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  {
   "cells": [
    {
     "cell_type": "markdown",
     "metadata": {},
     "source": [
      "# Chapter 1: Introductory Concepts"
     ]
    },
    {
     "cell_type": "markdown",
     "metadata": {},
     "source": [
      "## Example 1.1, Page number 23"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "# Express absolute viscosity of fluids in SI units and calculate the kinematic viscosity\n",
      "\n",
      "# Import required modules\n",
      "\n",
      "import numpy as np\n",
      "from prettytable import PrettyTable\n",
      "\n",
      "# Given\n",
      "\n",
      "mu = np.array([1,0.018,100]) # Absolute viscosities in centipoise\n",
      "mu_poise = np.array([1,0.018,100])/100 # Absolute viscosities in poise\n",
      "rho = np.array([1.0,0.0012,0.930]) # Densities in gm/cm^3\n",
      "mu_SI = mu/1000  # Absolute viscosities in SI units\n",
      "rho_SI = rho*1000 # Densities in SI units\n",
      "nu = mu_poise/rho # Kinematic viscosities in Stokes\n",
      "nu_SI = mu_SI/rho_SI # Kinematic viscosities in SI units\n",
      "\n",
      "# Tabulate results\n",
      "\n",
      "table = PrettyTable([\"Property\", \"Water\", \"Air\", \"Lube Oil\"])\n",
      "table.add_row(['Absolute Viscosity mu','  ','  ','  '])\n",
      "table.add_row([\"centipoise cP\",mu[0],mu[1],mu[2]])\n",
      "table.add_row([\"SI units (Ns/m^2)\",mu_SI[0],mu_SI[1],mu_SI[2]])\n",
      "table.add_row([' ',' ',' ',' '])\n",
      "table.add_row(['Mass Density rho','  ','  ','  '])\n",
      "table.add_row([\"g/cm^3\",rho[0],rho[1],rho[2]])\n",
      "table.add_row([\"SI units (kg/m^3)\",rho_SI[0],rho_SI[1],rho_SI[2]])\n",
      "table.add_row([' ',' ',' ',' '])\n",
      "table.add_row(['Kinematic Viscosity nu','  ','  ','  '])\n",
      "table.add_row([\"St\",nu[0],nu[1],round(nu[2],2)])\n",
      "table.add_row([\"SI units (m^2/s)\",nu_SI[0],nu_SI[1],\"{:.2e}\".format(nu_SI[2])])\n",
      "print table"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "+------------------------+--------+---------+----------+\n",
        "|        Property        | Water  |   Air   | Lube Oil |\n",
        "+------------------------+--------+---------+----------+\n",
        "| Absolute Viscosity mu  |        |         |          |\n",
        "|     centipoise cP      |  1.0   |  0.018  |  100.0   |\n",
        "|   SI units (Ns/m^2)    | 0.001  | 1.8e-05 |   0.1    |\n",
        "|                        |        |         |          |\n",
        "|    Mass Density rho    |        |         |          |\n",
        "|         g/cm^3         |  1.0   |  0.0012 |   0.93   |\n",
        "|   SI units (kg/m^3)    | 1000.0 |   1.2   |  930.0   |\n",
        "|                        |        |         |          |\n",
        "| Kinematic Viscosity nu |        |         |          |\n",
        "|           St           |  0.01  |   0.15  |   1.08   |\n",
        "|    SI units (m^2/s)    | 1e-06  | 1.5e-05 | 1.08e-04 |\n",
        "+------------------------+--------+---------+----------+\n"
       ]
      }
     ],
     "prompt_number": 21
    },
    {
     "cell_type": "markdown",
     "metadata": {},
     "source": [
      "## Example 1.2, Page number 23"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "# Calculate the force and power required to maintain the velocity\n",
      "\n",
      "# Given\n",
      "A = 0.1 # Area of the flat plate (m^2)\n",
      "U = 0.3 # Velocity of the flat plate (m/s)\n",
      "mu = 0.001 # viscosity of the fluid separating the plates (m)\n",
      "du = U - 0 # relative velocity between the plates\n",
      "dy = 0.0001 # relative distance between the plates\n",
      "\n",
      "tau = mu*du/dy # Shear stress (N/m^2)\n",
      "F = tau * A # Shear force (N)\n",
      "Power = F * U # Power required (W)\n",
      "print(\"The force required to maintain the velocity is %.2f N.\" %F)\n",
      "print(\"The power required is %.2f W.\" %Power)"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The force required to maintain the velocity is 0.30 N.\n",
