path: root/Heat_Transfer_Principles_And_Applications_by_Dutta/ch4.ipynb

{
 "cells": [
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "# Chapter 4 : Forced Convection"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Example 4.2 Page No : 112"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 4,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "The rate of heat loss is 5328 W\n"
     ]
    }
   ],
   "source": [
    "# Variables\n",
    "l = 2.                      \t\t\t#m, length of flat surface\n",
    "T1 = 150.                   \t\t\t#C, surface temp.\n",
    "p = 1.                      \t\t\t#atm, pressure\n",
    "T2 = 30.                    \t\t\t#C, bulk air  temp.\n",
    "V = 12.                     \t\t\t#m/s, air velocity\n",
    "\n",
    "#Calculation\n",
    "Tf = (T1+T2)/2            \t\t\t#C, mean air film temp.\n",
    "mu = 2.131*10**-5            \t\t\t#m**2/s, vismath.cosity\n",
    "k = 0.031                   \t\t\t#W/m C, thermal conductivity\n",
    "rho = 0.962                 \t\t\t#kg/m**3, density of air\n",
    "cp = 1.01                   \t\t\t#kj/kg C, specific heat of air\n",
    "Pr = cp*10**3*mu/k                \t\t\t#Prandtl no.\n",
    "Remax = l*V*rho/mu          \t\t\t#maximum Reynold no.\n",
    "Re = 5.*10**5                 \t\t\t#Reynold no. during transition to turbulent flow   \n",
    "L_ = (Re*mu)/(V*rho)        \t\t\t#m,dismath.tance from the leading edge\n",
    "#for laminar flow heat transfer coefficient h, \n",
    "#h16.707*x**-(1/2)\n",
    "#(a)\n",
    "#h2 = 31.4*x**(-1/5)\n",
    "#b\n",
    "hav = 22.2\n",
    "#c\n",
    "Q = hav*l*p*(T1-T2)\n",
    "\n",
    "# Results\n",
    "print \"The rate of heat loss is %.0f W\"%(Q)\n",
    "\n",
    "# rounding off error."
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Example 4.3 Page No : 114"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 2,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "The steady state temprature is 230 C\n",
      "The recalculated value  is almost equal to previous one.\n"
     ]
    }
   ],
   "source": [
    "import math\n",
    "\n",
    "# Variables\n",
    "d = 7.24*10**-4                  \t\t\t#m, diameter of wire\n",
    "l = 1.                           \t\t\t#m, length of wire\n",
    "I = 8.3                         \t\t\t#A, current in a wire\n",
    "R = 2.625                       \t\t\t#ohm/m, electrical resistance\n",
    "V = 10.                          \t\t\t#m/s, air velocity\n",
    "Tb = 27.                         \t\t\t#C, bulk air temp.\n",
    "#the properties at bulk temp.\n",
    "mu = 1.983*10**-5                \t\t\t#m**2/s, vismath.cosity\n",
    "k = 0.02624                     \t\t\t#W/m C, thermal conductivity\n",
    "rho = 1.1774                    \t\t\t#kg/m**3, density of air\n",
    "cp = 1.0057                     \t\t\t#kj/kg C, specific heat of air\n",
    "\n",
    "# Calculations and Results\n",
    "Pr = cp*10**3*mu/k               \t\t\t#Prandtl no.\n",
    "Re = d*V*rho/mu                 \t\t\t# Reynold no.\n",
    "#from eq. 4.19, nusslet no.\n",
    "Nu = 0.3+(0.62*Re**(1./2)*Pr**(1./3)/(1+(0.4/Pr)**(2./3))**(1./4))*(1+(Re/(2.82*10**5))**(5./8))**(4./5)\n",
    "hav = Nu*k/d                    \t\t\t#W/m**2 C, average heat transfer coefficient\n",
    "Q = I**2*R                       \t\t\t#W, rate of electrical heat generation\n",
    "A = math.pi*d*l\n",
