path: root/Transport_Phenomena:_A_Unified_Approach/ch12.ipynb

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   "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",
      "\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",
      "\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",
      "\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",
      "\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",
      "\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",
      "\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",
      "# 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": {}
  }
 ]
}