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 "metadata": {
  "name": "",
  "signature": "sha256:1f6973f8c8d9996abb4c23a2c5352b16bfcd60086af954d2e49808504d3171d0"
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 "worksheets": [
  {
   "cells": [
    {
     "cell_type": "heading",
     "level": 1,
     "metadata": {},
     "source": [
      "CHapter12-Unsteady Flow"
     ]
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Ex1-pg557"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "import math\n",
      "#calculate Time for which flow into the tank continues after the power failure \n",
      "import scipy\n",
      "from scipy import integrate\n",
      "Q=0.05; ## m^3/s\n",
      "d=0.15; ## m^2\n",
      "h=8.; ## m\n",
      "g=9.81; ## m/s^2\n",
      "l=90.; ## m\n",
      "f=0.007;\n",
      "\n",
      "u1=Q/(math.pi/4.*d**2.);\n",
      "\n",
      "def function(u):\n",
      "\tfun=(1./((h*g/l)+(2.*f/d)*u**2))\n",
      "\treturn fun\n",
      "\n",
      "t=scipy.integrate.quad(function,u1,0)\n",
      "\n",
      "\n",
      "print(\"Time for which flow into the tank continues after the power failure is\" )\n",
      "print'%s %.1f %s'%(\" \",-t[0],\"s\")\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Time for which flow into the tank continues after the power failure is\n",
        "  2.6 s\n"
       ]
      }
     ],
     "prompt_number": 1
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Ex4-pg588"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "import math\n",
      "#calculate Estimate the height of tank required\n",
      "\n",
      "print(\"Estimate the height of tank required\")\n",
      "\n",
      "f=0.006;\n",
      "l=1400.; ## m\n",
      "g=9.81; ## m/s^2\n",
      "d1=0.75; ## m\n",
      "d2=3.; ## m\n",
      "Q=1.2; ## m^3/s\n",
      "a=20.; ## m\n",
      "\n",
      "K=4*f*l/(2*g*d1);\n",
      "\n",
      "## 2*K*Y = l*a/(g*A) = 8.919 s^2\n",
      "\n",
      "## Y=2*K*Y/2*K\n",
      "\n",
      "Y=8.919/(2*K);\n",
      "## When t=0\n",
      "\n",
      "u0=Q/(math.pi/4*d1**2);\n",
      "\n",
      "y0=K*u0**2;\n",
      "\n",
      "C=-Y/K/math.exp(y0/Y);\n",
      "\n",
      "## To determine the height of the surge tank, we consider the condition y = y_max when u = 0. \n",
      "\n",
      "## 0 = 1/K*(y_max+Y) + C*exp(y_max/Y)\n",
      "\n",
      "## From the above eqn we get\n",
      "\n",
      "y_max=-Y;\n",
      "\n",
      "H=a-y_max;\n",
      "print'%s %.1f %s'%(\"The minimum height of the surge tank =\",H,\"m\")\n",
      "\n",
      "\n",
      "print(\"The actual design height should exceed the minimum required, say 23 m\")"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Estimate the height of tank required\n",
        "The minimum height of the surge tank = 22.0 m\n",
        "The actual design height should exceed the minimum required, say 23 m\n"
       ]
      }
     ],
     "prompt_number": 2
    }
   ],
   "metadata": {}
  }
 ]
}