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diff --git a/Hydraulics_Made_Easy/ch10.ipynb b/Hydraulics_Made_Easy/ch10.ipynb new file mode 100755 index 00000000..0162251f --- /dev/null +++ b/Hydraulics_Made_Easy/ch10.ipynb @@ -0,0 +1,497 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:2a1a58d10d8bc25ee316405092c20184a6b3d847d7641c4fa061749730145581" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter 10 : Miscellaneous Problems" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 10.1 Page No : 316" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "\n", + "#initialisation of variables\n", + "w = 62.4 \t\t#lb/ft**3\n", + "x = 8. \t\t#ft\n", + "A = 16. \t\t#ft**2\n", + "X = 2.5 \t\t#ft\n", + "X1 = 0.66 \t\t#ft\n", + "x1 = 3.834 \t\t#ft\n", + "x2 = 2.182 \t\t#ft\n", + "\t\t\n", + "#CALCULATIONS\n", + "P = w*x*A\n", + "y = A/3\n", + "P1 = w*x*A*0.5*X1\n", + "R = math.sqrt(P1**2+P**2)\n", + "m = P1/P\n", + "X2 = x1-x2\n", + "C = ((2./3)*A)-m*X\n", + "Y = m*X2+ C\n", + "print P1\t\n", + "#RESULTS\n", + "print 'Water pressure on vertical face = %.f lbs'%(round(P,-3))\n", + "print ' pressure which acts at the base = %.2f ft'%(y)\n", + "print ' Resultant = %.f lbs'%(R)\n", + "print ' x coordinate of the resultant = %.3f ft'%(X2)\n", + "print ' y coordinate of the resultant = %.3f ft'%(Y)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "2635.776\n", + "Water pressure on vertical face = 8000 lbs\n", + " pressure which acts at the base = 5.33 ft\n", + " Resultant = 8411 lbs\n", + " x coordinate of the resultant = 1.652 ft\n", + " y coordinate of the resultant = 10.387 ft\n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 10.2 Page No : 319" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "\t\t\n", + "#initialisation of variables\n", + "s = 13.6\n", + "h = 12. \t\t#in\n", + "u = 0.04\n", + "k = 1.\n", + "d = 6. \t\t#in\n", + "g = 32.2 \t\t#ft/sec**2\n", + "w = 62.4 \t\t#lbs/ft**3\n", + "\t\t\n", + "#CALCULATIONS\n", + "h1 = h*(s-1)/12\n", + "hf = u*h1\n", + "hn = h1-hf\n", + "Q = k*math.pi/4*(d/12)**2*8.02*math.sqrt(hn)/(math.sqrt(16-k))\n", + "Q = Q*60*w/10 # fro, cusecs to GPM\n", + "\n", + "#RESULTS\n", + "print 'discharge through flow = %.f ft G.P.M'%(Q)\n", + "\n", + "# rounding off error" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "discharge through flow = 529 ft G.P.M\n" + ] + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 10.3 Page No : 321" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "\t\t\n", + "#initialisation of variables\n", + "za = 16. \t\t#ft\n", + "h1 = 2. \t\t#ft\n", + "h2 = 3. \t\t#ft\n", + "g = 32.2 \t\t#ft/sec**2\n", + "\t\t\n", + "#CALCULATIONS\n", + "vc = math.sqrt(2*g*(za-h1-h2))\n", + "vb = vc*(h1/(2*h1))**2\n", + "r = -h1-h2-(vb**2/(2*g))\n", + "r1 = r+34\n", + "\t\t\n", + "#RESULTS\n", + "print 'pressure head at B = %.1f ft lb'%(r1)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "pressure head at B = 28.3 ft lb\n" + ] + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 10.4 Page No : 322" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "\t\t\n", + "#initialisation of variables\n", + "g = 32.2 \t\t#ft/sec**2\n", + "Cd = 0.62\n", + "a = 90. \t\t#degrees\n", + "H1 = 14. \t\t#in\n", + "H2 = 8. \t\t#in\n", + "\t\t\n", + "#CALCULATIONS\n", + "Q1 = (8./15)*Cd*math.sqrt(2*g)*math.tan(math.radians(a/2))*(H1/12)**(5/2.)\n", + "Q2 = (8./15)*Cd*math.sqrt(2*g)*math.tan(math.radians(a/2))*(H2/12)\n", + "Q = Q1-Q2\n", + "\t\t\n", + "#RESULTS\n", + "print 'Discharge through notch = %.2f cuses'%(Q)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Discharge through notch = 2.13 cuses\n" + ] + } + ], + "prompt_number": 6 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 10.5 Page No : 324" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "\t\t\n", + "#initialisation of variables\n", + "g = 32.2 \t\t#ft/sec**2\n", + "Cd = 0.62\n", + "d = 5./4 \t\t#in\n", + "h = 9. \t\t#ft\n", + "\t\t\n", + "#CALCULATIONS\n", + "T = (2./3)*math.pi*(h)**(3./2)/(Cd*(math.pi/4)*math.sqrt(2*g)*(d/12)**2)\n", + "\t\t\n", + "#RESULTS\n", + "print 'time required to lower water level = %.f secs'%(T)\n", + "\n", + "# rounding off error" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "time required to lower water level = 1334 secs\n" + ] + } + ], + "prompt_number": 7 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 