path: root/Heat_Transfer_Principles_And_Applications_by_Dutta/ch8.ipynb

{
 "cells": [
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "# Chapter 8 : Heat Exchanger"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Example 8.1 Page No : 303"
   ]
  },
  {
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   "execution_count": 1,
   "metadata": {
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   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "the heat duty of the exchanger is 47000 kj/h\n",
      "the water flow rate is 1122 kg/h\n",
      "heat transfer coefficient based on inside area  is 3560 W/m**2 C \n",
      "heat transfer coefficient based on outside area  is 880.3 W/m**2 C \n",
      "overall heat transfer coefficient outside area basis is 663.1 W/m**2 C \n",
      "overall heat transfer coefficient inside area basis is 802.0 W/m**2 C \n",
      "The fouling factor is 0.000949 m**2 C/W\n"
     ]
    }
   ],
   "source": [
    "import math\n",
    "\n",
    "# Variables\n",
    "#for  Benzene\n",
    "Mb = 1000.              \t\t\t#Kg, mass of benzene\n",
    "T1 = 75.                \t\t\t#C initial temp. of benzene\n",
    "T2 = 50.               \t\t\t#C final temp. of benzene\n",
    "Cp1 = 1.88             \t\t\t#Kj/Kg C. specific heat of benzene\n",
    "mu1 = 0.37             \t\t\t#cP. vismath.cosity of benzene\n",
    "rho1 = 860.             \t\t\t#kg/m**3, density\n",
    "k1 = 0.154             \t\t\t#W/m K. thermal conductivity\n",
    "\n",
    "#for water\n",
    "Tav = 35.               \t\t\t#C av, temp.\n",
    "Cp2 = 4.187             \t\t\t#specific heat\n",
    "mu2 = 0.8              \t\t\t#cP. vismath.cosity\n",
    "k2 = 0.623             \t\t\t#W/m K. thermal conductivity\n",
    "T3 = 30.                \t\t\t#C. initial temp.\n",
    "T4 = 40.                \t\t\t#C final temp.\n",
    "\n",
    "#Calculation and Results\n",
    "#(a)\n",
    "HD = Mb*Cp1*(T1-T2)   \t\t\t#Kj/h, heat duty\n",
    "WR = HD/(Cp2*(T4-T3))  \t\t\t#kg/h Water rate\n",
    "print \"the heat duty of the exchanger is %.0f kj/h\"%(HD)\n",
    "print \"the water flow rate is %d kg/h\"%(WR)\n",
    "\n",
    "#(b)\n",
    "#tube side (water) calculations\n",
    "# Variables\n",
    "di1 = 21.            \t\t\t#mm, inner diameter of inner tube \n",
    "do1 = 25.4          \t\t\t#mm, outer dia. of inner tube\n",
    "t = 2.2            \t\t\t#mm/ wall thickness\n",
    "kw = 74.5          \t\t\t#W/m K. thermal conductivity of the wall\n",
    "di2 = 41.           \t\t\t#mm, inner diameter of outer pipe\n",
    "do2 = 48.           \t\t\t#mm, outer diameter of outer pipe\n",
    "\n",
    "FA1 = (math.pi/4)*(di1*10**-3)**2    \t\t\t#m**2, flow area\n",
    "FR1 = WR/1000.\n",
    "v1 = FR1/(FA1*3600)                      \t\t\t#m/s, velocity\n",
    "Re1 = (di1*10**-3)*v1*1000/(mu2*10**-3)  \t\t\t#Reynold no.\n",