    "dt = Q/(hav*A)                  \t\t\t#C,temp. difference\n",
    "T = dt+Tb                       \t\t\t#C, steady state temp.\n",
    "print \"The steady state temprature is %.0f C\"%(T)\n",
    "\n",
    "Tm = (T+Tb)/2                   \t\t\t#C, mean air film temp.\n",
    "#the properties at Tm temp.\n",
    "mu1 = 2.30*10**-5                  \t\t\t#m**2/s, vismath.cosity\n",
    "k1 = 0.0338                       \t\t\t#W/m C, thermal conductivity\n",
    "rho1 = 0.878                      \t\t\t#kg/m**3, density of air\n",
    "cp1 = 1.014                       \t\t\t#kj/kg C, specific heat of air\n",
    "Re1 = d*V*rho1/mu1                \t\t\t# Reynold no.\n",
    "Pr1 = (1.014*10**3*2.30*10**-5)/k1             \t\t\t#Prandtl no.\n",
    "#from eq. 4.19, nusslet no.\n",
    "Nu1 = 0.3+(0.62*Re1**(1./2)*Pr1**(1./3)/(1+(0.4/Pr1)**(2./3))**(1./4))*(1+(Re1/(2.82*10**5))**(5./8))**(4./5)\n",
    "hav1 = Nu1*k1/d                    \t\t\t#W/m**2 C, average heat transfer coefficient\n",
    "dt1 = Q/(hav1*A)                  \t\t\t#C,temp. difference\n",
    "T1 = dt1+Tb                       \t\t\t#C, steady state temp.\n",
    "print \"The recalculated value  is almost equal to previous one.\"\n"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Example 4.4 Page No : 116"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 3,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "initial  rate of melting of ice is 0.0109 g/s\n",
      "The required time is is 1665 s\n"
     ]
    }
   ],
   "source": [
    "import math\n",
    "# Variables\n",
    "di = 0.04                       \t\t\t#m, diameter of ice ball\n",
    "V = 2.                           \t\t\t#m/s, air velocity\n",
    "T1 = 25.                         \t\t\t#C, steam temp.\n",
    "T2 = 0.\n",
    "#the properties  of air\n",
    "mu = 1.69*10**-5                 \t\t\t#kg/ms, vismath.cosity\n",
    "k = 0.026                       \t\t\t#W/m C, thermal conductivity\n",
    "rho = 1.248                     \t\t\t#kg/m**3, density \n",
    "cp = 1.005                      \t\t\t#kj/kg C, specific heat \n",
    "#propertice of ice\n",
    "lamda = 334.                     \t\t\t#kj/kg, heat of fusion\n",
    "rhoice = 920.                    \t\t\t#kg/m**3 density of ice\n",
    "\n",
    "# Calculations and Results\n",
    "Pr = cp*10**3*mu/k               \t\t\t#Prandtl no.\n",
    "Re = di*V*rho/mu                \t\t\t# Reynold no.\n",
    "#from eq. 4.19, nusslet no.\n",
    "Nu = 2+(0.4*Re**0.5+0.06*Re**(2./3))*Pr**0.4\n",
    "hav = Nu*k/di                          \t\t\t#W/m**2 C, average heat transfer coefficient\n",
    "Ai = math.pi*di**2                   \t\t\t#initial area of sphere\n",
    "Qi = Ai*hav*(T1-T2)                 \t\t\t#W = J/s, initial rate of heat transfer\n",
    "Ri = Qi/lamda                       \t\t\t#initial  rate of melting of ice\n",
    "print \"initial  rate of melting of ice is %.4f g/s\"%(Ri)\n",
    "\n",
    "#(b)\n",
    "#mass of ice ball 4/3*math.pi*r**3\n",
    "#Rate of melting =  Rm =  -d/dt(m)\n",
    "#Rate of heat input required  = -lamda*Rate of melting\n",
    "#heat balance equation\n",
    "# -lamda*(Rm) = h*4*math.pi*r**2*dt\n",
    "#integrating and solving\n",
    "rf = ((di/2)**3/2.)**(1./3)\n",
    "#solving eq. 3\n",
    "t1 = 1.355*10**-4/(8.136*10**-8)\n",
    "print \"The required time is is %.0f s\"%(t1)\n",
