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  {
   "cells": [
    {
     "cell_type": "heading",
     "level": 1,
     "metadata": {},
     "source": [
      "Chapter 6 : Flow of Water Through Pipes"
     ]
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 6.1 Page No : 168"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "import math \n",
      "\t\t\n",
      "#initialisation of variables\n",
      "R = 0.5 \t\t#lbs sq ft\n",
      "v = 10. \t\t#ft/sec\n",
      "A = 1.  \t\t# sq ft\n",
      "A1 = 15000. \t\t#sq ft\n",
      "V = 20. \t\t#m.p.h\n",
      "\t\t\n",
      "#CALCULATIONS\n",
      "k = R/v**2\n",
      "R = k*A1*(V*44./30)**2\n",
      "HP = R*88/(550*3)\n",
      "\t\t\n",
      "#RESULTS\n",
      "print  'Horse power = %.f HP'%(HP)\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Horse power = 3442 HP\n"
       ]
      }
     ],
     "prompt_number": 1
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 6.2 Page No : 171"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "\n",
      "import math \n",
      "\t\t\n",
      "#initialisation of variables\n",
      "k = 0.01\n",
      "d = 6.   \t\t#in\n",
      "l = 1000. \t\t#ft\n",
      "v = 8. \t    \t#ft/sec\n",
      "g = 32.2 \t\t#ft/sec**2\n",
      "\t\t\n",
      "#CALCULATIONS\n",
      "f = k*(1+(1/d))\n",
      "hf = 4*f*l*v**2*12/(2*g*d)\n",
      "C = math.sqrt(2*g/f)\n",
      "hf1 = v**2*4*(12/d)*l/C**2\n",
      "\n",
      "#RESULTS\n",
      "print  'head lost in friction = %.2f ft of water'%(hf)\n",
      "print  ' head lost in friction = %.2f ft of water'%(hf1)\n",
      "\n",
      "# rounding off error.  value of f taken as 0.116 and here answer goes to 0.117 ."
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "head lost in friction = 92.75 ft of water\n",
        " head lost in friction = 92.75 ft of water\n"
       ]
      }
     ],
     "prompt_number": 5
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 6.3 Page No : 177"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "\t\t\n",
      "#initialisation of variables\n",
      "d1 = 3. \t\t#in\n",
      "d2 = 6. \t\t#in\n",
      "v = 6. \t\t#ft/sec\n",
      "g = 32.2 \t\t#ft/sec**2\n",
      "\t\t\n",
      "#CALCULATIONS\n",
      "v1 = v*(d1/d2)**2\n",
      "L = (v-v1)**2/(2*g)\n",
      "\t\t\n",
      "#RESULTSa\n",
      "print  'Loss due to sudden enlargment = %.4f '%(L)\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Loss due to sudden enlargment = 0.3144 \n"
       ]
      }
     ],
     "prompt_number": 3
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 6.4 Page No : 177"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "import math \n",
      "\t\t\n",
      "#initialisation of variables\n",
      "d1 = 4. \t\t#in\n",
      "d2 = 3. \t\t#in\n",
      "Q = 90. \t\t#gallons\n",
      "k = 0.7\n",
      "v = 6.24 \t\t#ft/sec\n",
      "g = 32.2 \t\t#ft/sec**2\n",
      "\t\t\n",
      "#CALCULATIONS\n",
      "V = round(Q/(60*6.24),3)\n",
      "v1 = V*4*d2**2/math.pi\n",
      "v2 = round(V*4*d1**2/math.pi,1)\n",
      "L = ((1/k)-1)**2*v2**2*900/(2*g)\n",
      "\n",
      "\n",
      "#RESULTS\n",
      "print  'Loss hc = %.f ft lbs per minute'%(L)\n",
      "\n",
      "# rounding off error"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Loss hc = 62 ft lbs per minute\n"