    "Pr1 = Cp2*1000*(mu2*10**-3)/k2          \t\t\t#Prandtl no.\n",
    "#umath.sing dittus boelter eq.\n",
    "Nu1 = 0.023*(Re1)**(0.8)*(Pr1)**(0.3)    \t\t\t#nusslet no.\n",
    "h1 = round(Nu1*k2/(di1*10**-3),-1)                \t\t\t#W/m**2 C, heat transfer coefficient\n",
    "\n",
    "#Outer side (benzene) calculation\n",
    "FA2 = (math.pi/4)*(di2*10**-3)**2-(math.pi/4)*(do1*10**-3)**2   \t\t\t#flow area\n",
    "wp = math.pi*(di2*10**-3+do1*10**-3)                      \t\t\t#wettwd perimeter\n",
    "dh = 4*FA2/wp                                       \t\t\t#hydrolic diameter\n",
    "bfr = Mb/rho1                                       \t\t\t#m**3/h benzene flow rate\n",
    "v2 = bfr/(FA2*3600)                                 \t\t\t#m/s, velocity\n",
    "Re2 = dh*v2*rho1/(mu1*10**-3)                        \t\t\t#Reynold no\n",
    "Pr2 = Cp1*10**3*(mu1*10**-3)/k1                       \t\t\t#Prandtl no.\n",
    "Nu2 = 0.023*(Re2)**(0.8)*(Pr2)**(0.4)                 \t\t\t#nusslet no.\n",
    "h2 = Nu2*k1/(dh)                              \t\t\t#W/m**2 C, heat transfer coefficient\n",
    "\n",
    "print \"heat transfer coefficient based on inside area  is %.0f W/m**2 C \"%(h1)\n",
    "print \"heat transfer coefficient based on outside area  is %.1f W/m**2 C \"%(h2)\n",
    "\n",
    "#Calculation of clean  overall heat transfer coefficient, outside area basis\n",
    "#from eq. 8.28\n",
    "# Variables\n",
    "l = 1.     \t\t\t#assume , length\n",
    "Ao = math.pi*do1*10**-3*l\n",
    "Ai = math.pi*di1*10**-3*l\n",
    "Am = (do1*10**-3-di1*10**-3)*math.pi*l/(math.log(do1*10**-3/(di1*10**-3)))\n",
    "\n",
    "#overall heat transfer coefficient\n",
    "Uo = 1/((1/h2)+(Ao/Am)*((do1*10**-3-di1*10**-3)/(2*kw))+(Ao/Ai)*(1/h1))\n",
    "Ui = Uo*Ao/Ai\n",
    "\n",
    "#Calculation of LMTD\n",
    "dt1 = T1-T4\n",
    "dt2 = T2-T3\n",
    "LMTD = (dt1-dt2)/math.log(dt1/dt2)    \t\t\t#math.log mean temp. difference correction factor\n",
    "Q = HD*1000/3600                 \t\t\t#W, heat required\n",
    "Ao_ = Q/(Uo*LMTD)                \t\t\t#m**@, required area\n",
    "len = Ao_/(math.pi*do1*10**(-3))        \t\t\t#m, tube length necessary\n",
    "\n",
    "#(c)\n",
    "la = 15.                          \t\t\t#m ,actual length\n",
    "Aht = (math.pi*do1*10**(-3)*la)\n",
    "Udo = Q/(Aht*LMTD)              \t\t\t#W/m**2 C, overall heat transfer coefficient with dirt factor\n",
    "#from eq. 8.2\n",
    "Rdo = (1/Udo)-(1/Uo)           \t\t\t#m**2 C/W\n",
    "print \"overall heat transfer coefficient outside area basis is %.1f W/m**2 C \"%(Uo)\n",
    "print \"overall heat transfer coefficient inside area basis is %.1f W/m**2 C \"%(Ui)\n",
    "print \"The fouling factor is %f m**2 C/W\"%(Rdo)\n",
    "\n",
    "# note : rounding off error. please check."
