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diff --git a/Transport_Phenomena/ch12.ipynb b/Transport_Phenomena/ch12.ipynb deleted file mode 100755 index b890667e..00000000 --- a/Transport_Phenomena/ch12.ipynb +++ /dev/null @@ -1,674 +0,0 @@ -{ - "metadata": { - "name": "", - "signature": "sha256:cf1ae634b67dbf9ac3cc91134d4c30acd13efd02af03447d121a71d1c5277a05" - }, - "nbformat": 3, - "nbformat_minor": 0, - "worksheets": [ - { - "cells": [ - { - "cell_type": "heading", - "level": 1, - "metadata": {}, - "source": [ - "Chapter 12 : Transport past immersed bodies" - ] - }, - { - "cell_type": "heading", - "level": 3, - "metadata": {}, - "source": [ - "Example 12.2 - Page No :562\n" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "\n", - "import math \n", - "\n", - "# Variables\n", - "p = 1.2047*0.06243; \t\t\t #[lb/ft**3]\n", - "mu = (18.17*10**-6)*(0.6720); \t #[lb/ft*sec]\n", - "v = mu/p;\n", - "x = 2.; \t\t\t #[ft]\n", - "U = 6.; \t\t\t #[ft/sec]\n", - "\n", - "# Calculation and Results\n", - "Nre = (x*U)/v;\n", - "print \"The Reynolds number is well within the laminar region %.3e Nre\"%Nre\n", - "del_ = 5*x*(Nre)**(-1./2);\n", - "C1 = 0.33206;\n", - "Cd = 2.*C1*(Nre)**(-1./2);\n", - "L2 = 2.; \t\t\t #[ft]\n", - "L1 = 1.; \t\t\t #[ft]\n", - "b = 1.;\n", - "F = ((2*(C1)*U*b))*((mu*p*U)**(1./2))*(((L2)**(1./2))-((L1)**(1./2)));\n", - "gc = 32.174;\n", - "F = F/gc;\n", - "print \" The value of F properly expressed in force units is F = %.3e lbf\"%(F);\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "The Reynolds number is well within the laminar region 7.391e+04 Nre\n", - " The value of F properly expressed in force units is F = 1.204e-04 lbf\n" - ] - } - ], - "prompt_number": 1 - }, - { - "cell_type": "heading", - "level": 3, - "metadata": {}, - "source": [ - "Example 12.3 - Page No :569\n" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "\n", - "import math \n", - "\n", - "# Variables\n", - "U = 3.; \t\t\t #[m/sec]\n", - "x1 = 1.; \t\t\t #[m]\n", - "x2 = 2.; \t\t\t #[m]\n", - "\n", - "# Calculations\n", - "p = 1./(1.001*10**-3); \t\t\t #[kg/m**3];\n", - "mu = 1.*10**-3; \t\t\t #[kg/m*sec]\n", - "Nre1 = (x1*U*p)/(mu);\n", - "Nre2 = (x2*p*U)/(mu);\n", - "tauw = (1./2)*(p*(U**2))*((2*math.log10(Nre1)-0.65)**(-2.3));\n", - "B = 1700.;\n", - "Cd = (0.455*(math.log10(Nre2))**-2.58)-(B/(Nre2));\n", - "Lb = 2.0;\n", - "F = (1./2)*(p*(U**2))*(Lb)*(Cd);\n", - "\n", - "Xc = round((5*10**5 * mu)/(U*p),3)\n", - "CDlaminar = round(4*.33206*(5*10**5)**(-1./2),5)\n", - "Flaminar= round(1./2*(p*U**2)*Xc*CDlaminar,3)\n", - "Cd = round(.455*((math.log10(Nre2))**-2.58),6)\n", - "Fturbulent1 = round(1./2*(p*U**2)*x2*Cd,2)\n", - "Fturbulent2 = round(1./2*(p*U**2)*Xc*.005106,3)\n", - "Factual = 1.411 + Fturbulent1 - Fturbulent2\n", - "\n", - "\n", - "# Results\n", - "print \" the drag on the plate is F = %f kg*m/sec**2 = %.1f N\"%(F,F);\n", - "print ' total drag on the plate Factual = %.2f N'%Factual\n", - "print \" the shear