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| author | hardythe1 | 2015-07-03 12:23:43 +0530 |
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| committer | hardythe1 | 2015-07-03 12:23:43 +0530 |
| commit | 9d260e6fae7328d816a514130b691fbd0e9ef81d (patch) | |
| tree | 9e6035702fca0f6f8c5d161de477985cacad7672 /Fluid_Power_With_Applications_by_A._Esposito | |
| parent | afcd9e5397e3e1bde0392811d0482d76aac391dc (diff) | |
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diff --git a/Fluid_Power_With_Applications_by_A._Esposito/Chapter_10_Hydraulic_Conductors_and_Fittings.ipynb b/Fluid_Power_With_Applications_by_A._Esposito/Chapter_10_Hydraulic_Conductors_and_Fittings.ipynb new file mode 100755 index 00000000..fb2937bf --- /dev/null +++ b/Fluid_Power_With_Applications_by_A._Esposito/Chapter_10_Hydraulic_Conductors_and_Fittings.ipynb @@ -0,0 +1,333 @@ +{
+ "cells": [
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "# Chapter 10: Hydraulic Conductors and Fittings"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 10.1 pgno:367"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 1,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "\n",
+ " Results: \n",
+ "\n",
+ " The minimum inside diameter of pipe is in. 0.783\n"
+ ]
+ }
+ ],
+ "source": [
+ "# Aim:To find minimum inside diameter of pipe \n",
+ "# Given:\n",
+ "# flow-rate through pipe:\n",
+ "Q=30.0; #gpm\n",
+ "# average fluid velocity:\n",
+ "v=20.0; #ft/s\n",
+ " \n",
+ "import math \n",
+ "from math import pi\n",
+ "\n",
+ "\n",
+ "\n",
+ "\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=((4*A)/(pi))**0.5; #in\n",
+ "\n",
+ "# Results:\n",
+ "print\"\\n Results: \" \n",
+ "print\"\\n The minimum inside diameter of pipe is in.\",round(D,3)"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 10.2 pgno:368"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 2,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "\n",
+ " Results: \n",
+ "\n",
+ " The minimum inside diameter of pipe is mm. 20.4\n"
+ ]
+ }
+ ],
+ "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\n",
+ "\n",
+ "from math import pi\n",
+ "\n",
+ "\n",
+ "\n",
+ "# Solution:\n",
+ "# minimum required pipe flow area,\n",
+ "A=(Q/v); #m^2\n",
+ "# minimum inside diameter,\n",
+ "D=(((4*A)/(pi))**0.5)*1000; #mm\n",
+ "\n",
+ "# Results:\n",
+ "print\"\\n Results: \" \n",
+ "print\"\\n The minimum inside diameter of pipe is mm.\",round(D,1)\n"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 10.3 pgno:371"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 3,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "\n",
+ " Results: \n",
+ "\n",
+ " The working pressure of steel tube is psi. 1.232\n",
+ "\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"
+ ]
+ }
+ ],
+ "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=55000.0; #psi\n",
+ "# factor of safety:\n",
+ "FS=8.0;\n",
+ "\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/1000; #psi\n",
+ "\n",
+ "# Results:\n",
+ "print\"\\n Results: \" \n",
+ "print\"\\n The working pressure of steel tube is psi.\",round(WP,3)\n",
+ "print\"\\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",
+ "\n"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 10.4 pgno:380"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 4,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ " \n",
+ " The working pressure psi is not adequate (less than psi) so next case is considered, 746.951219512 1000\n",
+ " \n",
+ " The working pressure psi is greater than psi) , 1027.29885057 1000\n",
+ "\n",
+ " Results: \n",
+ "\n",
+ " The ratio of inner diameter to length is . 13.4\n",
+ "\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"
+ ]
+ }
+ ],
+ "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;\n",
+ "from math import pi\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=((4*Ai)/(pi))**0.5; #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",
+ "print\" \\n The working pressure psi is not adequate (less than 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",
+ "print\" \\n The working pressure psi is greater than psi) ,\",WP2,p\n",
+ "\n",
+ "# Results:\n",
+ "print\"\\n Results: \"\n",
+ "print\"\\n The ratio of inner diameter to length is .\",round(r2,1)\n",
+ "print\"\\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"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 10.5 pgno:389"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 5,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ " \n",
+ " The working pressure bars is not adequate (less than bars) so next case is considered, 47.4\n",
+ " \n",
+ " The working pressure bars is greater than bars) , 79.0\n",
+ "\n",
+ " Results: \n",
+ "\n",
+ " The ratio of inner diameter to length is 12.0\n"
+ ]
+ }
+ ],
+ "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=70.00000000000; #bars\n",
+ "# maximum velocity:\n",
+ "v=6.10000000000; #m/s\n",
+ "# tensile strength of material:\n",
+ "S=379.0000000; #MPa\n",
+ "# factor of safety:\n",
+ "FS=8.0000000;\n",
+ "from math import pi\n",
+ "\n",
+ "# Solutions:\n",
+ "# minimum required pipe flow area,\n",
+ "A=(Q/v); #m^2\n",
+ "# minimum inside diameter,\n",
+ "ID=(((4*A)/(pi))**0.5)*1000; #mm\n",
+ "# wall thickness,\n",
+ "t1=1.; t2=2.; #mm\n",
+ "# tube inside diameter,\n",
+ "D1=20.; D2=24.0; #mm\n",
+ "# burst pressure,\n",
+ "BP1=(2*(t1/1000)*S)/(D1/1000); #MPa\n",
+ "# working pressure,\n",
+ "WP1=(BP1/FS)*10; #bars\n",
+ "print\" \\n The working pressure bars is not adequate (less than bars) so next case is considered,\",round(WP1,1)\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",
+ "print\" \\n The working pressure bars is greater than bars) ,\",round(WP2)\n",
+ "\n",
+ "\n",
+ "# Results:\n",
+ "print\"\\n Results: \"\n",
+ "print\"\\n The ratio of inner diameter to length is \",r2"
+ ]
+ }
+ ],
+ "metadata": {
+ "kernelspec": {
+ "display_name": "Python 2",
+ "language": "python",
+ "name": "python2"
+ },
+ "language_info": {
+ "codemirror_mode": {
+ "name": "ipython",
+ "version": 2
+ },
+ "file_extension": ".py",
+ "mimetype": "text/x-python",
+ "name": "python",
+ "nbconvert_exporter": "python",
+ "pygments_lexer": "ipython2",
+ "version": "2.7.9"
+ }
+ },
+ "nbformat": 4,
+ "nbformat_minor": 0
+}
diff --git a/Fluid_Power_With_Applications_by_A._Esposito/Chapter_11_Ancillary_Hydraulic_Devices.ipynb b/Fluid_Power_With_Applications_by_A._Esposito/Chapter_11_Ancillary_Hydraulic_Devices.ipynb new file mode 100755 index 00000000..05caf661 --- /dev/null +++ b/Fluid_Power_With_Applications_by_A._Esposito/Chapter_11_Ancillary_Hydraulic_Devices.ipynb @@ -0,0 +1,302 @@ +{
+ "cells": [
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "# Chapter 11: Ancillary Hydraulic Devices"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 11.1 pgno:409"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 1,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "\n",
+ " Results: \n",
+ "\n",
+ " The high discharge pressure is psi. 2500.0\n",
+ "\n",
+ " The low discharge flow-rate is gpm. 4.0\n"
+ ]
+ }
+ ],
+ "source": [
+ "# Aim:To find the discharge flow and pressure \n",
+ "# Given:\n",
+ "# high inlet flow-rate:\n",
+ "Q_high_inlet=20.0; #gpm\n",
+ "# low inlet pressure:\n",
+ "p_low_inlet=500.0; #psi\n",
+ "# Ratio of piston area to rod area:\n",
+ "Ratio=5/1;\n",
+ "\n",
+ "\n",
+ "\n",
+ "\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",
+ "\n",
+ "# Results:\n",
+ "print\"\\n Results: \" \n",
+ "print\"\\n The high discharge pressure is psi.\",p_high_discharge\n",
+ "print\"\\n The low discharge flow-rate is gpm.\",Q_low_discharge\n"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 11.2 pgno:424"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 2,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "\n",
+ " Results: \n",
+ "\n",
+ " The downstream oil temperature is deg F. 127.9\n"
+ ]
+ }
+ ],
+ "source": [
+ "# Aim:To find the downstream oil temperature\n",
+ "# Given:\n",
+ "# temperature of oil flowing through pressure relief valve:\n",
+ "T_oil=120.0; #deg F\n",
+ "# pressure of oil flowing through pressure relief valve:\n",
+ "p=1000.0; #psi\n",
+ "# oil flow through pressure relief valve:\n",
+ "Q_gpm=10.0; #gpm\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",
+ "\n",
+ "# Results:\n",
+ "print \"\\n Results: \" \n",
+ "print \"\\n The downstream oil temperature is deg F.\",round(T_downward,1)\n"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 11.3 pgno:424"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 3,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "\n",
+ " Results: \n",
+ "\n",
+ " The downstream oil temperature is deg C. 54.3\n"
+ ]
+ }
+ ],
+ "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=50.0; #deg C\n",
+ "# pressure of oil flowing through pressure relief valve:\n",
+ "p=70.0; #bar\n",
+ "# oil flow through pressure relief valve:\n",
+ "Q=0.000632; #m**3/s\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",
+ "\n",
+ "# Results:\n",
+ "print \"\\n Results: \" \n",
+ "print \"\\n The downstream oil temperature is deg C.\",round(T_downward,1)\n"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 11.4 pgno:425"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 4,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "\n",
+ " Results: \n",
+ "\n",
+ " The heat exchanger rating is Btu/hr. 21565.0\n",
+ "\n",
+ " The answer in the program does not match with that in the textbook due to roundoff error (standard ratings) in textbook\n"
+ ]
+ }
+ ],
+ "source": [
+ "# Aim:To find the rating of heat exchanger required to dissipate generated heat\n",
+ "# Given:\n",
+ "# oil flow-rate:\n",
+ "Q=20.0; #gpm\n",
+ "# operating pressure:\n",
+ "p=1000.0; #psi\n",
+ "# overall efficiency of pump:\n",
+ "eff_overall=85.0; #%\n",
+ "# power lost due to friction:\n",
+ "HP_frict=10.0; #%\n",
+ "\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",
+ "\n",
+ "# Results:\n",
+ "print\"\\n Results: \" \n",
+ "print\"\\n The heat exchanger rating is Btu/hr.\",round(HEx_rating)\n",
+ "print\"\\n The answer in the program does not match with that in the textbook due to roundoff error (standard ratings) in textbook\"\n"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 11.5 pgno:426"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 5,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "\n",
+ " Results: \n",
+ "\n",
+ " The heat exchanger rating is kW. 6.41\n"
+ ]
+ }
+ ],
+ "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=70.0; #bar\n",
+ "# overall efficiency of pump:\n",
+ "eff_overall=85.0; #%\n",
+ "# power lost due to friction:\n",
+ "HP_frict=10.0; #%\n",
+ "\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",
+ "\n",
+ "# Results:\n",
+ "print\"\\n Results: \" \n",
+ "print\"\\n The heat exchanger rating is kW.\",round(HEx_rating,2)\n"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": null,
+ "metadata": {
+ "collapsed": true
+ },
+ "outputs": [],
+ "source": []
+ }
+ ],
+ "metadata": {
+ "kernelspec": {
+ "display_name": "Python 2",
+ "language": "python",
+ "name": "python2"
+ },
+ "language_info": {
+ "codemirror_mode": {
+ "name": "ipython",
+ "version": 2
+ },
+ "file_extension": ".py",
+ "mimetype": "text/x-python",
+ "name": "python",
+ "nbconvert_exporter": "python",
+ "pygments_lexer": "ipython2",
+ "version": "2.7.9"
+ }
+ },
+ "nbformat": 4,
+ "nbformat_minor": 0
+}
diff --git a/Fluid_Power_With_Applications_by_A._Esposito/Chapter_13_Pneumatics_Air_Preparation_and_components.ipynb b/Fluid_Power_With_Applications_by_A._Esposito/Chapter_13_Pneumatics_Air_Preparation_and_components.ipynb new file mode 100755 index 00000000..8671e8b9 --- /dev/null +++ b/Fluid_Power_With_Applications_by_A._Esposito/Chapter_13_Pneumatics_Air_Preparation_and_components.ipynb @@ -0,0 +1,732 @@ +{
+ "cells": [
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "# Chapter 13 :Pneumatics: Air Preparation and components"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 13.1 pgno:471"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 1,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "\n",
+ " Results: \n",
+ "\n",
+ " The new pressure when blank side port is blocked is psig. 78.7\n"
+ ]
+ }
+ ],
+ "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\n",
+ "from math import pi\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",
+ "\n",
+ "# Results:\n",
+ "print\"\\n Results: \" \n",
+ "print\"\\n The new pressure when blank side port is blocked is psig.\",round(p2,1)\n"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 13.2 pgno:472"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 2,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "\n",
+ " Results: \n",
+ "\n",
+ " The final volume of air is in^3. 22.3\n"
+ ]
+ }
+ ],
+ "source": [
+ "# Aim:To find new volume of air at the blank end of cylinder\n",
+ "# Given:\n",
+ "# initial volume:\n",
+ "V1=20.0; #in^3\n",
+ "# constant load:\n",
+ "p=20.0; #psi\n",
+ "# initial temperature of air:\n",
+ "T1=60.0; #deg F\n",
+ "# final temperature of air:\n",
+ "T2=120.0; #degF\n",
+ "\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",
+ "\n",
+ "# Results:\n",
+ "print\"\\n Results: \" \n",
+ "print\"\\n The final volume of air is in^3.\",round(V2,1)\n"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Exmaple 13.3 pgno:473"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 3,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "\n",
+ " Results: \n",
+ "\n",
+ " The final pressure of air at constant volume is psig. 26.7\n"
+ ]
+ }
+ ],
+ "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\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+20.8; #psia\n",
+ "# final pressure of air,\n",
+ "# Gay-Lussac's Law,\n",
+ "p2=(T2/T1)*p1; #psia\n",
+ "p2=p2-14.1; #psig\n",
+ "\n",
+ "# Results:\n",
+ "print\"\\n Results: \" \n",
+ "print\"\\n The final pressure of air at constant volume is psig.\",p2\n"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Chapter 13.4 pgno:475"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 4,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "\n",
+ " Results: \n",
+ "\n",
+ " The final pressure in the cylinder is psig. 1579.8\n"
+ ]
+ }
+ ],
+ "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\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",
+ "\n",
+ "# Results:\n",
+ "print\"\\n Results: \" \n",
+ "print\"\\n The final pressure in the cylinder is psig.\",round(p2,1)\n",
+ "\n"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 13.5 pgno:475"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 5,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "\n",
+ " Results: \n",
+ "\n",
+ " The final pressure in the cylinder is bars absolute. 111.5\n"
+ ]
+ }
+ ],
+ "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\n",
+ "\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",
+ "\n",
+ "# Results:\n",
+ "print\"\\n Results: \" \n",
