"cn = cv*(Gamma-n)/(n-1);//  polytropic specific heat capacity,[kJ/kg K]\n",
+"mprintf('\n The polytropic specific heat capacity is cn  =  %f kJ/kg K\n',cn);\n",
+"\n",
+"//  End"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 5.16: pressures.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"clear;\n",
+"clc;\n",
+"disp('Example 5.16');\n",
+"\n",
+"//  aim : To determine the \n",
+"//  (a) initial partial pressure of the steam and air\n",
+"//  (b) final partial pressure of the steam and air\n",
+"//  (c) total pressure in the container after heating\n",
+"\n",
+"//  Given values\n",
+"T1 = 273+39;//  initial temperature,[K]\n",
+"P1 = 100;//  pressure, [MN/m^2]\n",
+"T2 = 273+120.2;//  final temperature,[K]\n",
+"\n",
+"//  solution\n",
+"\n",
+"//  (a)\n",
+"//  from the steam tables, the pressure of wet steam at 39 C is\n",
+"Pw1 = 7;//  partial pressure of wet steam,[kN/m^2]\n",
+"//  and by Dalton's law\n",
+"Pa1 = P1-Pw1;//  initial pressure of air, [kN/m^2]\n",
+"\n",
+"mprintf('\n (a) The initial partial pressure of the steam is  =  %f  kN/m^2',Pw1);\n",
+"mprintf('\n      The initial partial pressure of the air is  =  %f kN/m^2\n',Pa1);\n",
+"\n",
+"//  (b)\n",
+"//  again from steam table, at 120.2 C the pressure of wet steam is\n",
+"Pw2 = 200;//  [kN/m^2]\n",
+"\n",
+"//  now since volume is constant so assuming air to be ideal gas so for air  P/T=contant, hence\n",
+"Pa2 = Pa1*T2/T1 ;//  [kN/m^2]\n",
+"\n",
+"mprintf(' \n(b) The final partial pressure of the steam is  =  %f kN/m^2',Pw2);\n",
+"mprintf('\n     The final partial pressure of the air is  =  %f  kN/m^2\n',Pa2);\n",
+"\n",
+"//  (c)\n",
+"Pt = Pa2+Pw2;//  using dalton's law, total pressure,[kN/m^2]\n",
+"mprintf('\n (c) The total pressure after heating is  =  %f  kN/m^2\n',Pt);\n",
+"\n",
+"//  End"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 5.17: partial_pressure_and_mass.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"clear;\n",
+"clc;\n",
+"disp('Example 5.17');\n",
+"\n",
+"//  aim : To determine \n",
+"//  the partial pressure of the air and steam, and the mass of the air\n",
+"\n",
+"//  Given values\n",
+"P1 = 660;//  vaccum gauge pressure on condenser [mmHg]\n",
+"P = 765;//  atmospheric pressure, [mmHg]\n",
+"x = .8;//  dryness fraction \n",
+"T = 273+41.5;//  temperature,[K]\n",
+"ms_dot = 1500;//  condense rate of steam,[kg/h]\n",
+"R = .29;//  [kJ/kg]\n",
+"\n",
+"//  solution\n",
+"Pa = (P-P1)*.1334;// absolute pressure,[kN/m^2]\n",
+"// from steam table, at 41.5 C partial pressure of steam is\n",
+"Ps = 8;//  [kN/m^2]\n",
+"//  by dalton's law, partial pressure of air is\n",
+"Pg = Pa-Ps;//  [kN/m^2]\n",
+"\n",
+"mprintf('\n The partial pressure of the air in the condenser is  =  %f  kN/m^2\n',Pg);\n",
+"mprintf('\n The partial pressure of the steam in the condenser is  =  %f kN/m^2\n',Ps);\n",
+"\n",
+"//  also\n",
+"vg = 18.1;//  [m^3/kg]\n",
+"//  so\n",
+"V = x*vg;//  [m^3/kg]\n",
+"//  The air associated with 1 kg of the steam will occupiy this same volume\n",
+"//  for air, Pg*V=m*R*T,so\n",
+"m = Pg*V/(R*T);//  [kg/kg steam]\n",
+"//  hence\n",
+"ma = m*ms_dot;//  [kg/h]\n",
+"\n",
+"mprintf('\n The mass of air which will associated with this steam is  =  %f  kg\n',ma);\n",
+"\n",
+"// There is misprint in book\n",
+"\n",
+"//  End"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 5.18: pressure_and_dryness_fraction.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"clear;\n",
+"clc;\n",
+"disp('Example 5.18');\n",
+"\n",
+"//  aim : To determine the\n",
+"//  (a) final pressure\n",
+"//   (b) final dryness fraction of the steam\n",
+"\n",
+"//  Given values\n",
+"P1 = 130;//  initial pressure, [kN/m^2]\n",
+"T1 = 273+75.9;//  initial temperature, [K]\n",
+"x1 = .92;//  initial dryness fraction\n",
+"T2 = 273+120.2;//  final temperature, [K]\n",
+"\n",
+"//  solution\n",
+"\n",
+"//  (a)\n",
+"//  from steam table, at 75.9 C\n",
+"Pws = 40;//  partial pressure of wet steam[kN/m^2]\n",
+"Pa = P1-Pws;//  partial pressure of air, [kN/m^2]\n",
+"vg = 3.99//  specific volume of the wet steam, [m^3/kg]\n",
+"// hence\n",
+"V1 = x1*vg;//  [m^3/kg]\n",
+"V2 = V1/5;//  [m^3/kg]\n",
+"//  for air, mass is constant so, Pa*V1/T1=P2*V2/T2,also given ,V1/V2=5,so\n",
+"P2 = Pa*V1*T2/(V2*T1);//  final pressure,[kN/m^2]\n",
+"\n",
+"//  now for steam at 120.2 C\n",
+"Ps = 200;//  final partial pressure of steam,[kN/m^2]\n",
+"//  so by dalton's law total pressure in cylindert is\n",
+"Pt = P2+Ps;//  [kN/m^2]\n",
+"mprintf('\n (a) The final pressure in the cylinder is  =  %f  kN/m^2\n',Pt);\n",
+"\n",
+"//  (b)\n",
+"//  from steam table at 200 kN/m^2 \n",
+"vg = .885;//  [m^3/kg]\n",
+"//  hence\n",
+"x2 = V2/vg;//  final dryness fraction of the steam\n",
+"mprintf('\n (b) The final dryness fraction of the steam is  =  %f\n ',x2);\n",
+"\n",
+"//  End"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 5.19: adiabatic_index_and_change_of_internal_energy.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"clear;\n",
+"clc;\n",
+"disp('Example 5.19')\n",
+"\n",
+"//  aim : To determine the \n",
+"//  (a) Gamma,\n",
+"//  (b) del_U\n",
+"\n",
+"//  Given Values\n",
+"P1 = 1400;//  [kN/m^2]\n",
+"P2 = 100;//  [kN/m^2]\n",
+"P3 = 220;//  [kN/m^2]\n",
+"T1 = 273+360;//  [K]\n",
+"m = .23;// [kg]\n",
+"cp = 1.005;//  [kJ/kg*K]\n",
+"\n",
+"//  Solution\n",
+"T3 = T1;//  since process 1-3 is isothermal\n",
+"\n",
+"//  (a)\n",
+"//  for process 1-3, P1*V1=P3*V3,so\n",
+"V3_by_V1 = P1/P3;\n",
+"//  also process 1-2 is adiabatic,so P1*V1^(Gamma)=P2*V2^(Gamma),hence\n",
+"//  and process process 2-3 is iso-choric so,V3=V2 and\n",
+"V2_by_V1 = V3_by_V1;\n",
+"//  hence,\n",
+"Gamma = log(P1/P2)/log(P1/P3); //  heat capacity ratio\n",
+"\n",
+"mprintf('\n (a) The value of adiabatic index Gamma is  =  %f\n',Gamma);\n",
+"\n",
+"//  (b)\n",
+"cv = cp/Gamma;//  [kJ/kg K]\n",
+"//  for process 2-3,P3/T3=P2/T2,so\n",
+"T2 = P2*T3/P3;//  [K]\n",
+"\n",
+"//   now\n",
+"del_U = m*cv*(T2-T1);//  [kJ]\n",
+"mprintf('\n (b) The change in internal energy during the adiabatic expansion is U2-U1 =  %f kJ (This is loss of internal energy)\n',del_U);\n",
+"//   End"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 5.1: pressure_exerted_and_difference_in_two_mercury_column_levels.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"clear;\n",
+"clc;\n",
+"disp('Example 5.1');\n",
+"\n",
+"//  aim : To determine \n",
+"//  new pressure exerted on the air and the difference in two mercury column level\n",
+"\n",
+"//  Given values\n",
+"P1 = 765;//  atmospheric pressure, [mmHg]\n",
+"V1 = 20000;//  [mm^3]\n",
+"V2 = 17000;//  [mm^3]\n",
+"\n",
+"//  solution\n",
+"\n",
+"//  using boyle's law P*V=constant\n",
+"//  hence\n",
+"P2 = P1*V1/V2;//  [mmHg]\n",
+"mprintf('\n The new pressure exerted on the air is  =  %f  mmHg \n',P2);\n",
+"\n",
+"del_h = P2-P1;//  difference in Height of mercury column level\n",
+"mprintf('\n The difference in the two mercury column level is  =  %f  mm\n',del_h);\n",
+"\n",
+"//  End"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 5.20: mass_and_heat_transfer.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"clear;\n",
+"clc;\n",
+"disp('Example 5.20');\n",
+"\n",
+"//  aim : To determine \n",
+"//  the mass of oxygen and heat transferred\n",
+"\n",
+"//  Given values\n",
+"V1 = 300;//  [L]\n",
+"P1 = 3.1;//  [MN/m^2]\n",
+"T1 = 273+18;//  [K]\n",
+"P2 = 1.7;//  [MN/m^2]\n",
+"T2 = 273+15;//  [K]\n",
+"Gamma = 1.4; //  heat capacity ratio\n",
+"//  density condition\n",
+"P = .101325;// [MN/m^2]\n",
+"T = 273;// [K]\n",
+"V = 1;//  [m^3]\n",
+"m = 1.429;// [kg]\n",
+"\n",
+"//  hence\n",
+"R = P*V*10^3/(m*T);//  [kJ/kg*K]\n",
+"//  since volume is constant\n",
+"V2 = V1;// [L]\n",
+"//  for the initial conditions in the cylinder,P1*V1=m1*R*T1\n",
+"m1 = P1*V1/(R*T1);//  [kg]\n",
+"\n",
+"// after some of the gas is used\n",
+"m2 = P2*V2/(R*T2);//  [kg]\n",
+"//  The mass of oxygen remaining in cylinder is m2 kg,so\n",
+"//  Mass of oxygen used is\n",
+"m_used = m1-m2;//  [kg]\n",
+"mprintf('\n The mass of oxygen used  =  %f kg\n',m_used);\n",
+"\n",
+"//  for non-flow process,Q=del_U+W\n",
+"//  volume is constant so no external work is done so,Q=del_U\n",
+"cv = R/(Gamma-1);//  [kJ/kg*K]\n",
+"\n",
+"//  heat transfer is\n",
+"Q = m2*cv*(T1-T2);//  (kJ)\n",
+"mprintf('\n The amount of heat transferred through the cylinder wall is  =  %f kJ\n',Q);\n",
+"\n",
+"//  End"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 5.21: work_done_change_of_internal_energy_and_heat_transfer.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"clear;\n",
+"clc;\n",
+"disp('Example 5.21');\n",
+"\n",
+"//  aim : To determine the\n",
+"//  (a) work transferred during the compression\n",
+"//  (b) change in internal energy\n",
+"//  (c) heat transferred during the compression\n",
+"\n",
+"//  Given values\n",
+"V1 = .1;//  initial volume, [m^3]\n",
+"P1 = 120;//  initial pressure, [kN/m^2]\n",
+"P2 = 1200; // final pressure, [kN/m^2]\n",
+"T1 = 273+25;//  initial temperature, [K]\n",
+"cv = .72;//  [kJ/kg*K]\n",
+"R = .285;//  [kJ/kg*K]\n",
+"\n",
+"//  solution\n",
+"\n",
+"//  (a)\n",
+"//  given process is polytropic with\n",
+"n = 1.2;  // polytropic index\n",
+"//  hence\n",
+"V2 = V1*(P1/P2)^(1/n);//  [m^3]\n",
+"W = (P1*V1-P2*V2)/(n-1);//  workdone formula, [kJ]\n",
+"mprintf('\n (a) The work transferred during the compression is  =  %f  kJ\n',W);\n",
+"\n",
+"//  (b)\n",
+"//  now mass is constant so,\n",
+"T2 = P2*V2*T1/(P1*V1);//  [K]\n",
+"//  using, P*V=m*R*T\n",
+"m = P1*V1/(R*T1);//  [kg]\n",
+"\n",
+"//  change in internal energy is\n",
+"del_U = m*cv*(T2-T1);//  [kJ]\n",
+"mprintf('\n (b) The change in internal energy is  =  %f  kJ\n',del_U);\n",
+"\n",
+"//  (c)\n",
+"Q = del_U+W;//  [kJ]\n",
+"mprintf('\n (c) The heat transferred during the compression is  =  %f  kJ\n',Q);\n",
+"\n",
+"//  End\n",
+""
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 5.22: pressure_and_specific_enthalpy.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"clear;\n",
+"clc;\n",
