{
 "cells": [
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Chapter 4 : Fluid Dynamics"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "### Example 4.1 Page no 159"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 1,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "Difference in pressure in section 1 and 2 =  1157.0 kN/m**2\n"
     ]
    }
   ],
   "source": [
    "# Difference in pressure at top and bottom\n",
    "\n",
    "from math import *\n",
    "\n",
    "from __future__ import division\n",
    "\n",
    "# Given\n",
    "\n",
    "d1 = 0.1                     # diameter in m\n",
    "\n",
    "d2 = 0.05                     # diameter in m\n",
    "\n",
    "Q = 0.1                      # discharge in m**3/s\n",
    "\n",
    "A1 = pi*d1**2/4\n",
    "\n",
    "A2 = pi*d2**2/4\n",
    "\n",
    "gma =9810                   # specific weight\n",
    "\n",
    "z= 6                        # difference in the height\n",
    "\n",
    "g = 9.81\n",
    "\n",
    "# Solution\n",
    "\n",
    "V1 = Q/A1                    # velocity at section 1\n",
    "\n",
    "V2 = Q/A2                    # velocity at section 2\n",
    "\n",
    "dP = gma*((V2**2/(2*g))-(V1**2/(2*g))-z)/1000\n",
    "\n",
    "print \"Difference in pressure in section 1 and 2 = \",round(dP,1),\"kN/m**2\""
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "### Example 4.2 Page no 160"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 2,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "Actual discharge = 2.0 l/s\n"
     ]
    }
   ],
   "source": [
    "# Actual discharge\n",
    "\n",
    "from math import *\n",
    "\n",
    "from __future__ import division\n",
    "\n",
    "# Given\n",
    "\n",
    "d = 2.5                 # diameter in cm\n",
    "\n",
    "h =200                        # head in cm\n",
    "\n",
    "Cd = 0.65                   # coefficient of discharge\n",
    "\n",
    "A  =pi*d**2/4\n",
    "\n",
    "g = 9.81                  # acceleration due to gravity in m/s**2 \n",
    "\n",
    "# Solution\n",
    "\n",
    "Q = Cd*A*sqrt(2*g*h)/100\n",
    "\n",
    "print \"Actual discharge =\",round(Q,2),\"l/s\"\n",
    "\n"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "### Example 4.3 Page no 162"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 4,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "Discharge through the orifice = 2.69 m**3/s\n"
     ]
    }
   ],
   "source": [
    "# Discharge through the orifice\n",
    "\n",
    "from __future__ import division\n",
    "\n",
    "from math import *\n",
    "\n",
    "from scipy import integrate\n",
    "\n",
    "import numpy as np\n",
    "\n",
    "# Given\n",
    "\n",
    "H1 = 3               # height in m\n",
    "\n",
    "H2 = 4               # height in m\n",
    "\n",
    "b = 0.5              # width in m\n",
    "\n",
    "Cd = 0.65            # co-efficient of discharge \n",
    "\n",
    "g = 9.81             # acceleration due to grvity in m/s**2\n",
    "\n",
    "# Solution\n",
    "\n",
    "q = lambda h: h**(1/2)\n",
    "    \n",
    "Q,err = integrate.quad(q, H1, H2)\n",
    "\n",
    "Qt = Cd*b*sqrt(2*g)*Q\n",
    "\n",
    "print \"Discharge through the orifice =\",round(Qt,2),\"m**3/s\"\n"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "### Example 4.4 Page no 163"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 6,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "Total Discharge = 1.95 m**3/s\n"
     ]
    }
   ],
   "source": [
    "# discharge through orifice\n",
    "    \n",
    "from math import *\n",
    "\n",
    "from scipy import integrate\n",
    "\n",
    "from __future__ import division\n",
    "\n",
    "import numpy as np\n",
    "\n",
    "# Given\n",
    "\n",
    "b = 1                  # bredth of the tank\n",
    "\n",
    "d = 0.5                # depth of the tank\n",
    "\n",
    "h1 = 0.2                 # height of the orifice in m\n",
    "\n",
    "Cd = 0.6               # coefficient of discharge\n",
    "\n",
    "H1 = 2                 # height in m\n",
    "\n",
    "H2 = 2+h1               # height in m\n",
    "\n",
    "g = 9.81               # acceleration due to gravity in m/s**2\n",
    "\n",
    "A = 1*0.3              # area of submerged section in m**2\n",
    "\n",
    "# Solution\n",
    "\n",
    "q = lambda h: h**(1/2)\n",
    "    \n",
    "Q,err = integrate.quad(q, H1, H2)\n",
    "\n",
    "Q1 = Cd*b*sqrt(2*g)*(Q) # Flow through area 1\n",
    "\n",
    "Q2 = Cd*sqrt(2*g*H2)*A\n",
    "\n",
    "Td = Q1+Q2\n",
    "\n",
    "print \"Total Discharge =\",round(Td,2),\"m**3/s\"\n"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "### Example 4.5 Page no 165"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 7,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "Actual discharge = 0.000454 m**3/s\n"
     ]
    }
   ],
   "source": [
    "# Determine flow rate of water\n",
    "\n",
    "from math import *\n",
    "\n",
    "from __future__ import division\n",
    "\n",
    "# Given\n",
