{
 "cells": [
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Chapter 9 : Turbulent flow in Pipes"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "### Example 9.1 Page no 308"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 1,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "It is a laminar flow\n",
      "(a) Head loss due to flow for glycerine = 17.1 m \n",
      "The flow is turbulent\n",
      "(a) Head loss due to flow for water = 5.42 m \n"
     ]
    }
   ],
   "source": [
    "# Determine Head loss\n",
    "\n",
    "from math import *\n",
    "\n",
    "# Given\n",
    "\n",
    "S = 1.26                # specific gravity\n",
    "\n",
    "mu = 0.826              # kinematic viscosity in Ns/m**2\n",
    "\n",
    "# for water\n",
    "\n",
    "rho = 998               # density of water in kg/m**3\n",
    "\n",
    "mu1 = 1.005*10**-3      # viscosity in Ns/m**2\n",
    "\n",
    "# for glycerine\n",
    "\n",
    "rho1 = S*rho            # density of glycerine in kg/m**3\n",
    "\n",
    "Q = 0.1                 # discharge in m**3/s\n",
    "\n",
    "d1 = 0.2                # diameter in m\n",
    "\n",
    "A = pi*d1**2/4          # area in m**2\n",
    "\n",
    "g = 9.81                # acceleration due to gravity in m/s**2\n",
    "\n",
    "l =100                  # length of the pipe\n",
    "\n",
    "# Solution\n",
    "\n",
    "V = Q/A\n",
    "\n",
    "R = rho1*V*d1/mu\n",
    "\n",
    "print \"It is a laminar flow\"\n",
    "\n",
    "f = 64/R               # friction factor\n",
    "\n",
    "Hf = f*l*V**2/(2*g*d1) # head loss due to friction\n",
    "\n",
    "print \"(a) Head loss due to flow for glycerine =\",round(Hf,1),\"m \"\n",
    "\n",
    "R1 = rho*V*d1/mu1\n",
    "\n",
    "print \"The flow is turbulent\"\n",
    "\n",
    "e = 0.025\n",
    "\n",
    "r = e/(d1*100)\n",
    "\n",
    "f = 0.021\n",
    "\n",
    "hf = f*l*V**2/(2*g*d1)\n",
    "\n",
    "print \"(a) Head loss due to flow for water =\",round(hf,2),\"m \"\n"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "### Example 9.2 Page no 311"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 2,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "3.1853324563\n",
      "Discharge = 0.1 m**3/s\n"
     ]
    }
   ],
   "source": [
    "# Discharge of water\n",
    "\n",
    "from math import *\n",
    "\n",
    "# Given\n",
    "\n",
    "# for water\n",
    "\n",
    "nu = 1.007*10**-6            # viscosity in m**2/s\n",
    "\n",
    "e = 0.025                    # for cast iron in cm\n",
    "\n",
    "L = 100                      # length of the pipe in m\n",
    "\n",
    "D = 0.2                      # diameter in m\n",
    "\n",
    "hf = 5.43                    # head loss due to friction\n",
    "\n",
    "r = e/(D*100)\n",
    "\n",
    "g = 9.81                     # acceleration due to gravity in m/s**2\n",
    "\n",
    "# Solution\n",
    "\n",
    "A = sqrt(2*g*D*hf/L)\n",
    "\n",
    "B = D/nu\n",
    "\n",
    "f = 0.021                   # from moodys diagram\n",
    "\n",
    "V = A/sqrt(f)\n",
    "\n",
    "print V\n",
    "\n",
    "R = B*f\n",
    "\n",
    "A = pi*D**2/4\n",
    "\n",
    "Q = A*V\n",
    "\n",
    "print \"Discharge =\",round(Q,2),\"m**3/s\"\n",
    "\n"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "### Example 9.3 Page no 314"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 3,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "Hence the convergence is attained, D= 0.2 m\n"
     ]
    }
   ],
   "source": [
    "# Size of the case iron\n",
    "\n",
    "from math import *\n",
    "\n",
    "# Given\n",
    "\n",
    "Q =0.1                          # discharge in m**3/s\n",
    "\n",
    "hf = 5.43                       # friction loss head in m\n",
    "\n",
    "L = 100                         # length of pipe\n",
    "\n",
    "nu = 1.00*10**-6                # viscosity in m**2/s\n",
    "\n",
    "e = 0.025                       # for cast iron in cm\n",
    "\n",
