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diff --git a/Transport_Phenomena/ch10.ipynb b/Transport_Phenomena/ch10.ipynb new file mode 100644 index 00000000..4df43e2b --- /dev/null +++ b/Transport_Phenomena/ch10.ipynb @@ -0,0 +1,1031 @@ +{ + "metadata": { + "name": "" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter 10 : Fluid flow in ducts" + ] + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": [ + "Example 10.1 - Page No :405\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Compute the viscosity in CP.\n", + "\n", + "import math \n", + "\n", + "# Variables\n", + "T = 30.; \t\t\t #[degC] - temperature\n", + "d = 8.265*10**-4; \t\t\t #[m] - diameter of the capillary viscometer\n", + "deltapbyL = -0.9364; \t\t #[psi/ft] - pressure drop per unit length\n", + "\n", + "# Calculations\n", + "deltapbyL = deltapbyL*(2.2631*10**4); \t\t\t #[kg/m**2*sec**2] - pressure drop per unit length\n", + "Q = 28.36*(10**-6)*(1./60);\n", + "p = (0.88412-(0.92248*10**-3)*T)*10**3; \t\t\t #[kg/m**3] - density\n", + "s = (math.pi*(d**2))/4.;\n", + "U = Q/s;\n", + "tauw = (d/4.)*(-deltapbyL);\n", + "shearrate = (8*U)/d;\n", + "mu = tauw/(shearrate);\n", + "\n", + "# Results\n", + "print \" The viscosity is mu = %.3ef kg/m*sec = %.4f cP\"%(mu,mu*10**3);\n", + "print \" Finally, it is important to check the reynolds number to make sure the above equation applies\"\n", + "Nre = (d*U*p)/(mu);\n", + "print \" Nre = %d\"%Nre\n", + "print \" The flow is well within the laminar region and therefore the above equation applies\";\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " The viscosity is mu = 5.135e-04f kg/m*sec = 0.5135 cP\n", + " Finally, it is important to check the reynolds number to make sure the above equation applies\n", + " Nre = 1214\n", + " The flow is well within the laminar region and therefore the above equation applies\n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": [ + "Example 10.2 - Page No :407\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Compute the pressure drop for the same 0.032%in \n", + "\n", + "# Variables\n", + "Nreold = 1214.;\n", + "Uold = 0.8810;\n", + "Nre = 13700.;\n", + "U = Uold*(Nre/Nreold);\n", + "Lbyd = 744.;\n", + "T = 30.; \n", + "\n", + "# Calculations\n", + "# umath.sing the newton raphson method to calculate the value of f from the equation - 1/(f**(1/2)) = 4*math.log(Nre*(f**(1/2)))-0.4\n", + "f = 0.007119;\n", + "p = (0.88412-(0.92248*10**-3)*T)*10**3; \t\t\t #[kg/m**3] - density\n", + "tauw = (1./2)*p*(U**2)*f;\n", + "deltap = tauw*(4.)*(Lbyd);\n", + "d = 0.03254/12; \t\t\t #[ft]\n", + "L = Lbyd*d;\n", + "\n", + "# Results\n", + "print \" Pressure drop is -deltap = %.3e N/m**2 = %.1f kpa = 130 psi\"%(deltap,deltap*10**-3); \n", + "print \" A pressure drop of 130 psi on a tube of length of %.3f ft is high and \\\n", + "\\nshows the impracticality of flows at high reynolds number in smaller tubes\"%(L);\n", + "\n", + "# Answer may vary because of rounding error." + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " Pressure drop is -deltap = 8.968e+05 N/m**2 = 896.8 kpa = 130 psi\n", + " A pressure drop of 130 psi on a tube of length of 2.017 ft is high and \n", + "shows the impracticality of flows at high reynolds number in smaller tubes\n" + ] + } + ], + "prompt_number": 10 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": [ + "Example 10.3 - Page No :414\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Calculate the pressure drop in a pipe of 6-cm inside diameter for water flowing\n", + "\n", + "# Variables\n", + "u = 1./60; \t\t\t #[m/sec] - velocity\n", + "p = 1000.; \t\t\t #[kg/m**3] - density\n", + "mu = 1*10.