Merge pull request #1 from prashantsinalkar/masterHEADmaster

Initial commit

13 files changed, 7838 insertions, 0 deletions

diff --git a/Fluid_Power_With_Applications_by_A_Esposito/10-HYDRAULIC_CONDUCTORS_AND_FITTINGS.ipynb b/Fluid_Power_With_Applications_by_A_Esposito/10-HYDRAULIC_CONDUCTORS_AND_FITTINGS.ipynb
new file mode 100644
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--- /dev/null
+++ b/Fluid_Power_With_Applications_by_A_Esposito/10-HYDRAULIC_CONDUCTORS_AND_FITTINGS.ipynb
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+{
+"cells": [
+ {
+		   "cell_type": "markdown",
+	   "metadata": {},
+	   "source": [
+       "# Chapter 10: HYDRAULIC CONDUCTORS AND FITTINGS"
+	   ]
+	},
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 10.1_a: find_minimum_inside_diameter_of_pipe.sci"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"// Aim:To find minimum inside diameter of pipe  \n",
+"// Given:\n",
+"// flow-rate through pipe:\n",
+"Q=30; //gpm\n",
+"// average fluid velocity:\n",
+"v=20; //ft/s"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 10.1_b: SOLUTION_minimum_inside_diameter_of_pipe.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"clc;\n",
+"pathname=get_absolute_file_path('10_1_soln.sce')\n",
+"filename=pathname+filesep()+'10_1_data.sci'\n",
+"exec(filename)\n",
+"// Solution:\n",
+"// flow-rate in ft^3/s,\n",
+"Q_fps=Q/449; //ft^3/s\n",
+"// minimum required pipe flow area,\n",
+"A=(Q_fps/v)*144; //in^2\n",
+"// minimum inside diameter,\n",
+"D=sqrt((4*A)/(%pi)); //in\n",
+"// Results:\n",
+"printf('\n  Results:  ')   \n",
+"printf('\n The minimum inside diameter of pipe is %.3f in.',D)"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 10.2_a: find_minimum_inside_diameter_in_SI.sci"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"// Aim:To find minimum inside diameter of pipe in Metric units  \n",
+"// Given:\n",
+"// flow-rate through pipe:\n",
+"Q=0.002; //m^3/s\n",
+"// average fluid velocity:\n",
+"v=6.1; //m/s"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 10.2_b: SOLUTION_minimum_inside_diameter_in_SI.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"clc;\n",
+"pathname=get_absolute_file_path('10_2_soln.sce')\n",
+"filename=pathname+filesep()+'10_2_data.sci'\n",
+"exec(filename)\n",
+"// Solution:\n",
+"// minimum required pipe flow area,\n",
+"A=(Q/v); //m^2\n",
+"// minimum inside diameter,\n",
+"D=sqrt((4*A)/(%pi))*1000; //mm\n",
+"// Results:\n",
+"printf('\n  Results:  ')   \n",
+"printf('\n The minimum inside diameter of pipe is %.1f mm.',D)"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 10.3_a: find_safe_working_pressure_of_tube.sci"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"// Aim:To find safe working pressure for the tube \n",
+"// Given:\n",
+"// outside diameter of steel tube:\n",
+"Do=1.250; //in\n",
+"// inside diameter of steel tube:\n",
+"Di=1.060; //in\n",
+"// tensile strength of steel tube:\n",
+"S=55000; //psi\n",
+"// factor of safety:\n",
+"FS=8;"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 10.3_b: SOLUTION_safe_working_pressure_of_tube.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"clc;\n",
+"pathname=get_absolute_file_path('10_3_soln.sce')\n",
+"filename=pathname+filesep()+'10_3_data.sci'\n",
+"exec(filename)\n",
+"// Solution:\n",
+"// wall thickness,\n",
+"t=(Do-Di)/2; //in\n",
+"// burst pressure,\n",
+"BP=(2*t*S)/Di; //psi\n",
+"// working pressure,\n",
+"WP=BP/FS; //psi\n",
+"// Results:\n",
+"printf('\n  Results:  ')   \n",
+"printf('\n The working pressure of steel tube is %.0f psi.',WP)\n",
+"printf('\n The answer in the program is different than that in textbook. It may be due to no.s of significant digit in data and calculation')"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 10.4_a: select_proper_size_steel_tube.sci"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"// Aim:Refer Example 10-4 for Problem Description\n",
+"// Given:\n",
+"// flow-rate:\n",
+"Q=30; //gpm\n",
+"// operating pressure:\n",
+"p=1000; //psi\n",
+"// maximum velocity:\n",
+"v=20; //ft/s\n",
+"// tensile strength of material:\n",
+"S=55000; //psi\n",
+"// factor of safety:\n",
+"FS=8;"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 10.4_b: SOLUTION_proper_size_steel_tube.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"clc;\n",
+"pathname=get_absolute_file_path('10_4_soln.sce')\n",
+"filename=pathname+filesep()+'10_4_data.sci'\n",
+"exec(filename)\n",
+"// Solutions:\n",
+"// flow-rate,\n",
+"Q=Q/449; //ft^3/s\n",
+"// minimum required pipe flow area,\n",
+"Ai=(Q/v)*144; //in^2\n",
+"// minimum inside diameter,\n",
+"Di=sqrt((4*Ai)/(%pi)); //in\n",
+"// wall thickness,\n",
+"t1=0.049; t2=0.065; //in\n",
+"// tube inside diameter,\n",
+"D1=0.902; D2=0.870; //in\n",
+"// burst pressure,\n",
+"BP1=(2*t1*S)/D1; //psi\n",
+"// working pressure,\n",
+"WP1=BP1/FS; //psi\n",
+"printf(' \n The working pressure %.0f psi is not adequate (less than %.0f psi) so next case is considered,',WP1,p)\n",
+"// burst pressure,\n",
+"BP2=(2*t2*S)/D2; //psi\n",
+"// working pressure,\n",
+"WP2=BP2/FS; //psi\n",
+"// ratio of inner diameter to thickness,\n",
+"r2=D2/t2;\n",
+"printf(' \n The working pressure %.0f psi is greater than %.0f psi) ,',WP2,p)\n",
+"// Results:\n",
+"printf('\n  Results:  ')\n",
+"printf('\n The ratio of inner diameter to length is %.1f.',r2)\n",
+"printf('\n The answer in the program is different than that in textbook. It may be due to no.s of significant digit in data and calculation')"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 10.5_a: select_proper_size_steel_tube_SI.sci"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"// Aim:Refer Example 10-5 for Problem Description\n",
+"// Given:\n",
+"// flow-rate:\n",
+"Q=0.00190; //m^3/s\n",
+"// operating pressure:\n",
+"p=70; //bars\n",
+"// maximum velocity:\n",
+"v=6.1; //m/s\n",
+"// tensile strength of material:\n",
+"S=379; //MPa\n",
+"// factor of safety:\n",
+"FS=8;"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 10.5_b: SOLUTION_proper_size_steel_tube_SI.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"clc;\n",
+"pathname=get_absolute_file_path('10_5_soln.sce')\n",
+"filename=pathname+filesep()+'10_5_data.sci'\n",
+"exec(filename)\n",
+"// Solutions:\n",
+"// minimum required pipe flow area,\n",
+"A=(Q/v); //m^2\n",
+"// minimum inside diameter,\n",
+"ID=sqrt((4*A)/(%pi))*1000; //mm\n",
+"// wall thickness,\n",
+"t1=1; t2=2; //mm\n",
+"// tube inside diameter,\n",
+"D1=20; D2=24; //mm\n",
+"// burst pressure,\n",
+"BP1=(2*(t1/1000)*S)/(D1/1000); //MPa\n",
+"// working pressure,\n",
+"WP1=(BP1/FS)*10; //bars\n",
+"printf(' \n The working pressure %.0f bars is not adequate (less than %.0f bars) so next case is considered,',WP1,p)\n",
+"// burst pressure,\n",
+"BP2=(2*(t2/1000)*S)/(D2/1000); //MPa\n",
+"// working pressure,\n",
+"WP2=(BP2/FS)*10;; //MPa\n",
+"// ratio of inner diameter to thickness,\n",
+"r2=D2/t2;\n",
+"printf(' \n The working pressure %.0f bars is greater than %.0f bars) ,',WP2,p)\n",
+"// Results:\n",
+"printf('\n  Results:  ')\n",
+"printf('\n The ratio of inner diameter to length is %.1f.',r2)"
+   ]
+   }
+],
+"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/Fluid_Power_With_Applications_by_A_Esposito/11-ANCILLARY_HYDRAULIC_DEVICES.ipynb b/Fluid_Power_With_Applications_by_A_Esposito/11-ANCILLARY_HYDRAULIC_DEVICES.ipynb
new file mode 100644
index 0000000..98cdfeb
--- /dev/null
+++ b/Fluid_Power_With_Applications_by_A_Esposito/11-ANCILLARY_HYDRAULIC_DEVICES.ipynb
@@ -0,0 +1,337 @@
+{
+"cells": [
+ {
+		   "cell_type": "markdown",
+	   "metadata": {},
+	   "source": [
+       "# Chapter 11: ANCILLARY HYDRAULIC DEVICES"
+	   ]
+	},
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 11.1_a: find_the_discharge_flow_and_pressure.sci"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"// Aim:To find the discharge flow and pressure \n",
+"// Given:\n",
+"// high inlet flow-rate:\n",
+"Q_high_inlet=20; //gpm\n",
+"// low inlet pressure:\n",
+"p_low_inlet=500; //psi\n",
+"// Ratio of piston area to rod area:\n",
+"Ratio=5/1;"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 11.1_b: SOLUTION_the_discharge_flow_and_pressure.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"clc;\n",
+"pathname=get_absolute_file_path('11_1_soln.sce')\n",
+"filename=pathname+filesep()+'11_1_data.sci'\n",
+"exec(filename)\n",
+"// Solution:\n",
+"// high discharge pressure,\n",
+"p_high_discharge=Ratio*p_low_inlet; //psi\n",
+"// low discharge flow-rate,\n",
+"Q_low_discharge=Q_high_inlet/Ratio; //gpm\n",
+"// Results:\n",
+"printf('\n  Results:  ')   \n",
+"printf('\n The high discharge pressure is %.0f psi.',p_high_discharge)\n",
+"printf('\n The low discharge flow-rate is %.0f gpm.',Q_low_discharge)"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 11.2_a: determine_the_downstream_oil_temperature.sci"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"// Aim:To find the downstream oil temperature\n",
+"// Given:\n",
+"// temperature of oil flowing through pressure relief valve:\n",
+"T_oil=120; //deg F\n",
+"// pressure of oil flowing through pressure relief valve:\n",
+"p=1000; //psi\n",
+"// oil flow through pressure relief valve:\n",
+"Q_gpm=10; //gpm"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 11.2_b: SOLUTION_the_downstream_oil_temperature.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"clc;\n",
+"pathname=get_absolute_file_path('11_2_soln.sce')\n",
+"filename=pathname+filesep()+'11_2_data.sci'\n",
+"exec(filename)\n",
+"// Solution:\n",
+"// heat generation rate,\n",
+"HP=(p*Q_gpm)/1714; //HP\n",
+"// heat generation rate in Btu/min,\n",
+"HP_btu=HP*42.4; //Btu/min\n",
+"// oil flow-rate in lb/min,\n",
+"Q_lb=7.42*Q_gpm; //lb/min\n",
+"// temperature increase,\n",
+"T_increase=HP_btu/(0.42*Q_lb); //deg F\n",
+"// downward oil temperature,\n",
+"T_downward=T_oil+T_increase; //deg F\n",
+"// Results:\n",
+"printf('\n  Results:  ')   \n",
+"printf('\n The downstream oil temperature is %.1f deg F.',T_downward)"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 11.3_a: determine_downstream_oil_temperature_in_SI.sci"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"// Aim:To find the downstream oil temperature in SI Unit\n",
+"// Given:\n",
+"// temperature of oil flowing through pressure relief valve:\n",
+"T_oil=50; //deg C\n",
+"// pressure of oil flowing through pressure relief valve:\n",
+"p=70; //bar\n",
+"// oil flow through pressure relief valve:\n",
+"Q=0.000632; //m^3/s"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 11.3_b: SOLUTION_downstream_oil_temperature_in_SI.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"clc;\n",
+"pathname=get_absolute_file_path('11_3_soln.sce')\n",
+"filename=pathname+filesep()+'11_3_data.sci'\n",
+"exec(filename)\n",
+"// Solution:\n",
+"// heat generation rate,\n",
+"kW=((p*10^5)*Q)/1000; //kW\n",
+"// oil flow-rate,\n",
+"Q_kg_s=895*Q; //kg/s\n",
+"// temperature increase,\n",
+"T_increase=kW/(1.8*Q_kg_s); //deg C\n",
+"// downward oil temperature,\n",
+"T_downward=T_oil+T_increase; //deg C\n",
+"// Results:\n",
+"printf('\n  Results:  ')   \n",
+"printf('\n The downstream oil temperature is %.1f deg C.',T_downward)"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 11.4_a: find_heat_exchanger_rating_of_system.sci"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"// Aim:To find the rating of heat exchanger required to dissipate generated heat\n",
+"// Given:\n",
+"// oil flow-rate:\n",
+"Q=20; //gpm\n",
+"// operating pressure:\n",
+"p=1000; //psi\n",
+"// overall efficiency of pump:\n",
+"eff_overall=85; //%\n",
+"// power lost due to friction:\n",
+"HP_frict=10; //%"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 11.4_b: SOLUTION_heat_exchanger_rating_of_system.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"clc;\n",
+"pathname=get_absolute_file_path('11_4_soln.sce')\n",
+"filename=pathname+filesep()+'11_4_data.sci'\n",
+"exec(filename)\n",
+"// Solution:\n",
+"// pump power loss,\n",
+"pump_HP_loss=((1/(eff_overall/100))-1)*((p*Q)/1714); //HP\n",
+"// PRV average HP loss,\n",
+"PRV_loss=0.5*((p*Q)/1714); //HP\n",
+"// line average HP loss,\n",
+"line_loss=(HP_frict/100)*PRV_loss; //HP\n",
+"// total average loss,\n",
+"total_loss=pump_HP_loss+PRV_loss+line_loss; //HP\n",
+"// heat exchanger rating,\n",
+"HEx_rating=total_loss*2544; //Btu/hr\n",
+"// Results:\n",
+"printf('\n  Results:  ')   \n",
+"printf('\n The heat exchanger rating is %.0f Btu/hr.',HEx_rating)\n",
+"printf('\n The answer in the program does not match with that in the textbook due to roundoff error (standard ratings) in textbook')"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 11.5_a: find_heat_exchanger_rating_in_SI.sci"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"// Aim:To find the rating of heat exchanger required to dissipate generated heat in SI unit\n",
+"// Given:\n",
+"// oil flow-rate:\n",
+"Q=0.00126; //m^3/s\n",
+"// operating pressure:\n",
+"p=70; //bar\n",
+"// overall efficiency of pump:\n",
+"eff_overall=85; //%\n",
+"// power lost due to friction:\n",
+"HP_frict=10; //%"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 11.5_b: SOLUTION_heat_exchanger_rating_in_SI.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"clc;\n",
+"pathname=get_absolute_file_path('11_5_soln.sce')\n",
+"filename=pathname+filesep()+'11_5_data.sci'\n",
+"exec(filename)\n",
+"// Solution:\n",
+"// pump power loss,\n",
+"pump_loss=((1/(eff_overall/100))-1)*((p*10^5*Q)/1000); //kW\n",
+"// PRV average HP loss,\n",
+"PRV_loss=0.5*((p*10^5*Q)/1000); //kW\n",
+"// line average HP loss,\n",
+"line_loss=(HP_frict/100)*PRV_loss; //kW\n",
+"// total average loss,\n",
+"HEx_rating=pump_loss+PRV_loss+line_loss; //kW\n",
+"// Results:\n",
+"printf('\n  Results:  ')   \n",
+"printf('\n The heat exchanger rating is %.2f kW.',HEx_rating)"
+   ]
+   }
+],
+"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/Fluid_Power_With_Applications_by_A_Esposito/13-PNEUMATICS_AIR_PREPARATION_AND_COMPONENTS.ipynb b/Fluid_Power_With_Applications_by_A_Esposito/13-PNEUMATICS_AIR_PREPARATION_AND_COMPONENTS.ipynb
new file mode 100644
index 0000000..2215f78
--- /dev/null
+++ b/Fluid_Power_With_Applications_by_A_Esposito/13-PNEUMATICS_AIR_PREPARATION_AND_COMPONENTS.ipynb
@@ -0,0 +1,843 @@
+{
+"cells": [
+ {
+		   "cell_type": "markdown",
+	   "metadata": {},
+	   "source": [
+       "# Chapter 13: PNEUMATICS AIR PREPARATION AND COMPONENTS"
+	   ]
+	},
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 13.10_a: determine_air_maximum_flowrate_in_scfm.sci"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"// Aim:To determine maximum flow-rate in units of scfm of air\n",
+"// Given:\n",
+"// upstream temperature:\n",
+"T1=80; //deg F\n",
+"// upstream pressure:\n",
+"p1=80; //psi\n",
+"// flow capacity constant:\n",
+"Cv=7.4; \n",
+"// diameter of orifice:\n",
+"d=0.5; //in"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 13.10_b: SOLUTION_air_maximum_flowrate_in_scfm.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"clc;\n",
+"pathname=get_absolute_file_path('13_10_soln.sce')\n",
+"filename=pathname+filesep()+'13_10_data.sci'\n",
+"exec(filename)\n",
+"// Solution:\n",
+"// upstream temperature in Rankine,\n",
+"T1=T1+460; //deg R\n",
+"// absolute upstream pressure,\n",
+"p1=p1+14.7; //psia\n",
+"// for maximum flow-rate,\n",
+"// absolute downstream pressure,\n",
+"p2=0.53*p1; //psia\n",
+"// volume flow-rate,\n",
+"Q=floor(22.7*Cv*sqrt(((p1-p2)*p2)/T1)); //scfm\n",
+"// Results:\n",
+"printf('\n  Results:  ')   \n",
+"printf('\n The maximum flow-rate is %.0f scfm of air.',Q)"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 13.11_a: determine_flow_capacity_constant_of_system.sci"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"// Aim:To determine size valve (Cv) for pneumatically powered impact tool\n",
+"// Given:\n",
+"// volume flow-rate of air:\n",
+"Q=50; //scfm\n",
+"// downstream pressure:\n",
+"p2=100; //psi\n",
+"// pressure drop across valve:\n",
+"del_p=12; //psi\n",
+"// upstream air temperature:\n",
+"T1=80; //deg F"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 13.11_b: SOLUTION_flow_capacity_constant_of_system.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"clc;\n",
+"pathname=get_absolute_file_path('13_11_soln.sce')\n",
+"filename=pathname+filesep()+'13_11_data.sci'\n",
+"exec(filename)\n",
+"// Solution:\n",
+"// upstream temperature in Rankine,\n",
+"T1=T1+460; //deg R\n",
+"// absolute downstream pressure,\n",
+"p2=p2+14.7; //psia\n",
+"// flow capacity constant,\n",
+"Cv=(Q/22.7)*sqrt(T1/(p2*del_p));\n",
+"// Results:\n",
+"printf('\n  Results:  ')   \n",
+"printf('\n The flow capacity constant is %.2f.',Cv)"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 13.12_a: determine_air_consumption_rate_in_scfm.sci"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"// Aim:To determine the air-consumption rate in scfm\n",
+"// Given:\n",
+"// piston diameter of pneumatic cylinder:\n",
+"d=1.75; //in\n",
+"// stroke length of cylinder:\n",
+"L=6; //in\n",
+"// number of cycles per minute:\n",
+"N=30; //cycles/min\n",
+"// atmospheric temperature:\n",
+"T1=68; //deg F\n",
+"// atmospheric pressure:\n",
+"p1=14.7; //psia\n",
+"// temperature of air in pneumatic cylinder:\n",
+"T2=80; //deg F\n",
+"// pneumatic cylinder pressure:\n",
+"p2=100; //psig"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 13.12_b: SOLUTION_air_consumption_rate_in_scfm.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"clc;\n",
+"pathname=get_absolute_file_path('13_12_soln.sce')\n",
+"filename=pathname+filesep()+'13_12_data.sci'\n",
+"exec(filename)\n",
+"// Solution:\n",
+"// atmospheric temperature in deg Rankine,\n",
+"T1=T1+460; //deg R\n",
+"// temperature of air in deg Rankine in pneumatic cylinder,\n",
+"T2=T2+460; //deg R\n",
+"// absolute pneumatic cylinder pressure,\n",
+"p2=p2+14.7; //psia\n",
+"// the volume per minute of air consumed by cylinder,\n",
+"Q2=(%pi/4)*(d/12)^2*(L/12)*N; //ft^3/min\n",
+"// air consumption rate,\n",
+"Q1=Q2*(p2/p1)*(T1/T2); //scfm\n",
+"// Results:\n",
+"printf('\n  Results:  ')   \n",
+"printf('\n The air consumption rate in scfm is %.2f.',Q1)"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 13.13_a: find_reciprocation_rate_of_pneumatic_cylinder.sci"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"// Aim:To determine the piston reciprocation rate\n",
+"// Given:\n",
+"// piston diameter of pneumatic cylinder:\n",
+"d=44.5; //mm\n",
+"// stroke length of cylinder:\n",
+"L=152; //mm\n",
+"// atmospheric temperature:\n",
+"T1=20; //deg C\n",
+"// atmospheric pressure:\n",
+"p1=101; //kPa\n",
+"// temperature of air in pneumatic cylinder:\n",
+"T2=27; //deg C\n",
+"// pneumatic cylinder pressure:\n",
+"p2=687; //kPa\n",
+"// air consumption rate:\n",
+"Q1=0.0555; //m^3/min"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 13.13_b: SOLUTION_reciprocation_rate_of_pneumatic_cylinder.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"clc;\n",
+"pathname=get_absolute_file_path('13_13_soln.sce')\n",
+"filename=pathname+filesep()+'13_13_data.sci'\n",
+"exec(filename)\n",
+"// Solution:\n",
+"// atmospheric temperature in kelvin,\n",
+"T1=T1+273; //K\n",
+"// temperature of air in kelvin in pneumatic cylinder,\n",
+"T2=T2+273; //K\n",
+"// absolute pneumatic cylinder pressure,\n",
+"p2=p2+101; //kPa abs\n",
+"// flow-rate of air consumed by cylinder,\n",
+"Q2=Q1*(p1/p2)*(T2/T1); //m^3/min\n",
+"// reciprocation rate of piston,\n",
+"N=floor(Q2/((%pi/4)*(d/1000)^2*(L/1000))); //cycles/min\n",
+"// Results:\n",
+"printf('\n  Results:  ')   \n",
+"printf('\n The reciprocation rate of piston is %.0f cycles/min.',N)"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 13.1_a: find_final_pressure_at_constant_temperature.sci"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"// Aim:To find new pressure in cylinder when its blank end is blocked\n",
+"// Given:\n",
+"// diameter of pneumatic piston:\n",
+"D=2; //in\n",
+"// length of retraction of piston:\n",
+"l_ret=4; //in\n",
+"// blank side pressure:\n",
+"p1=20; //psig\n",
+"// volume of cylinder for extension stroke:\n",
+"V1=20; //in^3"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 13.1_b: SOLUTION_final_pressure_at_constant_temperature.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"clc;\n",
+"pathname=get_absolute_file_path('13_1_soln.sce')\n",
+"filename=pathname+filesep()+'13_1_data.sci'\n",
+"exec(filename)\n",
+"// Solution:\n",
+"// volume of cylinder during retraction stroke,\n",
+"V2=(V1-((%pi * D^2 * l_ret)/4)); //in^3\n",
+"// absolute pressure on blank side,\n",
+"p1=p1+14.7; //psia\n",
+"// new pressure when blank side port is blocked,\n",
+"// Boyle's Law,\n",
+"p2=(p1*V1)/V2; //psia\n",
+"p2=p2-14.7; //psig\n",
+"// Results:\n",
+"printf('\n  Results:  ')   \n",
+"printf('\n The new pressure when blank side port is blocked is %.1f psig.',p2)"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 13.2_a: find_final_volume_at_constant_pressure.sci"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"// Aim:To find new volume of air at the blank end of cylinder\n",
+"// Given:\n",
+"// initial volume:\n",
+"V1=20; //in^3\n",
+"// constant load:\n",
+"p=20; //psi\n",
+"// initial temperature of air:\n",
+"T1=60; //deg F\n",
+"// final temperature of air:\n",
+"T2=120; //degF"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 13.2_b: SOLUTION_final_volume_at_constant_pressure.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"clc;\n",
+"pathname=get_absolute_file_path('13_2_soln.sce')\n",
+"filename=pathname+filesep()+'13_2_data.sci'\n",
+"exec(filename)\n",
+"// Solution:\n",
+"// initial temperature of air in Rankine,\n",
+"T1=T1+460; //deg R\n",
+"// final temperature of air in Rankine,\n",
+"T2=T2+460; //deg R\n",
+"// final volume of air,\n",
+"// Charle's Law,\n",
+"V2=(T2/T1)*V1; //in^3\n",
+"// Results:\n",
+"printf('\n  Results:  ')   \n",
+"printf('\n The final volume of air is %.1f in^3.',V2)"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 13.3_a: find_final_pressure_at_constant_volume.sci"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"// Aim:To find new pressure in cylinder when it is at locked position\n",
