diff options
Diffstat (limited to 'Fluid_Mechanics/Chapter9.ipynb')
| -rwxr-xr-x | Fluid_Mechanics/Chapter9.ipynb | 799 |
1 files changed, 799 insertions, 0 deletions
diff --git a/Fluid_Mechanics/Chapter9.ipynb b/Fluid_Mechanics/Chapter9.ipynb new file mode 100755 index 00000000..503ee70d --- /dev/null +++ b/Fluid_Mechanics/Chapter9.ipynb @@ -0,0 +1,799 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:81c0a87082bc2a398c522de3be6bcb735589adaa60c31ff7051c7c79b83652d0" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Chapter 9 : Turbulent flow in Pipes" + ] + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": [ + "Example 9.1 Page no 308" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "from math import *\n", + "\n", + "\n", + "S = 1.26 # specific gravity\n", + "\n", + "mu = 0.826 # kinematic viscosity in Ns/m**2\n", + "\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", + "\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", + "\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" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "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" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": [ + "Example 9.2 Page no 311" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "from math import *\n", + "\n", + "\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", + "\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" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "3.1853324563\n", + "Discharge = 0.1 m**3/s\n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": [ + "Example 9.3 Page no 314" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "from math import *\n", + "\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", + "\n", + "A = 8*L*Q**2/(hf*g*pi**2)\n", + "\n", + "B = 4*Q/(pi*nu)\n", + "\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", + "\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" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Hence the convergence is attained, D= 0.2 m\n" + ] + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": [ + "Example 9.4 Page no 318" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "from math import *\n", + "\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", + "\n", + "Hf = S*L\n", + "\n", + "print \"Head loss =\",round(Hf,0),\"ft\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Head loss = 2.0 ft\n" + ] + } + ], + "prompt_number": 4 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": [ + "Example 9.5 Page no 319" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "from math import *\n", + "\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", + "\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\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Head loss of water = 3.2 m\n" + ] + } + ], + "prompt_number": 6 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": [ + "Example 9.6 Page no 319" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "from math import *\n", + "\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", + "\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\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Head loss of water = 0.1 m**3/s\n" + ] + } + ], + "prompt_number": 8 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": [ + "Example 9.7 Page no 320" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "from math import *\n", + "\n", + "\n", + "eps = 0.025*10**-2 # for cast iron epsilon = 0.0025 cm\n", + "\n", + "\n", + "\n", + "D = 0.2 # value in m\n", + "\n", + "g = 9.81\n", + "\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\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "K = 0.432 from trial and error\n" + ] + } + ], + "prompt_number": 10 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": [ + "Example 9.8 Page no 326" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "from math import *\n", + "\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", + "\n", + "nu = 1.007*10**-6 # viscosity in m**2/s\n", + "\n", + "e = 0.025\n", + "\n", + "r = e/(10*d)\n", + "\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" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Total head loss = 83.7 m\n" + ] + } + ], + "prompt_number": 11 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": [ + "Example 9.9 Page no 328" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "from math import *\n", + "\n", + "\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", + "\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", + "\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\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Discharge through the pipe = 0.067 m**3/s\n" + ] + } + ], + "prompt_number": 12 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": [ + "Example 9.10 Page no 332" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "from math import *\n", + "\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", + "\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\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Replacement of the flow system = 94.3 m\n" + ] + } + ], + "prompt_number": 13 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": [ + "Example 9.11 Page no 335" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "from math import *\n", + "\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", + "\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", + "\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", + "\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", + "\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\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "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" + ] + } + ], + "prompt_number": 14 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": [ + "Example 9.14 Page no 349" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "from math import *\n", + "\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", + "\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\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Minimum depth = 17.6 ft\n" + ] + } + ], + "prompt_number": 15 + }, + { + "cell_type": "code", + "collapsed": false, + "input": [], + "language": "python", + "metadata": {}, + "outputs": [] + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file |