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+{
+ "metadata": {
+ "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 6 : Flow Through Open Channels"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 3,
+ "metadata": {},
+ "source": [
+ "Example 6.1 page no : 196"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# find the value of C and K\n",
+ "\n",
+ "import math \n",
+ "#initialisation of variables\n",
+ "i = 0.000146\n",
+ "v = 2.8 \t\t\t#ft/sec\n",
+ "m = 7. \t\t\t#ft\n",
+ "#CALCULAIONS\n",
+ "C = v/math.sqrt(m*i)\n",
+ "K = (157.6-C)*math.sqrt(m)/C\n",
+ "#RESULTS\n",
+ "print 'C = %.3f '%(C)\n",
+ "print 'K = %.3f '%(K)\n",
+ "\n",
+ "# answers may vary because of rounding error"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "C = 87.586 \n",
+ "K = 2.115 \n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 3,
+ "metadata": {},
+ "source": [
+ "Example 6.2 pageno : 197"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# find value of C \n",
+ "\n",
+ "import math \n",
+ "#initialisation of variables\n",
+ "b = 10. \t\t\t#ft\n",
+ "n = 1.\n",
+ "i = 1./1000\n",
+ "d = 1.5 \t\t\t#ft\n",
+ "C = 110.\n",
+ "w = 62.4 \t\t\t#lb/ft**3\n",
+ "#CALCULATIONS\n",
+ "L = math.sqrt(2*d**2)\n",
+ "P = b+2*L\n",
+ "A = d*(b+n*d)\n",
+ "m = A/P\n",
+ "v = round(C*math.sqrt(m*i),2)\n",
+ "A_v = round(A*v)\n",
+ "Q = A_v*w*60*60*24/10\n",
+ "#RESULTS\n",
+ "print 'Discharge = %.2e gal/day '%(Q)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Discharge = 3.56e+07 gal/day \n"
+ ]
+ }
+ ],
+ "prompt_number": 9
+ },
+ {
+ "cell_type": "heading",
+ "level": 3,
+ "metadata": {},
+ "source": [
+ "Example 6.3 page no : 197"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# find the discharge\n",
+ "\n",
+ "import math \n",
+ "#initialisation of variables\n",
+ "b = 10. \t\t\t#ft\n",
+ "n = 2.\n",
+ "d = 3.5 \t\t\t#ft\n",
+ "i = 1./625\n",
+ "#CALCULATIONS\n",
+ "A = d*(b+(d/n))\n",
+ "L = math.sqrt(d**2+(d/2)**2)\n",
+ "P = b+2*L\n",
+ "m = A/P\n",
+ "v = 1.486*m**(2./3)*i**0.5/0.03\n",
+ "Q = A*v\n",
+ "#RESULTS\n",
+ "print 'Discharge = %.1f cuses '%(Q)\n",
+ "\n",
+ "# Note : answer may vary because of rounding error."
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Discharge = 142.3 cuses \n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ },
+ {
+ "cell_type": "heading",
+ "level": 3,
+ "metadata": {},
+ "source": [
+ "Example 6.4 pageno : 198"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# find the discharge of a channel\n",
+ "import math \n",
+ "#initialisation of variables\n",
+ "d = 3. \t\t\t#ft\n",
+ "i = 1./4500\n",
+ "C = 80.\n",
+ "#CALCULATIONS\n",
+ "A = 0.5*(math.pi*d**2/4)\n",
+ "P = math.pi*d/2\n",
+ "m = A/P\n",
+ "v = C*math.sqrt(m*i)\n",
+ "Q = v*A\n",
+ "#RESULTS\n",
+ "print 'Discharge = %.2f cuses '%(Q)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Discharge = 3.65 cuses \n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "heading",
+ "level": 3,
+ "metadata": {},
+ "source": [
+ "Example 6.5 page no : 198\n"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# find diameter\n",
+ "\n",
+ "import math \n",
+ "#initialisation of variables\n",
+ "A = 2500. \t\t\t#acres\n",
+ "n = 20.\n",
+ "Q = 40. \t\t\t#gal/head\n",
+ "C = 130.\n",
+ "i = 1./3000\n",
+ "p = 7. \t\t\t#per cent\n",
+ "w = 62.4 \t\t\t#lb/ft**3\n",
+ "#CALCULATIONS\n",
+ "Q1 = Q*50000*p/(60*100*60*w)\n",
+ "Q2 = Q1+(A*4840*9/(12*24*60*60))\n",
+ "d = (Q2*8*math.sqrt(4/i)/(math.pi*C))**0.4\n",
+ "#RESULTS\n",
+ "print 'Diameter = %.3f ft '%(d)\n",
+ "\n",
+ "# answer is different because of rounding error."