+"// Given:\n",
+"// initial pressure:\n",
+"p1=20; //psig\n",
+"// initial temperature of air:\n",
+"T1=60; //deg F\n",
+"// final temperature of air:\n",
+"T2=160; //degF"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 13.3_b: SOLUTION_final_pressure_at_constant_volume.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"clc;\n",
+"pathname=get_absolute_file_path('13_3_soln.sce')\n",
+"filename=pathname+filesep()+'13_3_data.sci'\n",
+"exec(filename)\n",
+"// Solution:\n",
+"// initial temperature of air in Rankine,\n",
+"T1=T1+460; //deg R\n",
+"// final temperature of air in Rankine,\n",
+"T2=T2+460; //deg R\n",
+"// absolute initial pressure,\n",
+"p1=p1+14.7; //psia\n",
+"// final pressure of air,\n",
+"// Gay-Lussac's Law,\n",
+"p2=(T2/T1)*p1; //psia\n",
+"p2=p2-14.7; //psig\n",
+"// Results:\n",
+"printf('\n  Results:  ')   \n",
+"printf('\n The final pressure of air at constant volume is %.1f psig.',p2)"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 13.4_a: find_final_pressure_general_gas_law.sci"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"// Aim:To find final pressure in the cylinder\n",
+"// Given:\n",
+"// initial gas pressure:\n",
+"p1=1000; //psig\n",
+"// initial volume of cylinder:\n",
+"V1=2000; //in^3\n",
+"// initial temperature of cylinder:\n",
+"T1=100; //deg F\n",
+"// final volume of cylinder:\n",
+"V2=1500; //in^3\n",
+"// final temperature of cylinder:\n",
+"T2=200; //deg F"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 13.4_b: SOLUTION_final_pressure_general_gas_law.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"clc;\n",
+"pathname=get_absolute_file_path('13_4_soln.sce')\n",
+"filename=pathname+filesep()+'13_4_data.sci'\n",
+"exec(filename)\n",
+"// Solution:\n",
+"// final pressure in the cylinder,\n",
+"// General Gas Law, \n",
+"p2=((p1+14.7)*V1*(T2+460))/(V2*(T1+460))-14.7; //psig\n",
+"// Results:\n",
+"printf('\n  Results:  ')   \n",
+"printf('\n The final pressure in the cylinder is %.1f psig.',p2)"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 13.5_a: find_final_pressure_general_law_SI.sci"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"// Aim:To find final pressure in the cylinder in SI units\n",
+"// Given:\n",
+"// initial gas pressure:\n",
+"p1=70; //bar\n",
+"// initial volume of cylinder:\n",
+"V1=12900; //cm^3\n",
+"// initial temperature of cylinder:\n",
+"T1=37.8; //deg C\n",
+"// final volume of cylinder:\n",
+"V2=9680; //cm^3\n",
+"// final temperature of cylinder:\n",
+"T2=93.3; //deg C"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 13.5_b: SOLUTION_final_pressure_general_law_SI.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"clc;\n",
+"pathname=get_absolute_file_path('13_5_soln.sce')\n",
+"filename=pathname+filesep()+'13_5_data.sci'\n",
+"exec(filename)\n",
+"// Solution:\n",
+"// final pressure in the cylinder in absolute units,\n",
+"// General Gas Law, \n",
+"p2=(((p1+1)*10^5*V1*(T2+273))/(V2*(T1+273)))/10^5; //bars\n",
+"// Results:\n",
+"printf('\n  Results:  ')   \n",
+"printf('\n The final pressure in the cylinder is %.1f bars absolute.',p2)"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 13.6_a: how_much_air_compressor_must_provide.sci"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"// Aim:To find how many cfm of free air compressor must be provided\n",
+"// Given:\n",
+"// flow-rate of air from receiver:\n",
+"Q2=30; //cfm\n",
+"// temperature of air from receiver:\n",
+"T2=90; //deg F\n",
+"// pressure of air coming from receiver:\n",
+"p2=125; //psig\n",
+"// atmospheric temperature:\n",
+"T1=70; //deg F\n",
+"// atmospheric pressure:\n",
+"p1=14.7; //psig "
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 13.6_b: SOLUTION_air_compressor_must_provide.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"clc;\n",
+"pathname=get_absolute_file_path('13_6_soln.sce')\n",
+"filename=pathname+filesep()+'13_6_data.sci'\n",
+"exec(filename)\n",
+"// Solution:\n",
+"// cfm of free air compressor must be provided,\n",
+"Q1=Q2*((p2+14.7)/14.7)*((T1+460)/(T2+460)); //cfm\n",
+"// Results:\n",
+"printf('\n  Results:  ')   \n",
+"printf('\n The cfm of free air compressor must be provided is %.0f cfm of free air.',Q1)"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 13.7_a: find_receiver_size_for_pneumatic_system.sci"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"// Aim:Refer Example 13-7 for Problem Description\n",
+"// Given:\n",
+"// maximum pressure level in receiver:\n",
+"p_max=100; //psi\n",
+"// minimum pressure level in receiver:\n",
+"p_min=80; //psi\n",
+"// time that receiver can supply required amount of air:\n",
+"t=6; //min\n",
+"// consumption rate of pneumatic system:\n",
+"Qr=20; //scfm\n",
+"// output flow-rate of compressor:\n",
+"Qc=5; //scfm"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 13.7_b: SOLUTION_receiver_size_for_pneumatic_system.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"clc;\n",
+"pathname=get_absolute_file_path('13_7_soln.sce')\n",
+"filename=pathname+filesep()+'13_7_data.sci'\n",
+"exec(filename)\n",
+"// Solution:\n",
+"// required size of a receiver before compressor resumes operation,\n",
+"Vr=((14.7*t*(Qr-0))/(p_max-p_min))*7.48; //gal\n",
+"// required size of a receiver when compressor is running,\n",
+"Vr_run=((14.7*t*(Qr-Qc))/(p_max-p_min))*7.48; //gal\n",
+"// Results:\n",
+"printf('\n  Results:  ')   \n",
+"printf('\n The required size of a receiver before compressor resumes operation is %.0f gal.',Vr)\n",
+"printf('\n The required size of a receiver when compressor is running %.0f gal.',Vr_run)"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 13.8_a: determine_actual_power_required_for_compressor.sci"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"// Aim:To determine actual power required to drive a compressor\n",
+"// Given:\n",
+"// input flow-rate of air through compressor:\n",
+"Q=100; //scfm\n",
+"// inlet atmospheric pressure:\n",
+"p_in=14.7; //psig\n",
+"// outlet pressure:\n",
+"p_out=114.7; //psig\n",
+"// overall efficiency of compressor:\n",
+"eff=75; //%"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 13.8_b: SOLUTION_actual_power_required_for_compressor.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"clc;\n",
+"pathname=get_absolute_file_path('13_8_soln.sce')\n",
+"filename=pathname+filesep()+'13_8_data.sci'\n",
+"exec(filename)\n",
+"// Solution:\n",
+"// theoretical horsepower,\n",
+"HP_theo=((p_in*Q)/65.4)*((p_out/p_in)^0.286-1); //HP\n",
+"// actual horsepower,\n",
+"HP_act=HP_theo/(eff/100); //HP\n",
+"// Results:\n",
+"printf('\n  Results:  ')   \n",
+"printf('\n The actual power required to drive a compressor is %.0f HP.',HP_act)"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 13.9_a: find_moisture_received_by_pneumatic_system.sci"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"// Aim:Refer Example 13-9 for Problem Description\n",
+"// Given:\n",
+"// output flow-rate of compressor:\n",
+"Qc=100; //scfm\n",
+"// pressure at compressor outlet:\n",
+"p_out=100; //psig\n",
+"// temperature of saturated air at compressor inlet:\n",
+"T_in=80; //deg F\n",
+"// operation time of compressor per day:\n",
+"t=8; //hr/day"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 13.9_b: SOLUTION_moisture_received_by_pneumatic_system.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"clc;\n",
+"pathname=get_absolute_file_path('13_9_soln.sce')\n",
+"filename=pathname+filesep()+'13_9_data.sci'\n",
+"exec(filename)\n",
+"// Solution:\n",
+"// from fig 13-29,\n",
+"// entering moistue content at 80 deg F,\n",
+"moist_in=1.58/1000; //lb/ft^3\n",
+"// moisture rate which enters the compressor,\n",
+"moist_rate=moist_in*Qc; //lb/min\n",
+"// number of gallons/day received by pneumatic system,\n",
+"gal_per_day=(moist_rate*60*t)/8.34; //gal/day\n",
+"// moisture received by pneumatic system if aftercooler is installed,\n",
+"// from fig 13-29,\n",
+"moist_after=(1-((1.58-0.2)/1.58))*gal_per_day; //gal/day\n",
+"// moisture received by pneumatic system if air dryer is installed,\n",
+"// from fig 13-29,\n",
+"moist_air_dryer=(1-((1.58-0.05)/1.58))*gal_per_day; //gal/day\n",
+"// Results:\n",
+"printf('\n  Results:  ')   \n",
+"printf('\n The number of gallons/day received by pneumatic system is %.2f gal/day.',gal_per_day)\n",
+"printf('\n The moisture received by pneumatic system if aftercooler is installed is %.2f gal/day.',moist_after)\n",
+"printf('\n The moisture received by pneumatic system if air dryer is installed is %.2f gal/day.',moist_air_dryer)"
+   ]
+   }
+],
+"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/Fluid_Power_With_Applications_by_A_Esposito/14-PNEUMATICS_CIRCUITS_AND_APPLICATIONS.ipynb b/Fluid_Power_With_Applications_by_A_Esposito/14-PNEUMATICS_CIRCUITS_AND_APPLICATIONS.ipynb
new file mode 100644
index 0000000..dea66f3
--- /dev/null
+++ b/Fluid_Power_With_Applications_by_A_Esposito/14-PNEUMATICS_CIRCUITS_AND_APPLICATIONS.ipynb
@@ -0,0 +1,722 @@
+{
+"cells": [
+ {
+		   "cell_type": "markdown",
+	   "metadata": {},
+	   "source": [
+       "# Chapter 14: PNEUMATICS CIRCUITS AND APPLICATIONS"
+	   ]
+	},
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 14.10_a: calculate_required_size_of_accumulator_SI.sci"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"// Aim:Refer Example 14-7 for Problem Description\n",
+"// Given:\n",
+"// diamter of hydraulic cylinder:\n",
+"D=152; //mm\n",
+"// cylinder extension:\n",
+"L=2.54; //m\n",
+"// duration of cylinder extension:\n",
+"t=10; //s\n",
+"// time between crushing stroke:\n",
+"t_crush=5; //min\n",
+"// gas precharge pressure:\n",
+"p1=84; //bars abs\n",
+"// gas charge pressure when pump is turned on:\n",
+"p2=210; //bars abs\n",
+"// minimum pressure required to actuate load:\n",
+"p3=126; //bars abs"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 14.10_b: SOLUTION_required_size_of_accumulator_SI.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"clc;\n",
+"pathname=get_absolute_file_path('14_10_soln.sce')\n",
+"filename=pathname+filesep()+'14_10_data.sci'\n",
+"exec(filename)\n",
+"// Solutions:\n",
+"// volume of hydraulic cylinder,\n",
+"V=(%pi/4)*L*((D/1000)^2); //m^3\n",
+"// volume of cylinder in charged position,\n",
+"V2=V/((p2/p3)-1); //m^3\n",
+"// volume of cylinder in final position,\n",
+"V3=(p2/p3)*V2; //m^3\n",
+"// required size of accumulator,\n",
+"V1=floor(((p2*V2)/p1)*1000); //L\n",
+"// pump flow-rate with accumulator,\n",
+"Q_pump_acc=(2*V*1000)/(t_crush*60); //L/s\n",
+"// pump hydraulic power with accumulator,\n",
+"kW_pump_acc=(Q_pump_acc*10^-3*p2*10^5)/1000; //kW\n",
+"// pump flow-rate without accumulator,\n",
+"Q_pump_no_acc=V/t; //L/s\n",
+"// pump hydraulic power without accumulator,\n",
+"kW_pump_no_acc=(Q_pump_no_acc*10^-3*p3*10^5); //kW\n",
+"// Results:\n",
+"printf('\n  Results:  ')\n",
+"printf('\n The required size of accumulator is %.0f L.',V1)\n",
+"printf('\n The pump hydraulic horsepower with accumulator is %.2f kW.',kW_pump_acc)\n",
+"printf('\n The pump hydraulic horsepower without accumulator is %.1f kW.',kW_pump_no_acc)"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 14.1_a: find_pressure_loss_for_given_pipe.sci"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"// Aim:To find pressure loss for a 250 ft length of pipe\n",
+"// Given:\n",
+"// flow-rate:\n",
+"Q=100; //scfm\n",
+"// receiver pressure:\n",
+"p2=150; //psi\n",
+"// atmospheric pressure:\n",
+"p1=14.7; //psi\n",
+"// length of pipe:\n",
+"L=250; //ft"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 14.1_b: SOLUTION_pressure_loss_for_given_pipe.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"clc;\n",
+"pathname=get_absolute_file_path('14_1_soln.sce')\n",
+"filename=pathname+filesep()+'14_1_data.sci'\n",
+"exec(filename)\n",
+"// Solution:\n",
+"// compression ratio,\n",
+"CR=(p2+p1)/p1;\n",
+"// from fig 14-3,\n",
+"// inside diameter raised to 5.31,\n",
+"k=1.2892; //in\n",
+"// experimentally determined coefficient,\n",
+"c=0.1025/(1)^0.31;\n",
+"// pressure loss,\n",
+"p_f=(c*L*Q^2)/(3600*CR*k); //psi\n",
+"// Results:\n",
+"printf('\n  Results:  ')   \n",
+"printf('\n The pressure loss for a 250 ft length of pipe is %.2f psi.',p_f)"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 14.2_a: find_pressure_loss_with_pipe_valves.sci"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"// Aim:To find pressure loss for a  pipe with valves\n",
+"// Given:\n",
+"// experimentally determined coefficient:\n",
+"c=0.1025;\n",
+"// compression ratio:\n",
+"CR=11.2;\n",
+"// receiver pressure:\n",
+"p2=150; //psi\n",
+"// atmospheric pressure:\n",
+"p1=14.7; //psi\n",
+"// length of pipe:\n",
+"L=250; //ft"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 14.2_b: SOLUTION_pressure_loss_with_pipe_valves.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"clc;\n",
+"pathname=get_absolute_file_path('14_2_soln.sce')\n",
+"filename=pathname+filesep()+'14_2_data.sci'\n",
+"exec(filename)\n",
+"// Solution:\n",
+"// from fig 14-3,\n",
+"// inside diameter raised to 5.31,\n",
+"k=1.2892; //in\n",
+"// length of pipe along with valves,\n",
+"L=L+(2*0.56)+(3*29.4)+(5*1.5)+(4*2.6)+(6*1.23); //ft\n",
+"// pressure loss,\n",
+"p_f=(c*L*Q^2)/(3600*CR*k); //psi\n",
+"// Results:\n",
+"printf('\n  Results:  ')   \n",
+"printf('\n The pressure loss for a 250 ft length of pipe is %.2f psi.',p_f)"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 14.3_a: determine_cost_of_electricity_per_year.sci"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"// Aim:Refer Example 14-3 for Problem Description\n",
+"// Given:\n",
+"// air flow-rate:\n",
+"Q=270; //scfm\n",
+"// pressure at which compressor delivers air:\n",
+"p_out=100; //psig\n",
+"// overall efficiency of compressor:\n",
+"eff_o=75; //%\n",
+"// pressure at which compressor delivers air taking friction in account:\n",
+"p_out1=115; //psig\n",
+"// efficiency of electric motor driving compressor:\n",
+"eff_mot=92; //%\n",
+"// operating time of compressor:\n",
+"t=3000; //hr/year  \n",
+"// cost of electricity per watt:\n",
+"cost_per_wat=0.11; //$/kWh"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 14.3_b: SOLUTION_cost_of_electricity_per_year.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"clc;\n",
+"pathname=get_absolute_file_path('14_3_soln.sce')\n",
+"filename=pathname+filesep()+'14_3_data.sci'\n",
+"exec(filename)\n",
+"// Solutions:\n",
+"// inlet pressure,\n",
+"p_in=14.7; //psi\n",
+"// actual horsepower at 100 psig,\n",
+"act_HP=((p_in*Q)/(65.4*(eff_o/100)))*(((p_out+14.7)/p_in)^0.286-1); //HP\n",
+"// actual horsepower at 115 psig,\n",
+"act_HP1=((p_in*Q)/(65.4*(eff_o/100)))*(((p_out1+14.7)/p_in)^0.286-1); //HP\n",
+"// actual power at 100 psig in kW,\n",
+"act_kW=act_HP*0.746; //kW\n",
+"// electric power required to drive electric motor at 100 psig,\n",
+"elect_kW=act_kW/(eff_mot/100); //kW\n",
+"// cost of electricity per year at 100 psig,\n",
+"yearly_cost=elect_kW*t*cost_per_wat; //$/yr\n",
+"// actual power at 115 psig in kW,\n",
+"act_kW1=act_HP1*0.746; //kW\n",
+"// electric power required to drive electric motor at 115 psig,\n",
+"elect_kW1=act_kW1/(eff_mot/100); //kW\n",
+"// cost of electricity per year at 115 psig,\n",
+"yearly_cost1=elect_kW1*t*cost_per_wat; //$/yr\n",
+"// Results:\n",
+"printf('\n  Results:  ')\n",
+"printf('\n The actual HP required to drive the compressor at 100 psig is %.1f HP.',act_HP)\n",
+"printf('\n The actual HP required to drive the compressor at 115 psig is %.1f HP.',act_HP1)\n",
+"printf('\n The cost of electricity per year at 100 psig is %.0f $.',yearly_cost)\n",
+"printf('\n The cost of electricity per year at 115 psig is %.0f $.',yearly_cost1)\n",
+"printf('\n The answer in the program does not match with that in the textbook due to roundoff error (standard electric ratings)')"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 14.4_a: determine_cost_of_leakage_per_year.sci"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"// Aim:To determine the yearly cost of leakage of pneumatic system\n",
+"// Given:\n",
+"// air flow-rate:\n",
+"Q=270; //scfm\n",
+"// air flow-rate leakage:\n",
+"Q_leak=70; //scfm\n",
+"// // electric power required to drive electric motor at 100 psig:\n",
+"elect_kW=52.3; //kW\n",
+"// cost of electricity per watt:\n",
+"cost_per_wat=0.11; //$/kWh"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 14.4_b: SOLUTION_cost_of_leakage_per_year.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"clc;\n",
+"pathname=get_absolute_file_path('14_4_soln.sce')\n",
+"filename=pathname+filesep()+'14_4_data.sci'\n",
+"exec(filename)\n",
+"// Solutions:\n",
+"// electric power required to compensate for leakage,\n",
+"power_rate=(Q_leak/Q)*elect_kW; //kW\n",
+"// rounding off the above answer\n",
+"power_rate=fix(power_rate)+(fix(round((power_rate-fix(power_rate))*10))/10); //kW\n",
+"// cost of electricity per year at 100 psig,\n",
+"yearly_leak=power_rate*24*365*cost_per_wat; //$/yr\n",
+"// Results:\n",
+"printf('\n  Results:  ')\n",
+"printf('\n The cost of electricity for leakage per year at 100 psig is %.0f $.',yearly_leak)\n",
+"printf('\n The answer in the program does not match with that in the textbook due to roundoff error(standard electric ratings)')"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 14.5_a: how_heavy_object_can_be_lifted.sci"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"// Aim:Refer Example 14-5 for Problem Description\n",
+"// Given:\n",
+"// diameter of suction cup lip outer circle:\n",
+"Do=6; //in\n",
+"// diameter of suction cup inner lip circle:\n",
+"Di=5; //in\n",
+"// atmospheric pressure:\n",
+"p_atm=14.7; //psi\n",
+"// suction pressure:\n",
+"p_suc=-10; //psi"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 14.5_b: SOLUTION_heavy_object_can_be_lifted.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"clc;\n",
+"pathname=get_absolute_file_path('14_5_soln.sce')\n",
+"filename=pathname+filesep()+'14_5_data.sci'\n",
+"exec(filename)\n",
+"// Solution:\n",
+"// suction pressure in absolute,\n",
+"p_suc_abs=p_suc+p_atm; //psia\n",
+"// maximum weight that suction cup can lift,\n",
+"F=ceil((p_atm*(%pi/4)*Do^2)-(p_suc_abs*(%pi/4)*Di^2)); //lb\n",
+"// maximum weight suction cup can lift with perfect vaccum,\n",
+"W=p_atm*(%pi/4)*Do^2; //lb\n",
+"// Results:\n",
+"printf('\n  Results:  ')   \n",
+"printf('\n The maximum weight that suction cup can lift is %.0f lb.',F)\n",
+"printf('\n The maximum weight that suction cup can lift with perfect vacuum is %.0f lb.',W)"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 14.6_a: determine_time_for_achieving_vacuum_pressure.sci"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"// Aim:To determine the time required to achieve the desired vacuum pressure\n",
+"// Given:\n",
+"// total volume of space in the suction cup:\n",
+"V=6; //ft^3\n",
+"// flow-rate produced by vacuum pump:\n",
+"Q=4; //scfm\n",
+"// desired suction pressure:\n",
+"p_vacuum=6; //in Hg abs\n",
+"// atmospheric pressure:\n",
+"p_atm=30; //in Hg abs"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 14.6_b: SOLUTION_time_for_achieving_vacuum_pressure.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"clc;\n",
+"pathname=get_absolute_file_path('14_6_soln.sce')\n",
+"filename=pathname+filesep()+'14_6_data.sci'\n",
+"exec(filename)\n",
+"// Solutions:\n",
+"// time required to achieve the desired vacuum pressure,\n",
+"t=(V/Q)*log(p_atm/p_vacuum); //min\n",
+"// time required to achieve perfect vacuum pressure,\n",
+"t1=(V/Q)*log(p_atm/0.5); //min\n",
+"// Results:\n",
+"printf('\n  Results:  ')\n",
+"printf('\n The time required to achieve the desired vacuum pressure is %.2f min.',t)\n",
+"printf('\n The time required to achieve perfect vacuum pressure is %.2f min.',t1)"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 14.7_a: calculate_required_size_of_the_accumulator.sci"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"// Aim:Refer Example 14-7 for Problem Description\n",
+"// Given:\n",
+"// diamter of hydraulic cylinder:\n",
+"D=6; //in\n",
+"// cylinder extension:\n",
+"L=100; //in\n",
+"// duration of cylinder extension:\n",
+"t=10; //s\n",
+"// time between crushing stroke:\n",
+"t_crush=5; //min\n",
+"// gas precharge pressure:\n",
+"p1=1200; //psia\n",
+"// gas charge pressure when pump is turned on:\n",
+"p2=3000; //psia\n",
+"// minimum pressure required to actuate load:\n",
+"p3=1800; //psia"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 14.7_b: SOLUTION_required_size_of_the_accumulator.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"clc;\n",
+"pathname=get_absolute_file_path('14_7_soln.sce')\n",
+"filename=pathname+filesep()+'14_7_data.sci'\n",
+"exec(filename)\n",
+"// Solutions:\n",
+"// volume of hydraulic cylinder,\n",
+"V=(%pi/4)*L*(D^2); //in^3\n",
+"// volume of cylinder in charged position,\n",
+"V2=V/((p2/p3)-1); //in^3\n",
+"// volume of cylinder in final position,\n",
+"V3=(p2/p3)*V2; //in^3\n",
+"// required size of accumulator,\n",
+"V1=((p2*V2)/p1)/231; //gal\n",
+"// rounding off the above answer,\n",
+"V1=fix(V1)+(fix(floor((V1-fix(V1))*10))/10); //gal\n",
+"// pump flow-rate with accumulator,\n",
+"Q_pump_acc=((2*V)/231)/t_crush; //gpm\n",
+"// rounding off the above answer\n",
+"Q_pump_acc=fix(Q_pump_acc)+(fix(ceil((Q_pump_acc-fix(Q_pump_acc))*100))/100); //gpm\n",
+"// pump hydraulic power with accumulator,\n",
+"HP_pump_acc=(Q_pump_acc*p2)/1714; //HP\n",
+"// pump flow-rate without accumulator,\n",
+"Q_pump_no_acc=(V/231)/(t/60); //gpm\n",
+"// pump hydraulic power without accumulator,\n",
+"HP_pump_no_acc=(Q_pump_no_acc*p3)/1714; //HP\n",
+"// Results:\n",
+"printf('\n  Results:  ')\n",
+"printf('\n The required size of accumulator is %.1f gal.',V1)\n",
+"printf('\n The pump hydraulic horsepower with accumulator is %.2f HP.',HP_pump_acc)\n",
+"printf('\n The pump hydraulic horsepower without accumulator is %.1f HP.',HP_pump_no_acc)\n",
