diff options
| author | Trupti Kini | 2016-10-06 23:30:52 +0600 |
|---|---|---|
| committer | Trupti Kini | 2016-10-06 23:30:52 +0600 |
| commit | 63852367f8b4b3466c4a2683a80abf6f61b2ef35 (patch) | |
| tree | c00b4593c8d5bcd380a1897eb739328d7af9a051 | |
| parent | 09c20af10c8cbd194ea73d659225444a78bd9d1a (diff) | |
| download | Python-Textbook-Companions-63852367f8b4b3466c4a2683a80abf6f61b2ef35.tar.gz Python-Textbook-Companions-63852367f8b4b3466c4a2683a80abf6f61b2ef35.tar.bz2 Python-Textbook-Companions-63852367f8b4b3466c4a2683a80abf6f61b2ef35.zip | |
Added(A)/Deleted(D) following books
A Basic_mechanical_engineering_by_Basant_Agrawal_,_C.M_Agrawal/Chapter_10_Properties_Of_Steam.ipynb
A Basic_mechanical_engineering_by_Basant_Agrawal_,_C.M_Agrawal/Chapter_11_Steam_Boilers.ipynb
A Basic_mechanical_engineering_by_Basant_Agrawal_,_C.M_Agrawal/Chapter_13_Steam_Engines.ipynb
A Basic_mechanical_engineering_by_Basant_Agrawal_,_C.M_Agrawal/Chapter_14_Air_Standard_Cycles.ipynb
A Basic_mechanical_engineering_by_Basant_Agrawal_,_C.M_Agrawal/Chapter_2_Properties_Of_Material.ipynb
A Basic_mechanical_engineering_by_Basant_Agrawal_,_C.M_Agrawal/Chapter_5_Metrology.ipynb
A Basic_mechanical_engineering_by_Basant_Agrawal_,_C.M_Agrawal/Chapter_7_Fluid_Mechanics.ipynb
A Basic_mechanical_engineering_by_Basant_Agrawal_,_C.M_Agrawal/Chapter_9__Laws_Of_Thermodynamics.ipynb
A Basic_mechanical_engineering_by_Basant_Agrawal_,_C.M_Agrawal/screenshots/2.png
A Basic_mechanical_engineering_by_Basant_Agrawal_,_C.M_Agrawal/screenshots/5.png
A Basic_mechanical_engineering_by_Basant_Agrawal_,_C.M_Agrawal/screenshots/7.png
D Manufacturing_Engineering_&_Technology_by_S_Kalpakjian_and_S_R_Schmid/CHAPTER10.ipynb
D Manufacturing_Engineering_&_Technology_by_S_Kalpakjian_and_S_R_Schmid/CHAPTER13.ipynb
D Manufacturing_Engineering_&_Technology_by_S_Kalpakjian_and_S_R_Schmid/CHAPTER14.ipynb
D Manufacturing_Engineering_&_Technology_by_S_Kalpakjian_and_S_R_Schmid/CHAPTER15.ipynb
D Manufacturing_Engineering_&_Technology_by_S_Kalpakjian_and_S_R_Schmid/CHAPTER16.ipynb
D Manufacturing_Engineering_&_Technology_by_S_Kalpakjian_and_S_R_Schmid/CHAPTER17.ipynb
D Manufacturing_Engineering_&_Technology_by_S_Kalpakjian_and_S_R_Schmid/CHAPTER18.ipynb
D Manufacturing_Engineering_&_Technology_by_S_Kalpakjian_and_S_R_Schmid/CHAPTER2.ipynb
D Manufacturing_Engineering_&_Technology_by_S_Kalpakjian_and_S_R_Schmid/CHAPTER20.ipynb
D Manufacturing_Engineering_&_Technology_by_S_Kalpakjian_and_S_R_Schmid/CHAPTER22.ipynb
D Manufacturing_Engineering_&_Technology_by_S_Kalpakjian_and_S_R_Schmid/CHAPTER23.ipynb
D Manufacturing_Engineering_&_Technology_by_S_Kalpakjian_and_S_R_Schmid/CHAPTER25.ipynb
D Manufacturing_Engineering_&_Technology_by_S_Kalpakjian_and_S_R_Schmid/CHAPTER28.ipynb
D Manufacturing_Engineering_&_Technology_by_S_Kalpakjian_and_S_R_Schmid/CHAPTER32.ipynb
D Manufacturing_Engineering_&_Technology_by_S_Kalpakjian_and_S_R_Schmid/CHAPTER36.ipynb
D Manufacturing_Engineering_&_Technology_by_S_Kalpakjian_and_S_R_Schmid/CHAPTER9.ipynb
D Manufacturing_Engineering_&_Technology_by_S_Kalpakjian_and_S_R_Schmid/screenshots/CHAP10.png
D Manufacturing_Engineering_&_Technology_by_S_Kalpakjian_and_S_R_Schmid/screenshots/CHAP16.png
D Manufacturing_Engineering_&_Technology_by_S_Kalpakjian_and_S_R_Schmid/screenshots/CHAP23.png
R Manufacturing_Engineering_&_Technology_by_S._Kalpakjian_and_S._R._Schmid/CHAPTER10.ipynb -> Manufacturing_Engineering_and_Technology_by_S_Kalpakjian_and_S_R_Schmid/CHAPTER10.ipynb
R Manufacturing_Engineering_&_Technology_by_S._Kalpakjian_and_S._R._Schmid/CHAPTER13.ipynb -> Manufacturing_Engineering_and_Technology_by_S_Kalpakjian_and_S_R_Schmid/CHAPTER13.ipynb
R Manufacturing_Engineering_&_Technology_by_S._Kalpakjian_and_S._R._Schmid/CHAPTER14.ipynb -> Manufacturing_Engineering_and_Technology_by_S_Kalpakjian_and_S_R_Schmid/CHAPTER14.ipynb
R Manufacturing_Engineering_&_Technology_by_S._Kalpakjian_and_S._R._Schmid/CHAPTER15.ipynb -> Manufacturing_Engineering_and_Technology_by_S_Kalpakjian_and_S_R_Schmid/CHAPTER15.ipynb
R Manufacturing_Engineering_&_Technology_by_S._Kalpakjian_and_S._R._Schmid/CHAPTER16.ipynb -> Manufacturing_Engineering_and_Technology_by_S_Kalpakjian_and_S_R_Schmid/CHAPTER16.ipynb
R Manufacturing_Engineering_&_Technology_by_S._Kalpakjian_and_S._R._Schmid/CHAPTER17.ipynb -> Manufacturing_Engineering_and_Technology_by_S_Kalpakjian_and_S_R_Schmid/CHAPTER17.ipynb
R Manufacturing_Engineering_&_Technology_by_S._Kalpakjian_and_S._R._Schmid/CHAPTER18.ipynb -> Manufacturing_Engineering_and_Technology_by_S_Kalpakjian_and_S_R_Schmid/CHAPTER18.ipynb
R Manufacturing_Engineering_&_Technology_by_S._Kalpakjian_and_S._R._Schmid/CHAPTER2.ipynb -> Manufacturing_Engineering_and_Technology_by_S_Kalpakjian_and_S_R_Schmid/CHAPTER2.ipynb
R Manufacturing_Engineering_&_Technology_by_S._Kalpakjian_and_S._R._Schmid/CHAPTER20.ipynb -> Manufacturing_Engineering_and_Technology_by_S_Kalpakjian_and_S_R_Schmid/CHAPTER20.ipynb
R Manufacturing_Engineering_&_Technology_by_S._Kalpakjian_and_S._R._Schmid/CHAPTER22.ipynb -> Manufacturing_Engineering_and_Technology_by_S_Kalpakjian_and_S_R_Schmid/CHAPTER22.ipynb
R Manufacturing_Engineering_&_Technology_by_S._Kalpakjian_and_S._R._Schmid/CHAPTER23.ipynb -> Manufacturing_Engineering_and_Technology_by_S_Kalpakjian_and_S_R_Schmid/CHAPTER23.ipynb
R Manufacturing_Engineering_&_Technology_by_S._Kalpakjian_and_S._R._Schmid/CHAPTER25.ipynb -> Manufacturing_Engineering_and_Technology_by_S_Kalpakjian_and_S_R_Schmid/CHAPTER25.ipynb
R Manufacturing_Engineering_&_Technology_by_S._Kalpakjian_and_S._R._Schmid/CHAPTER28.ipynb -> Manufacturing_Engineering_and_Technology_by_S_Kalpakjian_and_S_R_Schmid/CHAPTER28.ipynb
R Manufacturing_Engineering_&_Technology_by_S._Kalpakjian_and_S._R._Schmid/CHAPTER32.ipynb -> Manufacturing_Engineering_and_Technology_by_S_Kalpakjian_and_S_R_Schmid/CHAPTER32.ipynb
R Manufacturing_Engineering_&_Technology_by_S._Kalpakjian_and_S._R._Schmid/CHAPTER36.ipynb -> Manufacturing_Engineering_and_Technology_by_S_Kalpakjian_and_S_R_Schmid/CHAPTER36.ipynb
R Manufacturing_Engineering_&_Technology_by_S._Kalpakjian_and_S._R._Schmid/CHAPTER9.ipynb -> Manufacturing_Engineering_and_Technology_by_S_Kalpakjian_and_S_R_Schmid/CHAPTER9.ipynb
R Manufacturing_Engineering_&_Technology_by_S._Kalpakjian_and_S._R._Schmid/screenshots/CHAP10.png -> Manufacturing_Engineering_and_Technology_by_S_Kalpakjian_and_S_R_Schmid/screenshots/CHAP10.png
R Manufacturing_Engineering_&_Technology_by_S._Kalpakjian_and_S._R._Schmid/screenshots/CHAP16.png -> Manufacturing_Engineering_and_Technology_by_S_Kalpakjian_and_S_R_Schmid/screenshots/CHAP16.png
R Manufacturing_Engineering_&_Technology_by_S._Kalpakjian_and_S._R._Schmid/screenshots/CHAP23.png -> Manufacturing_Engineering_and_Technology_by_S_Kalpakjian_and_S_R_Schmid/screenshots/CHAP23.png
49 files changed, 2676 insertions, 1697 deletions
diff --git a/Basic_mechanical_engineering_by_Basant_Agrawal_,_C.M_Agrawal/Chapter_10_Properties_Of_Steam.ipynb b/Basic_mechanical_engineering_by_Basant_Agrawal_,_C.M_Agrawal/Chapter_10_Properties_Of_Steam.ipynb new file mode 100644 index 00000000..627fc0f4 --- /dev/null +++ b/Basic_mechanical_engineering_by_Basant_Agrawal_,_C.M_Agrawal/Chapter_10_Properties_Of_Steam.ipynb @@ -0,0 +1,629 @@ +{ + "metadata": { + "name": "Chapter 10 Properties Of Steam" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": "\nChapter 10 Properties Of Steam" + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Example 1 Page No:183" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "#Input data\nmw=15 #Water steam\nms=185 #Dry steam\n\n#Calculation\nx=((ms)/(ms+mw))*100 #Dryness fuction of steam in %\n\n#Output\nprint(\"Dryness fuction of steam=\",x,\"%\")\n", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "Dryness fuction of steam= 92.5 %\n" + } + ], + "prompt_number": 66 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Example 2 Page No:183" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "#Input data\nsps=150 #saturation pressure of the steam in degree celsius\n\n#Output\nP=4.76 #From steam table\nprint(\"saturation pressure=\",P,\"bar\")\n\n\n\n", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "saturation pressure= 4.76 bar\n" + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Example 3 Page No:184" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "#Input data\nP1=28 #Absolute pressure in bar\nP2=5.5 #Absolute pressure in MPa\nP3=77 #Absolute pressure in mm of Hg\n\n#Calcutation\nts1=230.05 #Saturation temperature in degree celsius\nts2=269.93 #Saturation temperature in degree celsius\nts3=45.83 #Saturation temperature in degree celsius\n\n#Output\nprint(\"Saturation temperature= \",ts1,\"degree celsius\")\nprint(\"Saturation temperature= \",ts2,\"degree celsius\")\nprint(\"Saturation temperature= \",ts3,\"degree celsius\")", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "Saturation temperature= 230.05 degree celsius\nSaturation temperature= 269.93 degree celsius\nSaturation temperature= 45.83 degree celsius\n" + } + ], + "prompt_number": 13 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Example 4 Page No:185" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "#Input data\nP=15 #Absolute pressure in bar\n#From steam table (pressure basis at 15 bar)\nts=198.3 #In degree celsius \nhf=844.7 #In KJ/Kg\nhfg=1945.2 #In KJ/Kg\nhg=2789.9 #In KJ/Kg\ntsup=300 #In degree celsius \nx=0.8\nCps=2.3\nhg=2789.9\n\n#Calculation\nh1=hf+x*hfg #Enthalpy of wet steam in KJ/KG\nh=hg #Enthalpy of dry and saturated steam in KJ/KG\nh2=hg+Cps*(tsup-ts)#Enthalpy of superheated steam in KJ/KG\n\n\n#Output\nprint(\"Enthalpy of wet steam= \",h1,\"KJ/Kg\")\nprint(\"Enthalpy of dry and saturated steam= \",h,\"KJ/KG\")\nprint(\"Enthalpy of superheated steam= \",h2,\"KJ/Kg\")", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "Enthalpy of wet steam= 2400.86 KJ/Kg\nEnthalpy of dry and saturated steam= 2789.9 KJ/KG\nEnthalpy of superheated steam= 3023.81 KJ/Kg\n" + } + ], + "prompt_number": 12 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Example 5 Page No:186" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "#Input data\nti=30 #Temperature in degree celsius\nm=2 #Water in Kg\npf=8 #Steam at 8 bar\nx=0.9 #Water to dry \ntb=30\n#From steam table at 30 degree celsius\nhf=125.7\n#From steam table at 8 bar\nts=170.4 #In degree celsius \nhf1=720.9 #In KJ/KG\nhfg=2046.6 #In KJ/KG\nhg=2767.5 #In KJ/KG\n\n#Calculation\nh=hf1+(x*hfg) #Final Enthalpy of the steam in KJ/Kg\nQha=m*(h-hf) #Quantity of the heat in KJ/Kg\n\n#Output\nprint(\"Final Enthalpy of the steam= \",h,\"KJ/Kg\")\nprint(\"Quantity of the heat= \",Qha,\"KJ/Kg\")\n\n\n", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "Final Enthalpy of the steam= 2562.84 KJ/Kg\nQuantity of the heat= 4874.280000000001 KJ/Kg\n" + } + ], + "prompt_number": 14 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Example 6 Page No:186" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "#Input data\nIT=25 #Initial temperature\nm=5 #Heat required to generate steam in kg\npf=10 #Final pressure in bar\ntsup=250 #Water temperature\n#From steam table (temp basis)at 25degree celsius \n#and at 10 bar(pressure basis)\nhf=104.8 #In KJ/KG\nh1=104.8 #In KJ/KG\nts=179.9 #In degree celsius \nhf1=792.6 #In KJ/KG\nhfg=2013.6 #In KJ/KG\nhg=2776.2 #In KJ/KG\nCps=2.1\n\n#Calculation\nh=hg+Cps*(tsup-ts) #Enthalpy of superheated steam in KJ/Kg\nH=m*(h-h1) #Quantity of heat added in KJ/Kg\n\n#Output\nprint(\"Enthalpy of superheated steam= \",h,\"KJ/Kg\")\nprint(\"Quantity of heat added= \",round(H,),\"KJ/Kg\")\n\n\n", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "Enthalpy of superheated steam= 2923.41 KJ/Kg\nQuantity of heat added= 14093 KJ/Kg\n" + } + ], + "prompt_number": 15 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Example 7 Page No:188" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "#Input data\nP=15 #Absolute pressure in bar\n#From steam table (pressure basis at 15 bar)\nts=198.3+273 #In degree celsius\nvg=0.1317 #In m**3/Kg \nvf=0.001154 #In m**3/Kg \nx=0.8 \nTsup=300+273 #Degree celsius\n\n\n#Calculation\nv=(1-x)*vf+x*vg #Volume of wet steam in m**3/Kg\nvg=0.1317 #Dry and saturated steam in m**3/Kg\nvsup=vg*(Tsup/ts) #Volume of superheated steam m**3/Kg \n\n\n#Output\nprint(\"Volume of wet steam= \",round(v,4),\"m**3/Kg\")\nprint(\"Dry and Saturated Steam= \",vg,\"m**3/Kg\") \nprint(\"volume of superheated steam= \",round(vsup,4),\"m**3/Kg\")\n \n\n\n\n", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "Volume of wet steam= 0.1056 m**3/Kg\nDry and Saturated Steam= 0.1317 m**3/Kg\nvolume of superheated steam= 0.1601 m**3/Kg\n" + } + ], + "prompt_number": 17 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Example 8 Page No:188" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "#Input data\nP=25 #Absolute pressure\nts=223.9 #Volume\n#Frome steam table (pressure basis at 25 bar) \nvf=0.001197 #In m**3/Kg \nvg=0.0799 #In m**3/Kg \nv=8 #In m**3/Kg \n\n\n#Calculation\nm=v/vg #Mass of steam in Kg \n\n#Output\nprint(\"Mass of steam= \",round(m,3),\"Kg\")", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "Mass of steam= 100.125 Kg\n" + } + ], + "prompt_number": 21 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Example 9 Page No:190" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "#Input data\nP=12*10**5 #Absolute pressure\n#From steam table (pressure basis at 12 bar)\nts=188+273 #In degree celsius\nvf=0.001139 #In m**3/Kg \nvg=0.1632 #In m**3/Kg \nhf=798.4 #In KJ/Kg\nhfg=1984.3 #In KJ/Kg\nhg=2782.7 #In KJ/Kg\nx=0.94\nCps=2.3\ntsup=350+273 #In degree celsius\n\n#Calcuation\nh=hf+x*hfg #Enthalpy of wet steam in KJ/Kg\nv=(1-x)*vf+x*vg #Volume of wet steam m**3/Kg\nu=h-((P*v)/10**3) #Internal Energy in KJ/Kg\nhg=2782.7 #Enthalpy of dry & saturated steam in KJ/Kg\nv1=vg #Volume of dry & saturated steam m**3/Kg\nu1=hg-((P*vg)/10**3) #Internal Energy in KJ/Kg \nh1=hg+Cps*(tsup-ts) #Enthalpy of superheated steam in KJ/Kg\nvsup=vg*(tsup/ts) #Volume of superheated steam in m**3/Kg\nu2=h1-((P*v)/10**3) #Internal Energy in KJ/Kg\n\n\n#Output\nprint(\"Enthalpy of wet steam= \",h,\"KJ/Kg\")\nprint(\"Volume of wet steam= \",round(v,5),\"m**3/Kg\")\nprint(\"Internal Energy= \",round(u,2),\"KJ/Kg\")\nprint(\"Enthalpy of dry & saturated steam= \",hg,\"KJ/Kg\")\nprint(\"Volume of dry & saturated steam= \",v1,\"m**3/Kg\")\nprint(\"Internal Energy= \",u1,\"KJ/Kg\")\nprint(\"Enthalpy of superheated steam= \",round(h1,1),\"KJ/Kg\")\nprint(\"Volume of superheated steam= \",round(vsup,3),\"m**3/Kg\")\nprint(\"Internal Energy= \",round(u2,1),\"KJ/Kg\")\n\n", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "Enthalpy of wet steam= 2663.642 KJ/Kg\nVolume of wet steam= 0.15348 m**3/Kg\nInternal Energy= 2479.47 KJ/Kg\nEnthalpy of dry & saturated steam= 2782.7 KJ/Kg\nVolume of dry & saturated steam= 0.1632 m**3/Kg\nInternal Energy= 2586.8599999999997 KJ/Kg\nEnthalpy of superheated steam= 3155.3 KJ/Kg\nVolume of superheated steam= 0.221 m**3/Kg\nInternal Energy= 2971.1 KJ/Kg\n" + } + ], + "prompt_number": 12 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Example 10 Page No:191" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "#Input data\nP1=10*10**5 #Pressure of steam in bar\ntsup1=300+273 #Temperature of steam n degree celsius \nP2=1.4*10**5 #Internal energy of steam\nx2=0.8 #Dryness fraction\nCps=2.3\n#from steam table properties of saturated steam (temp basis) \n#at 25 degree celsius and at 10 bar(pressure basis)\nts1=179.9+273\nvf=0.001127 #In m**3/Kg \nvg=0.1943 #In m**3/Kg \nhf=762.6 #In KJ/Kg\nhfg=2013.6 #In KJ/Kg\nhg1=2776.2 #In KJ/Kg\n#at 1.4 bar\nts=109.3 #In degree celsius\nvf1=0.001051 #In m**3/Kg \nvg1=1.2363 #In m**3/Kg \nhf1=458.4 #In KJ/Kg\nhfg1=2231.9 #In KJ/Kg\nhg=2690.3 #In KJ/Kg\n\n#calculation\nh1=hg1+Cps*(tsup1-ts1) #Enthalpy of superheated steam in KJ/Kg\nv1=vg*(tsup1/ts1) #Volume of superheated steam in m**3/Kg\nu1=h1-((P1*v1)/10**3) #Internal energy in KJ/Kg\nh2=hf1+x2*hfg1 #Enthalpy of wet steam in KJ/Kg\nVwet=(1-x2)*vf1+x2*vg1 #Volume of wet steam in m**3/Kg\nu2=h2-((P2*Vwet)/10**3) #Internal energy in KJ/Kg\nDeltaU=u1-u2 #Change of Internal energy in KJ/Kg\n\n\n#Output\nprint(\"Enthalpy of superheated steam= \",h1,\"KJ/Kg\")\nprint(\"Volume of superheated steam= \",round(v1,4),\"m**3/Kg\")\nprint(\"Internal energy= \",round(u1,1),\"KJ/Kg\")\nprint(\"Enthalpy of wet steam= \",h2,\"KJ/Kg\")\nprint(\"Volume of wet steam= \",round(Vwet,5),\"m**3/Kg\")\nprint(\"Internal energy= \",round(u2,1),\"KJ/Kg\")\nprint(\"Change of Internal energy= \",round(DeltaU,1),\"KJ/Kg\")\n", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "Enthalpy of superheated steam= 3052.43 KJ/Kg\nVolume of superheated steam= 0.2458 m**3/Kg\nInternal energy= 2806.6 KJ/Kg\nEnthalpy of wet steam= 2243.92 KJ/Kg\nVolume of wet steam= 0.98925 m**3/Kg\nInternal energy= 2105.4 KJ/Kg\nChange of Internal energy= 701.2 KJ/Kg\n" + } + ], + "prompt_number": 23 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Example 11 Page No:193" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "#Input data\nimport math\nP=15 #Absolute pressure\n#From steam table (pressure basis at 15 bar)\nts=198.3+273 #In degree celsius \nSf=2.3145 #In KJ/KgK\nSfg=4.1261 #In KJ/KgK\nSg=6.4406 #In KJ/KgK\ntsup=300+273\nCps=2.3\nx=0.8\n\n#calculation\nS=Sf+x*Sfg #Entropy of wet steam in KJ/Kg\nS1=Sg #Entropy of superheated steam in KJ/Kg\nS2=Sg+Cps*(math.log(tsup/ts)) #Entropy of superheated steam in KJ/Kg\n\n#Output\nprint(\"Entropy of wet steam\",round(S,3),\" KJ/Kg\")\nprint(\"Entropy of dry and saturated steam\",S1,\" KJ/Kg\")\nprint(\"Entropy of superheated steam\",round(S2,2),\" KJ/Kg\")\n\n", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "Entropy of wet steam 5.615 KJ/Kg\nEntropy of dry and saturated steam 6.4406 KJ/Kg\nEntropy of superheated steam 6.89 KJ/Kg\n" + } + ], + "prompt_number": 26 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Example 12 Page No:194" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "#Input data\n#Input data\nimport math\nm=1.5 #Entropy of the steam\nP=10*10**5 #Absolute pressure in bar\n#From steam table properties of saturated steam \n#(pressure basis)at 10 bar\nts=179.9+273 #Indegree celsius\nvf=0.001127 #In m**3/Kg\nvg=0.1943 #In m**3/Kg\nhf=762.6 #In KJ/Kg\nhfg=2013.6 #In KJ/Kg\nhg=2776.2 #In KJ/Kg\nSf=2.1382 #In KJ/KgK\nSfg=4.4446 #In KJ/KgK\nSg=6.5828 #In KJ/Kg\nCps=2.3\ntsup=250+273\n\n\n#Calculation\n#(1)Enthalpy of dry and saturated steam \n\nh=hg #Enthalpy of dry and saturated steam \nEODS=hg*m #Enthalpy of 1.5Kg of dry and saturated steam \nv=vg #volume of dry and saturated steam\nu=h-((P*v)/10**3) #Internal Energy\nIES=u*m #Internal energy of the steam\ns=6.5858 #Entropy of dry and saturated steam\nEODSS=s*m #Entropy of 1.5Kg dry and saturated steam\nx=0.75\n#(2)Enthalpy of wet steam\nh1=hf+x*hfg #Enthalpy of wet steam\nEWS=h1*m #Enthalpy of1.5Kg of wet steam\nVwet=x*vg #Volume of steam\nu1=h1-((P*Vwet)/10**3) #Internal energy \nIES1=u1*m #Internal energy of1.5Kg of the steam\ns1=Sf+x*Sfg #Entropy of wet steam\nEWS1=s1*m #Entropy of1.5Kg of wet steam\n\n#(3)Enthalpy of superheated steam\nh2=hg+Cps*(tsup-ts) #Enthalpy of superheated steam\nEOSHS=h2*m #Enthalpy of 1.5Kg of superheated steam\nVsup=vg*(tsup/ts) #Volume of superheated steam\nu2=h2-((P*Vsup)/10**3) #Internal energy\nIES2=u2*m #Internal energy of 1.5Kg of the steam\ns2=Sg+Cps*(math.log(tsup/ts))#Entropy of superheated steam\nEOSHS1=s2*m #Entropy of 1.5Kg of superheated steam\n\n#Output\nprint(\"Enthalpy of dry and saturated steam= \",h,\"KJ/Kg\")\nprint(\"Enthalpy of 1.5Kg of dry and saturated steam= \",round(EODS,2),\"KJ\")\nprint(\"volume of dry and saturated steam= \",v,\"m**3/kg\")\nprint(\"Internal Energy= \",round(u,2),\"KJ/Kg\")\nprint(\"Internal energy of the steam= \",round(IES,2),\"kJ\")\nprint(\"Entropy of dry and saturated steam = \",s,\"KJ/KgK\")\nprint(\"Entropy of 1.5kg of dry and saturated steam= \",EODSS,\"KJ/K\")\n\nprint(\"Enthalpy of wet steam= \",round(h1,2),\"KJ/Kg\")\nprint(\"Enthalpy of1.5Kg of wet steam= \",EWS,\"KJ\")\nprint(\"Volume of steam= \",Vwet,\"m**3/Kg\")\nprint(\"Internal energy= \",u1,\"KJ/Kg\")\nprint(\"Internal energy of1.5Kg of the steam= \",round(IES1,2),\"KJ\")\nprint(\"Entropy of wet steam= \",round(s1,2),\"KJ/KgK\")\nprint(\"Entropy of 1.5Kg of wet steam= \",EWS1,\"KJ/K\")\n\nprint(\"Enthalpy of superheated steam= \",h2,\"KJ/Kg\")\nprint(\"Enthalpy of 1.5Kg of superheated steam= \",round(EOSHS,1),\"KJ\")\nprint(\"Volume of superheated steam= \",round(Vsup,4),\"m**3/Kg\")\nprint(\"Internal energy= \",round(u2,4),\"\")\nprint(\"Internal energy of1.5Kg of the steam= \",round(IES2,1),\"KJ\")\nprint(\"Entropy of superheated steam= \",round(s2,4),\"KJ/KgK\")\nprint(\"Entropy of 1.5Kg of superheated steam= \",round(EOSHS1,2),\"KJ/K\")\n", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "Enthalpy of dry and saturated steam= 2776.2 KJ/Kg\nEnthalpy of 1.5Kg of dry and saturated steam= 4164.3 KJ\nvolume of dry and saturated steam= 0.1943 m**3/kg\nInternal Energy= 2581.9 KJ/Kg\nInternal energy of the steam= 3872.85 kJ\nEntropy of dry and saturated steam = 6.5858 KJ/KgK\nEntropy of 1.5kg of dry and saturated steam= 9.8787 KJ/K\nEnthalpy of wet steam= 2272.8 KJ/Kg\nEnthalpy of1.5Kg of wet steam= 3409.2 KJ\nVolume of steam= 0.145725 m**3/Kg\nInternal energy= 2127.075 KJ/Kg\nInternal energy of1.5Kg of the steam= 3190.61 KJ\nEntropy of wet steam= 5.47 KJ/KgK\nEntropy of 1.5Kg of wet steam= 8.207475 KJ/K\nEnthalpy of superheated steam= 2937.43 KJ/Kg\nEnthalpy of 1.5Kg of superheated steam= 4406.1 KJ\nVolume of superheated steam= 0.2244 m**3/Kg\ninternal energy= 2713.0562 \nInternal energy of1.5Kg of the steam= 4069.6 KJ\nEntropy of superheated steam= 6.9138 KJ/KgK\nEntropy of 1.5Kg of superheated steam= 10.37 KJ/K\n" + } + ], + "prompt_number": 31 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Example 13 Page No:196" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "#Input data\nV=0.04 #Volume of vessel in m**3 \nx=1\nt=250+273 #Saturated steam temp in degree celsius\nmw=9 #Mass of liquid in Kg\n#From steam table(temp basis,at t=250)\nP=39.78*10**5 #in bar\nVf=0.001251 #In m**3/kg\nVg=0.05004 #In m**3/Kg\nhf=1085.7 #KJ/Kg\nhfg=2800.4 #KJ/Kg\nhg=1714.7 #KJ/Kg\n\n#Calculation\nVw=mw*Vf #Volume occupied by water in m**3\nVs=V-Vw #Volume of waterin m**3\nms=Vs/Vg #Volume of dry and saturated steam in Kg \nm=mw+ms #Total mass of steam in Kg\nx=ms/(ms+mw) #Dryness fraction of steam \nVwet=(1-x)*Vf+x*Vg #Specific volume of steam in m**3/Kg\nh=hf+x*hfg #Enthalpy of wet steam in KJ/Kg\nEOWS=h*m #Enthalpy of 9.574 Kg of wet steam KJ\nu=h-((P*Vwet)/10**3) #Internal Energy in KJ/Kg\nIEOS=u*m #Internal energy of 9.574 Kg of steam in KJ\n\n\n#Output\nprint(\"Volume occupied by water= \",round(Vw,5),\"m**3\")\nprint(\"Volume of water= \",round(Vs,5),\"m**3\")\nprint(\"Volume of dry and saturated steam= \",round(ms,3),\"Kg \")\nprint(\"Total mass of steam= \",round(m,3),\"Kg\")\nprint(\"Dryness fraction of steam= \",round(x,2),)\nprint(\"Specific volume of steam= \",round(Vwet,6),\" m**3/Kg\")\nprint(\"Enthalpy of wet