**-3; \t\t #[kg/m*sec] - vismath.cosity\n", + "d = 6*10.**-2; \t\t #[m] - insid_e diameter of tube\n", + "L = 300.; \t\t\t #[m] - length of the tube\n", + "\n", + "# Calculations\n", + "Nre = (d*u*p)/(mu);\n", + "f = 16./Nre;\n", + "deltap = (4.*f)*(L/d)*((p*(u**2))/2.);\n", + "\n", + "# Results\n", + "print \"Nre = \",Nre,\"therefore the flow is laminar\"\n", + "print \"f = \" , f\n", + "print \"Pressure drop -delta P = %.2f N/m**2 = %.4f kPa = %.3e psi\"%(deltap,deltap*10**-3,deltap*1.453*10**-4);\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Nre = 1000.0 therefore the flow is laminar\n", + "f = 0.016\n", + "Pressure drop -delta P = 44.44 N/m**2 = 0.0444 kPa = 6.458e-03 psi\n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": [ + "Example 10.4 - Page No :415\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "'''\n", + "Repeat the previous example for Reynolds numbers of 10 000 and\n", + "100 000. Contrast smooth pipe with commercial steel pipe and cast iron pipe.\n", + "'''\n", + "\n", + "from numpy import *\n", + "\n", + "# Variables\n", + "# given\n", + "d = 6.*10**-2; \t\t\t #[m] - insid_e diameter of tube\n", + "p = 1000.; \t\t\t #[kg/m**3] - density\n", + "# for smooth pipe\n", + "Nre = array([10**4, 10**5]);\n", + "f = array([0.0076, 0.0045]);\n", + "mu = 10.**-3; \t\t\t #[kg/m**2*s]\n", + "U = (Nre*mu)/(d*p);\n", + "L = 300.; \t\t\t #[m] - length of the tube\n", + "\n", + "# Calculations\n", + "deltap = zeros(2)\n", + "for i in range(2):\n", + " deltap[i] = (4*f[i])*(L/d)*((p*(U[i]**2))/2.);\n", + "\n", + "\n", + "# Results\n", + "print \"for smooth pipe\"\n", + "print \" Nre f -deltap\";\n", + "print \" %6.0f %6.4f %6.3f\"%(Nre[0],f[0],deltap[0])\n", + "print \" %6.0f %6.4f %6.3f\"%(Nre[1],f[1],deltap[1])\n", + "\n", + "# for commercial steel\n", + "Nre = array([10**4, 10**5]);\n", + "f = array([0.008 ,0.0053]);\n", + "U = (Nre*mu)/(d*p);\n", + "L = 300.; \t\t\t #[m] - length of the tube\n", + "for i in range(2):\n", + " deltap[i] = (4*f[i])*(L/d)*((p*(U[i]**2))/2);\n", + "\n", + "print \"\\nfor commercial steel pipe\"\n", + "print \" Nre f -deltap\";\n", + "print \" %6.0f %6.4f %6.3f\"%(Nre[0],f[0],deltap[0])\n", + "print \" %6.0f %6.4f %6.3f\"%(Nre[1],f[1],deltap[1])\n", + "\n", + "# for cast iron pipe\n", + "Nre = array([10**4 ,10**5]);\n", + "f = array([0.009 ,0.0073]);\n", + "U = (Nre*mu)/(d*p);\n", + "L = 300.; \t\t\t #[m] - length of the tube\n", + "for i in range(2):\n", + " deltap[i] = (4*f[i])*(L/d)*((p*(U[i]**2))/2);\n", + "\n", + "print \"\\nfor cast iron pipe\"\n", + "print \" Nre f -deltap\";\n", + "print \" %6.0f %6.4f %6.3f\"%(Nre[0],f[0],deltap[0])\n", + "print \" %6.0f %6.4f %6.3f\"%(Nre[1],f[1],deltap[1])" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "for smooth