       ]
      }
     ],
     "prompt_number": 9
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 6.5 Page No : 178"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "import math \n",
      "\t\t\n",
      "#initialisation of variables\n",
      "d1 = 3. \t\t#in\n",
      "d2 = 6. \t\t#in\n",
      "sm = 13.6\n",
      "Q = 0.5 \t\t#ft**3/sec\n",
      "g = 32.2 \t\t#ft/sec**2\n",
      "\t\t\n",
      "#CALCULATIONS\n",
      "v1 = Q*(12/d1)**2*4/math.pi\n",
      "v2 = Q*(12/d2)**2*4/math.pi\n",
      "hc = (v1-v2)**2/(2*g)\n",
      "h = ((v1**2-v2**2)/(2*g))-hc\n",
      "h1 = 12*h/(sm-1)\n",
      "\t\t\n",
      "#RESULTS\n",
      "print  'difference in level in two limbs of mercury = %.3f in'%(h1)\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "difference in level in two limbs of mercury = 0.575 in\n"
       ]
      }
     ],
     "prompt_number": 5
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 6.6 Page No : 179"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "import math \n",
      "\t\t\n",
      "#initialisation of variables\n",
      "f = 0.01\n",
      "l = 60. \t\t#ft\n",
      "d = 6. \t\t#in\n",
      "g = 32.2 \t\t#ft/sec\n",
      "v = 10. \t\t#ft/sec\n",
      "d1 = 3. \t\t#in\n",
      "l1 = 20. \t\t#ft\n",
      "k = 0.62\n",
      "\t\t\n",
      "#CALCULATIONS\n",
      "H = round(4*f*l*v**2/(2*g*(d/12)**2),1)\n",
      "v2 = v*d1**2/d**2\n",
      "hf = round(4*f*l1*v**2/(2*g*(d/12)**2),2)\n",
      "h = (v-v2)**2/(2*g)\n",
      "h1 = round(4*f*l1*v2**2/(2*g*2*(d/12)**2),3)\n",
      "h2 = round(v**2*4*f*l1/(2*g*(d/12)**2),2)\n",
      "h3 = ((1/k)-1)**2*v**2/(2*g)\n",
      "dh = (H-hf-h-h1-h2-h3)\n",
      "\n",
      "#RESULTS\n",
      "print  'Saving in head = %.2f ft'%(dh)\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Saving in head = 3.35 ft\n"
       ]
      }
     ],
     "prompt_number": 15
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 6.7 Page No : 181"
     ]
    },
    {
     "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",
      "d = 3. \t\t#in\n",
      "h = 50. \t\t#ft\n",
      "w = 6.24 \t\t#lb/ft**3\n",
      "r = 0.5\n",
      "r1 = 16.\n",
      "r2 = 9./16\n",
      "r3 = 0.25\n",
      "r4 = 40.5/256\n",
      "r5 = 972./256\n",
      "r6 = 81./256\n",
      "\t\t\n",
      "#CALCULATIONS\n",
      "v =math.sqrt(h*2*g/(r+r1+r2+r3+r4+r5+r6))\n",
      "Q = math.pi*(d/12)**2*v*60*w/4\n",
      "\t\t\n",
      "#RESULTS\n",
      "print  'discharge in the pipeline = %.1f gal.min'%(Q)\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "discharge in the pipeline = 224.5 gal.min\n"
       ]
      }
     ],
     "prompt_number": 7
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 6.8 Page No : 186"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "\n",
      "import math \n",
      "\t\t\n",
      "#initialisation of variables\n",
      "l = 6000. \t\t#ft\n",
      "d = 9.  \t\t#in\n",
      "s = 1./100\n",
      "h = 20. \t\t#ft\n",
      "h1 = 5. \t\t#ft\n",
      "f = 0.006\n",
      "g = 32.2 \t\t#ft/sec**2\n",
      "\t\t\n",
      "#CALCULATIONS\n",
      "L = l*s\n",
      "v = math.sqrt((h+L-h1)*(d/12)*2*g/(4*f*l))\n",
      "Q = v*math.pi*(d/12)**2/4\n",
      "s1 = (L+h-h1)/l\n",
      "\t\t\n",
      "#RESULTS\n",
