{
 "cells": [
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "# Chapter14 - Turbulent Flow in Pipe"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Example 14.1 page no 148"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 21,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "Example 14.1 page no 148\n",
      "\n",
      "\n",
      "\n",
      " Reynolds no R_e = 9769.23 \n"
     ]
    }
   ],
   "source": [
    "print \"Example 14.1 page no 148\\n\\n\" # a liquid flow through a tube\n",
    "meu=0.78e-2#viscosity of liquid,g/cm*s\n",
    "rho=1.50#density,g/cm**3\n",
    "D=2.54#diameter,cm\n",
    "v=20#flow velocity\n",
    "R_e=D*v*rho/meu#reynolds no\n",
    "print \"\\n Reynolds no R_e = %.2f \"%(R_e)#"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Example 14.2 page no 148"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 22,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "Example 14.2 page no 148\n",
      "\n",
      "\n",
      "\n",
      " velocity v = 0.28 ft/s\n"
     ]
    }
   ],
   "source": [
    "from __future__ import division\n",
    "print \"Example 14.2 page no 148\\n\\n\" # a fluid is moving through a cylinder in laminar flow\n",
    "meu=6.9216e-4#viscosity of fluid,lb/ft*s\n",
    "rho=62.4#density,lb/ft**3\n",
    "D=1/12#diameter,ft\n",
    "R_e=2100#reynolds no\n",
    "v=R_e*meu/(D*rho)#minimum velocity at which turbulance will appear\n",
    "print \"\\n velocity v = %.2f ft/s\"%(v)#"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Example 14.3 page no 152"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 23,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "Example 14.3 page no 152\n",
      "\n",
      "\n",
      "\n",
      " fanning friction factor f_a=0.01 \n",
      "\n",
      " friction factor f_b1=0.01 \n",
      "\n",
      " friction factor f_b2=0.01 \n",
      "\n",
      " friction factor f_c=0.01 \n",
      "\n",
      " friction factor f_d=0.01 \n",
      "\n",
      " friction factor f_e=0.01\n",
      "\n",
      " average friction f_av=0.01 \n"
     ]
    }
   ],
   "source": [
    "from math import log10\n",
    "print \"Example 14.3 page no 152\\n\\n\" # calculate the friction factor by using different equation's\n",
    "R_e=14080#reynolds no\n",
    "K_r=0.004#relative roughness (a) by PAT proposed equation\n",
    "f_a=0.0015+(8*(R_e)**0.30)**-1\n",
    "print \"\\n fanning friction factor f_a=%0.2f \"%(f_a)# equation for 5000<R_e>50000\n",
    "f_b1=0.0786/(R_e)**0.25 \n",
    "print \"\\n friction factor f_b1=%0.2f \"%(f_b1)#  equation for 30000<R_e>1000000\n",
    "f_b2=0.046/(R_e)**0.20\n",
    "print \"\\n friction factor f_b2=%0.2f \"%(f_b2)#  equation for the completely turbulent region \n",
    "f_c=1/(4*(1.14-2*log10(K_r))**2)\n",
    "print \"\\n friction factor f_c=%0.2f \"%(f_c)#  equation given by jain \n",
    "f_d=1/(2.28-4*log10(K_r+21.25/(R_e**.9)))**2\n",
    "print \"\\n friction factor f_d=%0.2f \"%(f_d)#\n",
    "f_e=0.0085 #from figur 14.2\n",
    "print \"\\n friction factor f_e=%0.2f\"%(f_e)#\n",
    "f_av=(f_a+f_b1+f_b2+f_c+f_d+f_e)/6\n",
    "print \"\\n average friction f_av=%0.2f \"%(f_av)#"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Example 14.4 page no 154"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 24,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "Example 14.4 page no 154\n",
      "\n",
      "\n",
      "\n",