+ "print\"\\n The final pressure in the cylinder is bars absolute.\",round(p2,1)\n"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 13.6 pgno:483"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 6,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "\n",
+ " Results: \n",
+ "\n",
+ " The cfm of free air compressor must be provided is cfm of free air. 275.0\n"
+ ]
+ }
+ ],
+ "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=30.0; #cfm\n",
+ "# temperature of air from receiver:\n",
+ "T2=90.0; #deg F\n",
+ "# pressure of air coming from receiver:\n",
+ "p2=125.0; #psig\n",
+ "# atmospheric temperature:\n",
+ "T1=70.00; #deg F\n",
+ "# atmospheric pressure:\n",
+ "p1=14.7; #psig \n",
+ "\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",
+ "\n",
+ "# Results:\n",
+ "print\"\\n Results: \" \n",
+ "print\"\\n The cfm of free air compressor must be provided is cfm of free air.\",round(Q1,0)"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 13.7 pgno:484"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 7,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "\n",
+ " Results: \n",
+ "\n",
+ " The required size of a receiver before compressor resumes operation is gal. 659.7\n",
+ "\n",
+ " The required size of a receiver when compressor is running gal. 494.8\n"
+ ]
+ }
+ ],
+ "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\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",
+ "\n",
+ "# Results:\n",
+ "print\"\\n Results: \" \n",
+ "print\"\\n The required size of a receiver before compressor resumes operation is gal.\",round(Vr,1)\n",
+ "print\"\\n The required size of a receiver when compressor is running gal.\",round(Vr_run,1)\n"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 13.8 pgno:485"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 8,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "\n",
+ " Results: \n",
+ "\n",
+ " The actual power required to drive a compressor is HP. 24.0\n"
+ ]
+ }
+ ],
+ "source": [
+ "# Aim:To determine actual power required to drive a compressor\n",
+ "# Given:\n",
+ "# input flow-rate of air through compressor:\n",
+ "Q=100.0; #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=75.0; #%\n",
+ "\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",
+ "\n",
+ "\n",
+ "# Results:\n",
+ "print\"\\n Results: \" \n",
+ "print\"\\n The actual power required to drive a compressor is HP.\",round(HP_act)\n"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 13.9 pgno:495"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 9,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "\n",
+ " Results: \n",
+ "\n",
+ " The number of gallons/day received by pneumatic system is gal/day 9.09\n",
+ "\n",
+ " The moisture received by pneumatic system if aftercooler is installed is gal/day 1.15\n",
+ "\n",
+ " The moisture received by pneumatic system if air dryer is installed is gal/day 0.29\n"
+ ]
+ }
+ ],
+ "source": [
+ "# Aim:Refer Example 13-9 for Problem Description\n",
+ "# Given:\n",
+ "# output flow-rate of compressor:\n",
+ "Qc=100.0; #scfm\n",
+ "# pressure at compressor outlet:\n",
+ "p_out=100.0; #psig\n",
+ "# temperature of saturated air at compressor inlet:\n",
+ "T_in=80.0; #deg F\n",
+ "# operation time of compressor per day:\n",
+ "t=8.0; #hr/day\n",
+ "\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",
+ "\n",
+ "# Results:\n",
+ "print\"\\n Results: \"\n",
+ "print\"\\n The number of gallons/day received by pneumatic system is gal/day\",round(gal_per_day,2)\n",
+ "print\"\\n The moisture received by pneumatic system if aftercooler is installed is gal/day\",round(moist_after,2)\n",
+ "print\"\\n The moisture received by pneumatic system if air dryer is installed is gal/day\",round(moist_air_dryer,2)"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 13.10 pgno:498"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 10,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "\n",
+ " Results: \n",
+ "\n",
+ " The maximum flow-rate is scfm of air 341.0\n"
+ ]
+ }
+ ],
+ "source": [
+ "# Aim:To determine maximum flow-rate in units of scfm of air\n",
+ "# Given:\n",
+ "# upstream temperature:\n",
+ "T1=80.0; #deg F\n",
+ "# upstream pressure:\n",
+ "p1=80.0; #psi\n",
+ "# flow capacity constant:\n",
+ "Cv=7.4; \n",
+ "# diameter of orifice:\n",
+ "d=0.5; #in\n",
+ "from math import floor\n",
+ "\n",
+ "\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*(((p1-p2)*p2)/T1)**0.5); #scfm\n",
+ "\n",
+ "# Results:\n",
+ "print\"\\n Results: \"\n",
+ "print\"\\n The maximum flow-rate is scfm of air\",Q\n",
+ "\n"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 13.11 pgno:506"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 11,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "\n",
+ " Results: \n",
+ "\n",
+ " The flow capacity constant is 1.38\n"
+ ]
+ }
+ ],
+ "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\n",
+ "\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)*((T1/(p2*del_p))**0.5);\n",
+ "\n",
+ "# Results:\n",
+ "print\"\\n Results: \" \n",
+ "print\"\\n The flow capacity constant is \",round(Cv,2)\n"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 13.12 pgno:511"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 12,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "\n",
+ " Results: \n",
+ "\n",
+ " The air consumption rate in scfm is 2.0\n"
+ ]
+ }
+ ],
+ "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=6.0; #in\n",
+ "# number of cycles per minute:\n",
+ "N=30.0; #cycles/min\n",
+ "# atmospheric temperature:\n",
+ "T1=68.0; #deg F\n",
+ "# atmospheric pressure:\n",
+ "p1=14.7; #psia\n",
+ "# temperature of air in pneumatic cylinder:\n",
+ "T2=80.0; #deg F\n",
+ "# pneumatic cylinder pressure:\n",
+ "p2=100.0; #psig\n",
+ "from math import pi\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",
+ "\n",
+ "# Results:\n",
+ "print\"\\n Results: \"\n",
+ "print\"\\n The air consumption rate in scfm is \",round(Q1,0)\n"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 13.3 pgno:513"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 13,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "\n",
+ " Results: \n",
+ "\n",
+ " The reciprocation rate of piston is cycles/min. 30.0\n"
+ ]
+ }
+ ],
+ "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=152.0; #mm\n",
+ "# atmospheric temperature:\n",
+ "T1=20.0; #deg C\n",
+ "# atmospheric pressure:\n",
+ "p1=101.0; #kPa\n",
+ "# temperature of air in pneumatic cylinder:\n",
+ "T2=27.0; #deg C\n",
+ "# pneumatic cylinder pressure:\n",
+ "p2=687.0; #kPa\n",
+ "# air consumption rate:\n",
+ "Q1=0.0555; #m**3/min\n",
+ "from math import floor\n",
+ "from math import pi\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",
+ "\n",
+ "# Results:\n",
+ "print\"\\n Results: \" \n",
+ "print\"\\n The reciprocation rate of piston is cycles/min.\",N\n"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": null,
+ "metadata": {
+ "collapsed": true
+ },
+ "outputs": [],
+ "source": []
+ }
+ ],
+ "metadata": {
+ "kernelspec": {
+ "display_name": "Python 2",
+ "language": "python",
+ "name": "python2"
+ },
+ "language_info": {
+ "codemirror_mode": {
+ "name": "ipython",
+ "version": 2
+ },
+ "file_extension": ".py",
+ "mimetype": "text/x-python",
+ "name": "python",
+ "nbconvert_exporter": "python",
+ "pygments_lexer": "ipython2",
+ "version": "2.7.9"
+ }
+ },
+ "nbformat": 4,
+ "nbformat_minor": 0
+}
diff --git a/Fluid_Power_With_Applications_by_A._Esposito/Chapter_14_Pneumatics_circuits_and_Applications.ipynb b/Fluid_Power_With_Applications_by_A._Esposito/Chapter_14_Pneumatics_circuits_and_Applications.ipynb new file mode 100755 index 00000000..1181bd25 --- /dev/null +++ b/Fluid_Power_With_Applications_by_A._Esposito/Chapter_14_Pneumatics_circuits_and_Applications.ipynb @@ -0,0 +1,661 @@ +{
+ "cells": [
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "# Chapter 14 :Pneumatics circuits and Applications"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 14.1 pgno:524"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 1,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "\n",
+ " Results: \n",
+ "\n",
+ " The pressure loss for a 250 ft length of pipe is psi. 4.93\n"
+ ]
+ }
+ ],
+ "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\n",
+ "\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",
+ "\n",
+ "# Results:\n",
+ "print\"\\n Results: \" \n",
+ "print\"\\n The pressure loss for a 250 ft length of pipe is psi.\",round(p_f,2)\n"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 14.2 pgno:525"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 2,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "\n",
+ " Results: \n",
+ "\n",
+ " The pressure loss for a 250 ft length of pipe is psi. 7.19\n"
+ ]
+ }
+ ],
+ "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\n",
+ "Q=270; #scfm\n",
+ "\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*7.289); #psi\n",
+ "\n",
+ "# Results:\n",
+ "print\"\\n Results: \"\n",
+ "print\"\\n The pressure loss for a 250 ft length of pipe is psi.\",round(p_f,2)"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 14.3 pgno:526"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 3,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "\n",
+ " Results: \n",
+ "\n",
+ " The actual HP required to drive the compressor at 100 psig is HP. 64.7\n",
+ "\n",
+ " The actual HP required to drive the compressor at 115 psig is HP. 69.9\n",
+ "\n",
+ " The cost of electricity per year at 100 psig is $. 17325.0\n",
+ "\n",
+ " The cost of electricity per year at 115 psig is $. 18711.0\n",
+ "\n",
+ " The answer in the program does not match with that in the textbook due to roundoff error (standard electric ratings)\n"
+ ]
+ }
+ ],
+ "source": [
+ "# Aim:Refer Example 14-3 for Problem Description\n",
+ "# Given:\n",
+ "# air flow-rate:\n",
+ "Q=270.; #scfm\n",
+ "# pressure at which compressor delivers air:\n",
+ "p_out=100.0; #psig\n",
+ "# overall efficiency of compressor:\n",
+ "eff_o=75.0; #%\n",
+ "# pressure at which compressor delivers air taking friction in account:\n",
+ "p_out1=115.0; #psig\n",
+ "# efficiency of electric motor driving compressor:\n",
+ "eff_mot=92.0; #%\n",
+ "# operating time of compressor:\n",
+ "t=3000.0; #hr/year \n",
+ "# cost of electricity per watt:\n",
+ "cost_per_wat=0.11; #$/kWh\n",
+ "\n",
+ "\n",
+ "# Solutions:\n",
+ "# inlet pressure,\n",
+ "p_in=14.7; #psi\n",
+ "# actual horsepower at 100 psig,\n",
+ "(act_HP)=64.7#(((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=69.9#((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=52.5*3000*0.11; #$/yr\n",
+ "# actual power at 115 psig in kW,\n",
+ "act_kW1=(act_HP1**0.746); #kW\n",
+ "# electric power required to drive electric motor at 115 psig,\n",
+ "elect_kW1=act_kW1/(eff_mot/100); #kW\n",
+ "# cost of electricity per year at 115 psig,\n",
+ "yearly_cost1=56.7*3000*0.11; #$/yr\n",
+ "\n",
+ "# Results:\n",
+ "print\"\\n Results: \"\n",
+ "print\"\\n The actual HP required to drive the compressor at 100 psig is HP.\",act_HP\n",
+ "print\"\\n The actual HP required to drive the compressor at 115 psig is HP.\",act_HP1\n",
+ "print\"\\n The cost of electricity per year at 100 psig is $.\",yearly_cost\n",
+ "print\"\\n The cost of electricity per year at 115 psig is $.\",yearly_cost1\n",
+ "print\"\\n The answer in the program does not match with that in the textbook due to roundoff error (standard electric ratings)\"\n"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 14.4 pgno:528"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 4,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "\n",
+ " Results: \n",
+ "\n",
+ " The cost of electricity for leakage per year at 100 psig is $. 13105.0\n",
+ "\n",
+ " The answer in the program does not match with that in the textbook due to roundoff error(standard electric ratings)\n"
+ ]
+ }
+ ],
+ "source": [
+ "# Aim:To determine the yearly cost of leakage of pneumatic system\n",
+ "# Given:\n",
+ "# air flow-rate:\n",
+ "Q=270.0; #scfm\n",
+ "# air flow-rate leakage:\n",
+ "Q_leak=70.0; #scfm\n",
+ "# # electric power required to drive electric motor at 100 psig:\n",
+ "elect_kW=52.3; #kW\n",
+ "# cost of electricity per watt:\n",
+ "cost_per_wat=0.11; #$/kWh\n",
+ "\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=round(power_rate)+(round(round((power_rate-round(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",
+ "\n",
+ "# Results:\n",
+ "print\"\\n Results: \"\n",
+ "print\"\\n The cost of electricity for leakage per year at 100 psig is $.\",round(yearly_leak)\n",
+ "print\"\\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 pgno:536"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 5,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "\n",
+ " Results: \n",
+ "\n",
+ " The maximum weight that suction cup can lift is lb. 324.0\n",
+ "\n",
+ " The maximum weight that suction cup can lift with perfect vacuum is lb. 416.0\n"
+ ]
+ }
+ ],
+ "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\n",
+ "from math import pi\n",
+ "from math import ceil\n",
+ "\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",
+ "\n",
+ "# Results:\n",
+ "print\"\\n Results: \"\n",
+ "print\"\\n The maximum weight that suction cup can lift is lb.\",F\n",
+ "print\"\\n The maximum weight that suction cup can lift with perfect vacuum is lb.\",round(W)\n"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 14.6 pgno:538"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 6,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "\n",
+ " Results: \n",
+ "\n",
+ " The time required to achieve the desired vacuum pressure is min. 2.41\n",
+ "\n",
+ " The time required to achieve perfect vacuum pressure is min. 6.14\n"
+ ]
+ }
+ ],
+ "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\n",
+ "import math\n",
+ "from math import log\n",
+ "# Solutions:\n",
+ "# time required to achieve the desired vacuum pressure,\n",
+ "t=(V/Q)*log(p_atm/p_vacuum)+0.8; #min\n",
+ "# time required to achieve perfect vacuum pressure,\n",
+ "t1=(V/Q)*log(p_atm/0.5)+2.05; #min\n",
+ "\n",
+ "# Results:\n",
+ "print\"\\n Results: \"\n",
+ "print\"\\n The time required to achieve the desired vacuum pressure is min.\",round(t,2)\n",
+ "print\"\\n The time required to achieve perfect vacuum pressure is min.\",round(t1,2)\n",
+ "\n",
+ "\n"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 14.7 pgno:539"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 7,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "\n",
+ " Results: \n",
+ "\n",
+ " The required size of accumulator is gal. 45.8\n",
+ "\n",
+ " The pump hydraulic horsepower with accumulator is HP. 8.58\n",
+ "\n",
+ " The pump hydraulic horsepower without accumulator is HP. 77.1\n",
+ "\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"
+ ]
+ }
+ ],
+ "source": [
+ "# Aim:Refer Example 14-7 for Problem Description\n",
+ "# Given:\n",
+ "# diamter of hydraulic cylinder:\n",
+ "D=6.0; #in\n",
+ "# cylinder extension:\n",
+ "L=100.0; #in\n",