        "The power required is 0.09 W.\n"
       ]
      }
     ],
     "prompt_number": 23
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 1.3, Page number 24"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "# Determine the torque and power\n",
      "\n",
      "# Import required modules\n",
      "\n",
      "import math \n",
      "import sympy\n",
      "\n",
      "# Given\n",
      "l = 0.10 # length of the shaft (m)\n",
      "d = 0.05 # diameter of the shaft (m)\n",
      "D = 0.051 # diameter of the concentric bearing (m)\n",
      "N = 500 # Rotational speed of the shaft (rpm)\n",
      "mu = 0.1 # Viscosity of the lubricating oil (Ns/m^2)\n",
      "theta = sympy.Symbol('theta')\n",
      "\n",
      "u = round(math.pi*d*N/60,2) # Peripheral speed of the shaft (m/s)\n",
      "du = u - 0 \n",
      "dy = (D-d)/2\n",
      "tau = round(mu*du/dy,0) # Shear stress (N/m^2)\n",
      "T = sympy.integrate(tau*d/2*d/2*l,(theta,0,2*math.pi)) # Torque required to turn the shaft (Nm)\n",
      "omega = u/(d/2) # Angular speed of the shaft\n",
      "Power = round(T,3)*omega # Power required to turn the shaft (W)\n",
      "\n",
      "print(\"The power required to turn the shaft is %1.3f W.\" %Power)\n",
      "# Wrong rounding-off of Torque T in textbook. Hence, the difference in value of power. Textbook answer Power = 5.387 W"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The power required to turn the shaft is 5.397 W.\n"
       ]
      }
     ],
     "prompt_number": 8
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 1.4, Page number 25"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "# The power dissipated in the bearing\n",
      "\n",
      "# Import required modules\n",
      "import sympy\n",
      "\n",
      "# Given\n",
      "R = 0.1/2 # Radius of the bearing (m)\n",
      "mu = 0.08 # Viscosity of oil film (Ns/m^2)\n",
      "dy = 0.0015 # separation distance (m)\n",
      "N = 100 # Rotational speed of the bearing (rpm)\n",
      "r = sympy.Symbol('r')\n",
      "theta = sympy.Symbol('theta')\n",
      "\n",
      "omega = round(2*math.pi*100/60,2) # Angular velocity of the bearing (rad/s)\n",
      "u = r*omega # Linear velocity of the bearing (m/s)\n",
      "du = u - 0 # Relative velocity \n",
      "tau = mu * du/dy # Shear stress (N/m^2)\n",
      "T = sympy.integrate(tau*r*r,(theta,0,2*math.pi),(r,0,R)) # Total torque on the shaft (Nm)\n",
      "Power = round(T,5)*omega # Power dissipated (W)\n",
      "print(\"The power dissipated by the bearing is %.4f W.\" %Power)"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The power dissipated by the bearing is 0.0574 W.\n"
       ]
      }
     ],
     "prompt_number": 14
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 1.5, Page number 26"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "# Determine shear stress for r/R ratios and calculate drag force per meter length of the pipe\n",
      "\n",
      "# Import required modules\n",
      "import sympy\n",
      "import math\n",
      "from tabulate import tabulate\n",
      "\n",
      "# Given\n",
      "r = sympy.Symbol('r') # Radial distance for the point\n",
      "R = sympy.Symbol('R') # Radial distance for the wall\n",
      "U = 10 # Centreline velocity (m/s)\n",
      "mu = 0.002 # Viscosity (Ns/m^2)\n",
      "r_R = [0.0,0.2,0.5,0.8,1.0] # r/R ratios\n",
      "u = U*(1-(r/R)**2) # Expression for velocity in a pipe-flow\n",
      "y = R-r # Distance from the wall\n",
      "\n",
      "du = sympy.diff(u,r) # Derivative of 'u' expression\n",
      "dy = sympy.diff(y,r) \n",
      "tau = mu*du/dy # Newton's law of viscosity (N/m^2)\n",
      "F = 2*math.pi*R*tau # Drag force (N)\n",
      "\n",
      "# Substitution of r/R ratios\n",
      "table = []\n",
      "for i, r_R in enumerate(r_R):      \n",
      "    table.append([r_R,round(tau.subs([(R,1.0/2.0),(r,r_R*1.0/2.0)]),4),\n",
      "          round(F.subs([(R,1.0/2.0),(r,r_R*1.0/2.0)]),4)])\n",
      "print tabulate(table, headers=['r/R', 'Shear stress, tau (N/m^2)', 'Drag force, F (N)'],tablefmt='grid',numalign=\"center\")\n",