+"printf('\n The answer in the program is different than that in textbook. It may be due to no.s of significant digit in data and calculation')"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 14.8_a: find_electricity_cost_per_year_SI.sci"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"// Aim:To calculate the cost of electricity per year in Metric Unit\n",
+"// Given:\n",
+"// air flow-rate:\n",
+"Q=7.65; //m^3/min\n",
+"// pressure at which compressor delivers air:\n",
+"p_out=687; //kPa gage\n",
+"// efficiency of compressor:\n",
+"eff_o=75; //%\n",
+"// efficiency of electric motor driving compressor:\n",
+"eff_mot=92; //%\n",
+"// operating time of compressor per year:\n",
+"t=3000; //hr  \n",
+"// cost of electricity:\n",
+"cost_per_wat=0.11; //$/kWh"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 14.8_b: SOLUTION_electricity_cost_per_year_SI.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"clc;\n",
+"pathname=get_absolute_file_path('14_8_soln.sce')\n",
+"filename=pathname+filesep()+'14_8_data.sci'\n",
+"exec(filename)\n",
+"// Solutions:\n",
+"// inlet pressure,\n",
+"p_in=101; //kPa\n",
+"// actual power,\n",
+"act_kW=((p_in*Q)/(17.1*(eff_o/100)))*(((p_out+101)/p_in)^0.286-1); //kW\n",
+"// electric power required to drive electric motor,\n",
+"elect_kW=act_kW/(eff_mot/100); //kW\n",
+"// rounding off the above answer\n",
+"elect_kW=fix(elect_kW)+(fix(round((elect_kW-fix(elect_kW))*10))/10); //kW\n",
+"// cost of electricity,\n",
+"yearly_cost=elect_kW*t*cost_per_wat; //$/yr\n",
+"// Results:\n",
+"printf('\n  Results:  ')\n",
+"printf('\n The cost of electricity per year is %.0f $.',yearly_cost)\n",
+"printf('\n The answer in the program does not match with that in the textbook due to roundoff error (standard electric ratings)')"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 14.9_a: what_flowrate_vacuum_pump_must_deliver.sci"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"// Aim:To find the flow-rate to be delivered by vacuum pump\n",
+"// Given:\n",
+"// lip outside diameter of suction cup:\n",
+"Do=100; //mm\n",
+"// lip inside diameter of suction cup:\n",
+"Di=80; //mm\n",
+"// weight of steel sheets:\n",
+"F=1000; //N\n",
+"// numbers of suction cups:\n",
+"N=4; \n",
+"// total volume of space inside the suction cup:\n",
+"V=0.15; //m^3\n",
+"// factor of safety:\n",
+"f=2;\n",
+"// time required to produce desired vacuum pressure:\n",
+"t=1; //min"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 14.9_b: SOLUTION_flowrate_vacuum_pump_must_deliver.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"clc;\n",
+"pathname=get_absolute_file_path('14_9_soln.sce')\n",
+"filename=pathname+filesep()+'14_9_data.sci'\n",
+"exec(filename)\n",
+"// Solutions:\n",
+"// atmospheric pressure,\n",
+"p_atm=101000; //Pa\n",
+"// lip outside area of suction cup,\n",
+"Ao=(%pi/4)*(Do/1000)^2; //m^2\n",
+"// lip inside area of suction cup,\n",
+"Ai=(%pi/4)*(Di/1000)^2; //m^2\n",
+"// required vacuum pressure,\n",
+"p=((p_atm*Ao)-((F*f)/N))/Ai; //Pa abs\n",
+"// flow-rate to be delivered by vacuum pump,\n",
+"Q=(V/t)*log(p_atm/p); //m^3/min\n",
+"// rounding off the above answer\n",
+"Q=fix(Q)+(fix(ceil((Q-fix(Q))*10000))/10000); //m^3/min\n",
+"// Results:\n",
+"printf('\n  Results:  ')\n",
+"printf('\n The flow-rate of air to be delivered by vacuum pump is %.4f m^3/min.',Q)"
+   ]
+   }
+],
+"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/Fluid_Power_With_Applications_by_A_Esposito/17-ADVANCED_ELECTRICAL_CONTROLS_FOR_FLUID_POWER_SYSTEMS.ipynb b/Fluid_Power_With_Applications_by_A_Esposito/17-ADVANCED_ELECTRICAL_CONTROLS_FOR_FLUID_POWER_SYSTEMS.ipynb
new file mode 100644
index 0000000..3f1ae72
--- /dev/null
+++ b/Fluid_Power_With_Applications_by_A_Esposito/17-ADVANCED_ELECTRICAL_CONTROLS_FOR_FLUID_POWER_SYSTEMS.ipynb
@@ -0,0 +1,234 @@
+{
+"cells": [
+ {
+		   "cell_type": "markdown",
+	   "metadata": {},
+	   "source": [
+       "# Chapter 17: ADVANCED ELECTRICAL CONTROLS FOR FLUID POWER SYSTEMS"
+	   ]
+	},
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 17.1_a: determine_system_accuracy_of_electrohydraulic_system.sci"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"// Aim:To determine the system accuracy of electrohydraulic servo system\n",
+"// Given:\n",
+"// servo valve gain:\n",
+"G_SV=0.15; //(in^3/s)/mA\n",
+"// cylinder gain:\n",
+"G_cyl=0.20; //in/in^3\n",
+"// feedback transducer gain:\n",
+"H=4; //V/in\n",
+"// weight of load:\n",
+"W=1000; //lb\n",
+"// mass of load:\n",
+"M=2.59; //lb.(s^2)/in\n",
+"// volume of oil under compression:\n",
+"V=50; //in^3\n",
+"// system deadband:\n",
+"SD=4; //mA\n",
+"// bulk modulus of oil:\n",
+"beta1=175000; //lb/in^2\n",
+"// cylinder piston area:\n",
+"A=5; //in^2"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 17.1_b: SOLUTION_system_accuracy_of_electrohydraulic_system.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"clc;\n",
+"pathname=get_absolute_file_path('17_1_soln.sce')\n",
+"filename=pathname+filesep()+'17_1_data.sci'\n",
+"exec(filename)\n",
+"// Solutions:\n",
+"// natural frequency of the oil,\n",
+"om_H=A*sqrt((2*beta1)/(V*M)); //rad/s\n",
+"// value of open-loop gain,\n",
+"open_loop=om_H/3; ///s\n",
+"// amplifier gain,\n",
+"G_A=open_loop/(G_SV*G_cyl*H); //mA/V\n",
+"// repeatable error,\n",
+"RE=SD/(G_A*H); //in\n",
+"// Results:\n",
+"printf('\n  Results:  ')\n",
+"printf('\n The repeatable error of system is %.5f in.',RE)"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 17.2_a: determine_system_accuracy_in_SI.sci"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"// Aim:To determine the system accuracy of in SI units\n",
+"// Given:\n",
+"// servo valve gain:\n",
+"G_SV=2.46; //(cm^3/s)/mA\n",
+"// cylinder gain:\n",
+"G_cyl=0.031; //cm/cm^3\n",
+"// feedback transducer gain:\n",
+"H=4; //V/cm\n",
+"// mass of load:\n",
+"M=450; //kg\n",
+"// volume of oil:\n",
+"V=819; //cm^3\n",
+"// system deadband:\n",
+"SD=4; //mA\n",
+"// bulk modulus of oil:\n",
+"beta1=1200; //MPa\n",
+"// cylinder piston area:\n",
+"A=32.3; //cm^2"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 17.2_b: SOLUTION_system_accuracy_in_SI.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"clc;\n",
+"pathname=get_absolute_file_path('17_2_soln.sce')\n",
+"filename=pathname+filesep()+'17_2_data.sci'\n",
+"exec(filename)\n",
+"// Solutions:\n",
+"// natural frequency of the oil,\n",
+"om_H=(A*10^-4)*sqrt((2*beta1*10^6)/(V*10^-6*M)); //rad/s\n",
+"// value of open-loop gain,\n",
+"open_loop=om_H/3; ///s\n",
+"// amplifier gain,\n",
+"G_A=open_loop/(G_SV*G_cyl*H); //mA/V\n",
+"// repeatable error,\n",
+"RE=SD/(G_A*H); //cm\n",
+"// rounding off the above answer,\n",
+"RE=fix(RE)+(fix(ceil((RE-fix(RE))*100000))/100000); //cm\n",
+"// Results:\n",
+"printf('\n  Results:  ')\n",
+"printf('\n The repeatable error of system is %.5f cm.',RE)"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 17.3_a: find_maximum_tracking_error.sci"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"// Aim:Refer Example 14-3 for Problem Description\n",
+"// Given:\n",
+"// servo valve current saturation:\n",
+"I=300; //mA\n",
+"// amplifier gain:\n",
+"G_A=724; //mA/V\n",
+"// feedback transducer gain:\n",
+"H=4; //V/in\n",
+"// feedback transducer gain in metric units\n",
+"H1=1.57; //V/cm"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 17.3_b: SOLUTION_maximum_tracking_error.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"clc;\n",
+"pathname=get_absolute_file_path('17_3_soln.sce')\n",
+"filename=pathname+filesep()+'17_3_data.sci'\n",
+"exec(filename)\n",
+"// Solutions:\n",
+"// tracking error,\n",
+"TE=I/(G_A*H); //in\n",
+"// tracking error,\n",
+"TE1=I/(G_A*H1); //cm\n",
+"// Results:\n",
+"printf('\n  Results:  ')\n",
+"printf('\n The tracking error of system is %.3f in.',TE)\n",
+"printf('\n The tracking error of system in SI Unit is %.3f cm.',TE1)"
+   ]
+   }
+],
+"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/Fluid_Power_With_Applications_by_A_Esposito/2-PHYSICAL_PROPERTIES_OF_HYDRAULIC_FLUIDS.ipynb b/Fluid_Power_With_Applications_by_A_Esposito/2-PHYSICAL_PROPERTIES_OF_HYDRAULIC_FLUIDS.ipynb
new file mode 100644
index 0000000..c277ed5
--- /dev/null
+++ b/Fluid_Power_With_Applications_by_A_Esposito/2-PHYSICAL_PROPERTIES_OF_HYDRAULIC_FLUIDS.ipynb
@@ -0,0 +1,1028 @@
+{
+"cells": [
+ {
+		   "cell_type": "markdown",
+	   "metadata": {},
+	   "source": [
+       "# Chapter 2: PHYSICAL PROPERTIES OF HYDRAULIC FLUIDS"
+	   ]
+	},
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 2.10_a: when_fahrenheit_and_celsius_temperature_equals.sci"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"// Aim:To find Temperature at which Fahrenheit and Celsius values are equal \n",
+"// Given:\n",
+"// T(degF) = T(degC)  //Eqn - 1"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 2.10_b: SOLUTION_fahrenheit_and_celsius_temp_equals.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"clc;\n",
+"pathname=get_absolute_file_path('2_10_soln.sce')\n",
+"filename=pathname+filesep()+'2_10_data.sci'\n",
+"exec(filename)\n",
+"// Solution:\n",
+"// We know that,\n",
+"// T(degF)=((1.8*T(degC))+32)  //Eqn - 2         \n",
+"// From Eqn 1 and 2\n",
+"// ((1.8*T(degC))+32)= T(degC)\n",
+"// (1-1.8)*T(degC)=32\n",
+"// -0.8*T(degC)=32\n",
+"TdegC=-32/0.8;\n",
+"// Results:\n",
+"printf('\n  Results:  ')\n",
+"printf('\n The temp at which Fahrenheit and Celsius values are equal is %0.1f deg.',TdegC)"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 2.11_a: find_change_in_volume_of_oil.sci"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"// Aim:To find change in volume of the oil\n",
+"// Given:\n",
+"// Volume of original oil:\n",
+"V=10; //in^3\n",
+"// Initial Pressure:\n",
+"P1=100; //psi\n",
+"// Final pressure:\n",
+"P2=2000; //psi\n",
+"// Bulk Modullus:\n",
+"betaa=250000; //psi"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 2.11_b: SOLUTION_change_in_volume_of_oil.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"clc;\n",
+"pathname=get_absolute_file_path('2_11_soln.sce')\n",
+"filename=pathname+filesep()+'2_11_data.sci'\n",
+"exec(filename)\n",
+"// Solution:\n",
+"// Change in pressure,\n",
+"delP=P2-P1; //psi\n",
+"// Change in volume,\n",
+"delV=-((V*delP)/betaa); //in^3 ,- sign indicates oil is being compressed\n",
+"// Results:\n",
+"printf('\n  Results:  ')\n",
+"printf('\n The change in volume of oil is %.3f in^3.',delV)"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 2.12_a: find_viscosity_of_oil.sci"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"// Aim:To find viscosity of oil in centistokes and centipoise\n",
+"// Given:\n",
+"// viscosity of oil:\n",
+"nu=230; //SUS at\n",
+"t=150; //degF.\n",
+"// specific gravity of oil:\n",
+"gamma1=0.9;"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 2.12_b: SOLUTION_viscosity_of_oil.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"clc;\n",
+"pathname=get_absolute_file_path('2_12_soln.sce')\n",
+"filename=pathname+filesep()+'2_12_data.sci'\n",
+"exec(filename)\n",
+"// Solution:\n",
+"// kinematic viscosity of oil in centistokes,\n",
+"nu_cs=((0.220*nu)-(135/t)); //centistokes\n",
+"// absolute viscosity of oil in centipoise,\n",
+"mu_cp=(gamma1*nu_cs); //centipoise\n",
+"// Results:\n",
+"printf('\n  Results:  ')\n",
+"printf('\n The viscosity of oil in centistokes is %0.0f cS.',nu_cs)\n",
+"printf('\n The viscosity of oil in centipoise is %0.0f cP.',mu_cp)"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 2.13_a: find_kinematic_and_absolute_viscosities.sci"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"// Aim:To find kinematic and absolute viscosity of oil in cS and cP respectively\n",
+"// Given:\n",
+"// Density of oil:\n",
+"Den=0.89; //g/cm^3\n",
+"// Time flow:\n",
+"t=250; //s\n",
+"// Calibration constant:\n",
+"cc=0.100;"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 2.13_b: SOLUTION_kinematic_and_absolute_viscosities.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"clc;\n",
+"pathname=get_absolute_file_path('2_13_soln.sce')\n",
+"filename=pathname+filesep()+'2_13_data.sci'\n",
+"exec(filename)\n",
+"// Solution:\n",
+"// kinematic viscosity of oil in centistokes,\n",
+"nu_cs=(t*cc); //centistokes\n",
+"// absolute viscosity of oil in centipoise,\n",
+"SG=Den;\n",
+"mu_cp=(SG*nu_cs); //centipoise\n",
+"// rounding off the above answer\n",
+"mu_cp=fix(mu_cp)+(fix(round((mu_cp-fix(mu_cp))*10))/10); //centipoise\n",
+"// Results:\n",
+"printf('\n  Results:  ')\n",
+"printf('\n The viscosity of oil in centistokes is %0.1f cS.',nu_cs)\n",
+"printf('\n The viscosity of oil in centipoise is %0.1f cP.',mu_cp)"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 2.14_a: find_viscosity_of_oil_at_100F.sci"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"// Aim:To find viscosity of oil at 100 degF in SUS\n",
+"// Given:\n",
+"// Viscosity Index:\n",
+"VI=80;\n",
+"// viscosity of O-VI oil at 100 degF:\n",
+"L=400; //SUS\n",
+"// viscosity of 100-VI oil at 100 degF:\n",
+"H=150; //SUS"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 2.14_b: SOLUTION_viscosity_of_oil_at_100F.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"clc;\n",
+"pathname=get_absolute_file_path('2_14_soln.sce')\n",
+"filename=pathname+filesep()+'2_14_data.sci'\n",
+"exec(filename)\n",
+"// Solution:\n",
+"// viscosity of sample oil at 100 degF,\n",
+"U=L-(((L-H)*VI)/100); //SUS\n",
+"// Results:\n",
+"printf('\n  Results:  ')\n",
+"printf('\n The viscosity of sample oil at 100 degF is %0.0f SUS.',U)"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 2.15_a: find_pressure_on_skin_diver_SI.sci"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"// Aim:To find pressure on the skin diver in SI units\n",
+"// Given:\n",
+"// Depth of Water Body:\n",
+"H=18.3; //m"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 2.15_b: SOLUTION_pressure_on_skin_diver_SI.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"clc;\n",
+"pathname=get_absolute_file_path('2_15_soln.sce')\n",
+"filename=pathname+filesep()+'2_15_data.sci'\n",
+"exec(filename)\n",
+"// Solution:\n",
+"// specific Weight of water,\n",
+"gamma1=9800; //N/m^3        \n",
+"// we know pressure,\n",
+"// p=(specific weight of liquid * liquid column height)\n",
+"p=(gamma1*H); //Pa\n",
+"pK=p/1000; //kPa\n",
+"// Results:\n",
+"printf('\n  Results:  ')\n",
+"printf('\n The pressure on skin diver is %.0f kPa.',pK)"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 2.16_a: convert_gage_to_absolute_pressure_SI.sci"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"// Aim:To convert given pressure into absolute pressure\n",
+"// Given:\n",
+"// Gage Pressure:\n",
+"Pg=-34000; //Pa"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 2.16_b: SOLUTION_gage_to_absolute_pressure_SI.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"clc;\n",
+"pathname=get_absolute_file_path('2_16_soln.sce')\n",
+"filename=pathname+filesep()+'2_16_data.sci'\n",
+"exec(filename)\n",
+"// Solution:\n",
+"// Atmospheric Pressure,\n",
+"Po=101000; //Pa           \n",
+"// Absolute Pressure(Pa) =Gage Pressure + Atmospheric Pressure\n",
+"Pa=Pg+Po; //Pa\n",
+"// Results:\n",
+"printf('\n  Results:  ')\n",
+"printf('\n The Absolute Pressure is %0.0f Pa.',Pa)"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 2.17_a: find_oil_volume_change_in_SI.sci"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"// Aim:To find % change in volume of the oil\n",
+"// Given:\n",
+"// Volume of original oil:V=164 //cm^3\n",
+"// Initial Pressure:\n",
+"P1=687; //kPa\n",
+"// Final pressure:\n",
+"P2=13740; //kPa\n",
+"// Bulk Modullus:\n",
+"betaa=1718; //MPa"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 2.17_b: SOLUTION_oil_volume_change_in_SI.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"clc;\n",
+"pathname=get_absolute_file_path('2_17_soln.sce')\n",
+"filename=pathname+filesep()+'2_17_data.sci'\n",
+"exec(filename)\n",
+"// Solution:\n",
+"// Change in pressure,\n",
+"delP=P2-P1; //kPa \n",
+"betaa=betaa*1000; //kPA\n",
+"// % Change in volume,\n",
+"delV=-(delP/betaa)*100; //% ,- sign indicates oil is being compressed\n",
+"// Results:\n",
+"printf('\n  Results:  ')\n",
+"printf('\n The Percentage change in volume of oil is %.3f.',delV)"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 2.18_a: find_absolute_viscosity_in_SI.sci"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"// Aim:To find absolute viscosity of oil in Ns/m^2 and cP\n",
+"// Given:\n",
+"// Area of moving plate surface in contact with oil:\n",
+"A=1; //m^2\n",
+"// Force applied to the moving plate:\n",
+"F=10; //N \n",
+"// velocity of the moving plate:\n",
+"v=1; //m/s\n",
+"// oil film thickness:\n",
+"y=5; //mm\n",
+"y=5*0.001; //m"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 2.18_b: SOLUTION_absolute_viscosity_in_SI.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"clc;\n",
+"pathname=get_absolute_file_path('2_18_soln.sce')\n",
+"filename=pathname+filesep()+'2_18_data.sci'\n",
+"exec(filename)\n",
+"// Solution:\n",
+"// absolute viscosity of oil,\n",
+"mu=(F/A)/(v/y); //Ns/m^2\n",
+"// absolute viscosity of oil in cP,\n",
+"mu_P=(F*100000*y*100)/(v*100*A*10000); //poise\n",
+"mu_cP=mu_P*100; //centipoise\n",
+"// Results:\n",
+"printf('\n  Results:  ')\n",
+"printf('\n The viscosity of oil is %0.2f Ns/m^2.',mu)\n",
+"printf('\n The viscosity of oil is %0.2f cP.',mu_cP)"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 2.1_a: find_weight_of_a_body.sci"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"// Aim:To Find Weight of Body\n",
+"// Given:\n",
+"// Mass of the Body:\n",
+"m=4; //slugs"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 2.1_b: SOLUTION_weight_of_a_body.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"clc;\n",
+"pathname=get_absolute_file_path('2_1_soln.sce')\n",
+"filename=pathname+filesep()+'2_1_data.sci'\n",
+"exec(filename)\n",
+"// Solutions:\n",
+"// we know acceleration due to gravity,\n",
+"g=32.2; //ft/s^2\n",
+"W=(m*g);\n",
+"// Results:\n",
+"printf('\n  Results:  ')\n",
+"printf('\n The weight of Body is %.0f lb.',W)"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 2.2_a: find_specific_weight_of_body.sci"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"// Aim:To find the specific weight of a body\n",
+"// Given:\n",
+"// Weigth of the Body:\n",
+"W=129; //lb\n",
+"// Volume of the Body:\n",
+"V=1.8; //ft^3"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 2.2_b: SOLUTION_specific_weight_of_body.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"clc;\n",
+"pathname=get_absolute_file_path('2_2_soln.sce')\n",
+"filename=pathname+filesep()+'2_2_data.sci'\n",
+"exec(filename)\n",
+"// Solution:\n",
+"// we know specific weight,\n",
+"// gamma=(Weigth of the Body/Volume of the Body)\n",
+"gamma1=(W/V); //lb/ft^3\n",
+"// rounding off the above answer\n",
+"gamma1=fix(gamma1)+(fix((gamma1-fix(gamma1))*10)/10); //lb/ft^3\n",
+"   \n",
+"// Results:\n",
+"printf('\n  Results:  ')\n",
+"printf('\n The specific weight of Body is %.1f lb/ft^3.',gamma1)"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 2.3_a: find_specific_gravity_of_air.sci"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"// Aim:To find the specific gravity of air at 68 degF\n",
+"// Given:\n",
+"// specific weight of air at 68 degF:\n",
+"gamma_air=0.0752; //lb/ft^3"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 2.3_b: SOLUTION_specific_gravity_of_air.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"clc;\n",
+"pathname=get_absolute_file_path('2_3_soln.sce')\n",
+"filename=pathname+filesep()+'2_3_data.sci'\n",
+"exec(filename)\n",
+"// Solution:\n",
+"// we know,\n",
+"// specific gravity of air=(specific weight of air/specific weight of water)\n",
+"// also we know,specific weight of water at 68 degF,\n",
+"gamma_water=62.4; //lb/ft^3\n",
+"SG_air=gamma_air/gamma_water;\n",
+"// Results:\n",
+"printf('\n  Results:  ')\n",
+"printf('\n The specific gravity of air %0.5f.',SG_air)"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 2.4_a: find_density_of_body.sci"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"// Aim:To find Density of body of Example 2-1 and 2-2\n",
+"// Given:\n",
+"// mass of the Body:\n",
+"m=4; //slugs\n",
+"// Volume of the Body:\n",
+"V=1.8; //ft^3"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 2.4_b: SOLUTION_density_of_body.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"clc;\n",
+"pathname=get_absolute_file_path('2_4_soln.sce')\n",
+"filename=pathname+filesep()+'2_4_data.sci'\n",
+"exec(filename)\n",
+"// Solution:\n",
+"// we know density,\n",
+"// rho1=(mass of the Body/Volume of the Body)\n",
+"rho1=(m/V); //slugs/ft^3\n",
+"// also density,rho2=(specific weight/acceleration due to gravity)\n",
+"g=32.2; //ft/s^2\n",
+"gamma1=71.6; //lb/ft^3\n",
+"rho2=(gamma1/g); //slugs/ft^3\n",
+"// Results:\n",
+"printf('\n  Results:  ')\n",
+"printf('\n The Density of Body is %.2f slugs/ft^3.',rho1)\n",
+"printf('\n The Density of Body is %.2f slugs/ft^3.',rho2)"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 2.5_a: find_pressure_on_skin_diver.sci"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"// Aim:To find pressure on the skin diver\n",
+"// Given:\n",
+"// Depth of Water Body:\n",
+"H=60; //ft"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 2.5_b: SOLUTION_pressure_on_skin_diver.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"clc;\n",
+"pathname=get_absolute_file_path('2_5_soln.sce')\n",
+"filename=pathname+filesep()+'2_5_data.sci'\n",
+"exec(filename)\n",
+"// Solution:\n",
+"// specific Weight of water,\n",
+"gamma1=0.0361; //lb/in^3 \n",
+"// Conversion: \n",
+"// 1 feet = 12 inches\n",
+"// 1 lb/in^2 = 1 psi            \n",
+"// we know pressure,\n",
+"// p=(specific weight of liquid * liquid column height)\n",
+"p=(gamma1*H*12); //psi\n",
+"// Results:\n",
+"printf('\n  Results:  ')\n",
+"printf('\n The pressure on skin diver is %.1f psi.',p)"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 2.6_a: find_height_of_barometer_tube.sci"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"// Aim:To find tube height of a Barometer\n",
+"// Given:\n",
+"// liquid used is Water instead of Mercury."