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Diameter = 8.754 ft \n"
+ ]
+ }
+ ],
+ "prompt_number": 6
+ },
+ {
+ "cell_type": "heading",
+ "level": 3,
+ "metadata": {},
+ "source": [
+ "Example 6.6 page no : 199"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# find depth of water\n",
+ "\n",
+ "import math \n",
+ "\n",
+ "#initialisation of variables\n",
+ "Qt = 150000. \t\t\t#cuses\n",
+ "i = 1/10000.\n",
+ "n1 = 1.\n",
+ "n2 = 2./3\n",
+ "d1 = 30. \t\t\t#ft\n",
+ "C1 = 100.\n",
+ "C2 = 75.\n",
+ "b1 = 600. \t\t\t#ft\n",
+ "b2 = 2000. \t\t\t#ft\n",
+ "r = 2.\n",
+ "A1 = (b1+d1)*d1\n",
+ "P1 = b1+(2*d1*math.sqrt(2))\n",
+ "m1 = A1/P1\n",
+ "v1 = C1*math.sqrt(m1*i)\n",
+ "Q1 = A1*v1\n",
+ "Q2 = Qt-Q1\n",
+ "v2 = v1/2\n",
+ "A2 = Q2/v2\n",
+ "d2 = (-b2+math.sqrt(b2**2+4*1.5*A2))/(2*1.5)\n",
+ "#RESULTS\n",
+ "print 'depth of water = %.f ft '%(d2)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "depth of water = 10 ft \n"
+ ]
+ }
+ ],
+ "prompt_number": 7
+ },
+ {
+ "cell_type": "heading",
+ "level": 3,
+ "metadata": {},
+ "source": [
+ "Example 6.7 page no : 200"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# find height of dam\n",
+ "\n",
+ "import math \n",
+ "#initialisation of variables\n",
+ "d = 3. \t\t\t#ft\n",
+ "i = 1./1000\n",
+ "C = 65.\n",
+ "Cd = 0.56\n",
+ "g = 32.2 \t\t\t#ft/sec**2\n",
+ "h1 = 7.5 \t\t\t#ft\n",
+ "h2 = 3. \t\t\t#ft\n",
+ "#CALCULATIONS\n",
+ "m = d\n",
+ "v = C*math.sqrt(m*i)\n",
+ "Q = v*d\n",
+ "H = (Q*d/(2*math.sqrt(2*g)*Cd))**(2./3)\n",
+ "h = h1+h2-H\n",
+ "#RESULTS\n",
+ "print 'Height of dam = %.2f ft '%(h)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Height of dam = 8.17 ft \n"
+ ]
+ }
+ ],
+ "prompt_number": 8
+ },
+ {
+ "cell_type": "heading",
+ "level": 3,
+ "metadata": {},
+ "source": [
+ "Example 6.8 page no : 207"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# find depth, bottom width and top width\n",
+ "\n",
+ "import math \n",
+ "#initialisation of variables\n",
+ "Q =100.\t\t\t#cuses\n",
+ "v = 2. \t\t\t#/ft/sec\n",
+ "n = 1.5\n",
+ "A = 50. \t\t\t#ft**2\n",
+ "C = 120.\n",
+ "#CALCULATIONS\n",
+ "d = math.sqrt((Q/v)/(2*math.sqrt(n**2+1)-n))\n",
+ "m = A/d\n",
+ "h1 = m-n*d\n",
+ "h2 = m+n*d\n",
+ "i = (v/C)**2*(2/d)\n",
+ "#RSULTS\n",
+ "print 'Depth = %.2f ft '%(d)\n",
+ "print ' Bottom width = %.2f ft '%(h1)\n",
+ "print ' Top width = %.2f ft '%(h2)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Depth = 4.87 ft \n",
+ " Bottom width = 2.95 ft \n",
+ " Top width = 17.57 ft \n"
+ ]
+ }
+ ],
+ "prompt_number": 9
+ },
+ {
+ "cell_type": "heading",
+ "level": 3,
+ "metadata": {},
+ "source": [
+ "Example 6.9 page no : 208"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# find depth, bottom width, and top width\n",
+ "import math \n",
+ "#initialisation of variables\n",
+ "Q = 1100. \t\t\t#cuses\n",
+ "i = 1/1800.\n",
+ "C = 95.\n",
+ "n = 1.5\n",
+ "#CALCULATIONS\n",
+ "d = ((Q*math.sqrt(3600)/C)/(n+0.6))**0.4\n",
+ "b = 0.6*d\n",
+ "ht = b+2*(n*d)\n",
+ "#RESULTS\n",
+ "print 'Depth = %.1f ft '%(d)\n",
