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-{
- "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
generated by cgit (git 2.34.1) at 2024-12-20 10:39:33 +0000