    "\n"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Example 4.5 Page No : 121"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 4,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "the required contact time is 1.43 s\n"
     ]
    }
   ],
   "source": [
    "from scipy.integrate import quad \n",
    "# Variables\n",
    "Vo = 0.5                        \t\t\t#m/s air velocity\n",
    "T1 = 800.                        \t\t\t#C, initial temp.\n",
    "T2 = 550.                        \t\t\t#C, final temp.\n",
    "Tam = 500.                       \t\t\t#C, air mean temp.\n",
    "P = 1.2                         \t\t\t#atm, pressure\n",
    "#the properties of solid particles.\n",
    "dp = 0.65*10**-3                 \t\t\t#m, average particle diameter\n",
    "cps = 0.196                     \t\t\t#kcal/kg C, specific heat\n",
    "rhos = 2550.                     \t\t\t#kg/m**3, density \n",
    "#Properties of air\n",
    "mu = 3.6*10**-5                  \t\t\t#kg/ms, vismath.cosity\n",
    "k = 0.05                        \t\t\t#kcal/hm C, thermal conductivity\n",
    "rho = 0.545                     \t\t\t#kg/m**3, density of air\n",
    "cp = 0.263                      \t\t\t#kcal/kg C, specific heat of air\n",
    "\n",
    "#calculation\n",
    "Pr = cp*mu*3600/k                    \t\t\t#Prandtl no.\n",
    "Redp = dp*Vo*rho/mu             \t\t\t# Reynold no.\n",
    "#from eq. 4.29(b) heat transfer coefficient\n",
    "h = (k/dp)*(2+0.6*(Redp)**(1./2)*(Pr)**(1./3))\n",
    "Tg = 500                        \t\t\t#C, gas temp.\n",
    "#from heat balance equation\n",
    "# -(dTs/dt) = 6h/(dp*rhos*cps)*(Ts-Tg)\n",
    "\n",
    "def f2(Ts): \n",
    "    return (1/(Ts-Tg))\n",
    "\n",
    "t = (dp*rhos*cps/(6*h))* quad(f2,550,800)[0]\n",
    "\n",
    "# Results\n",
    "print \"the required contact time is %.2f s\"%(t*3600)\n"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Example 4.6 Page No : 126"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 5,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "the required rate of flow of water is 1053 kg/h \n",
      "the overall heat transfer coefficient is 300 W/m**2 C\n"
     ]
    }
   ],
   "source": [
    "import math\n",
    "\n",
    "# Variables\n",
    "mo_ = 1000.           \t\t\t#kg/h, cooling rate of oil\n",
    "cpo = 2.05           \t\t\t#kj/kg C, specific heat of oil\n",
    "T1 = 70.              \t\t\t#C, initial temp. of oil\n",
    "T2 = 40.              \t\t\t#C,  temp. of oil after cooling\n",
    "cpw = 4.17           \t\t\t#kj/kg C, specific heat of water\n",
    "T3 = 42.              \t\t\t#C, initial temp. of water\n",
    "T4 = 28.              \t\t\t#C,  temp. of oil after cooling\n",
    "A = 3.                \t\t\t#m**2, heat exchange area\n",
    "\n",
    "# Calculation and Results\n",
    "mw_ = mo_*cpo*(T1-T2)/(cpw*(T3-T4))\n",
    "print \"the required rate of flow of water is %.0f kg/h \"%(mw_)\n",
    "Q = mo_*cpo*(T1-T2)/3600       \t\t\t#kw, heat duty\n",
    "dt1 = T1-T3                  \t\t\t#C, hot end temp. difference\n",
    "dt2 = T2-T4                  \t\t\t#C, cold end temp. difference\n",
    "LMTD = (dt1-dt2)/(math.log(dt1/dt2))  \t\t\t#math.log mean temp. difference\n",
    "dtm = LMTD\n",
    "U = Q*10**3/(A*dtm)\n",
    "print \"the overall heat transfer coefficient is %.0f W/m**2 C\"%(round(U,-1))\n"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Example 4.7 Page No : 126"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 6,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "The inlet temprature is  Ti = 26 C\n",