+ "print ' Bottom width = %.2f ft '%(b)\n",
+ "print ' Top width = %.1f ft '%(ht)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Depth = 10.2 ft \n",
+ " Bottom width = 6.11 ft \n",
+ " Top width = 36.7 ft \n"
+ ]
+ }
+ ],
+ "prompt_number": 20
+ },
+ {
+ "cell_type": "heading",
+ "level": 3,
+ "metadata": {},
+ "source": [
+ "Example 6.10 pageno : 209"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# find depth of channel\n",
+ "\n",
+ "import math \n",
+ "#initialisation of variables\n",
+ "\n",
+ "n = 1.5\n",
+ "Q = 800. \t\t\t#cuses\n",
+ "i = 2.5/5280\n",
+ "n1 = 9.24\n",
+ "r = 0.6\n",
+ "k = 1.49\n",
+ "#CALCULATIONS\n",
+ "d = (k*10**7*4/n1)**(1/8.)\n",
+ "#RESULTS\n",
+ "print 'Depth of channel = %.1f ft '%(d)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Depth of channel = 7.1 ft \n"
+ ]
+ }
+ ],
+ "prompt_number": 21
+ },
+ {
+ "cell_type": "heading",
+ "level": 3,
+ "metadata": {},
+ "source": [
+ "Example 6.11 page no : 210"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# find discharge\n",
+ "\n",
+ "import math \n",
+ "#initialisation of variables\n",
+ "d = 8. \t\t\t#ft\n",
+ "i = 1./1200\n",
+ "C = 90.\n",
+ "a = 308. \t\t\t#degrees\n",
+ "#CALCULATIONS\n",
+ "h = 0.95*d\n",
+ "A = (d/2)**2*(a*(math.pi/180)-math.sin(math.radians(a)))/2\n",
+ "m = 0.29*d\n",
+ "Q = A*C*math.sqrt(m*i)\n",
+ "#RESULTS\n",
+ "print 'Discharge = %.f cuses '%(Q)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Discharge = 195 cuses \n"
+ ]
+ }
+ ],
+ "prompt_number": 22
+ },
+ {
+ "cell_type": "heading",
+ "level": 3,
+ "metadata": {},
+ "source": [
+ "Example 6.12 page no : 213"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# find specific energy\n",
+ "\n",
+ "import math \n",
+ "#initialisation of variables\n",
+ "v = 5. \t\t\t#ft/sec\n",
+ "Q = 500. \t\t\t#cuses\n",
+ "w = 25. \t\t\t#ft\n",
+ "g = 32.2 \t\t\t#ft/sec**2\n",
+ "#CALCULATIONS\n",
+ "h = (Q/v)/w\n",
+ "E = h+(v**2/(2*g))\n",
+ "he = round((400*2/64.4)**(1./3),2)\n",
+ "ve = round(20./2.32,2)\n",
+ "Emin = he + (ve**2 / (g*2))\n",
+ "\n",
+ "#RESULTS\n",
+ "print 'Specific energy = %.2f ft '%(E)\n",
+ "print \"Critical Velocity of flow = %.2f ft/sec\"%ve\n",
+ "print 'Minimum energy Emin = %.2f ft'%Emin"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Specific energy = 4.39 ft \n",
+ "Critical Velocity of flow = 8.62 ft/sec\n",
+ "Minimum energy Emin = 3.47 ft\n"
+ ]
+ }
+ ],
+ "prompt_number": 16
+ },
+ {
+ "cell_type": "heading",
+ "level": 3,
+ "metadata": {},
+ "source": [
+ "Example 6.13 page no : 216"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# find slope of free water\n",
+ "\n",
+ "import math \n",
+ "#initialisation of variables\n",
+ "\n",
+ "i = 1./5000\n",
+ "C = 100.\n",
+ "b = 50. \t\t\t#ft\n",
+ "h = 10. \t\t\t#ft\n",
+ "Q = 1000. \t\t\t#cuses\n",
+ "g = 32.2 \t\t\t#ft/sec**2\n",
+ "#CALCULATIONS\n",
+ "f = 2.*g/C**2\n",
+ "m = (b*h)/(b+2*h)\n",
+ "v = Q/(b*h)\n",
+ "r = (i-(f*4/(2*g*m)))/(1-(2**2/(g*h)))\n",
+ "s = i-r\n",
+ "#RESULTS\n",
+ "print 'Slope = %.6f '%(s)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Slope = 0.000054 \n"