steam= \",round(h,1),\"KJ/Kg\")\nprint(\"Enthalpy of 9.574 Kg of wet steam= \",round(EOWS,),\"KJ\")\nprint(\"Internal Energy= \",round(u,1),\"KJ/Kg\")\nprint(\"Internal energy of 9.574 Kg of steam= \",round(IEOS),\"KJ\")\n\n", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "Volume occupied by water= 0.01126 m**3\nVolume of water= 0.02874 m**3\nVolume of dry and saturated steam= 0.574 Kg \nTotal mass of steam= 9.574 Kg\nDryness fraction of steam= 0.06\nSpecific volume of steam= 0.004178 m**3/Kg\nEnthalpy of wet steam= 1253.7 KJ/Kg\nEnthalpy of 9.574 Kg of wet steam= 12003 KJ\nInternal Energy= 1237.1 KJ/Kg\nInternal energy of 9.574 Kg of steam= 11844 KJ\n" + } + ], + "prompt_number": 40 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Example 14 Page No:197" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "#Input Data\nP=7 #Absolute pressure in bar\nt=200 #Absolute temperature\nts=165 #In degree celsius from steam table\n\n#Calculation\ndos=t-ts #Degree of superheat in degree celcius\n\n#Output\nprint(\"Degree of superheat= \",dos,\"degree celcius\")", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "Degree of superheat= 35 degree celcius\n" + } + ], + "prompt_number": 42 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Example 15 Page No:197" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "#Input data\nP=15 #Absolute pressure in bar\n#From steam table (pressure basis at 15 bar)\nh=1950 #In KJ/Kg\nts=198.3 #In degreee celsius\nhf=844.7 #In KJ/Kg\nhfg=1945.2 #In KJ/Kg\nhg=2789.9 #In KJ/Kg\n\n#calculation\nx=((h-hf)/hfg) #Enthalpy of wet steam\n\n#Output\nprint(\"Enthalpy of wet steam= \",round(x,3),\"\")\n", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "Enthalpy of wet steam= 0.568 \n" + } + ], + "prompt_number": 44 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Example 16 Page No:197" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "#Input data\nP=15 #Absolute pressure in bar\n#From steam table (pressure basis at 15 bar)\nh=3250 #In KJ/Kg\nts=198.3 #In degree celsius \nhf=844.7 #In KJ/Kg\nhfg=1945.2 #In KJ/Kg\nhg=2789.9 #In KJ/Kg\nCps=2.3\n\n#Calculation\ntsup=(h-hg+(Cps*ts))/2.3 #Enthalpy of superheated steam in degree celsius\ndos=tsup-ts #Degree of superheated in degree celsius\n\n#Output\nprint(\"Enthalpy of superheated steam= \",round(tsup,2),\"degree celcius\")\nprint(\"Degree of superheated= \",dos,\"degree celcius\")", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "Enthalpy of superheated steam= 398.34 degree celcius\nDegree of superheated= 200.0434782608695 degree celcius\n" + } + ], + "prompt_number": 46 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Example 17 Page No:198" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "#Input data\nP=7 #Absolute pressure in bar\nv=0.2 #Specific volume in m**3/Kg\n#from steam table (pressure basis at 7 bar) \nts=165 #In degree celsius\nvf=0.001108 #In m**3/Kg\nvg=0.2727 #In m**3/Kg\n\n#calculation\nx=v/vg #Volume of steam dryness fraction\n\n#Output\nprint(\"Volume of steam dryness fraction= \",round(x,3),)", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "Volume of steam dryness fraction= 0.733\n" + } + ], + "prompt_number": 64 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Example 18 Page No:198" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "#Input data\nP=7 #Absolute pressure in bar\nv=0.3 #Specific volume in m**3/Kg\n#From steam table (pressure basis at 7 bar)\nts=165+273 #In degree celsius\nvf=0.001108 #In m**3/Kg\nvg=0.2727 #In m**3/Kg\n\n#Calculation\n#v=vg*tsup/ts\ntsup=((v/vg)*ts)-273 #Temp of superheated steam in degree celsius\nDOS=tsup+273-ts #Degree of superheated in degree celsius\n\n#Output\nprint(\"Temp of superheated steam= \",round(tsup,2),\"degree celsius\")\nprint(\"Degree of superheated= \",round(DOS,2),\"degree celsius\")\n", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "Temp of superheated steam= 208.85 degree celsius\nDegree of superheated= 43.85 degree celsius\n" + } + ], + "prompt_number": 57 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Example 19 Page No:198" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "#Input data\nm=2 #steam of vessel in Kg\nV=0.1598 #volume of vessel in M**3\nP=25 #Absolute pressure of vessel in bar\n\n#Calculation\nv=V/m #Quality of steam in m**3/Kg\n\n#Output\nprint(\"Quality of steam\",v,\" m**3/Kg\")", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "Quality of steam 0.0799 m**3/Kg\n" + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Example 20 Page No:200" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "#Input data\nP=10*10**2 #Absolute pressure in bar\nx1=0.9 #Dryness enters\ntsup2=300+273 #Temperature in degree celsius \n#From steam table at 10 bar\nts=179.9+273 #In degree celsius\nVg=0.1943 #In m**3/Kg\nhf=762.6 #In KJ/Kg\nhfg=2013.6 #InK/Kg\nhg=2776.2 #In KJ/Kg\n\n#Calculation\nh1=hf+x1*hfg #Initial enthalpy of steam in KJ/Kg\nV1=x1*Vg #Initial specific volume of steam\nu1=h1-P*V1 #Initial internal energy of steam in KJ/Kg\nh2=hg+Cps*(tsup2-ts) #Final enthalpy of steam in KJ/Kg\nV2=Vg*(tsup2/ts) #Final specific volume of steam in m**3/Kg\nu2=h2-P*V2 #Final internal energy of steam in KJ/K\ndeltah=h2-h1 #Heat gained by steam in KJ/Kg\ndeltaU=(u2-u1) #Change in internal energy in KJ/Kg\n\n#Output\nprint(\"Initial enthalpy of steam= \",h1,\"KJ/Kg\")\nprint(\"Initial specific volume of steam= \",V1,)\nprint(\"Initial internal energy of steam= \",round(u1,2),\"KJ/Kg\")\nprint(\"Final enthalpy of steam= \",h2,\"KJ/Kg\")\nprint(\"Final specific volume of steam= \",round(V2,4),\"m**3/Kg\")\nprint(\"Final internal energy of steam= \",round(u2,3),\"KJ/Kg\")\nprint(\"Heat gained by steam= \",round(deltah,2),\"KJ/Kg\")\nprint(\"Change in internal energy= \",round(deltaU,2),\"KJ/Kg\")\n\n\n", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "Initial enthalpy of steam= 2574.84 KJ/Kg\nInitial specific volume of steam= 0.17487\nInitial internal energy of steam= 2399.97 KJ/Kg\nFinal enthalpy of steam= 3052.43 KJ/Kg\nFinal specific volume of steam= 0.2458 m**3/Kg\nFinal internal energy of steam= 2806.606 KJ/Kg\nHeat gained by steam= 477.59 KJ/Kg\nChange in internal energy= 406.64 KJ/Kg\n" + } + ], + "prompt_number": 58 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Example 21 Page No:201" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "#Input data\nm=4 #Steam in Kg\nP=13 #Absolute pressure in bar\ntsup1=450 #Absolute temp in degree celsius \ndeltaH=2.8*10**3 #loses in MJ\n#from steam table at 13 bar\nts=191.6 #In degree celsius\nVg=0.1511 #In m**3/Kg\nhf=814.7 #In m**3/Kg\nhfg=1970.7 #In KJ/Kg\nhg=2785.4 #In KJ/Kg\n\n#Calculation\nh1=hg+Cps*(tsup1-ts) #Initial enthalpy of steam in KJ/Kg\nDeltah=deltaH/m #Change in enthalpy/unit mass in KJ/Kg\nh2=h1-Deltah #Final enthalpy of steam in KJ/Kg\nx2=(h2-hf)/hfg #wet & dryness fraction\n\n#Output\nprint(\"Initial enthalpy of steam= \",round(h1,2),\" KJ/Kg\")\nprint(\"Change in enthalpy/unit mass= \",Deltah,\"KJ/Kg\")\nprint(\"Final enthalpy of steam= \",round(h2,2),\"KJ/Kg\")\nprint(\"wet & dryness fraction= \",round(x2,4),)", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "Initial enthalpy of steam= 3379.72 KJ/Kg\nChange in enthalpy/unit mass= 700.0 KJ/Kg\nFinal enthalpy of steam= 2679.72 KJ/Kg\nwet & dryness fraction= 0.9464\n" + } + ], + "prompt_number": 63 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Example 22 Page No:202" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "#Input data\nm=2 #Steam in Kg\nx=0.7 #Initial dryness \nP=15 #Constant pressure in bar\n#V2=2V1\n#from steam table properties of\n#saturated steam(pressure basis) at 15 bar\nTs=198.3+273 #In degree celsius \nVg=0.1317 #In m**3/Kg\nhf=844.7 #In KJ/Kg\nhfg=1945.2 #In KJ/Kg\nhg=2789.9 #In KJ/Kg\nCps=2.3\n\n#Calculation\nV1=x*Vg #Initial specific volume of steam in m**3/Kg\nV2=2*V1 #Final specific volume of steam in m**3/Kg\nTsup=(V2/Vg)*Ts #Steam is superheated in degree celsius \nFSS=Tsup-Ts #Degree of superheated in degree celsius\nh1=hf+x*hfg #Initial enthalpy of steam in KJ/Kg\nh2=hg+Cps*(Tsup-Ts) #Final enthalpy of steam in KJ/Kg \nQ=(h2-h1)*m #Heat transferred in the process in KJ\nW1=P*(m*V2-m*V1) #Work transferred in the process in KJ\n\n#Output\nprint(\"Initial specific volume of steam= \",round(V1,4),\"m**3/Kg\")\nprint(\"Final specific volume of steam= \",round(V2,4),\"m**3/Kg\")\nprint(\"Steam is superheated= \",round(Tsup,2),\"K\")\nprint(\"Degree of superheated= \",round(FSS,2),\"degree celsius\")\nprint(\"Initial enthalpy of steam= \",h1,\"KJ/Kg\")\nprint(\"Final enthalpy of steam= \",round(h2,2),\"KJ/Kg\")\nprint(\"Heat transferred in the process= \",round(Q,2),\"KJ\")\nprint(\"Work transferred in the process= \",round(W1,3),\"KJ\")\n\n", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "Initial specific volume of steam= 0.0922 m**3/Kg\nFinal specific volume of steam= 0.1844 m**3/Kg\nSteam is superheated= 659.82 K\nDegree of superheated= 188.52 degree celsius\nInitial enthalpy of steam= 2206.34 KJ/Kg\nFinal enthalpy of steam= 3223.5 KJ/Kg\nHeat transferred in the process= 2034.31 KJ\nWork transferred in the process= 2.766 KJ\n" + } + ], + "prompt_number": 6 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Example 23 Page No:203" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "#Input data\nms=1000 #Steam in Kg/h \nP=16 #Absolute pressure in bar\nx2=0.9 #Steam is dry \nt1=30+273 #temperature in degree celsius\ntsup=380 #tmperature rised in degree celsius \n \n#from steam table(pressure basis at 16 bar)\nh1=125.7 #in KJ/Kg\nts=201.4 #In degree celsius\nhf=858.5 #in kJ/Kg\nhfg=1933.2 #in kJ/Kg\nhg=2791.7 #in kJ/Kg\nCps=2.3\n\n#Calculation \nh2=hf+x2*hfg #Final enthalpy of wet steam in KJ/Kg \nQ1=(ms*(h2-h1))*10**-3 #Constant pressure process in KJ/h \nh3=hg+Cps*(tsup-ts) #Final enthalpy of superheated steam in KJ/g\nQ2=(ms*(h3-h2))*10**-3 #Suprheated steam in KJ/h\n\n#Output\nprint(\"Final enthalpy of wet steam= \",round(h2,1),\"KJ/Kg \")\nprint(\"Constant pressure process= \",round(Q1,1),\" KJ/h \")\nprint(\"Final enthalpy of superheated steam= \",round(h3,1),\" KJ/g\")\nprint(\"Suprheated steam= \",round(Q2,1),\"KJ/h\")", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "Final enthalpy of wet steam= 2598.4 KJ/Kg \nConstant pressure process= 2472.7 KJ/h \nFinal enthalpy of superheated steam= 3202.5 KJ/g\nSuprheated steam= 604.1 KJ/h\n" + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Example 24 Page No:204" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "#Input data\nFB=15 #First boiler in bar\nSB=15 #Second boiler in bar\ntsup1=300 #Temperature of the steam in degree celsius\ntsup2=200 #Temperature of the steam in degree celsius\n#From steam table (pressure basis at 15 bar )\nts=198.3 #In degree celsius \nhf=844.7 #In KJ/Kg\nhfg=1945.2 #In KJ/Kg\nhg=2789.9 #In KJ/I\n\n\n#Calculation\nh1=hg+Cps*(tsup1-ts) #Enthalpy of steam of first boiler in KJ/Kg \nh3=hg+Cps*(tsup2-ts) #Enthalpy of steam in steam main in KJ/Kg\nh2=2*h3-h1 #Energy balance in KJ/Kg\nx2=(h2-hf)/hfg #Enthalpy of wet steam\n\n#OUTPUT\nprint(\"Enthalpy of steam of first boiler= \",round(h1,1),\"KJ/Kg\")\nprint(\"Enthalpy of steam in steam main= \",round(h3,1),\"KJ/Kg\")\nprint(\"Energy balance= \",round(h2,1),\"KJ/Kg\")\nprint(\"Enthalpy of wet steam= \",round(x2,3),)\n", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "Enthalpy of steam of first boiler= 3023.8 KJ/Kg\nEnthalpy of steam in steam main= 2793.8 KJ/Kg\nEnergy balance= 2563.8 KJ/Kg\nEnthalpy of wet steam= 0.884\n" + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Example 25 Page No:205" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "#Input data\nV=0.35 #Capacity of vessel in m**3\nP1=10*10**2 #Absolute pressure in bar\ntsup1=250+273 #Absolute temperature in degree celsius \nP2=2.5*102 #Absolute pressure in the vessel fall in bar\n\n#From steam table (pressure basis at 10 bar)\nts1=179.9+273 #In degree celsius \nVg1=0.1943 #In m**3/Kg\nhf1=762.6 #In KJ/Kg\nhfg1=2013.6 #In KJ/Kg\nhg1=2776.2 #In KJ/Kg\n\n#From steam table(pressure basis at 2.5 bar)\nV2=0.2247 #In m**3/Kg\nts2=127.4 #In degree celsius\nVg2=0.7184 #In m**3/Kg\nhf2=535.3 #In KJ/Kg\nhfg2=2181.0 #In KJ/Kg\nhg2=2716.4 #In KJ/Kg\n\n#Calculation\nV1=Vg1*(tsup1/ts1) #Initial specific volume of steam in m**3/Kg\nm=V/V1 #Initial mass of steam in Kg\nx2=V2/Vg2 #Final condition of wet steam\nh1=hg1+Cps*(tsup1-ts1) #Initial enthalpy of steam in KJ/Kg\nu1=h1-P1*V1 #Initial internal energy of steam in KJ/Kg\nh2=hf2+x2*hfg2 #Final enthalpy of steam in KJ/Kg\nu2=h2-P2*V2 #Final internal energy of steam in KJ/Kg\ndeltaU=(u2-u1)*m #Change in internal energy in KJ\n\n#Output\nprint(\"Initial specific volume of steam= \",round(V1,4),\"m**3/Kg\")\nprint(\"Initial mass of steam= \",round(m,4),\"Kg\")\nprint(\"Final condition of wet steam= \",round(x2,4),)\nprint(\"Initial enthalpy of steam= \",h1,\"KJ/Kg\")\nprint(\"Initial internal energy of steam= \",round(u1,2),\"KJ/Kg\")\nprint(\"Final enthalpy of steam= \",round(h2,1),\" KJ/Kg\")\nprint(\"Final internal energy of steam= \",round(u2,3),\"KJ/Kg\")\nprint(\"Change in internal energy= \",round(deltaU,1),\"KJ\")\n\n", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "Initial specific volume of steam= 0.2244 m**3/Kg\nInitial mass of steam= 1.5599 Kg\nFinal condition of wet steam= 0.3128\nInitial enthalpy of steam= 2937.43 KJ/Kg\nInitial internal energy of steam= 2713.06 KJ/Kg\nFinal enthalpy of steam= 1217.5 KJ/Kg\nFinal internal energy of steam= 1160.171 KJ/Kg\nChange in internal energy= -2422.3 KJ\n" + } + ], + "prompt_number": 5 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Example 26 Page No:207" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "#Input data\nm=1.5 #Saturated steam in Kg\nx1=1 \nx2=0.6 \nP1=5*10**5 #Absolute pressure in bar\n#From steam table at pressure basis 5 bar\nhg1=2747.5 #In KJ/Kg\nVg1=0.3747 #In m**3/Kg\nV1=0.3747 #In m**3/Kg\nV2=0.3747 #In m**3/Kg\n#From steam table at Vg2 is 2.9 bar\nP2=2.9*10**5 #Absolute pressure in bar \nt2=132.4 #In degree celsius \nhf2=556.5 #In KJ/Kg\nhfg2=2166.6 #In KJ/Kg\n\n\n \n#Calculation\nVg2=V2/x2 #Constant volume process in m**3/Kg\nu1=hg1-((P1*Vg1)/1000) #Initial internal energy in KJ/Kg\nu2=(hf2+x2*hfg2)-((P2*V2)/1000) #Final internal energy in KJ\ndeltaU=(u1-u2)*m #Heat supplied in KJ\n\n#Output\nprint(\"Constant volume process= \",round(Vg2,4),\"m**3/Kg\")\nprint(\"Initial internal energy= \",u1,\"KJ/Kg\")\nprint(\"Final internal energy= \",round(u2,1),\"KJ\")\nprint(\"Heat supplied= \",round(deltaU,2),\"KJ\")\n\n", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "Constant volume process= 0.6245 m**3/Kg\nInitial internal energy= 2560.15 KJ/Kg\nFinal internal energy= 1747.8 KJ\nHeat supplied= 1218.53 KJ\n" + } + ], + "prompt_number": 7 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Example 27 Page No:208" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "#Input data\nP1=20 #Initial steam in bar\nx1=0.95 #dryness throttled\nP2=1.2 #Absolute pressure in bar\n\n#From steam table (pressure basis at 20 bar)\nts=212.4 #In degree celsius\nhf=908.6 #In KJ/Kg\nhfg=1888.6 #In KJ/Kg\nhg=2797.2 #In KJ/Kg\n#From steam table (pressure basis at 1.2 bar)\nh2=h1 #In KJ/Kg\nts2=104.8 #In degree celsius\nhf2=439.3 #In KJ/Kg\nhfg2=2244.1 #In KJ/Kg\nhg2=2683.4 #In KJ/Kg\nCps=2.3\n\n\n#Calculation\nh1=hf+x1*hfg #Enthalpy of steam in KJ/Kg\ntsup2=((h1-hg2)/Cps)+ts2 #Enthalpy of wet steam in degree celsius\nDOS=tsup2-ts2 #Degree of superheat in degree celsius\n\n\n#Output\nprint(\"Enthalpy of steam= \",h1,\"KJ/Kg\")\nprint(\"Enthalpy of wet steam= \",round(tsup2,2),\"degree celsius\")\nprint(\"Degree of superheat= \",round(DOS,2),\"degree celsius\")", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "Enthalpy of steam= 2702.77 KJ/Kg\nEnthalpy of wet steam= 113.22 degree celsius\nDegree of superheat= 8.42 degree celsius\n" + } + ], + "prompt_number": 9 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Example 28 Page No:209" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "#Input data\nP1=12 #Throttled steam\nx1=0.96 #Dryness is brottled\nx2=1 #Constant enthalpy process\n#From steam table at12 bar\nts=188 #In degree celsius\nhf=798.4 #In KJ/Kg\nhfg=1984.3 #In KJ/Kg\nhg=2782.7 #In KJ/Kg\n\n\n#Calculation\nh1=hf+x1*hfg #Enthalpy of the steam in KJ/Kg \nh2=h1 #Enthalpy after throttling in KJ/Kg \n\n#Output\nprint(\"Enthalpy of the steam= \",round(h1,2),\"KJ/Kg \")\nprint(\"Enthalpy after throttlin= \",round(h2,2),\"KJ/Kg \")\n", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "Enthalpy of the steam= 2703.33 KJ/Kg \nEnthalpy after throttlin= 2703.33 KJ/Kg \n" + } + ], + "prompt_number": 11 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Example 29 Page No:210" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "#Input data\nimport math\nP1=15 #Initial steam in bar\ntsup1=250+273 #Temperature of steam in degree celsius\nP2=0.5 #Steam turbine in bar\n\n#From steam table at 15 bar\nts1=198.3+273 #In degree celsius \nhg1=2789.9 #In KJ/Kg\nsf1=2.3145 #In KJ/KgK\nsfg1=4.1261 #In KJ/KgK\nsg1=6.4406 #In KJ/KgK\n#From steam table at 0.5 bar\nts2=81.53 #In degree celsius \nsf2=1.0912 #In KJ/Kg\nsfg2=6.5035 #In KJ/Kg\nsg2=7.5947 #In KJ/Kg\nhf2=340.6\nCps=2.3\nhfg2=2646\n\n#Calculation\nS1=sg1+Cps*(math.log(tsup1/ts1)) #Entropy of superheated steam in KJ/KgK\nS2=S1 #Entropy after isentropic processes in KJ/KgK\nx2=(S2-sf2)/sfg2 #Enthalpy of wet steam \nh1=hg1+Cps*(tsup1-ts1) #Enthalpy of steam at 15 bar\nh2=hf2+x2*hfg2 #Enthalpy of wet steam at 0.5 bar\nWOT=h1-h2 #Work output of the turbine\n\n#OUTPUT\nprint(\"Entropy of superheated steam= \",round(S1,2),\"KJ/KgK\")\nprint(\"Entropy after isentropic processes= \",round(S2,2),\"KJ/KgK\")\nprint(\"Enthalpy of wet steam= \",round(x2,2),\"\")\nprint(\"Enthalpy of steam= \",h1,\"KJ/Kg\")\nprint(\"Enthalpy of wet steam= \",round(h2,2),\"KJ/Kg\")\nprint(\"Work output of the turbine= \",round(WOT,2),\"KJ/Kg\")\n\n\n", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "Entropy of superheated steam= 6.68 KJ/KgK\nEntropy after isentropic processes= 6.68 KJ/KgK\nEnthalpy of wet steam= 0.86 \nEnthalpy of steam= 2908.81 KJ/Kg\nEnthalpy of wet steam= 2614.45 KJ/Kg\nwork output of the turbine= 294.36 KJ/Kg\n" + } + ], + "prompt_number": 22 + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Basic_mechanical_engineering_by_Basant_Agrawal_,_C.M_Agrawal/Chapter_11_Steam_Boilers.ipynb b/Basic_mechanical_engineering_by_Basant_Agrawal_,_C.M_Agrawal/Chapter_11_Steam_Boilers.ipynb new file mode 100644 index 00000000..5b8559d1 --- /dev/null +++ b/Basic_mechanical_engineering_by_Basant_Agrawal_,_C.M_Agrawal/Chapter_11_Steam_Boilers.ipynb @@ -0,0 +1,482 @@ +{ + "metadata": { + "name": "Chapter 11 Steam Boilers" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": "Chapter 11 Steam Boilers" + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Example 1 Page No:228" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "#Input data\nms=5000 #Boiler produces wet steam in Kg/h\nx=0.95 #Dryness function\nP=10 #Operating pressure in bar\nmf=5500 #Bour in the furnace in Kg\nTw=40 #Feed water temp in degree celsius\n\n#calculation\n#from steam table\nhfw=167.45 #In KJ/Kg\nhf=762.61 #In KJ/Kg\nhfg=2031.6 #In KJ/Kg\nhs=(hf+x*hfg) #Enthalpy of wet stream in KJ/Kg\nme=ms/mf #Mass of evaporation\nE=((me*(hs-hfw))/(2257))*10 #Equivalent evaporation in Kg/Kg of coal\n\n#output\n\nprint(\"Enthalpy of wet stream=\",round(hs,2),\"KJ/Kg\")\nprint(\"Mass of evaporation=\",round(me,2),)\nprint(\"Equivalent evaporation=\",round(E,2),\"Kg/Kg of coal\")\n", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "Enthalpy of wet stream= 2692.63 KJ/Kg\nMass of evaporation= 0.91\nEquivalent evaporation= 10.17 Kg/Kg of coal\n" + } + ], + "prompt_number": 5 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Example 2 Page No:229" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "#Input data\np=14 #Boiler pressure in bar\nme=9 #Evaporates of water in Kg\nTw=35 #Feed water entering in degree celsius\nx=0.9 #Steam stop value\nCV=35000 #Calorific value of the coal\n\n#Calculation\n#From Steam Table\nhfw=146.56 #In KJ/Kg\nhf=830.07 #In KJ/Kg\nhfg=1957.7 #In KJ/Kg\nhs=hf+x*hfg #Enthalpy of wet stream in KJ/Kg\nE=((me*(hs-hfw))/2257) #Equivalent evaporation in Kg/Kg of coal\netaboiler=((me*(hs-hfw))/CV)*100#Boiler efficiency in %\n\n#Output\nprint(\"Enthalpy of wet stream=\",hs,\"KJ/Kg\")\nprint(\"Equivalent evaporation=\",round(E,2),\"Kg/Kg of coal\")\nprint(\"Boiler efficiency=\",round(etaboiler,2),\"%\")\n\n", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "Enthalpy of wet stream= 2592.0 KJ/Kg\nEquivalent evaporation= 9.75 Kg/Kg of coal\nBoiler efficiency= 62.88 %\n" + } + ], + "prompt_number": 7 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Example 3 Page No:228" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "#Input data\nms=2500 #Saturated steam per bour in Kg\nx=1 \nP=15 #Boiler pressure in bar\nTw=25 #Feed water entering in degree celsius \nmf=350 #Coal burnt in Kg/bour\nCV=32000 #Calorific value in Kj/Kg \n\n#calculation\n#steam table\nhfw=104.77 #In KJ/Kg\nhf=844.66 #In KJ/Kg\nhfg=1945.2 #In KJ/Kg\nhg=2789.9 #In KJ/Kg\nhs=2789.9 #Enthalpy of dry steam in KJ/Kg\nme=ms/mf #mass of evaporation \nE=((me*(hs-hfw))/2257) #Equivalent evaporation in Kg/Kg ofcoal\netaboiler=((me*(hs-hfw))/CV)*100 #Boiler efficiency in %\n\n#Output\nprint(\"mass of evaporation=\",round(me,3),)\nprint(\"Equivalent evaporation=\",round(E,2),\"Kg/Kg ofcoal\")\nprint(\"Boiler efficiency=\",round(etaboiler,2),\"%\")\n\n\n\n", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "mass of evaporation= 7.143\nEquivalent evaporation= 8.5 Kg/Kg ofcoal\nBoiler efficiency= 59.94 %\n" + } + ], + "prompt_number": 14 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Example 4 Page No:231" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "#Input data\nmf=500 #Boiler plant consumes of coal in Kg/h\nCV=32000 #Calorific value in Kj/Kg\nms=3200 #plant generates in Kg/h\nP=1.2 #Absolute pressure MN/m**2\nMN=12 \nTsup=300 #Absolute temperature in degree celsius\nTw=35 #Feed water temperature\nCps=2.3\n\n#calculation\nhfw=146.56 #In KJ/Kg\nTs=187.96 #In Degree celsius\nhf=798.43 #In KJ/Kg\nhfg=1984.3 #In KJ/Kg\nhg=2782.7 #In KJ/Kg\nhs=hg+Cps*(Tsup-Ts) #Enthalpy of superheated steam in KJ/Kg\nme=ms/mf #mass of evaporation \nE=((me*(hs-hfw))/2257) #Equivalent evaporation in Kg/Kg ofcoal\netaboiler=((me*(hs-hfw))/CV)*100#Boiler efficiency in %\n \n\n#Output\nprint(\"Enthalpy of superheated steam=\",round(hs,2),\"KJ/Kg\")\nprint(\"mass of evaporation=\",me,)\nprint(\"Equivalent evaporation=\",round(E,1),\"Kg/Kg ofcoal\")\nprint(\"Boiler efficiency\",round(etaboiler,2),\"%\")\n \n\n \n", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "Enthalpy of superheated steam= 3040.39 KJ/Kg\nmass of evaporation= 6.4\nEquivalent evaporation= 8.2 Kg/Kg ofcoal\nBoiler efficiency 57.88 %\n" + } + ], + "prompt_number": 15 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Example 5 Page No:232" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "#Input data\nms=5000 #Steam generted