stress is %.f N/m^2\"%tauw\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - " the drag on the plate is F = 26.801111 kg*m/sec**2 = 26.8 N\n", - " total drag on the plate Factual = 26.93 N\n", - " the shear stress is 14 N/m^2\n" - ] - } - ], - "prompt_number": 2 - }, - { - "cell_type": "heading", - "level": 3, - "metadata": {}, - "source": [ - "Example 12.5 - Page No :576\n" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "\n", - "\n", - "# Variables \n", - "T = 290.; \t\t\t #[K] - temperature of flowing water\n", - "U = 3.; \t\t\t #[m/sec] - free stream velocity\n", - "Tfs = 285.; \t\t\t #[K] - temperature of free stream\n", - "vr = 10.**-3; \t\t\t #[m**3/kg] - volume per unit mass\n", - "p = 1./vr; \t\t\t #[kg/m**3] - density of water at Tfs\n", - "mu = 1225.*10**-6; \t #[N*sec/m**2]\n", - "k = 0.590; \t\t\t #[W/m*K]\n", - "Npr = 8.70;\n", - "\n", - "# Calculation and Results\n", - "# (a) The length of laminar boundary\n", - "Nre = 5.*10**5;\n", - "xc = (Nre)*(mu/(p*U));\n", - "print \" a) The length of laminar boundary is xc = %.4f m\"%(xc);\n", - "# (b) Thickness of the momentum boundary layer and thermal boundary layer\n", - "del_ = 5*xc*((Nre)**(-1./2));\n", - "del_h = del_*((Npr)**(-1./3));\n", - "print \" b) The thickness of momentum boundary layer is del_ = %.3e m \\n The \\\n", - " thickness of the hydryodynamic layer is del_h = %.3e m\"%(del_,del_h);\n", - "\n", - "# (c) Local heat transfer coefficient\n", - "x = 0.2042; \t\t\t #[ft]\n", - "hx = ((0.33206*k)/(x))*((Nre)**(1./2))*((Npr)**(1./3));\n", - "print \" c) The local heat transfer coefficient is h = %.0f W/m**2*K \\\n", - " = %.0f Btu/hr*ft**2*degF\"%(hx,hx*0.17611);\n", - "\n", - "# (d) Mean heat transfer coefficient\n", - "hm = 2*hx;\n", - "print \" d) The mean heat transfer coefficient is h = %.0f W/m**2*K \\\n", - " = %.0f Btu/hr*ft**2*degF\"%(hm,round(hm*0.17611,1));\n", - "\n", - "# Answer may vary because of rounding error.\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - " a) The length of laminar boundary is xc = 0.2042 m\n", - " b) The thickness of momentum boundary layer is del_ = 1.444e-03 m \n", - " The thickness of the hydryodynamic layer is del_h = 7.019e-04 m\n", - " c) The local heat transfer coefficient is h = 1395 W/m**2*K = 246 Btu/hr*ft**2*degF\n", - " d) The mean heat transfer coefficient is h = 2791 W/m**2*K = 492 Btu/hr*ft**2*degF\n" - ] - } - ], - "prompt_number": 8 - }, - { - "cell_type": "heading", - "level": 3, - "metadata": {}, - "source": [ - "Example 12.10 - Page No :590\n" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "\n", - "# Variables\n", - "T = 293.15; \t\t\t #[K]\n", - "pp = 999.; \t\t\t #[kg/m**3] - density of water\n", - "mu = 0.01817*10**-3; \t #[kg/m*sec] - viscosity of air\n", - "p = 1.205; \t\t\t #[kg/m**3] - density of air\n", - "d = 5*10**-6; \t\t\t #[m] - particle diameter\n", - "g = 9.80; \t\t\t #[m/sec**2]\n", - "\n", - "# Calculations\n", - "rp = d/2;\n", - "Ut = ((2*g*(rp**2))*(pp-p))/(9*mu);\n", - "Nre = (d*Ut*p)/(mu);\n", - "Fp = 6*math.pi*mu*rp*Ut;\n", - "\n", - "# Results\n", - "print \" The drag force is Fp = %.2e N\"%(Fp);" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - " The drag force is Fp = 6.40e-13 N\n" - ] - } - ], - "prompt_number": 9 - }, - { - "cell_type": "heading", - "level": 3, - "metadata": {}, - "source": [ - "Example 12.11 - Page No :591\n" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "\n", - "# Variables\n", - "T = 293.15; \t\t\t #[K]\n", - "pp = 999.; \t\t\t #[kg/m**3] - density of water\n", - "mu = 0.01817*10**-3; \t #[kg/m*sec] - viscosity of air\n", - "p = 1.205; \t\t\t #[kg/m**3] - density of air\n", - "d = 5*10**-6; \t\t\t #[m] - particle diameter\n", - "g = 9.80; \t\t\t #[m/sec**2]\n", - "\n", - "# Calculations\n", - "rp = d/2;\n", - "Ut = ((2*g*(rp**2))*(pp-p))/(9*mu);\n", - "Nre = (d*Ut*p)/(mu);\n", - "t = ((-2*(rp**2)*pp))/(9*mu)*(math.log(1-0.99));\n", - "\n", - "# Results\n", - "print \" Time for the drop of water in previous Example from an initial \\\n", - " velocity of zero to 0.99*Ut is \\n t = %.3e sec\"%(t);\n", - "print \" In other words, the drop accelerates almost instantaneously to its terminal velocity\"\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - " Time for the drop of water in previous Example from an initial velocity of zero to 0.99*Ut is \n", - " t = 3.517e-04 sec\n", - " In other words, the drop accelerates almost instantaneously to its terminal velocity\n" - ] - } - ], - "prompt_number": 11 - }, - { - "cell_type": "heading", - "level": 3, - "metadata": {}, - "source": [ - "Example 12.12 - Page No : 594\n" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "\n", - "# Variables\n", - "pp = 1.13*10**4; \t\t\t #[kg/m**3] - density of lead particle\n", - "p = 1.22; \t\t\t #[kg/m**3] - density of air\n", - "g = 9.80; \t\t\t #[m/sec**2] - acceleration due to gravity\n", - "d = 2*10**-3; \t\t\t #[m] - diameter of particle\n", - "mu = 1.81*10**-5; \t\t\t #[kg/m*sec] - viscosity of air\n", - "\n", - "# Calculations\n", - "# let us assume\n", - "Cd = 0.44;\n", - "Ut = ((4*d*g*(pp-p))/(3*p*Cd))**(1./2);\n", - "Nre = (Ut*d*p)/(mu);\n", - "\n", - "# from fig 12,16 value of Cd is\n", - "Cd = 0.4;\n", - "Ut = ((4*d*g*(pp-p))/(3*p*Cd))**(1./2);\n", - "Nre = (Ut*d*p)/(mu);\n", - "\n", - "# Results\n", - "# Within the readibility of the chart Cd is unchanged and therefore the above obtained Cd is the final answer\n", - "\n", - "print \" The terminal velocity is Ut = %.2f m/sec\"%(Ut);\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - " The terminal velocity is Ut = 24.60 m/sec\n" - ] - } - ], - "prompt_number": 23 - }, - { - "cell_type": "heading", - "level": 3, - "metadata": {}, - "source": [ - "Example 12.13 - Page No :595\n" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "\n", - "# Variables\n", - "distance = 1./12; \t #[ft]\n", - "time = 60.; \t\t #[sec]\n", - "Ut = distance/time;\n", - "mu = 1.68; \t\t #[lb/ft*sec] - viscosity \n", - "pp = 58.; \t\t\t #[lb/ft**3] - density of sphere\n", - "p = 50.; \t\t\t #[lb/ft**3] - density of polymer solution\n", - "g = 32.; \t\t\t #[ft/sec] - acceleration due to gravity\n", - "\n", - "# Calculations\n", - "rp = ((9*mu)*(Ut)*((2*g)**(-1))*((pp-p)**(-1)))**(1./2);\n", - "Nre = (rp*2*Ut*p)/(mu);\n", - "\n", - "# Results\n", - "print \" The required particle diameter would be about %.2f inch\"%(rp*2*12);\n", - "print \"Nre = %.2e\"%Nre\n", - "print \" This reynolds number is well within the stokes law region ; thus the design is reasonable\"\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - " The required particle diameter would be about 0.15 inch\n", - "Nre = 5.29e-04\n", - " This reynolds number is well within the stokes law region ; thus the design is reasonable\n" - ] - } - ], - "prompt_number": 14 - }, - { - "cell_type": "heading", - "level": 3, - "metadata": {}, - "source": [ - "Example 12.14 - Page No :616\n" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "\n", - "# Variables\n", - "T = 842.; \t\t\t #[degF] - temperature\n", - "P = 14.6; \t\t\t #[psia] - pressure\n", - "p = 0.487; \t\t #[kg/m**3] - density of air\n", - "mu = 3.431*10**-5; \t #[kg/m*sec] - viscosity of air\n", - "k = 0.05379; \t\t\t #[W/m*K] - thermal conductivity\n", - "Npr = 0.7025; \t\t\t #prandtl no.\n", - "\n", - "# Calculation and Results\n", - "# (a) static void_ fraction\n", - "mcoal = 15.*2000; \t #[lb] - mass of coal\n", - "pcoal = 94.; \t\t #[lbm/ft**3] - density of coal\n", - "d = 10.; \t\t\t #[ft]\n", - "L = 7.; \t\t\t #[ft]\n", - "area = ((math.pi*(d**2))/4);\n", - "Vcoal = mcoal/pcoal;\n", - "Vtotal = area*L;\n", - "e = (Vtotal-Vcoal)/(Vtotal);\n", - "print \"(a) The void_ fraction is E = %.2f\"%e\n", - "\n", - "# (b) minimum void_ fraction and bed height\n", - "d = 200.; \t\t\t #[um] - particle diameter\n", - "Emf = 1-0.356*((math.log10(d))-1);\n", - "\n", - "# this value seems to be a lottle low and therefore 0.58 will be used\n", - "Emf = 0.58;\n", - "Lmf = ((L)*(1-e))/(1-Emf);\n", - "print \" b) The bed height is Lmf = %.3f ft\"%(Lmf);\n", - "\n", - "# (c) Minimum fluid_ization velocity\n", - "P1 = 20.; \t\t\t #[psia]\n", - "P2 = 14.696; \t\t\t #[psia]\n", - "p1 = (p*P1)/(P2);\n", - "\n", - "# the archimid_es no. is\n", - "g = 9.78; \t\t\t #[m/sec**2]\n", - "Nar = p1*g*((d*10**-6)**3)*(1506-p1)*((1./(mu)**2));\n", - "C1 = 27.2;\n", - "C2 = 0.0408;\n", - "Nremf = (((C1**2)+C2*Nar)**(1./2))-C1;\n", - "Umf = (Nremf*mu)/((d*10**-6)*p1);\n", - "print \" c) The minimum fluid_ization velocity is Umf = %.4f %% m/sec\"%(Umf);\n", - "\n", - "# (d) Minimum pressure\n", - "del_tapmf = (1506-p1)*(g)*(1-Emf)*((Lmf*12*2.54)/(100))+p1*g*Lmf;\n", - "print \" d) The minimum pressure drop for fluid_ization is -del_tapmf = %.3e Pa\"%(del_tapmf);\n", - "\n", - "# (e) Particle settling velocity\n", - "Cd = 0.44;\n", - "Ut = (((8*((d*10**-6)/2)*g)*(1506-p1))/(3*p1*Cd))**(1./2);\n", - "Nrep = (Ut*d*10**-6*p1)/(mu);\n", - "print \"Nrep = %.2f\"%Nrep\n", - "Ut = ((5.923/18.5)*(((d*10**-6)*p1)/(mu))**(0.6))**(1./(2-0.6))\n", - "print \" e) The particle settling velocity is Ut = %.5f m/sec\"%(Ut);\n", - "\n", - "# (f) Bed to wall heat transfer coefficient\n", - "Nrefb = (d*10**-6)*2.5*Umf*p1*(1./mu);\n", - "Nnufb = 0.6*Npr*((Nrefb)**(0.3));\n", - "hw = Nnufb*(k/(d*10**-6));\n", - "print \" f) The bed to wall heat transfer coefficient is hw = %.1f W/m**2*K\"%(hw);\n", - "\n", - "# Answer may vary because of rounding error." - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "(a) The void_ fraction is E = 0.42\n", - " b) The bed height is Lmf = 9.675 ft\n", - " c) The minimum fluid_ization velocity is Umf = 0.0129 % m/sec\n", - " d) The minimum pressure drop for fluid_ization is -del_tapmf = 1.830e+04 Pa\n", - "Nrep = 14.18\n", - " e) The particle settling velocity is Ut = 0.79114 m/sec\n", - " f) The bed to wall heat transfer coefficient is hw = 60.6 W/m**2*K\n" - ] - } - ], - "prompt_number": 22 - }, - { - "cell_type": "heading", - "level": 3, - "metadata": {}, - "source": [ - "Example 12.15 - Page No :618\n" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "\n", - "\n", - "# Variables\n", - "pp = 249.6; \t\t #[lb/ft**3] - density of catalyst\n", - "p = 58.; \t\t\t #[lb/ft**3] - density of liquid\n", - "g = 32.174; \t\t #[ft/sec**2]\n", - "gc = 32.174;\n", - "Lmf = 5.; \t\t\t #[ft] - height of bed\n", - "mu = 6.72*10**-3; \t #[lbm/ft*sec] - viscosity of liquid\n", - "dp = 0.0157/12; \t #[ft] - diameter of particle\n", - "emf = 0.45;\n", - "\n", - "# Calculations\n", - "del_tapmf = (pp-p)*(g/gc)*(1-emf)*(Lmf);\n", - "Nar = (p*g*dp**3)*(pp-p)*(1./(mu)**2);\n", - "C1 = 27.2;\n", - "C2 = 0.0408;\n", - "Nremf = (((C1**2)+C2*Nar)**(1./2))-C1;\n", - "Umf = Nremf*(mu/(dp*p));\n", - "\n", - "# Results\n", - "print \" Minimum fluidization velocity is Umf = %.2e ft/sec\"%(Umf);\n", - "\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - " Minimum fluidization velocity is Umf = 1.18e-03 ft/sec\n" - ] - } - ], - "prompt_number": 23 - }, - { - "cell_type": "heading", - "level": 3, - "metadata": {}, - "source": [ - "Example 12.16 - Page No :624\n" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "\n", - "# Variables\n", - "d = 24.*10**-6; \t\t\t #[m] - diameter of wire\n", - "T = 415.; \t\t\t #[K] - operating temperature of hot wire anemometer\n", - "P = 0.1; \t\t\t #[W] - power consumption\n", - "L = 250.*d;\n", - "Tair = 385.; \t\t\t #[K] - temperature of air in duct\n", - "A = math.pi*d*L;\n", - "Tfilm = (T+Tair)/2.;\n", - "\n", - "# properties of air at Tfilm\n", - "p = 0.8825; \t\t\t #[kg/m**3]\n", - "mu = 2.294*10**-5; \t\t\t #[kg/m*s]\n", - "cpf = 1013.; \t\t\t #[J*kg/K]\n", - "kf = 0.03305; \t\t\t #[W/m*K]\n", - "Npr = 0.703;\n", - "\n", - "# Calculations\n", - "h = P/(A*(T-Tair));\n", - "Nnu = (h*d)/kf;\n", - "def func(x):\n", - " return Nnu-0.3-((0.62*(x**(1./2))*(Npr**(1./3)))/((1+((0.4/Npr)**(2./3)))**(1./4)))*((1+((x/(2.82*(10**5)))**(5./8)))**(4./5));\n", - "\n", - "# on