From 37d315828bbfc0f5cabee669d2b9dd8cd17b5154 Mon Sep 17 00:00:00 2001 From: hardythe1 Date: Wed, 17 Jun 2015 11:14:34 +0530 Subject: add books --- sample_notebooks/kumargugloth/Chapter1.ipynb | 348 +++++++++++++++++++++++++++ 1 file changed, 348 insertions(+) create mode 100755 sample_notebooks/kumargugloth/Chapter1.ipynb (limited to 'sample_notebooks/kumargugloth') diff --git a/sample_notebooks/kumargugloth/Chapter1.ipynb b/sample_notebooks/kumargugloth/Chapter1.ipynb new file mode 100755 index 00000000..df9ba4d0 --- /dev/null +++ b/sample_notebooks/kumargugloth/Chapter1.ipynb @@ -0,0 +1,348 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:84e452258bd05b64c16351467c4970051f4494cb47d7a832df03bdce07abddb8" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter1-Introduction" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Ex1-pg9" + ] + }, + { + "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" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "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" + ] + } + ], + "prompt_number": 4 + }, + { + "cell_type": "heading", + "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" + ] + }, + { + "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" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "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" + ] + } + ], + "prompt_number": 6 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Ex6-pg12" + ] + }, + { + "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" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "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" + ] + } + ], + "prompt_number": 7 + } + ], + "metadata": {} + } + ] +} \ No newline at end of file -- cgit