      "The outlet temprature is To = 107 C\n"
     ]
    }
   ],
   "source": [
    "from scipy.optimize import fsolve \n",
    "import math\n",
    "\n",
    "# Variables\n",
    "Q = 38700.             \t\t\t#kcal/h, heat duty\n",
    "W = 2000.              \t\t\t#kg/h gas flow rate\n",
    "cp = 0.239            \t\t\t#kcal/kg C, specific heat of nitrogen\n",
    "A = 10.                \t\t\t#m**2 ,heat exchanger area\n",
    "U = 70.                \t\t\t#kcal/hm**2 C, overall heat transfer coefficient\n",
    "n = 0.63              \t\t\t#fin efficiency\n",
    "\n",
    "#Calculation\n",
    "dt = Q/(W*cp)         \t\t\t#C, temp. difference\n",
    "#To-Ti = dt.........................(i)\n",
    "dtm = Q/(U*A*n)\n",
    "#(To-Ti)/(math.log((160-Ti)/(160-To))) = 87.8........(2)\n",
    "#solving  1  and 2\n",
    "def f(To): \n",
    "    return (To-(To-dt))/(math.log((160-(To-dt))/(160-To)))-87.8\n",
    "\n",
    "To = fsolve(f,100)\n",
    "Ti = To-dt\n",
    "\n",
    "# Results\n",
    "print \"The inlet temprature is  Ti = %.0f C\"%(Ti)\n",
    "print \"The outlet temprature is To = %.0f C\"%(To)\n",
    "\n",
    "# note : answers are slightly different because of fsolve function of python."
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Example 4.8 Page No : 127"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 7,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "The outlet eater temp. is 109.8 C\n"
     ]
    }
   ],
   "source": [
    "import math\n",
    "\n",
    "# Variables\n",
    "V = 1.8                  \t\t\t#m/s, velocity of hot water\n",
    "T1 = 110.                 \t\t\t#C, initial temp.\n",
    "l = 15.                   \t\t\t#m, length  of pipe\n",
    "t = 0.02                 \t\t\t#m, thickness of insulation\n",
    "kc = 0.12                \t\t\t#W/mC,thermal conductivity of insulating layer\n",
    "ho = 10.                  \t\t\t#Wm**2 C, outside film coefficient\n",
    "T2 = 20.                  \t\t\t#C, ambient temp.\n",
    "#the properties of water at 110 C\n",
    "mu = 2.55*10**-4          \t\t\t#m**2/s, vismath.cosity\n",
    "k = 0.685                \t\t\t#W/m C, thermal conductivity\n",
    "rho = 950.                \t\t\t#kg/m**3, density of air\n",
    "cp = 4.23                \t\t\t#kj/kg C, specific heat of air\n",
    "di = 0.035               \t\t\t#m, actual internal dia. of pipe\n",
    "ri = di/2.                \t\t\t#m,internal radius\n",
    "t1 = 0.0036              \t\t\t#m, actual thickness of 1-1/4 schedule 40 pipe\n",
    "ro = ri+t1               \t\t\t#m, outer radius of pipe\n",
    "r_ = ro+t                \t\t\t#m, outer radius of insulation\n",
    "kw = 43.                  \t\t\t#W/mC, thermal conductivity of steel\n",
    "\n",
    "#calculation\n",
    "Pr = cp*10**3*mu/k        \t\t\t#Prandtl no.\n",
    "Re = di*V*rho/mu         \t\t\t# Reynold no.\n",
    "#from eq. 4.9,  Nusslet no.\n",
    "Nu = 0.023*(Re)**0.88*Pr**0.3\n",
    "hi = Nu*k/di             \t\t\t#W/m**2 C, average heat transfer coefficient\n",
    "#the overall coefficient inside area basis Ui\n",
    "Ui = 1./(1/hi+(ri*math.log(ro/ri))/kw+(ri*math.log(r_/ro))/kc+ri/(r_*ho)) \n",
    "Ai = math.pi*di*l           \t\t\t#m**2, inside area basis\n",
    "W = math.pi*ri**2*V*rho      \t\t\t#kg/s, water flow rate\n",