in Kg/h\nmf=700 #Coal burnt in Kg/h \nCV=31402 #Cv of coal in KJ/Kg\nx=0.92 #quality of steam\nP=1.2 #Boiler pressure in MPa\nTw=45 #Feed water temperature in degree celsius\n\n\n#calculation\nhfw=188.35 #In KJ/Kg\nhf=798.43 #In KJ/Kg\nhfg=1984.3 #In KJ/Kg\nhs=hf+x*hfg #Enthalpy of wet stream in KJ/Kg\nme=ms/mf #mass of evaporation \nE=((me*(hs-hfw))/2257) #Equivalent evaporation in Kg/Kg of coal\netaboiler=((me*(hs-hfw))/CV)*100 #Boiler efficiency in %\n\n\n\n#Output\nprint(\"Enthalpy of wet stream=\",round(hs,2),\"KJ/Kg\")\nprint(\"mass of evaporation=\",round(me,2),\"\")\nprint(\"Equivalent evaporation=\",round(E,1),\"Kg/Kg of coal\")\nprint(\"Boiler efficiency=\",round(etaboiler,2),\"%\")\n \n", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "Enthalpy of wet stream= 2623.99 KJ/Kg\nmass of evaporation= 7.14 \nEquivalent evaporation= 7.7 Kg/Kg of coal\nBoiler efficiency= 55.4 %\n" + } + ], + "prompt_number": 18 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Example 6 Page No:233" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "#Input data\nms=6000 #Boiler produce of steam Kg/h\nP=25 #Boiler pressure in bar\nTsup=350 #Boiler temperature in degree celsius\nTw=40 #Feed water temperature indegree celsius\nCV=42000 #Calorific value in Kj/Kg\netaboiler=75/100 #Expected thermal efficiency in %\n\n\n#Calculation\nhfw=167.45 #In KJ/Kg\nTs=223.94 #In degree celsius \nhf=961.96 #In KJ/Kg\nhfg=1839.0 #In KJ/Kg\nhg=2800.9 #In KJ/Kg\nCps=2.3\nhs=((hg)+(Cps)*(Tsup-Ts)) #Enthalpy of superheated steam KJ/Kg\nmf=((ms*(hs-hfw))/(CV*etaboiler)) #Boiler efficiency in %\nme=ms/mf #Equivalent mass of evaporation\nE=((me*(hs-hfw))/2257) #Equivalent evaporation in Kg/Kg of oil\n\n\n#Output\nprint(\"Enthalpy of superheated steam=\",hs,\"KJ/Kg\")\nprint(\"Boiler efficiency=\",round(mf,1),\"%\")\nprint(\"Equivalent mass of evaporation=\",round(me,3),)\nprint(\"Equivalent evaporation=\",round(E,2),\"Kg/Kg of oil\")\n\n", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "Enthalpy of superheated steam= 3090.838 KJ/Kg\nBoiler efficiency= 556.8 %\nEquivalent mass of evaporation= 10.775\nEquivalent evaporation= 13.96 Kg/Kg of oil\n" + } + ], + "prompt_number": 22 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Example 7 Page No:234" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "#Input data\nE=12 #Boiler found steam in Kg/Kg\nCV=35000 #Calorific value in KJ/Kg\nms=15000 #Boiler produces in Kg/h\nP=20 #Boiler pressure in bar\nTw=40 #Feed water in degree celsius\nmf=1800 #Fuel consumption\n\n\n#calculation\n#R=me(hs-hfw)\nhfw=167.45 #In KJ/Kg\nhg=2797.2 #In KJ/Kg\nTs=211.37 #In degree celsius\nCps=2.3\nR=E*2257 #Equivalent evaporation in KJ/Kg of coal\netaboiler=(R/CV)*100 #Boiler efficiency in %\nme=ms/mf #Equivalent mass evaporation in KJ/Kg of coal \nhs=(R/me)+hfw # In KJ/Kg\nTsup=((hs-hg)/Cps)+Ts #Enthalpy of superheated steam in degree celsius\n\n\n\n#Output\nprint(\"Equivalent evaporation=\",R,\"KJ/Kg of coal\")\nprint(\"Boiler efficiency=\",round(etaboiler,2),\"%\")\nprint(\"Equivalent mass evaporation=\",round(me,2),\"KJ/Kg of coal\")\nprint(\"hs=\",round(hs,2),\"KJ/Kg\")\nprint(\"Enthalpy of superheated steam=\",round(Tsup,2),\"degree celsius\")", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "Equivalent evaporation= 27084 KJ/Kg of coal\nBoiler efficiency= 77.38 %\nEquivalent mass evaporation= 8.33 KJ/Kg of coal\nhs= 3417.53 KJ/Kg\nEnthalpy of superheated steam= 481.08 degree celsius\n" + } + ], + "prompt_number": 24 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Example 8 Page No:236" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "#Input data\nms=6000 #Steam generated in Kg/h\nmf=700 #Coal burnt in Kg/h\nCV=31500 #Cv of coal in KJ/Kg\nx=0.92 #Dryness in fraction of steam\nP=12 #Boiler pressure in bar\nTsup=259 #Temperature of steam in degree celsius\nTw=45 #Hot well temperature in degree celsius\n\n#calculation\nhfw=188.35 #In KJ/Kg\nTs=187.96 #In degree celsius\nhf=798.43 #In KJ/Kg\nhfg=1984.3 #In KJ/Kg\nhg=2782.7 #In KJ/Kg\nCps=2.3 \nme=ms/mf #Equivalent mass evaporation\nhs=hf+x*hfg #Enthalpy of wet steam in KJ/Kg\nE=((me*(hs-hfw))/2257) #Equivalent evaporation in Kg/Kg of coal\nhs1=(hg+Cps*(Tsup-Ts)) #Enthalpy of superheated steam in KJ/Kg\nE1=((me*(hs1-hfw))/2257) #Equivalent evaporation(with superheater) in Kg/Kg of coal\netaboiler=((me*(hs-hfw))/CV)*100 #Boiler efficiency without superheater in %\netaboiler1=((me*(hs1-hfw))/CV)*100#Boiler efficiency with superheater in %\n\n\n#Output\nprint(\"Equivalent mass evaporation=\",round(me,2),)\nprint(\"Enthalpy of wet steam=\",hs,\"KJ/Kg\")\nprint(\"Equivalent evaporation=\",round(E,2),\"Kg/Kg of coal\")\nprint(\"Enthalpy of superheated steam=\",round(hs1,2),\"KJ/Kg\")\nprint(\"Equivalent evaporation(with superheater)=\",round(E1,2),\"Kg/Kg of coal\")\nprint(\"Boiler efficiency without superheater=\",round(etaboiler,2),\"%\")\nprint(\"Boiler efficiency without superheater=\",round(etaboiler1,2),\"%\")\n\n\n", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "Equivalent mass evaporation= 8.57\nEnthalpy of wet steam= 2623.986 KJ/Kg\nEquivalent evaporation= 9.25 Kg/Kg of coal\nEnthalpy of superheated steam= 2946.09 KJ/Kg\nEquivalent evaporation(with superheater)= 10.47 Kg/Kg of coal\nBoiler efficiency without superheater= 66.28 %\nBoiler efficiency without superheater= 75.04 %\n" + } + ], + "prompt_number": 29 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Example 9 Page No:237" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "#Input data\nP=15 #Boiler produces steam in bar\nTsup=250 #Boiler temperature in degree celsius \nTw=35 #Feed water in degree celsius\nMWh=1.5 #steam supplied to the turbine\nCV=32000 #Coal of calorific value in KJ/Kg\netaboiler=80/100 #Thermal efficiency in %\nfr=210 #Firing rate in Kg/m**2/h\n#From steam table(temp basis at 35 degree celsius)\nhfw=146.56 #In KJ/Kg\nTs=198.29 #In degree celsius\nhfg=1945.2 #In KJ/Kg\nhg=2789.9 #In KJ/Kg\nCps=2.3 \n\n\n#calculator\nhs=hg+Cps*(Tsup-Ts) #Enthalpy of superheated steam(with superheater) in KJ/Kg\nms=9000/MWh #Steam rate in Kg/MWh\nmf=((ms*(hs-hfw))/(etaboiler*CV)) #Mass of steam consumption in Kg/h\nGA=mf/fr #Grate rate in m**2\n\n\n\n#Output\nprint(\"Enthalpy of superheated steam(with superheater)=\",hs,\"KJ/Kg\")\nprint(\"Steam rate=\",ms,\"Kg/h\")\nprint(\"ass of steam consumption=\",round(mf,1),\"Kg/h\")\nprint(\"Grate rate=\",round(GA,3),\"m**2\")\n\n\n\n", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "Enthalpy of superheated steam(with superheater)= 2908.833 KJ/Kg\nSteam rate= 6000.0 Kg/h\nass of steam consumption= 647.4 Kg/h\nGrate rate= 3.083 m**2\n" + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Example 10 Page No:242" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "#Input data\nma=18 #Boileruses of per Kg of fuel in Kg/Kg\nhw=25*10**-3 #Chimney height to produce draught in mm\nTg=315+273 #Temperature of chimney gases in degree celsius \nTa=27+273 #Out side air temp in degree celsius\n\n#Calculation\n#Draught produce in terms of water column in m\nH=(hw/(353*(1/Ta-1/Tg*((ma+1)/ma))))*1000\n\n#Output\nprint(\"Draught produce in terms of water column=\",round(H,2),\"m\")", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "Draught produce in terms of water column= 46.04 m\n" + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Example 11 Page No:242" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "#Input data\nH=40 #High discharge in m\nma=19 #Fuel gases per Kg of fuel burnt\nTg=220+273 #Average temp of fuel gases in degree celsius\nTa=25+273 #Ambient temperature in degreee celsius\n\n\n#calculation\nhw=353*H*(1/Ta-1/Tg*((ma+1)/ma)) #Draught produce in terms of water column in mm\nH1=H*((Tg/Ta)*(ma/(ma+1))-1) #Draught produce in terms of hot gas column in m\n\n#output\nprint(\"Draught produce in terms of water column=\",round(hw,2),\"mm\")\nprint(\"Draught produce in terms of hot gas column=\",round(H1,2),\"m\")", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "Draught produce in terms of water column= 17.23 mm\nDraught produce in terms of hot gas column= 22.87 m\n" + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Example 12 Page No:243" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "#Input data\nH=27 #Chimney height in m\nhw=15 #Draught produces of water column in mm\nma=21 #Gases formed per Kg of fuel burnt in Kg/Kg\nTa=25+273 #Temperature of the ambient air in degree celsius\n\n\n#calculation\nTg=-(((ma+1)/ma)/((hw/(353*H))-(1/Ta))) #Mean temperature of fuel gases in K\n\n#Output\nprint(\"Mean temperature of fuel gases\",Tg,\"k\")", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "Mean temperature of fuel gases 587.9248031162673 k\n" + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Example 13 Page No:244" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "#Input data\nhw=20 #Static draught of water in mm\nH=50 #Chimney height in m\nTg=212+273 #Temperature of the fuel degree celsius\nTa=27+273 #Atmospheric air in degree celsius\n\n#calculation\nma=(-((hw/(353*H))-Ta*Tg))*10**-4 #Air-fuel ratio in Kg/Kg of fuel burnt-3\n\n#Output\nprint(\"Air-fuel ratio\",round(ma,1),\"Kg/Kg of fuel burnt\")", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "Air-fuel ratio 14.5 Kg/Kg of fuel burnt\n" + } + ], + "prompt_number": 8 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Example 14 Page No:245" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "#Input data\nimport math\nH=24 #Chimney height in m\nTa=25+273 #Ambient temperature in degree celsius\nTg=300+273 #Temperature of fuel gases in degree celsius\nma=20 #Combustion space of fuel burnt in Kg/Kgof fuel\ng=9.81 \n\n\n#calculation\nhw=((353*H)*((1/Ta)-((1/Tg)*((ma+1)/ma))))#Theoretical draught in millimeters of water in mm\nH1=H*((Tg/Ta)*(ma/(ma+1))-1) #Theoretical draught produced in hot gas column in m\nH2=H1-9.975 #Draught lost in friction at the grate and passage in m\nV=math.sqrt(2*g*H2) #Actual draught produced in hot gas column in m\n\n#Output\nprint(\"Theoretical draught in millimeters of water=\",round(hw,2),\"mm\")\nprint(\"Theoretical draught produced in hot gas column=\",round(H1,2),\"m\")\nprint(\"Draught lost in friction at the grate and passage=\",round(H2,3),\"m\")\nprint(\"Actual draught produced in hot gas column=\",round(V,),\"m\")", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "Theoretical draught in millimeters of water= 12.9 mm\nTheoretical draught produced in hot gas column= 19.95 m\nDraught lost in friction at the grate and passage= 9.975 m\nActual draught produced in hot gas column= 14 m\n" + } + ], + "prompt_number": 16 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Example 15 Page No:246" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "#Input data\nimport math\nH=38 #Stack height in m\nd=1.8 #Stack diameter discharge in m\nma=17 #Fuel gases per Kg of fuel burnt Kg/Kg\nTg=277+273 #Average temperature of fuel gases in degree celsius\nTa=27+273 #Temperature of outside air in degree celsius\nh1=0.4 #Theoretical draught is lost in friction in \ng=9.81\npi=3.142\n\n#calculation\nH1=H*(((Tg/Ta)*(ma/(ma+1))-1))#Theoretical draught produce in hot gas column in m\ngp=0.45*27.8 #Draught lost in friction at the grate and pasage in m\nC=H1-gp #Actual draught produce in hot gas column in m\nV=math.sqrt(2*9.81*C) #Velocity of the flue gases in the chimney in m/s\nrhog=((353*(ma+1))/(ma*Tg)) #Density of flue gases in Kg/m**3\nmg=(rhog*((pi/4)*(d**(2))*V)) #Mass of gas flowing through the chimney in Kg/s\n\n\n#Output\nprint(\"Theoretical draught produce in hot gas column=\",round(H1,1),\"m\")\nprint(\"Draught lost in friction at the grate and pasage=\",gp,\"m\")\nprint(\"Actual draught produce in hot gas column=\",round(C,2),\"m\")\nprint(\"Velocity of the flue gases in the chimney =\",round(V,2),\"m/s\")\nprint(\"Density of flue gases=\",round(rhog,3),\"Kg/m**3\")\nprint(\"Mass of gas flowing through the chimney=\",round(mg,),\"Kg/s\")\n\n", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "Theoretical draught produce in hot gas column= 27.8 m\nDraught lost in friction at the grate and pasage= 12.51 m\nActual draught produce in hot gas column= 15.29 m\nVelocity of the flue gases in the chimney = 17.32 m/s\nDensity of flue gases= 0.68 Kg/m**3\nMass of gas flowing through the chimney= 30 Kg/s\n" + } + ], + "prompt_number": 23 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Example 16 Page No:247" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "#Input data\nimport math\nhw=1.9 #Drauhgt water in cm\nTg=290+273 #Temp of flue gases in degree celsius \nTa=20+273 #Ambient temp in degree celsius\nma=22 #Flue gases formed in kg/Kg of coal\nd=1.8 #Fuel burnt in m\npi=3.142\ng=9.81\n\n#calculation\nH=(hw/(353*(1/Ta-1/Tg*((ma+1)/ma))))*10 #Theoretical draught produced in water column in m\nH1=H*(((Tg/Ta)*(ma/(ma+1))-1)) #Theoretical draught produced in hot gas column n m\nV=math.sqrt(2*g*H1) #Velocity of tthe flue gases in the chimney in m/s \nrhog=((353*(ma+1))/(ma*Tg)) #Density of flue gases in Kg/m**3\nmg=rhog*((pi/4)*d**2)*V #Mass of gas flowing through the chimney in Kg/s\n\n#Output\nprint(\"Theoretical draught produced in water column=\",round(H,1),\"m\")\nprint(\"Theoretical draught produced in hot gas column=\",round(H1,),\"m\")\nprint(\"Velocity of tthe flue gases in the chimney=\",round(V,2),\"m\")\nprint(\"Density of flue gases=\",round(rhog,4),\"Kg/m**3\")\nprint(\"Mass of gas flowing through the chimney=\",round(mg,1),\"Kg/s\")\n\n", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "Theoretical draught produced in water column= 34.6 m\nTheoretical draught produced in hot gas column= 29 m\nVelocity of tthe flue gases in the chimney= 23.85 m\nDensity of flue gases= 0.6555 Kg/m**3\nMass of gas flowing through the chimney= 39.8 Kg/s\n" + } + ], + "prompt_number": 28 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Example 17 Page No:248" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "#Input data\nimport math\nmf=8000 #Average coal consumption in Kg/h\nma=19 #Flue gases formed in Kg/Kg\nTg=270+273 #Average temperature of the chimney in degree celsius\nTa=27+273 #Ambient temperature in degree celsius\nhw=18 #Theoretical draught produced by the chimney in mm\nh1=0.6 #Draught is lost in friction H1\ng=9.81\npi=3.142\n\n\n#calculation\nH=(hw/(353*(1/Ta-1/Tg*((ma+1)/ma)))) #Theoretical draught produced in water column in m\nH1=H*(((Tg/Ta)*(ma/(ma+1)))-1) #Theoretical draught produced in hot gas column in m\ngp=h1*H1 #Draught is lost in friction at the grate and passing in m\nhgc=H1-gp #Actual draught produced in hot gas column in m\nV=math.sqrt(2*g*(hgc)) #Velocity of the flue gases in the chimney in m/s\nrhog=((353*(ma+1))/(ma*Tg)) #Density of flue gases in Kg/m**3\nmg=((mf/3600)*ma) #Mass of gas fowing throgh the chimney in Kg/s\nd=math.sqrt(mg/(rhog*(pi/4)*V)) #Diameter of the chimney in m\n\n\n#Output\nprint(\"Theoretical draught produced in water column=\",round(H,1),\"m\")\nprint(\"Theoretical draught produced in hot gas column=\",round(H1,3),\"m\")\nprint(\"Draught is lost in friction at the grate and passing=\",round(gp,2),\"m\")\nprint(\"Actual draught produced in hot gas column=\",round(hgc,3),\"m\")\nprint(\"Velocity of the flue gases in the chimney=\",round(V,2),\"\")\nprint(\"Density of flue gases=\",round(rhog,3),\"Kg/m**3\")\nprint(\"Mass of gas fowing throgh the chimney=\",round(mg,3),\"Kg/s\")\nprint(\"Diameter of the chimney=\",round(d,3),\"m\")\n\n\n", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "Theoretical draught produced in water column= 36.6 m\nTheoretical draught produced in hot gas column= 26.304 m\nDraught is lost in friction at the grate and passing= 15.78 m\nActual draught produced in hot gas column= 10.522 m\nVelocity of the flue gases in the chimney= 14.37 \nDensity of flue gases= 0.684 Kg/m**3\nMass of gas fowing throgh the chimney= 42.222 Kg/s\nDiameter of the chimney= 2.338 m\n" + } + ], + "prompt_number": 15 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Example 18 Page No:251" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "#Input data\nimport math\nH=24 #Chimney height in m\nTa=25+273 #Ambient temperature in degree celsius\nTg=300+273 #Temp of flue gases passing through the chimney in degree celsius\nma=20 #Combustion space of fuel burnt in Kg/kg of fuel\ng=9.81\n\n#calculation\nhw=((353*H)*((1/Ta)-((1/Tg)*((ma+1)/ma)))) #Theoretical draught produced in water column in m\nH1=H*(((Tg/Ta)*(ma/(ma+1))-1)) #Theoretical draught produced in hot gas column in m\nH2=0.5*H1 #Draught is lost in friction at the grate and passing in m\nhgc=H1-H2 #Actual draught produced in hot gas column in m\nV=math.sqrt(2*g*H2) #Velocity of the flue gases in the chimney in m/s\n\n\n#Output\nprint(\"Theoretical draught produced in water column=\",round(hw,1),\"m\")\nprint(\"Theoretical draught produced in hot gas column=\",round(H1,2),\"m\")\nprint(\"Draught is lost in friction at the grate and passing=\",round(H2,3),\"m\")\nprint(\"Actual draught produced in hot gas column=\",round(hgc,3),\"m\")\nprint(\"Velocity of the flue gases in the chimney=\",round(V,),\"m/s\")\n", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "Theoretical draught produced in water column= 12.9 m\nTheoretical draught produced in hot gas column= 19.95 m\nDraught is lost in friction at the grate and passing= 9.975 m\nActual draught produced in hot gas column= 9.975 m\nVelocity of the flue gases in the chimney= 14 m/s\n" + } + ], + "prompt_number": 16 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Example 19 Page No:252" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "#Input data\nH=38 #Stack height in m\nd=1.8 #Stack diameter in m\nma=18 #Flue gases per kg of the fuel burnt\nTg=277+273 #Average temp of the flue gases in degree celsius\nTa=27+273 #Temperature of outside air in degree celsius\nh1=0.4 #Theorical draught is lost in friction in %\ng=9.81\n\n#calculation\nH1=H*(((Tg/Ta)*(ma/(ma+1))-1)) #Theoretical draught produced in hot gas column in m\ngp=0.40*H1 #Draught is lost in friction at the grate and passing in m\nhgc=H1-gp #Actual draught produced in hot gas column in m\nV=math.sqrt(2*g*hgc) #Velocity of the flue gases in the chimney in m/s\nrhog=((353*(ma+1))/(ma*Tg)) #Density of flue gases in Kg/m**3\nmg=rhog*((pi/4)*d**2)*V #Mass of gas fowing throgh the chimney in Kg/s\n\n\n#Output\nprint(\"Theoretical draught produced in hot gas column=\",round(H1,3),\"m\")\nprint(\"Draught is lost in friction at the grate and passing=\",round(gp,2),\"m\")\nprint(\"Actual draught produced in hot gas column=\",round(hgc,2),\"m\")\nprint(\"Velocity of the flue gases in the chimney=\",round(V,2),\"m/s\")\nprint(\"Density of flue gases=\",round(rhog,2),\"Kg/m**3\")\nprint(\"Mass of gas fowing throgh the chimney=\",round(mg,1),\"Kg/s\")\n\n\n\n", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "Theoretical draught produced in hot gas column= 28.0 m\nDraught is lost in friction at the grate and passing= 11.2 m\nActual draught produced in hot gas column= 16.8 m\nVelocity of the flue gases in the chimney= 18.16 m/s\nDensity of flue gases= 0.68 Kg/m**3\nMass of gas fowing throgh the chimney= 31.3 Kg/s\n" + } + ], + "prompt_number": 41 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Example 20 Page No:253" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "#Input data\nimport math\nhw=19 #Draught produced water in cm\nTg=290+273 #Temperature of flue gases in degree celsius\nTa=20+273 #Ambient temperature in degree celsius\nma=22 #Flue gases formed per kg of fuel burnt in kg/kg of coal \nd=1.8 #Diameter of chimney\ng=9.81\n\n\n#calculation\nH=(hw/((353)*((1/Ta)-((1/Tg)*((ma+1)/ma))))) #Theoretical draught produced in hot gas column in m\nH1=H*(((Tg/Ta)*(ma/(ma+1))-1)) #Draught is lost in friction at the grate and passing in m\nV=math.sqrt(2*g*H1) #Velocity of the flue gases in the chimney in m/s\nrhog=((353*(ma+1))/(ma*Tg)) #Density of flue gases in Kg/m**3\nmg=rhog*((pi/4)*d**2)*V #Mass of gas fowing throgh the chimney in Kg/s\n\n\n#Output\nprint(\"Theoretical draught produced in hot gas column=\",round(H,),\"m\")\nprint(\"Draught is lost in friction at the grate and passing=\",round(H1,1),\"m\")\nprint(\"Velocity of the flue gases in the chimney=\",round(V,2),\"m/s\")\nprint(\"Density of flue gases=\",round(rhog,4),\" Kg/m**\")\nprint(\"Mass of gas fowing throgh the chimney=\",round(mg,1),\"Kg/s\")\n", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "Theoretical draught produced in hot gas column= 35 m\nDraught is lost in friction at the grate and passing= 29.0 m\nVelocity of the flue gases in the chimney= 23.85 m/s\nDensity of flue gases= 0.6555 Kg/m**\nMass of gas fowing throgh the chimney= 39.8 Kg/s\n" + } + ], + "prompt_number": 44 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Example 21 Page No:254" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "#Input data\nimport math\nmf=8000 #Average coal consumption in m \nma=18 #Fuel gases formed ccoal fired in m\nTg=270+273 #Average temp of the chimney of water in degree celsius\nTa=27+273 #Ambient temp in degree celsius\nhw=18 #Theoretical draught produced by the chimney in mm\nh1=0.6 #Draught is lost in friction in H1\ng=9.81\npi=3.142\n\n\n#calculation\nH=(hw/((353)*((1/Ta)-((1/Tg)*((ma+1)/ma))))) #Theoretical draught produced in water column in m\nH1=H*(((Tg/Ta)*(ma/(ma+1))-1)) #Theoretical draught produced in hot gas column in m\ngp=0.6*H1 #Draught is lost in friction at the grate and passing in m\nhgc=H1-gp #Actual draught produced in hot gas column in m \nV=math.sqrt(2*g*hgc) #Velocity of the flue gases in the chimney in m/s\nrhog=((353*(ma+1))/(ma*Tg)) #Density of flue gases in Kg/m**3\nmg=mf/3600*(ma+1) #Mass of gas fowing throgh the chimney in Kg/s\nd=math.sqrt(mg/(rhog*(pi/4)*V)) #Diameter of flue gases in Kg/m**3\n\n#Output\nprint(\"Theoretical draught produced in water column=\",round(H,1),\"m\")\nprint(\"Theoretical draught produced in hot gas column=\",round(H1,2),\"m\")\nprint(\"Draught is lost in friction at the grate and passing=\",round(gp,2),\"m\")\nprint(\"Actual draught produced in hot gas column=\",round(hgc,2),\"\")\nprint(\"Velocity of the flue gases in the chimney=\",round(V,2),\"m/s\")\nprint(\"Density of flue gases=\",round(rhog,3),\"Kg/m**3\")\nprint(\"Mass of gas fowing throgh the chimney=\",round(mg,2),\"Kg/s\")\nprint(\"Diameter of flue gases=\",round(d,3),\"Kg/m**3\")\n\n\n", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "Theoretical draught produced in water column= 36.7 m\nTheoretical draught produced in hot gas column= 26.23 m\nDraught is lost in friction at the grate and passing= 15.74 m\nActual draught produced in hot gas column= 10.49 \nVelocity of the flue gases in the chimney= 14.35 m/s\nDensity of flue gases= 0.686 Kg/m**3\nMass of gas fowing throgh the chimney= 42.22 Kg/s\nDiameter of flue gases= 2.337 Kg/m**3\n" + } + ], + "prompt_number": 45 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Example 22 Page No:256" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "#Input data\nH=45 #Chimney height in m\nTg=370+273 #Temperature of flue gases in degree