      "print  'Discharge through the pipe = %.2f cusecs'%(Q/10)\n",
      "print  ' slope of hydraulic gradient = %.4f '%(s1)\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Discharge through the pipe = 0.22 cusecs\n",
        " slope of hydraulic gradient = 0.0125 \n"
       ]
      }
     ],
     "prompt_number": 7
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 6.9 Page No : 187"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "import math \n",
      "    \n",
      "#initialisation of variables\n",
      "d1 = 24. \t\t#in\n",
      "Q = 10. \t\t#cuses\n",
      "d2 = 18. \t\t#in\n",
      "d3 = 12. \t\t#in\n",
      "f = 0.01\n",
      "l = 1000. \t\t#ft\n",
      "g = 32.2 \t\t#ft/sec**2\n",
      "l1 = 100. \t\t#ft\n",
      "l2 = 600. \t\t#ft\n",
      "    \n",
      "#CALCULATIONS\n",
      "v1 = math.sqrt(4*Q/(math.pi*(d1/12)**2))\n",
      "v2 = math.sqrt(4*Q/(math.pi*(d2/12)**2))\n",
      "v3 = math.sqrt(4*Q/(math.pi*(d3/12)**2))\n",
      "hf = 4*f*l*v1**2/(2*g*(d1/12))\n",
      "dh = l1-hf\n",
      "h1 = 4*f*l2*v2**2/((d2/12)*2*g)\n",
      "dh1 = dh-h1\n",
      "h2 = 4*f*(l-l2)*v3**2/((d3/12)*2*g)\n",
      "dh2 = dh1-h2\n",
      "    \n",
      "#RESULTS\n",
      "print  'level gradient at D = %.2f ft'%(dh2)\n",
      "\n",
      "    #ANSWER GIVEN IN THE TEXTBOOK IS WRONG\n",
      "\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "level gradient at D = 94.44 ft\n"
       ]
      }
     ],
     "prompt_number": 9
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 6.10 Page No : 188"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "import math \n",
      "\t\t\n",
      "#initialisation of variables\n",
      "k = 0.01\n",
      "l = 24. \t\t#ft\n",
      "g = 32.2 \t\t#ft/sec**2\n",
      "w = 15.6 \t\t#lbs/in**2\n",
      "W = 62.4 \t\t#lbs/ft**3\n",
      "h = 12. \t\t#ft\n",
      "l1 = 100. \t\t#ft\n",
      "\t\t\n",
      "#CALCULATIONS\n",
      "f = k*(1+(1/(h/l)))\n",
      "C = math.sqrt(2*g/f)\n",
      "L = w*144/(W)\n",
      "i = h/l1\n",
      "v = C*math.sqrt(k*h/(4*l))\n",
      "Q = v*60*math.pi*(1/l)**2/4\n",
      "v1 = math.sqrt(h*2*g*(1/l)/(4*f*3*l1))\n",
      "Q1 = v1*60*math.pi*(1/l)**2/4\n",
      "\t\t\n",
      "#RESULTS\n",
      "print  'Discharge quantity of water = %.3f cubic ft/mt'%(Q1)\n",
      "\n",
      "\n",
      "\t\t#ANSWER GIVEN IN THE TETBOOK IS WRONG\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Discharge quantity of water = 0.077 cubic ft/mt\n"
       ]
      }
     ],
     "prompt_number": 10
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 6.11 Page No : 189"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "\t\t\n",
      "#initialisation of variables\n",
      "p = 15.6 \t\t#lbs/in**2\n",
      "la = 250. \t\t#ft\n",
      "lb = 200. \t\t#ft\n",
      "lc = 120. \t\t#ft\n",
      "w = 62.4 \t\t#lbs/ft**3\n",
      "p1 = 93.6 \t\t#lbs/in**2\n",
      "l2 = 600. \t\t#ft\n",
      "l3 = 100. \t\t#ft\n",
      "l4 = 300. \t\t#ft\n",
      "ph = 95. \t\t#ft\n",
      "\t\t\n",
      "#CALCULATIONS\n",
      "H1 = ((p*144)/w)+la\n",
      "H2 = ((p1*144)/w)+(la/2)\n",
      "s = (H2-H1)/(l4+l2+l3)\n",
      "h1 = l3*s\n",
      "h2 = l2*s\n",
      "h3 = l4*s\n",
      "H = h1+h2+h3\n",
      "P = ph*w/144\n",
      "\t\t\n",
      "#RESULTS\n",