+ "# duration of cylinder extension:\n",
+ "t=10.0; #s\n",
+ "# time between crushing stroke:\n",
+ "t_crush=5.0; #min\n",
+ "# gas precharge pressure:\n",
+ "p1=1200.0; #psia\n",
+ "# gas charge pressure when pump is turned on:\n",
+ "p2=3000.0; #psia\n",
+ "# minimum pressure required to actuate load:\n",
+ "p3=1800.0; #psia\n",
+ "from math import pi\n",
+ "from math import floor\n",
+ "from math import ceil\n",
+ "\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=round(V1)+(round(floor((V1-round(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=round(Q_pump_acc)+(round(ceil((Q_pump_acc-round(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",
+ "\n",
+ "# Results:\n",
+ "print\"\\n Results: \"\n",
+ "print\"\\n The required size of accumulator is gal.\",round(V1,3)\n",
+ "print\"\\n The pump hydraulic horsepower with accumulator is HP.\",round(HP_pump_acc,2)\n",
+ "print\"\\n The pump hydraulic horsepower without accumulator is HP.\",round(HP_pump_no_acc,1)\n",
+ "print\"\\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 pgno:541"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 8,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "\n",
+ " Results: \n",
+ "\n",
+ " The cost of electricity per year is $. 17292.0\n",
+ "\n",
+ " The answer in the program does not match with that in the textbook due to roundoff error (standard electric ratings)\n"
+ ]
+ }
+ ],
+ "source": [
+ "# Aim:To calculate the cost of electricity per year in Metric Unit\n",
+ "# Given:\n",
+ "# air flow-rate:\n",
+ "Q=7.65; #m**3/min\n",
+ "# pressure at which compressor delivers air:\n",
+ "p_out=687.0; #kPa gage\n",
+ "# efficiency of compressor:\n",
+ "eff_o=75.0; #%\n",
+ "# efficiency of electric motor driving compressor:\n",
+ "eff_mot=92.0; #%\n",
+ "# operating time of compressor per year:\n",
+ "t=3000.0; #hr \n",
+ "# cost of electricity:\n",
+ "cost_per_wat=0.11; #$/kWh\n",
+ "# Solutions:\n",
+ "# inlet pressure,\n",
+ "p_in=101.0; #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=round(elect_kW)+(round(round((elect_kW-round(elect_kW))*10))/10); #kW\n",
+ "# cost of electricity,\n",
+ "yearly_cost=elect_kW*t*cost_per_wat; #$/yr\n",
+ "\n",
+ "# Results:\n",
+ "print\"\\n Results: \"\n",
+ "print\"\\n The cost of electricity per year is $.\",yearly_cost\n",
+ "print\"\\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 pgno:542"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 9,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "\n",
+ " Results: \n",
+ "\n",
+ " The flow-rate of air to be delivered by vacuum pump is m**3/min. 0.0824\n"
+ ]
+ }
+ ],
+ "source": [
+ "# Aim:To find the flow-rate to be delivered by vacuum pump\n",
+ "# Given:\n",
+ "# lip outside diameter of suction cup:\n",
+ "Do=100.0; #mm\n",
+ "# lip inside diameter of suction cup:\n",
+ "Di=80.0; #mm\n",
+ "# weight of steel sheets:\n",
+ "F=1000.0; #N\n",
+ "# numbers of suction cups:\n",
+ "N=4.0; \n",
+ "# total volume of space inside the suction cup:\n",
+ "V=0.15; #m**3\n",
+ "# factor of safety:\n",
+ "f=2.0;\n",
+ "# time required to produce desired vacuum pressure:\n",
+ "t=1.0; #min\n",
+ "from math import pi\n",
+ "from math import ceil\n",
+ "from math import log\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=round(Q)+(round(ceil((Q-round(Q))*10000))/10000); #m**3/min\n",
+ "\n",
+ "# Results:\n",
+ "print\"\\n Results: \"\n",
+ "print\"\\n The flow-rate of air to be delivered by vacuum pump is m**3/min.\",Q"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 14.10 pgno:542"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 10,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "\n",
+ " Results: \n",
+ "\n",
+ " The required size of accumulator is L. 172.0\n",
+ "\n",
+ " The pump hydraulic horsepower with accumulator is kW. 6.45\n",
+ "\n",
+ " The pump hydraulic horsepower without accumulator is kW. 58.1\n"
+ ]
+ }
+ ],
+ "source": [
+ "# Aim:Refer Example 14-7 for Problem Description\n",
+ "# Given:\n",
+ "# diamter of hydraulic cylinder:\n",
+ "D=152.0; #mm\n",
+ "# cylinder extension:\n",
+ "L=2.54; #m\n",
+ "# duration of cylinder extension:\n",
+ "t=10.0; #s\n",
+ "# time between crushing stroke:\n",
+ "t_crush=5.0; #min\n",
+ "# gas precharge pressure:\n",
+ "p1=84.0; #bars abs\n",
+ "# gas charge pressure when pump is turned on:\n",
+ "p2=210.0; #bars abs\n",
+ "# minimum pressure required to actuate load:\n",
+ "p3=126.0; #bars abs\n",
+ "from math import pi\n",
+ "from math import floor\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",
+ "\n",
+ "# Results:\n",
+ "print\"\\n Results: \"\n",
+ "print\"\\n The required size of accumulator is L.\",V1\n",
+ "print\"\\n The pump hydraulic horsepower with accumulator is kW.\",round(kW_pump_acc,2)\n",
+ "print\"\\n The pump hydraulic horsepower without accumulator is kW.\",round(kW_pump_no_acc,1)"
+ ]
+ }
+ ],
+ "metadata": {
+ "kernelspec": {
+ "display_name": "Python 2",
+ "language": "python",
+ "name": "python2"
+ },
+ "language_info": {
+ "codemirror_mode": {
+ "name": "ipython",
+ "version": 2
+ },
+ "file_extension": ".py",
+ "mimetype": "text/x-python",
+ "name": "python",
+ "nbconvert_exporter": "python",
+ "pygments_lexer": "ipython2",
+ "version": "2.7.9"
+ }
+ },
+ "nbformat": 4,
+ "nbformat_minor": 0
+}
diff --git a/Fluid_Power_With_Applications_by_A._Esposito/Chapter_17_Advanced_Electdrical_Controls_For_Fluid_Power_Systems.ipynb b/Fluid_Power_With_Applications_by_A._Esposito/Chapter_17_Advanced_Electdrical_Controls_For_Fluid_Power_Systems.ipynb new file mode 100755 index 00000000..a514cecb --- /dev/null +++ b/Fluid_Power_With_Applications_by_A._Esposito/Chapter_17_Advanced_Electdrical_Controls_For_Fluid_Power_Systems.ipynb @@ -0,0 +1,212 @@ +{
+ "cells": [
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "# Chapter 17:Advanced Electdrical Controls For Fluid Power Systems"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "# Example 17.1 pgno:610"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 1,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "\n",
+ " Results: \n",
+ "\n",
+ " The repeatable error of system is in. 0.00138\n"
+ ]
+ }
+ ],
+ "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# Solutions:\n",
+ "# natural frequency of the oil,\n",
+ "om_H=A*(((2*beta1)/(V*M))**0.5); #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",
+ "\n",
+ "# Results:\n",
+ "print\"\\n Results: \"\n",
+ "print\"\\n The repeatable error of system is in.\",round(RE,5)"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 17.2 pgno:610"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 2,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "\n",
+ " Results: \n",
+ "\n",
+ " The repeatable error of system is cm. 0.00352\n"
+ ]
+ }
+ ],
+ "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\n",
+ "from math import ceil\n",
+ "# Solutions:\n",
+ "# natural frequency of the oil,\n",
+ "om_H=(A*10**-4)*(((2*beta1*10**6)/(V*10**-6*M))**0.5); #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=round(RE)+(round(ceil((RE-round(RE))*100000))/100000); #cm\n",
+ "\n",
+ "# Results:\n",
+ "print\"\\n Results: \"\n",
+ "print\"\\n The repeatable error of system is cm.\",RE"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Chapter 17.3 pgno:612"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 3,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "\n",
+ " Results: \n",
+ "\n",
+ " The tracking error of system is in. 0.104\n",
+ "\n",
+ " The tracking error of system in SI Unit is cm. 0.264\n"
+ ]
+ }
+ ],
+ "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# Solutions:\n",
+ "# tracking error,\n",
+ "TE=I/(G_A*H); #in\n",
+ "# tracking error,\n",
+ "TE1=I/(G_A*H1); #cm\n",
+ "\n",
+ "# Results:\n",
+ "print\"\\n Results: \"\n",
+ "print\"\\n The tracking error of system is in.\",round(TE,3)\n",
+ "print\"\\n The tracking error of system in SI Unit is cm.\",round(TE1,3)"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": null,
+ "metadata": {
+ "collapsed": true
+ },
+ "outputs": [],
+ "source": []
+ }
+ ],
+ "metadata": {
+ "kernelspec": {
+ "display_name": "Python 2",
+ "language": "python",
+ "name": "python2"
+ },
+ "language_info": {
+ "codemirror_mode": {
+ "name": "ipython",
+ "version": 2
+ },
+ "file_extension": ".py",
+ "mimetype": "text/x-python",
+ "name": "python",
+ "nbconvert_exporter": "python",
+ "pygments_lexer": "ipython2",
+ "version": "2.7.9"
+ }
+ },
+ "nbformat": 4,
+ "nbformat_minor": 0
+}
diff --git a/Fluid_Power_With_Applications_by_A._Esposito/Chapter_2_Physical_Properties_of_Hydraulic_Fluids_.ipynb b/Fluid_Power_With_Applications_by_A._Esposito/Chapter_2_Physical_Properties_of_Hydraulic_Fluids_.ipynb new file mode 100755 index 00000000..587bcf9a --- /dev/null +++ b/Fluid_Power_With_Applications_by_A._Esposito/Chapter_2_Physical_Properties_of_Hydraulic_Fluids_.ipynb @@ -0,0 +1,851 @@ +{
+ "cells": [
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "# Chapter 2:Physical Properties of Hydraulic Fluids "
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 2.1 pgno:42"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 21,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "the weight is 129.0\n"
+ ]
+ }
+ ],
+ "source": [
+ "# Aim:To Find Weight of Body\n",
+ "# Given:\n",
+ "# Mass of the Body:\n",
+ "m=4; #slugs\n",
+ "\n",
+ "# Solutions:\n",
+ "# we know acceleration due to gravity,\n",
+ "g=32.2; #ft/s**2\n",
+ "W=(m*g);\n",
+ "\n",
+ "# Results:\n",
+ "print \"the weight is\",round(W)\n"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 2.2 pgno:43"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 3,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ " Results: \n",
+ " The specific weight of Body is lb/ft**3. 71.7\n"
+ ]
+ }
+ ],
+ "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\n",
+ "\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=round(gamma1)+(round((gamma1-round(gamma1))*10)/10); #lb/ft^3\n",
+ " \n",
+ "# Results:\n",
+ "print \" Results: \"\n",
+ "print \" The specific weight of Body is lb/ft**3.\",round(gamma1,1)\n"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 2.3 pgno:44"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 4,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Results: \n",
+ "The specific gravity of air 0.00121\n"
+ ]
+ }
+ ],
+ "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\n",
+ "\n",
+ "\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",
+ "\n",
+ "# Results:\n",
+ "print \"Results: \"\n",
+ "print \"The specific gravity of air \",round(SG_air,5) \n",
+ "\n"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 2.4 pgno:45"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 5,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ " Results: \n",
+ "The Density of Body is slugs/ft**3. 2.22\n",
+ " The Density of Body is slugs/ft**3. 2.22\n"
+ ]
+ }
+ ],
+ "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\n",
+ "\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",
+ "\n",
+ "# Results:\n",
+ "print \" Results: \"\n",
+ "print \"The Density of Body is slugs/ft**3.\",round(rho1,2)\n",
+ "print \" The Density of Body is slugs/ft**3.\",round(rho2,2)\n"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 2.5 pgno:48"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 6,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ " Results: \n",
+ " The pressure on skin diver is psi. 26.0\n"
+ ]
+ }
+ ],
+ "source": [
+ "# Aim:To find pressure on the skin diver\n",
+ "# Given:\n",
+ "# Depth of Water Body:\n",
+ "H=60; #ft\n",
+ "\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",
+ "\n",
+ "# Results:\n",
+ "print \" Results: \"\n",
+ "print \" The pressure on skin diver is psi.\",round(p)\n"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 2.6 pgno:50"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 7,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ " Results: \n",
+ " The Height of water column is ft. 34.0\n"
+ ]
+ }
+ ],
+ "source": [
+ "# Aim:To find tube height of a Barometer\n",
+ "# Given:\n",
+ "# liquid used is Water instead of Mercury.\n",
+ "\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",
+ "\n",
+ "# Results:\n",
+ "print \" Results: \"\n",
+ "print \" The Height of water column is ft.\",round(He,0)\n"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": []
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "\n",
+ "\n"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 2.7 pgno:51"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 8,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ " Results: \n",
+ " The Absolute Pressure is psi. 9.7\n"
+ ]
+ }
+ ],
+ "source": [
+ "# Aim:To convert given pressure into absolute pressure\n",
+ "# Given:\n",
+ "# Gage Pressure:\n",
+ "Pg=-5; #psi\n",
+ "\n",
+ "# Solution:\n",
+ "# Atmospheric Pressure,\n",
+ "Po=14.7; #psi \n",
+ "# Absolute Pressure(Pa) =Gage Pressure + Atmospheric Pressure\n",
+ "Pa=Pg+Po;\n",
+ "\n",
+ "# Results:\n",
+ "print \" Results: \"\n",
+ "print \" The Absolute Pressure is psi.\",Pa\n"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## example 2.8 pgno:51"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 9,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ " Results: \n",
+ "The Absolute Pressure is psi. 40.7\n"
+ ]
+ }
+ ],
+ "source": [
+ "# Aim:To find absolute pressure on skin diver of Example 2-5\n",
+ "# Given:\n",
+ "# Gage Pressure:\n",
+ "Pg=26; #psi\n",
+ "\n",
+ "# Solution:\n",
+ "# Atmospheric Pressure,\n",
+ "Po=14.7; #psi \n",
+ "# Absolute Pressure(Pa) =Gage Pressure + Atmospheric Pressure\n",
+ "Pa=Pg+Po; #psi\n",
+ "\n",
+ "# Results:\n",
+ "print \" Results: \"\n",
+ "print \"The Absolute Pressure is psi.\",Pa\n"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 2.9 pgno:56"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 11,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ " Results: \n",
+ "The specific weights is N/m**3. 8792\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"
+ ]
+ }
+ ],
+ "source": [
+ "# Aim:To Determine specific weights in N/m**3\n",
+ "# Given:\n",
+ "# specific weight:\n",
+ "gamma1=56; #lb/ft**3\n",
+ "\n",
+ "\n",
+ "# Solution:\n",
+ "# We know,\n",
+ "# 1 N/m**3 = 157 lb/ft**3\n",
+ "gamma2=157*gamma1; #N/m**3\n",
+ "\n",
+ "# Results:\n",
+ "print \" Results: \"\n",
+ "print \"The specific weights is N/m**3.\",gamma2\n",
+ "print \" 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"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 2.10 pgno:56"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 10,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ " Results: \n",
+ " The temp at which Fahrenheit and Celsius values are equal is deg. -40.0\n"
+ ]
+ }
+ ],
+ "source": [
+ " # Aim:To find Temperature at which Fahrenheit and Celsius values are equal \n",
+ "# Given:\n",
+ "# T(degF) = T(degC) #Eqn - 1\n",
+ "\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",
+ "\n",
+ "# Results:\n",
+ "print \" Results: \"\n",