    "g = 9.81                        # acceleration due to gravity in m/s**2\n",
    "\n",
    "# Solution\n",
    "\n",
    "A = 8*L*Q**2/(hf*g*pi**2)\n",
    "\n",
    "B = 4*Q/(pi*nu)\n",
    "\n",
    "# for D = 0.172 ; f=0.01\n",
    "D = 0.172\n",
    "\n",
    "r = e/D\n",
    "\n",
    "Re = B/D\n",
    "\n",
    "f = 0.022                       # for Re and r\n",
    "\n",
    "# for D1=0.199 ; f=0.021\n",
    "\n",
    "D1 = 0.199\n",
    "\n",
    "r1 = e/D1\n",
    "\n",
    "R = B/D1\n",
    "\n",
    "f = 0.021                       # for R and r\n",
    "\n",
    "print \"Hence the convergence is attained, D=\",round(D1,1),\"m\"\n"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "### Example 9.4 Page no 318"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 4,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "Head loss = 2.0 ft\n"
     ]
    }
   ],
   "source": [
    "# Head loss \n",
    "\n",
    "from math import *\n",
    "\n",
    "# Given\n",
    "\n",
    "L = 500                 # length of the pipe in ft\n",
    "\n",
    "D= 9*2.54/100           # diameter in cm\n",
    "\n",
    "C = 100                 # constant\n",
    "\n",
    "S = 0.004\n",
    "\n",
    "# Solution\n",
    "\n",
    "Hf = S*L\n",
    "\n",
    "print \"Head loss =\",round(Hf,0),\"ft\"\n"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "### Example 9.5 Page no 319"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 6,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "Head loss of water = 3.2 m\n"
     ]
    }
   ],
   "source": [
    "# Head loss of water\n",
    "\n",
    "from math import *\n",
    "\n",
    "# Given\n",
    "\n",
    "Q = 0.1                  # water flow rate in m**3/s\n",
    "\n",
    "d = 30                  # diameter in m\n",
    "\n",
    "l = 500                  # length in m\n",
    "\n",
    "e = 0.025                # for cast iron\n",
    "\n",
    "g = 9.81                 # acceleration due to gravity in m/s**2\n",
    "\n",
    "# Solution\n",
    "\n",
    "r = log(d/e,10)\n",
    "\n",
    "K = (pi/4)*sqrt(2*g)*(2*r+1.14)*(0.3)**(2.5)\n",
    " \n",
    "S = (Q/K)**2\n",
    "\n",
    "hf = S*l\n",
    "\n",
    "print \"Head loss of water =\",round(hf,1),\"m\""
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "### Example 9.6 Page no 319"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 8,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "Head loss of water = 0.1 m**3/s\n"
     ]
    }
   ],
   "source": [
    "# Head loss by conveyance method\n",
    "\n",
    "from math import *\n",
    "\n",
    "# Given\n",
    "\n",
    "Q = 0.1                  # water flow rate in m**3/s\n",
    "\n",
    "d = 20                  # diameter in cm\n",
    "\n",
    "l = 500                  # length in m\n",
    "\n",
    "e = 0.025                # for cast iron\n",
    "\n",
    "g = 9.81                 # acceleration due to gravity in m/s**2\n",
    "\n",
    "S = 5.43 \n",
    "\n",
    "# Solution\n",
    "\n",
    "r = log(d/e,10)\n",
    "\n",
    "K = (pi/4)*sqrt(2*g)*(2*r+1.14)*(0.2)**2.5\n",
    "\n",
    "Q=K*sqrt(S/100)\n",
    "\n",
    "print \"Head loss of water =\",round(Q,2),\"m**3/s\""
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "### Example 9.7 Page no 320"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 10,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "K =  0.432  from trial and error\n"
     ]
    }
   ],
   "source": [
    "# Solve using the conveyence method\n",
    "\n",
    "from math import *\n",
    "\n",
    "# given\n",
    "\n",
    "eps = 0.025*10**-2           # for cast iron epsilon = 0.0025 cm\n",
    "\n",
    "# we get the value of K = 0.432 m**2/s\n",
    "# we need to do trial and error to find the value of D\n",
    "\n",
    "# we use the value of D = 0.2 m\n",
    "\n",
    "D = 0.2               # value in m\n",
    "\n",
    "g = 9.81\n",
    "\n",
    "# Solution\n",
    "\n",
    "K = (pi/4)*sqrt(2*g)*(2*log10(D/(eps))+1.14)*D**(2.5)\n",