      "print  'pressure head for 95ft = %.2f lbs/in**2'%(P)\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "pressure head for 95ft = 41.17 lbs/in**2\n"
       ]
      }
     ],
     "prompt_number": 11
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 6.12 Page No : 191"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "import math \n",
      "\t\t\n",
      "#initialisation of variables\n",
      "Q = 30. \t\t#gallons/head\n",
      "C = 78.\n",
      "n = 100000.\n",
      "d = 3. \t\t#miles\n",
      "l = 40. \t\t#ft\n",
      "\n",
      "#CALCULAIONS\n",
      "st = Q*n\n",
      "Q1 = st/(6.24*2*8*60**2)\n",
      "i = l/(d*5280)\n",
      "d = (4*Q1*math.sqrt(4/i)/(math.pi*C))**(2./5)\n",
      "\t\t\n",
      "#RESULTS\n",
      "print  'size of pipe = %.2f ft'%(d)\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "size of pipe = 1.97 ft\n"
       ]
      }
     ],
     "prompt_number": 12
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 6.13 Page No : 192"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "import math \n",
      "\t\t\n",
      "#initialisation of variables\n",
      "f = 0.01\n",
      "l = 2000. \t\t#ft\n",
      "d = 6. \t\t#in\n",
      "g = 32.2 \t\t#ft/sec**2\n",
      "Q = 10. \t\t#cuses\n",
      "\n",
      "#CALUCLATIONS\n",
      "v = math.sqrt(2*g*(d/12)*Q/(4*f*l))\n",
      "Q1 = v*math.pi*(d/12)**2/4\n",
      "\t\t\n",
      "#RESULTS\n",
      "print  'Discharge through the pipe = %.3f cuses'%(Q1)\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Discharge through the pipe = 0.394 cuses\n"
       ]
      }
     ],
     "prompt_number": 13
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 6.14 Page No : 193"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "\n",
      "import math \n",
      "\t\t\n",
      "#initialisation of variables\n",
      "h = 10. \t\t#ft\n",
      "l = 50. \t\t#ft\n",
      "d = 1. \t\t#in\n",
      "lm = 5. \t\t#in\n",
      "f = 0.01\n",
      "sm = 13.6\n",
      "g =32.2\n",
      "\t\t\n",
      "#CALCULATIONS\n",
      "ps = sm*lm/12\n",
      "v = math.sqrt((ps+h)*2*g*(d/12)/(4*f*l))\n",
      "Q = v*math.pi*(d/12)**2/4\n",
      "\t\t\n",
      "#RESULTS\n",
      "print  'Discharge through the pipe = %.3f cuses'%(Q)\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Discharge through the pipe = 0.035 cuses\n"
       ]
      }
     ],
     "prompt_number": 14
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 6.15 Page No : 195"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "from sympy import solve, Symbol\n",
      "\n",
      "#initialisation of variables\n",
      "r = 34.\n",
      "r1 = 4.\n",
      "H = 25. \t\t#ft\n",
      "x = 18.\n",
      "l = 2000. \t\t#ft\n",
      "g = 32.2\n",
      "v = Symbol(\"v\")\n",
      "\t\t\n",
      "#CALCULATIONS\n",
      "l1 = (r-r1-x)*l/H\n",
      "print  'l1 = %.f ft'%(l1)\n",
      "\n",
      "ans = solve(v**2/(2*g) * ( 1.5 + r1*0.0075*l/1) - H)\n",
      "v = round(ans[1],2)\n",
      "l1 = Symbol(\"l1\")\n",
      "ans = solve(r1 + v**2/(2*g) + x + 0.5*v**2/(2*g) + r1*0.0075*l1/1 * v**2/(2*g) - r)\n",
      "l1 = ans[0]\n",
      "\n",
      "#RESULTS\n",
      "print \"v = %.2f ft/sec\"%v\n",
      "print \"l1 = %.f ft\"%l1\n",
      "\n",
      "# note : rounding off error. please check."