      " equivalent diameter D_eq_a=3.33 in\n",
      "\n",
      " equivalent diameter D_eq_b=18.00 cm\n",
      "\n",
      " equivalent diameter D_eq_c=10.00 cm\n"
     ]
    }
   ],
   "source": [
    "from math import pi\n",
    "print \"Example 14.4 page no 154\\n\\n\" # for turbulent fluid flow in across section  (a) for a rectangle \n",
    "w=2#width of a rectangle,in\n",
    "h=10#height of rectangle,in\n",
    "S_a=h*w#cross sectional area\n",
    "P_a=2*h+2*w#perimeter of rectangle\n",
    "D_eq_a=4*S_a/P_a#equivalent diameter\n",
    "print \"\\n equivalent diameter D_eq_a=%0.2f in\"%(D_eq_a)# (b) for an annulus \n",
    "d_o=10#outer diameter of annulus\n",
    "d_i=8#inner diameter \n",
    "S_b=pi*(d_o**2-d_i**2)/4#cross sectional area\n",
    "P_b=pi*(d_o-d_i)#perimeter\n",
    "D_eq_b=(4*S_b)/(P_b)#eq. diameter\n",
    "print \"\\n equivalent diameter D_eq_b=%0.2f cm\"%(D_eq_b)# (c) for an half- full circle\n",
    "d_c=10#diameter of circle \n",
    "S_c=pi*d_c**2/8# cross sectional area\n",
    "P_c=pi*d_c/2#perimeter\n",
    "D_eq_c=4*S_c/P_c#eq. diameter\n",
    "print \"\\n equivalent diameter D_eq_c=%0.2f cm\"%(D_eq_c)# "
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Exampkle 14.5 page no 157"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 25,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "Example 14.5 page no 157\n",
      "\n",
      "\n",
      "\n",
      " pipe diameter D=0.29 ft\n",
      "\n",
      "D=0.69 \n",
      "\n",
      " flow velocity v=22.28 ft/s\n"
     ]
    }
   ],
   "source": [
    "print \"Example 14.5 page no 157\\n\\n\" # air is transported through a circular conduit \n",
    "MW=28.9#molecular weight of air \n",
    "R=10.73#gas constant\n",
    "T=500#temperature\n",
    "P=14.75#pressure,psia applying ideal gas law for density\n",
    "rho=P*MW/(R*T)#density \n",
    "rho=0.08#after round off\n",
    "meu=3.54e-7#viscosity of air at 40 degF assume flow is laminar\n",
    "q=8.33#flow rate ,ft**3/s\n",
    "L=800#length of pipe,ft\n",
    "P_1=.1#pressure at starting point\n",
    "P_2=.01#pressure at delivery point \n",
    "D=((128*meu*L*q)/(pi*(P_1-P_2)*144))**(1/4)#diameter\n",
    "print \"\\n pipe diameter D=%0.2f ft\"%(D)# check the flow type\n",
    "meu=1.14e-5\n",
    "R_e1=4*q*rho/(pi*D*meu)#reynolds no print \"\\n reynolds no R_e=%0.2f \"%(R_e)# from R_e we can conclude that laminar flow is not valid\n",
    "P_drop=12.96#pressure drop P_1-P2 in psf\n",
    "f=0.005#fanning friction factor\n",
    "g_c=32.174\n",
    "D=(32*rho*f*L*q**2/(g_c*pi**2*P_drop))**(0.2)#diamter from new assumption strat the second iteration with the newly calculated D\n",
    "k=0.00006/12#roughness factor\n",
    "K_r=k/D#relative roughness \n",
    "C_f=1.321224\n",
    "R_e_n=4*q*rho/(pi*D*meu)#new reynolds no print \"\\n new reynolds no R_e=%0.2f \"%(R_e)#\n",
    "f_n=0.0045#new fanning friction factor\n",
    "D=(((8*rho*f_n*L*q**2)/(g_c*pi**2*P_drop))**(0.2))*C_f#final calculated diameter because last diameter is same with this\n",
    "print \"\\nD=%0.2f \"%(D)# iteration may now be terminated\n",
    "S=pi*(D**2)/4#cross sectional area of pipe\n",
    "v=q/S#flow velocity\n",
    "print \"\\n flow velocity v=%0.2f ft/s\"%(v)##printing mistake in book in the value of meu in the formula of D is first time that's why this deviation in answer"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Example 14.6 page no 159"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 26,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "Example 14.6 page no. 159\n",
