{ "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 }