From 64d949698432e05f2a372d9edc859c5b9df1f438 Mon Sep 17 00:00:00 2001 From: kinitrupti Date: Fri, 12 May 2017 18:40:35 +0530 Subject: Revised list of TBCs --- sample_notebooks/kumargugloth/Chapter1.ipynb | 290 +++------------------ .../kumargugloth/Chapter1_wopEYRj.ipynb | 130 --------- 2 files changed, 36 insertions(+), 384 deletions(-) mode change 100755 => 100644 sample_notebooks/kumargugloth/Chapter1.ipynb delete mode 100644 sample_notebooks/kumargugloth/Chapter1_wopEYRj.ipynb (limited to 'sample_notebooks/kumargugloth') diff --git a/sample_notebooks/kumargugloth/Chapter1.ipynb b/sample_notebooks/kumargugloth/Chapter1.ipynb old mode 100755 new mode 100644 index df9ba4d0..fdfb0cb9 --- a/sample_notebooks/kumargugloth/Chapter1.ipynb +++ b/sample_notebooks/kumargugloth/Chapter1.ipynb @@ -1,7 +1,7 @@ { "metadata": { "name": "", - "signature": "sha256:84e452258bd05b64c16351467c4970051f4494cb47d7a832df03bdce07abddb8" + "signature": "sha256:281275d36b0e16d144d1212530d5ebac420ea6bfd258dbfe43c04ce417d0dbbc" }, "nbformat": 3, "nbformat_minor": 0, @@ -13,7 +13,7 @@ "level": 1, "metadata": {}, "source": [ - "Chapter1-Introduction" + "Chapter1-Atomic Weight " ] }, { @@ -21,34 +21,23 @@ "level": 2, "metadata": {}, "source": [ - "Ex1-pg9" + "Ex1-pg12" ] }, { "cell_type": "code", "collapsed": false, "input": [ - "\n", "import math\n", - " #determine\n", - "##This numerical is Ex 1_1E,page 9.\n", - "Pso=20.5\n", - "Psc=20.5*550##converting hp to fps system\n", - "Qo=385.\n", - "Qc=385./449.##converting gpm to ft^3/s\n", - "E=0.83\n", - "dp=E*Psc/(Qc*144.)\n", - "print\"%s %.2f %s \"%('The pressure rise is ',dp,' psi')\n", - "print(\"After rounding off,pressure rise is 75.8 psi\")\n", - "dpr=75.8\n", - "dHw=75.8*144/62.4##62.4 is accelaration due to gravity in fps system\n", - "print\"%s %.2f %s \"%(' The head of water is ',dHw,' ft of water')\n", - "print(\"After rounding off the value of head of water the answer is 175 ft of water.\")\n", - "dhwr=175##rounded off value of head of water\n", - "sg=0.72##specific gravity of oil\n", - "dHo=dhwr/sg\n", - "print\"%s %.2f %s \"%(' The head of oil is ',dHo,' ft of oil')\n", - "print(\"After rounding off the value of head of oil the answer is 243 ft of oil.\")\n" + "##Intitalisation of variables\n", + "#calculate the Molecular weight of carbon dioxide\n", + "dco= 1.9635 ##gms/lit\n", + "do= 1.4277 ##gms/lit\n", + "mo= 32. ##gms\n", + "##CALCULATIONS\n", + "mwt= dco*mo/do\n", + "##RESULTS\n", + "print'%s %.2f %s'% ('Molecular weight of carbon dioxide = ',mwt,'')\n" ], "language": "python", "metadata": {}, @@ -57,125 +46,7 @@ "output_type": "stream", "stream": "stdout", "text": [ - "The pressure rise is 75.79 psi \n", - "After rounding off,pressure rise is 75.8 psi\n", - " The head of water is 174.92 ft of water \n", - "After rounding off the value of head of water the answer is 175 ft of water.\n", - " The head of oil is 243.06 ft of oil \n", - "After rounding off the value of head of oil the answer is 243 ft of oil.\n" - ] - } - ], - "prompt_number": 1 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Ex2-pg10" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "#determine\n", - "##This numerical is Ex 1_1S,page 10.