      "\n",
      "\n",
      "\n",
      " R_e=106208.60 \n",
      "\n",
      " since R_e is more than 4000 flow is turbulent\n"
     ]
    }
   ],
   "source": [
    "print \"Example 14.6 page no. 159\\n\\n\" # ethyl alcohol is pumped through a horizontal tube\n",
    "rho=789#density .kg/m**3\n",
    "meu=1.1e-3#viscosity ,kg/m-s\n",
    "k=1.5e-6#roughness,m\n",
    "L=60#length of tube,m\n",
    "q=2.778e-3#flow rate \n",
    "g=9.807\n",
    "h_f=30#friction loss\n",
    "A=(L*q**2)/(g*h_f)\n",
    "A=1.574e-7\n",
    "D=0.66*(((k**1.25)*(A**4.75)+meu*(A**5.2)/(q*rho))**.04)\n",
    "D=0.0377 # calculate velocity of alcohol in the tube\n",
    "S=3.14*(D)**2/4#surface area\n",
    "v=q/S#velocity\n",
    "v=3.93#velocity\n",
    "neu=1.395e-6#dynamic viscosity\n",
    "R_e=D*v/neu#reynolds no \n",
    "print \"\\n R_e=%0.2f \"%(R_e)##printing mistake in book\n",
    "print \"\\n since R_e is more than 4000 flow is turbulent\" #"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Exanmple 14.7 page no 160"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 27,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "Example 14.7 page no 160\n",
      "\n",
      "\n",
      "\n",
      " average velocity v=2.37 m/s\n",
      "\n",
      " S=0.00 \n",
      "\n",
      " flow rate q=1244.02 m**3/s\n",
      "\n",
      " mass flow rate m_dot=1020094.94 kg/s\n",
      "\n",
      " v_max=2.91 m/s\n",
      "\n",
      " length L_c=1.36 m\n"
     ]
    }
   ],
   "source": [
    "print \"Example 14.7 page no 160\\n\\n\" # kerosene flow ina lng ,smooth ,horizontal pipe\n",
    "rho=820#density,kg/m**3\n",
    "D=0.0493#iside diameter of pipe by appendix A.5,m\n",
    "R_e=60000\n",
    "meu=0.0016#viscosity,kg/m.s\n",
    "v=(R_e*meu)/(D*rho)# flow average velocity\n",
    "print \"\\n average velocity v=%0.2f m/s\"%(v)#\n",
    "S=(pi/4)*D**2#cross sectional area\n",
    "print \"\\n S=%0.2f \"%(S)#\n",
    "q=v/S#flow rate \n",
    "print \"\\n flow rate q=%0.2f m**3/s\"%(q)##printing mistake in book\n",
    "m_dot=rho*q#mass flow rate\n",
    "print \"\\n mass flow rate m_dot=%0.2f kg/s\"%(m_dot)##printing mistake in book in the value of v\n",
    "n=7#seventh power apply\n",
    "v_max=v/(2*n**2/((n+1)*(2*n+1)))#maximum velocity\n",
    "print \"\\n v_max=%0.2f m/s\"%(v_max)# check the assumptioon of fully developed flow\n",
    "R_e=60000#reynolds no\n",
    "L_c=4.4*R_e**(1/6)*D#critical length\n",
    "print \"\\n length L_c=%0.2f m\"%(L_c)# since L_c <L th eassumption is valid"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Example 14.8 page no 161"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 28,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "\n",
      " Example 14.8 page no 161\n",
      "\n",
      "\n",
      "\n",
      " fanning friction factor f=0.01 \n",
      "\n",
      " h_f friction loss=1.07 m \n",
      "\n",
      " P_drop_a =0.09 atm\n"
     ]
    }
   ],
   "source": [
    "print \"\\n Example 14.8 page no 161\\n\\n\" # refer to example no 14.7\n",
    "rho=860#density\n",
    "R_e=60000#reynolds no\n",
    "f=.046/R_e**.2#fanning friction factor\n",
    "print \"\\n fanning friction factor f=%0.2f \"%(f)#\n",
    "L=9#length of tube\n",
    "v=2.38#velocity\n",
    "D=.0493#diameter of tube\n",
    "g=9.807\n",
