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  "name": ""
 },
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 "worksheets": [
  {
   "cells": [
    {
     "cell_type": "heading",
     "level": 1,
     "metadata": {},
     "source": [
      "Chapter 11 : Flow and Power Transmission through Pipe-lines"
     ]
    },
    {
     "cell_type": "heading",
     "level": 3,
     "metadata": {},
     "source": [
      "Example 11.2.1 page no : 307"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "\n",
      "import math \n",
      "#initialisation of variables\n",
      "Q= 0.5 \t\t\t#ft**3/sec\n",
      "d= 3. \t\t\t#in\n",
      "d1= 4. \t\t\t#in\n",
      "d2= 2. \t\t\t#in\n",
      "h= 12.7 \t\t\t#in\n",
      "g= 32.2 \t\t\t#ft/sec**2\n",
      "s= 13.6 \t\t\t#kg/m**3\n",
      "\t\t\t\n",
      "#CALCULATIONS\n",
      "k  = (s-1)*(h/2.)*2.*g*(math.pi/(Q*d**2*4))**2.+((d1/d)**4.-1)\n",
      "Cc = (d1/d2)**2./(math.sqrt(k)+1.)\n",
      "\t\t\t\n",
      "#CALCULATIONS\n",
      "print  ' contraction coefficient= %.3f '%(Cc)\n",
      "print  ' ANSWER GIVEN IN THE TEXTBOOK IS WRONG'\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        " contraction coefficient= 0.294 \n",
        " ANSWER GIVEN IN THE TEXTBOOK IS WRONG\n"
       ]
      }
     ],
     "prompt_number": 2
    },
    {
     "cell_type": "heading",
     "level": 3,
     "metadata": {},
     "source": [
      "Example 11.3.1 page no : 310"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "    \n",
      "import math \n",
      "#initialisation of variables\n",
      "Q= 400. \t\t\t#gallons\n",
      "d= 4. \t\t\t#in\n",
      "d1=6. \t\t\t#in\n",
      "C= 0.66\n",
      "g= 32.2 \t\t\t#ft/sec**2\n",
      "w=62.4\t\t\t#lbf/ft**3\n",
      "\t\t\t\n",
      "#CALCULATIONS\n",
      "u1= (Q/60.)*d1**2./(math.pi*6.23)\n",
      "u2= (d/d1)**2.*u1\n",
      "h= (u1-u2)**2./(2*g)\n",
      "w=62.4\t\t\t#lbf/ft**3\n",
      "p= (((u1**2-u2**2)/(2*g))-h)*w\n",
      "h1= ((1/C)-1.)**2*(u1**2/(2*g))\n",
      "p1= (((u1**2.-u2**2)/(2*g))+h1)*w\n",
      "p2= (u1**2.-u2**2)\n",
      "\t\t\t\n",
      "#RESULTS\n",
      "print  ' Loss of head due to the sudden enlargement= %.3f ft'%(h)\n",
      "print  '  difference in pressure = %.1f lbf/ft**2'%(p)\n",
      "print  '  difference in pressure = %.f lbf/ft**2'%round(p1,-1)\n",
      "print  '  difference in pressure = %d lbf/ft**2'%(p2)\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        " Loss of head due to the sudden enlargement= 0.721 ft\n",
        "  difference in pressure = 71.9 lbf/ft**2\n",
        "  difference in pressure = 160 lbf/ft**2\n",
        "  difference in pressure = 120 lbf/ft**2\n"
       ]
      }
     ],
     "prompt_number": 5
    },
    {
     "cell_type": "heading",
     "level": 3,
     "metadata": {},
     "source": [
      "Example 11.8.1 page no : 323"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "\n",
      "# variables\n",
      "f = .0075\n",
      "H = 96           # ft\n",
      "u1 = 25.         # ft\n",
      "g = 64.4        \n",
      "l1 = 5280        # ft\n",
      "\n",
      "\n",
      "# calculations\n",
      "d = (4*f*u1*(3*l1+(4*l1)/4))/(H*g)\n",
      "\n",
      "# results\n",
      "print \"Diameter of Pipe = %.2f ft\"%d"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Diameter of Pipe = 2.56 ft\n"