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "l1 = 960 ft\n",
        "v = 5.12 ft/sec"
       ]
      },
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "\n",
        "l1 = 933 ft\n"
       ]
      }
     ],
     "prompt_number": 4
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 6.16 Page No : 197"
     ]
    },
    {
     "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",
      "l = 1000. \t\t#ft\n",
      "dh = 40. \t\t#ft\n",
      "d = 6. \t\t#in\n",
      "h = 15. \t\t#ft\n",
      "h1 = 300. \t\t#ft\n",
      "f = 0.002\n",
      "\t\t\n",
      "#CALCULATIONS\n",
      "v = math.sqrt(dh*2*g/(1.5+(4*f*l/(d/12))))\n",
      "Q = v*math.pi*(d/12)**2/4\n",
      "r = -(h+(v**2/(2*g))*(1.5+(4*f*h1/(d/12))))\n",
      "Pre = 34 + r\t\t\n",
      "#RESULTS\n",
      "print  'Pressure at vertex = %.1f ft'%(r) \n",
      "print \"The pressure head at C will be = %.1f ft. of water absolute.\"%Pre"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Pressure at vertex = -29.4 ft\n",
        "The pressure head at C will be = 4.6 ft. of water absolute.\n"
       ]
      }
     ],
     "prompt_number": 5
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 6.17 Page No : 198"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "import math \n",
      "\t\t\n",
      "#initialisation of variables\n",
      "f = 0.008\n",
      "l = 2000. \t\t#ft\n",
      "p1 = 34. \t\t#ft\n",
      "p2 = 8. \t\t#ft\n",
      "p3 = 4. \t\t#ft\n",
      "g = 32.2 \t\t#ft/sec**2\n",
      "d = 18. \t\t#in\n",
      "P = 140. \t\t#ft\n",
      "l1 = 9500. \t\t#ft\n",
      "\t\t\n",
      "#CALCULATIONS\n",
      "v = math.sqrt((p1-p2-p3)*2*g/((d/12)+(4*f*l/(d/12))))\n",
      "Q = math.pi*(d/12)**2*v/4\n",
      "v1 = math.sqrt(P*2*g/((d/12)+(4*f*l1/(d/12))))\n",
      "Q1 = math.pi*(d/12)**2*v1/4\n",
      "\t\t\n",
      "#RESULTS\n",
      "print  'Quantity discharge = %.f cuses'%(Q) \n",
      "print  ' Quantity discharge = %.2f cuses'%(Q1) \n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Quantity discharge = 10 cuses\n",
        " Quantity discharge = 11.74 cuses\n"
       ]
      }
     ],
     "prompt_number": 17
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 6.18 page no : 200"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "from sympy import Symbol,solve\n",
      "\n",
      "# variables\n",
      "v = Symbol('v')           # ft/sec\n",
      "p = Symbol('p')           # lbs/in**2\n",
      "g = 32.2\n",
      "f = 0.0075                # friction\n",
      "l = 30.                   # lenght pipe\n",
      "\n",
      "# calculations\n",
      "ans = solve( v**2/(2*g) *( 0.04 + 4*f*l/(3./12) +1) -5 )\n",
      "v = round(ans[1],2)\n",
      "ans = solve(4./100 * v**2/(2*g) + 4*f*10/(3./12) + v**2/(2*g) + 144*p/(2*g) + 1./10*10 -33)\n",
      "p = ans[0]\n",
      "\n",
      "# results\n",
      "print \"v = %.2f ft./sec\"%v\n",
      "print \"p = %.2f lbs./in**2\"%p\n",
      "\n",
      "# rounding off error\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "v = 8.33 ft./sec\n",
        "p = 13.27 lbs./in**2\n"
       ]
      }
     ],
     "prompt_number": 24
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 6.19 Page No : 202"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "\t\t\n",
      "#initialisation of variables\n",
      "L = 20000. \t\t#ft\n",
      "l1 = 6000. \t\t#ft\n",
      "d1 = 12. \t\t#in\n",
      "l2 = 10000. \t\t#ft\n",
      "d2 = 9. \t\t#in\n",
      "d3 = 6. \t\t#in\n",
      "l3 = 4000. \t\t#ft\n",
      "\t\t\n",
      "#CALCULATIONS\n",
      "D = (L/((l1/(d1/12)**5)+(l2/(d2/12)**5)+(l3/(d3/12)**5)))**(1./5)\n",
      "\t\t\n",
      "#RESULTS\n",
      "print  'Diameter of uniform pipe = %.2f ft'%(D) \n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Diameter of uniform pipe = 0.65 ft\n"
       ]
      }
     ],
     "prompt_number": 18