    "h_f=4*f*(L*v**2)/(D*2*g)#friction loss \n",
    "print \"\\n h_f friction loss=%0.2f m \"%(h_f)# applying  bernoulli equation\n",
    "P_drop=rho*g*h_f#pressure drop in pa\n",
    "P_drop_a=P_drop/10**5#pressure drop in atm\n",
    "print \"\\n P_drop_a =%0.2f atm\"%(P_drop_a)#"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Example 14.9 page no 161"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 29,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      " Example 14.9 page no 161\n",
      "\n",
      "\n",
      "\n",
      " Force required to hold pipe F=16.58 N\n"
     ]
    }
   ],
   "source": [
    "print \" Example 14.9 page no 161\\n\\n\" # refer to example 14.7\n",
    "D=0.0493#diameter of tuube\n",
    "S=pi*D**2/4#cross sectional area\\\n",
    "P=8685#pressure\n",
    "F=P*S#force required to hold the pipe,direction is opposite the flow\n",
    "print \"\\n Force required to hold pipe F=%0.2f N\"%(F)# "
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Example 14.10 page no 163"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 30,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "Example 14.10 page no 163\n",
      "\n",
      "\n",
      "\n",
      " vz_bar=40.00\n",
      "\n",
      " vz_sqr=4.60\n",
      "\n",
      " intensity of turbulance I=0.05 \n"
     ]
    }
   ],
   "source": [
    "from math import sqrt\n",
    "print \"Example 14.10 page no 163\\n\\n\" # a fluid is moving in the turbulent flw through a pipe   a hot wire anemometer is inserted to measure the local velocity at a given point P in the system  following readings were recorded at equal time interval instantaneous velocities at subsequent time interval\n",
    "vz=[43.4,42.1,42,40.8,38.5,37,37.5,38,39,41.7]\n",
    "vz_bar=0#\n",
    "n=10#\n",
    "i = 0#\n",
    "sums=0#\n",
    "for i in range(0,10):\n",
    "    sums=sums+vz[i]#\n",
    "\n",
    "vz_bar=sums/n#\n",
    "print \"\\n vz_bar=%0.2f\"%(vz_bar)#\n",
    "sigma=0#\n",
    "for i in range(0,10):\n",
    "    sigma=sigma+(vz[i]-vz_bar)**2#\n",
    "    vz_sqr=sigma/10#\n",
    "\n",
    "print \"\\n vz_sqr=%0.2f\"%(vz_sqr)\n",
    "I = sqrt(vz_sqr)/vz_bar#intensity of turbulance\n",
    "print \"\\n intensity of turbulance I=%0.2f \"%(I)#"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Example 14.11 page no 164"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 31,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "Example 14.11 page no 164\n",
      "\n",
      "\n",
      "\n",
      " flow rate q_a=0.33 ft**3/min\n",
      " \n",
      " flow rate q_b=0.65 ft**3/min\n",
      "\n",
      " flow rate q_c=0.53 ft**3/min\n"
     ]
    }
   ],
   "source": [
    "print \"Example 14.11 page no 164\\n\\n\" # a fluid is flowing through a pipe\n",
    "D=2#inside diameter of pipe,in\n",
    "v_max=30#maximum velocity,ft/min\n",
    "A=(pi/4)*(D/12)**2#cross sectional area (a) for laminar flow \n",
    "v_a=(1/2)*v_max#average velocity\n",
    "q_a=v_a*A#volumatric flow rate\n",
    "print \"\\n flow rate q_a=%0.2f ft**3/min\"%(q_a)# (b) for plug flow \n",
    "v_b=v_max#average velocity \n",
    "q_b=v_b*A#volumatric flow rate\n",
    "print \" \\n flow rate q_b=%0.2f ft**3/min\"%(q_b)# (c)for turbulent flow\n",
    "v_c=(49/60)*v_max#average velocity\n",
    "q_c=v_c*A#volumatric flow rate\n",
    "print \"\\n flow rate q_c=%0.2f ft**3/min\"%(q_c)#"
   ]
  }
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