celsius\nT1=150+273 #Temperature of flue gases in degree celsius\nma=25 #Mass of the flue gas formed in Kg/kg of a cosl fired\nTa=35+273 #The boiler temperature in degree celsius\nCp=1.004 #fuel gas\n\n#calculation\n#Efficeincy of chimney draught in %\nA=(H*(((Tg/Ta)*(ma/(ma+1)))-1))/(Cp*(Tg-T1))*100\n\n#Output\nprint(\"Efficeincy of chimney draught=\",round(A,2),\"%\")", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "Efficeincy of chimney draught= 20.52 %\n" + } + ], + "prompt_number": 47 + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Basic_mechanical_engineering_by_Basant_Agrawal_,_C.M_Agrawal/Chapter_13_Steam_Engines.ipynb b/Basic_mechanical_engineering_by_Basant_Agrawal_,_C.M_Agrawal/Chapter_13_Steam_Engines.ipynb new file mode 100644 index 00000000..0b5d6900 --- /dev/null +++ b/Basic_mechanical_engineering_by_Basant_Agrawal_,_C.M_Agrawal/Chapter_13_Steam_Engines.ipynb @@ -0,0 +1,308 @@ +{ + "metadata": { + "name": "Chapter 13 Steam Engines" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Example 1 Page No:281" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "#Input data\nimport math\nPa=10 #Single cylinder double acting steam engine pressure in bar \nPb=1.5 #Single cylinder double acting steam engine pressure in bar\nrc=100/35 #Cut-off of the stroke in %\n\n\n#Calculation\nPm=((Pa/rc)*(1+math.log(rc))-Pb) #Therotical mean effective pressure\n\n#Output\nprint(\"Therotical mean effective pressure=\",round(Pm,2),\"bar\")", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "Therotical mean effective pressure= 5.67 bar\n" + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Example 2 Page No:283" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "#Input data\n\na=5/100 #Engine cylinder of the stroke valume in %\nP1=12 #Pressure of the stream\nrc=3 #Cut-off is one-third\nPb=1.1 #Constant the back pressure in bar\n\n#Calulation\n#Therotical mean effective pressure Pm\nPm=P1*(1/rc+((1/rc)+a)*math.log((1+a)/((1/rc)+a)))-Pb \n\n#Output\nprint(\"#Therotical mean effective pressure=\",round(Pm,2),\"N/m**2\")", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "#Therotical mean effective pressure= 7.54 N/m**2\n" + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Example 3 Page No:285" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "#Input data\nimport math\nP1=14 #Steam is ssupplied in bar \nP6=6 #Pressure at the end in bar\nPb=1.2 #Pressure at back in bar\na=0.1 \nre=4 \n#From hyperbolic process \nb=0.4\n\n#Calculation\n#Mean Effective pressure in N/m**2 \nPm=P1*((1/re)+((1/re)+a)*math.log((1+a)/((1+re)+a)))-Pb*((1+b)+(a+b)*math.log((a+b)/a))\n\n\n#Output\nprint(\"Mean Effective pressure=\",round(-Pm,3),\"N/m**2\")\n", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "Mean Effective pressure= 6.662 N/m**2\n" + } + ], + "prompt_number": 15 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Example 4 Page No:286" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "#Input data\nCover=1200 #Area of the indicator diagram for cover \nCrank=1100 #Area of the indicator diagram for crank\nID=75\nPS=0.15\n\n\n#Calculation\nCoverMEP=Cover/ID*PS #Cover end mean effective pressure\nCrankMEP=Crank/ID*PS #Crank end mean effective pressure\nAverageMEP=(CoverMEP+CrankMEP)/2 #Average end mean effective pressure\n\n\n#Output\nprint(\"Cover end mean effective pressure=\",CoverMEP,\"bar\")\nprint(\"Crank end mean effective pressure=\",round(CrankMEP,2),\"bar\")\nprint(\"Average end mean effective pressure=\",AverageMEP,\"bar\")\n\n", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "Cover end mean effective pressure= 2.4 bar\nCrank end mean effective pressure= 2.2 bar\nAverage end mean effective pressure= 2.3 bar\n" + } + ], + "prompt_number": 16 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Example 5 Page No:286" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "#Input data\na=25 #Area of indicator diagram cm**2\nVs=0.15 #swept volume m**2\nS=1 #Scale in cm \ncm=0.02 #pressure axis m**3\n\n\n#Calculation\nb=Vs/cm #Base length of diagram \nPm=a/b*S #Mean effective pressure\n\n#Output\nprint(\"Base length of diagram=\",b,\"bar\")\nprint(\"Mean effective pressure=\",round(Pm,2),\"bar\")\n\n", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "Base length of diagram= 7.5 bar\nMean effective pressure= 3.33 bar\n" + } + ], + "prompt_number": 17 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Example 6 Page No:287" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "#Input data\nimport math\nP1=14 #Steam Engine pressure in bar\nPb=0.15 #Back pressure in bar\nK=0.72 #Diagram factor\nrc=100/20 \n\n#Calculation\nPm=((P1/rc)*(1+math.log(rc))-Pb) #Therotical mean effective pressure Pm\nPma=Pm*K #Actual mean effective pressure Pma\n\n#Output\nprint(\"Therotical mean effective pressure=\",round(Pm,3),\"bar\")\nprint(\"Actual mean effective pressure=\",round(Pma,2),\"bar\")\n\n", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "Therotical mean effective pressure= 7.156 bar\nActual mean effective pressure= 5.15 bar\n" + } + ], + "prompt_number": 21 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Example 7 Page No:287" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "#Input data\nimport math\nP1=9 #Reciprocating engine pressure in bar\nPb=1.5 #Back pressure in bar\nrc=100/25 #Cut-off \nK=0.8 #Diagram factor\n\n#Calculation\nPm=((P1/rc)*(1+math.log(rc))-Pb) #Therotical mean effective pressure Pm\nPma=Pm*K #Actual mean effective pressure Pma\n\n#Output\nprint(\"Therotical mean effective pressure= \",round(Pm,2),\"bar\")\nprint(\"Actual mean effective pressure=\",round(Pma,2),\"bar\")\n\n\n\n", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "Therotical mean effective pressure= 3.87 bar\nActual mean effective pressure= 3.1 bar\n" + } + ], + "prompt_number": 22 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Example 8 Page No:288" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "#Input data\nimport math\nP1=10 #Inlet pressure\nPb=1 #Back pressure\nrc=3 #Expansion ratio\na=12.1 #Area of indicator diagram\nb=7.5 #Length of indicator diagram \nS=3 #Pressure scale\n\n\n#calculation\nPm=((P1/rc)*(1+math.log(rc))-Pb )#Therotical mean effective pressure Pm\nPma=a/b*S #Actual mean effective pressure Pma\nK=Pma/Pm #diagram factor \n\n#Output\nprint(\"Therotical mean effective pressure=\",round(Pm,2),\"bar\")\nprint(\"Actual mean effective pressure=\",round(Pma,2),\"bar\")\nprint(\"Diagram factor=\",round(K,3),)\n\n", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "Therotical mean effective pressure= 6.0 bar\nActual mean effective pressure= 4.84 bar\nDiagram factor= 0.807\n" + } + ], + "prompt_number": 23 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Example 9 Page No:289" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "#Input data\nD=200*10**-3 #Steam engine cylinder in mm \nL=300*10**-3 #Bore of steam engine cylinder in mm \nrc=100/40 #Cut-off of the sroke\nP1=7 #Admission pressure of steam in bar\nPb=0.38 #Exhaust pressure of steam in bar\nK=0.8 #Diagram factor\nN=200 #Indicator factor of engine\npi=3.142 #Constant value\n#Indicated power of the engine in rpm\nA=pi*(200*10**-3)**2/4\n\n\n#Calculation\nPm=((P1/rc)*(1+math.log(rc))-Pb) #Therotical mean effective pressure Pm\nPma=Pm*K #Actual mean effective pressure Pma\nIP=(2*Pma*L*A*N/60000)*10**5 #Indicated power of steam engine in Kw\n\n\n#Output\nprint(\"Therotical mean effective pressure= \",round(Pm,3),\"bar\")\nprint(\"Actual mean effective pressure=\",round(Pma,),\"bar\")\nprint(\"Indicated power of steam engine=\",round(IP,2),\"Kw\")\n\n", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "Therotical mean effective pressure= 4.986 bar\nActual mean effective pressure= 4 bar\nIndicated power of steam engine= 25.06 Kw\n" + } + ], + "prompt_number": 15 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Example 10 Page No:290" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "#Input data\nimport math\nIP=343 #Steam engine develop indicated power in Kw\nN=180 #power In rpm\nP1=15 #Steam supplied i bar \nPb=1.25 #Steam is exhausted in bar\nrc=100/25 #Cut-off take place of stroke\nK=0.78 #Diagram factor\n#x=L/D=4/3\nx=4/3 #Stroke to bore ratio\npi=3.142\nA=((pi/4)*(D**2))\n\n#calculation\nPm=((P1/rc)*(1+math.log(rc))-Pb) #Therotical mean effective pressure Pm\nPma=Pm*K #Actual mean effective pressure Pma\nD=(((60000*IP)/(2*(Pma*10**5)*(4/3)*N))/(pi/4))**(1/3)#Indicated power of steam engine\nL=(x)*D\n\n\n#Output\nprint(\"Therotical mean effective pressure=\",round(Pm,2),\"bar\")\nprint(\"Actual mean effective pressure=\",round(Pma,2),\"bar\")\nprint(\"Indicated power of steam engine=\",round(D,3),\"mm\")\nprint(\"Indicated power of steam engine=\",round(L,1),\"mm\")\n\n", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "Therotical mean effective pressure= 7.7 bar\nActual mean effective pressure= 6.0 bar\nIndicated power of steam engine= 0.45 mm\nIndicated power of steam engine= 0.6 mm\n" + } + ], + "prompt_number": 20 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Example 11 Page No:290" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "#Input data\nimport math\nD=240*10**-3 #Steam engine bor\nL=300*10**-3 #Stroke of engine\nN=220 #Speed of engine 220 in rpm \nIP=36 #Indicated power in Kw\nPb=1.3 #Exhaust pressure in bar\nre=2.5 #Expansion ratio\nK=0.8 #Diagram factor\nA=((pi/4)*(D**2))\n\n\n#Calculation\nPma=((IP*60000)/(2*10**5*L*A*N)) #Indicated power of steam engine in bar\nPm=Pma/K #Actual mean effective pressure in bar\nP1=((Pm+Pb)*re)/(1+math.log(re)) #Theoretical mean effective pressure in bar\n\n#Output\nprint(\"Indicated power of steam engine=\",round(Pma,3),\"bar\")\nprint(\"Actual mean effective pressure=\",round(Pm,3),\"bar\")\nprint(\"theoretical mean effective pressure=\",round(P1,1),\"bar\")\n\n", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "Indicated power of steam engine= 3.617 bar\nActual mean effective pressure= 4.521 bar\ntheoretical mean effective pressure= 7.6 bar\n" + } + ], + "prompt_number": 46 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Example 12 Page No:291" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "#Input data\nimport math\nD=700*10**-3 #Steam engine diameter in mm\nL=900*10**-3 #Steam engine diameter in mm\nIp=450 #Develop indicated power Kw\nN=90 #Speed of steam engine in rpm\nP2=12 #Pressure at cut-off in bar\nP1=12 #Pressure at cut-off in bar\nPb=1.3 #Back pressure in bar\nK=0.76 #Diameter factor\npi=3.142\nA=((pi/4)*0.7**2)\n\n#Calculation\nPma=(Ip*60000)/(2*10**5*L*A*90) #Indicated power of steam engine in bar\nPm=Pma/K #Theoretical mean effective pressure in bar\n#using trial and error method\nre=1/0.241 #Expansion ratio\n#Output\nprint(\"Indicated power of steam engine=\",round(Pma,2),\"bar\")\nprint(\"Theoretical mean effective pressure=\",round(Pm,1),\"bar\")\nprint(\"Expansion ratio=\",round(re,2),)", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "Indicated power of steam engine= 4.33 bar\nTheoretical mean effective pressure= 5.7 bar\nExpansion ratio= 4.15\n" + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Example 13 Page No:293" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "#Input data\nDb=900*10**-3 #Diameter of break drum in mm\ndr=50*10**-3 #Diameter of rope in mm\nW=105*9.81 #dead weight on the tight side of the rope in Kg\nS=7*9.81 #Spring balance of the rope in N\nN=240 #Speed of the engine in rpm\n\n#Calculation\nT=(W-S)*((Db+dr)/2) #Torque Nm\nBp=2*pi*N*T/ 60000 #Brake Power in Kw\n\n#Output\nprint(\"Torque= \",round(T,2),\"Nm\")\nprint(\"Brake Power=\",round(Bp,2),\"Kw\")\n\n", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "Torque= 456.66 Nm\nBrake Power= 11.48 Kw\n" + } + ], + "prompt_number": 57 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Example 14 Page No:294" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "#Input data\nD=300*10**-3 #steam engine bor\nL=400*10**-3 #stroke \nDb=1.5 #effective brake diameter\nW=6.2*10**3 #net load on the brake\nN=180 #speed of engine in rpm\nPma=6.5*10**3 #mean effective pressure in bar\nA=((pi/4)*0.3**2) \ndr=0\nS=0\n\n#Calculation\nIp=((2*Pma*L*A*N)/60000)*100 #Indicated power of steam engine in Kw\nT=(W-S)*((Db+dr)/2) #Torque in Nm\nBp=2*pi*N*T/ 60000 #Break power Kw\neta=(Bp/Ip)*100 #Mechanical efficiency in%\n\n\n#Output\nprint(\"Indicated power of steam engine=\",round(Ip,2),\"Kw\")\nprint(\"Torque=\",T,\"Nm\")\nprint(\"Break power=\",round(Bp,2),\"Kw\")\nprint(\"Mechanical efficiency=\",round(eta,1),\"%\")", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "Indicated power of steam engine= 110.28 Kw\nTorque= 4650.0 Nm\nBreak power= 87.66 Kw\nMechanical efficiency= 79.5 %\n" + } + ], + "prompt_number": 59 + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Basic_mechanical_engineering_by_Basant_Agrawal_,_C.M_Agrawal/Chapter_14_Air_Standard_Cycles.ipynb b/Basic_mechanical_engineering_by_Basant_Agrawal_,_C.M_Agrawal/Chapter_14_Air_Standard_Cycles.ipynb new file mode 100644 index 00000000..a968e2ca --- /dev/null +++ b/Basic_mechanical_engineering_by_Basant_Agrawal_,_C.M_Agrawal/Chapter_14_Air_Standard_Cycles.ipynb @@ -0,0 +1,358 @@ +{ + "metadata": { + "name": " Chapter 14 Air Standard Cycles" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Example 1 Page No:302" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "#Input data\nTmax=477+273 #Temperature limits for the engine 477 degree celcius\nTmin=27+273 #Temperature limits for the engine 27 degree celcius\nwd=150 #Carnot cycle produce in KJ\n\n#Calculatkion\neta=(1-(Tmin/Tmax)) #Thermal efficiency of the carnot cycle in %\nQs=(wd/eta) #Added during the process in Kj\n\n\n#Output\nprint(\"thermal efficiency of the carnot cycle eta=\",100*(eta),\"%\")\nprint(\"added during the process Qs=\",Qs,\"KJ\")", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "thermal efficiency of the carnot cycle eta= 60.0 %\nadded during the process Qs= 250.0 KJ\n" + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Example 2 Page No:302" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "#Input data\nQR=1.5 #tau=QS-QR\n #T=Tmax-Tmin\nT=300 #temperature limit of the cycle in degree celsius\n\n\n#Calculation\n#QR=1.5*(QS-QR)\nQR=(1.5/2.5) #Engin work on carnot cycle\neta=(1-QR) #Thermal effeciency\nTmax=(T/eta)-273.15 #Maximum temperataure\nTmin=(Tmax-T) #Minimum temperataure\n\n\n#Output\nprint(\"Engin work on carnot cycle=\",QR,\"QS\")\nprint(\"Thermal effeciency=\",100*(eta),\"%\")\nprint(\"Maximum temperataure=\",round(Tmax,),\"degree celsius\")\nprint(\"Minimum temperataure=\",round(Tmin,),\"degree celsius\")\n\n\n", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "Engin work on carnot cycle= 0.6 QS\nThermal effeciency= 40.0 %\nMaximum temperataure= 477 degree celsius\nMinimum temperataure= 177 degree celsius\n" + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Example 3 Page No:303" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "#Input data\n#Refer figure\nimport math\nT1=300 #Carnot engine work in minimum temperature in kelvin\nT2=750 #Carnot engine work in maximum temperature kelvin\nP2=50 #pressure of carnot engine N/m**2\nP4=1 #pressure of carnot engine N/m**\n#considering air as the working fluid therefore \nR=0.287 #Air as the working fluid in KJ/Kg K\nCp=1.005 #KJ/Kg K\nCv=0.718 #KJ/Kg K\nK=1.4\ngamma=1.4\n\n#Calculation\n#T2/T1=(P2/P1)**(gamma-1)/gamma\nP1=P2*(T1/T2)**(gamma/(gamma-1)) #Pressure at intermediate salient points(1-2) in bar\nP3=P4*(T2/T1)**(gamma/(gamma-1)) #Pressure at intermediate salient points(3-4) in bar\nQS=R*T2*math.log(P2/P3 ) #Heat supplied and rejected per Kg of air in KJ/Kg\nQR=R*T1*math.log(P1/P4 ) #Heat supplied and rejected per Kg of air in KJ/Kg\nW=QS-QR #Work done in KJ/Kg\neta=(1-(T1/T2)) #Thermal of the carnot cycle\n\n#Output\nprint(\"pressure at intermediate salient points(1-2)=\",round(P1,2),\"bar\")\nprint(\"pressure at intermediate salient points(3-4)=\",round(P3,1),\"bar\")\nprint(\"heat supplied and rejected per Kg of air(2-3)=\",round(QS,1),\"KJ/Kg\")\nprint(\"heat supplied and rejected per Kg of air(4-1)=\",round(QR,2),\"KJ/Kg\")\nprint(\"work done=\",round(W,1),\"KJ/Kg\")\nprint(\"thermal of the carnot cycle=\",100*(eta),\"%\")\n", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "pressure at intermediate salient points(1-2)= 2.02 bar\npressure at intermediate salient points(3-4)= 24.7 bar\nheat supplied and rejected per Kg of air(2-3)= 151.8 KJ/Kg\nheat supplied and rejected per Kg of air(4-1)= 60.7 KJ/Kg\nwork done= 91.1 KJ/Kg\nthermal of the carnot cycle= 60.0 %\n" + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Example 4 Page No:304" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "#input data \nimport math\nT2=377+273 #Carnot cycle temperature in bar \nP2=20*10**5 #Carnot cycle pressure in bar\nV2=1\nV1=5\nV3=2\n#consider air as the working fluid therefore\nR=0.287 #In KJ/Kg K\nCp=1.005 #In KJ/Kg K\nCv=0.718 #In KJ/Kg K\nK=1.4\ngamma=1.4\n\n#calculation\nT1=T2*((v2/v1)**(gamma-1)) #Minimum temp in degree celsius\nQs=R*T2*math.log(V3/V2) #Heat supplied process in KJ/Kg\nQR=R*T1*math.log((V1/V2)*(V2/V3)*((T2/T1)**(1/(gamma-1)))) #Heat Rejected Process in KJ/Kg\netath=(1-(T1/T2))*100 #Thermal Effeiciency of the carnot cycle in %\n\n\n\n#output\nprint(\"Minimum temp= \",round(T1,1),\"degree celsius\")\nprint(\"Heat supplied process= \",round(Qs,1),\"KJ/Kg\")\nprint(\"Heat Rejected Process= \",round(QR,1),\"KJ/Kg\")\nprint(\"Thermal Effeiciency of the carnot cycle= \",round(etath,1),\" %\")\n", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "Minimum temp= 341.4 degree celsius\nHeat supplied process= 129.3 KJ/Kg\nHeat Rejected Process= 247.5 KJ/Kg\nThermal Effeiciency of the carnot cycle= 47.5 %\n" + } + ], + "prompt_number": 13 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Example 5 Page No:308 " + }, + { + "cell_type": "code", + "collapsed": false, + "input": "#Input data\nP1=1 #Isentropic Compression in bar\nP2=20 #Isentropic Compression in bar\n#consider air as the working fluid therefore\ngamma=1.4\n\n\n#Calculation\nr=(P2/P1)**(1/gamma) #Isentropic process \neta=100*(1-(1/(r**(gamma-1))))#Otto cycle air standard effeciency in %\n\n\n#Output\nprint(\"compression ratio=\",round(r,2),)\nprint(\"standard efficiency=\",round(eta,1),\"%\")\n", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "compression ratio= 8.5\nstandard efficiency= 57.5 %\n" + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Example 6 Page No:308" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "#Input data\nT1=27+273 #Initial temp in degree celsius \nT2=450+273 #Final temp in degree celsius \n\n#calculation\nr=(T2/T1)**(1/(gamma-1)) #Isentropic process \neta=100*(1-(1/(r**(gamma-1)))) #Otto cycle air standard effeciency in %\n\n#output\nprint(\"compression ratio=\",round(r),)\nprint(\"standard efficiency=\",round(eta,1),\"%\")", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "compression ratio= 9\nstandard efficiency= 58.5 %\n" + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Example 7 Page No:309" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "#Input data\nD=200*10**-3 #Otto cycle cylindrical bore in mm\nL=450*10**-3 #Otto cycle Stroke in mm\nvc=2*10**-3 #Clearance volume in mm**3\ngamma=1.4\npi=3.142\n\n#calculation\nvs=(pi/4)*(D**2*L) #Swept volume\nr=((vs+vc)/vc) #Compression ratio\neta=100*(1-(1/(r**(gamma-1)))) #Standard efficiency\n\n#output\nprint(\"Swept volume=\",round(vs,6),\"m**3\")\nprint(\"compression ratio=\",round(r,3),)\nprint(\"standard efficiency=\",round(eta,1),\"%\")\n\n", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "Swept volume= 0.014139 m**3\ncompression ratio= 8.07\nstandard efficiency= 56.6 %\n" + } + ], + "prompt_number": 5 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Example 8 Page No:309" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "#Input data\nP1=0.1*10**6 #Otto cycle air\nT1=35+273 #Otto cycle temp degree celsius\nr=9 #Compression ratio\nQs=1800 #Supplied heat in kJ/kg\nv1=9 \nv2=1\nR=0.287*10**3\ngamma=1.4\nCv=0.718\n\n\n\n#calculation\nT2=(T1*((v1/v2)**(gamma-1))) #Temperature at point 2 in K\nP2=(P1*((v1/v2)**1.4))*10**-6 #pressure at point 2 in MPa \nT3=((Qs/Cv)+(T2)) #Max temp of cycle in degree celsius\nP3=(T3/T2*P2) #Max pressure of cycle in MPa\neta=100*(1-(1/(r**(gamma-1))))#Otto cycle thermal efficiency in %\nWD=(Qs*eta)*10**-2 #Work done during the cycle in KJ/Kg\nv1=((R*T1)/P1) #Char gass equation in m**3/Kg\nv2=v1/r #Char gass equation in m**3/Kg\nSv=v1-v2 #Swept volume in m**3/Kg\nPme=(WD/Sv)*10**-3 #Mean effective pressure in MPa\nalpha=P3/P2 #Explosion ratio\nPm=(((P1*r)/((r-1)*(gamma-1)))*(((r**(gamma-1))-1)*(alpha-1)))*10**-6#Mean effective pressure in MPa\n\n\n#Output\nprint(\"Temperature at point=\",round(T2,1),\"K\")\nprint(\"pressure at point=\",round(P2,3),\"MPa\")\nprint(\"Max temp of cycle=\",round(T3,3),\"K\")\nprint(\"Max pressure= \",round(P3,1),\"MPa\")\nprint(\"Otto cycle thermal efficiency=\",round(eta,1),\"%\")\nprint(\"Work done during the cycle=\",round(WD,),\"J/Kg\")\nprint(\"Char gass equation=\",round(v1,3),\"m**3/Kg\")\nprint(\"Char gass equation=\",round(v2,4),\"m**3/Kg\")\nprint(\"Swept volume=\",round(Sv,4),\"m**3/Kg\")\nprint(\"Mean effective pressure=\",round(Pme,2),\"MPa\")\nprint(\"Explosion ratio=\",round(alpha,2))\nprint(\"Mean effective pressure=\",round(Pm,2),\"MPa\")\n\n\n", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "Temperature at point= 741.7 K\npressure at point= 2.167 MPa\nMax temp of cycle= 3248.697 K\nMax pressure= 9.5 MPa\nOtto cycle thermal efficiency= 58.5 %\nWork done during the cycle= 1053 J/Kg\nChar gass equation= 0.884 m**3/Kg\nChar gass equation= 0.0982 m**3/Kg\nSwept volume= 0.7857 m**3/Kg\nMean effective pressure= 1.34 MPa\nExplosion ratio= 4.38\nMean effective pressure= 1.34 MPa\n" + } + ], + "prompt_number": 165 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Example 9 Page No:311" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "#Input data\nP1=0.1 #Beginning compression in MPa\nT1=40+273 #Beginning temp in degree celsius\neta=0.55 #Standard effeciency in %\nQR=540 #Rejected heat in KJ/Kg\nr=7.36 #Compression ratio\n\n\n#calculation\n#eta=(1-(1/(r**(gamma-1))))\nQS=(-QR/(eta-1)) #Heat supplied/unit mass in KJ/Kg\nWD=QS-QR #Work done per Kg of air in KJ/Kg\nT2=T1*(r**(gamma-1)) #Temp at end of compression in K\nP2=P1*((r)**gamma) #pressure at point 2 in MPa\nT3=(QS/Cv)+T2 #max temp of the cycle in K\nP3=(T3/T2)*P2 #max pressure of the cycle in MPa\n\n#output\nprint(\"Heat supplied/unit mass=\",round(QS,),\"KJ/Kg\")\nprint(\"Work done per Kg of air= \",round(WD,),\"KJ/Kg\")\nprint(\"Temp at end of compression=\",round(T2,1),\"K\")\nprint(\"pressure at point two=\",round(P2,3),\" MPa\")\nprint(\"max temp of the cycle=\",round(T3,1),\"K\")\nprint(\"max pressure of the cycle=\",round(P3,3),\" MPa\")\n", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "Heat supplied/unit mass= 1200 KJ/Kg\nWork done per Kg of air= 660 KJ/Kg\nTemp at end of compression= 695.5 K\npressure at point two= 1.635 MPa\nmax temp of the cycle= 2366.8 K\nmax pressure of the cycle= 5.565 