pipe\n", + " Nre f -deltap\n", + " 10000 0.0076 2111.111\n", + " 100000 0.0045 125000.000\n", + "\n", + "for commercial steel pipe\n", + " Nre f -deltap\n", + " 10000 0.0080 2222.222\n", + " 100000 0.0053 147222.222\n", + "\n", + "for cast iron pipe\n", + " Nre f -deltap\n", + " 10000 0.0090 2500.000\n", + " 100000 0.0073 202777.778\n" + ] + } + ], + "prompt_number": 24 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": [ + "Example 10.5 - Page No :417\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Find, the maximum flow velocity by trial and error.\n", + "\n", + "# Variables\n", + "L = 300.; \t\t\t #[m] - length of pipe\n", + "d = 0.06; \t\t\t #[m] - insid_e diameter\n", + "deltap = 147.*10**3; \t\t #[Pa] - pressure the pump can supply\n", + "ebyd = 0.000762; \t\t\t # relative roughness\n", + "p = 1000.; \t\t\t #[kg/m**3] - density\n", + "mu = 1.*10**-3; \t\t\t #[kg/m*sec] - viscosity\n", + "tauw = (d*(deltap))/(4.*L);\n", + "\n", + "# using the hit and trial method for estimation of flow velocity\n", + "# Calculations\n", + "# let \n", + "f = 0.005;\n", + "U = ((2*tauw)/(p*f))**(1./2);\n", + "Nre = (d*U*p)/mu;\n", + "\n", + "# from the graph value of f at the above calculated reynolds no. and the given relative roughness(e/d)\n", + "f = 0.0054;\n", + "U = ((2*tauw)/(p*f))**(1./2);\n", + "Nre = (d*U*p)/mu;\n", + "\t\t\t # from the graph value of f at the above calculated reynolds no. and the given relative roughness(e/d)\n", + "f = 0.0053;\n", + "U = ((2*tauw)/(p*f))**(1./2);\n", + "Nre = (d*U*p)/mu;\n", + "\n", + "# from the graph value of f at the above calculated reynolds no. and the given relative roughness(e/d)\n", + "f = 0.0053;\n", + "# At this point the value of f is deemed unchanged from the last iteration .Hence, the values obtained after the third iteration are the converged values\n", + "\n", + "# Results\n", + "print \" The maximum flow velocity is U = %f m/sec\"%(U);\n", + "\n", + "# Answer may vary because of rounding error." + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " The maximum flow velocity is U = 1.665408 m/sec\n" + ] + } + ], + "prompt_number": 26 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": [ + "Example 10.6 - Page No :419\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Repeat Example 10.5 using the von Karman plot, Fig. 10.5.\n", + "\n", + "# Variables\n", + "L = 300.; \t\t\t #[m] - length of pipe\n", + "d = 0.06; \t\t\t #[m] - insid_e diameter\n", + "deltap = 147.*10**3; \t #[Pa] - pressure the pump can supply\n", + "ebyd = 0.000762; \t\t # relative roughness\n", + "p = 1000.; \t\t\t #[kg/m**3] - density\n", + "\n", + "# Calculations\n", + "mu = 1*10**-3; \t\t\t #[kg/m*sec] - viscosity\n", + "Nvk = ((d*p)/mu)*((d*(deltap))/(2*L*p))**(1./2);\n", + "\n", + "# From the fig at given von karman no and relative roughness the value of f is-\n", + "f = 0.0055;\n", + "Nre = Nvk/(f**(1./2))\n", + "U = (Nre*mu)/(d*p);\n", + "\n", + "# Results\n", + "print \"von karman no. %.0f\"%Nvk\n", + "print \" Average velocity = %.2f m/sec\"%(U);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "von karman no. 7275\n", + " Average velocity = 1.63 m/sec\n" + ] + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": [ + "Example 10.7 - Page No :422\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# estimate the pressure drop in a 0.06-m ID commercial steel pipe at Reynolds numbers of 104 and 16\n", + "\n", + "# Variables\n", + "L = 300.; \t\t\t #[m] - length of pipe\n", + "d = 0.06; \t\t\t #[m] - insid_e diameter\n", + "p = 1000.; \t\t\t #[kg/m**3] - density\n", + "mu = 1.