       ]
      }
     ],
     "prompt_number": 3
    },
    {
     "cell_type": "heading",
     "level": 3,
     "metadata": {},
     "source": [
      "Example 11.8.2 page no : 324"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "\n",
      "import math\n",
      "#initialisation of variables\n",
      "l= 9. \t\t\t#ft\n",
      "g= 32.2 \t\t\t#ft/sec**2\n",
      "Q= 160.\n",
      "A= 21.*math.pi\n",
      "\t\t\t\n",
      "#CALCULATIONS\n",
      "d= 6./(l*2*g*(A/Q)**2-1.5)\n",
      "\t\t\t\n",
      "#RESULTS\n",
      "print  'diameter = %.2f ft'%(d)\n",
      "\n",
      "#ANSWER GIVEN IN THE TEXTBOOKIS WRONG\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "diameter = 0.06 ft\n"
       ]
      }
     ],
     "prompt_number": 6
    },
    {
     "cell_type": "heading",
     "level": 3,
     "metadata": {},
     "source": [
      "Example 11.8.3 pageno :325"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "\n",
      "import math \n",
      "#initialisation of variables\n",
      "a= 2.493\n",
      "b= 6.8\n",
      "c= -393./(4*2.493)\n",
      "d= 0.75 \t\t\t#ft\n",
      "\t\t\t\n",
      "#CALCULATIONS\n",
      "u2= (-b+math.sqrt(b**2-4*a*c))/(2*a)\n",
      "Q= math.pi*d**2*u2/4.\n",
      "\t\t\t\n",
      "#RESULTS\n",
      "print  ' Rate of flow = %.2f ft**3/sec'%(Q)\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        " Rate of flow = 1.25 ft**3/sec\n"
       ]
      }
     ],
     "prompt_number": 11
    },
    {
     "cell_type": "heading",
     "level": 3,
     "metadata": {},
     "source": [
      "Example 11.10.2 page no : 332"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "    \n",
      "import math \n",
      "#initialisation of variables\n",
      "g= 32.2 \t\t\t#ft/sec**2\n",
      "H= 100. \t\t\t#ft\n",
      "L= 1000. \t\t\t#ft\n",
      "h1= 0.03\n",
      "h= 0.05\n",
      "h2= 0.4\n",
      "d= 6. \t\t\t#in\n",
      "le= 1021. \t\t\t#ft\n",
      "\t\t\t\n",
      "#CALCULATIONS\n",
      "u1= math.sqrt((2*g*H)/(1+h+h2+(h1*L/0.5)))\n",
      "Q= math.pi*(d/12.)**2*u1/4.\n",
      "u2= math.sqrt((H*2*g)/(1+h+(1./16)*(1+h+h2+(h1*L/0.5))))\n",
      "Q1= math.pi*(d/24)**2*u2/4\n",
      "r= math.sqrt((d/12)/(2*h1*le))\n",
      "\t\t\t\n",
      "#RESULTS\n",
      "print  ' rate of discharge without a nozzle= %.2f ft**3/sec'%(Q)\n",
      "print  '  rate of discharge= %.2f ft**3/sec'%(Q1)\n",
      "print  '  diameter of nozzle= %.2f in'%(r)\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        " rate of discharge without a nozzle= 2.01 ft**3/sec\n",
        "  rate of discharge= 1.78 ft**3/sec\n",
        "  diameter of nozzle= 0.09 in\n"
       ]
      }
     ],
     "prompt_number": 12
    },
    {
     "cell_type": "heading",
     "level": 3,
     "metadata": {},
     "source": [
      "Example 11.10.3 pageno : 332"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "\n",
      "import math\n",
      "#initialisation of variables\n",
      "\n",
      "L= 1730. \t\t\t#ft\n",
      "l= 104. \t\t\t#ft\n",
      "hl= 234. \t\t\t#ft\n",
      "u= 2.3 \t\t\t#ft/sec\n",
      "g= 32.2 \t\t\t#ft/sec**2\n",
      "p1 = 750*144.\n",
      "w = 62.3\n",
      "\t\t\t\n",
      "#CALCULATIONS\n",
      "Q = round(3*math.pi/4.*1./4 *u,3)       # ft**3/sec\n",