+ ]
+ }
+ ],
+ "prompt_number": 24
+ },
+ {
+ "cell_type": "heading",
+ "level": 3,
+ "metadata": {},
+ "source": [
+ "Example 6.14 page no : 221"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# find height required\n",
+ "\n",
+ "import math \n",
+ "#Initialization of variables\n",
+ "B =48. \t\t\t#ft\n",
+ "D =5. \t\t\t#ft\n",
+ "f =0.005\n",
+ "i =1./1000\n",
+ "g =32.2\n",
+ "#calculations\n",
+ "C =math.sqrt(2*g/f)\n",
+ "m =B*D/(B+2*D)\n",
+ "V =C*math.sqrt(m*i)\n",
+ "Q =B*D*V\n",
+ "Dc =(Q**2 /(g*B**2))**(1./3)\n",
+ "d1 =2.25 \t\t\t#ft\n",
+ "Q1 =1*D*V\n",
+ "d2 =-d1/2 + math.sqrt(2*Q1**2 /(g*d1) + d1**2 /4)\n",
+ "hd =d2-d1\n",
+ "#results\n",
+ "print \"height required = %.1f ft\"%(hd)\n",
+ "#The answer is a bit different due to rounding off error in textbook\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "height required = 2.8 ft\n"
+ ]
+ }
+ ],
+ "prompt_number": 26
+ },
+ {
+ "cell_type": "heading",
+ "level": 3,
+ "metadata": {},
+ "source": [
+ "Example 6.15 page no : 222"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# find loss in energy\n",
+ "\n",
+ "import math \n",
+ "#initialisation of variables\n",
+ "Q = 360. \t\t\t#cfs\n",
+ "d1 = 1. \t\t\t#ft\n",
+ "B = 18. \t\t\t#ft\n",
+ "g = 32.2 \t\t\t#ft/sec**2\n",
+ "w1 = 624. \t\t\t#lb/ft**3\n",
+ "d2 =4.5 \t\t\t#ft\n",
+ "#CALCULATIONS\n",
+ "w = Q/B\n",
+ "v1 = w/d1\n",
+ "v2 = v1/d2\n",
+ "d2 = -0.5+math.sqrt((2*v1**2*d1/(g))+(d1**2./4))\n",
+ "El = (d1+(w**2/(2*g)))-(d2+(v2**2/(2*g)))\n",
+ "EL = round(w1*Q*El,-4)\n",
+ "#RESULTS\n",
+ "print 'loss in energy = %.f lb '%(EL)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "loss in energy = 540000 lb \n"
+ ]
+ }
+ ],
+ "prompt_number": 29
+ },
+ {
+ "cell_type": "heading",
+ "level": 3,
+ "metadata": {},
+ "source": [
+ "Example 6.16 page no : 223"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# find downstream depth \n",
+ "import math \n",
+ "#initialisation of variables\n",
+ "d1 = 4. \t\t\t#ft\n",
+ "v1 = 60. \t\t\t#ft/sec\n",
+ "g = 32.2 \t\t\t#ft/sec**2\n",
+ "#CALULATIONS\n",
+ "d2 = d1*(math.sqrt(1+8*v1**2/(g*d1))-1)/2.\n",
+ "#RESULTS\n",
+ "print 'd2 = %.f ft '%(d2)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "d2 = 28 ft \n"
+ ]
+ }
+ ],
+ "prompt_number": 30
+ },
+ {
+ "cell_type": "heading",
+ "level": 3,
+ "metadata": {},
+ "source": [
+ "Example 6.17 page no : 224"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# find approximate length\n",
+ "\n",
+ "import math \n",
+ "#initialisation of variables\n",
+ "b = 150. \t\t\t#ft\n",
+ "d = 12. \t\t\t#ft\n",
+ "N = 0.03\n",
+ "i = 1./10000\n",
+ "h = 10. \t\t\t#ft\n",
+ "g = 32.2 \t\t\t#ft/sec**2\n",
+ "#CALCULATIONS\n",
+ "A = b*d\n",
+ "P = b+2*d\n",
+ "m = A/P\n",
+ "v = m**(2/3.)*1.49*i**0.5/N\n",
+ "A1 = b*(h+d)\n",
+ "P1 = b+2*(h+d)\n",
+ "m1 = A1/P1\n",
+ "C1 = 1.49*m1**(1./6)/N\n",
+ "v1 = A*v/A1\n",
+ "s = (i-(v1**2/(C1**2*m1)))/(1-(v1**2/(g*(h+d))))\n",
+ "L = round(2*h/s,-3)\n",
+ "#RESULTS\n",