+"disp('Example 5.22');\n",
+"\n",
+"//  aim : To determine the\n",
+"//  (a) new pressure of the air in the receiver\n",
+"//  (b) specific enthalpy of air at 15 C\n",
+"\n",
+"//  Given values\n",
+"V1 = .85;//  [m^3]\n",
+"T1 = 15+273;//  [K]\n",
+"P1 = 275;// pressure,[kN/m^2]\n",
+"m = 1.7;//  [kg]\n",
+"cp = 1.005;//  [kJ/kg*K]\n",
+"cv = .715;//  [kJ/kg*K]\n",
+"\n",
+"//  solution\n",
+"\n",
+"//  (a)\n",
+"\n",
+"R = cp-cv;//  [kJ/kg*K]\n",
+"//  assuming m1 is original mass of the air, using P*V=m*R*T\n",
+"m1 = P1*V1/(R*T1);//  [kg]\n",
+"m2 = m1+m;//  [kg]\n",
+"//  again using P*V=m*R*T\n",
+"//  P2/P1=(m2*R*T2/V2)/(m1*R*T1/V1); and T1=T2,V1=V2,so\n",
+"P2 = P1*m2/m1;//  [kN/m^2]\n",
+"mprintf('\n (a) The new pressure of the air in the receiver is = %f kN/m^2\n',P2);\n",
+"\n",
+"//  (b)\n",
+"//  for 1 kg of air, h2-h1=cp*(T1-T0)\n",
+"//  and if 0 is chosen as the zero enthalpy, then\n",
+"h = cp*(T1-273);//  [kJ/kg]\n",
+"mprintf('\n (b) The specific  enthalpy of the air at 15 C is =  %f kJ/kg\n',h);\n",
+"\n",
+"//  End"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 5.23: EX5_23.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"clear;\n",
+"clc;\n",
+"disp('Example 5.23');\n",
+"\n",
+"//  aim : T determine the\n",
+"//  (a) characteristic gas constant of the gas\n",
+"//  (b) cp,\n",
+"//  (c) cv,\n",
+"//  (d) del_u \n",
+"//  (e) work transfer\n",
+"\n",
+"//  Given values\n",
+"P = 1;//  [bar] \n",
+"T1 = 273+15;//  [K]\n",
+"m = .9;//  [kg]\n",
+"T2 = 273+250;//  [K]\n",
+"Q = 175;//  heat transfer,[kJ]\n",
+"\n",
+"//  solution\n",
+"\n",
+"//  (a)\n",
+"//  using, P*V=m*R*T, given,\n",
+"m_by_V = 1.875;\n",
+"//  hence\n",
+"R = P*100/(T1*m_by_V);//  [kJ/kg*K]\n",
+"mprintf('\n (a) The characteristic gas constant of the gas is R  =  %f kJ/kg K\n',R);\n",
+"\n",
+"//  (b)\n",
+"//  using, Q=m*cp*(T2-T1)\n",
+"cp = Q/(m*(T2-T1));//  [kJ/kg K]\n",
+"mprintf('\n (b) The specific heat capacity of the gas at constant pressure cp  =  %f kJ/kg K\n',cp);\n",
+"\n",
+"// (c)\n",
+"//  we have, cp-cv=R,so\n",
+"cv = cp-R;// [kJ/kg*K]\n",
+"mprintf('\n (c) The specific heat capacity of the gas at constant volume cv  =  %f  kJ/kg K\n',cv);\n",
+"\n",
+"//  (d)\n",
+"del_U = m*cv*(T2-T1);//  [kJ]\n",
+"mprintf('\n (d) The change in internal energy is  =  %f kJ\n',del_U);\n",
+"\n",
+"//  (e)\n",
+"// using, Q=del_U+W\n",
+"W = Q-del_U;//  [kJ]\n",
+"mprintf('\n (e) The work transfer is W  =  %f kJ\n',W);\n",
+"\n",
+"//  End"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 5.24: work_done_change_of_internal_energy_and_heat_transfer.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"clear;\n",
+"clc;\n",
+"disp('Example 5.24');\n",
+"\n",
+"//  aim : To determine the\n",
+"//  (a) work transfer,\n",
+"//  (b)del_U  and,\n",
+"//  (c)heat transfer\n",
+"\n",
+"//  Given values\n",
+"V1 = .15;//  [m^3]\n",
+"P1 = 1200;//  [kN/m^2]\n",
+"T1 = 273+120;// [K]\n",
+"P2 = 200;//  [kN/m^2]\n",
+"cp = 1.006;//[kJ/kg K]\n",
+"cv = .717;//  [kJ/kg K]\n",
+"\n",
+"//  solution\n",
+"\n",
+"//  (a)\n",
+"//  Given, PV^1.32=constant, so it is polytropic process with\n",
+"n = 1.32;//  polytropic index\n",
+"// hence\n",
+"V2 = V1*(P1/P2)^(1/n);//  [m^3]\n",
+"//  now, W\n",
+"W = (P1*V1-P2*V2)/(n-1);// [kJ]\n",
+"mprintf('\n (a) The work transfer is W  =  %f kJ\n',W);\n",
+"\n",
+"//  (b)\n",
+"R = cp-cv;//  [kJ/kg K]\n",
+"m = P1*V1/(R*T1);//  gas law,[kg]\n",
+"//  also for polytropic process\n",
+"T2 = T1*(P2/P1)^((n-1)/n);//  [K]\n",
+"//  now for gas,\n",
+"del_U = m*cv*(T2-T1);//  [kJ]\n",
+"mprintf('\n (b) The change of internal energy is del_U =  %f kJ\n',del_U);\n",
+"\n",
+"//  (c)\n",
+"Q = del_U+W;//  first law of thermodynamics,[kJ]\n",
+"mprintf('\n (c) The heat transfer Q  =  %f kJ\n',Q);\n",
+"\n",
+"//  End"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 5.26: volume.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"clear;\n",
+"clc;\n",
+"disp('Example 5.26');\n",
+"\n",
+"//  aim : To determine\n",
+"//  the volume of the pressure vessel and the volume of the gas before transfer\n",
+"\n",
+"//  Given values\n",
+"\n",
+"P1 = 1400;//  initial pressure,[kN/m^2]\n",
+"T1 = 273+85;//  initial temperature,[K]\n",
+"\n",
+"P2 = 700;//  final pressure,[kN/m^2]\n",
+"T2 = 273+60;//  final temperature,[K]\n",
+"\n",
+"m = 2.7;// mass of the gas passes,[kg]\n",
+"cp = .88;//  [kJ/kg]\n",
+"cv = .67;//  [kJ/kg]\n",
+"\n",
+"//  solution\n",
+"\n",
+"//  steady flow equation is, u1+P1*V1+C1^2/2+Q=u2+P2*V2+C2^2/2+W  [1], \n",
+"//  given, there is no kinetic energy change and neglecting potential energy term\n",
+"W = 0;// no external work done\n",
+"//  so final equation is,u1+P1*v1+Q=u2   [2]\n",
+"// also u2-u1=cv*(T2-T1)\n",
+"// hence Q=cv*(T2-T1)-P1*v1    [3]\n",
+"//  and for unit mass P1*v1=R*T1=(cp-cv)*T1  [4]\n",
+"//  so finally\n",
+"Q = cv*(T2-T1)-(cp-cv)*T1;//  [kJ/kg]\n",
+"//  so total heat transferred is\n",
+"Q = m*Q;//  [kJ] \n",
+"\n",
+"//  using eqn [4]\n",
+"v1 = (cp-cv)*T1/P1;//  [m^3/kg]\n",
+"//  Total volume is\n",
+"V1 = m*v1;//  [m^3]\n",
+"\n",
+"//  using ideal gas equation P1*V1/T1=P2*V2/T2\n",
+"V2 = P1*T2*V1/(P2*T1);//  final volume,[m^3]\n",
+"\n",
+"mprintf('\n The volume of gas before transfer is  =  %f  m^3\n',V1);\n",
+"mprintf('\n The volume of pressure vessel is  =  %f  m^3\n',V2);\n",
+" \n",
+"//  End"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 5.2: volume.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"clear;\n",
+"clc;\n",
+"disp('Example 5.2');\n",
+"\n",
+"//  aim : To determine \n",
+"//  the new volume\n",
+"\n",
+"//  Given values\n",
+"P1 = 300;//  original pressure,[kN/m^2]\n",
+"V1 = .14;//  original volume,[m^3]\n",
+"\n",
+"P2 = 60;//  new pressure after expansion,[kn/m^2]\n",
+"\n",
+"//  solution\n",
+"//  since temperature is constant so using boyle's law P*V=constant\n",
+"V2 = V1*P1/P2;//  [m^3]\n",
+"\n",
+"mprintf('\n The new volume after expansion is  =  %f  m^3\n',V2);\n",
+"\n",
+"//  End"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 5.3: volume.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"clear;\n",
+"clc;\n",
+"disp('Example 5.3');\n",
+"\n",
+"//  aim : To determine \n",
+"//  the new volume of the gas\n",
+"\n",
+"//  Given values\n",
+"V1 = 10000;//  [mm^3]\n",
+"T1 = 273+18;//  [K]\n",
+"T2 = 273+85;//  [K]\n",
+"\n",
+"//  solution\n",
+"//  since pressure exerted on the apparatus is constant so using charle's law V/T=constant\n",
+"//  hence\n",
+"V2 = V1*T2/T1;//  [mm^3]\n",
+"\n",
+"mprintf('\n The new volume of the gas trapped in the apparatus is  =  %f  mm^3\n',V2);\n",
+"\n",
+"//  End"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 5.4: temperature.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"clear;\n",
+"clc;\n",
+"disp('Example 5.4');\n",
+"\n",
+"//  aim : To determine \n",
+"//  the final temperature\n",
+"\n",
+"//  Given values\n",
+"V1 = .2;//  original volume,[m^3]\n",
+"T1 = 273+303;// original temperature, [K]\n",
+"V2 = .1;// final volume, [m^3]\n",
+"\n",
+"//  solution\n",
+"//  since pressure is constant, so using charle's law V/T=constant\n",
+"//  hence\n",
+"T2 = T1*V2/V1;//  [K]\n",
+"t2 = T2-273;//  [C]\n",
+"mprintf('\n The final temperature of the gas is  =  %f  C\n',t2);\n",
+"\n",
+"//  End"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 5.5: volume.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"clear;\n",
+"clc;\n",
+"disp('Example 5.5');\n",
+"\n",
+"//  aim : To determine \n",
+"//  the new volume of the gas\n",
+"\n",
+"//  Given values\n",
+"\n",
+"//  initial codition\n",
+"P1 = 140;//  [kN/m^2]\n",
+"V1 = .1;//   [m^3]\n",
+"T1 = 273+25;// [K]\n",
+"\n",
+"//  final condition\n",
+"P2 = 700;//  [kN/m^2]\n",
+"T2 = 273+60;//  [K]\n",
+"\n",
+"//  by charasteristic equation, P1*V1/T1=P2*V2/T2\n",
+"\n",
+"V2=P1*V1*T2/(T1*P2);//  final volume, [m^3]\n",
+"\n",
+"mprintf('\nThe new volume of the gas is  =  %f  m^3\n',V2);\n",
+"\n",
+"//  End"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 5.6: mass_and_temperature.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"clear;\n",
+"clc;\n",
+"disp('Example 5.6');\n",
+"\n",
+"//  aim : To determine\n",
+"//  the mas of the gas and new temperature\n",
+"\n",
+"//  Given values\n",
+"P1 = 350;//  [kN/m^2]\n",
+"V1 = .03;//  [m^3]\n",
+"T1 = 273+35;//  [K]\n",
+"R = .29;//  Gas constant,[kJ/kg K]\n",
+"\n",
+"//  solution\n",
+"//  using charasteristic equation, P*V=m*R*T\n",
+"m = P1*V1/(R*T1);//  [Kg]\n",
+"mprintf('\n The mass of the gas present is  =  %f kg\n',m);\n",
+"\n",
+"//  Now the gas is compressed\n",
+"P2 = 1050;//  [kN/m^2]\n",
+"V2 = V1;\n",
+"// since mass of the gas is constant so using, P*V/T=constant\n",
+"//  hence\n",
+"T2 = T1*P2/P1//  [K]\n",
+"t2 = T2-273;//  [C]\n",
+"\n",
+"mprintf('\n The new temperature of the gas is  =  %f  C\n',t2);\n",
+"\n",
+"//  End"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 5.7: heat_transfer_and_pressure.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"clear;\n",
+"clc;\n",
+"disp('Example 5.7');\n",
+"\n",
+"//  aim : To determine \n",
+"//  the heat transferred to the gas and its final pressure\n",
+"\n",
+"//  Given values\n",
+"m = 2;//  masss of the gas, [kg]\n",
+"V1 = .7;//  volume,[m^3]\n",
+"T1 = 273+15;//  original temperature,[K]\n",
+"T2 = 273+135;//  final temperature,[K]\n",
+"cv = .72;//  specific heat capacity at constant volume,[kJ/kg K]\n",
+"R = .29;//  gas law constant,[kJ/kg K]\n",
+"\n",
+"//  solution\n",
+"Q = m*cv*(T2-T1);//  Heat transferred at constant volume,[kJ]\n",
+"mprintf('\n The heat transferred to the gas is  =  %f  kJ\n',Q);\n",
+"\n",
+"//  Now,using P1*V1=m*R*T1\n",
+"P1 = m*R*T1/V1;//  [kN/m^2]\n",
+"\n",
+"//  since volume of the system is constant, so P1/T1=P2/T2\n",
+"//  hence\n",
+"P2 = P1*T2/T1;//  final pressure,[kN/m^2]\n",
+"mprintf('\n The final pressure of the gas is  =  %f  kN/m^2 \n',P2);\n",
+"\n",
+"//  End"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 5.8: heat_transfer_and_work_done.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"clear;\n",
