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+{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:74db613da6801f860dadcdbf59265a8239e5864c8922257a769d0c0fa0f7c4e0"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Chapter 8 : Laminar Flow"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 3,
+ "metadata": {},
+ "source": [
+ "Example 8.1 Page no 286"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# Determine maximum velocity and shear stress\n",
+ "\n",
+ "# Given\n",
+ "\n",
+ "from math import *\n",
+ "\n",
+ "P1 = 200 # Pressure at inlet in kPa\n",
+ "\n",
+ "P2 = 260 # Pressure at outlet in kPa\n",
+ "\n",
+ "d = 0.004 # diameter in m\n",
+ "\n",
+ "L = 8 # length of pipe in meters\n",
+ "\n",
+ "z = 6 # height of the pipe from the ground\n",
+ "\n",
+ "g = 9.81 # acceleration due to gravity in m/s**2\n",
+ "\n",
+ "# properties of kerosene\n",
+ "\n",
+ "mu = 19.1*10**-4 # viscosity of kerosene at 20 deg C\n",
+ "\n",
+ "S = 0.81 # specific gravity of kerosene\n",
+ "\n",
+ "rho = 1000 # density in kg/m**3\n",
+ "\n",
+ "# Solution\n",
+ "\n",
+ "# calculating direction of flow\n",
+ "\n",
+ "p1 = (P1+g*z*S)*1000 # point 1\n",
+ "\n",
+ "p2 = (P2)*1000 # point 2\n",
+ "\n",
+ "# direction of flow is from point 1-2\n",
+ "\n",
+ "# shear stress\n",
+ "\n",
+ "Sp = -((p1-p2)/sqrt(L**2+z**2))\n",
+ "\n",
+ "r = d/2\n",
+ "\n",
+ "Tau_max = r*Sp/2\n",
+ "\n",
+ "print \"(a) Maximum shear stress =\",round(Tau_max,3),\"N/m**2\"\n",
+ "\n",
+ "# maximum velocity\n",
+ "\n",
+ "Vmax = r**2*Sp/(4*mu)\n",
+ "\n",
+ "print \"(b) Maximum velocity =\",round(Vmax,3),\"m/s\"\n",
+ "\n",
+ "# discharge\n",
+ "\n",
+ "Q = pi*r**4*Sp/(8*mu)\n",
+ "\n",
+ "print \"(c) Discharge = \",round(Q,7),\"m**3/s\"\n",
+ "\n",
+ "# calculate reynolds number\n",
+ "\n",
+ "V = Vmax/2\n",
+ "\n",
+ "R = rho*V*d*S/mu\n",
+ "\n",
+ "print \"Reynolds number =\",round(R,0),\"is less than 2000, the flow is laminar and the calculations are valid\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(a) Maximum shear stress = 1.232 N/m**2\n",
+ "(b) Maximum velocity = 0.645 m/s\n",
+ "(c) Discharge = 4.1e-06 m**3/s\n",
+ "Reynolds number = 547.0 is less than 2000, the flow is laminar and the calculations are valid\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 3,
+ "metadata": {},
+ "source": [
+ "Example no 8.2 Page no 289"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# Determine the head loss\n",
+ "\n",
+ "# Given\n",
+ "\n",
+ "d = 0.02 # diameter of the pipe in m\n",
+ "\n",
+ "l = 30 # length of the pipe in m\n",
+ "\n",
+ "v = 0.1 # velocity in m/s\n",
+ "\n",
+ "g = 9.81 # acceleration due to gravity in m/s**2\n",
+ "\n",
+ "# for water at 5 deg C\n",
+ "\n",
+ "nu = 1.54*10**-6 # kinematic viscosity of water in m**2/s\n",
+ "\n",
+ "# Solution\n",
+ "\n",
+ "R = v*d/nu\n",
+ "\n",
+ "print \"R = \",round(R,0),\"is lesss than 2000 , the flow is laminar\"\n",
+ "\n",
+ "f = 64/R # friction factor\n",
+ "\n",
+ "Hf = f*l*v**2/(2*g*d) # head loss due to friction\n",
+ "\n",
+ "H=Hf*100\n",
+ "\n",
+ "print \"Head loss = \",round(H,2),\"cm of water\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "R = 1299.0 is lesss than 2000 , the flow is laminar\n",
+ "Head loss = 3.77 cm of water\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 3,
+ "metadata": {},
+ "source": [
+ "Example 8.3 Page no 290"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# Horsepower required to pump 50 tons of oil\n",
+ "\n",
+ "# Given\n",
+ "\n",
+ "from math import *\n",
+ "\n",
+ "# oil properties\n",
+ "\n",
+ "S = 0.92 # specific gravity\n",
+ "\n",
+ "gma = S*62.4 # density in lbs/ft**3\n",
+ "\n",
+ "nu=0.0205 # viscosity in ft**2/s\n",
+ "\n",
+ "W = 50 # weight of oil\n",
+ "\n",
+ "d = 9 # diameter of the pipe in inches\n",
