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 "metadata": {
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  "signature": "sha256:84e452258bd05b64c16351467c4970051f4494cb47d7a832df03bdce07abddb8"
 },
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 "worksheets": [
  {
   "cells": [
    {
     "cell_type": "heading",
     "level": 1,
     "metadata": {},
     "source": [
      "Chapter1-Introduction"
     ]
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Ex1-pg9"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "\n",
      "import math\n",
      " #determine\n",
      "##This numerical is Ex 1_1E,page 9.\n",
      "Pso=20.5\n",
      "Psc=20.5*550##converting hp to fps system\n",
      "Qo=385.\n",
      "Qc=385./449.##converting gpm to ft^3/s\n",
      "E=0.83\n",
      "dp=E*Psc/(Qc*144.)\n",
      "print\"%s %.2f %s \"%('The pressure rise is ',dp,' psi')\n",
      "print(\"After rounding off,pressure rise is 75.8 psi\")\n",
      "dpr=75.8\n",
      "dHw=75.8*144/62.4##62.4 is accelaration due to gravity in fps system\n",
      "print\"%s %.2f %s \"%(' The head of water is ',dHw,' ft of water')\n",
      "print(\"After rounding off the value of head of water the answer is 175 ft of water.\")\n",
      "dhwr=175##rounded off value of head of water\n",
      "sg=0.72##specific gravity of oil\n",
      "dHo=dhwr/sg\n",
      "print\"%s %.2f %s \"%(' The head of oil is ',dHo,' ft of oil')\n",
      "print(\"After rounding off the value of head of oil the answer is 243 ft of oil.\")\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The pressure rise is  75.79  psi \n",
        "After rounding off,pressure rise is 75.8 psi\n",
        " The head of water is  174.92  ft of water \n",
        "After rounding off the value of head of water the answer is 175 ft of water.\n",
        " The head of oil is  243.06  ft of oil \n",
        "After rounding off the value of head of oil the answer is 243 ft of oil.\n"
       ]
      }
     ],
     "prompt_number": 1
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Ex2-pg10"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "import math \n",
      "#determine\n",
      "##This numerical is Ex 1_1S,page 10.\n",
      "E=0.83##efficiency\n",
      "Ps=15300.\n",
      "Q=87.4\n",
      "Qs=87.4/3600.##flow rate in meter cube per sec\n",
      "rho=998.\n",
      "g=9.81\n",
      "sg=0.72\n",
      "dp=E*Ps/Qs\n",
      "print\"%s %.2f %s \"%('\\n The change in pressure (dp)is ',dp,'')\n",
      "dpr=523000##rounded value of dp\n",
      "print(\"The rounded off value of dp is 523kPa.\")\n",
      "dHw=dpr/(rho*g)\n",
      "print\"%s %.2f %s \"%(' dHw is equal to ',dHw,' m of water')\n",
      "print(\"The rounded off value of dHw is 53.4 m of water.\")\n",
      "dHwr=53.4##rounded off value of dHw\n",
      "print(\"Thus we can determine head of oil.\")\n",
      "dHoil=dHwr/sg\n",
      "print\"%s %.2f %s \"%(' dHoil is given by ',dHoil,' m of oil')\n",
      "print(\"The rounded off value of dHoil is 74.2 m of oil.\")\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "\n",
        " The change in pressure (dp)is  523070.94  \n",
        "The rounded off value of dp is 523kPa.\n",
        " dHw is equal to  53.42  m of water \n",
        "The rounded off value of dHw is 53.4 m of water.\n",
        "Thus we can determine head of oil.\n",
        " dHoil is given by  74.17  m of oil \n",
        "The rounded off value of dHoil is 74.2 m of oil.\n"
       ]
      }
     ],
     "prompt_number": 2
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Ex3-pg10"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#determine\n",
      "##This numerical is Ex 1_2E,page 10.\n",
      "Q=12000.\n",
      "A=3.5\n",
      "rho_a=0.0762\n",
      "E=0.85\n",
      "r=2.5##resistance of duct system\n",
      "V=Q/(60.*A)\n",
      "print\"%s %.2f %s \"%('The air flow velocity at discharge is ',V,' ft/s')\n",
      "KE=(rho_a*(V**2))/(32.2*2)\n",
      "print\"%s %.2f %s \"%('\\n The product is ',KE,' lb/ft^2')\n",
      "##PE=KE\n",
      "Hv=KE/62.4\n",
      "print\"%s %.2f %s \"%('\\n The dynamic head is ',Hv,' ft')\n",
      "print(\"The value of dynamic head in inches of water is 0.74.\")\n",
      "Hvi=0.74##Head in inches\n",
      "Ht=r+Hvi\n",
      "print\"%s %.2f %s \"%('\\n The total head is ',Ht,' inches of water')\n",
      "p_tot=Ht*62.4\n",
      "Ps=Q*p_tot/(60.*12.*E)\n",
      "print\"%s %.2f %s \"%('\\n The shaft power is ',Ps,' ft-lb/s')\n",
      "print(\"The shaft power is 7.2 hp.\")\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The air flow velocity at discharge is  57.14  ft/s \n",
        "\n",
        " The product is  3.86  lb/ft^2 \n",