    "#from the relation b/w LMTD and rate of heat loss\n",
    "\n",
    "def f(To): \n",
    "    return (W*cp*10**3)/(Ui*Ai)*(T1-To)-((T1-To)/math.log((T1-T2)/(To-T2)))\n",
    "To = fsolve(f,100)\n",
    "\n",
    "# Results\n",
    "print \"The outlet eater temp. is %.1f C\"%(To)\n"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Example 4.9 Page No : 129"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 9,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "The exit water temp is 36 C\n"
     ]
    }
   ],
   "source": [
    "import math\n",
    "\n",
    "# Variables\n",
    "T1 = 28.              \t\t\t#C, inlet temp. \n",
    "T2 = 250.             \t\t\t#C,bulk temp.\n",
    "V = 10.               \t\t\t#m/s, gas velocity\n",
    "l = 20.               \t\t\t#m, length of pipe\n",
    "mw = 1.*3600          \t\t\t#kg/h, water flow rate\n",
    "di = 4.1*10**-2       \t\t\t#m, inlet diameter\n",
    "Tm = (T1+T2)/2       \t\t\t#C, mean temp.\n",
    "ro = 0.0484          \t\t\t#m, outside radius\n",
    "#properties of water\n",
    "mu = 8.6*10**-4       \t\t\t#kg/ms, vismath.cosity\n",
    "kw = 0.528           \t\t\t#kcal/h m C, thermal conductivity\n",
    "kw_ = 0.528*1.162    \t\t\t#W/ m C, thermal conductivity\n",
    "rho = 996.            \t\t\t#kg/m**3, density of air\n",
    "cp = 1*4.18          \t\t\t#kj/kg C, specific heat of air\n",
    "cp_ = 1.              \t\t\t#kcal/kg C\n",
    "#properties of flue gas\n",
    "mu1 = 2.33*10**-5     \t\t\t#kg/ms, vismath.cosity\n",
    "ka = 0.0292          \t\t\t#kcal/h m C, thermal conductivity\n",
    "rho1 = 0.891         \t\t\t#kg/m**3, density of air\n",
    "cp1 = 0.243          \t\t\t#kcal/kg C, specific heat of air\n",
    "Pr = 0.69\n",
    "\n",
    "#calculation\n",
    "A = math.pi/4*di**2       \t\t\t#m**2, cross section of pipe\n",
    "Vw = 1/(rho*A)       \t\t\t#m/s, velocity of warer\n",
    "Re = di*Vw*rho/mu    \t\t\t# Reynold no.\n",
    "Pr1 = cp*10**3*mu/kw_   \t\t\t#Prandtl no. for water\n",
    "Nu = 0.023*Re**0.8*Pr1**0.4               \t\t\t#Nusslet no.\n",
    "#water side heat transfer coefficient  hi\n",
    "hi = 206*kw/di\n",
    "#gas side heat transfer coefficient    ho\n",
    "a = 41              \t\t\t#mm, i.d. schedule\n",
    "Tw = 3.7            \t\t\t#mm, wall thickness\n",
    "do = a+2*Tw         \t\t\t#mm, outer diameter of pipe\n",
    "Re1 = do*10**-3*V*rho1/mu1    \t\t\t# Reynold no\n",
    "#from eq. 4.19, nusslet no.\n",
    "Nu1 = 0.3+(0.62*Re1**(1./2)*Pr**(1./3)/(1+(0.4/Pr)**(2./3))**(1/4.))*(1+(Re1/(2.82*10**5))**(5./8))**(4/5.)\n",
    "ho = (Nu1*ka/do)*10**3      \t\t\t#kcal/h m**2 C\n",
    "Uo = 1/(ro/(di/2*hi)+1/ho)   \t\t\t#kcal/h m**2 C, overall heat transfer coefficient\n",
    "\n",
    "#Heat balance\n",
    "A1 = math.pi*ro*l         \t\t\t#m62, outside area of pipe\n",
    "#from the formula of LMTD\n",
    "def f(T2_): \n",
    "    return mw*cp_*(T2_-T1)-Uo*A1*((T2_-T1)/math.log((T2-T1)/(T2-T2_)))\n",
    "T2_ = fsolve(f,1)\n",
    "\n",
    "# Results\n",
    "print \"The exit water temp is %.0f C\"%(T2_)\n"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Example 4.10 Page No : 131"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 10,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "The tube length is 123 m\n"
     ]
    }
   ],
   "source": [
    "import math\n",
    "# Variables\n",
    "dti = 0.0212        \t\t\t#m inner tube\n",
    "dto = 0.0254        \t\t\t#cm, outer tube\n",
