{ "metadata": { "name": "" }, "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", "Estimate the boundary layer thickness and the local drag coefficient at the trailing \n", "edge of the test section. Also estimate the force (drag) on the test section.\n", "'''\n", "\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", "calculate the shear stress\n", "at the wall (in units of N m-*) at a distance of 1 m from the leading edge of the plate.\n", "'''\n", "\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", "Find (a) the length of the laminar boundary layer, (b) the thickness of the\n", "momentum boundary layer and the thermal boundary layer at the transition\n", "point, (c) the local heat transfer coefficient at the transition point, and (d) the\n", "mean heat transfer coefficient between the leading edge of the plate and the\n", "transition point.\n", "'''\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", "Find the terminal velocity and drag force when a spherical water\n", "drop, 5 pm in diameter, falls through air at 20\u00b0C. Let g = 9.80 m s-\n", "'''\n", "\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", "Calculate the time for the drop of water in Example 12.10 to\n", "accelerate from an initial velocity of zero to 0.9!XJ,.\n", "'''\n", "\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": [ "# Determine its terminal velocity.\n", "\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", "Determine the appropriate diameter of the sphere to achieve\n", "the design criteria, if g = 32 ft s-*.\n", "'''\n", "\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", "Compute (a) the static void fraction, (b) the minimum void fraction and\n", "bed height for fluidization, (c) the settling velocity for a single particle in a static\n", "fluid, (d) the minimum pressure drop for fluidization, (e) the minimum velocity\n", "for entrainment, and (f) the heat transfer coefficient at the wall for a superficial\n", "velocity 2.5 times U,,.\n", "'''\n", "\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", "Calculate the\n", "pressure drop in psi and the superficial velocity (ft s-l) at the point of incipient\n", "'''\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": [ "# estimate the velocity in m s-i.\n", "\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": [ "# Calculate the exit temperature of the air if its properties can be estimated from the following correlations\n", "\n", "# Variables\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": {} } ] }