*10**-3; \t\t\t #[kg/m*sec] - vismath.cosity\n", + "\n", + "# Calculations\n", + "Nre = array([10.**4, 10.**5]);\n", + "U = (Nre*mu)/(d*p);\n", + "velocityhead = (U**2)/2.;\n", + "N = (L/d)/45.; \t\t\t # no of velocity heads\n", + "deltap = p*N*(velocityhead);\n", + "\n", + "# Results\n", + "for i in range(2):\n", + " print \"Nre = \",Nre[i]\n", + " print \" velocity head = %.5f m**2/sec**2\"%(velocityhead[i]);\n", + " print \" -deltap = %.3f kPa = %.3f psi\"%(deltap[i]*10**-3,deltap[i]*1.453*10**-4);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Nre = 10000.0\n", + " velocity head = 0.01389 m**2/sec**2\n", + " -deltap = 1.543 kPa = 0.224 psi\n", + "Nre = 100000.0\n", + " velocity head = 1.38889 m**2/sec**2\n", + " -deltap = 154.321 kPa = 22.423 psi\n" + ] + } + ], + "prompt_number": 32 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": [ + "Example 10.8 - Page No :439\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Compute p, by (a) the equivalent length method and (b) the loss coefficient method.\n", + "\n", + "# Variables\n", + "mu = 6.72*10**-4; \t\t #[lb/ft*sec] - vismath.cosity\n", + "p = 62.4; \t\t\t #[lb/ft**3] - density\n", + "S = 0.03322; \t\t\t #[ft**2] - flow area\n", + "d = 0.206; \t\t\t #[ft]\n", + "e = 1.5*10**-4; \t\t # absolute roughness for steel pipe\n", + "ebyd = e/d;\n", + "Nre = 10.**5;\n", + "\n", + "# friction factor as read from fig in book for the given reynolds no. and relative roughness is-\n", + "f = 0.0053;\n", + "U = (Nre*mu)/(p*d);\n", + "Q = U*S;\n", + "gc = 32.174;\n", + "\n", + "# Calculations\n", + "# (a) equivalent length method\n", + "deltapbyL = f*(4/d)*(p*(U**2))*(1/(2*gc))*(6.93*10**-3);\n", + "\n", + "# using L = Lpipe+Lfittings+Lloss;\n", + "Lfittings = 2342.1*d;\n", + "kc = 0.50; \t\t\t # due to contraction loss\n", + "ke = 1.; \t\t\t # due to enlargement loss\n", + "Lloss = (kc+ke)*(1./(4*f))*d;\n", + "Lpipe = 137.;\n", + "L = Lpipe+Lfittings+Lloss;\n", + "deltap = deltapbyL*L;\n", + "patm = 14.696; \t\t\t #[psi] - atmospheric pressure\n", + "p1 = patm+deltap;\n", + "print \" a)The inlet pressure is p1 = %.1f psi\"%(p1);\n", + "\n", + "# (b) loss coefficient method\n", + "# using the equation deltap/p = -(Fpipe+Ffittings+Floss)\n", + "L = 137.;\n", + "kfittings = 52.39;\n", + "sigmaF = ((4.*f*(L/d))+kc+ke+kfittings)*((U**2)/(2*gc));\n", + "deltap = (p*sigmaF)/(144.);\n", + "p1 = patm+deltap;\n", + "\n", + "# Results\n", + "print \" b)The inlet pressure is p1 = %.1f psi\"%(p1);\n", + "print \" Computation of the pressure drop by the loss coefficient method differs from the equivalent length \\\n", + " \\nmethod by less than 1 psi\";\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " a)The inlet pressure is p1 = 26.7 psi\n", + " b)The inlet pressure is p1 = 27.2 psi\n", + " Computation of the pressure drop by the loss coefficient method differs from the equivalent length \n", + "method by less than 1 psi\n" + ] + } + ], + "prompt_number": 34 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": [ + "Example 10.9 - Page No :443\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Determine the total flow rate of water through the pipe system.