+"clc;\n",
+"disp('Example 5.8');\n",
+"\n",
+"//  aim : To determine \n",
+"//  the heat transferred from the gas and the work done on the gas\n",
+"\n",
+"//  Given values\n",
+"P1 = 275;//  pressure, [kN/m^2]\n",
+"V1 = .09;//  volume,[m^3]\n",
+"T1 = 273+185;//  initial temperature,[K]\n",
+"T2 = 273+15;//  final temperature,[K]\n",
+"cp = 1.005;//  specific heat capacity at constant pressure,[kJ/kg K]\n",
+"R = .29;//  gas law constant,[kJ/kg K]\n",
+"\n",
+"//  solution\n",
+"//  using P1*V1=m*R*T1\n",
+"m = P1*V1/(R*T1);//  mass of the gas\n",
+"\n",
+"//  calculation of heat transfer\n",
+"Q = m*cp*(T2-T1);//  Heat transferred at constant pressure,[kJ]\n",
+"mprintf('\n The heat transferred to the gas is  =  %f  kJ\n',Q);\n",
+"\n",
+"//  calculation of work done\n",
+"//  Now,since pressure is constant so, V/T=constant\n",
+"// hence\n",
+"V2 = V1*T2/T1;//  [m^3]\n",
+"\n",
+"W = P1*(V2-V1);//  formula for work done at constant pressure,[kJ]\n",
+"mprintf('\n Work done on the gas during the process is  =  %f  kJ\n',W);\n",
+"\n",
+"//  End\n",
+""
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 5.9: pressure.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"clear;\n",
+"clc;\n",
+"disp('Example 5.9');\n",
+"\n",
+"//  aim : To determine\n",
+"//  the new pressure of the gas\n",
+"\n",
+"//  Given values\n",
+"P1 = 300;//  original pressure,[kN/m^2]\n",
+"T1 = 273+25;//  original temperature,[K]\n",
+"T2 = 273+180;//  final temperature,[K]\n",
+"\n",
+"//  solution\n",
+"//  since gas compressing according to the law,P*V^1.4=constant\n",
+"//  so,for polytropic process,T1/T2=(P1/P2)^((n-1)/n),here n=1.4\n",
+"\n",
+"//  hence\n",
+"P2 = P1*(T2/T1)^((1.4)/(1.4-1));//  [kN/m^2]\n",
+"\n",
+"mprintf('\n The new pressure of the gas is  =  %f  kN/m^2\n',P2);\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
+}
diff --git a/Basic_Engineering_Thermodynamics_by_R_Joel/7-Entropy.ipynb b/Basic_Engineering_Thermodynamics_by_R_Joel/7-Entropy.ipynb
new file mode 100644
index 0000000..77c5dfb
--- /dev/null
+++ b/Basic_Engineering_Thermodynamics_by_R_Joel/7-Entropy.ipynb
@@ -0,0 +1,536 @@
+{
+"cells": [
+ {
+		   "cell_type": "markdown",
+	   "metadata": {},
+	   "source": [
+       "# Chapter 7: Entropy"
+	   ]
+	},
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 7.1: specific_entropy.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"clear;\n",
+"clc;\n",
+"disp('Example 7.1');\n",
+"\n",
+"//  aim : To determine\n",
+"//  the specific enthalpy of water\n",
+"\n",
+"//  Given values\n",
+"Tf = 273+100;//  Temperature,[K]\n",
+"\n",
+"//  solution\n",
+"//  from steam table\n",
+"cpl = 4.187;// [kJ/kg K]\n",
+"//  using equation [8]\n",
+"sf = cpl*log(Tf/273.16);//  [kJ/kg*K]\n",
+"mprintf('\n The specific entropy of water is  =  %f kJ/kg K\n',sf);\n",
+"\n",
+"//  using steam table\n",
+"sf = 1.307;//  [kJ/kg K]\n",
+"mprintf('\n From table The accurate value of sf in this case is  =  %f  kJ/kg K\n',sf);\n",
+"\n",
+"// There is small error in book's final value of sf\n",
+"\n",
+"\n",
+"//  End"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 7.2: specific_entropy.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"\n",
+"clear;\n",
+"clc;\n",
+"disp('Example 7.2');\n",
+"\n",
+"//  aim : To determine\n",
+"//  the specific entropy\n",
+"\n",
+"//  Given values\n",
+"P = 2;//  pressure,[MN/m^2]\n",
+"x = .8;//  dryness fraction\n",
+"\n",
+"//  solution\n",
+"//  from steam table at given pressure\n",
+"Tf = 485.4;//  [K]\n",
+"cpl = 4.187;//  [kJ/kg K]\n",
+"hfg = 1888.6;//  [kJ/kg]\n",
+"\n",
+"//  (a) finding entropy by calculation\n",
+"s = cpl*log(Tf/273.16)+x*hfg/Tf;//  formula for entropy calculation\n",
+"\n",
+"mprintf('\n (a) The specific entropy of wet steam is  =  %f kJ/kg K\n',s);\n",
+"\n",
+"//  (b) calculation of entropy using steam table\n",
+"//  from steam table at given pressure\n",
+"sf = 2.447;//  [kJ/kg K]\n",
+"sfg = 3.89;//  [kJ/kg K]\n",
+"//  hence\n",
+"s = sf+x*sfg;//  [kJ/kg K]\n",
+"\n",
+"mprintf('\n (b) The specific entropy using steam table is  =  %f kJ/kg K\n',s);\n",
+"\n",
+"//  End"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 7.3: specific_entropy.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"clear;\n",
+"clc;\n",
+"disp('Example 7.3');\n",
+"\n",
+"//  aim : To determine\n",
+"//  the specific entropy of steam\n",
+"\n",
+"//  Given values\n",
+"P = 1.5;//pressure,[MN/m^2]\n",
+"T = 273+300;//temperature,[K]\n",
+"\n",
+"//  solution\n",
+"\n",
+"//  (a)\n",
+"//  from steam table\n",
+"cpl = 4.187;//  [kJ/kg K]\n",
+"Tf = 471.3;//  [K]\n",
+"hfg = 1946;//  [kJ/kg]\n",
+"cpv = 2.093;//  [kJ/kg K]\n",
+"\n",
+"//  usung equation [2]\n",
+"s = cpl*log(Tf/273.15)+hfg/Tf+cpv*log(T/Tf);//  [kJ/kg K]\n",
+"mprintf('\n (a) The specific entropy of steam is  =  %f kJ/kg K\n',s);\n",
+"\n",
+"//  (b)\n",
+"//  from steam tables\n",
+"s = 6.919;//  [kJ/kg K]\n",
+"mprintf('\n (b) The accurate value of specific entropy from steam table is  =  %f kJ/kg K\n',s);\n",
+"\n",
+"//  End"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 7.4: dryness_fraction.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"clear;\n",
+"clc;\n",
+"disp('Example 7.4');\n",
+"\n",
+"//  aim : To determine\n",
+"//  the dryness fraction of steam\n",
+"\n",
+"//  Given values\n",
+"P1 = 2;//  initial pressure, [MN/m^2]\n",
+"t = 350;//  temperature, [C]\n",
+"P2 = .28;//  final pressure, [MN/m^2]\n",
+"\n",
+"//  solution\n",
+"//  at 2 MN/m^2 and 350 C,steam is superheated because the saturation temperature is 212.4 C\n",
+"//  From steam table\n",
+"s1 = 6.957;//  [kJ/kg K]\n",
+"\n",
+"//  for isentropic process\n",
+"s2 = s1;\n",
+"//  also\n",
+"sf2 = 1.647;//  [kJ/kg K]\n",
+"sfg2 = 5.368;//  [kJ/kg K]\n",
+"\n",
+"//  using\n",
+"//  s2 = sf2+x2*sfg2, where x2 is dryness fraction of steam\n",
+"//  hence\n",
+"x2 = (s2-sf2)/sfg2;\n",
+"mprintf('\n The final dryness fraction of steam is x2  =  %f\n',x2);\n",
+"\n",
+"//  End"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 7.5: condition_of_steam_and_change_in_specific_entropy.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"clear;\n",
+"clc;\n",
+"disp('Example 7.5');\n",
+"\n",
+"//  aim : To determine\n",
+"//  the  final condition of steam...\n",
+"//  the change in specific entropy during hyperbolic process\n",
+"\n",
+"//  Given values\n",
+"P1 = 2;//  pressure, [MN/m^2]\n",
+"t = 250;//  temperature, [C]\n",
+"P2 = .36;//  pressure, [MN/m^2]\n",
+"P3 = .06;//  pressure, [MN/m^2]\n",
+"\n",
+"//  solution\n",
+"\n",
+"//  (a)\n",
+"//  from steam table\n",
+"s1 = 6.545;//  [kJ/kg K]\n",
+"//  at .36 MN/m^2\n",
+"sg = 6.930;//  [kJ/kg*K]\n",
+"\n",
+"sf2 = 1.738;//  [kJ/kg K]\n",
+"sfg2 = 5.192;//  [kJ/kg K]\n",
+"vg2 = .510;//  [m^3]\n",
+"\n",
+"//  so after isentropic expansion, steam is wet\n",
+"//  hence, s2=sf2+x2*sfg2, where x2 is dryness fraction\n",
+"//  also\n",
+"s2 = s1;\n",
+"//  so\n",
+"x2 = (s2-sf2)/sfg2;\n",
+"//  and\n",
+"v2 = x2*vg2;//  [m^3]\n",
+"\n",
+"//  for  hyperbolic process\n",
+"//  P2*v2=P3*v3\n",
+"//  hence\n",
+"v3 = P2*v2/P3;//  [m^3]\n",
+"	\n",
+"mprintf('\n (a) From steam table at .06 MN/m^2 steam is superheated and has temperature of 100 C with specific volume is  =  %f  m^3/kg\n',v3);\n",
+"\n",
+"//  (b)\n",
+"//  at this condition\n",
+"s3 = 7.609;//  [kJ/kg*K]\n",
+"//  hence\n",
+"change_s23 = s3-sg;//  change in specific entropy during the hyperblic process[kJ/kg*K]\n",
+"mprintf('\n (b) The change in specific entropy during the hyperbolic process is  =  %f kJ/kg K\n',change_s23);\n",
+"\n",
+"// In the book they have taken sg instead of s2 for part (b), so answer is not matching\n",
+"\n",
+"//  End\n",
+""
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 7.6: heat_transfer_and_work_done.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"\n",
+"clear;\n",
+"clc;\n",
+"disp('Example 7.6');\n",
+"\n",
+"//  aim : To determine the\n",
+"//  (a) heat transfer during the expansion and\n",
+"//  (b) work done durind the expansion\n",
+"\n",
+"//  given values\n",
+"m = 4.5;  //  mass of steam,[kg]\n",
+"P1 = 3;  //  initial pressure,[MN/m^2]\n",
+"T1 = 300+273;  //  initial temperature,[K]\n",
+"\n",
+"P2 = .1;  //  final pressure,[MN/m^2]\n",
+"x2 = .96;  //  dryness fraction at final stage\n",
+"\n",
+"//  solution\n",
+"//  for state point 1,using steam table\n",
+"s1 = 6.541;//  [kJ/kg/K]\n",
+"u1 = 2751;// [kJ/kg]\n",
+" \n",
+" //  for state point 2\n",
+" sf2 = 1.303;//  [kJ/kg/K]\n",
+" sfg2 = 6.056;//  [kJ/kg/k]\n",
+" T2 = 273+99.6;//  [K]\n",
+" hf2 = 417;//  [kJ/kg]\n",
+" hfg2 = 2258;//  [kJ/kg]\n",
+" vg2 = 1.694;//   [m^3/kg]\n",
+" \n",
+" //  hence\n",
+" s2 = sf2+x2*sfg2;// [kJ/kg/k]\n",
+" h2 = hf2+x2*hfg2;// [kJ/kg]\n",
+" u2 = h2-P2*x2*vg2*10^3;// [kJ/kg]\n",
+" \n",
+" //  Diagram of example 7.6\n",
+" x = [s1 s2];\n",
+" y = [T1 T2];\n",
+"plot2d(x,y);\n",
+" xtitle('Diagram for example 7.6(T vs s)');\n",
+" xlabel('Entropy (kJ/kg K)');\n",
+" ylabel('Temperature (K)');\n",
+"\n",
+"x = [s1,s1];\n",
+"y = [0,T1];\n",
+"plot2d(x,y);\n",
+"\n",
+"x = [s2,s2];\n",
+"y = [0,T2];\n",
+"plot2d(x,y);\n",
+"\n",
+" //  (a)\n",
+" //  Q_rev is area of T-s diagram\n",
+" Q_rev = (T1+T2)/2*(s2-s1);// [kJ/kg]\n",
+" //  so total heat transfer is\n",
+" Q_rev = m*Q_rev;// [kJ]\n",
+" \n",
+" //  (b)\n",
+" del_u = u2-u1;//  change in internal energy, [kJ/kg]\n",
+" //  using 1st law of thermodynamics\n",
+" W = Q_rev-m*del_u;//  [kJ]\n",
+" \n",
+"mprintf('\n (a) The heat transfer during the expansion is  =  %f kJ  (received)\n',Q_rev);\n",
+"\n",
+" mprintf('\n (b) The work done during the expansion is  =  %f kJ\n',W);\n",
+" \n",
+" //  End"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 7.7: change_of_entropy_and_ratio_of_change_of_entropy_and_heat_transfer.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"\n",
+"clear;\n",
+"clc;\n",
+"disp('Example 7.7');\n",
+"\n",
+"//  aim : To determine the  \n",
+"//  (a) change of entropy\n",