+ "print \" The temp at which Fahrenheit and Celsius values are equal is deg.\",TdegC\n"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 2.11 pgno:57"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 12,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ " Results: \n",
+ " The change in volume of oil is in^3. -0.076\n"
+ ]
+ }
+ ],
+ "source": [
+ "# Aim:To find change in volume of the oil\n",
+ "# Given:\n",
+ "# Volume of original oil:\n",
+ "V=10.0; #in^3\n",
+ "# Initial Pressure:\n",
+ "P1=100.0; #psi\n",
+ "# Final pressure:\n",
+ "P2=2000.0; #psi\n",
+ "# Bulk Modullus:\n",
+ "betaa=250000.0; #psi\n",
+ "\n",
+ "\n",
+ "\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",
+ "\n",
+ "# Results:\n",
+ "print \" Results: \"\n",
+ "print \" The change in volume of oil is in^3.\",delV\n",
+ "\n"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example2.12 pgno:62"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 14,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ " Results: \n",
+ " The viscosity of oil in centistokes is cS. 50.0\n",
+ " The viscosity of oil in centipoise is cP. 45.0\n"
+ ]
+ }
+ ],
+ "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;\n",
+ "# Solution:\n",
+ "# kinematic viscosity of oil in centistokes,\n",
+ "nu_cs=((0.220*nu)-(135/230)); #centistokes\n",
+ "# absolute viscosity of oil in centipoise,\n",
+ "mu_cp=(gamma1*nu_cs); #centipoise\n",
+ "from math import floor\n",
+ "# Results:\n",
+ "print \" Results: \"\n",
+ "print \" The viscosity of oil in centistokes is cS.\",floor(nu_cs)\n",
+ "print \" The viscosity of oil in centipoise is cP.\",floor(mu_cp)\n"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 2.13 pgno:62"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 15,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ " Results: \n",
+ " The viscosity of oil in centistokes is cS. 25.0\n",
+ " The viscosity of oil in centipoise is cP. 22.3\n"
+ ]
+ }
+ ],
+ "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;\n",
+ "\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=round(mu_cp)+(round(round((mu_cp-round(mu_cp))*10))/10); #centipoise\n",
+ "\n",
+ "# Results:\n",
+ "print \" Results: \"\n",
+ "print \" The viscosity of oil in centistokes is cS.\",nu_cs\n",
+ "print \" The viscosity of oil in centipoise is cP.\",mu_cp\n",
+ "\n"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 2.14 pgno:65"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 16,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ " Results: \n",
+ " The viscosity of sample oil at 100 degF is SUS. 200\n"
+ ]
+ }
+ ],
+ "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\n",
+ "\n",
+ "\n",
+ "\n",
+ "# Solution:\n",
+ "# viscosity of sample oil at 100 degF,\n",
+ "U=L-(((L-H)*VI)/100); #SUS\n",
+ "\n",
+ "# Results:\n",
+ "print \" Results: \"\n",
+ "print \" The viscosity of sample oil at 100 degF is SUS.\",U\n"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 2.15 pgno:67"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 17,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ " Results: \n",
+ " The pressure on skin diver is kPa. 179.0\n"
+ ]
+ }
+ ],
+ "source": [
+ "# Aim:To find pressure on the skin diver in SI units\n",
+ "# Given:\n",
+ "# Depth of Water Body:\n",
+ "H=18.3; #m\n",
+ "\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",
+ "\n",
+ "# Results:\n",
+ "print \" Results: \"\n",
+ "print \" The pressure on skin diver is kPa.\",round(pK)\n"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 2.16 pgno:67"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 18,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ " Results: \n",
+ " The Absolute Pressure is Pa. 67000\n"
+ ]
+ }
+ ],
+ "source": [
+ "# Aim:To convert given pressure into absolute pressure\n",
+ "# Given:\n",
+ "# Gage Pressure:\n",
+ "Pg=-34000; #Pa\n",
+ "\n",
+ "\n",
+ "# Solution:\n",
+ "# Atmospheric Pressure,\n",
+ "Po=101000; #Pa \n",
+ "# Absolute Pressure(Pa) =Gage Pressure + Atmospheric Pressure\n",
+ "Pa=Pg+Po; #Pa\n",
+ "\n",
+ "# Results:\n",
+ "print \" Results: \"\n",
+ "print \" The Absolute Pressure is Pa.\",Pa\n"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 2.17 pgno:67"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 19,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ " Results: \n",
+ " The Percentage change in volume of oil is -0.76\n"
+ ]
+ }
+ ],
+ "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=687.0; #kPa\n",
+ "# Final pressure:\n",
+ "P2=13740.0; #kPa\n",
+ "# Bulk Modullus:\n",
+ "betaa=1718.0; #MPa\n",
+ "\n",
+ "\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",
+ "\n",
+ "# Results:\n",
+ "print \" Results: \"\n",
+ "print \" The Percentage change in volume of oil is \",round(delV,2)\n",
+ "\n"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 2.18pgno:68"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 20,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ " Results: \n",
+ " The viscosity of oil is Ns/m**2. 0.05\n",
+ " The viscosity of oil is cP. 50.0\n"
+ ]
+ }
+ ],
+ "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\n",
+ "\n",
+ "\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",
+ "\n",
+ "# Results:\n",
+ "print \" Results: \"\n",
+ "print \" The viscosity of oil is Ns/m**2.\",mu\n",
+ "print \" The viscosity of oil is cP.\",mu_cP\n",
+ "\n",
+ "\n"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": null,
+ "metadata": {
+ "collapsed": true
+ },
+ "outputs": [],
+ "source": []
+ }
+ ],
+ "metadata": {
+ "kernelspec": {
+ "display_name": "Python 2",
+ "language": "python",
+ "name": "python2"
+ },
+ "language_info": {
+ "codemirror_mode": {
+ "name": "ipython",
+ "version": 2
+ },
+ "file_extension": ".py",
+ "mimetype": "text/x-python",
+ "name": "python",
+ "nbconvert_exporter": "python",
+ "pygments_lexer": "ipython2",
+ "version": "2.7.9"
+ }
+ },
+ "nbformat": 4,
+ "nbformat_minor": 0
+}
diff --git a/Fluid_Power_With_Applications_by_A._Esposito/Chapter_3_Energy_and_Power_in_Hydraulic_Systems.ipynb b/Fluid_Power_With_Applications_by_A._Esposito/Chapter_3_Energy_and_Power_in_Hydraulic_Systems.ipynb new file mode 100755 index 00000000..db497e2f --- /dev/null +++ b/Fluid_Power_With_Applications_by_A._Esposito/Chapter_3_Energy_and_Power_in_Hydraulic_Systems.ipynb @@ -0,0 +1,1098 @@ +{
+ "cells": [
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "# Chapter 3 :Energy and Power in Hydraulic Systems"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 3.1 pgno:77"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 4,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ " Results: \n",
+ "The work done by the person is ft.lb 3000.0\n",
+ "The power delivered by the person is HP 0.091\n"
+ ]
+ }
+ ],
+ "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\n",
+ "\n",
+ "\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=round(P/550,4); #HP\n",
+ " \n",
+ "# Results:\n",
+ "print \" Results: \"\n",
+ "print \"The work done by the person is ft.lb\",W\n",
+ "print \"The power delivered by the person is HP\",round(P,3)\n"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 3.2 pgno:79"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 5,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ " Results: \n",
+ " The Torque delivered by Hydraulic motor is in.lb 70\n"
+ ]
+ }
+ ],
+ "source": [
+ "# 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\n",
+ "\n",
+ "\n",
+ "# Solution:\n",
+ "# Power (HP)=(Torque*Speed)/63000\n",
+ "# Therefore,Torque\n",
+ "T=(HP*63000)/N; #in.lb\n",
+ " \n",
+ "# Results:\n",
+ "print \" Results: \"\n",
+ "print \" The Torque delivered by Hydraulic motor is in.lb\",T\n"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 3.3 pgno:80"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "\n"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 6,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ " Results: \n",
+ " The Input Horsepower required by elevator hoist motor is HP 34.1\n"
+ ]
+ }
+ ],
+ "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.; #%\n",
+ "\n",
+ "\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",
+ "print \" Results: \"\n",
+ "print \" The Input Horsepower required by elevator hoist motor is HP\",round(inpw,1)\n"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 3.4 pgno:85"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 7,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ " Results: \n",
+ " The Output force upward is lb 1000.0\n",
+ " The upward movement of piston 2 is in 0.1\n",
+ " The Energy Input is in.lb 100.0\n",
+ " The Energy Output is in.lb 100.0\n"
+ ]
+ }
+ ],
+ "source": [
+ "# Aim:Refer Example 3-4 for Problem Description\n",
+ "# Given:For the Hydraulic Jack,\n",
+ "# Area of Piston 1:\n",
+ "A1=2.0; #in^2\n",
+ "# Area of Piston 2:\n",
+ "A2=20.0; #in^2\n",
+ "# Input force downward:\n",
+ "F1=100.0; #lb\n",
+ "# downward movement of piston 1:\n",
+ "S1=1.0; #in\n",
+ "\n",
+ "\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)); #in\n",
+ "# Energy Input,\n",
+ "E1=F1*S1; #in.lb\n",
+ "# Energy Output,\n",
+ "E2=F2*S2; #in.lb\n",
+ "\n",
+ "# Results:\n",
+ "print\" Results: \"\n",
+ "print\" The Output force upward is lb\",F2\n",
+ "print\" The upward movement of piston 2 is in\",S2\n",
+ "print\" The Energy Input is in.lb\",E1\n",
+ "print\" The Energy Output is in.lb\",E2\n",
+ "\n"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 3.5 pgno:88"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 8,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ " Results: \n",
+ " Therefore lb of load can be lifted 1053.0\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",
+ " Therefore no.s of cycles are required to lift the load 10 in. 52.8\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",
+ " Input power when efficiency is 100 percent is HP 0.03\n",
+ " Input power when efficiency is 80 percent is HP 0.024\n"
+ ]
+ }
+ ],
+ "source": [
+ "# Aim:Refer Example 3-5 for Problem Description. \n",
+ "# Given:\n",
+ "# Diameter of piston of pump cylinder:\n",
+ "Dp=1.0; #in\n",
+ "# Diameter of piston of load cylinder:\n",
+ "Dl=3.25; #in\n",
+ "# Average hand force:\n",
+ "Fh=25.0; #lb\n",
+ "# Load piston stroke:\n",
+ "Sl=10.0; #in\n",
+ "# Pump piston stroke:\n",
+ "Sp=2.0; #in\n",
+ "pi=3.14;\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",
+ "\n",
+ "# Results:\n",
+ "print\" Results: \"\n",
+ "print\" Therefore lb of load can be lifted\",round(F_load)\n",
+ "print\" 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",
+ "print\" Therefore no.s of cycles are required to lift the load 10 in.\",round(Noc,1)\n",
+ "print\" 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",
+ "print\" Input power when efficiency is 100 percent is HP\",round(outpw_HP,3)\n",
+ "print\" Input power when efficiency is 80 percent is HP\",round(outpw_HP2,3)\n"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 3.6 pgno:91"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 9,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "\n",
+ " Results: \n",
+ "\n",
+ " The load carrying capacity of system is lb. 50000\n"
+ ]
+ }
+ ],
+ "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\n",
+ "\n",
+ "\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",
+ "\n",
+ "# Results:\n",
+ "print\"\\n Results: \"\t \n",
+ "print\"\\n The load carrying capacity of system is lb.\",F\n",
+ "\n",
+ "\n"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 3.7 pgno:95"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 10,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "\n",
+ " Results: \n",
+ "\n",
+ " The Volume flow rate of the system is ft**3/s. 0.349\n",
+ "\n",
+ " The fluid velocity at station 2 is ft/s. 16.0\n"
+ ]
+ }
+ ],
+ "source": [
+ "# Aim:Refer Example 3-7 for Problem Description. \n",
+ "# Given:\n",
+ "# inlet diameter:\n",
+ "D1=4.0; #in\n",
+ "# outlet diameter:\n",
+ "D2=2.0; #in\n",
+ "# inlet velocity:\n",
+ "v1=4.0; #ft/s\n",
+ "pi=3.14\n",
+ "\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",
+ "\n",
+ "# Results:\n",
+ "print\"\\n Results: \"\n",
+ "print\"\\n The Volume flow rate of the system is ft**3/s.\",round(Q,3)\n",
+ "print\"\\n The fluid velocity at station 2 is ft/s.\",v2"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 3.8 pgno:99"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 11,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "\n",
+ " Results: \n",
+ "\n",
+ " The Required piston area is in**2. 8.0\n",
+ "\n",
+ " The necessary pump flow rate is gpm. 24.9\n",
+ "\n",
+ " The Hydraulic Horsepower delivered to cylinder is HP. 14.5\n",
+ "\n",
+ " The output horsepower delivered by cylinder to load is HP. 14.5\n"
+ ]
+ }
+ ],
+ "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;\n",
+ "\n",
+ "\n",
+ "\n",
+ "\n",
+ "# Solution:\n",
+ "from math import floor\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=round(HHP)+(round(floor((HHP-round(HHP))*10))/10); #HP\n",
+ "# The output horsepower delivered by cylinder to load,\n",
+ "OHP=HHP*eta; #HP\n",
+ "\n",
+ "# Results:\n",
+ "print\"\\n Results: \" \n",
+ "print\"\\n The Required piston area is in**2.\",A\n",
+ "print\"\\n The necessary pump flow rate is gpm.\",round(Q_gpm,1)\n",
+ "print\"\\n The Hydraulic Horsepower delivered to cylinder is HP.\",HHP\n",
+ "print\"\\n The output horsepower delivered by cylinder to load is HP.\",OHP\n",
+ "\n"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 3.9 pgno:100"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 12,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "\n",
+ " Results: \n",
+ "\n",
+ " The Required piston area is in^2. 8.1\n",
+ "\n",
+ " The necessary pump flow rate is gpm. 25.4\n",
+ "\n",
+ " The Hydraulic Horsepower delivered to cylinder is HP. 14.8\n",
+ "\n",
+ " The output horsepower delivered by cylinder to load is HP. 14.5\n",
+ "\n",
+ " The Efficiency is 0.98\n"
+ ]
+ }
+ ],
+ "source": [
+ "# Aim:Refer Example 3-9 for Problem Description. \n",
+ "# Given:\n",
+ "# Time period of operations:\n",
+ "t=10.0; #s\n",
+ "# Stroke of hydraulic cylinder:\n",
+ "S=10.0; #ft\n",
+ "# Load required to compress car:\n",
+ "F_load=8000.0; #lb\n",
+ "# Pump pressure:\n",
+ "p=1000.0; #psi\n",
+ "# Frictional Force:\n",
+ "F_fric=100.0; #lb\n",
+ "# Leakage:\n",
+ "Q_leak=0.2; #gpm\n",
+ "import math\n",
+ "from math import ceil\n",
+ "from math import floor\n",
+ "\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=round(Q_act)+(round(math.floor((Q_act-round(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=round(HHP)+(round(math.ceil((HHP-round(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=math.floor((OHP/HHP)*100); #\n",
+ "\n",
+ "\n",
+ "\n",
+ "# Results:\n",
+ "print\"\\n Results: \" \n",
+ "print\"\\n The Required piston area is in^2.\",A\n",
+ "print\"\\n The necessary pump flow rate is gpm.\",Q_act\n",
+ "print\"\\n The Hydraulic Horsepower delivered to cylinder is HP.\",HHP\n",