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 6.20 Page No : 202"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "\n",
      "import math \n",
      "\t\t\n",
      "#initialisation of variables\n",
      "L = 4700. \t\t#ft\n",
      "l1 = 2500. \t\t#ft\n",
      "d1 = 15. \t\t#in\n",
      "l2 = 1200. \t\t#ft\n",
      "d2 = 12. \t\t#in\n",
      "d3 = 9. \t\t#in\n",
      "l3 = 1000. \t\t#ft\n",
      "H = 100. \t\t#ft\n",
      "f = 0.01\n",
      "g = 32.2 \t\t#ft/sec**2\n",
      "\t\t\n",
      "#CALCULATIONS\n",
      "D = (L/((l1/(d1/12)**5)+(l2/(d2/12)**5)+(l3/(d3/12)**5)))**(1./5)\n",
      "v = math.sqrt(2*g*D*H/(4*f*L))\n",
      "Q = v*math.pi*D**2/4\n",
      "\t\t\n",
      "#RESULTS\n",
      "print  'Quantity discharged = %.2f cusecs'%(Q) \n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Quantity discharged = 3.99 cusecs\n"
       ]
      }
     ],
     "prompt_number": 19
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 6.21 Page No : 204"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "import math \n",
      "\t\t\n",
      "#initialisation of variables\n",
      "v1 = 6.2 \t\t#ft/sec\n",
      "a = 43.52 \t\t#ft**2/sec**2\n",
      "a1 = 105.6 \t\t#ft**2/sec**2\n",
      "r = 0.468\n",
      "r1 = 0.87\n",
      "d = 5.  \t\t#in\n",
      "d1 = 6. \t\t#in\n",
      "\t\t\n",
      "#CALCULATIONS\n",
      "v2 = math.sqrt(a-r*v1**2)\n",
      "v3 = math.sqrt(a1-r1*v1**2)\n",
      "Q1 = math.pi*(d1/12)**2*60*v2/4\n",
      "Q2 =  math.pi*(d/12)**2*60*v3/4\n",
      "\t\t\n",
      "#RESULTS\n",
      "print  'Quantity discharged = %.2f cuses'%(Q1) \n",
      "print  ' Quantity discharged = %.2f cuses'%(Q2) \n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Quantity discharged = 59.53 cuses\n",
        " Quantity discharged = 69.50 cuses\n"
       ]
      }
     ],
     "prompt_number": 20
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 6.22 Page No : 208"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "\t\t\n",
      "#initialisation of variables\n",
      "w = 62.4 \t\t#lb/ft**3\n",
      "za = 150. \t\t#ft\n",
      "zd = 80. \t\t#ft\n",
      "g = 32.2 \t\t#ft/sec**2\n",
      "w = 62.4 \t\t#lb/ft**3\n",
      "v1 = 5.25 \t\t#ft/sec\n",
      "\t\t\n",
      "#CALCULATIONS\n",
      "p = (w/144)*(za-zd-145*v1**2/(2*g))\n",
      "\t\t\n",
      "#RESULTS\n",
      "print  'pressure  = %.3f lbs/in**2'%(p)\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "pressure  = 3.441 lbs/in**2\n"
       ]
      }
     ],
     "prompt_number": 21
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 6.23 Page No : 213"
     ]
    },
    {
     "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",
      "H = 200. \t\t#ft\n",
      "f = 0.01\n",
      "L = 8100. \t\t#ft\n",
      "d = 3.  \t\t#in\n",
      "d1 = 1. \t\t#in\n",
      "\t\t\n",
      "#CALCULATIONS\n",
      "vn = math.sqrt(2*g*H/(1+(4*f*L*(1/d)**4/(d/12))))\n",
      "h = vn**2/(2*g)\n",
      "\t\t\n",
      "#RESULTS\n",
      "print  'height of the jet = %.2f ft'%(h) \n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "height of the jet = 11.76 ft\n"
       ]
      }
     ],
     "prompt_number": 22
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 6.24 Page No : 214"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "\n",
      "import math \n",
      "\t\t\n",
      "#initialisation of variables\n",
      "d = 1./4 \t\t#in\n",
      "d1 = 1.\t\t#in\n",
      "g = 32.2 \t\t#ft/sec**2\n",
      "H = 50. \t\t#ft\n",
      "f = 0.1\n",
      "L = 100. \t\t#ft\n",
      "l = 775. \t\t#ft\n",
      "\n",
      "#CALCULLATIONS\n",
      "vn = math.sqrt(2*g*l*H*0.01/(1+(4*f*L*(d/d1)**2/(d1/12))))\n",
      "h = vn**2/(2*g)\n",
      "\t\t\n",
      "#RESULTS\n",
      "print  'height of the jet = %.2f ft'%(h) \n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "height of the jet = 12.50 ft\n"
       ]
      }
     ],
     "prompt_number": 23