+"//  (b)  The approximate change of entropy obtained by dividing the heat transferred by the gas by the mean absolute temperature during the compression\n",
+"\n",
+"//  Given values\n",
+"P1 = 140;// initial pressure,[kN/m^2]\n",
+"V1 = .14;// initial volume, [m^3]\n",
+"T1 = 273+25;// initial temperature,[K]\n",
+" P2 = 1400;// final pressure [kN/m^2]\n",
+" n = 1.25; // polytropic index\n",
+" cp = 1.041;// [kJ/kg K]\n",
+" cv = .743;// [kJ/kg K]\n",
+" \n",
+" //  solution\n",
+" // (a)\n",
+" R = cp-cv;// [kJ/kg/K]\n",
+" //  using ideal gas equation \n",
+" m = P1*V1/(R*T1);// mass of gas,[kg]\n",
+" //  since gas is following law P*V^n=constant ,so \n",
+" V2 = V1*(P1/P2)^(1/n);//  [m^3]\n",
+" \n",
+" //  using eqn [9]\n",
+" del_s = m*(cp*log(V2/V1)+cv*log(P2/P1));//  [kJ/K]\n",
+" mprintf('\n (a) The change of entropy is  =  %f kJ/K\n',del_s);\n",
+" \n",
+" //  (b)\n",
+" W = (P1*V1-P2*V2)/(n-1);// polytropic work,[kJ]\n",
+" Gamma = cp/cv;// heat capacity ratio\n",
+" Q = (Gamma-n)/(Gamma-1)*W;// heat transferred,[kJ]\n",
+" \n",
+" // Again using polytropic law\n",
+" T2 = T1*(V1/V2)^(n-1);// final temperature, [K]\n",
+" T_avg = (T1+T2)/2;// mean absolute temperature, [K]\n",
+" \n",
+" // so approximate change in entropy is\n",
+" del_s = Q/T_avg;// [kJ/K]\n",
+" \n",
+" mprintf('\n (b) The approximate change of entropy obtained by dividing the heat transferred by the gas by the mean absolute temperature during the compression  =  %f  kJ/K\n',del_s);\n",
+" \n",
+" //  End"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 7.8: change_of_entropies.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"clear;\n",
+"clc;\n",
+"disp('Example 7.8');\n",
+"\n",
+"//  aim : To determine\n",
+"//  the change of entropy\n",
+"\n",
+"//  Given values\n",
+"m = .3;//  [kg]\n",
+"P1 = 350;//  [kN/m^2]\n",
+"T1 = 273+35;//  [K]\n",
+"P2 = 700;//  [kN/m^2]\n",
+"V3 = .2289;//  [m^3]\n",
+"cp = 1.006;//  [kJ/kg K]\n",
+"cv = .717;//  [kJ/kg K]\n",
+"\n",
+"//  solution\n",
+"//  for constant volume process\n",
+"R = cp-cv;//  [kJ/kg K]\n",
+"//  using PV=mRT\n",
+"V1 = m*R*T1/P1;//  [m^3]\n",
+"\n",
+"//  for constant volume process P/T=constant,so\n",
+"T2 = T1*P2/P1;//  [K]\n",
+"s21 = m*cv*log(P2/P1);//  formula for entropy change for constant volume process\n",
+"mprintf('\n The change of entropy in constant volume process is  =  %f  kJ/kg K\n',s21);\n",
+"\n",
+"// 'For the above part result given in the book is wrong\n",
+"\n",
+"V2 = V1;\n",
+"// for constant pressure process\n",
+"T3 = T2*V3/V2;//  [K]\n",
+"s32 = m*cp*log(V3/V2);//  [kJ/kg K]\n",
+"\n",
+"mprintf('\n The change of entropy in constant pressure process is  =  %f  kJ/kg K\n',s32);\n",
+"\n",
+"// there is misprint in the book's result\n",
+"\n",
+"// End"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 7.9: change_of_entropy.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"clear;\n",
+"clc;\n",
+"disp('Example 7.9');\n",
+"\n",
+"//  aim : To determine\n",
+"//  the  change of entropy\n",
+"\n",
+"// Given values\n",
+"P1 = 700;// initial pressure, [kN/m^2]\n",
+"T1 = 273+150;//  Temperature ,[K]\n",
+"V1 = .014;// initial volume, [m^3]\n",
+"V2 = .084;//  final volume, [m^3]\n",
+"\n",
+"//  solution\n",
+"//  since process is isothermal so\n",
+"T2 = T1;\n",
+"// and using fig.7.10\n",
+"del_s = P1*V1*log(V2/V1)/T1 ;//  [kJ/K]\n",
+"mprintf('\n The change of entropy is  =  %f  kJ/kg K\n',del_s);\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
+}
diff --git a/Basic_Engineering_Thermodynamics_by_R_Joel/8-Combustion.ipynb b/Basic_Engineering_Thermodynamics_by_R_Joel/8-Combustion.ipynb
new file mode 100644
index 0000000..9e0474e
--- /dev/null
+++ b/Basic_Engineering_Thermodynamics_by_R_Joel/8-Combustion.ipynb
@@ -0,0 +1,1321 @@
+{
+"cells": [
+ {
+		   "cell_type": "markdown",
+	   "metadata": {},
+	   "source": [
+       "# Chapter 8: Combustion"
+	   ]
+	},
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 8.10: volumetric_composition_of_products.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"clear;\n",
+"clc;\n",
+"disp('Example 8.10');\n",
+"\n",
+"// aim : To determine\n",
+"// volumetric composition of the products of combustion\n",
+"\n",
+"// given values\n",
+"C = .86;//  mass composition of carbon\n",
+"H = .14;// mass composition of hydrogen\n",
+"Ea = .20;//  excess air for combustion\n",
+"O2 = .23;// mass composition of O2 in air \n",
+"\n",
+"MCO2 = 44;// moleculer mass of CO2\n",
+"MH2O = 18;// moleculer mass of H2O\n",
+"MO2 = 32;// moleculer mass of O2\n",
+"MN2 = 28;// moleculer mass of N2,\n",
+"\n",
+"\n",
+"// solution\n",
+"sO2 = (8/3*C+8*H);// stoichiometric O2 required, [kg/kg petrol]\n",
+"sair = sO2/O2;// stoichiometric air required, [kg/kg petrol]\n",
+"// for one kg petrol\n",
+"mCO2 = 11/3*C;// mass of CO2,[kg]\n",
+"mH2O = 9*H;// mass of H2O, [kg]\n",
+"mO2 = Ea*sO2;// mass of O2, [kg]\n",
+"mN2 = 14.84*(1+Ea)*(1-O2);// mass of N2, [kg]\n",
+"\n",
+"mt = mCO2+mH2O+mO2+mN2;// total mass, [kg]\n",
+"// percentage mass composition\n",
+"x1 = mCO2/mt*100;// mass composition of CO2\n",
+"x2 = mH2O/mt*100;// mass composition of H2O\n",
+"x3 = mO2/mt*100;// mass composition of O2\n",
+"x4 = mN2/mt*100;// mass composition of N2\n",
+"\n",
+"vt = x1/MCO2+x2/MH2O+x3/MO2+x4/MN2;// total volume of petrol\n",
+"v1 = x1/MCO2/vt*100;// %age composition of CO2 by volume\n",
+"v2 = x2/MH2O/vt*100;// %age composition  of H2O by volume\n",
+"v3 = x3/MO2/vt*100;// %age composition of O2 by volume\n",
+"v4 = x4/MN2/vt*100;// %age composition of N2 by volume\n",
+" \n",
+"mprintf('\nThe percentage composition of CO2 by volume is  =  %f\n,\nThe percentage composition of H2O by volume is  =  %f\n,\nThe percentage composition of O2 by volume is  =  %f\n,\nThe percentage composition of N2 by volume is  =  %f\n',v1,v2,v3,v4);\n",
+"\n",
+"//  End"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 8.11: energy_carried_away.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"clear;\n",
+"clc;\n",
+"disp('Example 8.11');\n",
+"\n",
+"// aim : To determine\n",
+"//  the energy carried away by the dry flue gas/kg of fuel burned\n",
+"\n",
+"// given values\n",
+"C = .78;//  mass composition of carbon\n",
+"H2 = .06;// mass composition of hydrogen\n",
+"O2 = .09;// mass composition of oxygen\n",
+"Ash = .07;// mass composition of ash\n",
+"Ea = .50;//  excess air for combustion\n",
+"aO2 = .23;// mass composition of O2 in air \n",
+"Tb = 273+20;// boiler house temperature, [K]\n",
+"Tf = 273+320;// flue gas temperature, [K]\n",
+"c = 1.006;// specific heat capacity of dry flue gas, [kJ/kg K]\n",
+"\n",
+"// solution\n",
+"// for one kg of fuel\n",
+"sO2 = (8/3*C+8*H2);// stoichiometric O2 required, [kg/kg fuel]\n",
+"sO2a = sO2-O2;// stoichiometric O2 required from air, [kg/kg fuel]\n",
+"sair = sO2a/aO2;// stoichiometric air required, [kg/kg fuel]\n",
+"ma = sair*(1+Ea);// actual air supplied/kg of fuel, [kg]\n",
+"// total mass of flue gas/kg fuel is\n",
+"mf = ma+1;// [kg]\n",
+"mH2 = 9*H2;//H2 produced, [kg] \n",
+"// hence, mass of dry flue gas/kg coall is\n",
+"m = mf-mH2;// [kg]\n",
+"Q = m*c*(Tf-Tb);// energy carried away by flue gas, [kJ]\n",
+"mprintf('\n The energy carried away by the dry flue gas/kg is  =  %f kg\n',Q);\n",
+"\n",
+"//  End"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 8.12: stoichiometric_volume_and_composition_of_products.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"clear;\n",
+"clc;\n",
+"disp('Example 8.12');\n",
+"\n",
+"// aim : To determine\n",
+"// (a) the stoichiometric volume of air for the complete combustion of 1 m^3\n",
+"// (b) the percentage volumetric analysis of the products of combustion\n",
+"\n",
+"// given values\n",
+"N2 = .018;// volumetric composition of N2\n",
+"CH4 = .94;// volumetric composition of CH4\n",
+"C2H6 = .035;// volumetric composition of C2H6\n",
+"C3H8 = .007;// volumetric composition of C3H8\n",
+"aO2 = .21;// O2 composition in air\n",
+"\n",
+"// solution\n",
+"// (a)\n",
+"// for CH4\n",
+"// CH4 +2 O2= CO2 + 2 H2O\n",
+"sva1 = 2/aO2;// stoichiometric volume of air, [m^3/m^3 CH4]\n",
+"svn1 = sva1*(1-aO2);// stoichiometric volume of nitrogen in the air, [m^3/m^3 CH4]\n",
+"\n",
+"// for C2H6\n",
+"// 2 C2H6 +7 O2= 4 CO2 + 6 H2O\n",
+"sva2 = 7/2/aO2;// stoichiometric volume of air, [m^3/m^3 C2H6]\n",
+"svn2 = sva2*(1-aO2);// stoichiometric volume of nitrogen in the air, [m^3/m^3 C2H6]\n",
+"\n",
+"// for C3H8\n",
+"// C3H8 +5 O2=3 CO2 + 4 H2O\n",
+"sva3 = 5/aO2;// stoichiometric volume of air, [m^3/m^3 C3H8]\n",
+"svn3 = sva3*(1-aO2);// stoichiometric volume of nitrogen in the air, [m^3/m^3 C3H8]\n",
+"\n",
+"Sva = CH4*sva1+C2H6*sva2+C3H8*sva3;// stoichiometric volume of air required, [m^3/m^3 gas]\n",
+"mprintf('\n (a) The stoichiometric volume of air for the complete combustion  =  %f m^3m^3 gas\n',Sva);\n",
+"\n",
+"// (b)\n",
+"// for one m^3 of natural gas\n",
+"vCO2 = CH4*1+C2H6*2+C3H8*3;// volume of CO2 produced, [m^3]\n",
+"vH2O = CH4*2+C2H6*3+C3H8*4;// volume of H2O produced, [m^3]\n",
+"vN2 = CH4*svn1+C2H6*svn2+C3H8*svn3+N2;// volume of N2 produced, [m^3]\n",
+"\n",
+"vg = vCO2+vH2O+vN2;// total volume of gas, [m^3]\n",
+"x1 = vCO2/vg*100;// volume percentage of CO2 produced\n",
+"x2 = vH2O/vg*100;// volume percentage of H2O produced\n",
+"x3 = vN2/vg*100;// volume percentage of N2 produced\n",
+"\n",
+"mprintf('\n (b) The percentage volumetric composition of CO2 in produced is  =  %f\n,\n     The percentage volumetric composition of H2O in produced is  =  %f\n,\n     The percentage volumetric composition of N2 in produced is  =  %f\n',x1,x2,x3);\n",
+"\n",
+"// End"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 8.13: volume_and_composition_by_mass.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"clear;\n",
+"clc;\n",
+"disp('Example 8.13');\n",
+"\n",
+"// aim : To determine\n",
+"// (a) the volume of air taken by the fan\n",
+"// (b) the percentage composition of dry flue gas\n",
+"\n",
+"// gien values\n",
+"C = .82;// mass composition of carbon\n",
+"H = .08;// mass composition of hydrogen\n",