\n", - "E=0.83##efficiency\n", - "Ps=15300.\n", - "Q=87.4\n", - "Qs=87.4/3600.##flow rate in meter cube per sec\n", - "rho=998.\n", - "g=9.81\n", - "sg=0.72\n", - "dp=E*Ps/Qs\n", - "print\"%s %.2f %s \"%('\\n The change in pressure (dp)is ',dp,'')\n", - "dpr=523000##rounded value of dp\n", - "print(\"The rounded off value of dp is 523kPa.\")\n", - "dHw=dpr/(rho*g)\n", - "print\"%s %.2f %s \"%(' dHw is equal to ',dHw,' m of water')\n", - "print(\"The rounded off value of dHw is 53.4 m of water.\")\n", - "dHwr=53.4##rounded off value of dHw\n", - "print(\"Thus we can determine head of oil.\")\n", - "dHoil=dHwr/sg\n", - "print\"%s %.2f %s \"%(' dHoil is given by ',dHoil,' m of oil')\n", - "print(\"The rounded off value of dHoil is 74.2 m of oil.\")\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "\n", - " The change in pressure (dp)is 523070.94 \n", - "The rounded off value of dp is 523kPa.\n", - " dHw is equal to 53.42 m of water \n", - "The rounded off value of dHw is 53.4 m of water.\n", - "Thus we can determine head of oil.\n", - " dHoil is given by 74.17 m of oil \n", - "The rounded off value of dHoil is 74.2 m of oil.\n" - ] - } - ], - "prompt_number": 2 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Ex3-pg10" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "#determine\n", - "##This numerical is Ex 1_2E,page 10.\n", - "Q=12000.\n", - "A=3.5\n", - "rho_a=0.0762\n", - "E=0.85\n", - "r=2.5##resistance of duct system\n", - "V=Q/(60.*A)\n", - "print\"%s %.2f %s \"%('The air flow velocity at discharge is ',V,' ft/s')\n", - "KE=(rho_a*(V**2))/(32.2*2)\n", - "print\"%s %.2f %s \"%('\\n The product is ',KE,' lb/ft^2')\n", - "##PE=KE\n", - "Hv=KE/62.4\n", - "print\"%s %.2f %s \"%('\\n The dynamic head is ',Hv,' ft')\n", - "print(\"The value of dynamic head in inches of water is 0.74.\")\n", - "Hvi=0.74##Head in inches\n", - "Ht=r+Hvi\n", - "print\"%s %.2f %s \"%('\\n The total head is ',Ht,' inches of water')\n", - "p_tot=Ht*62.4\n", - "Ps=Q*p_tot/(60.*12.*E)\n", - "print\"%s %.2f %s \"%('\\n The shaft power is ',Ps,' ft-lb/s')\n", - "print(\"The shaft power is 7.2 hp.\")\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "The air flow velocity at discharge is 57.14 ft/s \n", - "\n", - " The product is 3.86 lb/ft^2 \n", - "\n", - " The dynamic head is 0.06 ft \n", - "The value of dynamic head in inches of water is 0.74.\n", - "\n", - " The total head is 3.24 inches of water \n", - "\n", - " The shaft power is 3964.24 ft-lb/s \n", - "The shaft power is 7.2 hp.\n" + "Molecular weight of carbon dioxide = 44.01 \n" ] } ], @@ -186,91 +57,23 @@ "level": 2, "metadata": {}, "source": [ - "Ex4-pg11" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "##This numerical is Ex 1_2S,page 11.\n", - "Q=340.\n", - "A=0.325\n", - "V=Q/(60.*A)\n", - "print\"%s %.2f %s \"%('The air flow velocity at discharge is ',V,' m/s')\n", - "rho_a=1.22\n", - "Vr=17.4\n", - "Hd=(rho_a*(Vr**2))/2.\n", - "print\"%s %.2f %s \"%('\\n The dynamic pressure head is ',Hd,' Pa')\n", - "Hdr=184.7##rounded off value of Hd\n", - "rho_w=998.