+ "\n",
+ "g = 32.2 # acceleration due to gravity in ft/s**2\n",
+ "\n",
+ "# Solution\n",
+ "\n",
+ "Q = W*2000/(gma*3600) # discharge in ft**3/s\n",
+ "\n",
+ "A = pi*d**2/(4*144) # area of pipe\n",
+ "\n",
+ "V = Q*1000/(A) # velocity in ft/s\n",
+ "\n",
+ "R = V*0.75/(nu*1000) # Reynolds number\n",
+ "\n",
+ "print \"R =\",round(R,2),\"is less than 2000 and hence flow is laminar\"\n",
+ "\n",
+ "f = 64/R # friction factor\n",
+ "\n",
+ "Hf = (f*5280*(V/1000)**2)/(2*g*0.75)\n",
+ "\n",
+ "Hp = gma*Q*Hf/(550)\n",
+ "\n",
+ "print \"Horse power required to pump the oil = \",round(Hp,1)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "R = 40.07 is less than 2000 and hence flow is laminar\n",
+ "Horse power required to pump the oil = 10.6\n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ },
+ {
+ "cell_type": "heading",
+ "level": 3,
+ "metadata": {},
+ "source": [
+ "Example 8.4 Page no 291"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# Viscosity of the liquid in poise\n",
+ "\n",
+ "# Given\n",
+ "\n",
+ "from math import *\n",
+ "\n",
+ "V = 50 # Volume in m**3\n",
+ "\n",
+ "d = 5 # diameter in m\n",
+ "\n",
+ "d1 = 0.1 # diameter of bore\n",
+ "\n",
+ "l = 10 # length of the tube\n",
+ "\n",
+ "t = 20*60 # time in seconds\n",
+ "\n",
+ "rho = 0.88 # density in g/cm**3\n",
+ "\n",
+ "H1 = 5 # height from the base in m\n",
+ "\n",
+ "A = pi*d**2/4\n",
+ "\n",
+ "a = pi*d1**2/4\n",
+ "\n",
+ "# Solution\n",
+ "\n",
+ "# From derivation we obtain a equation for T\n",
+ "\n",
+ "H2 = H1-(V/A)\n",
+ "\n",
+ "mu = t*rho*a*(0.1)*98.1/(32*A*10*log(H1/H2))\n",
+ "\n",
+ "print \"Viscosity of the liquid =\",round(mu,4),\"poise\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Viscosity of the liquid = 0.0182 poise\n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "heading",
+ "level": 3,
+ "metadata": {},
+ "source": [
+ "Example 8.5 Page no 297"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# Velocity distribution; Discharge ; shear on the upper plate\n",
+ "\n",
+ "# Given\n",
+ "\n",
+ "from math import *\n",
+ "\n",
+ "# properties of kerosene oil at 20 deg C\n",
+ "\n",
+ "S = 0.81 # specific gravity of oil\n",
+ "\n",
+ "mu = 4*10**-5 # viscosity of oil in lb.s/ft**2\n",
+ "\n",
+ "gma = 62.4*S # density in lbs/ft**3\n",
+ "\n",
+ "p1 = 6.51 # pressure at point 1 in psia\n",
+ "\n",
+ "p2 = 8 # pressure at point 2 in psia\n",
+ "\n",
+ "h = 0.006 # distance between the plate in ft\n",
+ "\n",
+ "l = 4 # length of the plate in ft\n",
+ "\n",
+ "theta = pi/6 # angle of inclination\n",
+ "\n",
+ "# Solution\n",
+ "\n",
+ "# point 1\n",
+ "\n",
+ "P1 = p1*144 + gma*l*sin(theta)\n",
+ "\n",
+ "# point 2\n",
+ "\n",
+ "P2 = p2*144\n",
+ "\n",
+ "# flow is taking from poont 2-1\n",
+ "\n",
+ "Sp = (P2-P1)/4\n",
+ "\n",
+ "# equation for u = 2154.75*y-359125*y**2\n",
+ "\n",
+ "y = h\n",
+ "\n",
+ "# discharge per ft width\n",
+ "\n",
+ "q = (2154.75*y**2/2) - (359125*y**3/3)\n",
+ "\n",
+ "print \"Discharge q = \",round(q,3),\"per unit ft of the plate\"\n",
+ "\n",
+ "# to find shear at the top of the plate take du/dy = 0\n",
+ "\n",
+ "dV = 2154.75 - 718250*h\n",
+ "\n",
+ "# shear stress\n",
+ "\n",
+ "T = -mu*dV\n",
+ "\n",
+ "print \"Shear stress on the plate = \",round(T,3),\"lbs/ft**2 and resisting the motion of the plate\"\n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Discharge q = 0.013 per unit ft of the plate\n",
+ "Shear stress on the plate = 0.086 lbs/ft**2 and resisting the motion of the plate\n"
+ ]
+ }
+ ],
+ "prompt_number": 5
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [],
+ "language": "python",
+ "metadata": {},
+ "outputs": []
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+} \ No newline at end of file
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