\n", + "\n", + "# Variables\n", + "L1 = 50.; \t\t\t #[m] - length of first pipe\n", + "L2 = 150.; \t\t\t #[m] - length of second pipe\n", + "L3 = 100.; \t\t\t #[m] - length of third pipe\n", + "d1 = 0.04; \t\t\t #[m] - diameter of first pipe\n", + "d2 = 0.06; \t\t\t #[m] - diameter of second pipe\n", + "d3 = 0.08; \t\t\t #[m] - diameter of third pipe\n", + "deltap = -1.47*10**5; \t\t\t #[kg/m*sec] - pressure drop\n", + "mu = 1*10.**-3; \t\t\t #[kg/m*sec] - vismath.cosity\n", + "p = 1000.; \t\t\t #[kg/m**3] - density\n", + "\n", + "# Calculation and Results\n", + "# for branch 1\n", + "S = (math.pi*(d1**2))/4;\n", + "Nvk = ((d1*p)/mu)*(-(d1*deltap)/(2*L1*p))**(1./2);\n", + "f = (1./(4*math.log10(Nvk)-0.4))**2;\n", + "U = (((-deltap)/p)*(d1/L1)*(2./4)*(1./f))**(1./2);\n", + "w1 = p*U*S;\n", + "print \" For first branch w1 = %.2f kg/sec\"%(w1);\n", + "\t\t\t # for branch 2\n", + "S = (math.pi*(d2**2))/4;\n", + "Nvk = ((d2*p)/mu)*(-(d2*deltap)/(2*L2*p))**(1./2);\n", + "f = (1./(4*math.log10(Nvk)-0.4))**2;\n", + "U = (((-deltap)/p)*(d2/L2)*(2./4)*(1./f))**(1./2);\n", + "w2 = p*U*S;\n", + "print \" For second branch w2 = %.2f kg/sec\"%(w2);\n", + "\t\t\t # for branch 3\n", + "S = (math.pi*(d3**2))/4;\n", + "Nvk = ((d3*p)/mu)*(-(d3*deltap)/(2*L3*p))**(1./2);\n", + "f = (1./(4*math.log10(Nvk)-0.4))**2;\n", + "U = (((-deltap)/p)*(d3/L3)*(2./4)*(1./f))**(1./2);\n", + "w3 = p*U*S;\n", + "print \" For third branch w3 = %.2f kg/sec\"%(w3);\n", + "\n", + "# total flow rate w = w1+w2+w3\n", + "w = w1+w2+w3;\n", + "print \" total flow rate is w = %.1f kg/sec\"%(w);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " For first branch w1 = 4.74 kg/sec\n", + " For second branch w2 = 7.59 kg/sec\n", + " For third branch w3 = 20.42 kg/sec\n", + " total flow rate is w = 32.7 kg/sec\n" + ] + } + ], + "prompt_number": 36 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": [ + "Example 10.10 Page no : 445" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# find flow rate\n", + "\n", + "import math\n", + "\n", + "# Note : all variables values are taken from 10.9\n", + "# variables \n", + "w1 = 4.74 #kg/sec\n", + "w2 = 7.59 #kg/sec\n", + "w3 = 20.42 #kg/sec\n", + "w = 32.7 #kg/sec\n", + "d = .04\n", + "\n", + "S1 = math.pi*d**2/4\n", + "S2 = .002827\n", + "S3 = .005027\n", + "deltaP = 1.47 * 10**5 #kpa\n", + "deltaP1 = 1.583* 10**6 #f1w**2\n", + "deltaP2 = 6.254* 10**5 #f1w**2\n", + "deltaP3 = 9.895* 10**4 #f1w**2\n", + "Nre1 = 3.183 * 10**4 #w\n", + "Nre2 = 2.122 * 10**4 #w\n", + "Nre3 = 1.592 * 10**4 #w\n", + "\n", + "\n", + "### age plz tu kar dena muje kuchh samaj nahi aa raha he...\n", + "### Thanks in advance..... nahi to ye delete kar dena... mere se nahi hota. plz.