##density of water=rhow\n", - "g=9.81\n", - "H=0.0635\n", - "dp=rho_w*g*H##static pressure head\n", - "print\"%s %.2f %s \"%('\\n The static pressure head is ',dp,' Pa')\n", - "dpr=621.7\n", - "p_tot=Hdr+dpr\n", - "print\"%s %.2f %s \"%('\\n The total pressure head is ',p_tot,' Pa')\n", - "p_tot=806.4\n", - "E=0.85##efficiency\n", - "Ps=Q*p_tot/(60*E)\n", - "print\"%s %.2f %s \"%('\\n The shaft power is',Ps, 'W')\n", - "print(\"The shaft power is 5.376 kW.\")\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "The air flow velocity at discharge is 17.44 m/s \n", - "\n", - " The dynamic pressure head is 184.68 Pa \n", - "\n", - " The static pressure head is 621.69 Pa \n", - "\n", - " The total pressure head is 806.40 Pa \n", - "\n", - " The shaft power is 5376.00 W \n", - "The shaft power is 5.376 kW.\n" - ] - } - ], - "prompt_number": 5 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Ex5-pg11" + "Ex2-pg13" ] }, { "cell_type": "code", "collapsed": false, "input": [ - "#determine \n", "import math\n", - "##This numerical is Ex 1_3E,page 11.\n", - "H=295.##net head in ft\n", - "Q=148.##water flow rate\n", - "n=1800.##rpm\n", - "E=0.87##efficiency\n", - "a=62.4##product of density and accelaration due to gravity\n", - "omega=(n*2.*math.pi)/60.\n", - "dp=a*H\n", - "print\"%s %.2f %s \"%('The pressure is ',dp,' lb/ft^2')\n", - "Ps=E*Q*dp\n", - "print\"%s %.2f %s \"%('\\n Output power is equal to ',Ps,' lb-ft/s')\n", - "print(\"The output output power can also be written as 2.37*10^6 lb-ft/s\")\n", - "print(\"Output power in terms of horsepower is given by 4309hp.\")\n", - "Psr=2370000##rounded off value of Ps\n", - "Torque=Psr/omega\n", - "print\"%s %.2f %s \"%(' The output torque is ',Torque,' lb-ft.')\n", - "print(\"The output torque can also be written as 12.57*10^3 lb-ft\")\n", - "\n" + "##Intitalisation of variables\n", + "#calculate the atomic weight of lead\n", + "shl= 0.031 ##cal deg^-1 g^-1\n", + "ewlc= 103.605 ##gms\n", + "n= 2.\n", + "##CALCULATIONS\n", + "aw= n*ewlc\n", + "##RESULTS\n", + "print'%s %.2f %s'% ('Atomic weight of lead = ',aw,' gms')\n" ], "language": "python", "metadata": {}, @@ -279,49 +82,33 @@ "output_type": "stream", "stream": "stdout", "text": [ - "The pressure is 18408.00 lb/ft^2 \n", - "\n", - " Output power is equal to 2370214.08 lb-ft/s \n", - "The output output power can also be written as 2.37*10^6 lb-ft/s\n", - "Output power in terms of horsepower is given by 4309hp.\n", - " The output torque is 12573.24 lb-ft. \n", - "The output torque can also be written as 12.57*10^3 lb-ft\n" + "Atomic weight of lead = 207.21 gms\n" ] } ], - "prompt_number": 6 + "prompt_number": 3 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ - "Ex6-pg12" + "Ex3-pg13" ] }, { "cell_type": "code", "collapsed": false, "input": [ - "#determine c\n", "import math\n", - "##This numerical is Ex 1_3S,page 12.\n", - "H=90.\n", - "Q=4.2##water flow rate(in m^3/s)\n", - "n=1800.\n", - "E=0.87##efficiency\n", - "rho=998.\n", - "g=9.81\n", - "omega=(n*2.*math.pi)/60.\n", - "dp=rho*g*H\n", - "print\"%s %.2f %s \"%('The pressure is ',dp,' N/m^2')\n", - "Ps=E*Q*dp\n", - "print\"%s %.2f %s \"%('\\n Output power is equal to ',Ps,' N-m/s')\n", - "print(\"After rounding off the value of output power is 3220 kW.