solving the above function for x by umath.sing some root solver technique like Newton raphson method , we get\n", - "x = 107.7;\n", - "\t\t\t # or\n", - "Nre = 107.7;\n", - "y = func(x);\n", - "Um = (Nre*mu)/(d*p);\n", - "\n", - "# Results\n", - "print \" The velocity is Um = %.1f m/sec = %d ft/sec\"%(Um,Um*3.28);\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - " The velocity is Um = 116.6 m/sec = 382 ft/sec\n" - ] - } - ], - "prompt_number": 25 - }, - { - "cell_type": "heading", - "level": 3, - "metadata": {}, - "source": [ - "Example 12.17 - Page No :630\n" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "\n", - "dt = 0.75;\n", - "St = 1.5*dt;\n", - "Sl = 3.*dt;\n", - "Lw = 1.; \t\t\t #[m]\n", - "N = 12.;\n", - "Stotalarea = N*(St/12.)*Lw;\n", - "Sminarea = N*((St-dt)/12.)*Lw*0.3048;\n", - "\n", - "# properties of air at 293.15 K\n", - "p = 1.204; \t\t\t #[kg/m**3]\n", - "mu = 1.818*10**-5; \t #[kg/m*s]\n", - "cp = 1005.; \t\t\t #[J*kg/K];\n", - "k = 0.02560; \t\t\t #[J/s*m*K]\n", - "Npr = (cp*mu)/k;\n", - "U_inf = 7.; \t\t\t #[m/sec]\n", - "\n", - "# Calculations\n", - "Umax = U_inf*(St/(St-dt));\n", - "w = p*Umax*Sminarea;\n", - "C_tubes = 0.05983; \t\t\t #[m**2/m] - circumference of the tubes\n", - "N_tubes = 96.;\n", - "Atubes = N_tubes*C_tubes*Lw;\n", - "Tw = 328.15; \t\t\t #[K]\n", - "Tinf = 293.15; \t\t\t #[K]\n", - "Tin = 293.15; \t\t\t #[K]\n", - "Tout = 293.15; \t\t #[K]\n", - "u = 100.;\n", - "while u>10**-1:\n", - " T = (Tin+Tout)/2\n", - " Told = Tout;\n", - " p = -(0.208*(10**-3))+(353.044/T);\n", - " mu = -(9.810*(10**-6))+(1.6347*(10**-6)*(T**(1./2)));\n", - " cp = 989.85+(0.05*T);\n", - " k = 0.003975+7.378*(10**-5)*T;\n", - " Npr = (cp*mu)/k;\n", - " dt = 0.75*0.0254;\n", - " Gmax = w/Sminarea;\n", - " Nre = (dt*Gmax)/mu;\n", - " h = 0.27*(k/dt)*(Npr**0.36)*(Nre**0.63);\n", - " h = h*0.98;\n", - " del_taT = (h*Atubes*(Tw-Tinf))/(w*cp);\n", - " Tout = Tin+del_taT;\n", - " u = abs(Tout-Told);\n", - "\n", - "T = (Tin+Tout)/2\n", - "p = -(0.208*(10**-3))+(353.044/T);\n", - "mu = -(9.810*(10**-6))+(1.6347*(10**-6)*(T**(1./2)));\n", - "dt = 0.75;\n", - "dv = (4*(St*Sl-(math.pi*(dt**2)*(1./4))))/(math.pi*dt)*(0.09010/3.547);\n", - "de = dv;\n", - "Nre = (dv*24.72)/mu;\n", - "dv = dv/(0.09010/3.547);\n", - "ftb = 1.92*(Nre**(-0.145));\n", - "Zt = Sl;\n", - "Ltb = 8*Sl;\n", - "del_tap = (ftb*(24.72**2))/(2*p*(dv/Ltb)*((St/dv)**0.4)*((St/Zt)**0.6));\n", - "\n", - "# Results\n", - "print \" del_tap = %.0f kg/m*s = %.0f N/m**2 = %f psia\"%(del_tap,del_tap,round(del_tap*0.1614/1113,5))\n", - "print \" Exit temperature : %.2f K\"%T\n", - "# answer may slightly vary because of rounding error." - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - " del_tap = 1113 kg/m*s = 1113 N/m**2 = 0.161350 psia\n", - " Exit temperature : 299.87 K\n" - ] - } - ], - "prompt_number": 4 - } - ], - "metadata": {} - } - ] -}
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