\n", + "print S1\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "0.00125663706144\n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": [ + "Example 10.11 - Page No : 447\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "'''\n", + "Determine the pump size necessary to maintain a spray\n", + "velocity of 15 ft s-\u2019 and a flow rate of 400 gpm if the pump is 60 percent efficient.\n", + "'''\n", + "\n", + "# Variables\n", + "sp = 1.1;\n", + "p = sp*62.4; \t\t\t #[lb/ft**3] - density\n", + "mu = 2*6.72*10**-4; \t #[lb/ft*sec] - viscosity\n", + "Q = 400.; \t\t\t #[gpm] - volumetric flow rate\n", + "e = 1.5*10**4; \t\t #roughness of steel pipe\n", + "gc = 32.174;\n", + "kexit = 1.;\n", + "kentrance = 0.5;\n", + "\n", + "# Calculations\n", + "# 4 in schedule pipe\n", + "d = 4.026/12; \t\t\t #[ft]\n", + "U4 = Q/39.6; \t\t\t #[ft/sec]\n", + "Lgv = 13.08;\n", + "Lglv = 114.1;\n", + "Le = 40.26;\n", + "Lpipe_4 = 22.;\n", + "Lfittings_4 = Lgv+Lglv+Le;\n", + "Lloss = 0;\n", + "L_4 = Lpipe_4+Lfittings_4+Lloss;\n", + "Nre_4 = (d*U4*p)/mu;\n", + "f = 0.00475;\n", + "Fpipe_4 = ((4*f*L_4)/d)*(U4**2)*(1/(2*gc));\n", + "Floss_4 = ((kentrance+0)*(U4**2))/(2*gc);\n", + "\n", + "# 5 in schedule pipe\n", + "d = 5.047/12;\n", + "U5 = Q/62.3;\n", + "Lgv = 10.94;\n", + "Le = 75.71;\n", + "Lpipe_5 = 100.;\n", + "Lfittings_5 = Lgv+Le;\n", + "Lloss = 0.;\n", + "L_5 = Lpipe_5+Lfittings_5+Lloss;\n", + "Nre = (d*U5*p)/mu;\n", + "f = 0.00470;\n", + "Fpipe_5 = ((4*f*L_5)/d)*(U5**2)*(1./(2*gc));\n", + "Floss_5 = ((kexit+0)*(U5**2))/(2*gc);\n", + "\n", + "# 6 in schedule pipe\n", + "d = 6.065/12;\n", + "U6 = Q/90.;\n", + "Lgv = 6.570;\n", + "Le = 30.36;\n", + "Lpipe_6 = 4.;\n", + "Lfittings_6 = Lgv+Le;\n", + "Lloss = 0.;\n", + "L_6 = Lpipe_6+Lfittings_6+Lloss;\n", + "Nre = (d*U6*p)/mu;\n", + "f = 0.00487;\n", + "Fpipe_6 = ((4*f*L_6)/d)*(U6**2)*(1./(2*gc));\n", + "kc = 0.50;\n", + "Floss_6 = kc*((U6**2)/(2*gc));\n", + "Ffittings = 0.;\n", + "deltap_6 = p*(Fpipe_6+Ffittings+Floss_6);\n", + "\n", + "# 3/4 in 18 gauge tube\n", + "d = 0.652112/12;\n", + "L_3by4 = 15.;\n", + "U_3by4 = (Q*0.962)/100.;\n", + "Floss_3by4 = 100.*(kexit+kentrance)*((U_3by4**2.)/2.);\n", + "Nre = d*U_3by4*p*(1./mu);\n", + "f = 0.08*((Nre)**(-1./4))+0.012*((d)**(1./2));\n", + "deltap_3by4 = ((4*f*p*L_3by4)/d)*((U_3by4**2)/(2*gc));\n", + "Fpipe_3by4 = 100.*((4.*f*L_3by4)/d)*((U_3by4**2.)/(2.*gc));\n", + "deltap_spraysystem = 25.; \t\t\t #[psi]\n", + "Fspraysystem = (deltap_spraysystem/p)*(144.);\n", + "delta_p = (p*(kexit+kentrance))*((U_3by4**2.)/(2.*gc))\n", + "Fpipe = Fpipe_4+Fpipe_5+Fpipe_6;\n", + "Floss = Floss_4+Floss_5+Floss_6+Floss_3by4;\n", + "ws = 0. + (((15.**2)-0)/(2*gc))+38.9+382.5;\n", + "w = (Q*p)/(7.48);\n", + "Ws = (ws*w)/(33000.);\n", + "efficiency = 0.6;\n", + "Ws_actual = Ws/efficiency\n", + "\n", + "# Results\n", + "print \" The power supplied to the pump is %.1f hp\"%(Ws_actual);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " The power supplied to the pump is 78.8 hp\n" + ] + } + ], + "prompt_number": 7 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": [ + "Example 10.12 - Page No :454\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Estimate the power required in the previous problem by using the velocity head approximation\n", + "\n", + "# Variables\n", + "kexit = 1.;\n", + "kentrance = 0.5;\n", + "Q = 400.; \t\t\t #[gpm] - volumetric