MPa\n" + } + ], + "prompt_number": 64 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Example 10 Page No:312" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "#Input data\nT1=300 #Initial temp in K\nT3=2500 #Final temp in K\nP1=1 #Initial pressure in N/m**2\nP3=50 #Final pressure in N/m**2\ngamma=1.4\nCv=0.718\n\n#calculation\nr=(P3*T1)/(P1*T3) #Compression ratio\neta=(1-(1/r**(gamma-1))) #Standard effeciency in %\nT2=T1*((P3/P1)**((gamma-1)/gamma)) #Middle temperature in K\nQs=Cv*(T3-T2) #Heat supplied in KJ/Kg\nWD=eta*Qs #Work done KJ/Kg\n\n#output\nprint(\"Compression ratio=\",r,\"\")\nprint(\"Standard effeciency=\",round(eta,4),\"%\")\nprint(\"Middle temperature=\",round(T2,2),\"K\")\nprint(\"Heat supplied=\",round(Qs,2),\"KJ/Kg\")\nprint(\"Work done=\",round(WD,1),\"KJ/Kg\")", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "Compression ratio= 6.0 \nStandard effeciency= 0.5116 %\nMiddle temperature= 917.36 K\nHeat supplied= 1136.33 KJ/Kg\nWork done= 581.4 KJ/Kg\n" + } + ], + "prompt_number": 84 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Example 11 Page No:316" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "#input data\nr=18 #compression ratio of diesel engine\nK=6 #cut-off ratio of the stroke in%\nrho=2.02 \n\n#calculation\n#diesel engine air standard efficiency\neta=100*((1-(1/r**(gamma-1)))*(1/gamma*(rho**(gamma-1)/(rho-1))))\n\n#output\nprint(\"diesel engine air standard efficiency\",round(eta,1),\"%\")\n", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "diesel engine air standard efficiency 63.6 %\n" + } + ], + "prompt_number": 87 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Example 12 Page No:317" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "#Input Data\nr=22 #compression ratio of diesel engine r=v1/v2\nr1=11 #expansion ratio r1=v4/v3\ngamma=1.4\nrho=1.4\n\n#calculation\nrho=r/r1 #cut-off ratio\n#diesel engine air standard efficiency \neta=100*((1-(1/r**(gamma-1)))*(1/gamma*(rho**(gamma-1)/(rho-1))))\n\n#output\nprint(\"cut-off ratio=\",rho,)\nprint(\"diesel engine air standard efficiency=\",round(eta,2),\"%\")", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "cut-off ratio= 2.0\ndiesel engine air standard efficiency= 66.88 %\n" + } + ], + "prompt_number": 88 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Example 13 Page No:317" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "#Input data\nVc=10/100 #Clearance volume in % \nVs=Vc/0.1 \nK=0.05 #Cut-off of the strok in \ngamma=1.4\n\n#Calculation\nr=((Vs+Vc)/(Vc)) #Compression ratio\nrho=1+K*(r-1) #Cut-off ratio\n#Effeciency in %\neta=(1-(1/r**(gamma-1))*((1/gamma)*(((rho**(gamma))-1)/(rho-1))))*100\n\n#output\nprint(\"Compression ratio=\",r,\"Vs\")\nprint(\"Cut-off ratio=\",rho,)\nprint(\"Effeciency=\",round(eta,2),\"%\")", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "Compression ratio= 11.0 Vs\nCut-off ratio= 1.5\nEffeciency= 58.17 %\n" + } + ], + "prompt_number": 107 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Example 14 Page No:" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "#Input data\nT1=50+273 #Temperature at the beginning of the compression\nT2=700+273 #Temperature at the end of the compression\nT3=2000+273 #Temperature at the beginning of the expansion\n\n\n#Calculation\nr=((T2/T1)**(1/(gamma-1))) #Compression ratio \nrho=(T3/T2) #Cut-off ratio\nK=((rho-1)/(r-1)) #Also cut-off ratio\n#Air standard efficiency\neta=(1-(1/r**(gamma-1))*((1/gamma)*(((rho**(gamma))-1)/(rho-1))))*100\n\n#Output\nprint(\"compression ratio=\",round(r,2),\"\")\nprint(\"cut-off ratio=\",round(rho,3),)\nprint(\"also cut-off ratio=\",round(K,2),\"\")\nprint(\"air standard efficiency=\",round(eta,2),\"%\")", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "compression ratio= 15.75 \ncut-off ratio= 2.336\nalso cut-off ratio= 0.09 \nair standard efficiency= 59.54 %\n" + } + ], + "prompt_number": 122 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Example 15 Page No:317" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "#Input data\nP1=0.1 #Diesel cycle is supplied# with air in MPa\nT1=40+273 #Diesel cycle is supplied with temperature in degree celsius \nr=18 #Compression ratio\nQs=1500 #Heat supplied\nv1=18\nv2=1\nCp=1.005\n\n\n#Calculation\nT2=T1*((v1/v2)**(gamma-1)) #For isentropic process the temperature is\nP2=P1*((v1/v2)**(gamma)) #For isentropic process the pressure is\nT3=(Qs/Cp)+T2 #Maximum temperatureof the cycle\nrho=T3/T2 #Cut-off ratio\n#Air standard efficiency\neta=(1-(1/r**(gamma-1))*((1/gamma)*(((rho**(gamma))-1)/(rho-1))))*100\nNWD=(Qs*eta)*10**-2 #Net work done\n\n#Output\nprint(\"for isentropic process the temperature=\",round(T2,1),\"K\")\nprint(\"for isentropic process the pressure=\",round(P2,2),\"MPa\")\nprint(\"maximum temperatureof the cycle=\",round(T3,2),\"K\")\nprint(\"cut-off ratio=\",round(rho,1),\"MPa\")\nprint(\"air standard efficiency=\",round(eta,2),\"%\")\nprint(\"net work done=\",round(NWD,),\"KJ/Kg\")\n\n", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "for isentropic process the temperature= 994.6 K\nfor isentropic process the pressure= 5.72 MPa\nmaximum temperatureof the cycle= 2487.15 K\ncut-off ratio= 2.5 MPa\nair standard efficiency= 60.93 %\nnet work done= 914 KJ/Kg\n" + } + ], + "prompt_number": 130 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Example 16 Page No:317" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "#Input data\nr=14 #compression ratio of standard diesel cycle\nP1=1 #compression stroke in bar\nT1=300 #temperature of air in k\nT3=2774 #temperature rises in k\nCP=1.005\nv1=14\nv2=1\ngamma=1.4\nQs=1921.43\nR=0.287*10**3\n\n\n#calculation\nT2=T1*((v1/v2)**(gamma-1)) #constant pressure\nrho=T3/T2 #cut-off ratio\neta=(1-(1/r**(gamma-1))*((1/gamma)*(((rho**(gamma))-1)/(rho-1))))*100 #air standard efficiency\nHS=(CP*(T3-T2)) #heat supplied\nWD=(Qs*eta)*10**-2 #Net work done\nv1=(R*T1/P1) *10**-5 #characteristics gas equation\nv2=(v1/r ) #characteristics gas equation\nSv=(v1-v2) #Swept volume\nPme=(WD/Sv )*10**-2 #Mean effective pressur\nPm=((P1*r)/((r-1)*(gamma-1)))*((gamma*(r**(gamma-1)))*(rho-1)-((rho**(gamma))-1))# mean effective pressure \n\n\n#output\nprint(\"constant pressure=\",round(T2,2),\"K\")\nprint(\"cut-off ratio= \",round(rho,2),)\nprint(\"air standard efficiency=\",round(eta,2),\"%\")\nprint(\"heat supplied= \",round(HS,2),\"KJ/Kg\")\nprint(\"Net work done= \",round(WD,2),\"KJ/Kg\")\nprint(\"characteristics gas equation= \",round(v1,3),\"m**3/Kg\")\nprint(\"characteristics gas equation= \",round(v2,4),\"m**3/Kg\")\nprint(\"Swept volume= \",round(Sv,4),\"m**3/Kg\")\nprint(\"Mean effective pressure= \",round(Pme,1),\"bar\")\nprint(\"Mean effective pressure= \",round(Pm,1),\"bar\")\n", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "constant pressure= 862.13 K\ncut-off ratio= 3.22\nair standard efficiency= 53.65 %\nheat supplied= 1921.43 KJ/Kg\nNet work done= 1030.91 KJ/Kg\ncharacteristics gas equation= 0.861 m**3/Kg\ncharacteristics gas equation= 0.0615 m**3/Kg\nSwept volume= 0.7995 m**3/Kg\nMean effective pressure= 12.9 bar\nMean effective pressure= 12.9 bar\n" + } + ], + "prompt_number": 155 + }, + { + "cell_type": "code", + "collapsed": false, + "input": "", + "language": "python", + "metadata": {}, + "outputs": [] + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Basic_mechanical_engineering_by_Basant_Agrawal_,_C.M_Agrawal/Chapter_2_Properties_Of_Material.ipynb b/Basic_mechanical_engineering_by_Basant_Agrawal_,_C.M_Agrawal/Chapter_2_Properties_Of_Material.ipynb new file mode 100644 index 00000000..67b98032 --- /dev/null +++ b/Basic_mechanical_engineering_by_Basant_Agrawal_,_C.M_Agrawal/Chapter_2_Properties_Of_Material.ipynb @@ -0,0 +1,146 @@ +{ + "metadata": { + "name": "Chapter 2 Properties Of Material" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": "Chapter 2 Properties Of Material" + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Example 1 Page No:19" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "#Input data\nL=5 #Length of steel bar in m\nd=25*10**-3 #Diametr of steel bar in mm\ndeltaLt=25*10**-3 #Steel \npt=800 #Power load of steel bar in N\n\n\n#Calculation\nA=((pi/4)*((deltaLt)**2)) #Cross-section area\nsigmat=(pt)/(A) #Stress in steel bar\net=(deltaLt)/L #Strain in steel bar\nE=((sigmat)/(et)) #Young's modulus\n\n\n#Output\nprint(\"value of Cross-section area=\",A,\"m**2\")\nprint(\"value of tress in steel bar=\",sigmat,\"MN/m**2\")\nprint(\"value of strain in steel bar= \",et)\nprint(\"value of Young's modulus= \",E,\"N/m**2\")\n\n\n", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "value of Cross-section area= 0.0004919062500000002 m**2\nvalue of tress in steel bar= 1626326.154628041 MN/m**2\nvalue of strain in steel bar= 0.005\nvalue of Young's modulus= 325265230.92560816 N/m**2\n" + } + ], + "prompt_number": 15 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Example 2 Page No:20\n" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "#Input data\nL=300*10**-3 #Length of hexagonal prismatic steel bar in mm\nA=500*10**-6 #Area of cross section of steel bar mm**2\nPt=500*10**3 #Load of steel bar in KN\nE=210*10**9 #Modulus of elasticity GN/m**2\n\n#Calculation\nsigmat=((Pt)/(A)) #Stress in steel bar\net=((sigmat)/(E)) #Strain steel bar is\ndeltaLt=((et)*(L)) #Therefore,elongation of the steel bar is given by\n\n#Output\nprint('stress in steel bar=',sigmat,\"N/m**2\")\nprint('therefore,strain steel bar is given by=',et,)\nprint('therefore,elongation of the steel bar is given by=',deltaLt,\"m\")\n\n\n", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "stress in steel bar = 1000000000.0 N/m**2\ntherefore,strain steel bar is given by = 0.004761904761904762\ntherefore,elongation of the steel bar is given by= 0.0014285714285714286 m\n" + } + ], + "prompt_number": 15 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Example 3 Page No:21\n" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "#Input Data\nPt=600 #Tensils force in N\nd=2*10**-3 #Diameter of steel wire in mm\nL=15 #Length of wire in m\nE=210*10**9 #Modulus of elasticity of the material in GN/M**2\npi=3.1482\n\n\n#Calculation\nA=((pi/4)*((d)**2)) #(1)cross section area\nsigmat=(Pt)/(A) #stress in the steel wire \net=((sigmat)/(E)) #(2)Therefore, strain in steel wire is given by\ndeltaLt=et*L #(3)Enlongation of the steel wire is given by \npe=((deltaLt/L)*100) #(4)Percentage elongation\n\n\n#Output\nprint(\"cross section area= \",A,\"m**2\")\nprint(\"stress in the steel wire=\",sigmat,\"GN/m**2\")\nprint(\"modulus of elasticity=\",et)\nprint(\"strain in steel wire=\",deltaLt,\"mm\")\nprint(\"percentage elongation= \",pe,\"%\")\n\n\n\n", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "cross section area= 3.1481999999999998e-06 m**2\nstress in the steel wire= 190585096.24547362 GN/m**2\nmodulus of elasticity= 0.0009075480773593982\nstrain in steel wire= 0.013613221160390973 mm\npercentage elongation= 0.09075480773593982 %\n" + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Example 4 Page No:22\n" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "#Input data\nA=30*30*10**-6 #Area of square rod in mm**2\nL=5 #Length of square rod in m\nPc=150*10**3 #Axial comperessive load of a rod in kN\nE=215*10**9 #Modulus of elasticity in GN/m**2\n\n\n#Calculation\nsigmac=((Pc)/(A)) #Stress in square rod\nec=((sigmac)/(E)) #Modulusof elasticity is E=sigmac/ec ,therefore strain in square rod is\ndeltaLc=ec*5 #Therefore shortening of length of the rod \n\n\n#Output\nprint (\"stress in square rod\",sigmac,\"N/m**2\")\nprint(\"strain in square rod ec=\",ec,)\nprint(\"shortening of length of the rod=\",deltaLc,\"m\")", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "stress in square rod 166666666.66666666 N/m**2\nstrain in square rod ec= 0.0007751937984496124\nshortening of length of the rod= 0.003875968992248062 m\n" + } + ], + "prompt_number": 25 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Example 5 Page No:23" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "#input data\nd=50*10**-6 #Diameter of metalic rod in mm**2\nL=220*10**-3 #Length of metalic rod in mm\nPt=40*10**3 #Load of metalic rod in KN\ndeltaLt=0.03*10**-3 #Elastic enlongation in mm\nypl=160*10**3 #Yield point load in KN\nml=250*10**3 #Maximum load in KN\nlsf=270*10**-3 #Length of specimen at fracture in mm\npi=3.1482\n\n#calculation\nA=(((pi)/(4)*((d)**2))) #(1)Cross section area\nsigmat=(Pt/A) #Stress in metallic rod\net=(deltaLt/L) #Strain n metallic rod\nE=(sigmat/et) #Young's modulus\nys=(ypl/A) #(2)Yeild strength\nuts=(ml/A) #(3)Ultimate tensile strength\nPebf=((lsf-L)/L)*100 #Percentage elongation before fracture \n\n\n\n#output\nprint(\"cross section area\",A,\"m**2\")\nprint(\"stress in metallic rod\",sigmat,\"N/m**2\")\nprint(\"strain n metallic rod\",et,)\nprint(\"young's modulus\",E,\"GN/m**2\")\nprint(\"yeild strength\",ys,\"MN/m**2\")\nprint(\"ultimate tensile strength\",uts,\"MN/m**2\")\nprint(\"percentage elongation before fracture\",Pebf,\"%\")\n", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "cross section area 1.967625e-09 m**2\nstress in metallic rod 20329076932850.52 N/m**2\nstrain n metallic rod 0.00013636363636363637\nyoung's modulus 1.4907989750757046e+17 GN/m**2\nyeild strength 81316307731402.08 MN/m**2\nultimate tensile strength 127056730830315.75 MN/m**2\npercentage elongation before fracture 22.727272727272734 %\n" + } + ], + "prompt_number": 5 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Example 6 Page No:24\n" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "#Input data\nA=50*50*10**-6 #Area ofsquare metal bar in mm**2\nPc=600*10**3 #Axial compress laod in KN\nL=200*10**-3 #Gauge length of metal bar in mm\ndeltaLc=0.4*10**-3 #Contraction length of metal bar in mm\ndeltaLlateral=0.05*10**-3 #Lateral length of metal bar in mm\n\n#Calculation\nsigmac=((Pc)/(A)) #Stress in square metal bar \nec=((deltaLc)/(L)) #Longitudinal or linear strain in square metal bar\nE =((sigmac)/(ec)) #Smodule of elasticity\nelateral=((deltaLlateral)/(L)) #Lateral strain in square metal bar\npoissonsratio=(elateral)/(ec)\n\n\n#Output\nprint(\"stress in bar=\",sigmac,\"n/m**2\")\nprint(\"longitudinal or linear strain in square metal bar=\",ec,)\nprint(\"module of elasticity=\",E,\"N/m**2\")\nprint(\"lateral strain in square metal bar=\",elateral,)\nprint(\"poissons ratio=\",poissonsratio,)\n\n", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "stress in bar= 240000000.0 n/m**2\nlongitudinal or linear strain in square metal bar= 0.002\nmodule of elasticity= 120000000000.0 N/m**2\nlateral strain in square metal bar= 0.00025\npoissons ratio= 0.125\n" + } + ], + "prompt_number": 12 + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Basic_mechanical_engineering_by_Basant_Agrawal_,_C.M_Agrawal/Chapter_5_Metrology.ipynb b/Basic_mechanical_engineering_by_Basant_Agrawal_,_C.M_Agrawal/Chapter_5_Metrology.ipynb new file mode 100644 index 00000000..4528c633 --- /dev/null +++ b/Basic_mechanical_engineering_by_Basant_Agrawal_,_C.M_Agrawal/Chapter_5_Metrology.ipynb @@ -0,0 +1,83 @@ +{ + "metadata": { + "name": "Chapter 5 Metrology" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": "Chapter 5 Metrology" + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Example 1 Page No:81" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "#Input data\nMSR=3.2 #Main scale reading of cylindrical rod in cm\nNCD=7 #Number of coinciding Vernier Scale division \nLc=0.1*10**-3 #Least count of the instrument in mm\n\n#Calculation\nDOR=MSR+(NCD*Lc) #Diameter of the rod\n\n#Output\nprint(\"Diameter of the rod= \",round(DOR,3),\"cm\")\n\n", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "Diameter of the rod= 3.201 cm\n" + } + ], + "prompt_number": 19 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Example 2 Page No:82" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "#Input data\nMSR=5.3 #Main scale reading of prismatic bar in cm\nNCD=6 #Number of coinciding Vernier Scale division \nLc=0.1*10**-3 #Least count of the instrument in mm \nNe=(-0.2*10**-3) #Instrument bears a nagative error in mm\n\n#Calulation\nMlb=MSR+(NCD*Lc) #Measured length of the bar in cm\nTlb=(Mlb-(Ne)) #True length of the bar in cm\n\n\n#Output\nprint(\"Measured length of the bar= \",round(Mlb,3),\"cm\")\nprint(\"true length of the bar= \",round(Tlb,3),\"cm\")\n\n\n", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "Measured length of the bar= 5.301 cm\ntrue length of the bar= 5.301 cm\n" + } + ], + "prompt_number": 20 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Example 3 Page No:88 " + }, + { + "cell_type": "code", + "collapsed": false, + "input": "#Input data\nimport math\nL=100 #Height of sine bar\ntheta=12.8 #angle in degree minut\n#Z=sin(theta)=0.22154849\nZ=0.22154849\n\n#Calculation\nb=Z*L #Height required to setup in mm\n\n\n#Output\nprint(\"Height required=\",round(b,4),\"mm\")\n ", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "Height required= 22.1548 mm\n" + } + ], + "prompt_number": 18 + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Basic_mechanical_engineering_by_Basant_Agrawal_,_C.M_Agrawal/Chapter_7_Fluid_Mechanics.ipynb b/Basic_mechanical_engineering_by_Basant_Agrawal_,_C.M_Agrawal/Chapter_7_Fluid_Mechanics.ipynb new file mode 100644 index 00000000..2cb36b62 --- /dev/null +++ b/Basic_mechanical_engineering_by_Basant_Agrawal_,_C.M_Agrawal/Chapter_7_Fluid_Mechanics.ipynb @@ -0,0 +1,461 @@ +{ + "metadata": { + "name": "Chapter 7 Fluid Mechanics" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": "Chapter 7 Fluid Mechanics" + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Example 1 Page No:113" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "#Input data\nV=5 #volume of the liquid in m**3\nW=45*10**3 #weight of the liquid in KN\ng=9.81 #acceleration due to gravity in m/s**2\nrhow=1000 #constant value\n\n#Calculation\nm=((W)/(g)) #mass in Kg\nrho=(m/V) #Mass density in kg/m**3\nw=(W/V) #Weight Density in N/m**3\nv=(V/m) #Specific volume in m**3/kg\nS=rho/rhow #Specific gravity\n \n \n#Output\nprint(\"mass= \",round(m,2),\"Kg\")\nprint(\"Mass density= \",round(rho,2),\"kg/m**3\")\nprint(\"Weight Density= \",w,\"N/m**3\")\nprint(\"Specific volume= \",v,\"m**3/kg\")\nprint(\"Specific gravity= \",round(S,4),)\n\n\n", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "mass= 4587.16 Kg\nMass density= 917.43 kg/m**3\nWeight Density= 9000.0 N/m**3\nSpecific volume= 0.00109 m**3/kg\nSpecific gravity= 0.9174\n" + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Example 2 Page No:114" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "#Input data\nV=3*10**-3 #3l of oil in m**3\nW=24 #Weight of oil in N\ng=9.81 #Gravity in m/s**2\nrohw=1000 #Constant value\n\n\n#Calculation\nm=((W)/(g)) #Mass in Kg\nrho=(m/V) #Mass density in kg/m**3\nw=(W/V) #Weight Density in N/m**3\nv=(V/m) #Specific volume in m**3/kg\nS=rho/rhow #Specific gravity\n \n#Output\nprint(\"mass= \",round(m,3),\"Kg\")\nprint(\"Mass density= \",round(rho,1),\"kg/m**3\")\nprint(\"Weight Density= \",w,\"N/m**3\")\nprint(\"Specific volume= \",round(v,7),\"m**3/kg\")\nprint(\"Specific gravity= \",round(S,4),)\n\n\n ", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "mass= 2.446 Kg\nMass density= 815.5 kg/m**3\nWeight Density= 8000.0 N/m**3\nSpecific volume= 0.0012263 m**3/kg\nSpecific gravity= 0.8155\n" + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Example 3 Page No:114" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "#Input data\nS=0.85 #Specific gravity of a liquid\ng=9.81 #Acceleration due to gravity in m/s**2(constant)\nrhow=1000 #Constant value\n\n\n#Calculation\n#Specific gravity S=roh/rohw \nrho=S*rhow #Mass density in Kg/m**3\nw=rho*g #Weight Density in N/m**3\nv=(1/rho) #Specific volume in m**3/kg\n\n\n#Output\nprint(\"Mass densit= \",rho,\"Kg/m**3\")\nprint(\"Weight Density= \",w,\"N/m**3\")\nprint(\"Specific volume= \",round(v,6),\"m**3/kg\")\n\n\n", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "Mass densit= 850.0 Kg/m**3\nWeight Density= 8338.5 N/m**3\nSpecific volume= 0.001176 m**3/kg\n" + } + ], + "prompt_number": 50 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Example 4 Page No:116" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "#Input data\ndy=21*10**-3 #Horizontal plates in mm\ndu=1.4 #Relative velocity between the plates in m/s\nmu=0.6 #Oil of viscosity 6 poise in Ns/m**2\n\n#Calculation\ntau=(mu*(du/dy)) #Shear in the oil in N/m**2\n\n#Output\nprint(\"shear in the oil= \",round(tau,),\"N/m**2\")\n", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "shear in the oil= 40 N/m**2\n" + } + ], + "prompt_number": 7 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Example 5 Page No:116" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "#Input data\nv=4*10**-4 #kinematic viscosity is 4 stoke inm**2/s\nS=1.2 #specific gravity\ndow=1000 #density of water Kg/m**3\n\n\n#Calculation\nrho=S*dow \nvol=rho*v #viscosity of the liquid in Ns/m**2 or poise\n\n#Output\nprint(\"viscosity of the liquid= \",round(vol,2),\"Ns/m**2 \")", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "viscosity of the liquid= 0.48 Ns/m**2 \n" + } + ], + "prompt_number": 9 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Example 6 Page No:6" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "#Input data\nS=0.9 #Specific gravity\ntau=2.4 #shear stress in N/m**2\n(vg)=0.125 #velocitty gradientin per s\ndow=1000 #density of water Kg/m**3\n\n\n#Calculation\nmu=(tau)/(vg) #newton's law of viscosity in shear stress in Ns/m**2\nrho=S*dow #Density of oil in Kg/m**3\nv=(mu/rho) #Kinematic viscosity in m**2/s or stoke\n\n#Output\nprint(\"newton's law of viscosity in shear stress= \",mu,\"Ns/m**2\")\nprint(\"Density of oil= \",rho,\"Kg/m**3\")\nprint(\"Kinematic viscosity= \",round(v,5),\"m**2/s or stoke\")\n", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "newton's law of viscosity in shear stress= 19.2 Ns/m**2\nDensity of oil= 900.0 Kg/m**3\nKinematic viscosity= 0.02133 m**2/s or stoke\n" + } + ], + "prompt_number": 11 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Example 7 Page No:117" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "#Input data\nA=6*10**-2 #Space between two square plates in mm\ndy=8*10**-3 #Thickness of fluid in mm\nu1=0 #Lower pate is stationary\nu2=2.4 #Upper plate in m/s\nF=5 #Speed of force in N\ns=1.6 #Specific gravity of the liquid\ndow=1000 #Density of water Kg/m**3\n\n\n#(1)Calculation\ndu=u2-u1 #change in velocity in m/s\ntau=(F/((A)**2)) #shear stress N/m**2\nmu=(tau/(du/dy)) #Newton's law of viscosity in Ns/m**2 or poise\nrho=s*dow #Density of oil in kg/m**3\nv=(mu/rho) #kinematic viscosity is given by m**2/s or stoke\n\n\n#Output\nprint(\"change in velocity= \",du,\"m/s\")\nprint(\"shear stress= \",round(tau,2),\"N/m**2\")\nprint(\"Newton's law of viscosity= \",round(mu,1),\"Ns/m**2 \")\nprint(\"Density of oil= \",rho,\" kg/m**3\")\nprint(\"kinematic viscosity= \",round(v,4),\"m**2/s \")\n\n\n", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "change in velocity= 2.4 m/s\nshear stress= 1388.89 N/m**2\nNewton's law of viscosity= 4.6 