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 2.6_b: SOLUTION_height_of_barometer_tube.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"clc;\n",
+"pathname=get_absolute_file_path('2_6_soln.sce')\n",
+"filename=pathname+filesep()+'2_6_data.sci'\n",
+"exec(filename)\n",
+"// Solution:\n",
+"// specific Weight of water,\n",
+"gamma1=0.0361; //lb/in^3 \n",
+"// We also knows Atmospheric Pressure,\n",
+"p=14.7; //psi\n",
+"// Conversion: \n",
+"// 1 feet = 12 inches\n",
+"// 1 lb/in^2 = 1 psi            \n",
+"// we know pressure,\n",
+"// p=(specific weight of liquid * liquid column height)\n",
+"// Therefore,\n",
+"H=(p/gamma1); //in\n",
+"// He=Height in Feet.\n",
+"He=H*0.083; //ft\n",
+"// Results:\n",
+"printf('\n  Results:  ')\n",
+"printf('\n The Height of water column is %0.0f ft.',He)"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 2.7_a: convert_gage_to_absolute_pressure.sci"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"// Aim:To convert given pressure into absolute pressure\n",
+"// Given:\n",
+"// Gage Pressure:\n",
+"Pg=-5; //psi"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 2.7_b: SOLUTION_gage_to_absolute_pressure.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"clc;\n",
+"pathname=get_absolute_file_path('2_7_soln.sce')\n",
+"filename=pathname+filesep()+'2_7_data.sci'\n",
+"exec(filename)\n",
+"// Solution:\n",
+"// Atmospheric Pressure,\n",
+"Po=14.7; //psi           \n",
+"// Absolute Pressure(Pa) =Gage Pressure + Atmospheric Pressure\n",
+"Pa=Pg+Po;\n",
+"// Results:\n",
+"printf('\n  Results:  ')\n",
+"printf('\n The Absolute Pressure is %0.1f psi.',Pa)"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 2.8_a: find_absolute_pressure_on_skin_diver.sci"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"// Aim:To find absolute pressure on skin diver of Example 2-5\n",
+"// Given:\n",
+"// Gage Pressure:\n",
+"Pg=26; //psi"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 2.8_b: SOLUTION_absolute_pressure_on_skin_diver.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"clc;\n",
+"pathname=get_absolute_file_path('2_8_soln.sce')\n",
+"filename=pathname+filesep()+'2_8_data.sci'\n",
+"exec(filename)\n",
+"// Solution:\n",
+"// Atmospheric Pressure,\n",
+"Po=14.7; //psi           \n",
+"// Absolute Pressure(Pa) =Gage Pressure + Atmospheric Pressure\n",
+"Pa=Pg+Po; //psi\n",
+"// Results:\n",
+"printf('\n  Results:  ')\n",
+"printf('\n The Absolute Pressure is %0.1f psi.',Pa)"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 2.9_a: find_specific_weight_in_SI_system.sci"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"// Aim:To Determine specific weights in N/m^3\n",
+"// Given:\n",
+"// specific weight:\n",
+"gamma1=56; //lb/ft^3"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 2.9_b: SOLUTION_specific_weight_in_SI_system.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"clc;\n",
+"pathname=get_absolute_file_path('2_9_soln.sce')\n",
+"filename=pathname+filesep()+'2_9_data.sci'\n",
+"exec(filename)\n",
+"// Solution:\n",
+"// We know,\n",
+"// 1 N/m^3 = 157 lb/ft^3\n",
+"gamma2=157*gamma1; //N/m^3\n",
+"// Results:\n",
+"printf('\n  Results:  ')\n",
+"printf('\n The specific weights is %0.0f N/m^3.',gamma2)\n",
+"printf('\n The answer in the program is different than that in textbook. It may be due to no.s of significant digit in data and calculation')"
+   ]
+   }
+],
+"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/Fluid_Power_With_Applications_by_A_Esposito/3-ENERGY_AND_POWER_IN_HYDRAULIC_SYSTEMS.ipynb b/Fluid_Power_With_Applications_by_A_Esposito/3-ENERGY_AND_POWER_IN_HYDRAULIC_SYSTEMS.ipynb
new file mode 100644
index 0000000..e4eb9bf
--- /dev/null
+++ b/Fluid_Power_With_Applications_by_A_Esposito/3-ENERGY_AND_POWER_IN_HYDRAULIC_SYSTEMS.ipynb
@@ -0,0 +1,1168 @@
+{
+"cells": [
+ {
+		   "cell_type": "markdown",
+	   "metadata": {},
+	   "source": [
+       "# Chapter 3: ENERGY AND POWER IN HYDRAULIC SYSTEMS"
+	   ]
+	},
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 3.10_a: find_pressure_available_at_motor_inlet.sci"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"// Aim:Refer Example 3-10 for Problem Description. \n",
+"// Given:\n",
+"// Pump Power:\n",
+"HHP=5; //HP\n",
+"// Pump flow:\n",
+"Q=30; //gpm\n",
+"// Pipe Diameter:\n",
+"D=1; //in\n",
+"// specific gravity of oil:\n",
+"SG=0.9;\n",
+"// Pressure at Station 1:\n",
+"p1=0; //psig (It is atmospheric pressure.)\n",
+"// Head Loss due to friction between Station 1 and 2 of oil:\n",
+"Hl=30; //ft"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 3.10_b: SOLUTION_pressure_available_at_motor_inlet.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"clc;\n",
+"pathname=get_absolute_file_path('3_10_soln.sce')\n",
+"filename=pathname+filesep()+'3_10_data.sci'\n",
+"exec(filename)\n",
+"// Solution:\n",
+"// Acceleration due to gravity,\n",
+"g=32.2; //ft/s^2\n",
+"// Energy Equation between Station 1 and Station 2 is given by,\n",
+"// (Z1+P1+K1+Hp-Hm-Hl)=(Z2+P2+K2)\n",
+"// since, There is no Hydraulic motor between Station 1 and 2,\n",
+"// Therefore Motor Head,\n",
+"Hm=0; //ft\n",
+"// also, cross section of oil tank is very large, as a result oil is at rest,\n",
+"v1=0; //ft/s\n",
+"// Kinetic Energy Head at inlet,\n",
+"K1=(v1^2)/(2*g); //ft\n",
+"// Height of Station 1 from Datum,\n",
+"Z1=0; //ft\n",
+"// Height of Station 2 from Datum,\n",
+"Z2=20; //ft\n",
+"// Pressure Head at inlet,\n",
+"P1=p1/SG; //ft\n",
+"// Pump Head,\n",
+"Hp=ceil((3950*HHP)/(Q*SG)); //ft\n",
+"// Pump flow,\n",
+"Q_1=Q/449; //ft^3/s\n",
+"// Area of pipe,\n",
+"A=((%pi)*((D/12)^2))/4; //ft^2\n",
+"// Therefore, velocity in pipe,\n",
+"v2=Q_1/A; //ft/s\n",
+"// Kinetic Energy head at Station 2,\n",
+"K2=(v2^2)/(2*g); //ft\n",
+"// Therefore, Pressure Head at outlet,\n",
+"P2=Z1+P1+K1+Hp-Hm-Hl-Z2-K2; //ft\n",
+"// specific weight of oil,\n",
+"gamma1=SG*62.4; //lb/ft^3\n",
+"// Pressure available at inlet of hydraulic motor at station 2,\n",
+"p2=P2*gamma1; // lb/ft^2\n",
+"p2=floor(p2/144); //psi\n",
+"// Results:\n",
+"printf('\n  Results:  ')  \n",
+"printf('\n The Pressure available at inlet of hydraulic motor at Station 2 is %.0f psig.',p2)"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 3.11_a: find_jet_velocity_and_flow_rate.sci"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"// Aim:Refer Example 3-11 for Problem Description. \n",
+"// Given:\n",
+"// Fluid Head:\n",
+"h=36; //ft\n",
+"// Diameter of opening:\n",
+"d=2; //in\n",
+"// Frictional Head Losses:\n",
+"Hl=10; //ft"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 3.11_b: SOLUTION_jet_velocity_and_flow_rate.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"clc;\n",
+"pathname=get_absolute_file_path('3_11_soln.sce')\n",
+"filename=pathname+filesep()+'3_11_data.sci'\n",
+"exec(filename)\n",
+"// Solution:\n",
+"// Acceleration due to gravity,\n",
+"g=32.2; //ft/s^2\n",
+"// Assuming ideal fluid, Jet velocity,\n",
+"v2=sqrt(2*g*h); //ft/s\n",
+"// Area of the opening,\n",
+"A=(%pi/4)*((d/12)^2); //ft^2\n",
+"// flow rate,\n",
+"Q=A*v2; //ft^3/s\n",
+"Q_gpm=floor(449*Q); //gpm\n",
+"// Jet velocity considering friction losses,\n",
+"v2l=sqrt(64.4*(h-Hl)); //ft/s\n",
+"// since, flow rate is proportional to velocity,\n",
+"Ql=((v2l/v2)*Q_gpm); //gpm\n",
+"// Results:\n",
+"printf('\n  Results:  ')  \n",
+"printf('\n The Jet velocity is %.1f ft/s.',v2)\n",
+"printf('\n The answer in the program is different than that in textbook. It may be due to no.s of significant digit in data and calculation')\n",
+"printf('\n The Flow rate is %.0f gpm.',Q_gpm)\n",
+"printf('\n The Jet velocity considering friction losses is %.1f ft/s.',v2l)\n",
+"printf('\n The Flow rate considering friction losses is %.0f gpm.',Ql)\n",
+"printf('\n The answer in the program is different than that in textbook. It may be due to no.s of significant digit in data and calculation')"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 3.12_a: find_velocity_and_flowrate_through_siphon.sci"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"// Aim:Refer Example 3-12 for Problem Description. \n",
+"// Given:\n",
+"// Fluid Head:\n",
+"h=30; //ft\n",
+"// Frictional Head Losses:\n",
+"Hl=10; //ft\n",
+"// U-tube inside diameter:\n",
+"d=1; //in"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 3.12_b: SOLUTION_velocity_and_flowrate_through_siphon.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"clc;\n",
+"pathname=get_absolute_file_path('3_12_soln.sce')\n",
+"filename=pathname+filesep()+'3_12_data.sci'\n",
+"exec(filename)\n",
+"// Solution:\n",
+"// Acceleration due to gravity,\n",
+"g=32.2; //ft/s^2\n",
+"// Jet velocity through siphon,\n",
+"v2=sqrt(2*g*(h-Hl)); //ft/s\n",
+"// rounding off the above answer\n",
+"v2=fix(v2)+(fix(floor((v2-fix(v2))*10))/10); //ft/s\n",
+"// Area of the U tube,\n",
+"A=(%pi/4)*((d/12)^2); //ft^2\n",
+"// flow rate through siphon,\n",
+"Q=A*v2; //ft^3/s\n",
+"Q_gpm=449*Q; //gpm\n",
+"// rounding off the above answer\n",
+"Q_gpm=fix(Q_gpm)+(fix(floor((Q_gpm-fix(Q_gpm))*10))/10); //gpm\n",
+"// Results:\n",
+"printf('\n  Results:  ')  \n",
+"printf('\n The velocity through siphon is %.1f ft/s.',v2)\n",
+"printf('\n The Flow rate through siphon is %.1f gpm.',Q_gpm)"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 3.13_a: determine_force_and_displacement_for_piston2.sci"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"// Aim:Refer Example 3-13 for Problem Description\n",
+"// Given:For the Hydraulic Jack,\n",
+"// Area of Piston 1:\n",
+"A1=25; //cm^2\n",
+"// Area of Piston 2:\n",
+"A2=100; //cm^2\n",
+"// Input force downward:\n",
+"F1=200; //N\n",
+"// downward movement of piston 1:\n",
+"S1=5; //cm"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 3.13_b: SOLUTION_force_and_displacement_for_piston2.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"clc;\n",
+"pathname=get_absolute_file_path('3_13_soln.sce')\n",
+"filename=pathname+filesep()+'3_13_data.sci'\n",
+"exec(filename)\n",
+"// Solution:\n",
+"// Pascal law states, (F1*A1 = F2*A2) \n",
+"// Similarly, (S1*A1 = S2*A2)\n",
+"// Output force upward,\n",
+"F2=(A2/A1)*F1; //N\n",
+"// upward movement of piston 2,\n",
+"S2=(A1/A2)*S1; //cm\n",
+"// Results:\n",
+"printf('\n  Results:  ')   \n",
+"printf('\n The Output force upward is %.0f N',F2)\n",
+"printf('\n The upward movement of piston 2 is %.2f cm',S2)"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 3.14_a: find_velocity_of_oil_through_pipe.sci"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"// Aim:To Determine velocity through pipe. \n",
+"// Given:\n",
+"// Diameter of pipe:\n",
+"D=30; //mm\n",
+"// Flow through pipe:\n",
+"Q=60; //lpm"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 3.14_b: SOLUTION_velocity_of_oil_through_pipe.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"clc;\n",
+"pathname=get_absolute_file_path('3_14_soln.sce')\n",
+"filename=pathname+filesep()+'3_14_data.sci'\n",
+"exec(filename)\n",
+"// Solution:\n",
+"// Pump flow in m^3/s,\n",
+"Q_si=0.0000167*Q; //m^3/s\n",
+"// Diameter of pipe,\n",
+"D_m=D/1000; //m\n",
+"// Area of pipe,\n",
+"A=(%pi*(D_m^2))/4; //m^2\n",
+"// velocity,\n",
+"v=Q_si/A; //m/s\n",
+"// rounding off the above answer\n",
+"v=fix(v)+(fix(floor((v-fix(v))*100))/100); //m/s\n",
+"// Results:\n",
+"printf('\n  Results:  ')   \n",
+"printf('\n The velocity through pipe is %.2f m/s.',v)"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 3.15_a: find_hydraulic_power_delivered_by_pump.sci"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"// Aim:To Determine Hydraulic power delivered by pump. \n",
+"// Given:\n",
+"// Pump flow:\n",
+"Q=50; //lpm\n",
+"// Pressure delivered by pump:\n",
+"p=10000; //kPa"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 3.15_b: SOLUTION_hydraulic_power_delivered_by_pump.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"clc;\n",
+"pathname=get_absolute_file_path('3_15_soln.sce')\n",
+"filename=pathname+filesep()+'3_15_data.sci'\n",
+"exec(filename)\n",
+"// Solution:\n",
+"// Pump flow in m^3/s,\n",
+"Q_si=0.0000167*Q; //m^3/s\n",
+"// Hydraulic Power,\n",
+"HP=p*Q_si; //kW\n",
+"// Results:\n",
+"printf('\n  Results:  ')   \n",
+"printf('\n The Hydraulic power delivered by pump is %.2f kW.',HP)"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 3.16_a: find_torque_delivered_by_motor_SI.sci"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"// Aim:To determine torque delivered by hydraulic motor\n",
+"// Given:\n",
+"// Mechanical Output Power:\n",
+"OP=10; //kW\n",
+"// Speed of the Hydraulic motor: \n",
+"N=1450; //rpm"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 3.16_b: SOLUTION_torque_delivered_by_motor_SI.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"clc;\n",
+"pathname=get_absolute_file_path('3_16_soln.sce')\n",
+"filename=pathname+filesep()+'3_16_data.sci'\n",
+"exec(filename)\n",
+"// Solution:\n",
+"// Power(kW)=(Torque*Speed)/9550\n",
+"// Therefore,Torque\n",
+"T=(OP*9550)/N; //Nm\n",
+"   \n",
+"// Results:\n",
+"printf('\n  Results:  ')  \n",
+"printf('\n The Torque delivered by Hydraulic motor is %.1f Nm.',T)"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 3.17_a: find_pressure_at_hydraulicmotor_inlet_SI.sci"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"// Aim:Refer Example 3-17 for Problem Description. \n",
+"// Given:\n",
+"// Pump Power:\n",
+"HHP=3.73; //kW\n",
+"// Pump flow:\n",
+"Q=0.001896; //m^3/s\n",
+"// Pipe Diameter:\n",
+"D=0.0254; //m\n",
+"// specific gravity of oil:\n",
+"SG=0.9;\n",
+"// Pressure at Station 1:\n",
+"p1=0; //Pa (It is atmospheric pressure.)\n",
+"// Elevation Between Station 1 and 2:\n",
+"// Z=Z1-Z2\n",
+"Z=-6.096; //m   -ve sign indicates Station 2 is above Station 1\n",
+"// Head Loss due to friction between Station 1 and 2 of oil:\n",
+"Hl=9.144; //m"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 3.17_b: SOLUTION_pressure_at_hydraulicmotor_inlet_SI.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"clc;\n",
+"pathname=get_absolute_file_path('3_17_soln.sce')\n",
+"filename=pathname+filesep()+'3_17_data.sci'\n",
+"exec(filename)\n",
+"// Solution:\n",
+"// Acceleration due to gravity,\n",
+"g=9.81; //m/s^2\n",
+"// Energy Equation between Station 1 and Station 2 is given by,\n",
+"// (Z+P1+K1+Hp-Hm-Hl)=(P2+K2)\n",
+"// since, There is no Hydraulic motor between Station 1 and 2,\n",
+"// Therefore Motor Head,\n",
+"Hm=0; //m\n",
+"// also, cross section of oil tank is very large, as a result oil is at rest,\n",
+"v1=0; //m/s\n",
+"// Kinetic Energy Head at inlet,\n",
+"K1=(v1^2)/(2*g); //m\n",
+"// Pressure Head at inlet,\n",
+"P1=p1/SG; //m\n",
+"// specific weight of oil,\n",
+"gamma1=round(SG*9797); //N/m^3\n",
+"// Pump Power,\n",
+"W=HHP*1000; //W\n",
+"// Pump Head,\n",
+"Hp=(W/(Q*gamma1)); //m\n",
+"// Area of pipe,\n",
+"A=((%pi)*(D^2))/4; //m^2\n",
+"// Therefore, velocity in pipe,\n",
+"v2=Q/A; //m/s\n",
+"// Kinetic Energy head at Station 2,\n",
+"K2=(v2^2)/(2*g); //m\n",
+"// Therefore, Pressure Head at outlet,\n",
+"P2=Z+P1+K1+Hp-Hm-Hl-K2; //m\n",
+"// Pressure available at inlet of hydraulic motor at station 2,\n",
+"p2=floor((P2*gamma1)/1000); // kPa gage\n",
+"// Results:\n",
+"printf('\n  Results:  ')  \n",
+"printf('\n The Pressure available at inlet of hydraulic motor at Station 2 is %.0f kPa gage.',p2)"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 3.1_a: find_work_done_and_power_delivered.sci"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"// Aim:To find work done and power deliver\n",
+"// Given:\n",
+"// Force excerted by the person:\n",
+"F=30; //lb\n",
+"// Distance moved by hand truck: \n",
+"S=100; //ft\n",
+"// time taken:\n",
+"t=60; //s"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 3.1_b: SOLUTION_work_done_and_power_delivered.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"clc;\n",
+"pathname=get_absolute_file_path('3_1_soln.sce')\n",
+"filename=pathname+filesep()+'3_1_data.sci'\n",
+"exec(filename)\n",
+"// Solution:\n",
+"// we know,Work done=Force * Displacement,\n",
+"W=F*S; //ft.lb\n",
+"// Now,Power,\n",
+"P=W/t; //(ft.lb/s)\n",
+"P=P/550; //HP\n",
+"   \n",
+"// Results:\n",
+"printf('\n  Results:  ')   \n",
+"printf('\n The work done by the person is %.1f ft.lb',W)\n",
+"printf('\n The power delivered by the person is %.3f HP',P)"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 3.2_a: find_torque_delivered_by_hydraulic_motor.sci"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"clc;\n",
+"// Aim:To determine torque required by hydraulic motor\n",
+"// Given:\n",
+"// Power Supplied:\n",
+"HP=2; //HP\n",
+"// Speed of the Hydraulic motor: \n",
+"N=1800; //rpm"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 3.2_b: SOLUTION_torque_delivered_by_hydraulic_motor.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"clc;\n",
+"pathname=get_absolute_file_path('3_2_soln.sce')\n",
+"filename=pathname+filesep()+'3_2_data.sci'\n",
+"exec(filename)\n",
+"// Solution:\n",
+"// Power (HP)=(Torque*Speed)/63000\n",
+"// Therefore,Torque\n",
+"T=(HP*63000)/N; //in.lb\n",
+"   \n",
+"// Results:\n",
+"printf('\n  Results:  ')  \n",
+"printf('\n The Torque delivered by Hydraulic motor is %.1f in.lb',T)"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 3.3_a: find_input_horsepower_required_by_elevator.sci"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"// Aim: Refer Example 3-3 for Problem Description\n",
+"// Given:\n",
+"// Load to be raised:\n",
+"F=3000; //lb\n",
+"// Distance: \n",
+"S=50; //ft\n",
+"// time required:\n",
+"t=10; //s\n",
+"//efficiency of the system:\n",
+"eta=80; //%"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 3.3_b: SOLUTION_input_horsepower_required_by_elevator.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"clc;\n",
+"pathname=get_absolute_file_path('3_3_soln.sce')\n",
+"filename=pathname+filesep()+'3_3_data.sci'\n",
+"exec(filename)\n",
+"// Solution:\n",
+"// we know,output power=(Force * Displacement)/time,\n",
+"outpw=(F*S)/t; //ft.lb/s\n",
+"outpw_HP=outpw/550; //HP\n",
+"// Efficiency=output power/input power\n",
+"inpw=outpw_HP/(eta*0.01); //HP\n",
+"   \n",
+"// Results:\n",
+"printf('\n  Results:  ')    \n",
+"printf('\n The Input Horsepower required by elevator hoist motor is %.1f HP',inpw)"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 3.4_a: find_force_and_energy_for_jack.sci"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"// Aim:Refer Example 3-4 for Problem Description\n",
+"// Given:For the Hydraulic Jack,\n",
+"// Area of Piston 1:\n",
+"A1=2; //in^2\n",
+"// Area of Piston 2:\n",
+"A2=20; //in^2\n",