    "dpi = 0.035         \t\t\t#cm, outer pipe\n",
    "mo_ = 500.           \t\t\t#kh/h, cooling rate of oil\n",
    "To2 = 110.           \t\t\t#C, initial temo. of oil\n",
    "To1 = 70.            \t\t\t#C, temp. after cooling of oil\n",
    "Tw2 = 40.            \t\t\t#C, inlet temp. of water\n",
    "Tw1 = 29.            \t\t\t#C, outlet temp. of water\n",
    "#properties of oil\n",
    "cpo = 0.478         \t\t\t#kcal/kg C\n",
    "ko = 0.12           \t\t\t#kcal/h m C, thermal conductivity\n",
    "rho = 850.           \t\t\t#kg/m**3, density of oil\n",
    "#properties of water\n",
    "kw = 0.542          \t\t\t#kcal/h m C, thermal conductivity\n",
    "kw_ = (kw*1.162)    \t\t\t#kj/kg C\n",
    "muw = 7.1*10**-4     \t\t\t#kg/ms, vismath.cosity of water\n",
    "cpw = 1.             \t\t\t#kcal/kg C\n",
    "cpw_ = cpw*4.17     \t\t\t#kcal/kg C\n",
    "rhow = 1000.         \t\t\t#kg/m**3, density\n",
    "\n",
    "#calculation\n",
    "HL = mo_*cpo*(To2-To1)      \t\t\t#kcal/h, heat load of exchanger\n",
    "mw_ = HL/(cpw*(Tw2-Tw1))    \t\t\t#kg/h water flow rate\n",
    "mw_1 = mw_/(3600*10**3)      \t\t\t#m**3/s water flow rate\n",
    "A1 = (math.pi/4)*(dti)**2        \t\t\t#m**2, flow area of tube\n",
    "Vw = mw_1/A1                \t\t\t#m/s water velocity\n",
    "Rew = dti*Vw*rhow/muw       \t\t\t#Reynold no.\n",
    "Prw = cpw_*10**3*muw/kw_      \t\t\t#Prandtl no.\n",
    "Nuw = 0.023*Rew**0.8*Prw**0.4 \t\t\t#nusslet no.\n",
    "#water side heat transfer coefficient   hi\n",
    "hi = Nuw*kw/dti\n",
    "\n",
    "#oil side heat transfer coefficient\n",
    "A2 = math.pi/4*(dpi**2-dto**2)    \t\t\t#m**2, flow area of annulus\n",
    "Vo = mo_/(3600*rho*A2)      \t\t\t#m/s velocity of oil\n",
    "de = (dpi**2-dto**2)/dto      \t\t\t#m, equivalent dia of annulus\n",
    "Tmo = (To2+To1)/2           \t\t\t#C,mean oil temp.\n",
    "muoil = math.exp((5550./(Tmo+273))-19)  \t\t\t#kg/ms, vismath.cosity of oil\n",
    "Reo = de*Vo*rho/muoil\n",
    "Pro = cpo*muoil*3600/ko     \t\t\t#prandtl no. for oil    \n",
    "\n",
    "#assume (1st approximation)\n",
    "Nuo = 3.66\n",
    "ho = Nuo*ko/de             \t\t\t#kcal/h m**2 c\n",
    "L = 1                      \t\t\t#assume length of tube\n",
    "Ai = math.pi*dti*L\n",
    "Ao = math.pi*dto*L\n",
    "#overall heat transfer coefficient 1st approximation\n",
    "Uo = 1/(1/ho+Ao/(Ai*hi))\n",
    "LMTD = ((To2-Tw2)-(To1-Tw1))/(math.log((To2-Tw2)/(To1-Tw1)))\n",
    "Ao1 = HL/(Uo*LMTD)         \t\t\t                #m**2, heat transfer area\n",
    "Lt = Ao1/(math.pi*dto)          \t\t\t        #m, tube length\n",
    "#from eq.  4.8\n",
    "Nuo1 = 1.86*(Reo*Pro/(Lt/de))**(1./3)   \t\t\t#Nusslet no. \n",
    "ho1 = Nuo1*ko/de\n",
    "Tmw = (Tw1+Tw2)/2         \t\t\t#C, mean water temp.\n",
    "#balancing heat transfer rate of oil and water\n",
    "\n",
    "#average wall temp. Twall\n",
    "Twall = ((hi*dti*(-Tmw))-(ho1*dto*Tmo))/(-65.71216)\n",
    "#vismath.cosity of oil at this temp.\n",
    "muwall = math.exp((5550/(Twall+273))-19)  \t\t\t#kg/ms, vismath.cosity of oil\n",
    "#Nusslet no. \n",
    "Nuo2 = 1.86*(Reo*Pro/(Lt/de))**(1./3)*(muoil/muwall)**0.14\n",
    "ho2 = Nuo2*ko/de\n",
    "Uo2 = 1/((1/ho2)+(Ao/(Ai*hi)))\n",
    "Ao2 = HL/(Uo2*LMTD)\n",
    "Lt_ = Ao2/(math.pi*dto)\n",
    "\n",
    "# Results\n",
    "print \"The tube length is %d m\"%(Lt_)\n",
    "\n",
    "# rounding off error."