      "H1 = round(p1/w + (u**2)/(2*g),-1)      # ft\n",
      "Pi = round(w*Q*H1/550)                  # h.p.\n",
      "H2= L-l\n",
      "R= u**2/(2*g)\n",
      "H1= round(H2+R+hl)\n",
      "z = H2*100./H1\n",
      "\t\t\t\n",
      "#RESULTS\n",
      "print  ' efficiency of tramsmission = %.1f per cent'%(z)\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        " efficiency of tramsmission = 87.4 per cent\n"
       ]
      }
     ],
     "prompt_number": 15
    },
    {
     "cell_type": "heading",
     "level": 3,
     "metadata": {},
     "source": [
      "Example 11.11.1 page no : 335"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "\n",
      "import math \n",
      "#initialisation of variables\n",
      "g= 32.2 \t\t\t#ft/sec**2\n",
      "w= 62.3 \t\t\t#lbf/ft**3\n",
      "p= 40. \t\t\t#lbf/in**2\n",
      "k= 0.44\n",
      "d= 2. \t\t\t#in\n",
      "\t\t\t\n",
      "#CALCULATIONS\n",
      "Cu = math.sqrt(1/1.44)\n",
      "Q= (math.pi*(d/2)**2./144.)*math.sqrt(2*g*p*144/(w*0.981))\n",
      "P= w*Q**3*(144./math.pi)**2/(2*g*550)\n",
      "\t\t\t\n",
      "#RESULTS\n",
      "print ' Coefficient of velocity = %.3f'%Cu\n",
      "print  ' discharge rate = %.2f ft**3/min'%(Q)\n",
      "print  ' Power of jet = %.2f h.p'%(P)\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        " Coefficient of velocity = 0.833\n",
        " discharge rate = 1.70 ft**3/min\n",
        " Power of jet = 18.14 h.p\n"
       ]
      }
     ],
     "prompt_number": 14
    },
    {
     "cell_type": "heading",
     "level": 3,
     "metadata": {},
     "source": [
      "Example 11.11.2 page no : 336"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "    \n",
      "import math \n",
      "#initialisation of variables\n",
      "u= 80. \t\t\t#ft/sec\n",
      "d= 1.5 \t\t\t#in\n",
      "Cu= 0.97\n",
      "f= 0.007\n",
      "l= 150. \t\t\t#ft\n",
      "g= 32.2 \t\t\t#ft/sec**2\n",
      "n= 70. \t\t\t#percent\n",
      "Ho= -10. \t\t\t#ft\n",
      "w= 62.3 \t\t\t#lb/ft**3\n",
      "\t\t\t\n",
      "#CALCULATIONS\n",
      "Q=math.pi*d**2*u/(4*144.)\n",
      "ut= u/Cu\n",
      "H2= ut**2/(2*g)\n",
      "H1= 5*H2/4\n",
      "hf= H1/5\n",
      "D= ((4/math.pi)**2*4*f*l*Q**2*0.00237/(hf*2*g))**(1/5.)*12*(3.95/1.18)\n",
      "Ps= w*2*Q*(H1-Ho)*100/(n*550.)\n",
      "\t\t\t\n",
      "#RESULTS\n",
      "print  ' Diameter = %.2f in'%(D)\n",
      "print  '  shaft power of the pump = %.1f h.p'%(Ps)\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        " Diameter = 3.95 in\n",
        "  shaft power of the pump = 45.1 h.p\n"
       ]
      }
     ],
     "prompt_number": 15
    },
    {
     "cell_type": "heading",
     "level": 3,
     "metadata": {},
     "source": [
      "Example 11.13.1 pageno : 342"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "\n",
      "#initialisation of variables\n",
      "\n",
      "v= 10. \t\t\t#ft/sec\n",
      "g= 32. \t\t\t#ft/sec**2\n",
      "w= 62.3 \t\t\t#lbf/ft**3\n",
      "l= 200. \t\t\t#ft\n",
      "t= 0.5 \t\t\t#sec\n",
      "\t\t\t\n",
      "#CALCULATIONS\n",
      "dp= w*l*v/(g*t*144)\n",
      "\t\t\t\n",
      "#RESULTS\n",
      "print  ' Rise in pressure = %.1f lbf/in**2'%(dp)\n",
      "\n",
      "# Answer may vary because of rounding error."