+"// Input force downward:\n",
+"F1=100; //lb\n",
+"// downward movement of piston 1:\n",
+"S1=1; //in"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 3.4_b: SOLUTION_force_and_energy_for_jack.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"clc;\n",
+"pathname=get_absolute_file_path('3_4_soln.sce')\n",
+"filename=pathname+filesep()+'3_4_data.sci'\n",
+"exec(filename)\n",
+"// Solution:\n",
+"// Pascal law states, (F1*A1 = F2*A2) \n",
+"// Similarly, (S1*A1 = S2*A2)\n",
+"// Output force upward,\n",
+"F2=(A2/A1)*F1; //lb\n",
+"// upward movement of piston 2,\n",
+"S2=(A1/A2)*S1; //in\n",
+"// Energy Input,\n",
+"E1=F1*S1; //in.lb\n",
+"// Energy Output,\n",
+"E2=F2*S2; //in.lb\n",
+"// Results:\n",
+"printf('\n  Results:  ')   \n",
+"printf('\n The Output force upward is %.1f lb',F2)\n",
+"printf('\n The upward movement of piston 2 is %.1f in',S2)\n",
+"printf('\n The Energy Input is %.1f in.lb',E1)\n",
+"printf('\n The Energy Output is %.1f in.lb',E2)"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 3.5_a: what_is_output_horsepower.sci"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"// Aim:Refer Example 3-5 for Problem Description. \n",
+"// Given:\n",
+"// Diameter of piston of pump cylinder:\n",
+"Dp=1; //in\n",
+"// Diameter of piston of load cylinder:\n",
+"Dl=3.25; //in\n",
+"// Average hand force:\n",
+"Fh=25; //lb\n",
+"// Load piston stroke:\n",
+"Sl=10; //in\n",
+"// Pump piston stroke:\n",
+"Sp=2; //in"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 3.5_b: SOLUTION_output_horsepower.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"clc;\n",
+"pathname=get_absolute_file_path('3_5_soln.sce')\n",
+"filename=pathname+filesep()+'3_5_data.sci'\n",
+"exec(filename)\n",
+"// Solution:\n",
+"// Therfore, Force acting on rod of pump cylinder,\n",
+"F_rod=(8/2)*Fh; //lb\n",
+"// Area of piston of pump cylinder,\n",
+"Ap=(%pi/4)*Dp^2;//in^2\n",
+"// Area of piston of load cylinder,\n",
+"Al=(%pi/4)*Dl^2; //in^2\n",
+"// Pump cylinder discharge pressure,\n",
+"p=round(F_rod/Ap); //psi\n",
+"// Load carrying capacity,\n",
+"F_load=p*Al; //lb\n",
+"// Therefore, No.s of Cycles,\n",
+"Noc=(Al*Sl)/(Ap*Sp);\n",
+"// Output power,\n",
+"outpw=((F_load*(Sl/12))/Noc); //ft.lb/s\n",
+"outpw_HP=outpw/550; //HP\n",
+"// Assuming efficiency 80 %\n",
+"eta=0.8;\n",
+"outpw_HP2=eta*outpw_HP; //HP\n",
+"// Results:\n",
+"printf('\n  Results:  ') \n",
+"printf('\n Therefore %.0f lb of load can be lifted',F_load)\n",
+"printf('\n The answer in the program is different than that in textbook. It may be due to no.s of significant digit in data and calculation')\n",
+"printf('\n Therefore %.1f no.s of cycles are required to lift the load 10 in.',Noc)\n",
+"printf('\n The answer in the program is different than that in textbook. It may be due to no.s of significant digit in data and calculation')\n",
+"printf('\n Input power when efficiency is 100 percent is %.3f HP',outpw_HP)\n",
+"printf('\n Input power when efficiency is 80 percent is %.3f HP',outpw_HP2)"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 3.6_a: find_load_carrying_capacity_of_system.sci"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"// Aim:Refer Example 3-6 for Problem Description. \n",
+"// Given:\n",
+"// inlet air pressure:\n",
+"p1=100; //psi\n",
+"// air piston area:\n",
+"A1=20; //in^2\n",
+"// oil piston area:\n",
+"A2=1; //in^2\n",
+"// load piston area:\n",
+"A3=25; //in^2\n",
+"// load piston diameter:\n",
+"d3=5.64; //in"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 3.6_b: SOLUTION_load_carrying_capacity_of_system.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"clc;\n",
+"pathname=get_absolute_file_path('3_6_soln.sce')\n",
+"filename=pathname+filesep()+'3_6_data.sci'\n",
+"exec(filename)\n",
+"// Solution:\n",
+"// booster input force = booster output force\n",
+"// p1*A1 = p2*A2\n",
+"p2=(A1/A2)*p1; //psi\n",
+"// As per pascal law,\n",
+"p3=p2; // where p3=outlet pressure\n",
+"// Therefore load carrying capacity of system,\n",
+"F=p3*A3; //lb\n",
+"// Results:\n",
+"printf('\n  Results:  ')  \n",
+"printf('\n The load carrying capacity of system is %.0f lb.',F)"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 3.7_a: find_flow_rate_and_fluid_velocity.sci"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"// Aim:Refer Example 3-7 for Problem Description. \n",
+"// Given:\n",
+"// inlet diameter:\n",
+"D1=4; //in\n",
+"// outlet diameter:\n",
+"D2=2; //in\n",
+"// inlet velocity:\n",
+"v1=4; //ft/s"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 3.7_b: SOLUTION_flow_rate_and_fluid_velocity.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"clc;\n",
+"pathname=get_absolute_file_path('3_7_soln.sce')\n",
+"filename=pathname+filesep()+'3_7_data.sci'\n",
+"exec(filename)\n",
+"// Solution:\n",
+"// we know, Discharge=Area*Velocity\n",
+"A1=(%pi/4)*(D1/12)^2; //ft^2\n",
+"Q=A1*v1; //ft^3/s\n",
+"// Since, for hydraulic system, volume flow rate is always constant\n",
+"// we get,outlet velocity,\n",
+"v2=((D1/D2)^2)*v1; //ft/s\n",
+"// Results:\n",
+"printf('\n  Results:  ')  \n",
+"printf('\n The Volume flow rate of the system is %.3f ft^3/s.',Q)\n",
+"printf('\n The fluid velocity at station 2 is %.0f ft/s.',v2)"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 3.8_a: calculate_output_HP_delivered_by_cylinder.sci"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"// Aim:Refer Example 3-8 for Problem Description. \n",
+"// Given:\n",
+"// Time period of operations:\n",
+"t=10; //s\n",
+"// Stroke of hydraulic cylinder:\n",
+"S=10; //ft\n",
+"// Load required to compress car:\n",
+"F_load=8000; //lb\n",
+"// Pump pressure:\n",
+"p=1000; //psi\n",
+"// Efficiency of cylinder:100 %\n",
+"eta=1;"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 3.8_b: SOLUTION_output_HP_delivered_by_cylinder.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"clc;\n",
+"pathname=get_absolute_file_path('3_8_soln.sce')\n",
+"filename=pathname+filesep()+'3_8_data.sci'\n",
+"exec(filename)\n",
+"// Solution:\n",
+"// The required piston area,\n",
+"A=round(F_load/p); //in^2\n",
+"// The necessary pump flow rate,\n",
+"Q=((A/144)*S)/t; //ft^3/s\n",
+"Q_gpm=Q*449; //gpm\n",
+"// The Hydraulic Horsepower delivered to cylinder,\n",
+"HHP=(p*Q_gpm)/1714; //HP\n",
+"// rounding off the above answer\n",
+"HHP=fix(HHP)+(fix(floor((HHP-fix(HHP))*10))/10); //HP\n",
+"// The output horsepower delivered by cylinder to load,\n",
+"OHP=HHP*eta; //HP\n",
+"// Results:\n",
+"printf('\n  Results:  ')   \n",
+"printf('\n The Required piston area is %.0f in^2.',A)\n",
+"printf('\n The necessary pump flow rate is %.1f gpm.',Q_gpm)\n",
+"printf('\n The Hydraulic Horsepower delivered to cylinder is %.1f HP.',HHP)\n",
+"printf('\n The output horsepower delivered by cylinder to load is %.1f HP.',OHP)"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 3.9_a: calculate_efficiency_of_cylinder_assuming_leakage.sci"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"// Aim:Refer Example 3-9 for Problem Description. \n",
+"// Given:\n",
+"// Time period of operations:\n",
+"t=10; //s\n",
+"// Stroke of hydraulic cylinder:\n",
+"S=10; //ft\n",
+"// Load required to compress car:\n",
+"F_load=8000; //lb\n",
+"// Pump pressure:\n",
+"p=1000; //psi\n",
+"// Frictional Force:\n",
+"F_fric=100; //lb\n",
+"// Leakage:\n",
+"Q_leak=0.2; //gpm"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 3.9_b: SOLUTION_efficiency_of_cylinder_assuming_leakage.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"clc;\n",
+"pathname=get_absolute_file_path('3_9_soln.sce')\n",
+"filename=pathname+filesep()+'3_9_data.sci'\n",
+"exec(filename)\n",
+"// Solution:\n",
+"// The required piston area,\n",
+"A=(F_load+F_fric)/p; //in^2\n",
+"// The Theoretical pump flow rate,\n",
+"Q_theo=((A/144)*S)/t; //ft^3/s\n",
+"Q_gpm=(Q_theo*449); //gpm\n",
+"// The Actual pump flow rate,\n",
+"Q_act=Q_gpm+Q_leak; //gpm\n",
+"// rounding off the above answer\n",
+"Q_act=fix(Q_act)+(fix(floor((Q_act-fix(Q_act))*10))/10); //gpm\n",
+"// The Hydraulic Horsepower delivered to cylinder,\n",
+"HHP=(p*Q_gpm)/1714; //HP\n",
+"// rounding off the above answer\n",
+"HHP=fix(HHP)+(fix(ceil((HHP-fix(HHP))*10))/10); //HP\n",
+"// The output horsepower delivered by cylinder to load,\n",
+"OHP=(F_load*(S/t))/550; //HP\n",
+"// The Efficiency of System,\n",
+"eta=floor((OHP/HHP)*100); //%\n",
+"// Results:\n",
+"printf('\n  Results:  ')   \n",
+"printf('\n The Required piston area is %.2f in^2.',A)\n",
+"printf('\n The necessary pump flow rate is %.1f gpm.',Q_act)\n",
+"printf('\n The Hydraulic Horsepower delivered to cylinder is %.1f HP.',HHP)\n",
+"printf('\n The output horsepower delivered by cylinder to load is %.1f HP.',OHP)"
+   ]
+   }
+],
+"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/Fluid_Power_With_Applications_by_A_Esposito/4-FRICTIONAL_LOSSES_IN_HYDRAULIC_PIPELINES.ipynb b/Fluid_Power_With_Applications_by_A_Esposito/4-FRICTIONAL_LOSSES_IN_HYDRAULIC_PIPELINES.ipynb
new file mode 100644
index 0000000..cba24bd
--- /dev/null
+++ b/Fluid_Power_With_Applications_by_A_Esposito/4-FRICTIONAL_LOSSES_IN_HYDRAULIC_PIPELINES.ipynb
@@ -0,0 +1,718 @@
+{
+"cells": [
+ {
+		   "cell_type": "markdown",
+	   "metadata": {},
+	   "source": [
+       "# Chapter 4: FRICTIONAL LOSSES IN HYDRAULIC PIPELINES"
+	   ]
+	},
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 4.10_a: find_pressure_inlet_hydraulic_motor_SI.sci"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"// Aim:Refer Example 4-10 for Problem Description\n",
+"// Given:\n",
+"// Pump hydraulic power:\n",
+"HHP=3.73; //kW\n",
+"// Pump flow:\n",
+"Q=0.00190; //m^3/s\n",
+"// Inside Diameter of pipe:\n",
+"D=0.0254; //m\n",
+"// specific gravity of oil:\n",
+"SG_oil=0.9;\n",
+"// Kinematic viscosity of oil:\n",
+"nu=100; //cS\n",
+"// elevation between station 1 and 2:\n",
+"Z=-6.10; //m ,-ve sign indicates station 2 is above Station 1\n",
+"// Pressure at oil top surface level in hydraulic tank:\n",
+"p1=0; //Pa\n",
+"// Pump inlet pipe length:\n",
+"L1=1.53; //m\n",
+"// Pump outlet pipe length up to hydraulic motor:\n",
+"L2=4.88; //m"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 4.10_b: SOLUTION_pressure_inlet_hydraulic_motor_SI.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"clc;\n",
+"pathname=get_absolute_file_path('4_10_soln.sce')\n",
+"filename=pathname+filesep()+'4_10_data.sci'\n",
+"exec(filename)\n",
+"// Solution:\n",
+"// specific weight of oil,\n",
+"gamma1=SG_oil*9800; //N/m^3\n",
+"// acceleration due to gravity,\n",
+"g=9.80; //m/s^2\n",
+"// Since, There is no hydraulic motor,\n",
+"Hm=0; //m\n",
+"// oil in tank is at rest,\n",
+"v1=0; //m/s\n",
+"// velocity head at station 1,\n",
+"K1=(v1^2)/(2*g); //m\n",
+"// velocity through pipe,\n",
+"v2=Q/((%pi*(D^2))/4); //m/s\n",
+"// velocity head at station 2,\n",
+"K2=(v2^2)/(2*g); //m\n",
+"// Reynolds Number,\n",
+"N_R=((v2*D)/(nu/1000000));\n",
+"// friction factor,\n",
+"f=64/N_R;\n",
+"// From table of 'K factors of common valves and fittings',\n",
+"K=0.9;\n",
+"// equivalent length of standard elbow,\n",
+"Le_std_elbow=((K*(D/12))/f); //m\n",
+"// Total equivalent length,\n",
+"Le_tot=L1+L2+Le_std_elbow; //m\n",
+"// head loss due to friction between Station 1 and 2,\n",
+"H_L=((f*Le_tot*K2)/D); //m\n",
+"// Pump head,\n",
+"Hp=((1000*HHP)/(Q*gamma1)); //m\n",
+"// Pressure at station 2,\n",
+"p2=(Z+(p1/gamma1)+K1+Hp-Hm-H_L-K2); //m ,Modified Bernoulli equation\n",
+"p2=((p2*gamma1)/1000); //kPa\n",
+"// Results:\n",
+"printf('\n  Results:  ')\n",
+"printf('\n The Pressure available at the inlet to hydraulic motor is %.0f kPa.',p2)\n",
+"printf('\n The answer in the program is different than that in textbook. It may be due to no.s of significant digit in data and calculation')"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 4.1_a: find_reynolds_number_of_hydraulic_oil.sci"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"// Aim:To Find Reynolds number of oil \n",
+"// Given:\n",
+"// Kinematic viscosity of oil:\n",
+"nu=100; //cS\n",
+"// velocity of oil:\n",
+"v=10; //ft/s\n",
+"// Pipe diameter:\n",
+"D=1; //in"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 4.1_b: SOLUTION_reynolds_number_of_hydraulic_oil.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"clc;\n",
+"pathname=get_absolute_file_path('4_1_soln.sce')\n",
+"filename=pathname+filesep()+'4_1_data.sci'\n",
+"exec(filename)\n",
+"// Solution:\n",
+"// Reynolds Number,\n",
+"N_R=(7740*v*D)/nu;\n",
+"// Results:\n",
+"printf('\n  Results:  ')\n",
+"printf('\n The Reynolds number of given oil is %.0f.',N_R)"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 4.2_a: find_reynolds_number_of_oil_SI.sci"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"// Aim:To Find Reynolds number of oil \n",
+"// Given:\n",
+"// Kinematic viscosity of oil:\n",
+"nu=0.001; //m^2/s\n",
+"// velocity of oil:\n",
+"v=5; //m/s\n",
+"// Pipe diameter:\n",
+"D=50; //mm"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 4.2_b: SOLUTION_reynolds_number_of_oil_SI.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"clc;\n",
+"pathname=get_absolute_file_path('4_2_soln.sce')\n",
+"filename=pathname+filesep()+'4_2_data.sci'\n",
+"exec(filename)\n",
+"// Solution:\n",
+"// Reynolds Number,\n",
+"N_R=(v*(D/1000))/nu;\n",
+"// Results:\n",
+"printf('\n  Results:  ')\n",
+"printf('\n The Reynolds number of given oil is %.0f.',N_R)"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 4.3_a: find_head_loss_in_friction.sci"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"// Aim:Refer Example 4-3 for Problem Description \n",
+"// Given:\n",
+"// Kinematic viscosity of oil:\n",
+"nu=100; //cS\n",
+"// velocity of oil:\n",
+"v=10; //ft/s\n",
+"// Pipe diameter:\n",
+"D=1; //in\n",
+"// Length of pipe:\n",
+"L=100; //ft\n",
+"// specific gravity of oil:\n",
+"SG_oil=0.9;"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 4.3_b: SOLUTION_head_loss_in_friction.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"clc;\n",
+"pathname=get_absolute_file_path('4_3_soln.sce')\n",
+"filename=pathname+filesep()+'4_3_data.sci'\n",
+"exec(filename)\n",
+"// Solution:\n",
+"// acceleration due to gravity,\n",
+"g=32.2; //ft/s^2\n",
+"// Reynolds Number,\n",
+"N_R=(7740*v*D)/nu;\n",
+"// Head loss in pipe,\n",
+"H_L=round((64*L*(v^2))/(N_R*(D/12)*2*g)); //ft ,Hagen-Poiseuille Equation\n",
+"// Head loss in terms of psi,\n",
+"H_L=SG_oil*0.0361*12*H_L; //psi\n",
+"// Results:\n",
+"printf('\n  Results:  ')\n",
+"printf('\n The Head Loss due to friction in pipe is %.0f psi.',H_L)"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 4.4_a: find_head_loss_in_friction_SI.sci"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"// Aim:Refer Example 4-4 for Problem Description \n",
+"// Given:\n",
+"// Kinematic viscosity of oil:\n",
+"nu=0.001; //m^2/s\n",
+"// velocity of oil:\n",
+"v=5; //m/s\n",
+"// Pipe diameter:\n",
+"D=50; //mm\n",
+"// Length of pipe:\n",
+"L=50; //m\n",
+"// specific weigth of oil:\n",
+"gamma1=8800; //N/m^2"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 4.4_b: SOLUTION_head_loss_in_friction_SI.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"clc;\n",
+"pathname=get_absolute_file_path('4_4_soln.sce')\n",
+"filename=pathname+filesep()+'4_4_data.sci'\n",
+"exec(filename)\n",
+"// Solution:\n",
+"// acceleration due to gravity,\n",
+"g=9.80; //m/s^2\n",
+"// Reynolds Number,\n",
+"N_R=(v*(D/1000))/nu;\n",
+"// Head loss in pipe,\n",
+"H_L=floor((64*L*(v^2))/(N_R*(D/1000)*2*g)); //m ,Hagen-Poiseuille Equation\n",
+"// Head loss in terms of kPa,\n",
+"H_L1=(gamma1*H_L)/1000; //kPa\n",
+"// Results:\n",
+"printf('\n  Results:  ')\n",
+"printf('\n The Head Loss due to friction in pipe is %.0f m of oil.',H_L)"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 4.5_a: find_friction_factor_of_pipe.sci"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"// Aim:Refer Example 4-5 for Problem Description \n",
+"// Given:\n",
+"// Kinematic viscosity of oil:\n",
+"nu=50; //cS\n",
+"// Pipe diameter:\n",
+"D=1; //in\n",
+"// velocity of oil:\n",
+"v1=10; //ft/s\n",
+"v2=40; //ft/s"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 4.5_b: SOLUTION_friction_factor_of_pipe.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"clc;\n",
+"pathname=get_absolute_file_path('4_5_soln.sce')\n",
+"filename=pathname+filesep()+'4_5_data.sci'\n",
+"exec(filename)\n",
+"// Solution:\n",
+"// Reynolds Number in 1st case,\n",
+"N_R1=(7740*v1*D)/nu;\n",
+"// Using Moody diagram from fig 4-9,\n",
+"f1=0.042 ;\n",
+"// Reynolds Number in 2nd case,\n",
+"N_R2=(7740*v2*D)/nu;\n",
+"// relative roughness,\n",
+"rr=0.0018/D;\n",
+"// Using Moody diagram from fig 4-9,\n",
+"f2=0.036;\n",
+"// Results:\n",
+"printf('\n  Results:  ')\n",
+"printf('\n The friction factor in 1st case is %.3f.',f1)\n",
+"printf('\n The friction factor in 2nd case is %.3f.',f2)"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 4.6_a: find_head_loss_across_globe_valve.sci"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"// Aim:To Find Head Loss across valve \n",
+"// Given:\n",
+"// Diameter of globe valve:\n",
+"D=1; //in\n",
+"// specific gravity of oil:\n",
+"SG_oil=0.9;\n",
+"// flow rate:\n",
+"Q=30; //gpm"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 4.6_b: SOLUTION_head_loss_across_globe_valve.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"clc;\n",
+"pathname=get_absolute_file_path('4_6_soln.sce')\n",
+"filename=pathname+filesep()+'4_6_data.sci'\n",
+"exec(filename)\n",
+"// Solution:\n",
+"// fluid velocity,\n",
+"v=(Q/449)/((%pi*((D/12)^2))/4); //ft/s\n",
+"// rounding off the above answer\n",
+"v=fix(v)+(fix(floor((v-fix(v))*10))/10); //ft/s\n",
+"// From table of 'K factors of common valves and fittings',\n",
+"K=10;\n",
+"// acceleration due to gravity,\n",
+"g=32.2; //ft/s^2\n",
+"// Head Loss across globe valve,\n",
+"H_L=(K*(v^2))/(2*g); //ft\n",
+"// Pressure drop across Valve,\n",
+"delp=SG_oil*0.0361*12*H_L; //psi\n",
+"// Results:\n",
+"printf('\n  Results:  ')\n",
+"printf('\n The head loss across globe valve is %.1f ft of oil.',H_L)"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 4.7_a: find_head_loss_across_gate_valve.sci"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"// Aim:To Find Head Loss across valve \n",
+"// Given:\n",
+"// Diameter of gate valve:\n",
+"D=50; //mm\n",
+"// specific weight of oil:\n",
+"gamma1=8800; //N/m^2\n",
+"// kinemativ viscosity of oil:\n",
+"nu=0.001; //m^2/s\n",
+"// flow rate:\n",
+"Q=0.02; //m^3/s"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 4.7_b: SOLUTION_head_loss_across_gate_valve.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"clc;\n",