    "\n",
    "print \"K = \",round(K,3),\" from trial and error\""
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "### Example 9.8 Page no 326"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 11,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "Total head loss = 83.7 m\n"
     ]
    }
   ],
   "source": [
    "# Determine head loss\n",
    "\n",
    "from math import *\n",
    "\n",
    "# Given\n",
    "\n",
    "d = 0.1                # diameter of the pipe\n",
    "\n",
    "Q= 0.075               # discharge in m**3/s\n",
    "\n",
    "L = 30                 # length in m\n",
    "\n",
    "A = pi*d**2/4\n",
    "\n",
    "g = 9.81               # acceleration due to gravity in m/s**2\n",
    "\n",
    "# for water\n",
    "\n",
    "nu = 1.007*10**-6       # viscosity in m**2/s\n",
    "\n",
    "e = 0.025\n",
    "\n",
    "r = e/(10*d)\n",
    "\n",
    "# Solution\n",
    "\n",
    "V = Q/A\n",
    "\n",
    "Re = V*d/nu\n",
    "\n",
    "f = 0.025              # firction factor from moodys diagram\n",
    "\n",
    "hf = f*L*V**2/(2*g*d)  \n",
    "\n",
    "K= 0.5                 # contraction constant\n",
    "\n",
    "hc = K*V**2/(2*g)     \n",
    "\n",
    "K1 =10                 # loss of the globe valve\n",
    "\n",
    "hg = K1*V**2/(2*g)\n",
    "\n",
    "Th = hf+hc+hg\n",
    "\n",
    "print \"Total head loss =\",round(Th,1),\"m\"\n"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "### Example 9.9 Page no 328"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 12,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "Discharge through the pipe = 0.067 m**3/s\n"
     ]
    }
   ],
   "source": [
    "# discharge through the pipe\n",
    "\n",
    "from math import *\n",
    "\n",
    "# Given\n",
    "\n",
    "# for water\n",
    "\n",
    "nu = 1.007*10**-6                # viscosity in m**2/s\n",
    "\n",
    "d1 = 0.3                         # diameter of pipe 1 in m\n",
    "\n",
    "d2 = 0.15                        # diameter of pipe 2 in m\n",
    "\n",
    "d3 = 0.08                        # diameter of pipe 3 in m\n",
    "\n",
    "g = 9.81                         # acclelration due to gravity in m/s**2\n",
    "\n",
    "e = 0.025                        # for cast iron\n",
    "\n",
    "f1 = 0.019                       # foe e/d1\n",
    "\n",
    "f2 = 0.022                       # foe e/d2\n",
    "\n",
    "# Solution\n",
    "\n",
    "V3 = sqrt(2*g*100/((8.4*(f1)+268.85*(f2)+4.85)))\n",
    "\n",
    "V1 = (d3/d1)**2*V3\n",
    "\n",
    "V2 = (d3/d2)**2*V3\n",
    "\n",
    "# reynolds number for pipe BC\n",
    "\n",
    "R1 = V1*d1/nu\n",
    "\n",
    "R2 = V2*d2/nu\n",
    "\n",
    "Q = V3*pi*d3**2/4\n",
    "\n",
    "print \"Discharge through the pipe =\",round(Q,3),\"m**3/s\""
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "### Example 9.10 Page no 332"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 13,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "Replacement of the flow system = 94.3 m\n"
     ]
    }
   ],
   "source": [
    "# Replace the flow system\n",
    "\n",
    "from math import *\n",
    "\n",
    "# Given\n",
    "\n",
    "D = 0.2                      # diameter of pipe 1\n",
    "\n",
    "D1 = 0.15                    # diameter of pipe 2\n",
    "\n",
    "Q = 0.1                     # discharge in m**3/s\n",
    "\n",
    "nu = 1.007*10**-6           # viscosity in m**2/s\n",
    "\n",
    "e = 0.025                   # e for cast iron\n",
    "\n",
    "r = e/(100*D)                     \n",
    "\n",
    "# Solution\n",
    "\n",
    "V = Q/(pi*(0.2)**2/4)\n",
    "\n",
    "R = V*D/nu\n",
    "\n",
    "f = 0.021                 # from moodys law\n",
    "\n",
    "r2 = e/D1\n",
    "\n",
    "V1 = Q/(pi*D1**2/4)\n",
    "\n",
    "R1 = V*D1/nu\n",
    "\n",
    "f2 = 0.023                # from moodys law\n",
    "\n",
    "L2 = 28562*D1**5/f2\n",
    "\n",
    "print \"Replacement of the flow system =\",round(L2,2),\"m\""