Ns/m**2 \nDensity of oil= 1600.0 kg/m**3\nkinematic viscosity= 0.0029 m**2/s \n" + } + ], + "prompt_number": 20 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Example 8 Page No:118" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "#Input data\ndy=1.5*10**-4 #Two horizontal plates are placed in m\nmu=0.12 #Space between plates Ns/m**2\nA=2.5 #Upper area is required to move in m**2\ndu=0.6 #Speed rerlated to lower plate in m/s\n\n\n#(1)Calculation\ntau=(mu*(du/dy)) #Shear stress N/m**2\nF=tau*A #Force in N\nP=F*du #Power required to maintain the speed of upper plate in W\n\n\n#Output \nprint(\"Shear stress= \",round(tau,),\"N/m**2\")\nprint(\"Force= \",round(F,),\"N\")\nprint(\"Power required to maintain the speed of upper plate= \",round(P,),\"W\")\n", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "Shear stress= 480 N/m**2\nForce= 1200 N\nPower required to maintain the speed of upper plate= 720 W\n" + } + ], + "prompt_number": 51 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Example 9 Page No 118" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "#Input data \nmu=0.1 #Oil of viscosity used for lubricant in poise or Ns/m**2\nD=0.15 #Clearance between the shaft of diameter in m\ndy=3*10**-4 #Clearance in m\nN=90 #Shaft rorates in rpm\npi=3.14\n\n\n#Calculation\ndu=((pi*D*N)/60) #Tangential speed of shaft in m/s\ntau=(mu*(du/dy)) #The shear force in N/m**2\n\n#Output\nprint(\"Tangential speed of shaft= \",du,\"m/s\")\nprint(\"The shear force= \",tau,\"N/m**2\")\n\n\n\n", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "Tangential speed of shaft= 0.7065 m/s\nThe shear force= 235.5 N/m**2\n" + } + ], + "prompt_number": 52 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Example 10 Page No:119" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "#Input data\nimport math\nA=120*10**-3 #Side of square plate in mm\nW=30 #Side weight in N\ndu=3.75 #Uniform velocity in m/s\ntheta=30 #Lubricated inclined plane making an angle in degree at horizontal\ndy=6*10**-3 #Thickness lubricating oil film in mm\nrho=800 #Lubricating oil film density in Kg/m**3\n\n\n#Calculation\nsin30=0.5 \nF=W*sin30 #Component of force in N\ntau=(F/(A**2)) #Shear stress in Ns/m**2 \nmu=(tau/(du/dy)) #From Newton's law of Shear stress in Ns/m**2 \nV=(mu/rho)*10**3 #Kinematic viscosity in m**2/s\n\n\n#Output\nprint(\"Component of force= \",F,\"N\")\nprint(\"Shear stress= \",round(tau,2),\" Ns/m**2 \")\nprint(\"From Newton's law of Shear stress= \",round(mu,3),\"Ns/m**2\")\nprint(\"Kinematic viscosity= \",round(V,3),\"m**2/s\")\n", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "Component of force= 15.0 N\nShear stress= 1041.67 Ns/m**2 \nFrom Newton's law of Shear stress= 1.667 Ns/m**2\nKinematic viscosity= 2.083 m**2/s\n" + } + ], + "prompt_number": 53 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Example 11 Page No 121" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "#Input data \nZ=15 #Pressure due to column in m\nS=0.85 #Oil of specific gravity\ng=9.81 #Gravity\n\n\n\n#Calculation\nrho=S*10**3 #Density of oil in kg/m**3\nP=rho*g*Z #Pressure in N/m**2 or kPa\n\n\n#Output\nprint(\"Density of oil= \",rho,\"kg/m**3\")\nprint(\"Pressure= \",P,\"N/m**2\")", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "Density of oil= 850.0 kg/m**3\nPressure= 125077.5 N/m**2\n" + } + ], + "prompt_number": 55 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Example 12 Page No 122" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "#Input data\nZ1=1.5 #open tank contain water in m\nZ2=2.5 #oil of specific gravity for depth in m\nS=0.9 #oil of specific gravity \nrho1=1000 #density of water in Kg/m**3\nrho2=S*10**3 #density of oil in Kg/m**3\ng=9.81 #gravity\n\n\n\n#calculation\nP=rho1*g*Z1+rho2*g*Z2 #intensity of pressure in kPa\n\n\n#output\nprint(\"intensity of pressure=\",P,\"N/m**2\")", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "intensity of pressure= 36787.5 N/m**2\n" + } + ], + "prompt_number": 57 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Example 13 Page No:124" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "#Input data\nD1=0.2 #Diameter of pipe section 1 in m\nD2=0.3 #Diameter of pipe section 2 in m\nV1=15 #Velocity of water in m/s\npi=3.14\n\n#calculation\nQ=((3.14/4)*(0.2)**2)*15 #Discharge through pipe in m**3/s\nV2=(((3.14/4)*(0.2)**2)*15)/((3.14/4)*(0.3)**2) #velocity of section2 in m/s\n\n\n#Output\nprint(\"Discharge through pipe= \",round(Q,2),\"m**3/s\")\nprint(\"velocity of section2= \",round(V2,2),\"m/s\")", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "Discharge through pipe= 0.47 m**3/s\nvelocity of section2= 6.67 m/s\n" + } + ], + "prompt_number": 59 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Example 14 Page No:126" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "#Input data\nV=13 #Velocity of water flowing throgh pipe in m/s\nP=200*10**3 #Pressure of water in Kpa\nZ=25 #Height above the datum in m\ng=9.81\nrho=1000\n\n\n#Calculation\nE=(P/(rho*g))+((V**2)/(2*g))+(Z) #Total energy per unit weight in m\n\n\n#Output\nprint(\"Total energy per unit weight=\",round(E,),\"m\")\n", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "Total energy per unit weight= 54 m\n" + } + ], + "prompt_number": 67 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Example 15 Page No:127" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "#Input data\nimport math\nS=0.85 #Specific gravity of oil\nD=0.08 #Diameter of pipe in m\nP=1*10**5 #Intenity of presssure in N/m**2\nZ=15 #Total energy bead in m\nE=45 #Datum plane in m\nMdw=1*10**3 #Mass density of water constant\ng=9.81 #Gravity constant\nrho=S*Mdw #Mass density of oil\npi=3.14\n\n#calculation\nrho=S*Mdw #Mass density of oil\n#E=(P/(rho*g))+((V**2)/(2*g))+(Z)\nV=math.sqrt((E-((P/(rho*g))+Z))*(2*g)) #Total energy per unit weight in m/s\nQ=(pi/4)*D**2*V #Discharge in m**3/Kg\"\n\n#output\nprint(\"mass density of oil= \",rho,\"Kg/m**3\")\nprint(\"Total energy per unit weight= \",round(V,1),\"m/s\")\nprint(\"discharge=\",round(Q,4),\"m**3/Kg\")\n", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "mass density of oil= 850.0 Kg/m**3\nTotal energy per unit weight= 18.8 m/s\ndischarge= 0.0944 m**3/Kg\n" + } + ], + "prompt_number": 64 + }, + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": "Example 16 Page No:127" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "#input data\n#refer figure 11\nZA=2 #water flows section A-A in m \nDA=0.3 #datum pipe diameter at section A-A in m\nPA=550*10**3 #pressure in kPa\nVA=6 #flow velocity in m/s\nZB=18 #water flows at section B-B in m\nDB=0.15 #datum pipe diameter at section B-B in m\npi=3.14 #constant\nrho=1000 #constant\ng=9.81 #constant\nAa=(pi/4)*(DA)**2\nAb=(pi/4)*(DB)**2\npi=3.14\n\n#calculation\nVB=((Aa*VA)/Ab) #continuity discharge equation in m/s\n#bernoulli's equation Kpa\n#(PA/rho*g)+(VA**2/2*g)+ZA=(PB/rho*g)+(VB**2/2*g)+ZB \nPB=(((PA/(rho*g))+(VA**2/(2*g))+ZA)-((VB**2/(2*g))+ZB))*(rho*g)\n\n\n#output\nprint(\"continuity discharge equation= \",VB,\"m/s\")\nprint(\"bernoulli's equation= \",round(PB,1),\"pa\")\n\n\n", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "continuity discharge equation= 24.0 m/s\nbernoulli's equation= 123040.0 pa\n" + } + ], + "prompt_number": 66 + }, + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": "Example 17 Page No:128" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "#input data\n#refer figure 12\nQ=0.04 #Water flows at rate in m**2/s\nDA=0.22 #Pipe diameter at section A in m\nDB=0.12 #Pipe diameter at section B in m\nPA=400*10**3 #Intensity of pressure at setion A in kPa\nPB=150*10**3 #Intensity of pressure at setion B in kPa\npi=3.14 #Pi constant \ng=9.81 #Gravity constant\nrho=1000\n\n#calculation\nVA=Q/(pi/4*(DA)**2) #contuity equation for discharge\nVB=Q/(pi/4*(DB)**2) #bernoulli's equation for discharge\n#Z=ZB-ZA\nZ=(PA/(rho*g))+(VA**2/(2*g))-(PB/(rho*g))-(VB**2/(2*g))\n\n\n#output\nprint(\"contuity equation for discharge= \",round(VA,3),\"m**3\")\nprint(\"contuity equation for discharge= \",round(VB,3),\"m**3\")\nprint(\"bernoulli's equation for discharge= \",round(Z,2),\"m\")\n\n\n\n\n\n\n\n", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "contuity equation for discharge= 1.053 m**3\ncontuity equation for discharge= 3.539 m**3\nbernoulli's equation for discharge= 24.9 m\n" + } + ], + "prompt_number": 69 + }, + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": "Example 18 Page No:129" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "#Input data\nL=200 #length of pipe in m\nD1=1 #Diameter at high end in m\nD2=0.4 #Diameter at low end in m\nP1=50*10**3 #Pressure at high end in kPa\nQ=4000 #Rate of water flow l/min\nS=1 #Slope of pipe 1 in 100\nZ2=0 #Datum line is passing through the center of the low end,therefore\npi=3.14\n\n\n\n#calculation\nQ=(4000*10**-3)/60 #rate of water flow l/min in m**3/s\nZ1=1/100*L #slope of pipe 1 in 100 is in m\n#Q=A1*V1=A2V2 #continuity eqation ,discharge\nV1=Q/((pi/4)*(D1**2))#in m**3\nV2=Q/((pi/4)*(D2**2))#in m**3\n#bernoulli's equation \nP2=(((((P1/(rho*g))+(V1**2/(2*g))+Z1)-(V2**2/(2*g))-Z2))*(rho*g))*10**-3 \n\n\n#output\nprint(\"rate of water flow= \",round(Q,4),\"m**3/s\")\nprint(\"slope of pipe= \",Z1,\"m\")\nprint(\"continuity eqation ,discharge= \",round(V1,5),\"m**3\")\nprint(\"continuity eqation ,discharge= \",round(V2,4),\"m**3\")\nprint(\"bernoulli's equation for discharge= \",round(P2,2),\" Kpa\")\n", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "rate of water flow= 0.0667 m**3/s\nslope of pipe= 2.0 m\ncontinuity eqation ,discharge= 0.08493 m**3\ncontinuity eqation ,discharge= 0.5308 m**3\nbernoulli's equation for discharge= 69.48 Kpa\n" + } + ], + "prompt_number": 74 + }, + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": "Example 19 Page No:130" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "#Input data\nimport math\nL=36 #Length of pipe in m\nD1=0.15 #Diameter at upper side in m\nD2=0.3 #Diameter at lower side in m\nsin30=0.5\ntheta=math.sin(30) #Pipe slope upward at angle in degree\nV1=2 #Velocity of water at smaller section in m/s \npi=3.14 #Pi constant \nrho=1000 #Roh constant\ng=9.81 #Gravity constant\n\n\n#calculation\n#datum line is passing through the center of the low end,therefore\nZ1=0\nZ2=Z1+L*(0.5) #pipe inclined 30 degree,therefore in m\n#Q=A1*V1=A2*V2 continuity eqation ,discharge\nV2=(pi/4*(D1**2)*2)/(pi/4*(D2**2))\n#Z=P1-P2 bernoulli's equation \nZ=((((-V1**2)/(2*g))+((V2**2)/(2*g))-Z1+Z2)*(rho*g))*10**-3\n\n\n\n\n\n\n#output\nprint(\"pipe inclined 30 degree,therefore Z2=\",Z2,\"m\")\nprint(\"continuity eqation ,discharge V2=\",V2,\"m/s\")\nprint(\"#bernoulli's equation Z=\",round(Z,1),\"Kpa\")\n\n\n", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "pipe inclined 30 degree,therefore Z2= 18.0 m\ncontinuity eqation ,discharge V2= 0.5 m/s\n#bernoulli's equation Z= 174.7 Kpa\n" + } + ], + "prompt_number": 76 + }, + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": "Example 20 Page No:130-131" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "#Input data\nD1=0.25 #Diameter at inlet in m\nD2=0.175 #Diameter at outlet in m\nP1=450*10**3 #Intensity of pressure at inlet in kPa\nP2=200*10**3 #Intensity of pressure at outlet in kPa\npi=3.14 #pi constant \nrho=1000 #Roh constant\ng=9.81 #Gravity constant\nZ1=Z2\n\n#Calculation \n#A1*V1=A2*V2 Continuity eqation in V1\nV2=((pi/4)*(D1**2))/((pi/4)*(D2**2))\n#Z=V2**2-V1**2 Bernoulli's equation in m/s\nZ=-(((P2/(rho*g))-(P1/(rho*g)))*(2*g))\nX=Z/((V2**2)-1)\nV1=math.sqrt(X)\nQ=(pi/4)*(D1**2)*V1 #Flow rate Water in m**3/Kg\n\n\n#Output\nprint(\"Continuity eqation= \",round(V2,2),\"V1\")\nprint(\"Bernoulli's equation= \",Z,\"m/s\")\nprint(\"V1=\",round(V1,2),\"\")\nprint(\"Flow rate Water= \",round(Q,3),\"m**3/Kg\")\n\n\n\n", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "Continuity eqation= 2.04 V1\nBernoulli's equation= 500.0 m/s\nV1= 12.57 \nFlow rate Water= 0.617 m**3/Kg\n" + } + ], + "prompt_number": 120 + }, + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": "Example 21 Page No:131-132" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "#Input data\nL=300 #Length of pipe in m\nD1=0.9 #Diameter at higher end in m\nD2=0.6 #Diameter at lower end in m\nS=0.85 #Specific gravity \nQ=0.08 #Flow in l/s\nP1=40*10**3 #Pressure at higher end in kPa\npi=3.14 #pi constant \nrho=1000 #Roh constant\ng=9.81 #Gravity constant\nslop=1/50 #1 in 50\n\n\n#Calculation\n#Datum line is passing through the center of the low end,therefore\nZ2=0 \nZ1=slop*L\n#Q=A1*V1=A2*V2 Continuity eqation\nV1=Q/((pi/4)*(D1**2)) #Frome continuity eqation, discharge\nV2=Q/((pi/4)*(D2**2)) #Frome continuity eqation, discharge\n#Bernoulli's equation \nP2=(((((P1/(rho*S*g))+(V1**2/(2*g))+Z1)-(V2**2/(2*g))+Z2))*(S*rho*g))*10**-3\n\n\n\n#Output\nprint(\"Z1=\",Z1,\"m\")\nprint(\"continuity eqation, discharge V1=\",round(V1,5),\"m**3\")\nprint(\"continuity eqation, discharge V2=\",round(V2,5),\"m**3\")\nprint(\"bernoulli's equation= \",round(P2,),\"KPa\")\n\n", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "Z1= 6.0 m\ncontinuity eqation, discharge V1= 0.12582 m**3\ncontinuity eqation, discharge V2= 0.28309 m**3\nbernoulli's equation= 90 KPa\n" + } + ], + "prompt_number": 128 + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Basic_mechanical_engineering_by_Basant_Agrawal_,_C.M_Agrawal/Chapter_9__Laws_Of_Thermodynamics.ipynb b/Basic_mechanical_engineering_by_Basant_Agrawal_,_C.M_Agrawal/Chapter_9__Laws_Of_Thermodynamics.ipynb new file mode 100644 index 00000000..d459e09d --- /dev/null +++ b/Basic_mechanical_engineering_by_Basant_Agrawal_,_C.M_Agrawal/Chapter_9__Laws_Of_Thermodynamics.ipynb @@ -0,0 +1,209 @@ +{ + "metadata": { + "name": "Chapter 9 Laws Of Thermodynamics" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": " Chapter 9 Law Of Thermodynamics" + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Example 1 Page No:165" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "#Input data\nQab=720 #Heat transfer of 1st processes in KJ \nQbc=-80 #Heat transfer of 2nd processes in KJ\nQcd=40 #Heat transfer of 3rd processes in KJ\nQda=-640 #Heat transfer of 4th processes in KJ\nWab=-90 #Work transfer of 1st processes in KJ\nWbc=-50 #Work transfer of 2nd processes in KJ\nWcd=130 #Work transfer of 3rd processes in KJ\n\n\n#Calculation\n#From the 1st law of thermodynamic for close system undergoing a cycle.\n\n#Work interaction during the 4th processes \nWda=((Qab+Qbc+Qcd+Qda)-(Wab+Wbc+Wcd)) \n\n#Output\nprint(\"Work interaction during the 4th processes=\",Wda,\"KJ\")\n", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "Work interaction during the 4th processes= 50 KJ\n" + } + ], + "prompt_number": 5 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Example 2 Page No:166" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "#Input data\n #During compression\nW1=-9200 #Stroke work done by the piston in Nm\nNm1=-9.2 #Nm of work done\nQ1=-50 #Heat rejected during copression in KJ\n #During expansion\nW2=8400 #Stroke work done by the piston in Nm\nNm2=8.4 #Nm of work done\n\n#Calculation\n #Quantity of heat transferred\nQ2=-((Nm1+Nm2)+Q1) #-sign for indicate heat is transferred\n\n\n#Output\nprint(\"Quantity of heat transferred=\",Q2,\"KJ\")", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "Quantity of heat transferred= 50.8 KJ\n" + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Example 3 Page No:166" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "#input data\nW1=-20 #Work interaction to the fluid in KJ\nW2=42 #Work interaction from the fluid in KJ\nQ1=85 #Heat interaction to the fluid in KJ\nQ2=85 #Heat interaction to the fluid in KJ\nQ3=-50 #Heat interaction from the fluid in KJ\n\n#calculation\nW3=((Q1+Q2+Q3)-(W1+W2)) #Magnitude and direction of the third heat interation\n\n\n#output\nprint(\"Magnitude and direction of the third heat interation=\",W3,\"KJ\")", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "Magnitude and direction of the third heat interation= 98 KJ\n" + } + ], + "prompt_number": 51 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Example 4 Page No:168" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "#Input data\nQ=-2100 #Non flow process losses heat in KJ\ndeltaU=420 #Gain heat\n\n#Calculation\nW=Q-deltaU #Work done and compression process in KJ\n\n#Output\nprint(\"Work done and compression process=\",W,\"KJ\")", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "Work done and compression process= -2520 KJ\n" + } + ], + "prompt_number": 52 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Example 5 Page No:168" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "#Input data\nW=-2000 #Work input of panddle wheel in KJ\nQ=-6000 #Heat transferred to the surrounding from tank\n\n#Calculation\ndeltaU=Q-W #Change in interval energy\n\n#Output\nprint(\"change in interval energy drop=\",deltaU,\"KJ\")", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "change in interval energy drop= -4000 KJ\n" + } + ], + "prompt_number": 9 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Example 6 Page No:169" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "#Input data\nU1=520 #internal energy in KJ/Kg\nU2=350 #internal energy in KJ/Kg\nW=-80 #work done by the air in the cylinder KJ/kg\n\n#Calculation\ndeltaU=U2-U1\nQ=deltaU+W #Heat transferred during the process\n\n#Output\nprint(\"Heat transferred during the process=\",Q,\"KJ\")\n ", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "Heat transferred during the process= -250 KJ\n" + } + ], + "prompt_number": 12 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Example 7 Page No:169" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "#Input data\nW1=800 #Power of turbine shaft Kw\nW2=-5 #Work pump to feed in Kw \nQ1=2700 #Heat for steam generation KJ/Kg\nQ2=-1800 #Condenser rejected heat KJ/Kg\n\n#Calculation\nm=((W1+W2)/(Q1+Q2)) #Steam flow rate in Kg/h\n\n\n#Output\nprint(\"Steam flow rate=\",round(m,4),\"Kg/s\")\n", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "Steam flow rate= 0.8833 Kg/s\n" + } + ], + "prompt_number": 13 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Example 8 Page No:170" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "#input data\n#Data consistent with first law pf thermodynamics\nQabcd=-22 #In KJ\nN=150 #In Cycles/min\nQab=17580 #In KJ/min\nQbc=0 \nQcd=-3660 #In KJ/min\nWab=-8160 #In KJ/min\nWbc=4170 #In KJ/min \nDeltaUcd=-21630 #In KJ/min\n\n\n#calculation\nDeltaUab=Qab-Wab #In KJ/min\nDeltaUbc=Qbc-Wbc #In KJ/min \nWcd=Qcd-DeltaUcd #In KJ/min\nQabcd1=-220*150 #In KJ/min\nQda=((Qabcd1)-(Qab+Qbc+Qcd)) #In KJ/min\nWda=((Qabcd1)-(Wab+Wbc+Wcd)) #In KJ/min\nDeltaUabcd=0\nDeltaUda=((DeltaUabcd)-(DeltaUab+DeltaUbc+DeltaUcd)) #In KJ/min\nNWO=Qabcd1/60 #In KW\n\n\n#output\nprint(\"DeltaUab= \",DeltaUab,\"KJ/min\")\nprint(\"DeltaUbc= \",DeltaUbc,\"KJ/min\")\nprint(\"Wcd= \",Wcd,\"KJ/min\")\nprint(\"Qabcd1= \",Qabcd1,\"KJ/min\")\nprint(\"Qda= \",Qda,\"KJ/min\")\nprint(\"Wda= \",Wda,\"KJ/min\")\nprint(\"DeltaUabcd= \",DeltaUabcd,\"KJ/min\")\nprint(\"DeltaUda= \",DeltaUda,\"KJ/min\")\nprint(\"NWO=\",NWO,\"Kw\")\n", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "DeltaUab= 25740 KJ/min\nDeltaUbc= -4170 KJ/min\nWcd= 17970 KJ/min\nQabcd1= -33000 KJ/min\nQda= -46920 KJ/min\nWda= -46980 KJ/min\nDeltaUabcd= 0 KJ/min\nDeltaUda= 60 KJ/min\nNWO= -550.0 Kw\n" + } + ], + "prompt_number": 47 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Example 9 Page No:171" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "#Input data\nQab=-6500 #Heat transferred in 1st process KJ/min\nQbc=0 #Heat transferred in 2nd process \nQcd=-10200 #Heat transferred in 3rd process KJ/min\nQda=32600 #Heat transferred in 4th process KJ/min\nWab=-1050 #Heat transferred in 1st process KJ\nWbc=-3450 #Heat transferred in 2nd process KJ\nWcd=20400 #Heat transferred in 3rd process KJ\nWda=0 #Heat transferred in 4th process\n\n#Calculator\ndQ=Qab+Qbc+Qcd+Qda #Net heat transfer in 1st cycle\ndW=Wab+Wbc+Wcd+Wda #Net work done in 1st cycle\ndW1=dW/60 #Net work done in 1st cycle\nDeltaUab=Qab-Wab #ab process\nDeltaUbc=Qbc-Wbc #bc processes\nDeltaUcd=Qcd-Wcd #cd processes\nDeltaUda=Qda-Wda #dc processes\n\n#Output\nprint(\"Net heat transfer in 1st cycle= \",dQ,\"KJ/min\")\nprint(\"Net work done in 1st cycle= \",dW,\"KJ/min\")\nprint(\"Net work done in 1st cycle= \",dW1,\"KW\")\nprint(\"ab process= \",DeltaUab,\"KJ/min\")\nprint(\"bc processes= \",DeltaUbc,\"KJ/min\")\nprint(\"cd processes= \",DeltaUcd,\"KJ/min\")\nprint(\"dc processes= \",DeltaUda,\"KJ/min\")\n\n", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "Net heat transfer in 1st cycle= 15900 KJ/min\nNet work done in 1st cycle= 15900 KJ/min\nNet work done in 1st cycle= 265.0 KW\nab process= -5450 KJ/min\nbc processes= 3450 KJ/min\ncd processes= -30600 KJ/min\ndc processes= 32600 KJ/min\n" + } + ], + "prompt_number": 50 + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Basic_mechanical_engineering_by_Basant_Agrawal_,_C.M_Agrawal/screenshots/2.png b/Basic_mechanical_engineering_by_Basant_Agrawal_,_C.M_Agrawal/screenshots/2.png Binary files differnew file mode 100644 index 00000000..66dd064e --- /dev/null +++ b/Basic_mechanical_engineering_by_Basant_Agrawal_,_C.M_Agrawal/screenshots/2.png diff --git a/Basic_mechanical_engineering_by_Basant_Agrawal_,_C.M_Agrawal/screenshots/5.png b/Basic_mechanical_engineering_by_Basant_Agrawal_,_C.M_Agrawal/screenshots/5.png Binary files differnew file mode 100644 index 00000000..b667656a --- /dev/null +++ b/Basic_mechanical_engineering_by_Basant_Agrawal_,_C.M_Agrawal/screenshots/5.png diff --git a/Basic_mechanical_engineering_by_Basant_Agrawal_,_C.M_Agrawal/screenshots/7.png b/Basic_mechanical_engineering_by_Basant_Agrawal_,_C.M_Agrawal/screenshots/7.png Binary files differnew file mode 100644 index 00000000..5074ec29 --- /dev/null +++ b/Basic_mechanical_engineering_by_Basant_Agrawal_,_C.M_Agrawal/screenshots/7.png diff --git a/Manufacturing_Engineering_&_Technology_by_S_Kalpakjian_and_S_R_Schmid/CHAPTER10.ipynb b/Manufacturing_Engineering_&_Technology_by_S_Kalpakjian_and_S_R_Schmid/CHAPTER10.ipynb deleted file mode 100755 index 719b96fb..00000000 --- a/Manufacturing_Engineering_&_Technology_by_S_Kalpakjian_and_S_R_Schmid/CHAPTER10.ipynb +++ /dev/null @@ -1,88 +0,0 @@ -{
- "metadata": {
- "name": "",
- "signature": "sha256:7bab090be4ea3beabd54f01ab20d8f4629c694669924098a83be7720130d8118"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "CHAPTER 10 - Fundamentals of Metal Casting"
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "EXAMPLE 10.1 - PG NO. 252"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Example 10.1 \n",
- "#page no. 252\n",
- "# Given that\n",
- "#three metal piece being cast have the same volume but different shapes\n",
- "#shapes are sphere,cube,cylinder(height=diameter)\n",
- "\n",
- "\n",
- "\n",
- "print(\"\\n #solidification time for various shapes# \\n\")\n",
- "\n",
- "#solidification time is inversely proportional to the square of surface area\n",
- "\n",
- "#for sphere\n",
- "r=(3./(4.*3.14))**(1./3.)#radius of the sphere from volume of sphere v=(4*3.14*r**3)/3\n",
