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diff --git a/Problems_In_Hydraulics/ch4.ipynb b/Problems_In_Hydraulics/ch4.ipynb new file mode 100755 index 00000000..aefd97a7 --- /dev/null +++ b/Problems_In_Hydraulics/ch4.ipynb @@ -0,0 +1,291 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:f02a7c0105dc5d8192eae252809dedbf748fa59e0b9b78e4af3b1079610b7a69" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter 4 : Orifices and Notches" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 4.1 Page No : 51" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "import math \n", + "\n", + "#initialisation of variables\n", + "Cd= 0.98 # velocity\n", + "g= 32.2 \t#ft/sec**2\n", + "H= 2. \t#ft\n", + "\n", + "#CALCULATIONS\n", + "v= math.sqrt(2*g*H)\n", + "t= H/v\n", + "h= 0.5*g*t**2\n", + "\n", + "#RESULTS\n", + "print 'Vertical distance fallen in this ttime = %.3f ft'%(h) \n", + "\n", + "#Note : The answer given in textbook is wrong.\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Vertical distance fallen in this ttime = 0.500 ft\n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 4.2 Page No : 51" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "\n", + "#initialisation of variables\n", + "r= 53.4\n", + "T= 60. \t #F pressure of air\n", + "h= 29.7 \t#in of mercury\n", + "sm= 13.6\n", + "w= 62.4 \t#lb/ft**3\n", + "d= 1.5 \t#in diameter\n", + "Qin= 2. \t#cuses air\n", + "g=32.2 \t #ft/s**2\n", + "\n", + "#CALCULATIONS\n", + "W= h*sm*w/(r*(460+T)*12)\n", + "dP= 0.75*w/(12*W)\n", + "Q= math.sqrt(2*g*dP)*math.pi*d**2/(4*144)\n", + "W= Q*W*60\n", + "Cd= Qin/W\n", + "\n", + "#RESULTS\n", + "print 'coefficient of discharge = %.2f '%(Cd) \n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "coefficient of discharge = 0.62 \n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 4.3 Page No : 52" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "import math \n", + "\n", + "#initialisation of variables\n", + "H1= 34. \t#ft height\n", + "H2= 8. \t #ft head\n", + "H3= 7. \t#ft pressure head\n", + "g= 32.2 \t#ft/sec**2\n", + "d= 1.5 \t #in\n", + "\n", + "#CALCULATIONS\n", + "v2= math.sqrt(2*g*(H1+H2-H3))\n", + "Q= v2*math.pi*d**2/(4*144)\n", + "v3= (2*v2+math.sqrt(4*v2**2-4*6*(v2**2-H2*2*5*g)))/12\n", + "dr= math.sqrt(v2/v3)\n", + "\n", + "#RESULTS\n", + "print 'ratio of diameteres = %.1f '%(dr) \n", + "print \" Maximum discharge = %.3f cusec\"%(Q)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "ratio of diameteres = 1.6 \n", + " Maximum discharge = 0.583 cusec\n" + ] + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 4.4 Page No : 54" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "import math \n", + "\n", + "#initialisation of variables\n", + "Q1= 8./15 \t#cuses\n", + "Q2= 2./15 \t#cuses\n", + "\n", + "#CALCULATIONS\n", + "A= math.degrees(math.atan(Q2/Q1))\n", + "\n", + "#RESULTS\n", + "print 'Angle of inclination = %.2f degrees'%(A) \n", + "\n", + "# rounding off error" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Angle of inclination = 14.04 degrees\n" + ] + } + ], + "prompt_number": 4 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 4.5 Page No : 56" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "import math \n", + "\n", + "#initialisation of variables\n", + "g= 32.2 \t#ft/sec**2\n", + "\n", + "#CALCULATIONS\n", + "r= g**2/((math.sqrt(2))**2*g**2)\n", + "\n", + "#RESULTS\n", + "print 'coefficient of contraction = %.1f '%(r) \n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "coefficient of contraction = 0.5 \n" + ] + } + ], + "prompt_number": 6 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 4.6 Page No : 56" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "import math \n", + "\n", + "#initialisation of variables\n", + "B= 3. \t #ft long\n", + "H= 2. \t#ft depth of water\n", + "H1= 3.75 \t#ft \n", + "w= 4. \t#ft wide\n", + "g= 32.2 \t#ft/sec**2\n", + "\n", + "#CALCULATIONS\n", + "Q= 3.33*(B-(H1/5))*H**1.5\n", + "v= Q/(H*w)\n", + "kh= v**2/(2*g)\n", + "Q1= 3.33*(B-(H1/5)-kh)*(((H1/5)+kh)**1.5-kh**1.5)\n", + "\n", + "#RESULTS\n", + "print 'Discharge = %.2f cuses'%(Q1) \n", + "\n", + "\n", + "# NOte : ANSWER IN THE TEXTBOOK IS WRONG\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Discharge = 5.42 cuses\n" + ] + } + ], + "prompt_number": 5 + }, + { + "cell_type": "code", + "collapsed": false, + "input": [], + "language": "python", + "metadata": {}, + "outputs": [] + } + ], + "metadata": {} + } + ] +}
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