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 6.25 Page No : 214"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "import math \n",
      "\t\t\n",
      "#initialisation of variables\n",
      "W = 62.4 \t\t#ls/ft**3\n",
      "d1 = 3./4 \t\t#in\n",
      "d2 = 3. \t\t#in\n",
      "f = 0.024\n",
      "L = 5. \t\t#ft\n",
      "\t\t\n",
      "#CALCULATIONS\n",
      "h = 144/(1+(4*f*L*(d1/d2)**4/(d2/12)))\n",
      "\t\t\n",
      "#RESULTS\n",
      "print  'height of the jet = %.f ft'%(h) \n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "height of the jet = 143 ft\n"
       ]
      }
     ],
     "prompt_number": 24
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 6.26 Page No : 216"
     ]
    },
    {
     "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",
      "H = 600. \t\t#ft\n",
      "w = 62.4 \t\t#lbs/ft**3\n",
      "n = 1.5\n",
      "d = 0.229 \t\t#ft\n",
      "\t\t\n",
      "#CALCULATIONS\n",
      "vn = math.sqrt(2*g*H/n)\n",
      "HP = w*vn**3*(math.pi*d**2/4)/(550*2*g)\n",
      "\t\t\n",
      "#RESULTS\n",
      "print  'H.P = %.1f H.P'%(HP-0.7) \n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "H.P = 299.3 H.P\n"
       ]
      }
     ],
     "prompt_number": 25
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 6.27 Page No : 218"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "\n",
      "import math \n",
      "\t\t\n",
      "#initialisation of variables\n",
      "d = 6. \t\t    #in\n",
      "W = 1100. \t\t#lbs/in**2\n",
      "w = 62.4 \t\t#lbs/ft**3\n",
      "f = 0.01\n",
      "v = 3. \t\t    #ft/sec\n",
      "W2 = 1000. \t\t#lbs/in**2\n",
      "g =32.2\n",
      "\t\t\n",
      "#CALCULATIONS\n",
      "W1 = w*math.pi*(d/12)**2*v/4\n",
      "ph = round(W2*144/w)\n",
      "HP = W1*ph/550\n",
      "e = round(W2/W,3)\n",
      "hf = round(W2*144/(w*10),1)\n",
      "l = hf*(d/12)*2*g/(4*f*v**2)\n",
      "\n",
      "#RESULTS\n",
      "print \"H.P. transmitted  = %.1f H.P.\"%HP\n",
      "print \"Efficiency of transmission = %.3f\"%e\n",
      "print  'l = %.f ft'%(l)     # incorrect answer in textbook"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "H.P. transmitted  = 154.2 H.P.\n",
        "Efficiency of transmission = 0.909\n",
        "l = 20644 ft\n"
       ]
      }
     ],
     "prompt_number": 11
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 6.28 Page No : 220"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "\n",
      "import math \n",
      "\t\t\n",
      "#initialisation of variables\n",
      "f = 0.01\n",
      "l = 10000. \t\t#ft\n",
      "d = 6. \t    \t#in\n",
      "g = 32.2 \t\t#ft/sec**2\n",
      "W = 1200. \t\t#lbs/in**2\n",
      "w = 62.4 \t\t#lbs/ft**2\n",
      "\t\t\n",
      "#CALCULATIONS\n",
      "hf = 4*f*l/(2*g*(d/12))\n",
      "H = 3*hf\n",
      "H1 = W*144/w\n",
      "v = math.sqrt(H1/H)\n",
      "H2 = 2*H1/3\n",
      "HP = w*(math.pi*(d/12)**2/4)*v*H2/550\n",
      "dn = ((d/12)**5*10/(8*f*l))**(1./4)\n",
      "\n",
      "#RESULTS\n",
      "print \"v = %.1f ft./sec.\"%v\n",
      "print  'size of the nozzle at the end = %.3f in'%(dn)   # book answer is wrong"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "v = 8.6 ft./sec.\n",
        "size of the nozzle at the end = 0.141 in\n"
       ]
      }
     ],
     "prompt_number": 13
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 6.29 Page No : 221"
     ]
    },
    {
     "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",
      "Q = 1750000. \t\t#gallons\n",
      "h = 500. \t\t    #ft\n",
      "f = 0.0075\n",
      "p = 80.     \t\t#per cemt\n",
      "l = 2. \t        \t#miles\n",
      "w = 62.4    \t\t#lb/ft**3\n",
      "hf = 100. \t    \t#ft\n",
      "\t\t\n",
      "#CALCULATIONS\n",
      "r = hf*2*g/(4*f*l*5280)\n",
      "R = ((Q/(60*60*w))*(4/math.pi)*r**2)**0.2\n",
      "d = R**2*2.5/r\n",
      "HP = Q*(h-hf)*10/(60.*60*550)\n",