- "A=4*3.14*((r)**2)\n",
- "time1=1./(A)**2.\n",
- "print'%s %.6f %s' %(\"\\n the solidification time for the sphere is \",time1,\"C\")\n",
- "\n",
- "#for cube\n",
- "a=1#edge of the cube\n",
- "A=6*a**2\n",
- "time2=1./(A)**2\n",
- "print'%s %.6f %s' %(\"\\n the solidification time for the cube is \",time2,\"C\")\n",
- "\n",
- "#for cylinder\n",
- "#given height =diameter \n",
- "#radius=2*height\n",
- "r=(1./(2*3.14))**(1./3.)#radius of the cylinder from volume of the cylinder v=3.14*r**2*h\n",
- "A=(6*3.14*(r**2)) #area of the cylinder = (2*3.14*radius**2) + (2*3.14*radius*height)\n",
- "time3=1./(A)**2.\n",
- "print'%s %.6f %s' %(\"\\n the solidification time for the sphere is \",time3,\"C\")\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- " #solidification time for various shapes# \n",
- "\n",
- "\n",
- " the solidification time for the sphere is 0.042774 C\n",
- "\n",
- " the solidification time for the cube is 0.027778 C\n",
- "\n",
- " the solidification time for the sphere is 0.032643 C\n"
- ]
- }
- ],
- "prompt_number": 3
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Manufacturing_Engineering_&_Technology_by_S_Kalpakjian_and_S_R_Schmid/CHAPTER13.ipynb b/Manufacturing_Engineering_&_Technology_by_S_Kalpakjian_and_S_R_Schmid/CHAPTER13.ipynb deleted file mode 100755 index 151e498b..00000000 --- a/Manufacturing_Engineering_&_Technology_by_S_Kalpakjian_and_S_R_Schmid/CHAPTER13.ipynb +++ /dev/null @@ -1,84 +0,0 @@ -{
- "metadata": {
- "name": "",
- "signature": "sha256:7430f82497981f14c807b82dc97f1ffae56cea7bca1ef54c84ec5f6d9a82fb1c"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "CHAPTER 13 - Rolling of Metals"
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "EXAMPLE 13.1 - PG NO. 323"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Example 13.1\n",
- "#page no. 323\n",
- "# Given that\n",
- "import math\n",
- "w=9. #in inch width of thee strip\n",
- "ho=1. #in inch initial thickness of the strip\n",
- "hf=0.80 #in inch thickness of the strip after one pass\n",
- "r=12. #in inch roll radius\n",
- "N=100. #in rpm\n",
- "\n",
- "# Sample Problem on page no. 323\n",
- "\n",
- "print(\"\\n #Calculation of roll force and torque# \\n\")\n",
- "\n",
- "L=(r*(ho-hf))**(1./2.)\n",
- "\n",
- "E=math.log10(1./hf)#absolute value of true strain\n",
- "\n",
- "Y=26000. #in psi average stress from the data in the book \n",
- "F=L*w*Y # roll force\n",
- "F1=F*4.448/(10.**6.)#in mega newton\n",
- "print'%s %.2f %s' %(\"\\n\\nRoll force = \",F1+0.13,\"MN \")\n",
- "\n",
- "P=(2*3.14*F*L*N)/(33000.*12.)\n",
- "P1=P*7.457*(10.**2.)/(10.**3.)#in KW\n",
- "print'%s %d %s' %(\"\\n\\npower per roll = \",round(P1+41),\"KW\")\n",
- "\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- " #Calculation of roll force and torque# \n",
- "\n",
- "\n",
- "\n",
- "Roll force = 1.74 MN \n",
- "\n",
- "\n",
- "power per roll = 705 KW\n"
- ]
- }
- ],
- "prompt_number": 1
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Manufacturing_Engineering_&_Technology_by_S_Kalpakjian_and_S_R_Schmid/CHAPTER14.ipynb b/Manufacturing_Engineering_&_Technology_by_S_Kalpakjian_and_S_R_Schmid/CHAPTER14.ipynb deleted file mode 100755 index fb2e297e..00000000 --- a/Manufacturing_Engineering_&_Technology_by_S_Kalpakjian_and_S_R_Schmid/CHAPTER14.ipynb +++ /dev/null @@ -1,78 +0,0 @@ -{
- "metadata": {
- "name": "",
- "signature": "sha256:9c20d0a94b4cbce6b1960b4b814f748dc5e36a521148e77cc13a8657ef82f50b"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "CHAPTER 14 - Forging of Metals"
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "EXAMPLE 14.1 - PG NO. 344"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#example 14.1\n",
- "#page no. 344\n",
- "# Given that\n",
- "import math\n",
- "d=150.#in mm Diameter of the solid cylinder \n",
- "Hi=100. #in mm Height of the cylinder\n",
- "u=0.2 # Cofficient of friction\n",
- "\n",
- "# Sample Problem on page no. 344\n",
- "\n",
- "print(\"\\n # Calculation of forging force # \\n\")\n",
- "\n",
- "#cylinder is reduced in height by 50%\n",
- "Hf=100./2.\n",
- "#Volume before deformation= Volume after deformation\n",
- "r=math.sqrt((3.14*75**2*100)/(3.14*50.))#r is the final radius of the cylinder\n",
- "E=math.log(Hi/Hf)#absolute value of true strain\n",
- "#given that cylinder is made of 304 stainless steel\n",
- "Yf=1000. #in Mpa flow stress of the material from data in the book\n",
- "F = Yf*(10.**6.)*3.14*(r**2.)*10.**-6.*(1.+((2.*u*r)/(3.*Hf)))#Forging Force\n",
- "F1=F/(10.**6.)\n",
- "print'%s %d %s' %(\"\\n\\n Forging force = \",F1,\"MN\")\n",
- "\n",
- "\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- " # Calculation of forging force # \n",
- "\n",
- "\n",
- "\n",
- " Forging force = 45 MN\n"
- ]
- }
- ],
- "prompt_number": 1
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Manufacturing_Engineering_&_Technology_by_S_Kalpakjian_and_S_R_Schmid/CHAPTER15.ipynb b/Manufacturing_Engineering_&_Technology_by_S_Kalpakjian_and_S_R_Schmid/CHAPTER15.ipynb deleted file mode 100755 index 247f69a7..00000000 --- a/Manufacturing_Engineering_&_Technology_by_S_Kalpakjian_and_S_R_Schmid/CHAPTER15.ipynb +++ /dev/null @@ -1,72 +0,0 @@ -{
- "metadata": {
- "name": "",
- "signature": "sha256:11c648b7483b18a34763046366215f9df144424896a38077f7d1c80df90ae003"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "CHAPTER 15 -Extrusion and Drawing of Metals"
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "EXAMPLE 15.1 - PG NO.372"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#example 15.1\n",
- "#page no. 372\n",
- "# Given that\n",
- "import math\n",
- "di=5.#in inch Diameter of the round billet\n",
- "df=2.#in inch Diameter of the round billet after extrusion\n",
- "\n",
- "# Sample Problem on page no. 372\n",
- "\n",
- "print(\"\\n # Calculation of force in Hot Extrusion# \\n\")\n",
- "\n",
- "#As 70-30 Brass is given, so the value of the extrusion constant is 35000psi from the diagram given in the book\n",
- "k=35000.#in psi\n",
- "F=3.14*(di/2.)**2.*k*math.log((3.14*(di**2.))/(3.14*(df**2.)))\n",
- "F1=F*4.448/(10**6)\n",
- "print'%s %.6f %s' %(\"\\n\\n Extrusion force=\",F1,\"MN\")\n",
- "\n",
- "\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- " # Calculation of force in Hot Extrusion# \n",
- "\n",
- "\n",
- "\n",
- " Extrusion force= 5.598940 MN\n"
- ]
- }
- ],
- "prompt_number": 3
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Manufacturing_Engineering_&_Technology_by_S_Kalpakjian_and_S_R_Schmid/CHAPTER16.ipynb b/Manufacturing_Engineering_&_Technology_by_S_Kalpakjian_and_S_R_Schmid/CHAPTER16.ipynb deleted file mode 100755 index fb4b2ceb..00000000 --- a/Manufacturing_Engineering_&_Technology_by_S_Kalpakjian_and_S_R_Schmid/CHAPTER16.ipynb +++ /dev/null @@ -1,71 +0,0 @@ -{
- "metadata": {
- "name": "",
- "signature": "sha256:37b99b331f0cf2427d454d75229004d53c86c8f3b22d1edc37006f11ec00901a"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "CHAPTER 16 - Sheet Metal Forming Processes "
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "EXAMPLE 16.1 - PG NO. 396"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#example 16.1\n",
- "#page no. 396\n",
- "# Given that\n",
- "d=1.#in inch Diameter of the hole\n",
- "T=(1./8.)#in inch thickness of the sheet\n",
- "\n",
- "# Sample Problem on page no. 396\n",
- "\n",
- "print(\"\\n # Calculation of Punch Force# \\n\")\n",
- "\n",
- "UTS=140000.#in psi Ultimate Tensile Strength of the titanium alloy Ti-6Al-4V\n",
- "L=3.14*d#total length sheared which is the perimeter of the hole\n",
- "F=0.7*T*L*UTS\n",
- "F1=F*4.448/(10**6)\n",
- "print'%s %.6f %s' %(\"\\n\\n Extrusion force=\",F1,\"MN\")\n",
- "\n",
- "\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- " # Calculation of Punch Force# \n",
- "\n",
- "\n",
- "\n",
- " Extrusion force= 0.171092 MN\n"
- ]
- }
- ],
- "prompt_number": 1
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Manufacturing_Engineering_&_Technology_by_S_Kalpakjian_and_S_R_Schmid/CHAPTER17.ipynb b/Manufacturing_Engineering_&_Technology_by_S_Kalpakjian_and_S_R_Schmid/CHAPTER17.ipynb deleted file mode 100755 index e6d2e527..00000000 --- a/Manufacturing_Engineering_&_Technology_by_S_Kalpakjian_and_S_R_Schmid/CHAPTER17.ipynb +++ /dev/null @@ -1,85 +0,0 @@ -{
- "metadata": {
- "name": "",
- "signature": "sha256:f3d611ffad9a6eb77db0dafc2647d41da22e72400b02fe9943316083c1df665f"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "CHAPTER 17 - Processing of Powder Metals Ceramics, Glass and Superconductors"
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "EXAMPLE 17.1 - PG NO. 466"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#example 17.1 \n",
- "#page no. 466\n",
- "# Given that\n",
- "L=20#in mm Final length of the ceramic part\n",
- "#Linear shrinkage during drying and firing is 7% and 6% respectively\n",
- "Sd=0.070#Linear shrinkage during drying\n",
- "Sf=0.06#Linear shrinkage during firing\n",
- "\n",
- "# Sample Problem on page no. 466\n",
- "\n",
- "print(\"\\n # Dimensional changes during the shaping of ceramic components # \\n\")\n",
- "\n",
- "#part (a)\n",
- "\n",
- "Ld=L/(1.-Sf)#dried length\n",
- "Lo=(1.+Sd)*Ld#initial length\n",
- "print'%s %.6f %s' %(\"\\n\\nInitial Length=\",Lo,\"mm\")\n",
- "\n",
- "#Answer in the book is approximated to 22.77mm\n",
- "\n",
- "#part(b)\n",
- "\n",
- "Pf=0.03#Fired Porosity\n",
- "r = (1.-Pf)# Where r = Va/Vf\n",
- "R = 1./((1.-Sf)**3.)# Where R = Vd/Vf\n",
- "Pd = (1.-r/R)\n",
- "print'%s %d %s' %(\"\\n\\nDried porosity is \",Pd*100,\"%\")\n",
- "\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- " # Dimensional changes during the shaping of ceramic components # \n",
- "\n",
- "\n",
- "\n",
- "Initial Length= 22.765957 mm\n",
- "\n",
- "\n",
- "Dried porosity is 19 %\n"
- ]
- }
- ],
- "prompt_number": 1
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Manufacturing_Engineering_&_Technology_by_S_Kalpakjian_and_S_R_Schmid/CHAPTER18.ipynb b/Manufacturing_Engineering_&_Technology_by_S_Kalpakjian_and_S_R_Schmid/CHAPTER18.ipynb deleted file mode 100755 index a0571d20..00000000 --- a/Manufacturing_Engineering_&_Technology_by_S_Kalpakjian_and_S_R_Schmid/CHAPTER18.ipynb +++ /dev/null @@ -1,128 +0,0 @@ -{
- "metadata": {
- "name": "",
- "signature": "sha256:a21f94bb09c85281db9a114e59ae40fd5f2f40a3ccc78c3387df46257a6d865c"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "CHAPTER 18 - Forming and Shaping Plastics and Composite Materials"
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "EXAMPLE 18.1 - PG NO. 484"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#example 18.1\n",
- "#page no. 491\n",
- "# Given that\n",
- "W=400.#in mm Lateral(width) Dimension of a plastic shopping bag \n",
- "\n",
- "# Sample Problem on page no. 484\n",
- "\n",
- "print(\"\\n # Blown Film # \\n\")\n",
- "\n",
- "#part(a)\n",
- "\n",
- "P=2.*W#in mm Perimeter of bag\n",
- "D=P/3.14#in mm blown diameter calculated from Permeter=3.14*diameter\n",
- "#Given in this process, a tube is expanded to form 1.5 to 2.5 in times the extrusion die diameter, so take maximum value 2.5\n",
- "Dd=D/2.5#Extrusion die diameter\n",
- "print'%s %d %s' %(\"\\n\\n Extrusion Die Diameter =\",Dd,\"mm\")\n",
- "\n",
- "#Answer varies due to approximations\n",
- "\n",
- "#part(b) is theoritical\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- " # Blown Film # \n",
- "\n",
- "\n",
- "\n",
- " Extrusion Die Diameter = 101 mm\n"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "EXAMPLE 18.2 - PG NO. 488"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#example 18.2\n",
- "#page no. 488\n",
- "# Given that\n",
- "W=250.#in ton Weight of injection moulding machine\n",
- "d=4.5#in inch diameter of spur gear\n",
- "t=0.5#in inch thickness of spur gear\n",
- "#Gears have a fine tooth profile\n",
- "\n",
- "# Sample Problem on page no. 488\n",
- "\n",
- "print(\"\\n # Injection Molding of Parts # \\n\")\n",
- "\n",
- "#because of fine tooth profile pressure required in the mould cavity is assumed to be of the order 100MPa or 15Ksi\n",
- "\n",
- "p=15#inKsi\n",
- "A=(3.14*(d**2))/4#in inch^2 area of the gear\n",
- "F=A*15*1000\n",
- "n=(W*2000)/F #weight is converted into lb by multiplying it by 2000\n",
- "print'%s %d' %(\"\\n\\n Number of gears that can be injected =\",n)\n",
- "\n",
- "#print'%s %d %s' %(\"\\n\\n Force required is = \",A/10000,\"MN\" )\n",
- "\n",
- "# Second part of this question is theoritical\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- " # Injection Molding of Parts # \n",
- "\n",
- "\n",
- "\n",
- " Number of gears that can be injected = 2\n"
- ]
- }
- ],
- "prompt_number": 2
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Manufacturing_Engineering_&_Technology_by_S_Kalpakjian_and_S_R_Schmid/CHAPTER2.ipynb b/Manufacturing_Engineering_&_Technology_by_S_Kalpakjian_and_S_R_Schmid/CHAPTER2.ipynb deleted file mode 100755 index 40a2982b..00000000 --- a/Manufacturing_Engineering_&_Technology_by_S_Kalpakjian_and_S_R_Schmid/CHAPTER2.ipynb +++ /dev/null @@ -1,82 +0,0 @@ -{
- "metadata": {
- "name": "",
- "signature": "sha256:fda21ffdd3ec2f6f3990ab05fef8a23ff12faaa312bb1ada662f140fd61efe28"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "CHAPTER 2- Mechanical Behavior, Testing, and Manufacturing Properties of Materials"
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "EXAMPLE 2.1 - PG NO. 63"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#example 2.1,chapter 2, page 63\n",
- "\n",
- "# Given that\n",
- "#True stress=100000*(True strain)**0.5\n",
- "\n",
- "# Sample Problem on page no. 63\n",
- "import math\n",
- "print(\"\\n # Calculation of Ultimate Tensile Strength # \\n\")\n",
- "#from the data given\n",
- "n=0.5\n",
- "E=0.5\n",
- "K=100000.\n",
- "Truestress=K*((E)**n)\n",
- "#let An(area of neck)/Ao=t\n",
- "#from math.log(Ao/An)=n\n",
- "print'%s %.3f %s' %(\"true Ultimate Tensile Strength =\",Truestress,\"psi \\n\")\n",
- "t=math.exp(-n)\n",
- "print'%s %.7f %s' %(\"t =\",t,\"\\n\")\n",
- "UTS=Truestress*t#from the math.expression UTS= P/Ao where P(Maximum Load)=Truestress*An\n",
- "print'%s %.3f %s' %(\"Ultimate Tensile Strength =\",UTS,\"psi\")\n",
- "#answer in the book is approximated to 42850 psi \n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- " # Calculation of Ultimate Tensile Strength # \n",
- "\n",
- "true Ultimate Tensile Strength = 70710.678 psi \n",
- "\n",
- "t = 0.6065307 \n",
- "\n",
- "Ultimate Tensile Strength = 42888.194 psi\n"
- ]
- }
- ],
- "prompt_number": 1
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Manufacturing_Engineering_&_Technology_by_S_Kalpakjian_and_S_R_Schmid/CHAPTER20.ipynb b/Manufacturing_Engineering_&_Technology_by_S_Kalpakjian_and_S_R_Schmid/CHAPTER20.ipynb deleted file mode 100755 index 2872ff69..00000000 --- a/Manufacturing_Engineering_&_Technology_by_S_Kalpakjian_and_S_R_Schmid/CHAPTER20.ipynb +++ /dev/null @@ -1,129 +0,0 @@ -{
- "metadata": {
- "name": "",
- "signature": "sha256:28aeea19a74efb3bf099a391966ddf61856eae1ab0012ac6592d6c461a166282"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "CHAPTER 20 - Machining Processes used to Produce Round Shape"
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "EXAMPLE 20.1 - PG NO. 548"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#example 20.1\n",
- "#page no. 548\n",
- "import math\n",
- "# Given that\n",
- "to=0.005#in inch depth of cut\n",
- "V=400.#in ft/min cutting speed\n",
- "X=10.#in degree rake angle\n",
- "w=0.25#in inch width of cut\n",
- "tc=0.009#in inch chip thickness\n",
- "Fc=125.#in lb Cutting force\n",
- "Ft=50.#in lb thrust force\n",
- "\n",
- "# Sample Problem on page no. 548\n",
- "\n",
- "print(\"\\n # Relative Energies in cutting # \\n\")\n",
- "\n",
- "r=to/tc#cutting ratio\n",
- "R=math.sqrt((Ft**2.)+(Fc**2.))\n",
- "B=math.cos(math.degrees(Fc/R))+X#friction angle\n",
- "F=R*math.sin(math.degrees(B))\n",
- "P=((F*r)/Fc)*100.\n",
- "print'%s %d %s' %(\"\\n\\n Percentage of total energy going into overcoming friction =\",P-28.40367,\" pecrent\")\n",
- "\n",
- "#Answer in the book is approximated to 32 due to approximation in calculation of R and B\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- " # Relative Energies in cutting # \n",
- "\n",
- "\n",
- "\n",
- " Percentage of total energy going into overcoming friction = 31 pecrent\n"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "EXAMPLE 20.2 - PG NO. 555"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#example 20.2\n",
- "#page no. 555\n",
- "import numpy\n",
- "import math\n",
- "# Given that\n",
- "n=0.5#exponent that depends on tool and workpiece material\n",
- "C=400.#constant\n",
- "\n",
- "# Sample Problem on page no. 555\n",
- "\n",
- "print(\"\\n # Increasing tool life by Reducing the Cutting Speed # \\n\")\n",
- "\n",
- "V1=numpy.poly([0])\n",
- "r=0.5# it is the ratio of V2/V1 where V1 and V2 are the initial and final cutting speed of the tool\n",
- "#let t=T2/T1 where T1 and T2 are the initial and final tool life\n",
- "t=1./(r**(1./n))#from the relation V1*(T1**n)=V2*(T2**n)\n",
- "P=(t-1)*100\n",
- "print'%s %d %s' %(\"\\n\\n Percent increase in tool life =\",P,\"percent\")\n",
- "\n",
- "\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- " # Increasing tool life by Reducing the Cutting Speed # \n",
- "\n",
- "\n",
- "\n",
- " Percent increase in tool life = 300 percent\n"
- ]
- }
- ],
- "prompt_number": 2
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Manufacturing_Engineering_&_Technology_by_S_Kalpakjian_and_S_R_Schmid/CHAPTER22.ipynb b/Manufacturing_Engineering_&_Technology_by_S_Kalpakjian_and_S_R_Schmid/CHAPTER22.ipynb deleted file mode 100755 index a057d34b..00000000 --- a/Manufacturing_Engineering_&_Technology_by_S_Kalpakjian_and_S_R_Schmid/CHAPTER22.ipynb +++ /dev/null @@ -1,157 +0,0 @@ -{
- "metadata": {
- "name": "",
- "signature": "sha256:893d2b9b70668c0aef0dd9c06849e89a29e7b7b29867a2d8588481db5fed5a14"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "CHAPTER 22 - Machining Processes used to Produce Round Shape"
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "EXAMPLE 22.1 - PG NO. 600"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#example 22.1\n",
- "#page no. 600 \n",
- "# Given that\n",
- "l=6.#in inch Length of rod \n",
- "di=1./2.#in inch initial diameter of rod\n",
- "df=0.480#in inch final diameter of rod\n",
- "N=400.#in rpm spindle rotation\n",
- "Vt=8#in inch/minute axial speed of the tool\n",
- "\n",
- "# Sample Problem on page no. 600\n",
- "\n",
- "print(\"\\n # Material Removal Rate and Cutting Force in Turning # \\n\")\n",
- "\n",
- "V=3.14*di*N\n",
- "print'%s %d %s' %(\"\\n\\n Cutting speed=\",V,\" m/min\")\n",
- "\n",
- "v1=3.14*df*N#cutting speed from machined diameter\n",
- "d=(di-df)/2#depth of cut\n",
- "f=Vt/N#feed\n",
- "Davg=(di+df)/2.\n",
- "MRR=3.14*Davg*d*f*N \n",
- "print'%s %.6f %s' %(\"\\n\\n Material Removal Rate =\",MRR,\"=in^3/min\")\n",
- "\n",
- "t=l/(f*N)\n",
- "print'%s %.6f %s' %(\"\\n\\n Cutting time=\",t,\" min\")\n",
- "\n",
- "P=(4./2.73)*MRR#average value of stainless steel is taken as 4 ws/mm3 or 4/2.73 hpmin/mm3\n",
- "print'%s %.6f %s' %(\"\\n\\n Cutting power=\",P,\"hp\")\n",
- "\n",
- "Fc=((P*396000)/(N*2*3.14))/(Davg/2.)\n",
- "print'%s %d %s' %(\"\\n\\n Cutting force=\",Fc,\"lb\")\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- " # Material Removal Rate and Cutting Force in Turning # \n",
- "\n",
- "\n",
- "\n",
- " Cutting speed= 628 m/min\n",
- "\n",
- "\n",
- " Material Removal Rate = 0.123088 =in^3/min\n",
- "\n",
- "\n",
- " Cutting time= 0.750000 min\n",
- "\n",
- "\n",
- " Cutting power= 0.180349 hp\n",
- "\n",
- "\n",
- " Cutting force= 116 lb\n"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "EXAMPLE 22.2 - PG NO. 632"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#example 22.2\n",
- "#page no. 632\n",
- "# Given that \n",
- "d=10.#in mm diameter of drill bit\n",
- "f=0.2#in mm/rev feed\n",
- "N=800#in rpm spindle rotation\n",
- "\n",
- "# Sample Problem on page no. 632\n",
- "\n",
- "print(\"\\n # Material Removal Rate and Torque in Drilling # \\n\")\n",
- "\n",
- "MRR=(((3.14*(d**2))/4)*f*N)/60.\n",
- "print'%s %d %5s' %(\"\\n\\n Material Removal Rate \",MRR,\"=mm^3/sec\")\n",
- "\n",
- "\n",
- "#from the book data an average unit power of 0.5Ws/mm2 for magnesium is taken\n",
- "T=(MRR*0.5)/((N*2.*3.14)/60.)\n",
- "print'%s %.6f %s' %(\"\\n\\n Torque on the drill \",T,\"=Nm\")\n",
- "\n",
- "\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- " # Material Removal Rate and Torque in Drilling # \n",
- "\n",
- "\n",
- "\n",
- " Material Removal Rate 209 =mm^3/sec\n",
- "\n",
- "\n",
- " Torque on the drill 1.250000 =Nm\n"
- ]
- }
- ],
- "prompt_number": 3
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Manufacturing_Engineering_&_Technology_by_S_Kalpakjian_and_S_R_Schmid/CHAPTER23.ipynb b/Manufacturing_Engineering_&_Technology_by_S_Kalpakjian_and_S_R_Schmid/CHAPTER23.ipynb deleted file mode 100755 index ef186b9c..00000000 --- a/Manufacturing_Engineering_&_Technology_by_S_Kalpakjian_and_S_R_Schmid/CHAPTER23.ipynb +++ /dev/null @@ -1,175 +0,0 @@ -{
- "metadata": {
- "name": "",
- "signature": "sha256:c676368a76c6427cb17b90c9717661144a581e3ccee839bd5939f7f9199412f4"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "CHAPTER 23 - Machining Processes used to Produce Various Shapes"
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "EXAMPLE 23.1 - PG NO. 600"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Example 23.1\n",
- "#page no. 600\n",
- "# Given that\n",
- "import math\n",
- "l=12.#in inch Length of block\n",
- "w=4\n",
- "f=0.01#in inch/tooth feed \n",
- "d=0.125#in inch depth of cut\n",
- "D=2.#in inch diameter of cutter\n",
- "n=20.#no. of teeth\n",
- "N=100.#in rpm spindle rotation\n",
- "Vt=8.#in inch/minute axial speed of the tool\n",
- "\n",
- "# Sample Problem on page no. 600\n",
- "\n",
- "print(\"\\n # Material Removal Rate , Power required and Cutting Time in slab milling # \\n\")\n",
- "\n",
- "v=f*N*n\n",
- "MRR=w*d*v \n",
- "print'%s %d %s' %(\"\\n\\n Material Removal Rate = \",MRR,\" in^3/min\")\n",
- "\n",
- "#for annealed mild steel unit power is taken as 1.1 hp min/in3\n",
- "P=1.1*MRR\n",
- "print'%s %d %s' %(\"\\n\\n Cutting power=\",P,\"hp\")\n",
- "\n",
- "T=P*33000/(N*2*3.14)\n",
- "print'%s %d %s' %(\"\\n\\n Cutting torque=\",T,\"lb-ft\")\n",
- "\n",
- "lc=math.sqrt(d*D)\n",
- "t=(300.+12.2)/500.\n",
- "print'%s %.6f %s' %(\"\\n\\n Cutting time=\",t*60,\"sec\")\n",
- "\n",
- "#Answers vary due to aproximations \n",
- "\n",
- "\n",
- "\n",
- "\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- " # Material Removal Rate , Power required and Cutting Time in slab milling # \n",
- "\n",
- "\n",
- "\n",
- " Material Removal Rate = 10 in^3/min\n",
- "\n",
- "\n",
- " Cutting power= 11 hp\n",
- "\n",
- "\n",
- " Cutting torque= 578 lb-ft\n",
- "\n",
- "\n",
- " Cutting time= 37.464000 sec\n"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "EXAMPLE 23.2 - PG NO. 655"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#example 23.2\n",
- "#page no. 655\n",
- "# Given that\n",
- "l=500#in mm Length\n",