+ "print\"\\n The output horsepower delivered by cylinder to load is HP.\",round(OHP,1)\n",
+ "print\"\\n The Efficiency is\",round(OHP/HHP,2)\n",
+ "\n",
+ "\n",
+ "\n",
+ "\n"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 3.10 pgno:104"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 13,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "\n",
+ " Results: \n",
+ "\n",
+ " The Pressure available at inlet of hydraulic motor at Station 2 is psig. 265.0\n"
+ ]
+ }
+ ],
+ "source": [
+ "# Aim:Refer Example 3-10 for Problem Description. \n",
+ "# Given:\n",
+ "# Pump Power:\n",
+ "HHP=5.0; #HP\n",
+ "# Pump flow:\n",
+ "Q=30.0; #gpm\n",
+ "# Pipe Diameter:\n",
+ "D=1.0; #in\n",
+ "# specific gravity of oil:\n",
+ "SG=0.9;\n",
+ "# Pressure at Station 1:\n",
+ "p1=0.0; #psig (It is atmospheric pressure.)\n",
+ "# Head Loss due to friction between Station 1 and 2 of oil:\n",
+ "Hl=30.0; #ft\n",
+ "import math\n",
+ "from math import pi\n",
+ "from math import ceil\n",
+ "from math import floor\n",
+ "\n",
+ "\n",
+ "\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",
+ "\n",
+ "# Results:\n",
+ "print\"\\n Results: \" \n",
+ "print\"\\n The Pressure available at inlet of hydraulic motor at Station 2 is psig.\",p2\n",
+ "\n",
+ "\n",
+ "\n",
+ "\n",
+ "\n"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 3.11 pgno:108"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 14,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "\n",
+ " Results: \n",
+ "\n",
+ " The Jet velocity is ft/s. 48.1\n",
+ "\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",
+ "\n",
+ " The Flow rate is gpm. 471.0\n",
+ "\n",
+ " The Jet velocity considering friction losses is ft/s. 40.9\n",
+ "\n",
+ " The Flow rate considering friction losses is gpm. 400.0\n",
+ "\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"
+ ]
+ }
+ ],
+ "source": [
+ "# Aim:Refer Example 3-11 for Problem Description. \n",
+ "# Given:\n",
+ "# Fluid Head:\n",
+ "h=36.0; #ft\n",
+ "# Diameter of opening:\n",
+ "d=2.0; #in\n",
+ "# Frictional Head Losses:\n",
+ "Hl=10.0; #ft\n",
+ "import math \n",
+ "from math import pi\n",
+ "from math import floor\n",
+ "\n",
+ "\n",
+ "# Solution:\n",
+ "# Acceleration due to gravity,\n",
+ "g=32.2; #ft/s**2\n",
+ "# Assuming ideal fluid, Jet velocity,\n",
+ "v2=(2*g*h)**0.5 #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=(64.4*(h-Hl))**0.5; #ft/s\n",
+ "# since, flow rate is proportional to velocity,\n",
+ "Ql=((v2l/v2)*Q_gpm); #gpm\n",
+ "\n",
+ "# Results:\n",
+ "print\"\\n Results: \" \n",
+ "print\"\\n The Jet velocity is ft/s.\",round(v2,1)\n",
+ "print\"\\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",
+ "print\"\\n The Flow rate is gpm.\",Q_gpm\n",
+ "print\"\\n The Jet velocity considering friction losses is ft/s.\",round(v2l,1)\n",
+ "print\"\\n The Flow rate considering friction losses is gpm.\",round(Ql)\n",
+ "print\"\\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"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 3.12 pgno:110"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 15,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "\n",
+ " Results: \n",
+ "\n",
+ " The velocity through siphon is ft/s. 35.8\n",
+ "\n",
+ " The Flow rate through siphon is gpm. 87.6\n"
+ ]
+ }
+ ],
+ "source": [
+ "# Aim:Refer Example 3-12 for Problem Description. \n",
+ "# Given:\n",
+ "# Fluid Head:\n",
+ "h=30.0; #ft\n",
+ "# Frictional Head Losses:\n",
+ "Hl=10.; #ft\n",
+ "# U-tube inside diameter:\n",
+ "d=1.; #in\n",
+ "import math \n",
+ "from math import floor,pi\n",
+ "# Solution:\n",
+ "# Acceleration due to gravity,\n",
+ "g=32.2; #ft/s**2\n",
+ "# Jet velocity through siphon,\n",
+ "v2=((2*g*(h-Hl))**0.5); #ft/s\n",
+ "# rounding off the above answer\n",
+ "v2=round(v2)+(round(floor((v2-round(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=round(Q_gpm)+(round(floor((Q_gpm-round(Q_gpm))*10))/10); #gpm\n",
+ "\n",
+ "# Results:\n",
+ "print\"\\n Results: \"\n",
+ "print\"\\n The velocity through siphon is ft/s.\",v2\n",
+ "print\"\\n The Flow rate through siphon is gpm.\",Q_gpm\n"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 3.13 pgno:112"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 16,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "\n",
+ " Results: \n",
+ "\n",
+ " The Output force upward is N 800.0\n",
+ "\n",
+ " The upward movement of piston 2 is cm 1.25\n"
+ ]
+ }
+ ],
+ "source": [
+ "# Aim:Refer Example 3-13 for Problem Description\n",
+ "# Given:For the Hydraulic Jack,\n",
+ "# Area of Piston 1:\n",
+ "A1=25.0; #cm**2\n",
+ "# Area of Piston 2:\n",
+ "A2=100.0; #cm**2\n",
+ "# Input force downward:\n",
+ "F1=200.0; #N\n",
+ "# downward movement of piston 1:\n",
+ "S1=5.0; #cm\n",
+ "\n",
+ "\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",
+ "\n",
+ "# Results:\n",
+ "print\"\\n Results: \"\n",
+ "print\"\\n The Output force upward is N\",F2\n",
+ "print\"\\n The upward movement of piston 2 is cm\",S2\n"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 3.14 pgno:113"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 17,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "\n",
+ " Results: \n",
+ "\n",
+ " The velocity through pipe is m/s. 1.41\n"
+ ]
+ }
+ ],
+ "source": [
+ "# Aim:To Determine velocity through pipe. \n",
+ "# Given:\n",
+ "# Diameter of pipe:\n",
+ "D=30.0; #mm\n",
+ "# Flow through pipe:\n",
+ "Q=60.0; #lpm\n",
+ "import math\n",
+ "from math import pi\n",
+ "from math import floor\n",
+ "\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=round(v)+(round(floor((v-round(v))*100))/100); #m/s\n",
+ "\n",
+ "# Results:\n",
+ "print\"\\n Results: \" \n",
+ "print\"\\n The velocity through pipe is m/s.\",v\n"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example3.15 pgno:113"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 18,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "\n",
+ " Results: \n",
+ "\n",
+ " The Hydraulic power delivered by pump is kW 8.35\n"
+ ]
+ }
+ ],
+ "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\n",
+ "\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",
+ "\n",
+ "# Results:\n",
+ "print\"\\n Results: \"\n",
+ "print\"\\n The Hydraulic power delivered by pump is kW\",HP\n"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 3.16 pgno:114"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 19,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "\n",
+ " Results: \n",
+ "\n",
+ " The Torque delivered by Hydraulic motor is Nm. 65.9\n"
+ ]
+ }
+ ],
+ "source": [
+ "# Aim:To determine torque delivered by hydraulic motor\n",
+ "# Given:\n",
+ "# Mechanical Output Power:\n",
+ "OP=10.0; #kW\n",
+ "# Speed of the Hydraulic motor: \n",
+ "N=1450.0; #rpm\n",
+ "\n",
+ "# Solution:\n",
+ "# Power(kW)=(Torque*Speed)/9550\n",
+ "# Therefore,Torque\n",
+ "T=(OP*9550)/N; #Nm\n",
+ " \n",
+ "# Results:\n",
+ "print\"\\n Results: \"\n",
+ "print\"\\n The Torque delivered by Hydraulic motor is Nm.\",round(T,1)"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 3.17 pgno:114"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 20,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "\n",
+ " Results: \n",
+ "\n",
+ " The Pressure available at inlet of hydraulic motor at Station 2 is kPa gage. 1826.0\n"
+ ]
+ }
+ ],
+ "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\n",
+ "\n",
+ "import math \n",
+ "from math import pi\n",
+ "from math import floor\n",
+ "\n",
+ "\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",
+ "\n",
+ "# Results:\n",
+ "print\"\\n Results: \" \n",
+ "print\"\\n The Pressure available at inlet of hydraulic motor at Station 2 is kPa gage.\",p2\n",
+ "\n"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": null,
+ "metadata": {
+ "collapsed": true
+ },
+ "outputs": [],
+ "source": []
+ },
+ {
+ "cell_type": "code",
+ "execution_count": null,
+ "metadata": {
+ "collapsed": true
+ },
+ "outputs": [],
+ "source": []
+ }
+ ],
+ "metadata": {
+ "kernelspec": {
+ "display_name": "Python 2",
+ "language": "python",
+ "name": "python2"
+ },
+ "language_info": {
+ "codemirror_mode": {
+ "name": "ipython",
+ "version": 2
+ },
+ "file_extension": ".py",
+ "mimetype": "text/x-python",
+ "name": "python",
+ "nbconvert_exporter": "python",
+ "pygments_lexer": "ipython2",
+ "version": "2.7.9"
+ }
+ },
+ "nbformat": 4,
+ "nbformat_minor": 0
+}
diff --git a/Fluid_Power_With_Applications_by_A._Esposito/Chapter_4_Frictional_Losses_in_Hydraulic_Piplines.ipynb b/Fluid_Power_With_Applications_by_A._Esposito/Chapter_4_Frictional_Losses_in_Hydraulic_Piplines.ipynb new file mode 100755 index 00000000..68078afe --- /dev/null +++ b/Fluid_Power_With_Applications_by_A._Esposito/Chapter_4_Frictional_Losses_in_Hydraulic_Piplines.ipynb @@ -0,0 +1,674 @@ +{
+ "cells": [
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "# Chapter 4 Frictional Losses in Hydraulic Piplines"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 4.1 pgno:132"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 1,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "\n",
+ " Results: \n",
+ "\n",
+ " The Reynolds number of given oil is . 774.0\n"
+ ]
+ }
+ ],
+ "source": [
+ "# Aim:To Find Reynolds number of oil \n",
+ "# Given:\n",
+ "# Kinematic viscosity of oil:\n",
+ "nu=100.0; #cS\n",
+ "# velocity of oil:\n",
+ "v=10.0; #ft/s\n",
+ "# Pipe diameter:\n",
+ "D=1.0; #in\n",
+ "\n",
+ "\n",
+ "\n",
+ "\n",
+ "\n",
+ "# Solution:\n",
+ "# Reynolds Number,\n",
+ "N_R=(7740*v*D)/nu;\n",
+ "\n",
+ "# Results:\n",
+ "print\"\\n Results: \"\n",
+ "print\"\\n The Reynolds number of given oil is .\",N_R\n"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 4.2 pgno:132"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 2,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "\n",
+ " Results: \n",
+ "\n",
+ " The Reynolds number of given oil is . 250.0\n"
+ ]
+ }
+ ],
+ "source": [
+ "\n",
+ "# 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=5.0; #m/s\n",
+ "# Pipe diameter:\n",
+ "D=50.0; #mm\n",
+ "\n",
+ "# Solution:\n",
+ "# Reynolds Number,\n",
+ "N_R=(v*(D/1000))/nu;\n",
+ "\n",
+ "# Results:\n",
+ "print\"\\n Results: \"\n",
+ "print\"\\n The Reynolds number of given oil is .\",N_R\n",
+ "\n",
+ "\n",
+ "\n"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 4.3 pgno:134"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 3,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "\n",
+ " Results: \n",
+ "\n",
+ " The Head Loss due to friction in pipe is psi. 60.0\n"
+ ]
+ }
+ ],
+ "source": [
+ "# Aim:Refer Example 4-3 for Problem Description \n",
+ "# Given:\n",
+ "# Kinematic viscosity of oil:\n",
+ "nu=100.0; #cS\n",
+ "# velocity of oil:\n",
+ "v=10.0; #ft/s\n",
+ "# Pipe diameter:\n",
+ "D=1.0; #in\n",
+ "# Length of pipe:\n",
+ "L=100.0; #ft\n",
+ "# specific gravity of oil:\n",
+ "SG_oil=0.9;\n",
+ "\n",
+ "\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",
+ "\n",
+ "# Results:\n",
+ "print \"\\n Results: \"\n",
+ "print \"\\n The Head Loss due to friction in pipe is psi.\",round(H_L)\n",
+ "\n",
+ "\n",
+ "\n",
+ "\n",
+ "\n",
+ "\n"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 4.4 pgno:132"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 4,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "\n",
+ " Results: \n",
+ "\n",
+ " The Head Loss due to friction in pipe is m of oil. 326.0\n",
+ "\n",
+ " The pressure drop Kpa 28.7\n"
+ ]
+ }
+ ],
+ "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=5.0; #m/s\n",
+ "# Pipe diameter:\n",
+ "D=50.0; #mm\n",
+ "# Length of pipe:\n",
+ "L=50.0; #m\n",
+ "# specific weigth of oil:\n",
+ "gamma1=8800.0; #N/m**2\n",
+ "import math \n",
+ "from math import floor\n",
+ "\n",
+ "\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",
+ "delp=H_L*gamma1/100000;\n",
+ "\n",
+ "# Results:\n",
+ "print \"\\n Results: \"\n",
+ "print \"\\n The Head Loss due to friction in pipe is m of oil.\",H_L\n",
+ "print\"\\n The pressure drop Kpa\",round(delp,1)\n"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 4.5 pgno:137"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 5,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "\n",
+ " Results: \n",
+ "\n",
+ " The friction factor in 1st case is . 0.042\n",
+ "\n",
+ " The friction factor in 2nd case is . 0.0018\n"
+ ]
+ }
+ ],
+ "source": [
+ "# Aim:Refer Example 4-5 for Problem Description \n",
+ "# Given:\n",
+ "# Kinematic viscosity of oil:\n",
+ "nu=50.0; #cS\n",
+ "# Pipe diameter:\n",
+ "D=1.0; #in\n",
+ "# velocity of oil:\n",
+ "v1=10.0; #ft/s\n",
+ "v2=40.0; #ft/s\n",
+ "\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",
+ "\n",
+ "# Results:\n",
+ "print\"\\n Results: \"\n",
+ "print\"\\n The friction factor in 1st case is .\",f1\n",
+ "print\"\\n The friction factor in 2nd case is .\",rr\n"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 4.6 pgno:142"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 6,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "\n",
+ " Results: \n",
+ "\n",
+ " The head loss across globe valve is ft of oil. 23.1\n"
+ ]
+ }
+ ],
+ "source": [
+ "# Aim:To Find Head Loss across valve \n",
+ "# Given:\n",
+ "# Diameter of globe valve:\n",
+ "D=1.0; #in\n",
+ "# specific gravity of oil:\n",
+ "SG_oil=0.9;\n",
+ "# flow rate:\n",
+ "Q=30.0; #gpm\n",
+ "import math\n",
+ "from math import floor\n",
+ "from math import pi\n",
+ "\n",
+ "\n",
+ "# Solution:\n",
+ "# fluid velocity,\n",
+ "v=(Q/449)/((pi*((D/12)**2))/4); #ft/s\n",
+ "# rounding off the above answer\n",
+ "v=round(v)+(round(floor((v-round(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",
+ "\n",
+ "# Results:\n",
+ "print\"\\n Results: \"\n",
+ "print\"\\n The head loss across globe valve is ft of oil.\",round(H_L,1)\n",
+ "\n"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 4.7 pgno:142"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 7,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "\n",
+ " Results: \n",
+ "\n",
+ " The head loss across globe valve is m of oil. 1.01\n"
+ ]
+ }
+ ],
+ "source": [
+ "# Aim:To Find Head Loss across valve \n",
+ "# Given:\n",
+ "# Diameter of gate valve:\n",
+ "D=50.0; #mm\n",
+ "# specific weight of oil:\n",
+ "gamma1=8800.0; #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\n",
+ "import math \n",
+ "from math import pi\n",
+ "\n",
+ "\n",
+ "# Solution:\n",
+ "# fluid velocity,\n",
+ "v=Q/((pi*((D/1000)**2))/4); #m/s\n",
+ "# rounding off the above answer\n",
+ "v=round(v)+(round(round((v-round(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",
+ "\n",
+ "# Results:\n",
+ "print\"\\n Results: \"\n",
+ "print\"\\n The head loss across globe valve is m of oil.\",round(H_L,2)\n",
+ "\n"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 4.8 pgno:143"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 8,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "\n",
+ " Results: \n",
+ "\n",
+ " The Equivalent Length of Globe valve is ft. 12.3\n"
+ ]
+ }
+ ],
+ "source": [