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Example 4.11 Page No : 135"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 11,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      " the rate of heat transfer to water.is 6.93e+05 kcal/h\n"
     ]
    }
   ],
   "source": [
    "import math\n",
    "\n",
    "# Variables\n",
    "Ti = 260.                 \t\t\t#C, initial  temp.\n",
    "Ts = 70.                  \t\t\t#C, skin temp.\n",
    "St = 0.15                \t\t\t#m,space between tubes in equilateral triangular arrangement\n",
    "Sd = St                  \t\t\t#space between tubes\n",
    "mu = 4.43*10**-5                \t\t\t#m**2/s, momentum diffusity\n",
    "k = 0.0375                     \t\t\t#W/m C, thermal conductivity\n",
    "rho = 0.73                     \t\t\t#kg/m**3, density of air\n",
    "cp = 0.248                     \t\t\t#kj/kg C, specific heat of air\n",
    "V = 16.                         \t\t\t#m/s, velociity \n",
    "d = 0.06                       \t\t\t#m, outside diameter of tube\n",
    "Nt = 15.                        \t\t\t#no. of tubes in transverse row\n",
    "Nl = 14.                        \t\t\t#no. of tubes in longitudinal row\n",
    "N = Nl*Nt                      \t\t\t#total no. of tubes\n",
    "L = 1.                          \t\t\t#m, length\n",
    "#Calculation\n",
    "Sl = (math.sqrt(3)/2)*St\n",
    "Pr = cp*mu*3600*rho/k          \t\t\t#Prandtl no. of bulk air\n",
    "Pr = 0.62\n",
    "Prw = 0.70                     \t\t\t#Prandtl no. of air at wall temp. 70 C\n",
    "#from eq. 4.25\n",
    "Vmax = (St/(St-d))*V\n",
    "#from eq. 4.26\n",
    "Vmax1 = (St/(2*(St-d)))*V\n",
    "Redmax = d*Vmax/mu\n",
    "p = St/Sl                     \t\t\t#pitch ratio\n",
    "#from table 4.3\n",
    "m = 0.6\n",
    "C = 0.35*(St/Sl)**0.2\n",
    "h = round((k/d)*C*(36163)**m*(Pr)**(0.36)*(Pr/Prw)**(0.25))\n",
    "#from eq. 4.28\n",
    "dt = round(190*math.exp(-math.pi*d*N*h/(rho*V*3600*Nt*St*cp)))\n",
    "LMTD = ((Ti-Ts)-(dt))/math.log((Ti-Ts)/dt)\n",
    "A = round(math.pi*d*L*N,1)                \t\t\t#m**2, heat transfer area\n",
    "Q = h*A*LMTD\n",
    "\n",
    "# Results\n",
    "print \" the rate of heat transfer to water.is %.2e kcal/h\"%(Q)\n",
    "\n",
    "# Note : Value of LMTD is wrong in book please check."