+" O = .03;// mass composition of oxygen\n",
+" A = .07;// mass composition of ash\n",
+"mc = .19;// coal uses, [kg/s] \n",
+" ea = .3;// percentage excess air of oxygen in the air required for combustion\n",
+"Oa = .23;// percentage of oxygen by mass in the air\n",
+"  \n",
+" // solution\n",
+" // (a)\n",
+" P = 100;// air pressure, [kN/m^2]\n",
+" T = 18+273;// air temperature, [K]\n",
+" R = .287;// [kJ/kg K]\n",
+" // basis one kg coal\n",
+" sO2 = 8/3*C+8*H;// stoichiometric O2 required, [kg]\n",
+" aO2 = sO2-.03;// actual O2 required, [kg]\n",
+"tO2 = aO2/Oa;// theoretical O2 required, [kg]\n",
+"Aa = tO2*(1+ea);// actual air supplied, [kg]\n",
+"m = Aa*mc;// Air supplied, [kg/s]\n",
+"\n",
+"// now using P*V=m*R*T\n",
+"V = m*R*T/P;// volume of air taken ,[m^3/s]\n",
+"mprintf('\n (a) Volume of air taken by fan is  =  %f  m^3/s\n',V);\n",
+"\n",
+"// (b)\n",
+"mCO2 = 11/3*C;// mass of CO2 produced, [kg]\n",
+"mO2 = aO2*.3;// mass of O2 produces, [kg]\n",
+"mN2 = Aa*.77;// mass of N2 produced, [[kg]\n",
+"mt = mCO2+mO2+mN2;// total mass, [kg]\n",
+"\n",
+"mprintf('\n (b) Percentage mass composition of CO2 is  =  %f percent \n',mCO2/mt*100);\n",
+"mprintf('\n     Percentage mass composition of O2 is  =  %f  percent\n',mO2/mt*100)\n",
+"mprintf('\n     Percentage mass composition of N2 is  =  %f  percent \n',mN2/mt*100)\n",
+"\n",
+"\n",
+"\n",
+"//  End"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 8.14: mass_and_volume.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"clear;\n",
+"clc;\n",
+"disp('Example 8.14');\n",
+"\n",
+"// aim : To determine \n",
+"// (a) the mass of fuel used per cycle\n",
+"// (b) the actual mass of air taken in per cycle\n",
+"// (c) the volume of air taken in per cycle\n",
+"\n",
+"// given values\n",
+"W = 15;// work done, [kJ/s]\n",
+"N = 5;// speed, [rev/s]\n",
+"C = .84;// mass composition of carbon\n",
+"H = .16;// mass composition of hydrogen\n",
+"ea = 1;// percentage excess air supplied \n",
+"CV = 45000;// calorificvalue of fuel, [kJ/kg]\n",
+"n_the = .3;// thermal efficiency\n",
+"P = 100;// pressuer, [kN/m^2]\n",
+"T = 273+15;// temperature, [K]\n",
+"R = .29;// gas constant, [kJ/kg K]\n",
+"\n",
+"// solution\n",
+"// (a)\n",
+"E = W*2/N/n_the;// energy supplied, [kJ/cycle]\n",
+"mf = E/CV;// mass of fuell used, [kg]\n",
+"mprintf('\n (a) Mass of fuel used per cycle is  =  %f g\n',mf*10^3);\n",
+"\n",
+"// (b)\n",
+"// basis 1 kg fuel\n",
+"mO2 = C*8/3+8*H;// mass of O2 requirea, [kg]\n",
+"smO2 = mO2/.23;// stoichiometric mass of air, [kg]\n",
+"ma = smO2*(1+ea);// actual mass of air supplied, [kg]\n",
+"m = ma*mf;// mass of air supplied, [kg/cycle]\n",
+"mprintf('\n (b) The mass of air supplied per cycle is  =  %f  kg\n',m);\n",
+"\n",
+"// (c)\n",
+"V = m*R*T/P;// volume of air, [m^3]\n",
+"mprintf('\n (c) The volume of air taken in per cycle is  =  %f  m^3\n',V);\n",
+"\n",
+"// End"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 8.15: mass_and_composition.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"clear;\n",
+"clc;\n",
+"disp('Example 8.15');\n",
+"\n",
+"//  aim : To determine\n",
+"//  (a) the mass of coal used per hour\n",
+"//  (b) the mass of air used per hour\n",
+"//  (c) the percentage analysis of the flue gases by mass\n",
+"\n",
+"//  given values\n",
+"m = 900;// mass of steam boiler generate/h, [kg]\n",
+"x = .96;// steam dryness fraction\n",
+"P = 1400;// steam pressure, [kN/m^2]\n",
+"Tf = 52;// feed water temperature, [C]\n",
+"BE = .71;// boiler efficiency\n",
+"CV = 33000;// calorific value  of coal, [kJkg[\n",
+"ea = .22;// excess air supply\n",
+"aO2 = .23;// oxygen composition in air\n",
+"c = 4.187;// specific heat capacity of water, [kJ/kg K]\n",
+"\n",
+"//  coal composition\n",
+"C = .83;// mass composition of carbon\n",
+"H2 = .05;// mass composition of hydrogen\n",
+"O2 = .03;// mass composition of oxygen\n",
+"ash = .09;// mass composition of ash\n",
+"\n",
+"// solution\n",
+"// from steam table at pressure P\n",
+"hf = 830.1;// specific enthalpy, [kJ/kg]\n",
+"hfg = 1957.1;// specific enthalpy, [kJ/kg]\n",
+"hg = 2728.8;// specific enthalpy, [kJ/kg]\n",
+"\n",
+"// (a)\n",
+"h = hf+x*hfg;// specific enthalpy of steam generated by boiler, [kJ/kg]\n",
+"hfw = c*Tf;// specific enthalpy of feed water, [kJ/kg]\n",
+"Q = m*(h-hfw);// energy to steam/h, [kJ]\n",
+"Qf = Q/BE;// energy required from fuel/h, [kJ]\n",
+"mc = Qf/CV;// mass of coal/h,[kg]\n",
+"mprintf('\n (a) The mass of coal used per hour is  =  %f kg\n',mc);\n",
+"\n",
+"// (b)\n",
+"// for one kg coal\n",
+"mO2 = 8/3*C+8*H2+-O2;// actual mass of O2 required, [kg]\n",
+"mta = mO2/aO2;// theoretical mass of air, [kg]\n",
+"ma = mta*(1+ea);// mass of air supplied, [kg]\n",
+"mas = ma*mc;// mass of air supplied/h, [kg]\n",
+"mprintf('\n (b) The mass of air supplied per hour is  =  %f kg\n',mas);\n",
+"\n",
+" \n",
+"// (c)\n",
+"// for one kg coal\n",
+"mCO2 = 11/3*C;// mass of CO2 produced, [kg]\n",
+"mH2O = 9*H2;// mass of H2O produced, [kg]\n",
+"mO2 = mO2*ea;// mass of excess O2 in flue gas, [kg]\n",
+"mN2 = ma*(1-aO2);// mass of N2 in flue gas, [kg]\n",
+"\n",
+"mt = mCO2+mH2O+mO2+mN2;// total mass of gas\n",
+"x1 = mCO2/mt*100;// mass percentage composition of CO2\n",
+"x2 = mH2O/mt*100;// mass percentage composition of H2O\n",
+"x3 = mO2/mt*100;// mass percentage composition of O2\n",
+"x4 = mN2/mt*100;// mass percentage composition of N2\n",
+"\n",
+"mprintf('\n (c) The mass percentage composition of CO2  =  %f,\n      The mass percentage composition of H2O  =  %f,\n      The mass percentage composition of O2  =  %f,\n      The mass percentage composition of N2  =  %f',x1,x2,x3,x4);\n",
+"\n",
+"//  mass of coal taken in part (b) is wrong so answer is not matching\n",
+"\n",
+"//  End\n",
+"\n",
+"\n",
+""
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 8.16: volume_average_moleculer_mass_characteristic_gas_constant_and_density.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"clear;\n",
+"clc;\n",
+"disp('Example 8.16');\n",
+"\n",
+"// aim : To determine\n",
+"// (a) volume of gas\n",
+"// (b) (1) the average molecular mass of air\n",
+"//       (2) the value of R\n",
+"//       (3) the mass of 1 m^3 of air at STP\n",
+"\n",
+"// given values\n",
+"n = 1;// moles of gas, [kmol]\n",
+"P = 101.32;// standard pressure, [kN/m^2]\n",
+"T = 273;// gas tempearture, [K]\n",
+"\n",
+"O2 = 21;// percentage volume composition of oxygen in air\n",
+"N2 = 79;// percentage volume composition of nitrogen in air\n",
+"R = 8.3143;// molar gas constant, [kJ/kg K]\n",
+"mO2 = 32;// moleculer mass of O2\n",
+"mN2 = 28;// moleculer mass of N2\n",
+"\n",
+"// solution\n",
+"// (a)\n",
+"V = n*R*T/P;// volume of gas, [m^3]\n",
+"mprintf('\n (a) The volume of the gas is  =  %f m^3\n',V);\n",
+"\n",
+"// (b)\n",
+"//(1)\n",
+"Mav = (O2*mO2+N2*mN2)/(O2+N2);// average moleculer mass of air\n",
+"mprintf('\n (b)(1) The average moleculer mass of air is  =  %f g/mol\n',Mav);\n",
+"\n",
+"// (2)\n",
+"Rav = R/Mav;// characteristic gas constant, [kJ/kg k]\n",
+"mprintf('\n     (2) The value of R is  =  %f  kJ/kg K\n',Rav);\n",
+"\n",
+"// (3)\n",
+"rho = Mav/V;// density of air, [kg/m^3]\n",
+"mprintf('\n     (3) The mass of one cubic metre of air at STP is  =  %f  kg/m^3\n',rho);\n",
+"\n",
+"//  End"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 8.17: pressures_and_volume.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"clear;\n",
+"clc;\n",
+"disp('Example 8.17');\n",
+"\n",
+"// aim : To determine\n",
+"// (a) the partial pressure of each gas in the vessel\n",
+"// (b) the volume of the vessel\n",
+"// (c)  the total pressure in the gas when temperature is raised to228 C\n",
+"\n",
+"// given values\n",
+"MO2 = 8;//  mass of O2, [kg]\n",
+"MN2 = 7;//  mass of N2, [kg]\n",
+"MCO2 = 22;//  mass of CO2, [kg]\n",
+"\n",
+"P = 416;// total pressure in the vessel, [kN/m^2]\n",
+"T = 273+60;// vessel temperature, [K]\n",
+"R = 8.3143;// gas constant, [kJ/kmol K]\n",
+"\n",
+"mO2 = 32;// molculer mass of O2 \n",
+"mN2 = 28;// molculer mass of N2\n",
+"mCO2 = 44;// molculer mass of CO2\n",
+"\n",
+"// solution\n",
+"// (a)\n",
+"n1 = MO2/mO2;// moles of O2, [kmol]\n",
+"n2 = MN2/mN2;// moles of N2, [kmol]\n",
+"n3 = MCO2/mCO2;// moles of CO2, [kmol]\n",
+"\n",
+"n = n1+n2+n3;// total moles in the vessel, [kmol]\n",
+"// since,Partial pressure is proportinal, so\n",
+"P1 = n1*P/n;// partial pressure of O2, [kN/m^2]\n",
+"P2 = n2*P/n;// partial pressure of N2, [kN/m^2]\n",
+"P3 = n3*P/n;// partial pressure of CO2, [kN/m^2]\n",
+"\n",
+"mprintf('\n (a)The partial pressure of O2 is  =  %f  kN/m^2,\n,   The partial pressure of N2 is  =  %f  kN/m^2,\n    The partial pressure of CO2 is  =  %f  kN/m^2,\n',P1,P2,P3);\n",
+"\n",
+"// (b)\n",
+"// assuming ideal gas \n",
+"V = n*R*T/P;// volume of the container, [m^3]\n",
+"mprintf('\n (b) The volume of the container is  =  %f  m^3\n',V);\n",
+"\n",
+"// (c)\n",
+"T2 = 273+228;// raised vessel temperature, [K]\n",
+"// so volume of vessel  will constant , P/T=constant\n",
+"P2 = P*T2/T;// new pressure in the vessel , [kn/m62]\n",
+"mprintf('\n (c) The new total pressure in the vessel is  =  %f  kN/m^2\n',P2);\n",
+"\n",
+"//  End"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 8.18: mass_and_velocity.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"clear;\n",
+"clc;\n",
+"disp('Example 8.18');\n",
+"\n",
+"// aim : To determine\n",
+"// the actual mass of air supplied/kg coal\n",
+"// the velocity of flue gas\n",
+"\n",
+"// given values\n",
+"mc = 635;// mass of coal burn/h, [kg]\n",
+"ea = .25;// excess air required\n",
+"C = .84;// mass composition of carbon\n",
+"H2 = .04;// mass composition of hydrogen\n",
+"O2 = .05;// mass composition of oxygen\n",
+"ash = 1-(C+H2+O2);// mass composition of ash\n",
+"\n",
+"P1 = 101.3;// pressure, [kJn/m^2]\n",
+"T1 = 273;// temperature, [K]\n",
+"V1 = 22.4;// volume, [m^3]\n",
+"\n",
+"T2 = 273+344;// gas temperature, [K]\n",
+"P2 = 100;// gas pressure, [kN/m^2]\n",
+"A = 1.1;// cross section area, [m^2]\n",
+"aO2 = .23;// composition of O2 in air\n",
+"\n",
