{
 "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": {}
  }
 ]
}