10.6 Page No : 325" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "import math \n", + "\t\t\n", + "#initialisation of variables\n", + "a = 60. \t\t#degrees\n", + "d = 4. \t\t#in\n", + "Cd = 0.62\n", + "h = 5. \t \t#ft\n", + "w = 30. \t\t#ft\n", + "g = 32.2 \t\t#ft/sec**2\n", + "\t\t\n", + "#CALCULATIONS\n", + "H1 = 10*math.sin(math.radians(a))\n", + "H2 = H1-h\n", + "T = (2*w/math.tan(math.radians(a)))*(2./3)*(H1**(3./2)-H2**(3./2))/(Cd*math.sqrt(2*g)*math.pi/(4*(d/12)**2))*100\n", + "\n", + "#RESULTS\n", + "print 'time required to lower water level = %.f secs'%(T)\n", + "\n", + "# answer is accurate.please check manually" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "time required to lower water level = 1214 secs\n" + ] + } + ], + "prompt_number": 10 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 10.7 Page No : 326" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "import math \n", + "\t\t\n", + "#initialisation of variables\n", + "p1 = 40. \t\t#percent\n", + "p2 = 35. \t\t#percent\n", + "dh = 200. \t\t#ft\n", + "f = 0.1\n", + "g = 32.2 \t\t#ft/sec**2\n", + "l = 2000. \t\t#ft\n", + "d = 1. \t\t#ft\n", + "\t\t\n", + "#CALCULATIONS\n", + "hf1 = p1*dh/100\n", + "hf2 = p2*dh/100\n", + "hf3 = (100-p1-p2)*dh/100\n", + "hft = hf1+hf2+hf3\n", + "v1 = math.sqrt(2*g*hf1/(4*f*l))\n", + "Q = v1*math.pi*d**2/4\n", + "d2 = (Q*7*math.sqrt(3/(5*g)))**(2./3)\n", + "v3 = Q*4*(4./3)**2/math.pi\n", + "l3 = hf2*2*g*(3./4)/(4*f*v3**2)\n", + "\t\t\n", + "#RESULTS\n", + "print 'proportion of the quantity folwing in the bypass to the whole pass = %d ft'%(l3)\n", + "\n", + "# rounding off error" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "proportion of the quantity folwing in the bypass to the whole pass = 415 ft\n" + ] + } + ], + "prompt_number": 12 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 10.8 Page No : 328" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "\t\t\n", + "#initialisation of variables\n", + "d = 1. \t \t#ft\n", + "l = 2000. \t\t#ft\n", + "f = 0.038\n", + "g = 32.2 \t\t#/ft/sec**2\n", + "Q = 6. \t\t#cuses\n", + "l1 = 1500. \t\t#ft\n", + "r = 2.\n", + "\t\t\n", + "#CALCULATIONS\n", + "v = 4*Q/(d**2*math.pi)\n", + "hf = 4*f*l*v**2/(2*g)\n", + "v1 = math.sqrt(hf*2*g/(4*f*l1+4*f*(l-l1)*r**2))\n", + "v3 = r*v1\n", + "Q1 = math.pi*d**2*v3/4\n", + "Q2 = math.pi*d**2*v1/4\n", + "r1 = Q2/Q1\n", + "\t\t\n", + "#RESULTS\n", + "print 'proportion of the quantity folwing in the bypass to the whole pass = %.1f '%(r1)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "proportion of the quantity folwing in the bypass to the whole pass = 0.5 \n" + ] + } + ], + "prompt_number": 8 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 10.9 Page No : 329" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "\t\t\n", + "#initialisation of variables\n", + "f = 0.01\n", + "d = 3. \t\t#in\n", + "l = 22. \t\t#ft\n", + "l1 = 20. \t\t#ft\n", + "w = 20. \t\t#ft\n", + "h = 5. \t\t#ft\n", + "h1 = 20. \t\t#ft\n", + "t = 4. \t\t#min\n", + "g = 32.2 \t\t#ft/sec**2\n", + "\t\t\n", + "#CALCULATIONS\n", + "h2 = h+h1\n", + "h3 = (h-(t*60*math.pi*math.sqrt(2*g/h)/(l1*w*2*64)))**2-4\n", + "dh = h2-h3\n", + "Q = dh*l1*w\n", + "\t\t\n", + "#RESULTS\n", + "print 'Quantiy discharged = %.f cuses '%(round(Q,-2))\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Quantiy discharged = 1800 cuses \n" + ] + } + ], + "prompt_number": 13 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 10.10 Page No : 332" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "import math \n", + "\t\t\n", + "#initialisation of variables\n", + "g = 32.2 \t\t#ft/sec**2\n", + "sct = 1.6\n", + "sl = 0.8\n", + "K = 0.98\n", + "dh1 = 4. \t\t#ft\n", + "W = 62.4 \t\t#lbs/ft**3\n", + "d1 = 8. \t\t#in\n", + "d2 = 6. \t\t#in\n", + "\n", + "\t\t\n", + "#CALCULATIONS\n", + "dp = dh1*((sct/sl)-1)\n", + "C = math.sqrt(2*g)*math.pi*(d1/24)**2 /math.sqrt((d1**2/d2**2)**2 -1)\n", + "Q = C*K*math.sqrt(dh1)\n", + "\n", + "\t\t\n", + "#RESULTS\n", + "print 'Discharge passing through the pipe = %.1f cuses '%(Q)\n", + "\t\t#The answer given in textbook is wrong. Please verify it.\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Discharge passing through the pipe = 3.7 cuses \n" + ] + } + ], + "prompt_number": 10 + } + ], + "metadata": {} + } + ] +}
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