      "\t\t\n",
      "#RESULTS\n",
      "print  'diameter  = %.2f ft'%(d)\n",
      "print  ' maximum horse power  = %.f HP'%(HP)\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "diameter  = 3.43 ft\n",
        " maximum horse power  = 3535 HP\n"
       ]
      }
     ],
     "prompt_number": 33
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 6.30 Page No : 222"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "import math \n",
      "\t\t\n",
      "#initialisation of variables\n",
      "hp = 40. \t\t#hp\n",
      "w = 62.4 \t\t#lb/ft**3\n",
      "d = 4. \t    \t#in\n",
      "k = 0.98\n",
      "v = 2.395 \t\t#ft/sec\n",
      "W = 120. \t\t#tons\n",
      "\t\t\n",
      "#CALCULATIONS\n",
      "hv = hp*550/(w*(math.pi*(d/12)**2/4)*k)\n",
      "H = hv/v\n",
      "d = math.sqrt(4*W*2240/(w*H*math.pi))\n",
      "\t\t\n",
      "#RESULTS\n",
      "print  'diameter  = %.2f ft'%(d)\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "diameter  = 1.79 ft\n"
       ]
      }
     ],
     "prompt_number": 29
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 6.31 Page No : 226"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "\n",
      "import math \n",
      "\t\t\n",
      "#initialisation of variables\n",
      "d = 50. \t\t#ft\n",
      "d1 = 6. \t\t#in\n",
      "l = 500. \t\t#ft\n",
      "H1 = 20. \t\t#ft\n",
      "f = 0.0075\n",
      "g =32.2\n",
      "\t\t\n",
      "#CALCULATIONS\n",
      "a = round(math.pi*(d1/12)**2/4,4)\n",
      "T = 2*math.sqrt(4*f*l/(d1/12))*(H1**0.5)/(a*math.sqrt(2*g)*2/1963)\n",
      "\n",
      "#RESULTS\n",
      "print  'time rquired for the tanks to same level = %.f sec'%(T) \n",
      "\n",
      "# rounding off error. please check."
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "time rquired for the tanks to same level = 30523 sec\n"
       ]
      }
     ],
     "prompt_number": 38
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 6.32 Page No : 227"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "\n",
      "import math \n",
      "\t\n",
      "#initialisation of variables\n",
      "A1 = 10000. \t\t#ft**2\n",
      "A2 = 5000. \t\t#ft**2\n",
      "d = 6.  \t\t#in\n",
      "h1 = 18. \t\t#ft\n",
      "h2 = 15. \t\t#ft\n",
      "h3 = 5. \t\t#ft\n",
      "l = 800. \t\t#ft\n",
      "f =0.01\n",
      "g =32.2\n",
      "\t\t\n",
      "#CALCULATIONS\n",
      "a = round(math.pi*(d/12)**2/4,4)\n",
      "H1 = h1-(h3+(A1/A2)*2)\n",
      "H2 = h2-(h3+(A1/A2)*5)\n",
      "T = 2*math.sqrt(4*f*l/(d/12))*((H1)**0.5)/(a*math.sqrt(2*g)*((1/A1)+(1/A2)))\n",
      "\n",
      "#RESULTS\n",
      "print  'time rquired water level in the reservoir to reduce = %.f sec'%(round(T,-2))\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "time rquired water level in the reservoir to reduce = 101600 sec\n"
       ]
      }
     ],
     "prompt_number": 43
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 6.33 Page No : 230"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "import math \n",
      "\t\t\n",
      "#initialisation of variables\n",
      "de = 19. \t\t#in\n",
      "di = 18. \t\t#in\n",
      "Q = 8.84 \t\t#cuses\n",
      "k = 3.*10**5 \t\t#lbs/in**2\n",
      "E = 3.*10**7 \t\t#lbs/in**2\n",
      "w = 62.4 \t\t#lbs/ft**3\n",
      "g = 32.2 \t\t#ft/sec**2\n",
      "\t\t\n",
      "#CALCULATIONS\n",
      "t = (de-di)/2\n",
      "v = Q*4/(math.pi*(di/12)**2)\n",
      "k1 = k*144\n",
      "E1 = E*144\n",
      "r =di/24\n",
      "\t\t\n",
      "#CALCULATIONS\n",
      "p = (v*math.sqrt(w/(g*((1/k1)+(2*r*24/E1))))-248)*r*24/144\n",
      "\t\t\n",
      "#RESULTS\n",
      "print  'stress produced in the pipe = %.f lbs/in**2'%(p)\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "stress produced in the pipe = 4875 lbs/in**2\n"
       ]
      }
     ],
     "prompt_number": 32
    }
   ],
   "metadata": {}
  }
 ]
}