    "\n",
    "d1 = 2               # radius of pipe\n",
    "\n",
    "d2 = 1               # radius of throat\n",
    "\n",
    "D1 = 40\n",
    "\n",
    "D2 = 20\n",
    "\n",
    "A1 = pi*D1**2/4\n",
    "\n",
    "A2 = pi*D2**2/4\n",
    "\n",
    "Cd = 0.95\n",
    "\n",
    "# Solution\n",
    "\n",
    "V2 = sqrt(21582/0.9375)\n",
    "\n",
    "Q = 1.52*pi*(d1/100)**2/4\n",
    "\n",
    "Qa = Q*Cd\n",
    "\n",
    "print \"Actual discharge =\",round(Qa,6),\"m**3/s\""
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "### Example 4.6 Page no 166"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 8,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "velocity at the dept of 1 ft = 5.67 ft/s\n"
     ]
    }
   ],
   "source": [
    "# Velocity of stream point at the point of insertion\n",
    "\n",
    "from math import *\n",
    "\n",
    "from __future__ import division\n",
    "\n",
    "# Given\n",
    "\n",
    "dx = 0.5              # in ft\n",
    "\n",
    "K = 1                 # constant\n",
    "\n",
    "g = 32.2              # acceleration due to gravity in ft/s**2\n",
    "\n",
    "# solution\n",
    "\n",
    "V = sqrt(2*g*dx)\n",
    "\n",
    "print \"velocity at the dept of 1 ft =\",round(V,2),\"ft/s\""
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "### Example no 4.7 Page no 172"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 10,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "Discharge throught the system = 190.0 l/s\n"
     ]
    }
   ],
   "source": [
    "# Discharge throught the system\n",
    "\n",
    "from math import *\n",
    "\n",
    "from __future__ import division\n",
    "\n",
    "gma= 0.8      # specific weight\n",
    "\n",
    "V2 = 40       # velocity in m/s\n",
    "\n",
    "z1 =25        # height at point 1\n",
    "\n",
    "g = 9.81      # acceleration due to gravity in m/s**2\n",
    "\n",
    "d = 15        # diameter of the nozzle in cm\n",
    "\n",
    "# Solution\n",
    "\n",
    "V2 = sqrt(2*g*z1/4.25)\n",
    "\n",
    "A = pi*(d/100)**2/4\n",
    "\n",
    "Q = A*V2*1000\n",
    "\n",
    "print \"Discharge throught the system =\",round(Q,0),\"l/s\""
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "### Example 4.8 Page no 174"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 11,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "Power input =  32.5 kW\n"
     ]
    }
   ],
   "source": [
    "# Power input to the pump\n",
    "\n",
    "from math import *\n",
    "\n",
    "from __future__ import division\n",
    "\n",
    "\n",
    "# Given\n",
    "\n",
    "Eff = 0.8     # pump efficiency\n",
    "\n",
    "Hl = 30       # head loss in m\n",
    "\n",
    "D1 =6         # diameter in cm\n",
    "\n",
    "D2 = 2        # diameter in cm\n",
    "\n",
    "gma = 9810    # specific weight in N/m**3\n",
    "\n",
    "V2 = 40       # velocity in m/s\n",
    "\n",
    "P1 = -50      # pressure at point 1 in N/m**2\n",
    "\n",
    "z2 = 100      # height at point 2\n",
    "\n",
    "g = 9.8       # acceleration due to gravity in m/s**2\n",
    "\n",
    "z1 = 30       # height in m\n",
    "\n",
    "# Solution\n",
    "\n",
    "V1=(2/6)**2*V2\n",
    "\n",
    "Q = (pi*6**2/4)*V1*10**-4\n",
    "\n",
    "Hs = z2 + (V2**2/(2*g)) + z1 + (50/gma) -(V1**2/(2*g))\n",
    "\n",
    "P = gma*Q*Hs\n",
    "\n",
    "Pi = (P/Eff)/1000\n",
    "\n",
    "print \"Power input = \",round(Pi,1),\"kW\""
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "### Example 4.9 Page no 176"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 12,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "Elevation at height A = 492.39 m\n",
      "Elevation at height B = 576.23 m\n"
     ]
    }
   ],
   "source": [
    "# Pressure head at A and B\n",
    "\n",
    "from math import *\n",
    "\n",
    "# Given\n",
    "\n",
    "Q = 0.2           # discharge in m**3/s\n",
    "\n",
    "d1 = 0.25           # diameter of the pipe in m\n",
    "\n",
    "A = pi*d1**2/4    # area of the pipe\n",
    "\n",
    "za = 480          # height in m\n",
    "\n",
    "z1 = 500          # height in m\n",
    "\n",
    "z3 = 550          # elevation in m\n",
    "\n",
    "gma =9810         # specific weight in N/m**2\n",
    "\n",
    "g =9.81           # acceleration due to gravity in m/s**2\n",
    "\n",
    "# Solution\n",
    "\n",
    "V=Q/A           # Velocity of in m/s\n",
    "\n",
    "Hl1 = (0.02*100*V**2/(0.25*2*9.81))\n",
    "\n",
    "# pressure head at A\n",
    "\n",
    "Pa =(z1-za-(V**2/(2*g))-Hl1)\n",
    "\n",
    "El = za+Pa\n",
    "\n",
    "print \"Elevation at height A =\",round(El,2),\"m\"\n",
    "\n",
    "# pressure head at B\n",
    "\n",
    "hs = z3 - z1 + (0.02*(500/0.25)*(V**2/(2*g))) \n",
    "\n",
    "El2 = El+hs\n",
    "\n",
    "print \"Elevation at height B =\",round(El2,2),\"m\"\n",
    "\n",
    "\n"
   ]
  },
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   "execution_count": null,
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