      "# The Drag force printed in the textbook for the r/R = 0.8 is wrong"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "+-------+-----------------------------+---------------------+\n",
        "|  r/R  |  Shear stress, tau (N/m^2)  |  Drag force, F (N)  |\n",
        "+=======+=============================+=====================+\n",
        "|   0   |              0              |          0          |\n",
        "+-------+-----------------------------+---------------------+\n",
        "|  0.2  |            0.016            |       0.0503        |\n",
        "+-------+-----------------------------+---------------------+\n",
        "|  0.5  |            0.04             |       0.1257        |\n",
        "+-------+-----------------------------+---------------------+\n",
        "|  0.8  |            0.064            |       0.2011        |\n",
        "+-------+-----------------------------+---------------------+\n",
        "|   1   |            0.08             |       0.2513        |\n",
        "+-------+-----------------------------+---------------------+\n"
       ]
      }
     ],
     "prompt_number": 5
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 1.6, Page number 27"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "# Determine average thickness of the film\n",
      "\n",
      "# Import required modules\n",
      "import sympy\n",
      "\n",
      "# Given\n",
      "n = 800 # Normal reaction by the ice on the skater (N)\n",
      "f = 0.02 # Coefficient of friction between the skates and the ice \n",
      "u = 54*1000/3600 # Speed of the skater (m/s)\n",
      "A = 10e-4 # Skating area (m^2)\n",
      "mu = 0.001 # Viscosity of water (Ns/m^2)\n",
      "h = sympy.Symbol('h') # average thickness of the film\n",
      "\n",
      "F = f*n # Frictional reaction (N)\n",
      "du_dy = (u-0)/h # Velocity gradient\n",
      "tau = mu*du_dy # Shear stress (N/m^2)\n",
      "print('The average thickness of the film is %.3e m.'\n",
      "      %sympy.solve(sympy.Eq(tau*A,F),h)[0]) # Solve for h by equating drag force to frictional reaction"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The average thickness of the film is 9.375e-07 m.\n"
       ]
      }
     ],
     "prompt_number": 23
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 1.7, Page number 30"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "# Determine necessary increase of pressure\n",
      "\n",
      "# Given\n",
      "K = 2.07e6 # Bulk modulus of water (kN/m^2)\n",
      "gamma = 1.4 # Specific heat ratio\n",
      "p = 101.324 # Atmospheric pressure (kN/m^2)\n",
      "vol_red = 0.01 # Volume reduction \n",
      "\n",
      "# (a) At same temperature\n",
      "dp = vol_red * K # increase in pressure (kN/m^2)\n",
      "print('The increase in pressure required for water is %d kN/m^2.' %dp)\n",
      "# (b) isentropic compression of air\n",
      "K = gamma * p # Bulk modulus of air (kN/m^2)\n",
      "dp = vol_red * K # increase in pressure (kN/m^2)\n",
      "print('The increase in pressure required for air is %.2f kN/m^2.' %dp)"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The increase in pressure required for water is 20700 kN/m^2.\n",
        "The increase in pressure required for air is 1.42 kN/m^2.\n"
       ]
      }
     ],
     "prompt_number": 25
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 1.7, Page number 34"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "# Height of capillary rise\n",
      "\n",
      "# Import required modules\n",
      "import math\n",
      "\n",
      "# Given\n",
      "sigma = 0.0736 # Surface tension between water and glass (N/m)\n",
      "theta = 0 # Angle of contact\n",
      "d = 2e-3 # Diameter of the glass tube (m)\n",
      "g = 9.81 # Acceleration due to gravity (m/s^2)\n",
      "rho = 1000 # Density of water (kg/m^3)\n",
      "\n",
      "h = 4*sigma*math.cos(theta)/(rho*g*d) # height of capillary rise (m)\n",