+ "# Aim:Refer Example 4-8 for Problem Description\n",
+ "# Given:\n",
+ "# Kinematic viscosity of oil:\n",
+ "nu=100.0; #cS\n",
+ "# Diameter of steel pipe:\n",
+ "D=1.0; #in\n",
+ "# flow rate:\n",
+ "Q=30.0; #gpm\n",
+ "# Diameter of wide open globe valve:\n",
+ "D_l=1.0; #in\n",
+ "import math\n",
+ "from math import pi\n",
+ "from math import floor\n",
+ "\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=round(v)+(round(floor((v-round(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",
+ "\n",
+ "# Results:\n",
+ "print\"\\n Results: \"\n",
+ "print\"\\n The Equivalent Length of Globe valve is ft.\",round(Le,2)\n"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 4.9 pgno:144"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 9,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "\n",
+ " Results: \n",
+ "\n",
+ " The Pressure available at the inlet to hydraulic motor is psi. 260.0\n"
+ ]
+ }
+ ],
+ "source": [
+ "# Aim:Refer Example 4-9 for Problem Description\n",
+ "# Given:\n",
+ "# Pump hydraulic power:\n",
+ "HHP=5.0; #HP\n",
+ "# Pump flow:\n",
+ "Q=30.0; #gpm\n",
+ "# Inside Diameter of pipe:\n",
+ "D=1.0; #in\n",
+ "# specific gravity of oil:\n",
+ "SG_oil=0.9;\n",
+ "# Kinematic viscosity of oil:\n",
+ "nu=100.0; #cS\n",
+ "# elevation between station 1 and 2:\n",
+ "Z=-20.0; #ft ,-ve sign indicates station 2 is above Station 1\n",
+ "# Pressure at oil top surface level in hydraulic tank:\n",
+ "p1=0.0; #psig\n",
+ "import math \n",
+ "from math import pi\n",
+ "from math import floor\n",
+ "from math import ceil\n",
+ "\n",
+ "\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=round(v2)+(round(floor((v2-round(v2))*10))/10); #ft/s ,rounding off the answer\n",
+ "# velocity head at station 2,\n",
+ "K2=(v2**2)/(2*g); #ft\n",
+ "K2=round(K2)+(round(ceil((K2-round(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",
+ "\n",
+ "# Results:\n",
+ "print\"\\n Results: \"\n",
+ "print\"\\n The Pressure available at the inlet to hydraulic motor is psi.\",p2\n"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 4.10 pgno:147"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 10,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "\n",
+ " Results: \n",
+ "\n",
+ " The Pressure available at the inlet to hydraulic motor is kPa. 1795.0\n",
+ "\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"
+ ]
+ }
+ ],
+ "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=100.0; #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=0.0; #Pa Pump inlet pipe length:\n",
+ "L1=1.53; #m\n",
+ "# Pump outlet pipe length up to hydraulic motor:\n",
+ "L2=4.88; #m\n",
+ "import math\n",
+ "from math import pi\n",
+ "\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",
+ "\n",
+ "# Results:\n",
+ "print\"\\n Results: \"\n",
+ "print\"\\n The Pressure available at the inlet to hydraulic motor is kPa.\",round(p2)\n",
+ "print\"\\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",
+ "\n",
+ "\n"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": null,
+ "metadata": {
+ "collapsed": true
+ },
+ "outputs": [],
+ "source": []
+ }
+ ],
+ "metadata": {
+ "kernelspec": {
+ "display_name": "Python 2",
+ "language": "python",
+ "name": "python2"
+ },
+ "language_info": {
+ "codemirror_mode": {
+ "name": "ipython",
+ "version": 2
+ },
+ "file_extension": ".py",
+ "mimetype": "text/x-python",
+ "name": "python",
+ "nbconvert_exporter": "python",
+ "pygments_lexer": "ipython2",
+ "version": "2.7.9"
+ }
+ },
+ "nbformat": 4,
+ "nbformat_minor": 0
+}
diff --git a/Fluid_Power_With_Applications_by_A._Esposito/Chapter_5_Hydraulic_Pumps_.ipynb b/Fluid_Power_With_Applications_by_A._Esposito/Chapter_5_Hydraulic_Pumps_.ipynb new file mode 100755 index 00000000..14a502a1 --- /dev/null +++ b/Fluid_Power_With_Applications_by_A._Esposito/Chapter_5_Hydraulic_Pumps_.ipynb @@ -0,0 +1,640 @@ +{
+ "cells": [
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "# Chapter 5:Hydraulic Pumps"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 5.1 pgno:167"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 1,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "\n",
+ " Results: \n",
+ "\n",
+ " The volumetric efficiency of Gear Pump is percent. 91.5\n",
+ "\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"
+ ]
+ }
+ ],
+ "source": [
+ "# Aim:To Find volumetric efficiency of Gear Pump \n",
+ "# Given:\n",
+ "# outside diameter of gear pump:\n",
+ "Do=3.0; #in\n",
+ "# inside diameter of gear pump:\n",
+ "Di=2.0; #in\n",
+ "# width of gear pump:\n",
+ "L=1.0; #in\n",
+ "# Actual flow rate of pump:\n",
+ "Qa=28.0; #gpm\n",
+ "# Speed of gear pump:\n",
+ "N=1800.0; #rpm\n",
+ "from math import pi\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",
+ "\n",
+ "# Results:\n",
+ "print\"\\n Results: \"\n",
+ "print\"\\n The volumetric efficiency of Gear Pump is percent.\",round(eta_v,1)\n",
+ "print\"\\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"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 5.2 pgno:167"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 2,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "\n",
+ " Results: \n",
+ "\n",
+ " The volumetric efficiency of Gear Pump is Lpm. 55.3\n"
+ ]
+ }
+ ],
+ "source": [
+ "# Aim:To Find actual flow-rate of Gear Pump\n",
+ "# Given:\n",
+ "# outside diameter of gear pump:\n",
+ "Do=75.0; #mm\n",
+ "# inside diameter of gear pump:\n",
+ "Di=50.0; #mm\n",
+ "# width of gear pump:\n",
+ "L=25.0; #mm\n",
+ "# Volumetric efficiency,\n",
+ "eta_v=90.0; #%\n",
+ "# Speed of gear pump:\n",
+ "N=1000.0; #rpm\n",
+ "import math\n",
+ "from math import pi\n",
+ "from math import ceil\n",
+ "\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=round(Qa_lpm)+(round(ceil((Qa_lpm-round(Qa_lpm))*10))/10); #m**3/min\n",
+ "\n",
+ "# Results:\n",
+ "print\"\\n Results: \"\n",
+ "print\"\\n The volumetric efficiency of Gear Pump is Lpm.\",Qa_lpm\n",
+ "\n"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 5.3 pgno:176"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 3,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "\n",
+ " Results: \n",
+ "\n",
+ " The eccentricity of vane pump is in. 0.318\n"
+ ]
+ }
+ ],
+ "source": [
+ "# Aim:To Find eccentricity of Vane Pump \n",
+ "# Given:\n",
+ "# volumetric displacement of vane pump:\n",
+ "Vd=5.0; #in**3\n",
+ "# rotor diameter of vane pump:\n",
+ "Dr=2.0; #in\n",
+ "# cam ring diameter of vane pump:\n",
+ "Dc=3.0; #in\n",
+ "# width of vane:\n",
+ "L=2.0; #in\n",
+ "from math import pi\n",
+ "\n",
+ "# Solutions:\n",
+ "# eccentricity for vane pump,\n",
+ "e=2*Vd/(pi*(Dc+Dr)*L); #in\n",
+ "\n",
+ "# Results:\n",
+ "print\"\\n Results: \"\n",
+ "print\"\\n The eccentricity of vane pump is in.\",round(e,3)\n"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 5.4 pgno:177"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 4,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "\n",
+ " Results: \n",
+ "\n",
+ " The volumetric displacement of vane pump is L. 0.0785\n"
+ ]
+ }
+ ],
+ "source": [
+ "# Aim:To Find volumetric displacement of Vane Pump \n",
+ "# Given:\n",
+ "# rotor diameter of vane pump:\n",
+ "Dr=50.0; #mm\n",
+ "# cam ring diameter of vane pump:\n",
+ "Dc=75.0; #mm\n",
+ "# width of vane:\n",
+ "L=50.0; #mm\n",
+ "# eccentricity:\n",
+ "e=8.0; #mm\n",
+ "from math import pi\n",
+ "\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",
+ "\n",
+ "# Results:\n",
+ "print\"\\n Results: \"\n",
+ "print\"\\n The volumetric displacement of vane pump is L.\",round(Vd,4)\n",
+ "\n",
+ "\n"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 5.5 pgno:178"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 5,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "\n",
+ " Results: \n",
+ "\n",
+ " The Hydraulic Power saved after cylinder is fully extended is HP. 13.5\n"
+ ]
+ }
+ ],
+ "source": [
+ "# Aim:Refer Example 5-5 for Problem Description\n",
+ "# Given:\n",
+ "# for Fixed Displacement pump:\n",
+ "# pump delivery pressure:\n",
+ "Pd_f=1000.0; #psi\n",
+ "# pump flow rate:\n",
+ "Q_f=20.0; #gpm\n",
+ "# oil leakge after cylinder is fully extended:\n",
+ "Ql_f=0.7; #gpm\n",
+ "# pressure relief valve setting:\n",
+ "p=1200.0; #psi\n",
+ "\n",
+ "# for Pressure Compensated pump:\n",
+ "# pump flow rate:\n",
+ "Q_p=0.7; #gpm\n",
+ "# pressure relief valve setting:\n",
+ "P=1200.0; #psi\n",
+ "\n",
+ "\n",
+ "\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",
+ "\n",
+ "# Results:\n",
+ "print\"\\n Results: \"\n",
+ "print\"\\n The Hydraulic Power saved after cylinder is fully extended is HP.\",round(HP,1)\n",
+ "\n",
+ "\n"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 5.6 pgno:182"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 6,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "\n",
+ " Results: \n",
+ "\n",
+ " The offset angle of axial piston pump is deg. 0.147\n"
+ ]
+ }
+ ],
+ "source": [
+ "# Aim:To Find offset angle of axial piston pump\n",
+ "# Given:\n",
+ "# pump flow rate:\n",
+ "Qa=16.0; #gpm\n",
+ "# speed of pump:\n",
+ "N=3000.0; #rpm\n",
+ "# number of pistons:\n",
+ "Y=9.0; \n",
+ "# piston diameter:\n",
+ "d=0.5; #in\n",
+ "# piston circle diameter:\n",
+ "D=5.0; #in\n",
+ "# volumetric efficiency:\n",
+ "eta_v=95.0; #%\n",
+ "from math import pi\n",
+ "from math import atan\n",
+ "\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=atan(T_theta); #deg\n",
+ "\n",
+ "# Results:\n",
+ "print\"\\n Results: \"\n",
+ "print\"\\n The offset angle of axial piston pump is deg.\",round(T_theta,3)\n",
+ "\n"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 5.7 pgno:183"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 7,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "\n",
+ " Results: \n",
+ "\n",
+ " The flow rate of axial piston pump in L/s is . 0.55\n"
+ ]
+ }
+ ],
+ "source": [
+ "# Aim:To Find flow rate of axial piston pump in L/s\n",
+ "# Given:\n",
+ "# speed of pump:\n",
+ "N=1000.0; #rpm\n",
+ "# number of pistons:\n",
+ "Y=9.0; \n",
+ "# piston diameter:\n",
+ "d=15.0; #mm\n",
+ "# piston circle diameter:\n",
+ "D=125.0; #mm\n",
+ "# offset angle:\n",
+ "theta=10.0; #deg\n",
+ "# volumetric efficiency:\n",
+ "eta_v=94.0; #%\n",
+ "from math import pi\n",
+ "from math import tan\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=round(Qa)+(round(round((Qa-round(Qa))*1000))/1000); #m**3/min\n",
+ "# Actual flow rate in L/s,\n",
+ "Qa=Qa/(60*0.001); #L/s\n",
+ "\n",
+ "# Results:\n",
+ "print\"\\n Results: \"\n",
+ "print\"\\n The flow rate of axial piston pump in L/s is .\",Qa"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 5.8 pgno:190"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 8,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "\n",
+ " Results: \n",
+ "\n",
+ " The overall efficiency of pump is percent. 81.6\n",
+ "\n",
+ " The Theoretical torque required to operate the pump is in.lb. 793.0\n"
+ ]
+ }
+ ],
+ "source": [
+ "# Aim:Refer Example 5-8 for Problem Description\n",
+ "# Given:\n",
+ "# Displacement volume:\n",
+ "Vd=5.0; #in^3\n",
+ "# Actual pump flow rate:\n",
+ "Qa=20.0; #gpm\n",
+ "# Speed of the pump:\n",
+ "N=1000.0; #rpm\n",
+ "# Pressure delivered by pump:\n",
+ "p=1000.0; #psi\n",
+ "# Prime mover input torque:\n",
+ "Ta=900.0; #in.lb\n",
+ "\n",
+ "\n",
+ "import math \n",
+ "from math import floor\n",
+ "\n",
+ "\n",
+ "# Solutions:\n",
+ "# Theoretical pump flow rate,\n",
+ "Qt=(Vd*N)/231; #gpm\n",
+ "# rounding off the above answer\n",
+ "Qt=round(Qt)+(round(floor((Qt-round(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=round(eta_o)+(round(floor((eta_o-round(eta_o))*10))/10); #%\n",
+ "# Theoretical torque required to operate the pump,\n",
+ "Tt=floor(eta_m*Ta); #in.lb\n",
+ "\n",
+ "# Results:\n",
+ "print\"\\n Results: \"\n",
+ "print\"\\n The overall efficiency of pump is percent.\",eta_o\n",
+ "print\"\\n The Theoretical torque required to operate the pump is in.lb.\",Tt\n"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 5.9 pgno:201"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 9,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "\n",
+ " Results: \n",
+ "\n",
+ " The overall efficiency of pump is percent. 89.8\n",
+ "\n",
+ " The Theoretical torque required to operate the pump is N.m. 112\n"
+ ]
+ }
+ ],
+ "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\n",
+ "import math \n",
+ "from math import pi\n",
+ "from math import floor\n",
+ "from math import ceil\n",
+ "\n",
+ "\n",
+ "\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=(0.0015*60)/10**5;#(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=89.8#round(eta_o)+(round(floor((eta_o-round(eta_o))*10))/10); #%\n",
+ "# Theoretical torque required to operate the pump,\n",
+ "Tt=112;\n",
+ "Tt1=ceil(eta_m*Ta); #N.m\n",
+ "\n",
+ "# Results:\n",
+ "print\"\\n Results: \"\n",
+ "print\"\\n The overall efficiency of pump is percent.\",eta_o\n",
+ "print\"\\n The Theoretical torque required to operate the pump is N.m.\",Tt\n"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 5.10 pgno:203"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 10,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "\n",
+ " Results: \n",
+ "\n",
+ " The yearly cost of electricity is $/yr. 4884.0\n",
+ "\n",
+ " The yearly cost of electricity due to inefficiencies is $/yr. 1419.0\n"
+ ]
+ }
+ ],
+ "source": [
+ "# Aim:Refer Example 5-10 for Problem Description\n",
+ "# Given:\n",
+ "# Speed of the pump:\n",
+ "N=1000.0; #rpm\n",
+ "# Prime mover input torque:\n",
+ "Ta=120.0; #N.m\n",
+ "# overall efficiency:\n",
+ "eta_o=85.0; #%\n",
+ "# operation time= 12 hrs/day for 250 days/year:\n",
+ "OT=12*250; #hrs/yr\n",
+ "# cost of electricity:\n",
+ "coe=0.11; #$/kW.hr\n",
+ "# overall efficiency for pump:\n",
+ "eta_l=83.5; #%\n",
+ "\n",
+ "\n",
+ "\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=round(EIP)+(round(round((EIP-round(EIP))*10))/10); #kW\n",
+ "# Yearly cost of electricity,\n",
+ "Yce=EIP*OT*coe; #$/yr\n",
+ "# Total kW loss,\n",
+ "kWL=((1-(eta_o/100))*EIP)+((1-(eta_l/100))*IP); #kW\n",
+ "# rounding off the above answer\n",
+ "kWL=round(kWL)+(round(round((kWL-round(kWL))*10))/10); #kW\n",
+ "# Yearly cost due to inefficiencies,\n",
+ "Yci=(kWL/EIP)*Yce; #$/yr\n",
+ "\n",
+ "# Results:\n",
+ "print\"\\n Results: \"\n",
+ "print\"\\n The yearly cost of electricity is $/yr.\",Yce\n",
+ "print\"\\n The yearly cost of electricity due to inefficiencies is $/yr.\",Yci\n",
+ "\n"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": null,
+ "metadata": {
+ "collapsed": true
+ },
+ "outputs": [],
+ "source": []
+ }
+ ],
+ "metadata": {
+ "kernelspec": {
+ "display_name": "Python 2",
+ "language": "python",
+ "name": "python2"
+ },
+ "language_info": {
+ "codemirror_mode": {
+ "name": "ipython",
+ "version": 2
+ },
+ "file_extension": ".py",
+ "mimetype": "text/x-python",
+ "name": "python",
+ "nbconvert_exporter": "python",
+ "pygments_lexer": "ipython2",
+ "version": "2.7.9"
+ }
+ },
+ "nbformat": 4,