- "w=60#in mm width\n",
- "v=0.6#in m/min \n",
- "d=3#in mm depth of cut\n",
- "D=150#in mm diameter of cutter\n",
- "n=10#no. of inserts\n",
- "N=100#in rpm spindle rotation\n",
- "\n",
- "# Sample Problem on page no. 655\n",
- "\n",
- "print(\"\\n # Material Removal Rate , Power Required and Cutting Time in Face Milling # \\n\")\n",
- "\n",
- "MRR=w*d*v*1000. \n",
- "print'%s %d %s' %(\"\\n\\n Material Removal Rate = \",MRR,\"mm^3/min\")\n",
- "\n",
- "lc=D/2.\n",
- "t=((l+(2.*lc))/((v*1000.)/60.)) # velocity is converted into mm/sec\n",
- "t1=t/60.\n",
- "print'%s %.6f %s' %(\"\\n\\n Cutting time= \",t1,\"f min\")\n",
- "\n",
- "f=(v*1000.*60.)/(60.*N*n) # N is converted into rev/sec by dividing by 60 , velocity is converted into mm/sec\n",
- "print'%s %.6f %s' %(\"\\n\\n Feed per Tooth =\",f,\"mm/tooth\")\n",
- "\n",
- "#for high strength aluminium alloy unit power is taken as 1.1 W s/mm3\n",
- "P=(1.1*MRR)/60. # MRR is converted into mm3/sec by dividing by 60\n",
- "P1=P/(1000.)#in KW\n",
- "print'%s %.6f %s' %(\"\\n\\n Cutting power =\",P1,\"KW\")\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- " # Material Removal Rate , Power Required and Cutting Time in Face Milling # \n",
- "\n",
- "\n",
- "\n",
- " Material Removal Rate = 108000 mm^3/min\n",
- "\n",
- "\n",
- " Cutting time= 1.083333 f min\n",
- "\n",
- "\n",
- " Feed per Tooth = 0.600000 mm/tooth\n",
- "\n",
- "\n",
- " Cutting power = 1.980000 KW\n"
- ]
- }
- ],
- "prompt_number": 2
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Manufacturing_Engineering_&_Technology_by_S_Kalpakjian_and_S_R_Schmid/CHAPTER25.ipynb b/Manufacturing_Engineering_&_Technology_by_S_Kalpakjian_and_S_R_Schmid/CHAPTER25.ipynb deleted file mode 100755 index ca6edbc0..00000000 --- a/Manufacturing_Engineering_&_Technology_by_S_Kalpakjian_and_S_R_Schmid/CHAPTER25.ipynb +++ /dev/null @@ -1,146 +0,0 @@ -{
- "metadata": {
- "name": "",
- "signature": "sha256:64df0f62add5674f5bdb088af7b099c9cc8d5cd0f207eda3eced371fc280619b"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "CHAPTER 25 - Abrasive Machining and Finishing Operations"
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "EXAMPLE 25.1 - PG NO. 713"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#example 25.1\n",
- "#page no. 713 \n",
- "import math\n",
- "# Given that\n",
- "D=200#in mm Grinding Wheel diameter \n",
- "d=0.05#in mm depth of cut\n",
- "v=30#m/min workpiece velocity\n",
- "V=1800#in m/min wheel velocity\n",
- "\n",
- "# Sample Problem on page no. 713\n",
- "\n",
- "print(\"\\n # Chip Dimensions in Surface Grinding # \\n\")\n",
- "\n",
- "l=math.sqrt(D*d)\n",
- "l1=l/2.54*(10**-1)\n",
- "print'%s %.6f %s'%(\"\\n\\n Undeformed Chip Length =\",l1,\"mm\")\n",
- "\n",
- "#the answer in the book is approximated to 0.13 in\n",
- "\n",
- "#assume\n",
- "C=2.#in mm\n",
- "r=15.\n",
- "t=math.sqrt(((4*v)/(V*C*r))*math.sqrt(d/D))\n",
- "t1=t/2.54*(10**-1)\n",
- "print'%s %.6f %s' %(\"\\n\\n Undeformed chip Thickness =\",t1,\"in\")\n",
- "\n",
- "#the answer in the book is approximated to 0.00023in\n",
- "\n",
- "\n",
- "\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- " # Chip Dimensions in Surface Grinding # \n",
- "\n",
- "\n",
- "\n",
- " Undeformed Chip Length = 0.124499 mm\n",
- "\n",
- "\n",
- " Undeformed chip Thickness = 0.000233 in\n"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 25.2 - Pg no. 715"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#example 25.2\n",
- "#page no. 715\n",
- "# Given that\n",
- "D=10.#in inch Grinding Wheel diameter\n",
- "N=4000.#in rpm \n",
- "w=1.#in inch \n",
- "d=0.002#in inch depth of cut\n",
- "v=60.#inch/min feed rate of the workpiece\n",
- "\n",
- "# Sample Problem on page no. 715\n",
- "\n",
- "print(\"\\n # force in Surface Grinding # \\n\")\n",
- "\n",
- "Mrr=d*w*v#material removal rate\n",
- "#for low carbon steel , the specific energy is 15hp min/in3\n",
- "u=15.#in hp min/in3\n",
- "P=u*Mrr*396000.#in lb/min\n",
- "Fc = P/(2*3.14*N*(D/2.))\n",
- "\n",
- "print'%s %.6f %s' %(\"\\n\\n Cutting Force =\",Fc,\"lb\")\n",
- "\n",
- "\n",
- "Fn = Fc+(30./100.)*Fc\n",
- "\n",
- "print'%s %.6f %s' %(\"\\n\\n Thrust Force =\",Fn,\"lb\")\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- " # force in Surface Grinding # \n",
- "\n",
- "\n",
- "\n",
- " Cutting Force = 5.675159 lb\n",
- "\n",
- "\n",
- " Thrust Force = 7.377707 lb\n"
- ]
- }
- ],
- "prompt_number": 2
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Manufacturing_Engineering_&_Technology_by_S_Kalpakjian_and_S_R_Schmid/CHAPTER28.ipynb b/Manufacturing_Engineering_&_Technology_by_S_Kalpakjian_and_S_R_Schmid/CHAPTER28.ipynb deleted file mode 100755 index 69d0545d..00000000 --- a/Manufacturing_Engineering_&_Technology_by_S_Kalpakjian_and_S_R_Schmid/CHAPTER28.ipynb +++ /dev/null @@ -1,90 +0,0 @@ -{
- "metadata": {
- "name": "",
- "signature": "sha256:c23f3c48fbab89dd7aecad46afb711bd99aac090001a21cb6e6d1af8ece79b90"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "CHAPTER 28 - Solid-State Welding Processes"
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "EXAMPLE 28.1 - PG NO. 805"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#example 28.1\n",
- "#page no. 805\n",
- "\n",
- "# Given that\n",
- "t=1.#in mm thickness of chip\n",
- "I=5000.#in Ampere current\n",
- "T=0.1#in sec\n",
- "d=5.#in mm diameter of electrode\n",
- "\n",
- "\n",
- "# Sample Problem on page no. 805\n",
- "\n",
- "print(\"\\n # Heat Generated in Spot Welding # \\n\")\n",
- "\n",
- "#It is assumed in the book that effective restiance = 200 micro ohm\n",
- "R=200.*(10.**-6.)\n",
- "H=(I**2.)*R*T\n",
- "\n",
- "print'%s %d %s' %(\"\\n\\n Heat Generated =\",H,\"J\")\n",
- "\n",
- "# It is assumed in the book that \n",
- "V=30.#in mm3 volume\n",
- "D=0.008#in g/mm3 density\n",
- "M=D*V\n",
- "#Heat required to melt 1 g of steel is about 1400J\n",
- "m1=1400.*M\n",
- "print'%s %d %s' %(\"\\n\\n Heat Required to melt weld nugget =\",m1,\" J\")\n",
- "\n",
- "m2=H-m1\n",
- "print'%s %d %s' %(\"\\n\\n Heat Dissipitated into the metal surrounding the nugget =\",m2,\" J\")\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- " # Heat Generated in Spot Welding # \n",
- "\n",
- "\n",
- "\n",
- " Heat Generated = 500 J\n",
- "\n",
- "\n",
- " Heat Required to melt weld nugget = 336 J\n",
- "\n",
- "\n",
- " Heat Dissipitated into the metal surrounding the nugget = 164 J\n"
- ]
- }
- ],
- "prompt_number": 1
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Manufacturing_Engineering_&_Technology_by_S_Kalpakjian_and_S_R_Schmid/CHAPTER32.ipynb b/Manufacturing_Engineering_&_Technology_by_S_Kalpakjian_and_S_R_Schmid/CHAPTER32.ipynb deleted file mode 100755 index 42cb2166..00000000 --- a/Manufacturing_Engineering_&_Technology_by_S_Kalpakjian_and_S_R_Schmid/CHAPTER32.ipynb +++ /dev/null @@ -1,73 +0,0 @@ -{
- "metadata": {
- "name": "",
- "signature": "sha256:4743efe71abeffbe112fc3cd99d282bbc1e9ab7a7bbe1d11e47f5fb2ca018d40"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "CHAPTER 32 - Tribology Friction Wear and Lubrication"
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "EXAMPLE 32.1 - PG NO. 886"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#example 32.1\n",
- "#page no. 886\n",
- "import math\n",
- "# Given that\n",
- "hi=10.#in mm height of specimen\n",
- "ODi=30.#in mm outside diameter \n",
- "IDi=15.#in mm inside diameter \n",
- "ODf=38.#in mm outside diameter after deformaton\n",
- "#Specimen is reduced in thickness by 50%\n",
- "hf=(50./100.)*hi\n",
- "\n",
- "# Sample Problem on page no. 886\n",
- "\n",
- "print(\"\\n # Determination of Cofficient of Friction # \\n\")\n",
- "\n",
- "IDf=math.sqrt((ODf**2.)-((((ODi**2.)-(IDi**2.))*hi)/hf)) #new internal diameter calculated , by comparing the volume before and after deformation (3.14/4)*(ODi**2-IDi**2)*hi=(3.14/4)*(ODf**2-IDf**2)*hf\n",
- "ID=((IDi-IDf)/IDi)*100#change in internal diameter \n",
- "\n",
- "print'%s %d %s %s' %(\"\\n\\n With a 50 percent reduction in height and a \",ID,\"%\",\" reduction in internal diameter, from the book data Cofficient of Friction = 0.21\") \n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- " # Determination of Cofficient of Friction # \n",
- "\n",
- "\n",
- "\n",
- " With a 50 percent reduction in height and a 35 % reduction in internal diameter, from the book data Cofficient of Friction = 0.21\n"
- ]
- }
- ],
- "prompt_number": 2
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Manufacturing_Engineering_&_Technology_by_S_Kalpakjian_and_S_R_Schmid/CHAPTER36.ipynb b/Manufacturing_Engineering_&_Technology_by_S_Kalpakjian_and_S_R_Schmid/CHAPTER36.ipynb deleted file mode 100755 index 7835b3b6..00000000 --- a/Manufacturing_Engineering_&_Technology_by_S_Kalpakjian_and_S_R_Schmid/CHAPTER36.ipynb +++ /dev/null @@ -1,159 +0,0 @@ -{
- "metadata": {
- "name": "",
- "signature": "sha256:1a827ea66b627083accd4f2478b3a04f75476c8ea952df501a53503ee0d2378e"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "CHAPTER 36 - Quality Assurance, Testing, and Inspection"
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "EXAMPLE 36.1 - PG NO. 978"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#example 36.1\n",
- "#page no.978 \n",
- "# Given that\n",
- "T=2.6#in mm wall thickness\n",
- "USL=3.2#in mm upper specification limit \n",
- "LSL=2.#in mm lower specification limit \n",
- "Y=2.6#in mm mean\n",
- "s=0.2#in mm standard deviation\n",
- "C1=10.#in dollar shipping included cost\n",
- "C2=50000.#in dollars improvement cost\n",
- "n=10000.#sections of tube per month\n",
- "# Sample Problem on page no. 978\n",
- "\n",
- "print(\"\\n # Production of Polymer Tubing # \\n\")\n",
- "\n",
- "k=C1/(USL-T)**2.\n",
- "LossCost=k*(((Y-T)**2.)+(s**2.))\n",
- "#after improvement the variation is half\n",
- "s1=0.2/2.\n",
- "LossCost1=k*(((Y-T)**2.)+(s1**2.))\n",
- "print'%s %.6f %s' %(\"\\n\\n Taguchi Loss Function = $\",LossCost1,\" per unit \")\n",
- "#answer in the book is approximated to $0.28 per unit \n",
- "\n",
- "savings=(LossCost-LossCost1)*n\n",
- "paybackperiod=C2/savings\n",
- "print'%s %.6f %s' %(\"\\n\\n Payback Period = \",paybackperiod+0.02,\" months\")\n",
- "#answer in the book is 6.02 months due to approximation savings \n",
- "\n",
- "\n",
- "\n",
- "\n",
- "\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- " # Production of Polymer Tubing # \n",
- "\n",
- "\n",
- "\n",
- " Taguchi Loss Function = $ 0.277778 per unit \n",
- "\n",
- "\n",
- " Payback Period = 6.020000 months\n"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "EXAMPLE 36.2 - PG NO. 990"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "# Given that\n",
- "n=5# in inch sample size\n",
- "m=10# in inch number of samples\n",
- "# The table of the queston is given of page no.990 Table 36.3\n",
- "\n",
- "# Sample Problem on page no. 990\n",
- "\n",
- "print(\"\\n # Calculation of Control Limits and Standard Deviation# \\n\")\n",
- "avgx=44.296 #from the table 36.3 by adding values of mean of x\n",
- "x = avgx/m\n",
- "avgR=1.03 #from the table 36.3 by adding values of R\n",
- "R = avgR/m\n",
- "#from the data in the book \n",
- "A2=0.577\n",
- "D4=2.115\n",
- "D3=0\n",
- "UCLx = x+(A2*R)\n",
- "LCLx = x-(A2*R)\n",
- "print'%s %.6f %s %.6f %s' %(\"\\n\\n Control Limits for Averages are =\\n UCLx =\",UCLx,\"in \\n UCLy =\",LCLx,\"in\") \n",
- "\n",
- "UCLR =D3*R\n",
- "LCLR =D4*R\n",
- "\n",
- "print'%s %.6f %s %.6f %s' %(\"\\n\\n Control Limits for Ranges are =\\n UCLR =\",UCLR,\"in \\n UCLR =\",LCLR,\"in\") \n",
- "\n",
- "#from table\n",
- "d2=2.326\n",
- "sigma= R/d2\n",
- "print'%s %.6f %s' %(\"\\n\\n Standard Deviation =\",sigma,\" in\") \n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- " # Calculation of Control Limits and Standard Deviation# \n",
- "\n",
- "\n",
- "\n",
- " Control Limits for Averages are =\n",
- " UCLx = 4.489031 in \n",
- " UCLy = 4.370169 in\n",
- "\n",
- "\n",
- " Control Limits for Ranges are =\n",
- " UCLR = 0.000000 in \n",
- " UCLR = 0.217845 in\n",
- "\n",
- "\n",
- " Standard Deviation = 0.044282 in\n"
- ]
- }
- ],
- "prompt_number": 4
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Manufacturing_Engineering_&_Technology_by_S_Kalpakjian_and_S_R_Schmid/CHAPTER9.ipynb b/Manufacturing_Engineering_&_Technology_by_S_Kalpakjian_and_S_R_Schmid/CHAPTER9.ipynb deleted file mode 100755 index bd6d81cf..00000000 --- a/Manufacturing_Engineering_&_Technology_by_S_Kalpakjian_and_S_R_Schmid/CHAPTER9.ipynb +++ /dev/null @@ -1,80 +0,0 @@ -{
- "metadata": {
- "name": "",
- "signature": "sha256:a2aee11149701e9ff173d8fdcf2dfa928b2913e3dc5ef2bf971d98cb5fa0f495"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "CHAPTER 9 - Composite Materials: Structure, General\n",
- "Properties, and Applications"
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "EXAMPLE 9.1 - PG NO. 229"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#example 9.1\n",
- "#page no. 229\n",
- "# Given that\n",
- "x=0.2# Area fraction of the fibre in the composite \n",
- "Ef= 300. # Elastic modulus of the fibre in GPa\n",
- "Em= 100. # Elastic modulus of the matrix in GPa\n",
- "\n",
- "# Sample Problem on page no. 229\n",
- "\n",
- "print(\"\\n # application of reinforced plastics # \\n\")\n",
- "\n",
- "Ec = x*Ef + (1.-x)*Em\n",
- "print'%s %d %s' %(\"\\n\\n The Elastic Modulus of the composite is = \",Ec,\"GPa\")\n",
- "\n",
- "#Let Pf/Pm be r\n",
- "r=x*Ef/((1.-x)*Em) \n",
- " \n",
- "#Let Pc/Pf be R\n",
- "R=1.+(1./r) # from the relation Pc = Pf + Pm\n",
- "P=(1.*100.)/R\n",
- "print'%s %.6f %s' %(\"\\n\\n The Fraction of load supported by Fibre is =\",P,\"%\")\n",
- "# Answer in the book is approximated to 43 %\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- " # application of reinforced plastics # \n",
- "\n",
- "\n",
- "\n",
- " The Elastic Modulus of the composite is = 140 GPa\n",
- "\n",
- "\n",
- " The Fraction of load supported by Fibre is = 42.857143 %\n"
- ]
- }
- ],
- "prompt_number": 2
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Manufacturing_Engineering_&_Technology_by_S_Kalpakjian_and_S_R_Schmid/screenshots/CHAP10.png b/Manufacturing_Engineering_&_Technology_by_S_Kalpakjian_and_S_R_Schmid/screenshots/CHAP10.png Binary files differdeleted file mode 100755 index b55eef3c..00000000 --- a/Manufacturing_Engineering_&_Technology_by_S_Kalpakjian_and_S_R_Schmid/screenshots/CHAP10.png +++ /dev/null diff --git a/Manufacturing_Engineering_&_Technology_by_S_Kalpakjian_and_S_R_Schmid/screenshots/CHAP16.png b/Manufacturing_Engineering_&_Technology_by_S_Kalpakjian_and_S_R_Schmid/screenshots/CHAP16.png Binary files differdeleted file mode 100755 index 41bb30ad..00000000 --- a/Manufacturing_Engineering_&_Technology_by_S_Kalpakjian_and_S_R_Schmid/screenshots/CHAP16.png +++ /dev/null diff --git a/Manufacturing_Engineering_&_Technology_by_S_Kalpakjian_and_S_R_Schmid/screenshots/CHAP23.png b/Manufacturing_Engineering_&_Technology_by_S_Kalpakjian_and_S_R_Schmid/screenshots/CHAP23.png Binary files differdeleted file mode 100755 index e0522cdd..00000000 --- a/Manufacturing_Engineering_&_Technology_by_S_Kalpakjian_and_S_R_Schmid/screenshots/CHAP23.png +++ /dev/null diff --git a/Manufacturing_Engineering_&_Technology_by_S._Kalpakjian_and_S._R._Schmid/CHAPTER10.ipynb b/Manufacturing_Engineering_and_Technology_by_S_Kalpakjian_and_S_R_Schmid/CHAPTER10.ipynb index ced687b6..ced687b6 100755 --- a/Manufacturing_Engineering_&_Technology_by_S._Kalpakjian_and_S._R._Schmid/CHAPTER10.ipynb +++ b/Manufacturing_Engineering_and_Technology_by_S_Kalpakjian_and_S_R_Schmid/CHAPTER10.ipynb diff --git a/Manufacturing_Engineering_&_Technology_by_S._Kalpakjian_and_S._R._Schmid/CHAPTER13.ipynb b/Manufacturing_Engineering_and_Technology_by_S_Kalpakjian_and_S_R_Schmid/CHAPTER13.ipynb index 29fe03ce..29fe03ce 100755 --- a/Manufacturing_Engineering_&_Technology_by_S._Kalpakjian_and_S._R._Schmid/CHAPTER13.ipynb +++ b/Manufacturing_Engineering_and_Technology_by_S_Kalpakjian_and_S_R_Schmid/CHAPTER13.ipynb diff --git a/Manufacturing_Engineering_&_Technology_by_S._Kalpakjian_and_S._R._Schmid/CHAPTER14.ipynb b/Manufacturing_Engineering_and_Technology_by_S_Kalpakjian_and_S_R_Schmid/CHAPTER14.ipynb index 7107f78a..7107f78a 100755 --- a/Manufacturing_Engineering_&_Technology_by_S._Kalpakjian_and_S._R._Schmid/CHAPTER14.ipynb +++ b/Manufacturing_Engineering_and_Technology_by_S_Kalpakjian_and_S_R_Schmid/CHAPTER14.ipynb diff --git a/Manufacturing_Engineering_&_Technology_by_S._Kalpakjian_and_S._R._Schmid/CHAPTER15.ipynb b/Manufacturing_Engineering_and_Technology_by_S_Kalpakjian_and_S_R_Schmid/CHAPTER15.ipynb index c751ca22..c751ca22 100755 --- a/Manufacturing_Engineering_&_Technology_by_S._Kalpakjian_and_S._R._Schmid/CHAPTER15.ipynb +++ b/Manufacturing_Engineering_and_Technology_by_S_Kalpakjian_and_S_R_Schmid/CHAPTER15.ipynb diff --git a/Manufacturing_Engineering_&_Technology_by_S._Kalpakjian_and_S._R._Schmid/CHAPTER16.ipynb b/Manufacturing_Engineering_and_Technology_by_S_Kalpakjian_and_S_R_Schmid/CHAPTER16.ipynb index ced88066..ced88066 100755 --- a/Manufacturing_Engineering_&_Technology_by_S._Kalpakjian_and_S._R._Schmid/CHAPTER16.ipynb +++ b/Manufacturing_Engineering_and_Technology_by_S_Kalpakjian_and_S_R_Schmid/CHAPTER16.ipynb diff --git a/Manufacturing_Engineering_&_Technology_by_S._Kalpakjian_and_S._R._Schmid/CHAPTER17.ipynb b/Manufacturing_Engineering_and_Technology_by_S_Kalpakjian_and_S_R_Schmid/CHAPTER17.ipynb index 2318f215..2318f215 100755 --- a/Manufacturing_Engineering_&_Technology_by_S._Kalpakjian_and_S._R._Schmid/CHAPTER17.ipynb +++ b/Manufacturing_Engineering_and_Technology_by_S_Kalpakjian_and_S_R_Schmid/CHAPTER17.ipynb diff --git a/Manufacturing_Engineering_&_Technology_by_S._Kalpakjian_and_S._R._Schmid/CHAPTER18.ipynb b/Manufacturing_Engineering_and_Technology_by_S_Kalpakjian_and_S_R_Schmid/CHAPTER18.ipynb index 1ea3fcf0..1ea3fcf0 100755 --- a/Manufacturing_Engineering_&_Technology_by_S._Kalpakjian_and_S._R._Schmid/CHAPTER18.ipynb +++ b/Manufacturing_Engineering_and_Technology_by_S_Kalpakjian_and_S_R_Schmid/CHAPTER18.ipynb diff --git a/Manufacturing_Engineering_&_Technology_by_S._Kalpakjian_and_S._R._Schmid/CHAPTER2.ipynb b/Manufacturing_Engineering_and_Technology_by_S_Kalpakjian_and_S_R_Schmid/CHAPTER2.ipynb index 0f29dc44..0f29dc44 100755 --- a/Manufacturing_Engineering_&_Technology_by_S._Kalpakjian_and_S._R._Schmid/CHAPTER2.ipynb +++ b/Manufacturing_Engineering_and_Technology_by_S_Kalpakjian_and_S_R_Schmid/CHAPTER2.ipynb diff --git a/Manufacturing_Engineering_&_Technology_by_S._Kalpakjian_and_S._R._Schmid/CHAPTER20.ipynb b/Manufacturing_Engineering_and_Technology_by_S_Kalpakjian_and_S_R_Schmid/CHAPTER20.ipynb index 00e9c991..00e9c991 100755 --- a/Manufacturing_Engineering_&_Technology_by_S._Kalpakjian_and_S._R._Schmid/CHAPTER20.ipynb +++ b/Manufacturing_Engineering_and_Technology_by_S_Kalpakjian_and_S_R_Schmid/CHAPTER20.ipynb diff --git a/Manufacturing_Engineering_&_Technology_by_S._Kalpakjian_and_S._R._Schmid/CHAPTER22.ipynb b/Manufacturing_Engineering_and_Technology_by_S_Kalpakjian_and_S_R_Schmid/CHAPTER22.ipynb index 9975e7dd..9975e7dd 100755 --- a/Manufacturing_Engineering_&_Technology_by_S._Kalpakjian_and_S._R._Schmid/CHAPTER22.ipynb +++ b/Manufacturing_Engineering_and_Technology_by_S_Kalpakjian_and_S_R_Schmid/CHAPTER22.ipynb diff --git a/Manufacturing_Engineering_&_Technology_by_S._Kalpakjian_and_S._R._Schmid/CHAPTER23.ipynb b/Manufacturing_Engineering_and_Technology_by_S_Kalpakjian_and_S_R_Schmid/CHAPTER23.ipynb index 1a90ba0b..1a90ba0b 100755 --- a/Manufacturing_Engineering_&_Technology_by_S._Kalpakjian_and_S._R._Schmid/CHAPTER23.ipynb +++ b/Manufacturing_Engineering_and_Technology_by_S_Kalpakjian_and_S_R_Schmid/CHAPTER23.ipynb diff --git a/Manufacturing_Engineering_&_Technology_by_S._Kalpakjian_and_S._R._Schmid/CHAPTER25.ipynb b/Manufacturing_Engineering_and_Technology_by_S_Kalpakjian_and_S_R_Schmid/CHAPTER25.ipynb index d985de21..d985de21 100755 --- a/Manufacturing_Engineering_&_Technology_by_S._Kalpakjian_and_S._R._Schmid/CHAPTER25.ipynb +++ b/Manufacturing_Engineering_and_Technology_by_S_Kalpakjian_and_S_R_Schmid/CHAPTER25.ipynb diff --git a/Manufacturing_Engineering_&_Technology_by_S._Kalpakjian_and_S._R._Schmid/CHAPTER28.ipynb b/Manufacturing_Engineering_and_Technology_by_S_Kalpakjian_and_S_R_Schmid/CHAPTER28.ipynb index ca84e122..ca84e122 100755 --- a/Manufacturing_Engineering_&_Technology_by_S._Kalpakjian_and_S._R._Schmid/CHAPTER28.ipynb +++ b/Manufacturing_Engineering_and_Technology_by_S_Kalpakjian_and_S_R_Schmid/CHAPTER28.ipynb diff --git a/Manufacturing_Engineering_&_Technology_by_S._Kalpakjian_and_S._R._Schmid/CHAPTER32.ipynb b/Manufacturing_Engineering_and_Technology_by_S_Kalpakjian_and_S_R_Schmid/CHAPTER32.ipynb index 8e0a7bd9..8e0a7bd9 100755 --- a/Manufacturing_Engineering_&_Technology_by_S._Kalpakjian_and_S._R._Schmid/CHAPTER32.ipynb +++ b/Manufacturing_Engineering_and_Technology_by_S_Kalpakjian_and_S_R_Schmid/CHAPTER32.ipynb diff --git a/Manufacturing_Engineering_&_Technology_by_S._Kalpakjian_and_S._R._Schmid/CHAPTER36.ipynb b/Manufacturing_Engineering_and_Technology_by_S_Kalpakjian_and_S_R_Schmid/CHAPTER36.ipynb index 8ec75437..8ec75437 100755 --- a/Manufacturing_Engineering_&_Technology_by_S._Kalpakjian_and_S._R._Schmid/CHAPTER36.ipynb +++ b/Manufacturing_Engineering_and_Technology_by_S_Kalpakjian_and_S_R_Schmid/CHAPTER36.ipynb diff --git a/Manufacturing_Engineering_&_Technology_by_S._Kalpakjian_and_S._R._Schmid/CHAPTER9.ipynb b/Manufacturing_Engineering_and_Technology_by_S_Kalpakjian_and_S_R_Schmid/CHAPTER9.ipynb index 12734f12..12734f12 100755 --- a/Manufacturing_Engineering_&_Technology_by_S._Kalpakjian_and_S._R._Schmid/CHAPTER9.ipynb +++ b/Manufacturing_Engineering_and_Technology_by_S_Kalpakjian_and_S_R_Schmid/CHAPTER9.ipynb diff --git a/Manufacturing_Engineering_&_Technology_by_S._Kalpakjian_and_S._R._Schmid/screenshots/CHAP10.png b/Manufacturing_Engineering_and_Technology_by_S_Kalpakjian_and_S_R_Schmid/screenshots/CHAP10.png Binary files differindex b55eef3c..b55eef3c 100755 --- a/Manufacturing_Engineering_&_Technology_by_S._Kalpakjian_and_S._R._Schmid/screenshots/CHAP10.png +++ b/Manufacturing_Engineering_and_Technology_by_S_Kalpakjian_and_S_R_Schmid/screenshots/CHAP10.png diff --git a/Manufacturing_Engineering_&_Technology_by_S._Kalpakjian_and_S._R._Schmid/screenshots/CHAP16.png b/Manufacturing_Engineering_and_Technology_by_S_Kalpakjian_and_S_R_Schmid/screenshots/CHAP16.png Binary files differindex 41bb30ad..41bb30ad 100755 --- a/Manufacturing_Engineering_&_Technology_by_S._Kalpakjian_and_S._R._Schmid/screenshots/CHAP16.png +++ b/Manufacturing_Engineering_and_Technology_by_S_Kalpakjian_and_S_R_Schmid/screenshots/CHAP16.png diff --git a/Manufacturing_Engineering_&_Technology_by_S._Kalpakjian_and_S._R._Schmid/screenshots/CHAP23.png b/Manufacturing_Engineering_and_Technology_by_S_Kalpakjian_and_S_R_Schmid/screenshots/CHAP23.png Binary files differindex e0522cdd..e0522cdd 100755 --- a/Manufacturing_Engineering_&_Technology_by_S._Kalpakjian_and_S._R._Schmid/screenshots/CHAP23.png +++ b/Manufacturing_Engineering_and_Technology_by_S_Kalpakjian_and_S_R_Schmid/screenshots/CHAP23.png |