\")\n", - "Psr=3220000.##rounded off value of Ps\n", - "Torque=Psr/omega\n", - "print\"%s %.2f %s \"%(' The output torque is ',Torque,' N-m.')\n", - "print(\"After rounding off the output torque comes out to be 17.1*10^3 N-m.\")\n" + "##Intitalisation of variables\n", + "\n", + "ewt= 17.337 ##gms\n", + "n=3.\n", + "##CALCULATIONS\n", + "aw= ewt*n\n", + "##RESULTS\n", + "print'%s %.2f %s'% ('Atomic weight of chromium = ',aw,' gms')\n" ], "language": "python", "metadata": {}, @@ -330,16 +117,11 @@ "output_type": "stream", "stream": "stdout", "text": [ - "The pressure is 881134.20 N/m^2 \n", - "\n", - " Output power is equal to 3219664.37 N-m/s \n", - "After rounding off the value of output power is 3220 kW.\n", - " The output torque is 17082.63 N-m. \n", - "After rounding off the output torque comes out to be 17.1*10^3 N-m.\n" + "Atomic weight of chromium = 52.01 gms\n" ] } ], - "prompt_number": 7 + "prompt_number": 2 } ], "metadata": {} diff --git a/sample_notebooks/kumargugloth/Chapter1_wopEYRj.ipynb b/sample_notebooks/kumargugloth/Chapter1_wopEYRj.ipynb deleted file mode 100644 index fdfb0cb9..00000000 --- a/sample_notebooks/kumargugloth/Chapter1_wopEYRj.ipynb +++ /dev/null @@ -1,130 +0,0 @@ -{ - "metadata": { - "name": "", - "signature": "sha256:281275d36b0e16d144d1212530d5ebac420ea6bfd258dbfe43c04ce417d0dbbc" - }, - "nbformat": 3, - "nbformat_minor": 0, - "worksheets": [ - { - "cells": [ - { - "cell_type": "heading", - "level": 1, - "metadata": {}, - "source": [ - "Chapter1-Atomic Weight " - ] - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Ex1-pg12" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math\n", - "##Intitalisation of variables\n", - "#calculate the Molecular weight of carbon dioxide\n", - "dco= 1.9635 ##gms/lit\n", - "do= 1.4277 ##gms/lit\n", - "mo= 32. ##gms\n", - "##CALCULATIONS\n", - "mwt= dco*mo/do\n", - "##RESULTS\n", - "print'%s %.2f %s'% ('Molecular weight of carbon dioxide = ',mwt,'')\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Molecular weight of carbon dioxide = 44.01 \n" - ] - } - ], - "prompt_number": 4 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Ex2-pg13" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math\n", - "##Intitalisation of variables\n", - "#calculate the atomic weight of lead\n", - "shl= 0.031 ##cal deg^-1 g^-1\n", - "ewlc= 103.605 ##gms\n", - "n= 2.\n", - "##CALCULATIONS\n", - "aw= n*ewlc\n", - "##RESULTS\n", - "print'%s %.2f %s'% ('Atomic weight of lead = ',aw,' gms')\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Atomic weight of lead = 207.21 gms\n" - ] - } - ], - "prompt_number": 3 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Ex3-pg13" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math\n", - "##Intitalisation of variables\n", - "\n", - "ewt= 17.337 ##gms\n", - "n=3.\n", - "##CALCULATIONS\n", - "aw= ewt*n\n", - "##RESULTS\n", - "print'%s %.2f %s'% ('Atomic weight of chromium = ',aw,' gms')\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Atomic weight of chromium = 52.01 gms\n" - ] - } - ], - "prompt_number": 2 - } - ], - "metadata": {} - } - ] -} \ No newline at end of file -- cgit