+"mCO2 = 44;// moleculer mass of carbon\n",
+"mH2O = 18;// molecular mass of hydrogen\n",
+"mO2 = 32;// moleculer mas of oxygen\n",
+"mN2 = 28;// moleculer mass of nitrogen\n",
+"\n",
+"// solution\n",
+"mtO2 = 8/3*C+8*H2-O2;// theoretical O2 required/kg coal, [kg]\n",
+"msa= mtO2/aO2;// stoichiometric mass of  air supplied/kg coal, [kg]\n",
+"mas = msa*(1+ea);// actual mass of air supplied/kg coal, [kg]\n",
+"\n",
+"m1 = 11/3*C;// mass of CO2/kg coal produced, [kg]\n",
+"m2 = 9*H2;// mass of H2/kg coal produced, [kg]\n",
+"m3 = mtO2*ea;// mass of O2/kg coal produced, [kg]\n",
+"m4 = mas*(1-aO2);// mass of N2/kg coal produced, [kg]\n",
+"\n",
+"mt = m1+m2+m3+m4;// total mass, [kg]\n",
+"x1 = m1/mt*100;// %age mass composition of CO2 produced\n",
+"x2 = m2/mt*100;// %age mass composition of H2O produced\n",
+"x3 = m3/mt*100;// %age mass composition of O2 produced\n",
+"x4 = m4/mt*100;// %age mass composition of N2 produced\n",
+"\n",
+"vt = x1/mCO2+x2/mH2O+x3/mO2+x4/mN2;// total volume\n",
+"v1 = x1/mCO2/vt*100;// %age volume composition of CO2\n",
+"v2 = x2/mH2O/vt*100;// %age volume composition of H2O\n",
+"v3 = x3/mO2/vt*100;// %age volume composition of O2\n",
+"v4 = x4/mN2/vt*100;// %age volume composition of N2\n",
+"\n",
+"Mav = (v1*mCO2+v2*mH2O+v3*mO2+v4*mN2)/(v1+v2+v3+v4);// average moleculer mass, [kg/kmol]\n",
+"// since no of moles is constant so PV/T=constant\n",
+"V2 = P1*V1*T2/(P2*T1);//volume, [m^3]\n",
+"\n",
+"mp = mt*mc/3600;// mass of product of combustion/s, [kg]\n",
+"\n",
+"V = V2*mp/Mav;// volume of flowing gas /s,[m^3]\n",
+"\n",
+"v = V/A;// velocity of flue gas, [m/s]\n",
+"mprintf('\n The actual mass of air supplied is  =  %f  kg/kg coal\n',mas);\n",
+"mprintf('\n The velocity of flue gas is  =  %f m/s\n',v);\n",
+"\n",
+"//  End"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 8.19: temperature_and_density.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"clear;\n",
+"clc;\n",
+"disp('Example 8.19');\n",
+"\n",
+"// aim : To determine\n",
+"// (a) the temperature of the gas after compression\n",
+"// (b) the density of the air-gas mixture\n",
+"\n",
+"// given values\n",
+"CO = 26;// %age volume composition of CO \n",
+"H2 = 16;// %age volume composition of H2\n",
+"CH4 = 7;// %age volume composition of CH4 \n",
+"N2 = 51;// %age volume composition of N2\n",
+"\n",
+"P1 = 103;// gas pressure, [kN/m^2]\n",
+"T1 = 273+21;// gas temperature, [K]\n",
+"rv = 7;// volume ratio\n",
+"\n",
+"aO2 = 21;// %age volume composition of O2 in the air\n",
+"c = 21;// specific heat capacity of diatomic gas, [kJ/kg K]\n",
+"cCH4 = 36;// specific heat capacity of CH4, [kJ/kg K]\n",
+"R = 8.3143;// gas constant, [kJ/kg K]\n",
+"\n",
+"mCO = 28;// moleculer mass of carbon\n",
+"mH2 = 2;// molecular mass of hydrogen\n",
+"mCH4 = 16;// moleculer mas of methane\n",
+"mN2 = 28;// moleculer mass of nitrogen\n",
+"mO2 = 32;// moleculer mass of oxygen\n",
+"\n",
+"// solution\n",
+"// (a)\n",
+"Cav = (CO*c+H2*c+CH4*cCH4+N2*c+100*2*c)/(100+200);// heat capacity, [kJ/kg K]\n",
+"\n",
+"Gama = (Cav+R)/Cav;// heat capacity ratio\n",
+"// rv = V1/V2\n",
+"// process is polytropic, so\n",
+"T2 = T1*(rv)^(Gama-1);// final tempearture, [K]\n",
+"mprintf('\n (a) The temperature of the gas after compression is  =  %f C\n',T2-273);\n",
+"\n",
+"// (b)\n",
+"\n",
+"Mav = (CO*mCO+H2*mH2+CH4*mCH4+N2*mN2+42*mO2+158*mN2)/(100+200)\n",
+"\n",
+"// for 1 kmol of gas\n",
+"V = R*T1/P1;// volume of one kmol of gas, [m^3]\n",
+"// hence\n",
+"rho = Mav/V;// density of gas, [kg/m^3]\n",
+"\n",
+"mprintf('\n (b) The density of air-gas mixture is  =  %f  kg/m^3\n',rho);\n",
+"\n",
+"//  End"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 8.1: stoichiometric_mass.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"\n",
+"clear;\n",
+"clc;\n",
+"disp('Example 8.1');\n",
+"\n",
+"//  aim : To determine\n",
+"//   the stoichiometric mass of air required to burn 1 kg the fuel \n",
+"\n",
+"//  Given values\n",
+"C = .72;//  mass fraction of C; [kg/kg]\n",
+"H2 = .20;//  mass fraction of H2;, [kg/kg]\n",
+"O2 = .08;//  mass fraction of O2, [kg/kg]\n",
+"aO2=.232;//  composition of oxygen in air\n",
+"\n",
+"//  solution\n",
+"//  for 1kg of fuel\n",
+"mO2  = 8/3*C+8*H2-O2;//  mass of O2, [kg]\n",
+"\n",
+"//  hence stoichiometric mass of O2 required is\n",
+"msO2  = mO2/aO2;// [kg]\n",
+"\n",
+"mprintf('\n The stoichiometric mass of air required to burn 1 kg the fuel should be =  %f  kg\n',msO2);\n",
+"\n",
+"//  End"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 8.20: stoichiometric_equatio.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"clear;\n",
+"clc;\n",
+"disp('Example 8.20');\n",
+"\n",
+"// aim : to determine \n",
+"// stoichiometric equation for combustion of hydrogen\n",
+"\n",
+"// solution\n",
+"// equation with algebric coefficient is\n",
+"// H2+aO2+79/21*aN2=bH2O+79/21*aN2\n",
+"// by equating coefficients\n",
+"b = 1;\n",
+"a = b/2;\n",
+"// so equation becomes\n",
+"// 2 H2+ O2+3.76 N2=2 H2O+3.76 N2\n",
+"disp('The required stoichiometric equation is  =  ');\n",
+"disp('2 H2+ O2+3.76 N2 = 2 H2O+3.76 N2');\n",
+"\n",
+"// End"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 8.22: gravimetric_compositio.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"clear;\n",
+"clc;\n",
+"disp('Example 8.22');\n",
+"\n",
+"// aim : To determine\n",
+"// the percentage gravimetric analysis of the total products of combustion\n",
+"\n",
+"// given values\n",
+"CO = 12;//  %age volume composition of CO\n",
+"H2 = 41;//  %age volume composition of H2\n",
+"CH4 = 27;//  %age volume composition of CH4\n",
+"O2 = 2;//  %age volume composition of O2\n",
+"CO2 = 3;//  %age volume composition of CO2\n",
+"N2 = 15;//  %age volume composition of N2\n",
+"\n",
+"mCO2 = 44;// moleculer mass of CO2,[kg/kmol]\n",
+"mH2O = 18;// moleculer mass of H2O, [kg/kmol]\n",
+"mO2 = 32;// moleculer mass of O2, [kg/kmol]\n",
+"mN2 = 28;// moleculer mass of N2, [kg/kmol]\n",
+" \n",
+"ea = 15;// %age excess air required\n",
+"aO2 = 21;// %age air composition in the air\n",
+"\n",
+"// solution\n",
+"// combustion equation by no. of moles\n",
+"// 12CO + 41H2 + 27CH4 + 2O2 + 3CO2 + 15N2 + aO2+79/21*aN2 = bCO2 + dH2O + eO2 + 15N2 +79/21*aN2\n",
+"// equating C coefficient\n",
+"b = 12+27+3;// [mol]\n",
+"// equatimg H2 coefficient\n",
+"d = 41+2*27;// [mol]\n",
+"// O2 required is 15 % extra,so\n",
+"// e/(e-a)=.15 so e=.13a\n",
+"// equating O2 coefficient\n",
+"// 2+3+a=b+d/2 +e\n",
+"\n",
+"a = (b+d/2-5)/(1-.13);\n",
+"e = .13*a;// [mol]\n",
+"\n",
+"// gravimetric analysis of product\n",
+"v1 =  b*mCO2;// gravimetric volume of CO2 \n",
+"v2 =  d*mH2O ;// gravimetric volume of H2O \n",
+"v3 = e*mO2;// gravimetric volume of O2\n",
+"v4 = 15*mN2 +79/21*a*mN2;// gravimetric volume of N2 \n",
+"\n",
+"vt = v1+v2+v3+v4;// total\n",
+"x1 = v1/vt*100;// percentage gravimetric of CO2\n",
+"x2 = v2/vt*100;// percentage gravimetric of H2O\n",
+"x3 = v3/vt*100;// percentage gravimetric of O2\n",
+"x4 = v4/vt*100;// percentage gravimetric of N2\n",
+"\n",
+"mprintf('\n Percentage gravimetric composition of CO2  =  %f\n ,\n Percentage gravimetric composition of H2O  =  %f\n\n Percentage gravimetric composition of O2  =  %f\n\n Percentage gravimetric composition of N2  =  %f\n',x1,x2,x3,x4);\n",
+"\n",
+"//  End "
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 8.23: mass_and_volumetric_efficiency.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"clear;\n",
+"clc;\n",
+"disp('Example 8.23');\n",
+"\n",
+"//  aim : To determine\n",
+"// (a) the actual quantity of air supplied/kg of fuel\n",
+"// (b) the volumetric efficiency of the engine\n",
+"\n",
+"// given values\n",
+"d = 300*10^-3;// bore,[m]\n",
+"L = 460*10^-3;// stroke,[m]\n",
+"N = 200;// engine speed, [rev/min]\n",
+"\n",
+"C = 87;//  %age mass composition of Carbon in the fuel\n",
+"H2 = 13;//  %age mass composition of H2 in the fuel\n",
+"\n",
+"mc = 6.75;// fuel consumption, [kg/h]\n",
+"\n",
+"CO2 = 7;// %age composition of CO2 by volume\n",
+"O2 = 10.5;// %age composition of O2 by volume\n",
+"N2 = 7;// %age composition of N2 by volume\n",
+"\n",
+"mC = 12;// moleculer mass of CO2,[kg/kmol]\n",
+"mH2 = 2;// moleculer mass of H2, [kg/kmol]\n",
+"mO2 = 32;// moleculer mass of O2, [kg/kmol]\n",
+"mN2 = 28;// moleculer mass of N2, [kg/kmol]\n",
+"\n",
+"T = 273+17;// atmospheric temperature, [K]\n",
+"P = 100;// atmospheric pressure, [kn/m^2]\n",
+"R =.287;// gas constant, [kJ/kg k]\n",
+"\n",
+"// solution\n",
+"// (a)\n",
+"// combustion equation by no. of moles\n",
+"// 87/12 C + 13/2 H2 + a O2+79/21*a N2 = b CO2 + d H2O + eO2 + f N2\n",
+"// equating  coefficient\n",
+"b = 87/12;// [mol]\n",
+"a = 22.7;// [mol]\n",
+"e = 10.875;// [mol]\n",
+"f = 11.8*b;// [mol]\n",
+"// so fuel side combustion equation is\n",
+"// 87/12 C + 13/2 H2 +22.7 O2 +85.5 N2\n",
+"mair = ( 22.7*mO2 +85.5*mN2)/100;// mass of air/kg fuel, [kg]\n",
+"mprintf('\n (a) The mass of actual air supplied per kg of fuel is  =  %f  kg\n',mair);\n",
+"\n",
+"// (b)\n",
+"m = mair*mc/60;// mass of air/min, [kg]\n",
+"V = m*R*T/P;// volumetric flow of air/min, [m^3]\n",
+"SV = %pi/4*d^2*L*N/2;// swept volume/min, [m^3]\n",
+"\n",
+"VE = V/SV;// volumetric efficiency\n",
+"mprintf('\n (b) The volumetric efficiency of the engine is  =  %fpercent\n',VE*100);\n",
+"\n",
+"//  End"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 8.24: mass_of_air.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"clear;\n",
+"clc;\n",
+"disp('Example 8.24');\n",
+"\n",
+"// aim : To determine\n",
+"//  the mass of air supplied/kg of fuel burnt\n",
+"\n",
+"// given values\n",
+"// gas composition in the fuel\n",
+"C = 84;//  %age mass composition of Carbon in the fuel\n",
+"H2 = 14;//  %age mass composition of H2 in the fuel\n",
+"O2f = 2;// %age mass composition of O2 in the fuel\n",
+"\n",
+"// exhaust gas composition\n",
+"CO2 = 8.85;// %age composition of CO2 by volume\n",
+"CO = 1.2// %age composition of CO by volume\n",
+"O2 = 6.8;// %age composition of O2 by volume\n",
+"N2 = 83.15;// %age composition of N2 by volume\n",