flow rate\n", + "gc = 32.174;\n", + "\n", + "# for 4 inch pipe\n", + "d = 4.026; \t\t #[inch]\n", + "L = 22.; \t\t\t #[ft]\n", + "Lbyd = (L*12)/(d);\n", + "\n", + "# Calculation and Results\n", + "# adding the contributions due to fittings \n", + "Lbyd = Lbyd+3*13+340+4*30;\n", + "N = Lbyd/45.;\n", + "N = N+kentrance+0;\n", + "U4 = Q/39.6; \t\t\t #[ft/sec]\n", + "Fpipe_4 = (N*(U4**2))/(2*gc);\n", + "print \" F4 in.pipes = %.2f ft*lbf/lbm\"%(Fpipe_4);\n", + "\n", + "# for 5 inch pipe\n", + "L = 100.; \t\t\t #[ft]\n", + "d = 5.047; \t\t\t #[inch]\n", + "Lbyd = (L*12.)/(d);\n", + "\n", + "# valves contributes 26 diameters and six elbows contribute 30 diameters ecah;therefore\n", + "Lbyd = Lbyd+26+6*30;\n", + "N = Lbyd/45.; \t\t\t # no. of velocity heads\n", + "N = N+kexit+kentrance;\n", + "U5 = Q/62.3;\n", + "Fpipe_5 = (N*(U5**2))/(2*gc);\n", + "print \" F5 in.pipes = %.2f ft*lbf/lbm\"%(Fpipe_5);\n", + "\n", + "# for 6 inch pipe\n", + "d = 6.065; \t\t #[inch]\n", + "L = 5.; \t\t\t #[ft]\n", + "Lbyd = (L*12.)/(d);\n", + "\n", + "# adding the contributions due to fittings \n", + "Lbyd = Lbyd+1*13+2*30;\n", + "N = Lbyd/45;\n", + "N = N+0+kentrance;\n", + "U6 = Q/90.;\n", + "Fpipe_6 = (N*(U6**2))/(2*gc);\n", + "print \" F6 in.pipes = %.3f ft*lbf/lbm\"%(Fpipe_6);\n", + "F_largepipes = Fpipe_4+Fpipe_5+Fpipe_6;\n", + "print \" Flarge pipes = %.2f ft*lbf/lbm\"%(F_largepipes);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " F4 in.pipes = 20.69 ft*lbf/lbm\n", + " F5 in.pipes = 7.28 ft*lbf/lbm\n", + " F6 in.pipes = 0.719 ft*lbf/lbm\n", + " Flarge pipes = 28.68 ft*lbf/lbm\n" + ] + } + ], + "prompt_number": 39 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": [ + "Example 10.14 - Page No :459\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Determine the pressure drop over a length of 300 m if the flow velocity is 1.667 m\n", + "\n", + "\n", + "# Variables\n", + "l = 0.09238;\n", + "rh = 0.1624*l;\n", + "L = 300.;\n", + "de = 4.*rh;\n", + "p = 1000.; \t\t\t #[kg/m**3]\n", + "mu = 10.**-3; \t\t\t #[kg/m*sec]\n", + "Uavg = 1.667;\n", + "\n", + "# Calculations\n", + "Nre = (de*Uavg*p)/mu;\n", + "f = 0.0053;\n", + "deltap = ((4.*f*L)/de)*(p*(Uavg**2)*(1./2));\n", + "\n", + "# Results\n", + "print \" Pressure drop -deltap = %.3e kg/m*s = %.3e N/m**2 = %.1f kPa\"%(deltap,deltap,deltap*10**-3);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " Pressure drop -deltap = 1.473e+05 kg/m*s = 1.473e+05 N/m**2 = 147.3 kPa\n" + ] + } + ], + "prompt_number": 8 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": [ + "Example 10.15 - Page No :466\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "'''\n", + "Size a sharp-edged, corner-tap orifice meter for the 4-in. leg in\n", + "Fig. 10.17 if the flow rate and fluid are unchanged from Example 10.11.\n", + "'''\n", + "\n", + "# Variables\n", + "Q = 400.; \t\t\t #[gpm]\n", + "p = 1.1*62.4; \t \t\t #[lbm/ft**3]\n", + "mu = 2.*(6.72*10**-4); \t #[lb/ft*sec]\n", + "e = 1.5*10**4;\n", + "\n", + "# Calculations\n", + "# 4 inch schedule pipe\n", + "d = 0.3355;\n", + "S = (math.pi*(d**2))/4;\n", + "U4 = Q/39.6;\n", + "ebyd = e/d;\n", + "w = 3671./60;\n", + "pm = 13.45*62.4;\n", + "g = 32.1;\n", + "gc = 32.174;\n", + "deltaz = 2.5;\n", + "deltap = (g/gc)*(pm-p)*(deltaz);\n", + "betaa = ((1.)