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "### Example 9.11 Page no 335"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 14,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "Discharge through branch 1 = 1.35 m**3/s\n",
      "Discharge through branch 2 = 0.566 m**3/s\n",
      "Discharge through branch 3 = 0.082 m**3/s\n",
      "It is found that hf1=hf2=hf3 = 43.5 The distribution od discharge is correct\n"
     ]
    }
   ],
   "source": [
    "# Discharge through each branch\n",
    "\n",
    "from math import *\n",
    "\n",
    "# Given\n",
    "\n",
    "e = 0.025                # in cm\n",
    "\n",
    "nu = 1.007*10**-6        # viscosity in m**2/s\n",
    "\n",
    "Q1 = 0.5                 # discharge in m**3/s\n",
    "\n",
    "D1 = 50\n",
    "\n",
    "L1 = 500                 # length in m\n",
    "\n",
    "g = 9.81\n",
    "\n",
    "# Pipe 1\n",
    "\n",
    "r1 = e/D                 # r1 for pipe 1\n",
    "\n",
    "V1 = Q1/(pi*(0.5)**2/4)\n",
    "\n",
    "R = V*(0.5)/nu\n",
    "\n",
    "f1 = 0.018               # for the reynolds no\n",
    "\n",
    "hf1 = f*L1*V1**2/(2*g*D1)\n",
    "\n",
    "# pipe 2\n",
    "\n",
    "hf2 = hf1\n",
    "\n",
    "L2 =200                 # length in m\n",
    "\n",
    "D2 = 0.3                # diameter in m\n",
    "\n",
    "r2 = e/D2\n",
    "\n",
    "f2 =  0.02 \n",
    "\n",
    "V2 = 0.419/sqrt(f2) \n",
    "\n",
    "R2 = V2*D2/nu\n",
    "\n",
    "Q2 = V2*(pi*D2**2/4)\n",
    "\n",
    "#pipe 3\n",
    "\n",
    "hf3=hf1\n",
    "\n",
    "L3 = 300              # length of pipe 3 in m\n",
    "\n",
    "D3  =0.15             # diameter of pipe 3 in m\n",
    "\n",
    "r3 = e/D3         \n",
    "\n",
    "f = 0.022            # from moody's law\n",
    "\n",
    "V3 = 0.242/sqrt(f2)\n",
    "\n",
    "R3 = V3*D3/nu\n",
    "\n",
    "Q3 = V3*(pi*D3**2/4)\n",
    "\n",
    "Td = Q1+Q2+Q3\n",
    "\n",
    "q1 = Q1*(2.0/Td)\n",
    "\n",
    "q2 = Q2*(2.0/Td)\n",
    "\n",
    "q3 = Q3*(2.0/Td)\n",
    "\n",
    "print \"Discharge through branch 1 =\",round(q1,2),\"m**3/s\"\n",
    "\n",
    "print \"Discharge through branch 2 =\",round(q2,3),\"m**3/s\"\n",
    "\n",
    "print \"Discharge through branch 3 =\",round(q3,3),\"m**3/s\"\n",
    "\n",
    "# Actual head loss\n",
    "\n",
    "d = 0.5\n",
    "\n",
    "v1 = q1/(pi*(d)**2/4)\n",
    "\n",
    "R4 = v1*d/nu\n",
    "\n",
    "r4 = 0.0005                 # ratio of e/D\n",
    "\n",
    "f = 0.018\n",
    "\n",
    "Hf1 = f*L1*v1**2/(2*g*d)\n",
    "\n",
    "print \"It is found that hf1=hf2=hf3 =\",round(Hf1,1),\"The distribution od discharge is correct\""
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "### Example 9.14 Page no 349"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 15,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "Minimum depth = 17.6 ft\n"
     ]
    }
   ],
   "source": [
    "# Find minimum depth below the ridge\n",
    "\n",
    "from math import *\n",
    "\n",
    "# Given\n",
    "\n",
    "e = 0.00015             # from moody's chart\n",
    "\n",
    "D = 2                   # depth in ft\n",
    "\n",
    "r = e/D\n",
    "\n",
    "z1 = 100                # elevation in ft\n",
    "\n",
    "mu = 1.084*10**-5       # viscosity in Ns/ft**2\n",
    "\n",
    "p1 = 34                 # pressure head in ft\n",
    "\n",
    "p2 = 10                 # pressure head in ft\n",
    "\n",
    "g = 32.2                # acclelration due to gravity in ft/s**2\n",
    "\n",
    "L = 1000                # length in ft\n",
    "\n",
    "# Solution\n",
    "\n",
    "f = 0.011           # assume\n",
    "\n",
    "V = sqrt(100/(10000/(2*2*g)))/sqrt(f)\n",
    "\n",
    "R = V*D/mu\n",
    "\n",
    "V1 = 10.15\n",
    "\n",
    "f1 = 0.0125\n",
    "\n",
    "Q = V1*pi*D**2/4\n",
    "\n",
    "x = p1-p2-(V1**2/(2*g))-(f1*L*V1**2/(2*g*D))\n",
    "\n",
    "Dp = 30 - x\n",
    "\n",
    "print \"Minimum depth =\",round(Dp,2),\"ft\""
   ]
  },
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