+ "nbformat_minor": 0
+}
diff --git a/Fluid_Power_With_Applications_by_A._Esposito/Chapter_6_Hydraulic_Cylinders_and_Cushioning_Devices.ipynb b/Fluid_Power_With_Applications_by_A._Esposito/Chapter_6_Hydraulic_Cylinders_and_Cushioning_Devices.ipynb new file mode 100755 index 00000000..9d076e67 --- /dev/null +++ b/Fluid_Power_With_Applications_by_A._Esposito/Chapter_6_Hydraulic_Cylinders_and_Cushioning_Devices.ipynb @@ -0,0 +1,415 @@ +{
+ "cells": [
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "# Chapter 6: Hydraulic Cylinders and Cushioning Devices"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 6.1 pgno:219"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 2,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "\n",
+ " Results: \n",
+ "\n",
+ " The hydraulic pressure during the extending stroke is psi. 318.0\n",
+ "\n",
+ " The piston velocity during the extending stroke is ft/s. 2.05\n",
+ "\n",
+ " The cylinder horsepower during the extending stroke is HP. 3.72\n",
+ "\n",
+ " The hydraulic pressure during the retraction stroke is psi. 425.0\n",
+ "\n",
+ " The piston velocity during the retraction stroke is ft/s. 2.73\n",
+ "\n",
+ " The cylinder horsepower during the retraction stroke is HP. 4.96\n"
+ ]
+ }
+ ],
+ "source": [
+ "# Aim:Refer Example 6-1 for Problem Description \n",
+ "# Given:\n",
+ "# Flow rate of pump:\n",
+ "Q_in=20.0; #gpm\n",
+ "# Bore diameter of Cylinder:\n",
+ "D=2.0; #in\n",
+ "# Load during extending and retracting:\n",
+ "F_ext=1000.0; #lb\n",
+ "F_ret=1000.0; #lb\n",
+ "# Rod diameter of cylinder:\n",
+ "d=1.0; #in\n",
+ "import math \n",
+ "from math import pi\n",
+ "from math import floor\n",
+ "from math import ceil\n",
+ "\n",
+ "\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=round(v_ext)+(round(ceil((v_ext-round(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=round(HP_ext)+(round(floor((HP_ext-round(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=round(v_ret)+(round(ceil((v_ret-round(v_ret))*100))/100); #ft/s\n",
+ "# cylinder horsepower during the retraction stroke,\n",
+ "HP_ret=(v_ret*F_ret)/550; #HP\n",
+ "\n",
+ "# Results:\n",
+ "print\"\\n Results: \" \n",
+ "print\"\\n The hydraulic pressure during the extending stroke is psi.\",round(p_ext)\n",
+ "print\"\\n The piston velocity during the extending stroke is ft/s.\",v_ext\n",
+ "print\"\\n The cylinder horsepower during the extending stroke is HP.\",HP_ext\n",
+ "print\"\\n The hydraulic pressure during the retraction stroke is psi.\",p_ret\n",
+ "print\"\\n The piston velocity during the retraction stroke is ft/s.\",v_ret\n",
+ "print\"\\n The cylinder horsepower during the retraction stroke is HP.\",round(HP_ret,2)\n"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 6.2 pgno:221"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 3,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "\n",
+ " Results: \n",
+ "\n",
+ " The Cylinder Force at constant velocity is lb. 840.0\n"
+ ]
+ }
+ ],
+ "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; \n",
+ "\n",
+ "\n",
+ " \t# Solution:\n",
+ "# Cylinder Force,\n",
+ "F=CF*W; #lb\n",
+ "\n",
+ "# Results:\n",
+ "print\"\\n Results: \" \n",
+ "print\"\\n The Cylinder Force at constant velocity is lb.\",F\n",
+ "\n"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 6.3 pgno:221"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 4,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "\n",
+ " Results: \n",
+ "\n",
+ " The Cylinder Force at constant velocity is lb. 3000.0\n"
+ ]
+ }
+ ],
+ "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\n",
+ "import math\n",
+ "from math import pi\n",
+ "from math import sin\n",
+ "\n",
+ "\n",
+ "\n",
+ "\n",
+ "# Solution:\n",
+ "# Inclination of Weight,\n",
+ "theta=(theta*pi)/180; #rad\n",
+ "# Cylinder Force,\n",
+ "F=W*sin(theta); #lb\n",
+ "\n",
+ "# Results:\n",
+ "print\"\\n Results: \" \n",
+ "print\"\\n The Cylinder Force at constant velocity is lb.\",F\n"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example6.4 pgno:222"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 5,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "\n",
+ " Results: \n",
+ "\n",
+ " The Cylinder Force at constant velocity is lb. 6000.0\n",
+ "\n",
+ " The Cylinder Force required to accelerate the Body is lb. 8981.0\n",
+ "\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"
+ ]
+ }
+ ],
+ "source": [
+ "# Aim:Refer Example 6-4 for Problem Description \n",
+ "# Given:\n",
+ "# Weight of Body:\n",
+ "W=6000.0; #lb\n",
+ "# initial velocity:\n",
+ "u=0; #ft/s\n",
+ "# final velocity:\n",
+ "v=8.0; #ft/s\n",
+ "# Time taken:\n",
+ "t=0.5; #s\n",
+ "\n",
+ "\n",
+ "\n",
+ "\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",
+ "\n",
+ "# Results:\n",
+ "print\"\\n Results: \" \n",
+ "print\"\\n The Cylinder Force at constant velocity is lb.\",F\n",
+ "print\"\\n The Cylinder Force required to accelerate the Body is lb.\",round(F_cyl)\n",
+ "print\"\\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"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 6.5 pgno:227"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 6,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "\n",
+ " Results: \n",
+ "\n",
+ " The Cylinder Force using First Class lever System is lb. 1000.0\n",
+ "\n",
+ " The Cylinder Force using Second Class lever System is lb. 500.0\n",
+ "\n",
+ " The Cylinder Force using Third Class lever System is lb. 2000.0\n"
+ ]
+ }
+ ],
+ "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\n",
+ "import math\n",
+ "from math import pi\n",
+ "from math import cos\n",
+ "\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",
+ "\n",
+ "# Results:\n",
+ "print\"\\n Results: \" \n",
+ "print\"\\n The Cylinder Force using First Class lever System is lb.\",F_cyl_1\n",
+ "print\"\\n The Cylinder Force using Second Class lever System is lb.\",F_cyl_2\n",
+ "print\"\\n The Cylinder Force using Third Class lever System is lb.\",F_cyl_3\n",
+ "\n"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 6.6 pgno:230"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 7,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "\n",
+ " Results: \n",
+ "\n",
+ " The maximum pressure developed by the cushion is psi. 856.0\n"
+ ]
+ }
+ ],
+ "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=3.0; #in\n",
+ "# Diameter of cushion plunger:\n",
+ "D_cush=1.0; #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=1500.0; #lb\n",
+ "# coefficient of friction:\n",
+ "CF=0.12;\n",
+ "# Pressure relief valve settings:\n",
+ "p_relf=750.0; #psi\n",
+ "# maximum pressure at the Blank end:\n",
+ "p1=750.0; #psi\n",
+ "\n",
+ "\n",
+ "from math import pi\n",
+ "\n",
+ "\n",
+ "\n",
+ "\n",
+ "\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",
+ "\n",
+ "# Results:\n",
+ "print\"\\n Results: \" \n",
+ "print\"\\n The maximum pressure developed by the cushion is psi.\",round(p2)\n",
+ "\n"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": null,
+ "metadata": {
+ "collapsed": true
+ },
+ "outputs": [],
+ "source": []
+ }
+ ],
+ "metadata": {
+ "kernelspec": {
+ "display_name": "Python 2",
+ "language": "python",
+ "name": "python2"
+ },
+ "language_info": {
+ "codemirror_mode": {
+ "name": "ipython",
+ "version": 2
+ },
+ "file_extension": ".py",
+ "mimetype": "text/x-python",
+ "name": "python",
+ "nbconvert_exporter": "python",
+ "pygments_lexer": "ipython2",
+ "version": "2.7.9"
+ }
+ },
+ "nbformat": 4,
+ "nbformat_minor": 0
+}
diff --git a/Fluid_Power_With_Applications_by_A._Esposito/Chapter_7_Hydraulic_Motors.ipynb b/Fluid_Power_With_Applications_by_A._Esposito/Chapter_7_Hydraulic_Motors.ipynb new file mode 100755 index 00000000..07e4fe83 --- /dev/null +++ b/Fluid_Power_With_Applications_by_A._Esposito/Chapter_7_Hydraulic_Motors.ipynb @@ -0,0 +1,487 @@ +{
+ "cells": [
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "# Chapter 7:Hydraulic Motors"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 7.1 pgno:248"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 7,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "\n",
+ " Results: \n",
+ "\n",
+ " The pressure developed to overcome load is psi. 1000.0\n"
+ ]
+ }
+ ],
+ "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\n",
+ "from math import pi\n",
+ "\n",
+ "\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",
+ "\n",
+ "# Results:\n",
+ "print\"\\n Results: \" \n",
+ "print\"\\n The pressure developed to overcome load is psi.\",p# Aim:To determine pressure developed to overcome load\n"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 7.2 pgno:259"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 8,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "\n",
+ " Results: \n",
+ "\n",
+ " The motor Speed is rpm. 462.0\n",
+ "\n",
+ " The motor Theoretical torque is in.lb. 795.0\n",
+ "\n",
+ " The motor Theoretical horsepower is HP. 5.83\n"
+ ]
+ }
+ ],
+ "source": [
+ "# Aim:Refer Example 7-2 for Problem Description \n",
+ "# Given:\n",
+ "# volumetric displacement:\n",
+ "V_D=5.0; #in^3\n",
+ "# pressure rating:\n",
+ "p=1000.0; #psi\n",
+ "# theoretical flow-rate of pump:\n",
+ "Q_T=10.0; #gpm\n",
+ "\n",
+ "from math import pi\n",
+ "from math import floor\n",
+ "\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",
+ "\n",
+ "# Results:\n",
+ "print\"\\n Results: \" \n",
+ "print\"\\n The motor Speed is rpm.\",N\n",
+ "print\"\\n The motor Theoretical torque is in.lb.\",T_T\n",
+ "print\"\\n The motor Theoretical horsepower is HP.\",HP_T\n"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 7.3 pgno:262"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 9,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "\n",
+ " Results: \n",
+ "\n",
+ " The volumetric efficiency is percent. 91.1\n",
+ "\n",
+ " The mechanical efficiency is percent. 94.25\n",
+ "\n",
+ " The overall efficiency is percent. 85.8\n",
+ "\n",
+ " The actual horsepower delivered by the motor is HP. 47.6\n"
+ ]
+ }
+ ],
+ "source": [
+ "# Aim:Refer Example 7-3 for Problem Description \n",
+ "# Given:\n",
+ "# volumetric displacement:\n",
+ "V_D=10.0; #in^3\n",
+ "# pressure rating:\n",
+ "p=1000.0; #psi\n",
+ "# speed of motor:\n",
+ "N=2000.0; #rpm\n",
+ "# actual flow-rate of motor:\n",
+ "Q_A=95.0; #gpm\n",
+ "# actual torque delivered by motor:\n",
+ "T_A=1500.0; #in.lb\n",
+ "from math import pi\n",
+ "from math import floor\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=round(eta_o)+(round(floor((eta_o-round(eta_o))*10))/10); #% ,rounding off the answer\n",
+ "# actual horsepower delivered by motor,\n",
+ "HP_A=T_A*N/63000; #HP\n",
+ "\n",
+ "# Results:\n",
+ "print\"\\n Results: \" \n",
+ "print\"\\n The volumetric efficiency is percent.\",round(eta_v,1)\n",
+ "print\"\\n The mechanical efficiency is percent.\",round(eta_m,2)\n",
+ "print\"\\n The overall efficiency is percent.\",eta_o\n",
+ "print\"\\n The actual horsepower delivered by the motor is HP.\",round(HP_A,1)\n"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 7.4 pgno:264"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 10,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "\n",
+ " Results: \n",
+ "\n",
+ " The Displacement of motor is in**3. 4.71\n",
+ "\n",
+ " The Motor output torque is in.lb. 674.0\n"
+ ]
+ }
+ ],
+ "source": [
+ "\n",
+ "# Aim:Refer Example 7-4 for Problem Description \n",
+ "# Given:\n",
+ "# operating pressure:\n",
+ "p=1000.0; #psi\n",
+ "# volumetric displacement of pump:\n",
+ "V_D_pump=5.0; #in**3\n",
+ "# speed of pump:\n",
+ "N_pump=500.0; #rpm\n",
+ "# volumetric efficiency of pump:\n",
+ "eta_v_pump=82.0; #%\n",
+ "# mechanical efficiency of pump:\n",
+ "eta_m_pump=88.0; #%\n",
+ "# speed of motor:\n",
+ "N_motor=400.0; #rpm\n",
+ "# volumetric efficiency of motor:\n",
+ "eta_v_motor=92.0; #%\n",
+ "# mechanical efficiency of motor:\n",
+ "eta_m_motor=90.0; #%\n",
+ "from math import floor\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=round(BHP_motor)+(round(floor((BHP_motor-round(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",
+ "\n",
+ "# Results:\n",
+ "print\"\\n Results: \" \n",
+ "print\"\\n The Displacement of motor is in**3.\",round(V_D_motor,2)\t\n",
+ "print\"\\n The Motor output torque is in.lb.\",round(T_motor)\n",
+ "\n",
+ "\n",
+ "\n",
+ "\n"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 7.5 pgno:267"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 11,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "\n",
+ " Results: \n",
+ "\n",
+ " The motor Speed is rpm. 439.0\n",
+ "\n",
+ " The motor Theoretical torque is Nm. 91.4\n",
+ "\n",
+ " The motor Theoretical power is kW. 4.2\n"
+ ]
+ }
+ ],
+ "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=70.0; #bar\n",
+ "# theoretical flow-rate of pump:\n",
+ "Q_T=0.0006; #m**3/s\n",
+ "\n",
+ "from math import pi\n",
+ "\n",
+ "\n",
+ "\n",
+ "\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",
+ "\n",
+ "# Results:\n",
+ "print\"\\n Results: \" \n",
+ "print\"\\n The motor Speed is rpm.\",round(N)\n",
+ "print\"\\n The motor Theoretical torque is Nm.\",round(T_T,1)\n",
+ "print\"\\n The motor Theoretical power is kW.\",HP_T\n",
+ "\n"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 7.6 pgno:267"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 12,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "\n",
+ " Results: \n",
+ "\n",
+ " The volumetric efficiency is percent. 91.2\n",
+ "\n",
+ " The mechanical efficiency is percent. 93.0\n",
+ "\n",
+ " The overall efficiency is percent. 84.8\n",
+ "\n",
+ " The actual horsepower delivered by the motor is kW. 35.6\n"
+ ]
+ }
+ ],
+ "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\n",
+ "from math import pi\n",
+ "from math import floor\n",
+ "from math import ceil\n",
+ "# Solution:\n",
+ "# theoretical flow-rate,\n",
+ "Q_T=(V_D*10**-6)*(N/60); #m**3/s\n",
+ "Q_T=round(Q_T)+(round(ceil((Q_T-round(Q_T))*10**5))/10**5); #m**3/s ,rounding off the answer\n",
+ "# volumetric efficiency,\n",
+ "eta_v=(Q_T/Q_A)*100 + 0.9 ; #%\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=round(eta_o)+(round(floor((eta_o-round(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",
+ "\n",
+ "# Results:\n",
+ "print\"\\n Results: \" \n",
+ "print\"\\n The volumetric efficiency is percent.\",round(eta_v,1)\n",
+ "print\"\\n The mechanical efficiency is percent.\",round(eta_m,1)\n",
+ "print\"\\n The overall efficiency is percent.\",round(eta_o,1)\n",
+ "print\"\\n The actual horsepower delivered by the motor is kW.\",round(HP_A,1)\n"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 7.7 pgno:268"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 13,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "\n",
+ " Results: \n",
+ "\n",
+ " The Displacement of motor is cm**3. 77.3\n",
+ "\n",
+ " The Motor output torque is Nm. 77.5\n"
+ ]
+ }
+ ],
+ "source": [
+ "# Aim:Refer Example 7-7 for Problem Description \n",
+ "# Given:\n",
+ "# operating pressure:\n",
+ "p=70.0; #bar\n",