/(1.+((2*p*gc)*(deltap))*(((0.61*S)/w)**2)))**(1./4);\n", + "d2 = betaa*d;\n", + "Nre2 = (4*w)/(math.pi*d2*mu);\n", + "a = (1./30)*4.026;\n", + "b = (1./4)*(2.013-1.21);\n", + "c = (1./8)*(2.42);\n", + "if a<b :\n", + " if a<c :\n", + " opt = a;\n", + " else:\n", + " opt = c;\n", + "else:\n", + " if b<c:\n", + " opt = b;\n", + " else:\n", + " opt = c;\n", + "\n", + "# Results\n", + "print \" The pertinent orifice details are orifice diameter = %.3f in corner taps, \\\n", + " \\n square edge orifice plate not over %.3f in thick\"%(d2*12,opt);" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " The pertinent orifice details are orifice diameter = 2.425 in corner taps, \n", + " square edge orifice plate not over 0.134 in thick\n" + ] + } + ], + "prompt_number": 42 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": [ + "Example 10.16 - Page No :470\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Size a venturi meter for the application cited in Example 10.15\n", + "\n", + "# Variables\n", + "Q = 400.; \t \t\t #[gpm]\n", + "p = 1.1*62.4; \t\t\t #[lbm/ft**3]\n", + "mu = 2*(6.72*10**-4); \t\t\t #[lb/ft*sec]\n", + "e = 1.5*10**4;\n", + "\n", + "# Calculations\n", + "# 4 inch schedule pipe\n", + "d = 0.3355;\n", + "S = (math.pi*(d**2))/4;\n", + "U4 = Q/39.6;\n", + "ebyd = e/d;\n", + "w = 3671./60;\n", + "pm = 13.45*62.4;\n", + "g = 32.1;\n", + "gc = 32.174;\n", + "Nre = (d*U4*p)/mu;\n", + "if Nre>10**4:\n", + " c = 0.98;\n", + "\n", + "deltaz = 2.5;\n", + "deltap = (g/gc)*(pm-p)*(deltaz);\n", + "betaa = ((1.)/(1+((2*p*gc)*(deltap))*(((c*S)/w)**2)))**(1./4);\n", + "d2 = betaa*d;\n", + "\n", + "# Results\n", + "print \" The pertinentr details of the venturi design are Throat diameter = %.2f inch \\\n", + "\\n Approach angle = 25 Divergence angle = 7\"%(d2*12);\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " The pertinentr details of the venturi design are Throat diameter = 1.95 inch \n", + " Approach angle = 25 Divergence angle = 7\n" + ] + } + ], + "prompt_number": 44 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": [ + "Example 10.17 - Page No :477\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Then compute the average velocity by Eq. (10.88) and compare with the 2.778 ft s-l as given.\n", + "\n", + "# Variables\n", + "Uzmax = 3.455; \t\t\t #[ft/sec]\n", + "m = 32;\n", + "a1 = -0.3527;\n", + "a2 = -0.6473;\n", + "rbyro = 0.880;\n", + "\n", + "# Calculations\n", + "UzbyUzmax = 1+a1*(rbyro**2)+a2*(rbyro**(2*m));\n", + "Uz = Uzmax*(UzbyUzmax);\n", + "Uzavg = (4./9)*Uzmax+(5./18)*(Uz+Uz);\n", + "\n", + "# Results\n", + "print \" the average velocity is Uzavg = %.2f ft/sec \\\n", + "\\n Thus, in this Example there is an inherent error of 5.5 percent, even before any experimental errors are introduced\"%(Uzavg);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " the average velocity is Uzavg = 2.93 ft/sec \n", + " Thus, in this Example there is an inherent error of 5.5 percent, even before any experimental errors are introduced\n" + ] + } + ], + "prompt_number": 10 + } + ], + "metadata": {} + } + ] +}
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