From 476705d693c7122d34f9b049fa79b935405c9b49 Mon Sep 17 00:00:00 2001 From: prashantsinalkar Date: Tue, 14 Apr 2020 10:19:27 +0530 Subject: Initial commit --- .../10-HYDRAULIC_CONDUCTORS_AND_FITTINGS.ipynb | 357 ++++++ .../11-ANCILLARY_HYDRAULIC_DEVICES.ipynb | 337 ++++++ ...PNEUMATICS_AIR_PREPARATION_AND_COMPONENTS.ipynb | 843 ++++++++++++++ .../14-PNEUMATICS_CIRCUITS_AND_APPLICATIONS.ipynb | 722 ++++++++++++ ...ECTRICAL_CONTROLS_FOR_FLUID_POWER_SYSTEMS.ipynb | 234 ++++ ...2-PHYSICAL_PROPERTIES_OF_HYDRAULIC_FLUIDS.ipynb | 1028 +++++++++++++++++ .../3-ENERGY_AND_POWER_IN_HYDRAULIC_SYSTEMS.ipynb | 1168 ++++++++++++++++++++ ...-FRICTIONAL_LOSSES_IN_HYDRAULIC_PIPELINES.ipynb | 718 ++++++++++++ .../5-HYDRAULIC_PUMPS.ipynb | 690 ++++++++++++ ...YDRAULIC_CYLINDERS_AND_CUSHIONING_DEVICES.ipynb | 419 +++++++ .../7-HYDRAULIC_MOTORS.ipynb | 510 +++++++++ .../8-HYDRAULIC_VALVES.ipynb | 383 +++++++ .../9-HYDRAULIC_CIRCUIT_DESIGN_AND_ANALYSIS.ipynb | 429 +++++++ 13 files changed, 7838 insertions(+) create mode 100644 Fluid_Power_With_Applications_by_A_Esposito/10-HYDRAULIC_CONDUCTORS_AND_FITTINGS.ipynb create mode 100644 Fluid_Power_With_Applications_by_A_Esposito/11-ANCILLARY_HYDRAULIC_DEVICES.ipynb create mode 100644 Fluid_Power_With_Applications_by_A_Esposito/13-PNEUMATICS_AIR_PREPARATION_AND_COMPONENTS.ipynb create mode 100644 Fluid_Power_With_Applications_by_A_Esposito/14-PNEUMATICS_CIRCUITS_AND_APPLICATIONS.ipynb create mode 100644 Fluid_Power_With_Applications_by_A_Esposito/17-ADVANCED_ELECTRICAL_CONTROLS_FOR_FLUID_POWER_SYSTEMS.ipynb create mode 100644 Fluid_Power_With_Applications_by_A_Esposito/2-PHYSICAL_PROPERTIES_OF_HYDRAULIC_FLUIDS.ipynb create mode 100644 Fluid_Power_With_Applications_by_A_Esposito/3-ENERGY_AND_POWER_IN_HYDRAULIC_SYSTEMS.ipynb create mode 100644 Fluid_Power_With_Applications_by_A_Esposito/4-FRICTIONAL_LOSSES_IN_HYDRAULIC_PIPELINES.ipynb create mode 100644 Fluid_Power_With_Applications_by_A_Esposito/5-HYDRAULIC_PUMPS.ipynb create mode 100644 Fluid_Power_With_Applications_by_A_Esposito/6-HYDRAULIC_CYLINDERS_AND_CUSHIONING_DEVICES.ipynb create mode 100644 Fluid_Power_With_Applications_by_A_Esposito/7-HYDRAULIC_MOTORS.ipynb create mode 100644 Fluid_Power_With_Applications_by_A_Esposito/8-HYDRAULIC_VALVES.ipynb create mode 100644 Fluid_Power_With_Applications_by_A_Esposito/9-HYDRAULIC_CIRCUIT_DESIGN_AND_ANALYSIS.ipynb (limited to 'Fluid_Power_With_Applications_by_A_Esposito') 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 index 0000000..ca5dee8 --- /dev/null +++ b/Fluid_Power_With_Applications_by_A_Esposito/10-HYDRAULIC_CONDUCTORS_AND_FITTINGS.ipynb @@ -0,0 +1,357 @@ +{ +"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 +} -- cgit