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        " Rise in pressure = 54.1 lbf/in**2\n"
       ]
      }
     ],
     "prompt_number": 16
    },
    {
     "cell_type": "heading",
     "level": 3,
     "metadata": {},
     "source": [
      "Example 11.14.1 page no : 343"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "\n",
      "import math \n",
      "#initialisation of variables\n",
      "w= 62.3 \t\t\t#lb/ft**3\n",
      "g= 32.2\t\t\t#ft/se**2\n",
      "k= 3.*10.**5 \t\t\t#lbf/in**2\n",
      "u= 10.\t\t\t#ft/sec\n",
      "\t\t\t\n",
      "#CALCULATIONS\n",
      "P= u*math.sqrt(w*k/g)/12.\n",
      "\t\t\t\n",
      "#RESULTS\n",
      "print  ' Rise in pressure = %.f lbf/in**2'%(P)\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        " Rise in pressure = 635 lbf/in**2\n"
       ]
      }
     ],
     "prompt_number": 17
    },
    {
     "cell_type": "heading",
     "level": 3,
     "metadata": {},
     "source": [
      "Example 11.14.2 page no: 344"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "\n",
      "import math \n",
      "#initialisation of variables\n",
      "g= 32.2 \t\t\t#ft/sec**2\n",
      "w= 62.3 \t\t\t#lb/ft**3\n",
      "k= 3.*10.**5 \t\t\t#lbf/in**2\n",
      "\t\t\t\n",
      "#CALCULATIONS\n",
      "v= math.sqrt(k*g*144/w)\n",
      "\t\t\t\n",
      "#RESULTS\n",
      "print  ' velocity of sound in the fluid = %.f ft/sec'%(v)\n",
      "\n",
      "# answer may vary because of rounding error"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        " velocity of sound in the fluid = 4725 ft/sec\n"
       ]
      }
     ],
     "prompt_number": 18
    },
    {
     "cell_type": "heading",
     "level": 3,
     "metadata": {},
     "source": [
      "Example 11.14.4 pageno : 347"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "\n",
      "import math \n",
      "#initialisation of variables\n",
      "w= 62.3 \t\t\t#lb/ft**3\n",
      "d= 6. \t\t\t#in\n",
      "t= 5./8 \t\t\t#in\n",
      "k= 3.*10**5 \t\t\t#lbf/in**2\n",
      "E= 18.*10**6 \t\t\t#lbf/in**2\n",
      "M= 3. \t\t\t#tonf\n",
      "\t\t\t\n",
      "#CALCULATIONS\n",
      "u= math.sqrt(((M*2240)**2/w)*(t*2/d)*32.2*114*((t*2/(d*k))+(2/E)))\n",
      "Q= (math.pi*(d/2)**2/144)*u\n",
      "\t\t\t\n",
      "#RESULTS\n",
      "print  ' maximum permissible flow = %.2f ft**3/sec'%(Q)\n",
      "\n",
      "# answer may vary because of rounding error."
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        " maximum permissible flow = 4.15 ft**3/sec\n"
       ]
      }
     ],
     "prompt_number": 19
    }
   ],
   "metadata": {}
  }
 ]
}