+"pathname=get_absolute_file_path('4_7_soln.sce')\n",
+"filename=pathname+filesep()+'4_7_data.sci'\n",
+"exec(filename)\n",
+"// Solution:\n",
+"// fluid velocity,\n",
+"v=Q/((%pi*((D/1000)^2))/4); //m/s\n",
+"// rounding off the above answer\n",
+"v=fix(v)+(fix(round((v-fix(v))*10))/10); //m/s\n",
+"// From table of 'K factors of common valves and fittings',\n",
+"K=0.19;\n",
+"// acceleration due to gravity,\n",
+"g=9.80; //m/s^2\n",
+"// Head Loss across globe valve,\n",
+"H_L=(K*(v^2))/(2*g); //m\n",
+"// Pressure drop across Valve,\n",
+"delp=(gamma1*H_L)/1000; //kPa\n",
+"// Results:\n",
+"printf('\n  Results:  ')\n",
+"printf('\n The head loss across globe valve is %.2f m of oil.',H_L)"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 4.8_a: find_equivalent_length_of_globe_valve.sci"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"// Aim:Refer Example 4-8 for Problem Description\n",
+"// Given:\n",
+"// Kinematic viscosity of oil:\n",
+"nu=100; //cS\n",
+"// Diameter of steel pipe:\n",
+"D=1; //in\n",
+"// flow rate:\n",
+"Q=30; //gpm\n",
+"// Diameter of wide open globe valve:\n",
+"D_l=1; //in"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 4.8_b: SOLUTION_equivalent_length_of_globe_valve.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"clc;\n",
+"pathname=get_absolute_file_path('4_8_soln.sce')\n",
+"filename=pathname+filesep()+'4_8_data.sci'\n",
+"exec(filename)\n",
+"// Solution:\n",
+"// velocity through steel pipes,\n",
+"v=(Q/449)/((%pi*((D/12)^2))/4); //ft/s\n",
+"// rounding off the above answer\n",
+"v=fix(v)+(fix(floor((v-fix(v))*10))/10); //ft/s\n",
+"// Reynolds Number,\n",
+"N_R=(7740*v*D)/nu;\n",
+"// friction factor,\n",
+"f=64/N_R;\n",
+"// From table of 'K factors of common valves and fittings',\n",
+"K=10;\n",
+"// Equivalent Length,\n",
+"Le=(K*(D_l/12))/f; //ft\n",
+"// Results:\n",
+"printf('\n  Results:  ')\n",
+"printf('\n The Equivalent Length of Globe valve is %.1f ft.',Le)"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 4.9_a: find_pressure_at_inlet_of_hydraulicmotor.sci"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"// Aim:Refer Example 4-9 for Problem Description\n",
+"// Given:\n",
+"// Pump hydraulic power:\n",
+"HHP=5; //HP\n",
+"// Pump flow:\n",
+"Q=30; //gpm\n",
+"// Inside Diameter of pipe:\n",
+"D=1; //in\n",
+"// specific gravity of oil:\n",
+"SG_oil=0.9;\n",
+"// Kinematic viscosity of oil:\n",
+"nu=100; //cS\n",
+"// elevation between station 1 and 2:\n",
+"Z=-20; //ft ,-ve sign indicates station 2 is above Station 1\n",
+"// Pressure at oil top surface level in hydraulic tank:\n",
+"p1=0; //psig"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 4.9_b: SOLUTION_pressure_at_inlet_of_hydraulicmotor.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"clc;\n",
+"pathname=get_absolute_file_path('4_9_soln.sce')\n",
+"filename=pathname+filesep()+'4_9_data.sci'\n",
+"exec(filename)\n",
+"// Solution:\n",
+"// specific weight of oil,\n",
+"gamma1=SG_oil*62.4; //lb/ft^3\n",
+"// acceleration due to gravity,\n",
+"g=32.2; //ft/s^2\n",
+"// Since, There is no hydraulic motor,\n",
+"Hm=0; //ft\n",
+"// oil in tank is at rest,\n",
+"v1=0; //ft/s\n",
+"// velocity head at station 1,\n",
+"K1=(v1^2)/(2*g); //ft\n",
+"// velocity through pipe,\n",
+"v2=(Q/449)/((%pi*((D/12)^2))/4); //ft/s\n",
+"v2=fix(v2)+(fix(floor((v2-fix(v2))*10))/10); //ft/s ,rounding off the answer\n",
+"// velocity head at station 2,\n",
+"K2=(v2^2)/(2*g); //ft\n",
+"K2=fix(K2)+(fix(ceil((K2-fix(K2))*10))/10); //ft ,rounding off the answer\n",
+"// Reynolds Number,\n",
+"N_R=round((7740*v2*D)/nu);\n",
+"// friction factor,\n",
+"f=64/N_R;\n",
+"// From table of 'K factors of common valves and fittings',\n",
+"K=0.9;\n",
+"// equivalent length of standard elbow,\n",
+"Le_std_elbow=((K*(D/12))/f); //ft\n",
+"// Total equivalent length,\n",
+"Le_tot=21+Le_std_elbow; //ft\n",
+"// head loss due to friction between Station 1 and 2,\n",
+"H_L=round((f*Le_tot*K2)/(D/12)); //ft\n",
+"// Pump head,\n",
+"Hp=ceil((3950*HHP)/(Q*SG_oil)); //ft\n",
+"// Pressure at station 2,\n",
+"p2=round(Z+(p1/gamma1)+K1+Hp-Hm-H_L-K2); //ft ,Modified Bernoulli equation\n",
+"p2=round((p2*gamma1)/144); //psi\n",
+"// Pressure increase across the pump,\n",
+"delp=ceil((gamma1*Hp)/144); \n",
+"// Results:\n",
+"printf('\n  Results:  ')\n",
+"printf('\n The Pressure available at the inlet to hydraulic motor is %.0f psi.',p2)"
+   ]
+   }
+],
+"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/Fluid_Power_With_Applications_by_A_Esposito/5-HYDRAULIC_PUMPS.ipynb b/Fluid_Power_With_Applications_by_A_Esposito/5-HYDRAULIC_PUMPS.ipynb
new file mode 100644
index 0000000..f364263
--- /dev/null
+++ b/Fluid_Power_With_Applications_by_A_Esposito/5-HYDRAULIC_PUMPS.ipynb
@@ -0,0 +1,690 @@
+{
+"cells": [
+ {
+		   "cell_type": "markdown",
+	   "metadata": {},
+	   "source": [
+       "# Chapter 5: HYDRAULIC PUMPS"
+	   ]
+	},
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 5.10_a: find_yearly_cost_of_electricity.sci"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"// Aim:Refer Example 5-10 for Problem Description\n",
+"// Given:\n",
+"// Speed of the pump:\n",
+"N=1000; //rpm\n",
+"// Prime mover input torque:\n",
+"Ta=120; //N.m\n",
+"// overall efficiency:\n",
+"eta_o=85; //%\n",
+"// operation time= 12 hrs/day for 250 days/year:\n",
+"OT=12*250; //hrs/yr\n",
+"// cost of electricity:\n",
+"coe=0.11; //$/kW.hr\n",
+"// overall efficiency for pump:\n",
+"eta_l=83.5; //%"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 5.10_b: SOLUTION_yearly_cost_of_electricity.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"clc;\n",
+"pathname=get_absolute_file_path('5_10_soln.sce')\n",
+"filename=pathname+filesep()+'5_10_data.sci'\n",
+"exec(filename)\n",
+"// Solutions:\n",
+"// Pump input power,\n",
+"IP=Ta*N/9550; //kW\n",
+"// Electric motor input power,\n",
+"EIP=IP/(eta_o/100); //kW\n",
+"// rounding off the above answer\n",
+"EIP=fix(EIP)+(fix(round((EIP-fix(EIP))*10))/10); //kW\n",
+"// Yearly cost of electricity,\n",
+"Yce=EIP*OT*coe; //$/yr\n",
+"// Total kW loss,\n",
+"kWL=((1-(eta_o/100))*EIP)+((1-(eta_l/100))*IP); //kW\n",
+"// rounding off the above answer\n",
+"kWL=fix(kWL)+(fix(round((kWL-fix(kWL))*10))/10); //kW\n",
+"// Yearly cost due to inefficiencies,\n",
+"Yci=(kWL/EIP)*Yce; //$/yr\n",
+"// Results:\n",
+"printf('\n  Results:  ')\n",
+"printf('\n The yearly cost of electricity is %.0f $/yr.',Yce)\n",
+"printf('\n The yearly cost of electricity due to inefficiencies is %.0f $/yr.',Yci)"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 5.1_a: find_volumetric_efficiency_of_gear_pump.sci"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"// Aim:To Find volumetric efficiency of Gear Pump \n",
+"// Given:\n",
+"// outside diameter of gear pump:\n",
+"Do=3; //in\n",
+"// inside diameter of gear pump:\n",
+"Di=2; //in\n",
+"// width of gear pump:\n",
+"L=1; //in\n",
+"// Actual flow rate of pump:\n",
+"Qa=28; //gpm\n",
+"// Speed of gear pump:\n",
+"N=1800; //rpm"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 5.1_b: SOLUTION_volumetric_efficiency_of_gear_pump.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"clc;\n",
+"pathname=get_absolute_file_path('5_1_soln.sce')\n",
+"filename=pathname+filesep()+'5_1_data.sci'\n",
+"exec(filename)\n",
+"// Solutions:\n",
+"// Volumetric Displacementis is given by,\n",
+"Vd=(%pi/4)*((Do^2)-(Di^2))*L; //in^3\n",
+"// Theoretical Flow rate,\n",
+"Qt=(Vd*N)/231; //gpm\n",
+"// Volumetric efficiency,\n",
+"eta_v=(Qa/Qt)*100; //%\n",
+"// Results:\n",
+"printf('\n  Results:  ')\n",
+"printf('\n The volumetric efficiency of Gear Pump  is %.1f percent.',eta_v)\n",
+"printf('\n The answer in the program is different than that in textbook. It may be due to no.s of significant digit in data and calculation')"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 5.2_a: find_actual_flowrate_of_gear_pump.sci"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"// Aim:To Find actual flow-rate of Gear Pump\n",
+"// Given:\n",
+"// outside diameter of gear pump:\n",
+"Do=75; //mm\n",
+"// inside diameter of gear pump:\n",
+"Di=50; //mm\n",
+"// width of gear pump:\n",
+"L=25; //mm\n",
+"// Volumetric efficiency,\n",
+"eta_v=90; //%\n",
+"// Speed of gear pump:\n",
+"N=1000; //rpm"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 5.2_b: SOLUTION_actual_flowrate_of_gear_pump.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"clc;\n",
+"pathname=get_absolute_file_path('5_2_soln.sce')\n",
+"filename=pathname+filesep()+'5_2_data.sci'\n",
+"exec(filename)\n",
+"// Solutions:\n",
+"// Volumetric Displacementis is given by,\n",
+"Vd=(%pi/4)*(((Do/1000)^2)-((Di/1000)^2))*(L/1000); //m^3/rev\n",
+"// Actual Flow-rate,\n",
+"Qa=Vd*N*(eta_v/100); //m^3/min\n",
+"Qa_lpm=Qa*1000; //Lpm\n",
+"// rounding off the above answer\n",
+"Qa_lpm=fix(Qa_lpm)+(fix(ceil((Qa_lpm-fix(Qa_lpm))*10))/10); //m^3/min\n",
+"// Results:\n",
+"printf('\n  Results:  ')\n",
+"printf('\n The volumetric efficiency of Gear Pump  is %.1f Lpm.',Qa_lpm)"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 5.3_a: find_eccentricity_of_vane_pump.sci"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"// Aim:To Find eccentricity of Vane Pump \n",
+"// Given:\n",
+"// volumetric displacement of vane pump:\n",
+"Vd=5; //in^3\n",
+"// rotor diameter of vane pump:\n",
+"Dr=2; //in\n",
+"// cam ring diameter of vane pump:\n",
+"Dc=3; //in\n",
+"// width of vane:\n",
+"L=2; //in"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 5.3_b: SOLUTION_eccentricity_of_vane_pump.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"clc;\n",
+"pathname=get_absolute_file_path('5_3_soln.sce')\n",
+"filename=pathname+filesep()+'5_3_data.sci'\n",
+"exec(filename)\n",
+"// Solutions:\n",
+"// eccentricity for vane pump,\n",
+"e=2*Vd/(%pi*(Dc+Dr)*L); //in\n",
+"// Results:\n",
+"printf('\n  Results:  ')\n",
+"printf('\n The eccentricity of vane pump  is %.3f in.',e)"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 5.4_a: find_volumetric_displacement_of_vane_pump.sci"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"// Aim:To Find volumetric displacement of Vane Pump \n",
+"// Given:\n",
+"// rotor diameter of vane pump:\n",
+"Dr=50; //mm\n",
+"// cam ring diameter of vane pump:\n",
+"Dc=75; //mm\n",
+"// width of vane:\n",
+"L=50; //mm\n",
+"// eccentricity:\n",
+"e=8; //mm"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 5.4_b: SOLUTION_volumetric_displacement_of_vane_pump.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"clc;\n",
+"pathname=get_absolute_file_path('5_4_soln.sce')\n",
+"filename=pathname+filesep()+'5_4_data.sci'\n",
+"exec(filename)\n",
+"// Solutions:\n",
+"// volumetric displacement of pump,\n",
+"Vd=(%pi*((Dc/1000)+(Dr/1000))*(e/1000)*(L/1000))/2; //m^3\n",
+"// since,1m^3 = 1000L\n",
+"Vd=1000*Vd; //L\n",
+"// Results:\n",
+"printf('\n  Results:  ')\n",
+"printf('\n The volumetric displacement of vane pump  is %.4f L.',Vd)"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 5.5_a: find_power_pres_compensated_pump_saved.sci"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"// Aim:Refer Example 5-5 for Problem Description\n",
+"// Given:\n",
+"// for Fixed Displacement pump:\n",
+"// pump delivery pressure:\n",
+"Pd_f=1000; //psi\n",
+"// pump flow rate:\n",
+"Q_f=20; //gpm\n",
+"// oil leakge after cylinder is fully extended:\n",
+"Ql_f=0.7; //gpm\n",
+"// pressure relief valve setting:\n",
+"p=1200; //psi\n",
+"// for Pressure Compensated pump:\n",
+"// pump flow rate:\n",
+"Q_p=0.7; //gpm\n",
+"// pressure relief valve setting:\n",
+"P=1200; //psi"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 5.5_b: SOLUTION_power_pres_compensated_pump_saved.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"clc;\n",
+"pathname=get_absolute_file_path('5_5_soln.sce')\n",
+"filename=pathname+filesep()+'5_5_data.sci'\n",
+"exec(filename)\n",
+"// Solutions:\n",
+"// Hydraulic Power lost in Fixed Displacemnt pump,\n",
+"HP_f=(p*Q_f)/1714; //HP\n",
+"// Hydraulic Power lost in Pressure Compensated pump,\n",
+"HP_p=(P*Q_p)/1714; //HP\n",
+"// Therefore, Hydraulic Power saved,\n",
+"HP=HP_f-HP_p; //HP\n",
+"// Results:\n",
+"printf('\n  Results:  ')\n",
+"printf('\n The Hydraulic Power saved after cylinder is fully extended is %.2f HP.',HP)"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 5.6_a: find_offset_angle_of_piston_pump.sci"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"// Aim:To Find offset angle of axial piston pump\n",
+"// Given:\n",
+"// pump flow rate:\n",
+"Qa=16; //gpm\n",
+"// speed of pump:\n",
+"N=3000; //rpm\n",
+"// number of pistons:\n",
+"Y=9; \n",
+"// piston diameter:\n",
+"d=0.5; //in\n",
+"// piston circle diameter:\n",
+"D=5; //in\n",
+"// volumetric efficiency:\n",
+"eta_v=95; //%"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 5.6_b: SOLUTION_offset_angle_of_piston_pump.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"clc;\n",
+"pathname=get_absolute_file_path('5_6_soln.sce')\n",
+"filename=pathname+filesep()+'5_6_data.sci'\n",
+"exec(filename)\n",
+"// Solutions:\n",
+"// Theoretical flow rate,\n",
+"Qt=Qa/(eta_v/100); //gpm\n",
+"// Area of piston,\n",
+"A=(%pi/4)*(d^2); //in^2\n",
+"// tan of offset angle,\n",
+"T_theta=(231*Qt)/(D*A*N*Y); \n",
+"// offset angle,\n",
+"theta=atand(T_theta); //deg\n",
+"// Results:\n",
+"printf('\n  Results:  ')\n",
+"printf('\n The offset angle of axial piston pump is %.1f deg.',theta)"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 5.7_a: find_flowrate_of_axial_piston_pump.sci"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"// Aim:To Find flow rate of axial piston pump in L/s\n",
+"// Given:\n",
+"// speed of pump:\n",
+"N=1000; //rpm\n",
+"// number of pistons:\n",
+"Y=9; \n",
+"// piston diameter:\n",
+"d=15; //mm\n",
+"// piston circle diameter:\n",
+"D=125; //mm\n",
+"// offset angle:\n",
+"theta=10; //deg\n",
+"// volumetric efficiency:\n",
+"eta_v=94; //%"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 5.7_b: SOLUTION_flowrate_of_axial_piston_pump.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"clc;\n",
+"pathname=get_absolute_file_path('5_7_soln.sce')\n",
+"filename=pathname+filesep()+'5_7_data.sci'\n",
+"exec(filename)\n",
+"// Solutions:\n",
+"// Area of piston,\n",
+"A=(%pi/4)*((d/1000)^2); //m^2\n",
+"// offset angle,\n",
+"theta=(theta*%pi)/180; //rad\n",
+"// Theoretical flow rate,\n",
+"Qt=(D/1000)*A*N*Y*tan(theta); //m^3/min\n",
+"// Actual flow rate,\n",
+"Qa=Qt*(eta_v/100); //m^3/min\n",
+"// rounding off the above answer\n",
+"Qa=fix(Qa)+(fix(round((Qa-fix(Qa))*1000))/1000); //m^3/min\n",
+"// Actual flow rate in L/s,\n",
+"Qa=Qa/(60*0.001); //L/s\n",
+"// Results:\n",
+"printf('\n  Results:  ')\n",
+"printf('\n The flow rate of axial piston pump in L/s is %.3f L/s.',Qa)"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 5.8_a: find_theoretical_torque_required_by_pump.sci"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"// Aim:Refer Example 5-8 for Problem Description\n",
+"// Given:\n",
+"// Displacement volume:\n",
+"Vd=5; //in^3\n",
+"// Actual pump flow rate:\n",
+"Qa=20; //gpm\n",
+"// Speed of the pump:\n",
+"N=1000; //rpm\n",
+"// Pressure delivered by pump:\n",
+"p=1000; //psi\n",
+"// Prime mover input torque:\n",
+"Ta=900; //in.lb"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 5.8_b: SOLUTION_theoretical_torque_required_by_pump.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"clc;\n",
+"pathname=get_absolute_file_path('5_8_soln.sce')\n",
+"filename=pathname+filesep()+'5_8_data.sci'\n",
+"exec(filename)\n",
+"// Solutions:\n",
+"// Theoretical pump flow rate,\n",
+"Qt=(Vd*N)/231; //gpm\n",
+"// rounding off the above answer\n",
+"Qt=fix(Qt)+(fix(floor((Qt-fix(Qt))*10))/10); //gpm\n",
+"// Therefore,volumetric efficiency,\n",
+"eta_v=(Qa/Qt);\n",
+"// Now, mechanical efficiency,\n",
+"eta_m=((p*Qt)/1714)/((Ta*N)/63000);\n",
+"// overall Efficiency,\n",
+"eta_o=eta_v*eta_m*100; //%\n",
+"// rounding off the above answer\n",
+"eta_o=fix(eta_o)+(fix(floor((eta_o-fix(eta_o))*10))/10); //%\n",
+"// Theoretical torque required to operate the pump,\n",
+"Tt=floor(eta_m*Ta); //in.lb\n",
+"// Results:\n",
+"printf('\n  Results:  ')\n",
+"printf('\n The overall efficiency of pump is %.1f percent.',eta_o)\n",
+"printf('\n The Theoretical torque required to operate the pump  is %.0f in.lb.',Tt)"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 5.9_a: find_theoretical_torque_required_in_SI.sci"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"// Aim:Refer Example 5-9 for Problem Description\n",
+"// Given:\n",
+"// Displacement volume:\n",
+"Vd=100; //cm^3\n",
+"// Actual pump flow rate:\n",
+"Qa=0.0015; //m^3/s\n",
+"// Speed of the pump:\n",
+"N=1000; //rpm\n",
+"// Pressure delivered by pump:\n",
+"p=70; //bars\n",
+"// Prime mover input torque:\n",
+"Ta=120; //N.m"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 5.9_b: SOLUTION_theoretical_torque_required_in_SI.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"clc;\n",
+"pathname=get_absolute_file_path('5_9_soln.sce')\n",
+"filename=pathname+filesep()+'5_9_data.sci'\n",
+"exec(filename)\n",
+"// Solutions:\n",
+"// volumetric displacement in m^3/rev,\n",
+"Vd=100/(10^6); //m^3/rev\n",
+"// Speed of pump in rps,\n",
+"N=N/60; //rps\n",
+"// Theoretical pump flow rate,\n",
+"Qt=Vd*N; //m^3/s\n",
+"// Therefore,volumetric efficiency,\n",
+"eta_v=(Qa/Qt);\n",
+"// Now, mechanical efficiency,\n",
+"eta_m=(p*10^5*Qt)/(Ta*N*2*(%pi));\n",
+"// overall Efficiency,\n",
+"eta_o=eta_v*eta_m*100; //%\n",
+"// rounding off the above answer\n",
+"eta_o=fix(eta_o)+(fix(floor((eta_o-fix(eta_o))*10))/10); //%\n",
+"// Theoretical torque required to operate the pump,\n",
+"Tt=ceil(eta_m*Ta); //N.m\n",
+"// Results:\n",
+"printf('\n  Results:  ')\n",
+"printf('\n The overall efficiency of pump is %.1f percent.',eta_o)\n",
+"printf('\n The Theoretical torque required to operate the pump  is %.0f N.m.',Tt)"
+   ]
+   }
+],
+"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/Fluid_Power_With_Applications_by_A_Esposito/6-HYDRAULIC_CYLINDERS_AND_CUSHIONING_DEVICES.ipynb b/Fluid_Power_With_Applications_by_A_Esposito/6-HYDRAULIC_CYLINDERS_AND_CUSHIONING_DEVICES.ipynb
new file mode 100644
index 0000000..33a8f10
--- /dev/null
+++ b/Fluid_Power_With_Applications_by_A_Esposito/6-HYDRAULIC_CYLINDERS_AND_CUSHIONING_DEVICES.ipynb
@@ -0,0 +1,419 @@
+{
+"cells": [
+ {
+		   "cell_type": "markdown",
+	   "metadata": {},
+	   "source": [
+       "# Chapter 6: HYDRAULIC CYLINDERS AND CUSHIONING DEVICES"
+	   ]
+	},