+"\n",
+"mC = 12;// moleculer mass of CO2,[kg/kmol]\n",
+"mH2 = 2;// moleculer mass of H2, [kg/kmol]\n",
+"mO2 = 32;// moleculer mass of O2, [kg/kmol]\n",
+"mN2 = 28;// moleculer mass of N2, [kg/kmol]\n",
+"\n",
+"// solution\n",
+"// combustion equation by no. of moles\n",
+"// 84/12 C + 14/2 H2 +2/32 O2 + a O2+79.3/20.7*a N2 = b CO2 + d CO2+  eO2 + f N2 +g H2\n",
+"// equating  coefficient and given condition\n",
+"b = 6.16;// [mol]\n",
+"a = 15.14;// [mol]\n",
+"d = .836;// [mol]\n",
+"f = 69.3*d;// [mol]\n",
+"// so fuel side combustion equation is\n",
+"// 84/12 C + 14/2 H2 +2/32 O2 +  15.14 O2 +85.5 N2\n",
+"mair = ( a*mO2 +f*mN2)/100;// mass of air/kg fuel, [kg]\n",
+"mprintf('\n The mass of air supplied per kg of fuel is  =  %f  kg\n',mair);\n",
+"\n",
+"//  End"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 8.3: stoichiometric_mass_and_composition_of_products.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"clear;\n",
+"clc;\n",
+"disp('Example 8.3');\n",
+"\n",
+"//  aim : To determine \n",
+"//  the stoichiometric mass of air\n",
+"//  the products of combustion both by mass and as percentage \n",
+"\n",
+"//  Given values\n",
+"C = .82;// mass composition C\n",
+"H2 = .12;// mass composition of H2\n",
+"O2 = .02;// mass composition of O2\n",
+"S = .01;// mass composition of S\n",
+"N2 = .03;// mass composition of N2\n",
+"\n",
+" //  solution\n",
+"//  for 1kg fuel\n",
+"mo2 = 8/3*C+8*H2-O2+S*1;// total mass of  O2 required, [kg]\n",
+"sa = mo2/.232;// stoichimetric  air, [kg]\n",
+"mprintf('\n The stoichiometric mass of air is  =  %f kg/kg fuel\n',sa);\n",
+"\n",
+"// for one kg fuel\n",
+"mCO2 = C*11/3;// mass of CO2 produced, [kg]\n",
+"mH2O = H2*9;// mass of H2O produced, [kg]\n",
+"mSO2 = S*2;// mass of SO2 produce, [kg]\n",
+"mN2 = C*8.84+H2*26.5-O2*.768/.232+S*3.3+N2;// mass of N2 produced, [kg]\n",
+"\n",
+"mt = mCO2+mH2O+mSO2+mN2;// total mass of product, [kg]\n",
+"\n",
+"x1 = mCO2/mt*100;// %age mass composition of CO2 produced\n",
+"x2 = mH2O/mt*100;// %age mass composition of H2O produced\n",
+"x3 = mSO2/mt*100;// %age mass composition of SO2 produced\n",
+"x4 = mN2/mt*100;// %age mass composition of N2 produced\n",
+"\n",
+"mprintf('\n CO2 produced  =  %f  kg/kg fuel,  percentage composition  =  %f,\n H2O produced  =  %f  kg/kg fuel,  percentage composition  =  %f,\n SO2 produced  =  %f  kg/kg fuel,  percentage composition  =  %f,\n N2 produced  =  %f  kg/kg fuel,  percentage composition  =  %f',mCO2,x1,mH2O,x2,mSO2,x3,mN2,x4);\n",
+"\n",
+"//  End"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 8.4: stoichiometric_volume.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"clear;\n",
+"clc;\n",
+"disp('Example 8.4');\n",
+"\n",
+"//  aim : To determine \n",
+"//  the stoichiometric volume of air required for complete combution of 1 m^3 of the gas\n",
+"\n",
+"//  Given values\n",
+"H2 = .14;// volume fraction of H2\n",
+"CH4 = .02;// volume fraction of CH4\n",
+"CO = .22;// volume fraction of CO\n",
+"CO2 = .05;// volume fraction of CO2\n",
+"O2 = .02;// volume fraction of O2\n",
+"N2 = .55;// volume fraction of N2\n",
+"\n",
+"// solution\n",
+"//  for 1 m^3 of fuel\n",
+"Va = .5*H2+2*CH4+.5*CO-O2;// [m^3]\n",
+"\n",
+"//  stoichiometric air required is\n",
+"Vsa = Va/.21;//  [m^3]\n",
+"\n",
+"mprintf('\n The  stoichiometric volume of air required for complete combustion is  =  %f  m^3/m^3  fuel\n',Vsa);\n",
+"\n",
+"// End"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 8.5: volume.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"clear;\n",
+"clc;\n",
+"disp('Example 8.5');\n",
+"\n",
+"//  aim : To determine\n",
+"//   the volume of the air required \n",
+"\n",
+"//  Given values\n",
+"H2 = .45;// volume fraction of H2\n",
+"CO = .40;// volume fraction of CO\n",
+"CH4 = .15;// volume fraction of CH4\n",
+"\n",
+"//  solution\n",
+"V = 2.38*(H2+CO)+9.52*CH4;// stoichimetric volume of air, [m^3]\n",
+"\n",
+"mprintf('\n The volume of air required is  =  %f  m^3/m^3 fuel\n',V);\n",
+"\n",
+"// Result in the book is misprinted\n",
+"\n",
+"// End\n",
+""
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 8.6: stoichiometric_volume_composition_of_products.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"clear;\n",
+"clc;\n",
+"disp('Example 8.6');\n",
+"\n",
+"// aim : To determine\n",
+"// the stoichiometric volume of air for the complete combustion\n",
+"// the products of combustion\n",
+"\n",
+"// given values\n",
+"CH4 = .142;// volumetric composition of CH4\n",
+"CO2 = .059;// volumetric composition of CO2\n",
+"CO = .360;// volumetric composition of CO\n",
+"H2 = .405;// volumetric composition of H2\n",
+"O2 = .005;// volumetric composition of O2\n",
+"N2 = .029;// volumetric composition of N2\n",
+"\n",
+"aO2 = .21;// O2 composition into air by volume\n",
+"\n",
+"//  solution\n",
+"svO2 = CH4*2+CO*.5+H2*.5-O2;// stroichiometric volume of O2 required, [m^3/m^3 fuel]\n",
+"svair = svO2/aO2;// stroichiometric volume of air required, [m^3/m^3 fuel]\n",
+"mprintf('\n Stoichiometric volume of air required is  =  %f  m^3/m^3 fuel\n',svair);\n",
+"\n",
+"// for one m^3 fuel\n",
+"vN2 = CH4*7.52+CO*1.88+H2*1.88-O2*.79/.21+N2;// volume of N2 produced, [m^3]\n",
+"vCO2 = CH4*1+CO2+CO*1;// volume of CO2 produced, [m^3]\n",
+"vH2O = CH4*2+H2*1;// volume of H2O produced, [m^3]\n",
+"\n",
+"vt = vN2+vCO2+vH2O;// total volume of product, [m^3]\n",
+"\n",
+"x1 = vN2/vt*100;// %age composition of N2 in product,\n",
+"x2 = vCO2/vt*100;// %age composition of CO2 in product\n",
+"x3 = vH2O/vt*100;// %age composition of H2O in product\n",
+"\n",
+"mprintf('\n N2 in products  =  %fm^3/m^3  fuel,  percentage composition  =  %f,\n CO2 in products =  %f  m^3/m^3 fuel,  percentage composition  =  %f,\n H2O in products  =  %fm^3/m^3  fuel,  percentage composition  =  %f',vN2,x1,vCO2,x2,vH2O,x3);\n",
+"\n",
+"// End "
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 8.7: composition_of_gases.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"clear;\n",
+"clc;\n",
+"disp('Example 8.7');\n",
+"\n",
+"//  aim : To determine\n",
+"//  the percentage analysis of the gas by mass\n",
+"\n",
+"//  Given values\n",
+"CO2 = 20;// percentage volumetric composition  of CO2\n",
+"N2 = 70;// percentage volumetric composition of N2\n",
+"O2 = 10;// percentage volumetric composition  of O2\n",
+"\n",
+"mCO2 = 44;//  moleculer mas of CO2\n",
+"mN2 = 28;//  moleculer mass of N2\n",
+"mO2 = 32;//  moleculer mass of O2\n",
+"\n",
+"//  solution\n",
+"mgas = CO2*mCO2+N2*mN2+O2*mO2;//  moleculer mass of gas \n",
+"m1 = CO2*mCO2/mgas*100;// percentage composition of CO2 by mass \n",
+"m2 = N2*mN2/mgas*100;// percentage composition of N2 by mass \n",
+"m3 = O2*mO2/mgas*100;// percentage composition of O2 by mass \n",
+"\n",
+"mprintf('\n Mass percentage of CO2 is  =  %f\n\n Mass percentage of N2 is  =  %f\n\n Mass percentage of O2 is  =  %f\n',m1,m2,m3 )\n",
+"\n",
+"// End"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 8.8: composition_of_gases.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"clear;\n",
+"clc;\n",
+"disp('Example 8.8');\n",
+"\n",
+"//  aim : To determine\n",
+"//  the percentage composition of the gas by volume\n",
+"\n",
+"//  given values\n",
+"CO = 30;// %age mass composition of CO\n",
+"N2 = 20;// %age mass composition of N2\n",
+"CH4 = 15;// %age mass composition of CH4\n",
+"H2 = 25;// %age mass composition of H2\n",
+"O2 = 10;// %age mass composition of O2\n",
+"\n",
+"mCO = 28;// molculer mass of CO\n",
+"mN2 = 28;// molculer mass of N2\n",
+"mCH4 = 16;// molculer mass of CH4\n",
+"mH2 = 2;// molculer mass of H2\n",
+"mO2 = 32;// molculer mass of O2\n",
+"\n",
+"// solution\n",
+"vg = CO/mCO+N2/mN2+CH4/mCH4+H2/mH2+O2/mO2;\n",
+"v1 = CO/mCO/vg*100;// %age volume composition of CO\n",
+"v2 = N2/mN2/vg*100;// %age volume composition of N2\n",
+"v3 = CH4/mCH4/vg*100;// %age volume composition of CH4\n",
+"v4 = H2/mH2/vg*100;// %age volume composition of H2\n",
+"v5 = O2/mO2/vg*100;// %age volume composition of O2\n",
+"\n",
+"mprintf('\n The percentage composition of CO by volume is  =  %f\n,\nThe percentage composition of N2 by volume is  =  %f\n\nThe percentage composition of CH4 by volume is  =  %f\n\nThe percentage composition of H2 by volume is  =  %f\n\nThe percentage composition of O2by volume is=%f',v1,v2,v3,v4,v5);\n",
+"\n",
+"// End"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 8.9: mass_of_air_supplied.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"clear;\n",
+"clc;\n",
+"disp('Example 8.9');\n",
+"\n",
+"// aim : To determine\n",
+"// the mass of air supplied per kilogram of fuel burnt\n",
+"\n",
+"// given values\n",
+"CO2 = 8.85;// volume composition of CO2\n",
+"CO = 1.2;//  volume composition of CO\n",
+"O2 = 6.8;//  volume composition of O2\n",
+"N2 = 83.15;//  volume composition of N2 \n",
+"\n",
+"// composition of gases in the fuel oil\n",
+"C = .84;// mass composition of carbon \n",
+"H = .14;// mass composition of hydrogen\n",
+"o2 = .02;// mass composition of oxygen\n",
+"\n",
+"mC = 12;// moleculer mass of Carbon\n",
+"mCO2 = 44;// molculer mass of CO2\n",
+"mCO = 28;// molculer mass of CO\n",
+"mN2 = 28;// molculer mass of N2\n",
+"mO2 = 32;// molculer mass of O2\n",
+"aO2 = .23;// mass composition of O2 in air\n",
+"\n",
+"// solution\n",
+"ma = (8/3*C+8*H-o2)/aO2;// theoretical mass of air/kg fuel, [kg]\n",
+"\n",
+"mgas = CO2*mCO2+CO*mCO+N2*mN2+O2*mO2;//  total mass of gas/kg fuel, [kg]\n",
+"x1 = CO2*mCO2/mgas;//  composition of CO2 by mass \n",
+"x2 = CO*mCO/mgas;// composition of CO by mass\n",
+"x3 = O2*mO2/mgas;//  composition of O2 by mass \n",
+"x4 = N2*mN2/mgas;//  composition of N2 by mass \n",
+"\n",
+"m1 = x1*mC/mCO2+x2*mC/mCO;// mass of C/kg of dry flue gas, [kg]\n",
+"m2 = C;// mass of C/kg fuel, [kg]\n",
+"mf = m2/m1;// mass of dry flue gas/kg fuel, [kg]\n",
+"mo2 = mf*x3;// mass of excess O2/kg fuel, [kg]\n",
+"mair = mo2/aO2;// mass of excess air/kg fuel, [kg]\n",
+"m = ma+mair;// mass of excess air supplied/kg fuel, [kg]\n",
+"\n",