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Example 4.12 Page No : 140"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 12,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "Outlet temp. of water for one pass through the tubes is 51 C\n"
     ]
    }
   ],
   "source": [
    "import math\n",
    "\n",
    "# Variables\n",
    "W = 0.057                      \t\t\t#m**3/min/tube, flow rate of water\n",
    "W_ = W*16.66                   \t\t\t#kg/s. water flow rate\n",
    "di = 0.0212                    \t\t\t#m,inside diameter\n",
    "Ti = 32.                        \t\t\t#C, inlet water temp.\n",
    "Tw = 80.                        \t\t\t#C, wall temp.\n",
    "L = 3.                          \t\t\t#m, length of pip\n",
    "\n",
    "#Calculation\n",
    "V = (W/60)*(1/((math.pi/4)*di**2))   \t\t\t#m/s, water velocity\n",
    "#the properties of water at mean liquid temp..\n",
    "mu = 7.65*10**-4                 \t\t\t#m**2/s, vismath.cosity\n",
    "k = 0.623                       \t\t\t#W/m C, thermal conductivity\n",
    "rho = 995.                       \t\t\t#kg/m**3, density of air\n",
    "cp = 4.17                       \t\t\t#kj/kg C, specific heat of air\n",
    "\n",
    "Pr = cp*10**3*mu/k               \t\t\t#Prandtl no.\n",
    "Re = di*V*rho/mu                \t\t\t# Reynold no.\n",
    "#from eq. 4.19, nusslet no.\n",
    "#from dittus boelter eq.\n",
    "Nu = 0.023*Re**0.8*Pr**0.4        \t\t\t#Prandtl no.\n",
    "f = 0.0014+0.125*Re**-0.32       \t\t\t#friction factor\n",
    "#Reynold anamath.logy\n",
    "St = f/2                       \t\t\t#Smath.tanton no.\n",
    "Nu1 = Re*Pr*St\n",
    "#Prandtl anamath.logy\n",
    "St1 = (f/2)/(1+5*(Pr-1)*math.sqrt(f/2))\n",
    "Nu2 = St1*Re*Pr               \n",
    "#colburn analogy\n",
    "Nu3 = Re*Pr**(1./3)*(f/2)\n",
    "h = Nu3*k/(di)               \t\t\t#W/m**2 C av heat transfer coefficient\n",
    "#Q = W_*cp*10**3*(To-Ti) = h*A*LMTD\n",
    "A = math.pi*di*L       \t\t\t#m**2\n",
    "def f(To): \n",
    "    return W_*cp*10**3*(To-Ti)-h*A*((To-Ti)/math.log((Tw-Ti)/(Tw-To)))\n",
    "To = fsolve(f,1)\n",
    "#Revised calculation\n",
    "Tm = (Ti+To)/2               \t\t\t#C, mean liquid temp.\n",
    "#the properties of water at new mean liquid temp..\n",
    "mu1 = 6.2*10**-4                 \t\t\t#m**2/s, vismath.cosity\n",
    "k1 = 0.623                       \t\t\t#W/m C, thermal conductivity\n",
    "rho1 = 991.                       \t\t\t#kg/m**3, density of air\n",
    "cp1 = 4.17                       \t\t\t#kj/kg C, specific heat of air\n",
    "\n",
    "Pr1 = cp1*10**3*mu1/k1               \t\t\t#Prandtl no.\n",
    "Re1 = di*V*rho1/mu1                \t\t\t# Reynold no.\n",
    "#from dittus boelter eq.\n",
    "f1 = 0.0014+0.125*Re1**(-0.32)       \t\t\t#friction factor\n",
    "#colburn anamath.logy\n",
    "Nu4 = Re1*Pr1**(1./3)*(f1/2)\n",
    "h1 = Nu4*k1/(di)               \t\t\t#W/m**2 C av heat transfer coefficient\n",
    "def f(To_): \n",
    "    return W_*cp*10**3*(To_-Ti)-h1*A*((To_-Ti)/math.log((Tw-Ti)/(Tw-To_)))\n",
    "To_ = fsolve(f,1)\n",
    "\n",
    "print \"Outlet temp. of water for one pass through the tubes is %.0f C\"%(To_)\n",
    "\n"
   ]
  }
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