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 6.1_a: find_pressure_velocity_and_horsepower.sci"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"// Aim:Refer Example 6-1 for Problem Description \n",
+"// Given:\n",
+"// Flow rate of pump:\n",
+"Q_in=20; //gpm\n",
+"// Bore diameter of Cylinder:\n",
+"D=2; //in\n",
+"// Load during extending and retracting:\n",
+"F_ext=1000; //lb\n",
+"F_ret=1000; //lb\n",
+"// Rod diameter of cylinder:\n",
+"d=1; //in"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 6.1_b: SOLUTION_pressure_velocity_and_horsepower.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"clc;\n",
+"pathname=get_absolute_file_path('6_1_soln.sce')\n",
+"filename=pathname+filesep()+'6_1_data.sci'\n",
+"exec(filename)\n",
+"// Solution:\n",
+"// Area of blank end of piston,\n",
+"Ap=(%pi/4)*(D^2); //in^2\n",
+"// Area of rod end of piston,\n",
+"Ar=(%pi/4)*(d^2); //in^2\n",
+"// hydraulic pressure during the extending stroke,\n",
+"p_ext=F_ext/Ap; //psi\n",
+"// piston velocity during the extending stroke,\n",
+"v_ext=(Q_in/449)/(Ap/144); //ft/s\n",
+"// rounding off the above answer\n",
+"v_ext=fix(v_ext)+(fix(ceil((v_ext-fix(v_ext))*100))/100); //ft/s\n",
+"// cylinder horsepower during the extending stroke,\n",
+"HP_ext=(v_ext*F_ext)/550; //HP\n",
+"// rounding off the above answer\n",
+"HP_ext=fix(HP_ext)+(fix(floor((HP_ext-fix(HP_ext))*100))/100); //HP\n",
+"// hydraulic pressure during the retraction stroke,\n",
+"p_ret=ceil(F_ret/(Ap-Ar)); //psi\n",
+"// piston velocity during the retraction stroke,\n",
+"v_ret=(Q_in/449)/((Ap-Ar)/144); //ft/s;\n",
+"// rounding off the above answer\n",
+"v_ret=fix(v_ret)+(fix(ceil((v_ret-fix(v_ret))*100))/100); //ft/s\n",
+"// cylinder horsepower during the retraction stroke,\n",
+"HP_ret=(v_ret*F_ret)/550; //HP\n",
+"// Results:\n",
+"printf('\n  Results:  ')   \n",
+"printf('\n The hydraulic pressure during the extending stroke is %.0f psi.',p_ext)\n",
+"printf('\n The piston velocity during the extending stroke is %.2f ft/s.',v_ext)\n",
+"printf('\n The cylinder horsepower during the extending stroke is %.2f HP.',HP_ext)\n",
+"printf('\n The hydraulic pressure during the retraction stroke is %.0f psi.',p_ret)\n",
+"printf('\n The piston velocity during the retraction stroke is %.2f ft/s.',v_ret)\n",
+"printf('\n The cylinder horsepower during the retraction stroke is %.2f HP.',HP_ret)"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 6.2_a: find_cylinder_force_to_move_6000lb.sci"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"// Aim:Refer Example 6-2 for Problem Description \n",
+"// Given:\n",
+"// Weight of Body:\n",
+"W=6000; //lb\n",
+"// coefficient of friction between weight and horizontal support:\n",
+"CF=0.14; "
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 6.2_b: SOLUTION_cylinder_force_to_move_6000lb.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"clc;\n",
+"pathname=get_absolute_file_path('6_2_soln.sce')\n",
+"filename=pathname+filesep()+'6_2_data.sci'\n",
+"exec(filename)\n",
+"// Solution:\n",
+"// Cylinder Force,\n",
+"F=CF*W; //lb\n",
+"// Results:\n",
+"printf('\n  Results:  ')   \n",
+"printf('\n The Cylinder Force at constant velocity is %.0f lb.',F)"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 6.3_a: find_force_to_move_inclined_weight.sci"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"// Aim:Refer Example 6-3 for Problem Description \n",
+"// Given:\n",
+"// Weight of Body:\n",
+"W=6000; //lb\n",
+"// Inclination of Weight:\n",
+"theta=30; //deg"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 6.3_b: SOLUTION_force_to_move_inclined_weight.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"clc;\n",
+"pathname=get_absolute_file_path('6_3_soln.sce')\n",
+"filename=pathname+filesep()+'6_3_data.sci'\n",
+"exec(filename)\n",
+"// Solution:\n",
+"// Inclination of Weight,\n",
+"theta=(theta*%pi)/180; //rad\n",
+"// Cylinder Force,\n",
+"F=W*sin(theta); //lb\n",
+"// Results:\n",
+"printf('\n  Results:  ')   \n",
+"printf('\n The Cylinder Force at constant velocity is %.0f lb.',F)"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 6.4_a: find_cylinder_force_to_accelerate_weight.sci"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"// Aim:Refer Example 6-4 for Problem Description \n",
+"// Given:\n",
+"// Weight of Body:\n",
+"W=6000; //lb\n",
+"// initial velocity:\n",
+"u=0; //ft/s\n",
+"// final velocity:\n",
+"v=8; //ft/s\n",
+"// Time taken:\n",
+"t=0.5; //s"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 6.4_b: SOLUTION_cylinder_force_to_accelerate_weight.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"clc;\n",
+"pathname=get_absolute_file_path('6_4_soln.sce')\n",
+"filename=pathname+filesep()+'6_4_data.sci'\n",
+"exec(filename)\n",
+"// Solution:\n",
+"// For constant velocity,Cylinder Force,\n",
+"F=W; //lb\n",
+"// Rate of change of velocity,\n",
+"a=(v-u)/t; //ft/s^2\n",
+"// Force required to accelerate the weight,\n",
+"F_acc=(F/32.2)*a; //lb\n",
+"// Therefore, Cylinder Force,\n",
+"F_cyl=(F+F_acc); //lb\n",
+"// Results:\n",
+"printf('\n  Results:  ')   \n",
+"printf('\n The Cylinder Force at constant velocity is %.0f lb.',F)\n",
+"printf('\n The Cylinder Force required to accelerate the Body is %.0f lb.',F_cyl)\n",
+"printf('\n The answer in the program is different than that in textbook. It may be due to no.s of significant digit in data and calculation')"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 6.5_a: find_cylinder_force_using_lever_system.sci"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"// Aim:Refer Example 6-5 for Problem Description\n",
+"// Given:\n",
+"L1=10; //in\n",
+"L2=10; //in\n",
+"// Inclination of cylinder axis with vertical axis:\n",
+"phi=0; //deg\n",
+"// cylinder load:\n",
+"F_load=1000; //lb"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 6.5_b: SOLUTION_cylinder_force_using_lever_system.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"clc;\n",
+"pathname=get_absolute_file_path('6_5_soln.sce')\n",
+"filename=pathname+filesep()+'6_5_data.sci'\n",
+"exec(filename)\n",
+"// Solution:\n",
+"// Inclination of cylinder axis with vertical axis,\n",
+"phi=(phi*%pi)/180; //rad\n",
+"// cylinder force required to overcome load using First Class Lever Sytem,\n",
+"F_cyl_1=(L2*F_load)/(L1*cos(phi)); //lb\n",
+"// cylinder force required to overcome load using Second Class Lever Sytem,\n",
+"F_cyl_2=(L2*F_load)/((L1+L2)*cos(phi)); //lb\n",
+"// cylinder force required to overcome load using Third Class Lever Sytem,\n",
+"F_cyl_3=((L1+L2)*F_load)/(L2*cos(phi)); //lb\n",
+"// Results:\n",
+"printf('\n  Results:  ')   \n",
+"printf('\n The Cylinder Force using First Class lever System is %.0f lb.',F_cyl_1)\n",
+"printf('\n The Cylinder Force using Second Class lever System is %.0f lb.',F_cyl_2)\n",
+"printf('\n The Cylinder Force using Third Class lever System is %.0f lb.',F_cyl_3)"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 6.6_a: find_maximum_pressure_developed_by_cushion.sci"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"// Aim:Refer Example 6-6 for Problem Description \n",
+"// Given:\n",
+"// Flow rate of pump:\n",
+"Q_pump=18.2; //gpm\n",
+"// Diameter of blank end of piston:\n",
+"D=3; //in\n",
+"// Diameter of cushion plunger:\n",
+"D_cush=1; //in\n",
+"// Stroke of cushion plunger:\n",
+"L_cush=0.75; //in\n",
+"// Distance Piston decelerates at the end of extending stroke:\n",
+"L=0.75; //in\n",
+"// Weight of Body:\n",
+"W=1500; //lb\n",
+"// coefficient of friction:\n",
+"CF=0.12;\n",
+"// Pressure relief valve settings:\n",
+"p_relf=750; //psi\n",
+"// maximum pressure at the Blank end:\n",
+"p1=750; //psi"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 6.6_b: SOLUTION_maximum_pressure_developed_by_cushio.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"clc;\n",
+"pathname=get_absolute_file_path('6_6_soln.sce')\n",
+"filename=pathname+filesep()+'6_6_data.sci'\n",
+"exec(filename)\n",
+"// Solution:\n",
+"// Area of blank end of piston,\n",
+"A_piston=(%pi/4)*(D^2); //in^2\n",
+"// piston velocity prior to deceleration,\n",
+"v=(Q_pump/449)/(A_piston/144); //ft/s\n",
+"// deceleration of piston at the end of extending stroke,\n",
+"a=(v^2)/(2*(L/12)); //ft/s^2\n",
+"// Area of cushion plunger,\n",
+"A_cush=(%pi/4)*(D_cush^2); //in^2\n",
+"// maximum pressure developed by the cushion,\n",
+"p2=(((W*a)/32.2)+(p1*A_piston)-(CF*W))/(A_piston-A_cush); //psi\n",
+"// Results:\n",
+"printf('\n  Results:  ')   \n",
+"printf('\n The maximum pressure developed by the cushion is %.0f psi.',p2)"
+   ]
+   }
+],
+"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/Fluid_Power_With_Applications_by_A_Esposito/7-HYDRAULIC_MOTORS.ipynb b/Fluid_Power_With_Applications_by_A_Esposito/7-HYDRAULIC_MOTORS.ipynb
new file mode 100644
index 0000000..269d88e
--- /dev/null
+++ b/Fluid_Power_With_Applications_by_A_Esposito/7-HYDRAULIC_MOTORS.ipynb
@@ -0,0 +1,510 @@
+{
+"cells": [
+ {
+		   "cell_type": "markdown",
+	   "metadata": {},
+	   "source": [
+       "# Chapter 7: HYDRAULIC MOTORS"
+	   ]
+	},
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 7.1_a: find_pressure_developed_to_overcome_load.sci"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"// Aim:To determine pressure developed to overcome load\n",
+"// Given:\n",
+"// outer radius of rotor:\n",
+"R_R=0.5; //in\n",
+"// outer radius of vane:\n",
+"R_V=1.5; //in\n",
+"// width of vane:\n",
+"L=1; //in\n",
+"// Torque Load:\n",
+"T=1000; //in.lb"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 7.1_b: SOLUTION_pressure_developed_to_overcome_load.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"clc;\n",
+"pathname=get_absolute_file_path('7_1_soln.sce')\n",
+"filename=pathname+filesep()+'7_1_data.sci'\n",
+"exec(filename)\n",
+"// Solution:\n",
+"// volumetric displacement,\n",
+"V_D=%pi*((R_V^2)-(R_R^2))*L; //in^3\n",
+"// pressure developed to overcome load,\n",
+"p=2*%pi*T/V_D; //psi\n",
+"// Results:\n",
+"printf('\n  Results:  ')   \n",
+"printf('\n The pressure developed to overcome load is %.0f psi.',p)"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 7.2_a: determine_theoretical_horsepower_of_hydraulic_motor.sci"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"// Aim:Refer Example 7-2 for Problem Description \n",
+"// Given:\n",
+"// volumetric displacement:\n",
+"V_D=5; //in^3\n",
+"// pressure rating:\n",
+"p=1000; //psi\n",
+"// theoretical flow-rate of pump:\n",
+"Q_T=10; //gpm"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 7.2_b: SOLUTION_theoretical_horsepower_of_hydraulic_motor.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"clc;\n",
+"pathname=get_absolute_file_path('7_2_soln.sce')\n",
+"filename=pathname+filesep()+'7_2_data.sci'\n",
+"exec(filename)\n",
+"// Solution:\n",
+"// motor speed,\n",
+"N=231*Q_T/V_D; //rpm\n",
+"// Theoretical torque,\n",
+"T_T=floor(V_D*p/(2*%pi)); //in.lb\n",
+"// Theoretical horsepower,\n",
+"HP_T=T_T*N/63000; //HP\n",
+"// Results:\n",
+"printf('\n  Results:  ')   \n",
+"printf('\n The motor Speed is %.0f rpm.',N)\n",
+"printf('\n The motor Theoretical torque is %.0f in.lb.',T_T)\n",
+"printf('\n The motor Theoretical horsepower is %.2f HP.',HP_T)"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 7.3_a: find_actual_horsepower_delivered_by_motor.sci"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"// Aim:Refer Example 7-3 for Problem Description \n",
+"// Given:\n",
+"// volumetric displacement:\n",
+"V_D=10; //in^3\n",
+"// pressure rating:\n",
+"p=1000; //psi\n",
+"// speed of motor:\n",
+"N=2000; //rpm\n",
+"// actual flow-rate of motor:\n",
+"Q_A=95; //gpm\n",
+"// actual torque delivered by motor:\n",
+"T_A=1500; //in.lb"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 7.3_b: SOLUTION_actual_horsepower_delivered_by_motor.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"clc;\n",
+"pathname=get_absolute_file_path('7_3_soln.sce')\n",
+"filename=pathname+filesep()+'7_3_data.sci'\n",
+"exec(filename)\n",
+"// Solution:\n",
+"// theoretical flow-rate,\n",
+"Q_T=V_D*N/231; //gpm\n",
+"// volumetric efficiency,\n",
+"eta_v=(Q_T/Q_A)*100; //%\n",
+"// theoretical torque,\n",
+"T_T=(V_D*p/(2*%pi)); //in.lb\n",
+"// mechanical efficiency,\n",
+"eta_m=(T_A/T_T)*100; //%\n",
+"// overall efficiency,\n",
+"eta_o=(eta_v/100)*(eta_m/100)*100; //%\n",
+"eta_o=fix(eta_o)+(fix(floor((eta_o-fix(eta_o))*10))/10); //% ,rounding off the answer\n",
+"// actual horsepower delivered by motor,\n",
+"HP_A=T_A*N/63000; //HP\n",
+"// Results:\n",
+"printf('\n  Results:  ')   \n",
+"printf('\n The volumetric efficiency is %.1f percent.',eta_v)\n",
+"printf('\n The mechanical efficiency is %.1f percent.',eta_m)\n",
+"printf('\n The overall efficiency is %.1f percent.',eta_o)\n",
+"printf('\n The actual horsepower delivered by the motor is %.1f HP.',HP_A)"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 7.4_a: find_motor_displacement_and_output_torque.sci"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"// Aim:Refer Example 7-4 for Problem Description \n",
+"// Given:\n",
+"// operating pressure:\n",
+"p=1000; //psi\n",
+"// volumetric displacement of pump:\n",
+"V_D_pump=5; //in^3\n",
+"// speed of pump:\n",
+"N_pump=500; //rpm\n",
+"// volumetric efficiency of pump:\n",
+"eta_v_pump=82; //%\n",
+"// mechanical efficiency of pump:\n",
+"eta_m_pump=88; //%\n",
+"// speed of motor:\n",
+"N_motor=400; //rpm\n",
+"// volumetric efficiency of motor:\n",
+"eta_v_motor=92; //%\n",
+"// mechanical efficiency of motor:\n",
+"eta_m_motor=90; //%"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 7.4_b: SOLUTION_motor_displacement_and_output_torque.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"clc;\n",
+"pathname=get_absolute_file_path('7_4_soln.sce')\n",
+"filename=pathname+filesep()+'7_4_data.sci'\n",
+"exec(filename)\n",
+"// Solution:\n",
+"// pump theoretical flow-rate,\n",
+"Q_T_pump=V_D_pump*N_pump/231; //gpm\n",
+"// pump actual flow rate,\n",
+"Q_A_pump=Q_T_pump*(eta_v_pump/100); //gpm\n",
+"// motor theoretical flow-rate,\n",
+"Q_T_motor=Q_A_pump*(eta_v_motor/100); //gpm ,motor actual flow-rate = pump actual flow rate\n",
+"// motor displacement,\n",
+"V_D_motor=Q_T_motor*231/N_motor; //in^3\n",
+"// hydraulic HP delivered to motor,\n",
+"HHP_motor=p*Q_A_pump/1714; //HP\n",
+"// brake HP delivered by motor,\n",
+"BHP_motor=HHP_motor*(eta_v_motor/100)*(eta_m_motor/100); //HP\n",
+"BHP_motor=fix(BHP_motor)+(fix(floor((BHP_motor-fix(BHP_motor))*100))/100); //HP ,rounding off the answer\n",
+"// torque delivered by motor,\n",
+"T_motor=(BHP_motor*63000/N_motor); //in.lb\n",
+"// Results:\n",
+"printf('\n  Results:  ')   \n",
+"printf('\n The Displacement of motor is %.2f in^3.',V_D_motor)\n",
+"printf('\n The Motor output torque is %.0f in.lb.',T_motor)"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 7.5_a: find_motor_theoretical_power_in_SI.sci"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"// Aim:Refer Example 7-5 for Problem Description \n",
+"// Given:\n",
+"// volumetric displacement:\n",
+"V_D=0.082; //L\n",
+"// pressure rating:\n",
+"p=70; //bar\n",
+"// theoretical flow-rate of pump:\n",
+"Q_T=0.0006; //m^3/s"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 7.5_b: SOLUTION_motor_theoretical_power_in_SI.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"clc;\n",
+"pathname=get_absolute_file_path('7_5_soln.sce')\n",
+"filename=pathname+filesep()+'7_5_data.sci'\n",
+"exec(filename)\n",
+"// Solution:\n",
+"// motor speed,\n",
+"N=(Q_T*60)/(V_D*10^-3); //rpm\n",
+"// Theoretical torque,\n",
+"T_T=((V_D*10^-3)*(p*10^5))/(2*%pi); //Nm\n",
+"// Theoretical power,\n",
+"HP_T=T_T*N*2*%pi/(60*1000); //kW\n",
+"// Results:\n",
+"printf('\n  Results:  ')   \n",
+"printf('\n The motor Speed is %.0f rpm.',N)\n",
+"printf('\n The motor Theoretical torque is %.1f Nm.',T_T)\n",
+"printf('\n The motor Theoretical power is %.2f kW.',HP_T)"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 7.6_a: find_actual_KW_delivered_by_motor.sci"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"// Aim:Refer Example 7-6 for Problem Description \n",
+"// Given:\n",
+"// volumetric displacement:\n",
+"V_D=164; //cm^3\n",
+"// pressure rating:\n",
+"p=70; //bar\n",
+"// speed of motor:\n",
+"N=2000; //rpm\n",
+"// actual flow-rate of motor:\n",
+"Q_A=0.006; //m^3/s\n",
+"// actual torque delivered by motor:\n",
+"T_A=170; //Nm"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 7.6_b: SOLUTION_actual_KW_delivered_by_motor.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"clc;\n",
+"pathname=get_absolute_file_path('7_6_soln.sce')\n",
+"filename=pathname+filesep()+'7_6_data.sci'\n",
+"exec(filename)\n",
+"// Solution:\n",
+"// theoretical flow-rate,\n",
+"Q_T=(V_D*10^-6)*(N/60); //m^3/s\n",
+"Q_T=fix(Q_T)+(fix(ceil((Q_T-fix(Q_T))*10^5))/10^5); //m^3/s ,rounding off the answer\n",
+"// volumetric efficiency,\n",
+"eta_v=(Q_T/Q_A)*100; //%\n",
+"// theoretical torque,\n",
+"T_T=((V_D*10^-6)*(p*10^5))/(2*%pi); //Nm\n",
+"// mechanical efficiency,\n",
+"eta_m=(T_A/T_T)*100; //%\n",
+"// overall efficiency,\n",
+"eta_o=(eta_v/100)*(eta_m/100)*100; //%\n",
+"eta_o=fix(eta_o)+(fix(floor((eta_o-fix(eta_o))*10))/10); //% ,rounding off the answer\n",
+"// actual horsepower delivered by motor,\n",
+"HP_A=(T_A*N*2*%pi)/(60*1000); //kW\n",
+"// Results:\n",
+"printf('\n  Results:  ')   \n",
+"printf('\n The volumetric efficiency is %.1f percent.',eta_v)\n",
+"printf('\n The mechanical efficiency is %.1f percent.',eta_m)\n",
+"printf('\n The overall efficiency is %.1f percent.',eta_o)\n",
+"printf('\n The actual horsepower delivered by the motor is %.1f kW.',HP_A)"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 7.7_a: find_motor_output_torque_in_SI.sci"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"// Aim:Refer Example 7-7 for Problem Description \n",
+"// Given:\n",
+"// operating pressure:\n",
+"p=70; //bar\n",
+"// volumetric displacement of pump:\n",
+"V_D_pump=82; //cm^3\n",
+"// speed of pump:\n",
+"N_pump=500; //rpm\n",
+"// volumetric efficiency of pump:\n",
+"eta_v_pump=82; //%\n",
+"// mechanical efficiency of pump:\n",
+"eta_m_pump=88; //%\n",
+"// speed of motor:\n",
+"N_motor=400; //rpm\n",
+"// volumetric efficiency of motor:\n",
+"eta_v_motor=92; //%\n",
+"// mechanical efficiency of motor:\n",
+"eta_m_motor=90; //%"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 7.7_b: SOLUTION_motor_output_torque_in_SI.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"clc;\n",
+"pathname=get_absolute_file_path('7_7_soln.sce')\n",
+"filename=pathname+filesep()+'7_7_data.sci'\n",
+"exec(filename)\n",
+"// Solution:\n",
+"// pump theoretical flow-rate,\n",