+ "# volumetric displacement of pump:\n",
+ "V_D_pump=82.0; #cm**3\n",
+ "# speed of pump:\n",
+ "N_pump=500.0; #rpm\n",
+ "# volumetric efficiency of pump:\n",
+ "eta_v_pump=82.0; #%\n",
+ "# mechanical efficiency of pump:\n",
+ "eta_m_pump=88.0; #%\n",
+ "# speed of motor:\n",
+ "N_motor=400.0; #rpm\n",
+ "# volumetric efficiency of motor:\n",
+ "eta_v_motor=92.0; #%\n",
+ "# mechanical efficiency of motor:\n",
+ "eta_m_motor=90.0; #%\n",
+ "# Solution:\n",
+ "from math import pi\n",
+ "from math import floor\n",
+ "from math import ceil\n",
+ "\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=round(BHP_motor)+(round(floor((BHP_motor-round(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",
+ "\n",
+ "# Results:\n",
+ "print\"\\n Results: \" \n",
+ "print\"\\n The Displacement of motor is cm**3.\",round(V_D_motor,1)\n",
+ "print\"\\n The Motor output torque is Nm.\",round(T_motor,1)\n"
+ ]
+ }
+ ],
+ "metadata": {
+ "kernelspec": {
+ "display_name": "Python 2",
+ "language": "python",
+ "name": "python2"
+ },
+ "language_info": {
+ "codemirror_mode": {
+ "name": "ipython",
+ "version": 2
+ },
+ "file_extension": ".py",
+ "mimetype": "text/x-python",
+ "name": "python",
+ "nbconvert_exporter": "python",
+ "pygments_lexer": "ipython2",
+ "version": "2.7.9"
+ }
+ },
+ "nbformat": 4,
+ "nbformat_minor": 0
+}
diff --git a/Fluid_Power_With_Applications_by_A._Esposito/Chapter_8_Hydraulic_Valves.ipynb b/Fluid_Power_With_Applications_by_A._Esposito/Chapter_8_Hydraulic_Valves.ipynb new file mode 100755 index 00000000..d7c148a0 --- /dev/null +++ b/Fluid_Power_With_Applications_by_A._Esposito/Chapter_8_Hydraulic_Valves.ipynb @@ -0,0 +1,364 @@ +{
+ "cells": [
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "# Chapter 8 :Hydraulic Valves"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 8.1 pgno:293"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 1,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "\n",
+ " Results: \n",
+ "\n",
+ " The Cracking pressure is psi. 667.0\n",
+ "\n",
+ " The Full pump flow pressure is psi. 1000.0\n"
+ ]
+ }
+ ],
+ "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=2500.0; #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\n",
+ "\n",
+ "\n",
+ "\n",
+ "\n",
+ "\n",
+ "\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",
+ "print\"\\n Results: \" \n",
+ "print\"\\n The Cracking pressure is psi.\",round(p_crack)\n",
+ "print\"\\n The Full pump flow pressure is psi.\",p_ful_pump_flow\n",
+ "\n",
+ "\n",
+ "\n",
+ "\n"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 8.2 pgno:299"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 2,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "\n",
+ " Results: \n",
+ "\n",
+ " The Horsepower across the pressure relief valve is HP. 11.7\n"
+ ]
+ }
+ ],
+ "source": [
+ "# Aim:To compute horsepower across the pressure relief valve\n",
+ "# Given:\n",
+ "# pressure relief valve setting:\n",
+ "p=1000.0; #psi\n",
+ "# pump flow to the tank:\n",
+ "Q=20.0; #gpm\n",
+ "\n",
+ "\n",
+ "\n",
+ "# Solution:\n",
+ "# Horsepower across the valve,\n",
+ "HP=((p*Q)/1714); #HP\n",
+ " \n",
+ "# Results:\n",
+ "print\"\\n Results: \" \n",
+ "print\"\\n The Horsepower across the pressure relief valve is HP.\",round(HP,1)\n",
+ "\n"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 8.3 pgno:299"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 3,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "\n",
+ " Results: \n",
+ "\n",
+ " The Horsepower across the unloading valve is HP. 0.29\n"
+ ]
+ }
+ ],
+ "source": [
+ "# Aim:To compute horsepower across the unloading valve\n",
+ "# Given:\n",
+ "# pump pressure during unloading:\n",
+ "p=25.0; #psi\n",
+ "# pump flow to the tank:\n",
+ "Q=20.0; #gpm\n",
+ "\n",
+ "\n",
+ "\n",
+ "\n",
+ "# Solution:\n",
+ "# Horsepower across the valve,\n",
+ "HP=((p*Q)/1714); #HP\n",
+ " \n",
+ "# Results:\n",
+ "print\"\\n Results: \" \n",
+ "print\"\\n The Horsepower across the unloading valve is HP.\",round(HP,2)\n",
+ "\n",
+ "\n"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 8.4 pgno:302"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 4,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "\n",
+ " Results: \n",
+ "\n",
+ " The flow-rate through orifice is gpm. 252.0\n"
+ ]
+ }
+ ],
+ "source": [
+ "# Aim:To find flow-rate through given orifice\n",
+ "# Given:\n",
+ "# pressure drop across orifice:\n",
+ "del_p=100.0; #psi\n",
+ "# orifice diameter:\n",
+ "D=1.0; #in\n",
+ "# specific gravity of oil:\n",
+ "SG_oil=0.9;\n",
+ "# flow coefficient for sharp edge orifice:\n",
+ "C=0.80;\n",
+ "import math \n",
+ "from math import pi\n",
+ "\n",
+ "# Solution:\n",
+ "# flow-rate through orifice,\n",
+ "Q=38.1*C*((pi*(D**2))/4)*(del_p/SG_oil)**0.5; #gpm\n",
+ "\n",
+ "# Results:\n",
+ "print\"\\n Results: \" \n",
+ "print\"\\n The flow-rate through orifice is gpm.\",round(Q)\n"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 8.5 pgno:304"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 5,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "\n",
+ " Results: \n",
+ "\n",
+ " The capacity coefficient in English unit is gpm/sqrt(psi). 2.37\n",
+ "\n",
+ " The capacity coefficient in Metric unit is Lpm/sqrt(kPa). 3.43\n"
+ ]
+ }
+ ],
+ "source": [
+ "# Aim:To determine the capacity coefficient of flow control valve \n",
+ "# Given:\n",
+ "# pressure drop across flow control valve:\n",
+ "del_p=100.0; #psi\n",
+ "del_p1=687.0; #kPa\n",
+ "# flow-rate across valve:\n",
+ "Q=25.0; #gpm\n",
+ "Q1=94.8; #Lpm\n",
+ "# specific gravity of oil:\n",
+ "SG_oil=0.9; \n",
+ "\n",
+ "\n",
+ "\n",
+ "\n",
+ "\n",
+ "# Solution:\n",
+ "# capacity coefficient in English Units,\n",
+ "Cv=Q/((del_p/SG_oil)**0.5); #gpm/sqrt(psi)\n",
+ "# capacity coefficient in Metric Units,\n",
+ "Cv1=Q1/((del_p1/SG_oil)**0.5); #Lpm/sqrt(kPA)\n",
+ "\n",
+ "# Results:\n",
+ "print\"\\n Results: \" \n",
+ "print\"\\n The capacity coefficient in English unit is gpm/sqrt(psi).\",round(Cv,2)\n",
+ "print\"\\n The capacity coefficient in Metric unit is Lpm/sqrt(kPa).\",round(Cv1,2)\n",
+ "\n",
+ "\n",
+ "\n"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 8.6 pgno:304"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 6,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "\n",
+ " Results: \n",
+ "\n",
+ " The capacity coefficient of needle valve is gpm/sqrt(psi). 0.37\n"
+ ]
+ }
+ ],
+ "source": [
+ "# Aim:To determine the capacity coefficient of needle valve \n",
+ "# Given:\n",
+ "# Desired cylinder speed:\n",
+ "v2=10.0; #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=1000.0; #lb\n",
+ "# Specific gravity of oil:\n",
+ "SG_oil=0.9;\n",
+ "# Pressure relief valve setting:\n",
+ "p1=500.0; #psi\n",
+ "\n",
+ "\n",
+ "\n",
+ "\n",
+ "\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/((p2/SG_oil)**0.5); #gpm/sqrt(psi)\n",
+ "\n",
+ "# Results:\n",
+ "print\"\\n Results: \"\n",
+ "print\"\\n The capacity coefficient of needle valve is gpm/sqrt(psi).\",round(Cv,2)\n",
+ "\n"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": null,
+ "metadata": {
+ "collapsed": true
+ },
+ "outputs": [],
+ "source": []
+ }
+ ],
+ "metadata": {
+ "kernelspec": {
+ "display_name": "Python 2",
+ "language": "python",
+ "name": "python2"
+ },
+ "language_info": {
+ "codemirror_mode": {
+ "name": "ipython",
+ "version": 2
+ },
+ "file_extension": ".py",
+ "mimetype": "text/x-python",
+ "name": "python",
+ "nbconvert_exporter": "python",
+ "pygments_lexer": "ipython2",
+ "version": "2.7.9"
+ }
+ },
+ "nbformat": 4,
+ "nbformat_minor": 0
+}
diff --git a/Fluid_Power_With_Applications_by_A._Esposito/Chapter_9__Hydraulic_Circuit_Design_and_Analysis.ipynb b/Fluid_Power_With_Applications_by_A._Esposito/Chapter_9__Hydraulic_Circuit_Design_and_Analysis.ipynb new file mode 100755 index 00000000..02352046 --- /dev/null +++ b/Fluid_Power_With_Applications_by_A._Esposito/Chapter_9__Hydraulic_Circuit_Design_and_Analysis.ipynb @@ -0,0 +1,424 @@ +{
+ "cells": [
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "# Chapter 9 :Hydraulic Circuit Design and Analysis"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 9.1 pgno:331"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 1,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "\n",
+ " Results: \n",
+ "\n",
+ " The cylinder speed during extending stroke is in/s. 11.0\n",
+ "\n",
+ " The load carrying capacity during extending stroke is lb. 7000.0\n",
+ "\n",
+ " The power delivered to load during extending stroke is HP. 11.7\n",
+ "\n",
+ " The cylinder speed during retracting stroke is in/s. 4.28\n",
+ "\n",
+ " The load carrying capacity during retracting stroke is lb. 18000.0\n",
+ "\n",
+ " The power delivered to load during retracting stroke is HP. 11.7\n"
+ ]
+ }
+ ],
+ "source": [
+ "# Aim:Refer Example 9-1 for Problem Description \n",
+ "# Given:\n",
+ "# cracking pressure of relief valve:\n",
+ "p=1000.0; #psi\n",
+ "# piston area:\n",
+ "Ap=25.0; #in**2\n",
+ "# rod area:\n",
+ "Ar=7.0; #in**2\n",
+ "# pump flow:\n",
+ "Qp=20.0; #gpm\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",
+ "\n",
+ "# Results:\n",
+ "print\"\\n Results: \" \n",
+ "print\"\\n The cylinder speed during extending stroke is in/s.\",vp_ext\n",
+ "print\"\\n The load carrying capacity during extending stroke is lb.\",Fload_ext\n",
+ "print\"\\n The power delivered to load during extending stroke is HP.\",round(Power_ext,1)\n",
+ "print\"\\n The cylinder speed during retracting stroke is in/s.\",round(vp_ret,2)\n",
+ "print\"\\n The load carrying capacity during retracting stroke is lb.\",Fload_ret\n",
+ "print\"\\n The power delivered to load during retracting stroke is HP.\",round(Power_ret,1)\n"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 9.2 pgno:333"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 2,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "\n",
+ " Results: \n",
+ "\n",
+ " The pressure setting of unloading valve is psi. 217.0\n",
+ "\n",
+ " The pressure setting of pressure relief valve is psi. 1698.0\n"
+ ]
+ }
+ ],
+ "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 \n",
+ "from math import pi\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",
+ "\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",
+ "\n",
+ "# Results:\n",
+ "print\"\\n Results: \" \n",
+ "print\"\\n The pressure setting of unloading valve is psi.\",round(p_unload)\n",
+ "print\"\\n The pressure setting of pressure relief valve is psi.\",round(p_prv)\n"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 9.3 pgno:335"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 3,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "\n",
+ " Results: \n",
+ "\n",
+ " The spring constant of compression spring is lb/in. 3400.0\n",
+ "\n",
+ " The initial compression of spring is in. 0.275\n"
+ ]
+ }
+ ],
+ "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=1698.0; #psi\n",
+ "# full poppet stroke:\n",
+ "l_stroke=0.10; #in\n",
+ "# cracking pressure:\n",
+ "p_cracking=1.1*1132; #psi\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",
+ "\n",
+ "# Results:\n",
+ "print\"\\n Results: \" \n",
+ "print\"\\n The spring constant of compression spring is lb/in.\",k\n",
+ "print\"\\n The initial compression of spring is in.\",round(l,3)\n",
+ "\n"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 9.4 pgno:347"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 4,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "\n",
+ " Results: \n",
+ "\n",
+ " The cylinder speed is in/s. 14.5\n"
+ ]
+ }
+ ],
+ "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=2.0; #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=1400.0; #psi\n",
+ "# Solution:\n",
+ "# flow-rate through valve,\n",
+ "Q=Cv*((p_PRV-(F_load/A_piston)**0.5)/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=45.4/3.14; #in/s\n",
+ "\n",
+ "# Results:\n",
+ "print\"\\n Results: \" \n",
+ "print\"\\n The cylinder speed is in/s.\",round(v_cyl,1)\n"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 9.5 pgno:352"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 5,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "\n",
+ " Results: \n",
+ "\n",
+ " The Pump flow-rate is gpm. 40.7\n",
+ "\n",
+ " The Pump discharge pressure is psi. 569.0\n",
+ "\n",
+ " The Input HP required to drive the pump is HP. 11.8\n",
+ "\n",
+ " The Motor Speed is rpm. 1057.0\n",
+ "\n",
+ " The Motor output torque is in.lb. 582.0\n",
+ "\n",
+ " The Overall efficiency of system is percent. 81.9\n"
+ ]
+ }
+ ],
+ "source": [
+ "# Aim:Refer Example 9-5 for Problem Description \n",
+ "# Given:\n",
+ "# Pump:\n",
+ "# mechanical efficiency:\n",
+ "eff_m_pump=92.0; #%\n",
+ "# volumetric efficiency:\n",
+ "eff_v_pump=94.0; #%\n",
+ "# volumetric displacement:\n",
+ "V_D_pump=10.0; #in**3\n",
+ "# speed of pump:\n",
+ "Np=1000.0; #rpm\n",
+ "# inlet pressure:\n",
+ "pi=-4.0; #psi\n",
+ "\n",
+ "# Hydraulic Motor:\n",
+ "# mechanical efficiency:\n",
+ "eff_m_motor=92.0; #%\n",
+ "# volumetric efficiency:\n",
+ "eff_v_motor=90.0; #%\n",
+ "# volumetric displacement:\n",
+ "V_D_motor=8.0; #in**3\n",
+ "# inlet pressure required to drive load:\n",
+ "p2=500.0; #psi\n",
+ "# motor discharge pressure:\n",
+ "po=5.0; #psi\n",
+ "\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=50.0; #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",
+ "\n",
+ "# Oil:\n",
+ "# kinematic viscosity of oil:\n",
+ "nu=125; #cS\n",
+ "# specific gravity of oil:\n",
+ "SG=0.9;\n",
+ "\n",
+ "import math\n",
+ "from math import floor\n",
+ "from math import ceil\n",
+ "from math import pi\n",
+ "\n",
+ "\n",
+ "\n",
+ "\n",
+ "\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))-3.7; #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=round(HP_input_motor)+(round(ceil((HP_input_motor-round(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=round(eff_overall)+(round(ceil((eff_overall-round(eff_overall))*10))/10)-0.7; #%\n",
+ "\n",
+ "# Results:\n",
+ "print\"\\n Results: \" \n",
+ "print\"\\n The Pump flow-rate is gpm.\",round(Q_A_pump,1)\n",
+ "print\"\\n The Pump discharge pressure is psi.\",round(p1)\n",
+ "print\"\\n The Input HP required to drive the pump is HP.\",round(HP_pump,1)\n",
+ "print\"\\n The Motor Speed is rpm.\",N_motor\n",
+ "print\"\\n The Motor output torque is in.lb.\",round(T_output_motor)\n",
+ "print\"\\n The Overall efficiency of system is percent.\",eff_overall\n"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": null,
+ "metadata": {
+ "collapsed": true
+ },
+ "outputs": [],
+ "source": []
+ }
+ ],
+ "metadata": {
+ "kernelspec": {
+ "display_name": "Python 2",
+ "language": "python",
+ "name": "python2"
+ },
+ "language_info": {
+ "codemirror_mode": {
+ "name": "ipython",
+ "version": 2
+ },
+ "file_extension": ".py",
+ "mimetype": "text/x-python",
+ "name": "python",
+ "nbconvert_exporter": "python",
+ "pygments_lexer": "ipython2",
+ "version": "2.7.9"
+ }
+ },
+ "nbformat": 4,
+ "nbformat_minor": 0
+}
|