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+{

+ "metadata": {

+  "name": "",

+  "signature": "sha256:84e452258bd05b64c16351467c4970051f4494cb47d7a832df03bdce07abddb8"

+ },

+ "nbformat": 3,

+ "nbformat_minor": 0,

+ "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": {}

+  }

+ ]

+}
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