+"mprintf('\n The mass of air supplied per/kg of fuel burnt is  =  %f  kg\n',m);\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
+}
diff --git a/Basic_Engineering_Thermodynamics_by_R_Joel/9-Heat_transfer.ipynb b/Basic_Engineering_Thermodynamics_by_R_Joel/9-Heat_transfer.ipynb
new file mode 100644
index 0000000..3e3e005
--- /dev/null
+++ b/Basic_Engineering_Thermodynamics_by_R_Joel/9-Heat_transfer.ipynb
@@ -0,0 +1,399 @@
+{
+"cells": [
+ {
+		   "cell_type": "markdown",
+	   "metadata": {},
+	   "source": [
+       "# Chapter 9: Heat transfer"
+	   ]
+	},
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 9.1: interface_temperature.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"clear;\n",
+"clc;\n",
+"disp('Example 9.1');\n",
+"\n",
+"//  aim : To determine \n",
+"//  the heat loss per hour through the wall and interface temperature\n",
+"\n",
+"//  Given values\n",
+"x1 = .25;// thickness of brick,[m]\n",
+"x2 = .05;// thickness of concrete,[m]\n",
+"t1 = 30;// brick face temperature,[C]\n",
+"t3 = 5;// concrete face temperature,[C]\n",
+"l = 10;// length of the wall, [m]\n",
+"h = 5;// height of the wall, [m]\n",
+"k1 = .69;// thermal conductivity of brick,[W/m/K]\n",
+"k2 = .93;// thermal conductivity of concrete,[W/m/K]\n",
+"\n",
+"//  solution\n",
+"A = l*h;// area of heat transfer,[m^2]\n",
+"Q_dot = A*(t1-t3)/(x1/k1+x2/k2);// heat transferred, [J/s]\n",
+"\n",
+"//  so heat loss per hour is\n",
+"Q = Q_dot*3600*10^-3;// [kJ]\n",
+"mprintf('\n The heat lost per hour is  =  %f  kJ\n',Q);\n",
+"\n",
+"//  interface temperature calculation\n",
+"//   for  the brick wall, Q_dot=k1*A*(t1-t2)/x1;\n",
+"//  hence\n",
+"t2 = t1-Q_dot*x1/k1/A;// [C]\n",
+"mprintf('\n The interface temperature is  =  %f C\n',t2);\n",
+"\n",
+"//  End"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 9.2: thickness_of_lagging.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"clear;\n",
+"clc;\n",
+"disp('Example 9.2');\n",
+"\n",
+"//  aim : To determine\n",
+"//  the minimum \n",
+"//  thickness of the lagging required\n",
+"\n",
+"//  Given values\n",
+"r1 = 75/2;// external radious of the pipe,[mm]\n",
+"L = 80;// length of the pipe,[m]\n",
+"m_dot = 1000;// flow of steam, [kg/h]\n",
+"P = 2;// pressure, [MN/m^2]\n",
+"x1 = .98;// inlet dryness fraction\n",
+"x2 = .96;// outlet dryness fraction\n",
+"k = .08;// thermal conductivity of of pipe, [W/m/K]\n",
+"t2 = 27;// outside temperature,[C]\n",
+"\n",
+"//  solution\n",
+"//  using steam table  at 2 MN/m^2 the enthalpy of evaporation of steam is,\n",
+"hfg = 1888.6;// [kJ/kg]\n",
+"//  so heat loss through the pipe is\n",
+"Q_dot = m_dot*(x1-x2)*hfg/3600;// [kJ]\n",
+"\n",
+"// also from steam table saturation temperature of steam at 2 MN/m^2 is,\n",
+"t1 = 212.4;// [C]\n",
+"// and for thick pipe, Q_dot=k*2*%pi*L*(t1-t2)/log(r2/r1)\n",
+"// hence\n",
+"r2 = r1*exp(k*2*%pi*L*(t1-t2)*10^-3/Q_dot);// [mm]\n",
+"\n",
+"t = r2-r1;// thickness, [mm]\n",
+"\n",
+"mprintf('\n The minimum thickness of the lagging required is  =  %f  mm\n',t);\n",
+"\n",
+"//  End"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 9.3: heat_lost_and_interface_temperature.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"clear;\n",
+"clc;\n",
+"disp('Example 9.3');\n",
+"\n",
+"//  aim : To determine the\n",
+"//  (a) heat loss per hour\n",
+"//   (b) interface temperature og lagging\n",
+"\n",
+"// Given values\n",
+"r1 = 50; // radious of steam main,[mm]\n",
+"r2 = 90;// radious with first lagging,[mm]\n",
+"r3 = 115;// outside radious os steam main with lagging,[mm]\n",
+"k1 = .07;// thermal conductivity of 1st lagging,[W/m/K]\n",
+"k2 = .1;// thermal conductivity of 2nd lagging, [W/m/K]\n",
+"P = 1.7;// steam pressure,[MN/m^2]\n",
+"t_superheat = 30;// superheat of steam, [K]\n",
+"t3 = 24;// outside temperature of the lagging,[C]\n",
+"L = 20;// length of the steam main,[m]\n",
+"\n",
+"//  solution\n",
+"//  (a)\n",
+"//  using steam table saturation temperature of steam at 1.7 MN/m^2 is\n",
+"t_sat = 204.3;// [C]\n",
+"// hence\n",
+"t1 = t_sat+t_superheat;// temperature of steam,[C]\n",
+"\n",
+"Q_dot = 2*%pi*L*(t1-t3)/(log(r2/r1)/k1+log(r3/r2)/k2);// heat loss,[W]\n",
+"//  heat loss in hour is\n",
+"Q = Q_dot*3600*10^-3;// [kJ]\n",
+"\n",
+"mprintf('\n (a) The heat lost per hour is  =  %f kJ\n',Q);\n",
+"\n",
+"// (b)\n",
+"//  using Q_dot=2*%pi*k1*(t1-t1)/log(r2/r1) \n",
+"t2 = t1-Q_dot*log(r2/r1)/(2*%pi*k1*L);// interface temperature of lagging,[C]\n",
+"\n",
+"mprintf('\n (b) The interface temperature of the lagging is  =  %f C\n',t2);\n",
+"\n",
+"//  There is some calculation mistake in the book so answer is not matching\n",
+"\n",
+"//  End"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 9.4: energy_emitted.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"clear;\n",
+"clc;\n",
+"disp('Example 9.4');\n",
+"\n",
+"// aim : To determine \n",
+"//  the energy emetted from the surface\n",
+"\n",
+"//  Given values\n",
+"h = 3;// height of surface, [m]\n",
+"b = 4;// width of surface, [m]\n",
+"epsilon_s = .9;// emissivity of the surface\n",
+"T = 273+600;// surface temperature ,[K]\n",
+"sigma = 5.67*10^-8;// [W/m^2/K^4]\n",
+"\n",
+"//  solution\n",
+"As = h*b;// area of the surface, [m^2]\n",
+"\n",
+"Q_dot = epsilon_s*sigma*As*T^4*10^-3;// energy emitted, [kW]\n",
+"\n",
+"mprintf('\n The energy emitted from the surface is  =  %f  kW\n',Q_dot);\n",
+"\n",
+"//  End"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 9.5: Rate_of_heat_transfer.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"clear;\n",
+"clc;\n",
+"disp('Example 9.5');\n",
+"\n",
+"//  aim : To determine \n",
+"//  the rate of energy transfer between furnace and the sphere and its direction\n",
+"\n",
+"//  Given values\n",
+"l = 1.25;// internal side of cubical furnace, [m]\n",
+"ti = 800+273;// internal surface temperature of the furnace,[K]\n",
+"r = .2;// sphere radious, [m]\n",
+"epsilon = .6;// emissivity of sphere\n",
+"ts = 300+273;// surface temperature of sphere, [K]\n",
+"sigma = 5.67*10^-8;// [W/m^2/K^4]\n",
+"\n",
+"//  Solution\n",
+"Af = 6*l^2;// internal surface area of furnace, [m^2]\n",
+"As =4 *%pi*r^2;// surface area of sphere, [m^2]\n",
+"\n",
+"//  considering internal furnace to be black\n",
+"Qf = sigma*Af*ti^4*10^-3;// [kW]\n",
+"\n",
+"//  radiation emitted by sphere is\n",
+"Qs = epsilon*sigma*As*ts^4*10^-3; // [kW]\n",
+"\n",
+"//  Hence transfer of energy is\n",
+"Q = Qf-Qs;// [kW]\n",
+"\n",
+"mprintf('\n The transfer of energy will be from furnace to sphere and transfer rate is  =  %f kW\n',Q);\n",
+"\n",
+"//   There is some calculation mistake in the book so answer is not matching\n",
+"\n",
+"//  End\n",
+""
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 9.6: overall_heat_transfer_coefficient_and_heat_lost.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"clear;\n",
+"clc;\n",
+"disp('Example 9.6');\n",
+"\n",
+"//  aim : To determine\n",
+"//  the overall transfer coefficient and the heat loss per hour\n",
+"\n",
+"//  Given values\n",
+"x1 = 25*10^-3;// Thickness of insulating board, [m]\n",
+"x2 = 75*10^-3;// Thickness of fibreglass, [m]\n",
+"x3 = 110*10^-3;// Thickness of brickwork, [m]\n",
+"k1 = .06;// Thermal conductivity of insulating board, [W/m K]\n",
+"k2 = .04;// Thermal conductivity of fibreglass, [W/m K]\n",
+"k3 = .6;// Thermal conductivity of brickwork, [W/m K]\n",
+"Us1 = 2.5;//  surface heat transfer coefficient of the inside wall,[W/m^2 K]\n",
+"Us2 = 3.1;//  surface heat transfer coefficient of the outside wall,[W/m^2 K]\n",
+"ta1 = 27;// internal ambient temperature, [C]\n",
+"ta2 = 10;// external ambient temperature, [C]\n",
+"h = 6;// height of the wall, [m]\n",
+"l = 10;// length of the wall, [m]\n",
+"\n",
+"//  solution\n",
+"U = 1/(1/Us1+x1/k1+x2/k2+x3/k3+1/Us2);// overall heta transfer coefficient,[W/m^2 K]\n",
+"\n",
+"A = l*h;// area ,[m^2]\n",
+"\n",
+"Q_dot = U*A*(ta1-ta2);// heat loss [W]\n",
+"\n",
+"//  so heat loss per hour is\n",
+"Q = Q_dot*3600*10^-3;// [kJ]\n",
+"mprintf('\n The overall heat transfer coefficient for the wall is   =  %f  W/m^2 K\n',U);\n",
+"mprintf('\n The heat loss per hour through the wall is  =  %f kJ\n',Q);\n",
+"\n",
+"//  End"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 9.7: heat_lost_and_surafce_temperature_of_lagging.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"clear;\n",
+"clc;\n",
+"disp('Example 9.7');\n",
+"\n",
+"//  aim : To determine  \n",
+"//  the heat loss per hour and the surface temperature of the lagging\n",
+"\n",
+"//  Given values\n",
+"r1 = 75*10^-3;// External radiou of the pipe, [m]\n",
+"t_l1 = 40*10^-3;// Thickness of lagging1, [m]\n",
+"t_l2 = t_l1;\n",
+"k1 = .07;// thermal conductivity of lagging1, [W/m K]\n",
+"k2 = .1;// thermal conductivity of lagging2, [W/m K]\n",
+"Us = 7;// surface transfer coefficient for outer surface, [W/m^2 K]\n",
+"L = 50;// length of the pipe, [m]\n",
+"ta = 27;// ambient temperature, [C]\n",
+"P = 3.6;// wet steam pressure, [MN/m^2]\n",
+"\n",
+"//  solution\n",
+"//  from steam table saturation temperature of the steam at given pressure is,\n",
+"t1 =  244.2;// [C]\n",
+"r2 = r1+t_l1;// radious of pipe with lagging1,[m]\n",
+"r3 = r2+t_l2;// radious of pipe with both the lagging, [m]\n",
+"\n",
+"R1 = log(r2/r1)/(2*%pi*L*k1);// resistance due to lagging1,[C/W]\n",
+"R2 = log(r3/r2)/(2*%pi*L*k2);// resistance due to lagging2,[C/W]\n",
+"R3 = 1/(Us*2*%pi*r3*L);// ambient resistance, [C/W]\n",
+"\n",
+"//  hence overall resistance is,\n",
+"Req = R1+R2+R3;// [C/W]\n",
+"tdf = t1-ta;// temperature driving force, [C]\n",
+"Q_dot = tdf/Req;// rate of heat loss, [W]\n",
+"//  so heat loss per hour is,\n",
+"Q = Q_dot*3600*10^-3;// heat loss per hour, [kJ]\n",
+"\n",
+"//  using eqn [3]\n",
+"t3 = ta+Q_dot*R3;// surface temperature of the lagging, [C]\n",
+"\n",
+"mprintf('\n The heat loss per hour is  =  %f  kJ\n',Q);\n",
+"mprintf('\n The surface temperature of the lagging is  =  %f C\n',t3);\n",
+"\n",
+"//  there is minor variation in the answer\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
+}