        "\n",
        " The dynamic head is  0.06  ft \n",
        "The value of dynamic head in inches of water is 0.74.\n",
        "\n",
        " The total head is  3.24  inches of water \n",
        "\n",
        " The shaft power is  3964.24  ft-lb/s \n",
        "The shaft power is 7.2 hp.\n"
       ]
      }
     ],
     "prompt_number": 4
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Ex4-pg11"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "##This numerical is Ex 1_2S,page 11.\n",
      "Q=340.\n",
      "A=0.325\n",
      "V=Q/(60.*A)\n",
      "print\"%s %.2f %s \"%('The air flow velocity at discharge is ',V,' m/s')\n",
      "rho_a=1.22\n",
      "Vr=17.4\n",
      "Hd=(rho_a*(Vr**2))/2.\n",
      "print\"%s %.2f %s \"%('\\n The dynamic pressure head is ',Hd,' Pa')\n",
      "Hdr=184.7##rounded off value of Hd\n",
      "rho_w=998.##density of water=rhow\n",
      "g=9.81\n",
      "H=0.0635\n",
      "dp=rho_w*g*H##static pressure head\n",
      "print\"%s %.2f %s \"%('\\n The static pressure head is ',dp,' Pa')\n",
      "dpr=621.7\n",
      "p_tot=Hdr+dpr\n",
      "print\"%s %.2f %s \"%('\\n The total pressure head is ',p_tot,' Pa')\n",
      "p_tot=806.4\n",
      "E=0.85##efficiency\n",
      "Ps=Q*p_tot/(60*E)\n",
      "print\"%s %.2f %s \"%('\\n The shaft power is',Ps, 'W')\n",
      "print(\"The shaft power is 5.376 kW.\")\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The air flow velocity at discharge is  17.44  m/s \n",
        "\n",
        " The dynamic pressure head is  184.68  Pa \n",
        "\n",
        " The static pressure head is  621.69  Pa \n",
        "\n",
        " The total pressure head is  806.40  Pa \n",
        "\n",
        " The shaft power is 5376.00 W \n",
        "The shaft power is 5.376 kW.\n"
       ]
      }
     ],
     "prompt_number": 5
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Ex5-pg11"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#determine \n",
      "import math\n",
      "##This numerical is Ex 1_3E,page 11.\n",
      "H=295.##net head in ft\n",
      "Q=148.##water flow rate\n",
      "n=1800.##rpm\n",
      "E=0.87##efficiency\n",
      "a=62.4##product of density and accelaration due to gravity\n",
      "omega=(n*2.*math.pi)/60.\n",
      "dp=a*H\n",
      "print\"%s %.2f %s \"%('The pressure is ',dp,' lb/ft^2')\n",
      "Ps=E*Q*dp\n",
      "print\"%s %.2f %s \"%('\\n Output power is equal to ',Ps,' lb-ft/s')\n",
      "print(\"The output output power can also be written as 2.37*10^6 lb-ft/s\")\n",
      "print(\"Output power in terms of horsepower is given by 4309hp.\")\n",
      "Psr=2370000##rounded off value of Ps\n",
      "Torque=Psr/omega\n",
      "print\"%s %.2f %s \"%(' The output torque is ',Torque,' lb-ft.')\n",
      "print(\"The output torque can also be written as 12.57*10^3 lb-ft\")\n",
      "\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The pressure is  18408.00  lb/ft^2 \n",
        "\n",
        " Output power is equal to  2370214.08  lb-ft/s \n",
        "The output output power can also be written as 2.37*10^6 lb-ft/s\n",
        "Output power in terms of horsepower is given by 4309hp.\n",
        " The output torque is  12573.24  lb-ft. \n",
        "The output torque can also be written as 12.57*10^3 lb-ft\n"
       ]
      }
     ],
     "prompt_number": 6
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Ex6-pg12"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#determine c\n",
      "import math\n",
      "##This numerical is Ex 1_3S,page 12.\n",
      "H=90.\n",
      "Q=4.2##water flow rate(in m^3/s)\n",
      "n=1800.\n",
      "E=0.87##efficiency\n",
      "rho=998.\n",
      "g=9.81\n",
      "omega=(n*2.*math.pi)/60.\n",
      "dp=rho*g*H\n",
      "print\"%s %.2f %s \"%('The pressure is ',dp,' N/m^2')\n",
      "Ps=E*Q*dp\n",
      "print\"%s %.2f %s \"%('\\n Output power is equal to ',Ps,' N-m/s')\n",
      "print(\"After rounding off the value of output power is 3220 kW.\")\n",
      "Psr=3220000.##rounded off value of Ps\n",
      "Torque=Psr/omega\n",
      "print\"%s %.2f %s \"%(' The output torque is ',Torque,' N-m.')\n",
      "print(\"After rounding off the output torque comes out to be 17.1*10^3 N-m.\")\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The pressure is  881134.20  N/m^2 \n",
        "\n",
        " Output power is equal to  3219664.37  N-m/s \n",
        "After rounding off the value of output power is 3220 kW.\n",
        " The output torque is  17082.63  N-m. \n",
        "After rounding off the output torque comes out to be 17.1*10^3 N-m.\n"
       ]
      }
     ],
     "prompt_number": 7
    }
   ],
   "metadata": {}
  }
 ]
}