+"Q_T_pump=(V_D_pump*10^-6)*(N_pump/60); //m^3/s\n",
+"// pump actual flow rate,\n",
+"Q_A_pump=Q_T_pump*(eta_v_pump/100); //m^3/s\n",
+"// motor theoretical flow-rate,\n",
+"Q_T_motor=Q_A_pump*(eta_v_motor/100); //m^3/s ,motor actual flow-rate = pump actual-flow rate\n",
+"// motor displacement,\n",
+"V_D_motor=(Q_T_motor/(N_motor/60))*10^6; //cm^3\n",
+"// hydraulic HP delivered to motor,\n",
+"HHP_motor=(p*10^5)*Q_A_pump; //W\n",
+"// brake HP delivered by motor,\n",
+"BHP_motor=HHP_motor*(eta_v_motor/100)*(eta_m_motor/100); //W\n",
+"BHP_motor=fix(BHP_motor)+(fix(floor((BHP_motor-fix(BHP_motor))*100))/100); //W ,rounding off the answer\n",
+"// torque delivered by motor,\n",
+"T_motor=(BHP_motor/N_motor)*(60/(2*%pi)); //Nm\n",
+"// Results:\n",
+"printf('\n  Results:  ')   \n",
+"printf('\n The Displacement of motor is %.1f cm^3.',V_D_motor)\n",
+"printf('\n The Motor output torque is %.1f Nm.',T_motor)"
+   ]
+   }
+],
+"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/Fluid_Power_With_Applications_by_A_Esposito/8-HYDRAULIC_VALVES.ipynb b/Fluid_Power_With_Applications_by_A_Esposito/8-HYDRAULIC_VALVES.ipynb
new file mode 100644
index 0000000..9072260
--- /dev/null
+++ b/Fluid_Power_With_Applications_by_A_Esposito/8-HYDRAULIC_VALVES.ipynb
@@ -0,0 +1,383 @@
+{
+"cells": [
+ {
+		   "cell_type": "markdown",
+	   "metadata": {},
+	   "source": [
+       "# Chapter 8: HYDRAULIC VALVES"
+	   ]
+	},
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 8.1_a: determine_cracking_and_full_flow_pressure.sci"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"// Aim:Refer Example 8-1 for Problem Description \n",
+"// Given:\n",
+"// area of relief valve:\n",
+"A=0.75; //in^2\n",
+"// spring constant:\n",
+"k=2500; //lb/in\n",
+"// initial compressed length of spring:\n",
+"S=0.20; //in\n",
+"// poppet displacement to pass full pump flow:\n",
+"L=0.10; //in"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 8.1_b: SOLUTION_cracking_and_full_flow_pressure.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"clc;\n",
+"pathname=get_absolute_file_path('8_1_soln.sce')\n",
+"filename=pathname+filesep()+'8_1_data.sci'\n",
+"exec(filename)\n",
+"// Solution:\n",
+"// spring force excerted on poppet when it is fully closed,\n",
+"F=k*S; //lb\n",
+"// Cracking pressure,\n",
+"p_crack=F/A; //psi\n",
+"// spring force when poppet moves 0.10 in from its fully closed position,\n",
+"F_new=k*(L+S); //lb\n",
+"// Full pump flow pressure,\n",
+"p_ful_pump_flow=F_new/A; //psi\n",
+"  \n",
+"// Results:\n",
+"printf('\n  Results:  ')   \n",
+"printf('\n The Cracking pressure is %.0f psi.',p_crack)\n",
+"printf('\n The Full pump flow pressure  is %.0f psi.',p_ful_pump_flow)"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 8.2_a: compute_horsepower_across_pressure_relief_valve.sci"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"// Aim:To compute horsepower across the pressure relief valve\n",
+"// Given:\n",
+"// pressure relief valve setting:\n",
+"p=1000; //psi\n",
+"// pump flow to the tank:\n",
+"Q=20; //gpm"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 8.2_b: SOLUTION_horsepower_across_pressure_relief_valve.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"clc;\n",
+"pathname=get_absolute_file_path('8_2_soln.sce')\n",
+"filename=pathname+filesep()+'8_2_data.sci'\n",
+"exec(filename)\n",
+"// Solution:\n",
+"// Horsepower across the valve,\n",
+"HP=((p*Q)/1714); //HP\n",
+" \n",
+"// Results:\n",
+"printf('\n  Results:  ')   \n",
+"printf('\n The Horsepower across the pressure relief valve is %.1f HP.',HP)"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 8.3_a: compute_horsepower_across_unloading_valve.sci"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"// Aim:To compute horsepower across the unloading valve\n",
+"// Given:\n",
+"// pump pressure during unloading:\n",
+"p=25; //psi\n",
+"// pump flow to the tank:\n",
+"Q=20; //gpm"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 8.3_b: SOLUTION_horsepower_across_unloading_valve.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"clc;\n",
+"pathname=get_absolute_file_path('8_3_soln.sce')\n",
+"filename=pathname+filesep()+'8_3_data.sci'\n",
+"exec(filename)\n",
+"// Solution:\n",
+"// Horsepower across the valve,\n",
+"HP=((p*Q)/1714); //HP\n",
+" \n",
+"// Results:\n",
+"printf('\n  Results:  ')   \n",
+"printf('\n The Horsepower across the unloading valve is %.2f HP.',HP)"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 8.4_a: find_flow_rate_through_the_orifice.sci"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"// Aim:To find flow-rate through given orifice\n",
+"// Given:\n",
+"// pressure drop across orifice:\n",
+"del_p=100; //psi\n",
+"// orifice diameter:\n",
+"D=1; //in\n",
+"// specific gravity of oil:\n",
+"SG_oil=0.9;\n",
+"// flow coefficient for sharp edge orifice:\n",
+"C=0.80;\n",
+" "
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 8.4_b: SOLUTION_flow_rate_through_the_orifice.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"clc;\n",
+"pathname=get_absolute_file_path('8_4_soln.sce')\n",
+"filename=pathname+filesep()+'8_4_data.sci'\n",
+"exec(filename)\n",
+"// Solution:\n",
+"// flow-rate through orifice,\n",
+"Q=38.1*C*((%pi*(D^2))/4)*sqrt(del_p/SG_oil); //gpm\n",
+"// Results:\n",
+"printf('\n  Results:  ')   \n",
+"printf('\n The flow-rate through orifice is %.0f gpm.',Q)"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 8.5_a: determine_capacity_coefficient_of_flowcontrol_valve.sci"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"// Aim:To determine the capacity coefficient of flow control valve \n",
+"// Given:\n",
+"// pressure drop across flow control valve:\n",
+"del_p=100; //psi\n",
+"del_p1=687; //kPa\n",
+"// flow-rate across valve:\n",
+"Q=25; //gpm\n",
+"Q1=94.8; //Lpm\n",
+"// specific gravity of oil:\n",
+"SG_oil=0.9; "
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 8.5_b: SOLUTION_capacity_coefficient_of_flowcontrol_valve.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"clc;\n",
+"pathname=get_absolute_file_path('8_5_soln.sce')\n",
+"filename=pathname+filesep()+'8_5_data.sci'\n",
+"exec(filename)\n",
+"// Solution:\n",
+"// capacity coefficient in English Units,\n",
+"Cv=Q/sqrt(del_p/SG_oil); //gpm/sqrt(psi)\n",
+"// capacity coefficient in Metric Units,\n",
+"Cv1=Q1/sqrt(del_p1/SG_oil); //Lpm/sqrt(kPA)\n",
+"// Results:\n",
+"printf('\n  Results:  ')   \n",
+"printf('\n The capacity coefficient in English unit is %.2f gpm/sqrt(psi).',Cv)\n",
+"printf('\n The capacity coefficient in Metric unit is %.2f Lpm/sqrt(kPa).',Cv1)"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 8.6_a: determine_capacity_coefficient_of_needle_valve.sci"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"// Aim:To determine the capacity coefficient of needle valve \n",
+"// Given:\n",
+"// Desired cylinder speed:\n",
+"v2=10; //in/s\n",
+"// Cylinder piston area:\n",
+"A1=3.14; //in^2\n",
+"// Cylinder rod area:\n",
+"Ar=0.79; //in^2\n",
+"// Cylinder load:\n",
+"F_load=1000; //lb\n",
+"// Specific gravity of oil:\n",
+"SG_oil=0.9;\n",
+"// Pressure relief valve setting:\n",
+"p1=500; //psi"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 8.6_b: SOLUTION_capacity_coefficient_of_needle_valve.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"clc;\n",
+"pathname=get_absolute_file_path('8_6_soln.sce')\n",
+"filename=pathname+filesep()+'8_6_data.sci'\n",
+"exec(filename)\n",
+"// Solution:\n",
+"// annular area of cylinder,\n",
+"A2=A1-Ar; //in^2\n",
+"// back pressure in the rod end,\n",
+"p2=((p1*A1)-F_load)/A2; //psi\n",
+"// flow rate through needle valve based on desired cylinder speed,\n",
+"Q=(A2*v2*60)/231; //gpm\n",
+"// capacity coefficient of needle valve,\n",
+"Cv=Q/sqrt(p2/SG_oil); //gpm/sqrt(psi)\n",
+"// Results:\n",
+"printf('\n  Results:  ')   \n",
+"printf('\n The capacity coefficient of needle valve is %.2f gpm/sqrt(psi).',Cv)"
+   ]
+   }
+],
+"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/Fluid_Power_With_Applications_by_A_Esposito/9-HYDRAULIC_CIRCUIT_DESIGN_AND_ANALYSIS.ipynb b/Fluid_Power_With_Applications_by_A_Esposito/9-HYDRAULIC_CIRCUIT_DESIGN_AND_ANALYSIS.ipynb
new file mode 100644
index 0000000..7ed67f9
--- /dev/null
+++ b/Fluid_Power_With_Applications_by_A_Esposito/9-HYDRAULIC_CIRCUIT_DESIGN_AND_ANALYSIS.ipynb
@@ -0,0 +1,429 @@
+{
+"cells": [
+ {
+		   "cell_type": "markdown",
+	   "metadata": {},
+	   "source": [
+       "# Chapter 9: HYDRAULIC CIRCUIT DESIGN AND ANALYSIS"
+	   ]
+	},
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 9.1_a: determine_speed_power_for_regenerative_circuit.sci"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"// Aim:Refer Example 9-1 for Problem Description \n",
+"// Given:\n",
+"// cracking pressure of relief valve:\n",
+"p=1000; //psi\n",
+"// piston area:\n",
+"Ap=25; //in^2\n",
+"// rod area:\n",
+"Ar=7; //in^2\n",
+"// pump flow:\n",
+"Qp=20; //gpm"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 9.1_b: SOLUTION_speed_power_for_regenerative_circuit.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"clc;\n",
+"pathname=get_absolute_file_path('9_1_soln.sce')\n",
+"filename=pathname+filesep()+'9_1_data.sci'\n",
+"exec(filename)\n",
+"// Solution:\n",
+"// cylinder speed during extending stroke,\n",
+"vp_ext=(Qp*231)/(Ar*60); //in/s\n",
+"// load carrying capacity during extending stroke,\n",
+"Fload_ext=p*Ar; //lb\n",
+"// power delivered to load during extending stroke,\n",
+"Power_ext=(Fload_ext*vp_ext)/(550*12); //HP\n",
+"// cylinder speed during retracting stroke,\n",
+"vp_ret=(Qp*231)/((Ap-Ar)*60); //in/s\n",
+"// load carrying capacity during retracting stroke,\n",
+"Fload_ret=p*(Ap-Ar); //lb\n",
+"// power delivered to load during retracting stroke,\n",
+"Power_ret=(Fload_ext*vp_ext)/(550*12); //HP\n",
+"// Results:\n",
+"printf('\n  Results:  ')   \n",
+"printf('\n The cylinder speed during extending stroke is %.1f in/s.',vp_ext)\n",
+"printf('\n The load carrying capacity during extending stroke is %.0f lb.',Fload_ext)\n",
+"printf('\n The power delivered to load during extending stroke is %.1f HP.',Power_ext)\n",
+"printf('\n The cylinder speed during retracting stroke is %.2f in/s.',vp_ret)\n",
+"printf('\n The load carrying capacity during retracting stroke is %.0f lb.',Fload_ret)\n",
+"printf('\n The power delivered to load during retracting stroke is %.1f HP.',Power_ret)"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 9.2_a: find_unloading_relief_valve_pressure_settings.sci"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"// Aim:Refer Example 9-2 for Problem Description \n",
+"// Given:\n",
+"// force required for sheet metal punching operations:\n",
+"F_load=2000; //lb\n",
+"// piston diameter:\n",
+"Dp=1.5; //in\n",
+"// rod diameter:\n",
+"Dr=0.5; //in\n",
+"// frictional pressure loss in line from high-flow pump to blank end during rapid extension:\n",
+"p_loss1=100; //psi\n",
+"// frictional pressure loss in return line from rod end to oil tank during rapid extension:\n",
+"p_loss2=50; //psi "
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 9.2_b: SOLUTION_unloading_relief_valve_pressure_settings.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"clc;\n",
+"pathname=get_absolute_file_path('9_2_soln.sce')\n",
+"filename=pathname+filesep()+'9_2_data.sci'\n",
+"exec(filename)\n",
+"// Solution:\n",
+"// Unloading Valve:\n",
+"// load due to back pressure force on cylinder,\n",
+"F_back_pressure=(p_loss2*%pi*((Dp^2)-(Dr^2)))/4; //psi\n",
+"// back pressure force on cylinder,\n",
+"P_cyl_blank_end=F_back_pressure/((%pi*(Dp^2))/4); //psi\n",
+"// pressure setting of the unloading valve,\n",
+"p_unload=1.5*(P_cyl_blank_end+p_loss1); //psi\n",
+"// Pressure Relief Valve:\n",
+"// pressure to overcome punching operations,\n",
+"P_punching=F_load/((%pi*(Dp^2))/4); //psi\n",
+"// pressure setting of the pressure relief valve,\n",
+"p_prv=1.5*P_punching; //psi\n",
+"// Results:\n",
+"printf('\n  Results:  ')   \n",
+"printf('\n The pressure setting of unloading valve is %.0f psi.',p_unload)\n",
+"printf('\n The pressure setting of pressure relief valve is %.0f psi.',p_prv)"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 9.3_a: find_spring_constant_of_PRV_valve.sci"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"// Aim:Refer Example 9-3 for Problem Description \n",
+"// Given:\n",
+"// poppet area:\n",
+"A_poppet=0.75; //in^2\n",
+"// hydraulic pressure:\n",
+"p_hydraulic=1698; //psi\n",
+"// full poppet stroke:\n",
+"l_stroke=0.10; //in\n",
+"// cracking pressure:\n",
+"p_cracking=1.1*1132; //psi"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 9.3_b: SOLUTION_spring_constant_of_PRV_valve.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"clc;\n",
+"pathname=get_absolute_file_path('9_3_soln.sce')\n",
+"filename=pathname+filesep()+'9_3_data.sci'\n",
+"exec(filename)\n",
+"// Solution:\n",
+"// spring force at full pump flow pressure,\n",
+"F_spr_full=round(p_hydraulic*A_poppet); //lb\n",
+"// spring force at cracking pressure,\n",
+"F_spr_crack=round(p_cracking*A_poppet); //lb\n",
+"// spring constant of compression spring,\n",
+"k=(F_spr_full-F_spr_crack)/l_stroke; //lb/in\n",
+"// initial compression of spring,\n",
+"l=F_spr_crack/k; //in\n",
+"// Results:\n",
+"printf('\n  Results:  ')   \n",
+"printf('\n The spring constant of compression spring is %.0f lb/in.',k)\n",
+"printf('\n The initial compression of spring is %.3f in.',l)"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 9.4_a: determine_cylinder_speed_of_meterin_circuit.sci"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"// Aim:To determine cylinder speed for given meter-in circuit\n",
+"// Given:\n",
+"// valve capacity coefficient:\n",
+"Cv=1.0; //gpm/sqrt(psi)\n",
+"// cylinder piston diameter and area:\n",
+"D=2; //in\n",
+"A_piston=3.14; //in^2\n",
+"// cylinder load:\n",
+"F_load=4000; //lb\n",
+"// specific gravity of oil:\n",
+"SG=0.9;\n",
+"// pressure relief valve setting:\n",
+"p_PRV=1400; //psi"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 9.4_b: SOLUTION_cylinder_speed_of_meterin_circuit.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"clc;\n",
+"pathname=get_absolute_file_path('9_4_soln.sce')\n",
+"filename=pathname+filesep()+'9_4_data.sci'\n",
+"exec(filename)\n",
+"// Solution:\n",
+"// flow-rate through valve,\n",
+"Q=Cv*sqrt((p_PRV-(F_load/A_piston))/SG); //gpm\n",
+"// flow-rate through valve in in^3/s,\n",
+"Q=(Q*231)/60; //in^3/s\n",
+"// cylinder speed,\n",
+"v_cyl=Q/A_piston; //in/s\n",
+"// Results:\n",
+"printf('\n  Results:  ')   \n",
+"printf('\n The cylinder speed is %.1f in/s.',v_cyl)"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 9.5_a: find_overall_efficiency_of_given_system.sci"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"// Aim:Refer Example 9-5 for Problem Description \n",
+"// Given:\n",
+"// Pump:\n",
+"// mechanical efficiency:\n",
+"eff_m_pump=92; //%\n",
+"// volumetric efficiency:\n",
+"eff_v_pump=94; //%\n",
+"// volumetric displacement:\n",
+"V_D_pump=10; //in^3\n",
+"// speed of pump:\n",
+"Np=1000; //rpm\n",
+"// inlet pressure:\n",
+"pi=-4; //psi\n",
+"// Hydraulic Motor:\n",
+"// mechanical efficiency:\n",
+"eff_m_motor=92; //%\n",
+"// volumetric efficiency:\n",
+"eff_v_motor=90; //%\n",
+"// volumetric displacement:\n",
+"V_D_motor=8; //in^3\n",
+"// inlet pressure required to drive load:\n",
+"p2=500; //psi\n",
+"// motor discharge pressure:\n",
+"po=5; //psi\n",
+"// Pipe and Fittings:\n",
+"// inside diameter of pipe:\n",
+"D=1.040; //in\n",
+"// Length of pipe between station 1 and 2:\n",
+"L_pipe=50; //ft\n",
+"// K factor of standard 90 deg elbow:\n",
+"K_elbow=0.75;\n",
+"// K factor of check valve:\n",
+"K_check=4.0;\n",
+"// Oil:\n",
+"// kinematic viscosity of oil:\n",
+"nu=125; //cS\n",
+"// specific gravity of oil:\n",
+"SG=0.9;"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 9.5_b: SOLUTION_overall_efficiency_of_given_system.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"clc;\n",
+"pathname=get_absolute_file_path('9_5_soln.sce')\n",
+"filename=pathname+filesep()+'9_5_data.sci'\n",
+"exec(filename)\n",
+"// Solution:\n",
+"// acceleration due to gravity,\n",
+"g=32.2; //ft/s^2\n",
+"// pump's theoretical flow-rate,\n",
+"Q_T_pump=(V_D_pump*Np)/231; //gpm\n",
+"// pump's actual flow-rate,\n",
+"Q_A_pump=(Q_T_pump*eff_v_pump)/100; //gpm\n",
+"// velocity of oil,\n",
+"v=((Q_A_pump)/449)/((%pi*((D/12)^2))/4); //ft/s\n",
+"// Reynolds number,\n",
+"N_R=(7740*v*D)/nu; \n",
+"// friction factor,\n",
+"f=64/N_R; \n",
+"// equivalent length of 90 deg standard elbow,\n",
+"Le_elbow=(K_elbow*(D/12))/f; //ft\n",
+"// equivalent length of check valve,\n",
+"Le_check_valve=(K_check*(D/12))/f; //ft\n",
+"// total length of pipe,\n",
+"LeTOT=L_pipe+(2*Le_elbow)+Le_check_valve; //ft\n",
+"// head loss due to friction,\n",
+"H_L=(f*LeTOT*(v^2))/(2*g*(D/12)); //ft\n",
+"// head developed due to hydraulic motor and pump,\n",
+"Hp=0; //ft\n",
+"Hm=0; //ft\n",
+"// height difference between station 1 and station 2,\n",
+"Z=20; //ft\n",
+"// pump discharge pressure,\n",
+"p1=(((Z+H_L+Hm+Hp)*SG*62.4)/144)+p2; //psi\n",
+"// input HP required to drive pump,\n",
+"HP_pump=((p1-pi)*Q_A_pump)/(1714*(eff_m_pump/100)*(eff_v_pump/100)); //Hp\n",
+"// motor theoretical power,\n",
+"Q_T_motor=Q_A_pump*(eff_v_motor/100); //gpm\n",
+"// speed of motor,\n",
+"N_motor=floor((Q_T_motor*231)/V_D_motor); //rpm\n",
+"// motor input horsepower,\n",
+"HP_input_motor=((p2-po)*Q_A_pump)/1714; //HP\n",
+"// rounding off the above answer\n",
+"HP_input_motor=fix(HP_input_motor)+(fix(ceil((HP_input_motor-fix(HP_input_motor))*10))/10); //HP\n",
+"// motor output horsepower,\n",
+"HP_output_motor=(HP_input_motor*(eff_m_motor/100)*(eff_v_motor/100)); //HP\n",
+"// motor output torque,\n",
+"T_output_motor=(HP_output_motor*63000)/N_motor; //in.lb\n",
+"// overall efficiency of system,\n",
+"eff_overall=(HP_output_motor/HP_pump)*100; //%\n",
+"// rounding off the above answer\n",
+"eff_overall=fix(eff_overall)+(fix(ceil((eff_overall-fix(eff_overall))*10))/10); //%\n",
+"// Results:\n",
+"printf('\n  Results:  ')   \n",
+"printf('\n The Pump flow-rate is %.1f gpm.',Q_A_pump)\n",
+"printf('\n The Pump discharge pressure is %.0f psi.',p1)\n",
+"printf('\n The Input HP required to drive the pump is %.1f HP.',HP_pump)\n",
+"printf('\n The Motor Speed is %.0f rpm.',N_motor)\n",
+"printf('\n The Motor output torque is %.0f in.lb.',T_output_motor)\n",
+"printf('\n The Overall efficiency of system is %.1f percent.',eff_overall)"
+   ]
+   }
+],
+"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
+}