      "print('The water in the glass tube rises through a height of %0.3f m'%h)"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The water in the glass tube rises through a height of 0.015 m\n"
       ]
      }
     ],
     "prompt_number": 28
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 1.8, Page number 34"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "# Gauge pressure and absolute pressure within a droplet and a jet\n",
      "\n",
      "# Given\n",
      "d_droplet = 0.004 # Diamter of the droplet (m)\n",
      "d_jet = 0.004 # Diameter of the jet (m)\n",
      "sigma = 0.073 # Viscosity of water (Ns/m^2)\n",
      "P_atm = 101300 # Atmospheric pressure (N/m^2)\n",
      "\n",
      "# (a) For the droplet\n",
      "P_gauge = 4*sigma/d_droplet # Gauge pressure for droplet (N/m^2)\n",
      "P_abs = P_atm + P_gauge # Absolute pressure (N/m^2)\n",
      "print('The gauge pressure and absolute pressure within a droplet is %d N/m^2 and %.3f kN/m^2 respectively.' %(P_gauge,P_abs/1000))\n",
      "\n",
      "# (a) For the jet\n",
      "P_gauge = 2*sigma/d_jet # Gauge pressure for jet (N/m^2)\n",
      "P_abs = P_atm + P_gauge # Absolute pressure (N/m^2)\n",
      "print('The gauge pressure and absolute pressure within a jet is %.1f N/m^2 and %.2f kN/m^2 respectively.' %(P_gauge,P_abs/1000))"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The gauge pressure and absolute pressure within a droplet is 73 N/m^2 and 101.373 kN/m^2 respectively.\n",
        "The gauge pressure and absolute pressure within a jet is 36.5 N/m^2 and 101.34 kN/m^2 respectively.\n"
       ]
      }
     ],
     "prompt_number": 38
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 1.9, Page number 34"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "# Difference in level of the miniscii\n",
      "\n",
      "# Import required modules\n",
      "import sympy\n",
      "\n",
      "# Given\n",
      "d_1 = 1.0e-3 # Diameter of capillary (m)\n",
      "d_2 = 1.5e-3 # Diameter of another capillary (m)\n",
      "sigma = 0.0075 # Surface tension of water (Ns/m^2)\n",
      "g = 9.81 # Acceleration due to gravity (m/s^2)\n",
      "rho = 1000 # Density of water (kg/m^3)\n",
      "h = sympy.Symbol('h') # Difference in level of the miniscii (m)\n",
      "\n",
      "h = sympy.solve(sympy.Eq(math.pi*d_2*sigma-math.pi*d_1*sigma,math.pi*d_2**2*h*rho*g/4),h)[0]*1000 # Solve for h\n",
      "print('The difference in level of the miniscii is %.2f mm' %h)"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The difference in level of the miniscii is 0.68 mm\n"
       ]
      }
     ],
     "prompt_number": 51
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 1.10, Page number 38"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "# Shear stress calculation and estimate the viscosity\n",
      "\n",
      "# Import required modules\n",
      "import sympy\n",
      "\n",
      "# Given\n",
      "U_max = 0.2 # Maximum velocity (m/s)\n",
      "h = 0.01 # film thickness (m)\n",
      "mu = 0.5 # Viscosity of the non-Newtonian fluid (Ns/m^2)\n",
      "y = sympy.Symbol('y') \n",
      "u = sympy.Symbol('u') \n",
      "u = U_max * (2*(y/h)-(y/h)**3/3) # Expression for velocity\n",
      "\n",
      "# (a) Shear stress calculation\n",
      "du_dy = sympy.diff(u,y) # Velocity gradient\n",
      "tau = mu*(round(du_dy.subs(y,h)))**1.3 # Shear stress of the non-Newtonian fluid (N/m^2)\n",
      "print('The shear stress at the solid surface is %.2f N/m^2.' %tau)\n",
      "\n",
      "# (b) Estimation of the viscosity of the Newtonian fluid\n",
      "mu = sympy.Symbol('mu')\n",
      "mu = sympy.solve(sympy.Eq(round(tau,2),mu*round(du_dy.subs(y,h))))[0] # Solve for mu for the same shear stress using Newton's law of viscosity\n",
      "print('The viscosity of a Newtonian fluid to induce the same shear stress is %.2f Ns/m^2.' %mu)"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The shear stress at the solid surface is 24.56 N/m^2.\n",
        "The viscosity of a Newtonian fluid to induce the same shear stress is 1.23 Ns/m^2.\n"
       ]
      }
     ],
     "prompt_number": 69
    }
   ],
   "metadata": {}
  }
 ]
}