+ "print 'Length of back water = %.f ft '%(L)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Length of back water = 236000 ft \n"
+ ]
+ }
+ ],
+ "prompt_number": 32
+ },
+ {
+ "cell_type": "heading",
+ "level": 3,
+ "metadata": {},
+ "source": [
+ "Example 6.18 page no : 227"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# find discharge through the venturi flume \n",
+ "\n",
+ "import math \n",
+ "#initialisation of variables\n",
+ "b1 = 3.2 \t\t\t#ft\n",
+ "b2 = 1.3 \t\t\t#ft\n",
+ "h1 = 1.86 \t\t\t#ft\n",
+ "h2 = 1.63 \t\t\t#ft\n",
+ "g = 32.2 \t\t\t#ft/sec**2\n",
+ "#CALCULATIONS\n",
+ "a1 = b1*h1\n",
+ "a2 = b2*h2\n",
+ "Q = a1*a2*math.sqrt(2*g)*math.sqrt(h1-h2)/(math.sqrt(a1**2-a2**2))\n",
+ "#RESULTS\n",
+ "print 'Discharge = %.1f cuses '%(Q)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Discharge = 8.7 cuses \n"
+ ]
+ }
+ ],
+ "prompt_number": 33
+ },
+ {
+ "cell_type": "heading",
+ "level": 3,
+ "metadata": {},
+ "source": [
+ "Example 6.19 page no : 227"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# find discharge and depth of flow at the throat\n",
+ "\n",
+ "import math \n",
+ "#initialisation of variables\n",
+ "b1 = 4. \t\t\t#ft\n",
+ "b2 = 2. \t\t\t#ft\n",
+ "h1 = 2. \t\t\t#ft\n",
+ "g = 32.2 \t\t\t#ft/sec**2\n",
+ "#CALCULATIONS\n",
+ "Qmax = 3.09*b2*h1**1.5\n",
+ "v1 = Qmax/(b1*h1)\n",
+ "H = h1+(v1**2/(2*g))\n",
+ "Qmax2 = 3.09*b2*H**1.5\n",
+ "h2 = 2*H/3\n",
+ "#RESULTS\n",
+ "print 'Qmax = %.2f cfs '%(Qmax2)\n",
+ "print 'h2 = %.3f ft '%(h2)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Qmax = 18.46 cfs \n",
+ "h2 = 1.383 ft \n"
+ ]
+ }
+ ],
+ "prompt_number": 17
+ },
+ {
+ "cell_type": "heading",
+ "level": 3,
+ "metadata": {},
+ "source": [
+ "Example 6.20 pageno : 229"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "'''\n",
+ "find\n",
+ "a) discharge\n",
+ "b) depth of water\n",
+ "c) height of standing wave\n",
+ "d) depth of water in channel\n",
+ "'''\n",
+ "import math \n",
+ "#initialisation of variables\n",
+ "h1 = 8. \t\t\t#ft\n",
+ "b1 = 32. \t\t\t#ft\n",
+ "h = 1. \t\t\t#ft\n",
+ "b2 = 24. \t\t\t#ft\n",
+ "g = 32.2 \t\t\t#ft/sec**2\n",
+ "\n",
+ "#CALCULATIONS\n",
+ "H = h1-h\n",
+ "h = 9.65\n",
+ "Q = 3.09*H**1.5*b2\n",
+ "v1 = Q/(b1*h1)\n",
+ "Q1 = 3.09*(H+(v1**2/(2*g)))**1.5*b2\n",
+ "hc = (Q1**2/(g*b2**2))**(1./3)\n",
+ "d2 = -(hc/2)+math.sqrt(9*hc**2/2)\n",
+ "\n",
+ "#RESULTS\n",
+ "print 'Q = %.f cfs '%(Q1)\n",
+ "print 'hc = %.2f ft '%(hc)\n",
+ "print 'max depth = %.2f ft '%(d2)\n",
+ "print 'Maximum depth of water downstream %.2f ft'%(d2+1)\n",
+ "print 'h = %.2f ft'%h\n",
+ "\n",
+ "# answers may vary because of rounding error."
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Q = 1507 cfs \n",
+ "hc = 4.97 ft \n",
+ "max depth = 8.05 ft \n",
+ "Maximum depth of water downstream 9.05 ft\n",
+ "h = 9.65 ft\n"
+ ]
+ }
+ ],
+ "prompt_number": 20
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [],
+ "language": "python",
+ "metadata": {},
+ "outputs": []
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+} \ No newline at end of file
generated by cgit (git 2.34.1) at 2024-12-20 11:52:34 +0000