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Example 8.2 Page No : 309"
   ]
  },
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   "metadata": {
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   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "Tube side Pressure drop is 1.118e+04 N/m**2 \n",
      "Shell side Pressure drop is 120 N/m**2 \n"
     ]
    }
   ],
   "source": [
    "import math\n",
    "# Variables\n",
    "Cp = 50.          \t\t\t#tpd, plant capacity\n",
    "T1 = 135.         \t\t\t#C, Temp.\n",
    "T2 = 40.          \t\t\t#C temp.\n",
    "T3 = 30.          \t\t\t#C temp.\n",
    "dt1 = (T1-T2)    \t\t\t#C hot end temp. \n",
    "dt2 = (T2-T3)    \t\t\t#C cold end temp.\n",
    "#Properties of ethylbenzene\n",
    "rho1 = 840.       \t\t\t#kg/m**3, density\n",
    "cp1 = 2.093      \t\t\t#kj/kg K , specific heat\n",
    "T = 87.5         \t\t\t#C\n",
    "mu1 = math.exp(-6.106+1353/(T+273)+5.112*10**-3*(T+273)-4.552*10**-6*((T+273)**2))\n",
    "k1 = 0.2142-(3.44*10**-4)*(T+273)+(1.947*10**-7)*(T+273)**2\n",
    "k1_ = k1*0.86         \t\t\t#kcal/h m K\n",
    "#properties of water\n",
    "rho2 = 993.      \t\t\t#kg/m**3, density\n",
    "mu2 = 8*10.**-4   \t\t\t#kg/m s , vismath.cosity \n",
    "cp2 = 4.175     \t\t\t#kj/kg K , specific heat\n",
    "k2 = 0.623      \t\t\t#W/m K, thermal conductivity\n",
    "k2_ = k2*0.8603  \t\t\t#kcal/h m**2 K\n",
    "#Calculation\n",
    "#(i) Energy balance\n",
    "Cp = Cp*1000./24          \t\t\t#kg/h, plant capacity\n",
    "Cp = 2083.       \t\t\t#approx.\n",
    "HD = Cp*cp1*dt1      \t\t\t#kj/h, Heat duty \n",
    "HD_ = HD*0.238837        \t\t\t#kcal/h\n",
    "wfr = HD/(cp2*dt2)\n",
    "\n",
    "#(ii)\n",
    "mu1 = mu1      \t\t\t#cP, vismath.cosity of ethylbenzene\n",
    "k1 = k1        \t\t\t#W/m K, thermal conductivity of ethylbenzene\n",
    "\n",
    "#(iii)\n",
    "#LMTD calculation\n",
    "LMTD = (dt1-dt2)/math.log(dt1/dt2)\n",
    "#assume\n",
    "Udo = 350.             \t\t\t#kcal/h m**2 C, overall coefficient\n",
    "A = HD_/(Udo*LMTD)    \t\t\t#m**2, area required\n",
    "\n",
    "#(iv)\n",
    "id_ = 15.7            \t\t\t#mm, internal diameter of tube\n",
    "od = 19.              \t\t\t#mm, outer diameter of tube\n",
    "l = 3000.             \t\t\t#mm, length\n",
    "OSA = math.pi*(od*10**-3)*(l*10**-3)  \t\t\t#m**2. outer surface area\n",
    "n = A/OSA                       \t\t\t#no. of tubes required\n",
    "fa = n*(math.pi/4)*(id_*10**-3)**2     \t\t\t#m**2, flow arae\n",
    "lv = (wfr/1000)/(3600*fa)       \t\t\t#m/s, linear velocity\n",
    "\n",
    "#(v)\n",
    "n1 = 44.             \t\t\t#total no. of tubes that can be accomodated in a 10 inch shell\n",
    "np = 11.             \t\t\t#no. of tubes in each pass\n",
    "#(vi)\n",
    "bf = 0.15          \t\t\t#m, baffel spacing\n",
    "#(vii)\n",
    "#estimation of heat transfer coefficient\n",
    "#Tube side (water)\n",
    "fa1 = (math.pi/4)*(id_*10**-3)**2*np   \t\t\t#m**2, flow area\n",
    "v1 = (wfr/1000.)/(3600*fa1)      \t\t\t#m/s, velocity\n",
    "Re = (id_*10**-3)*v1*rho2/mu2     \t\t\t#Reynold no.\n",
    "#from fig . 8.11(a)\n",
    "jh = 85.                         \t\t\t#colburn factor\n",
    "#jh = (hi*di)/k*(cp*mu/k)**-1/3     \n",
    "#assume,   (cp*mu/k) = x\n",
    "hi = jh*(k2_/(id_*10**-3))*(cp2*1000*mu2/k2)**(1/3)  \t\t\t#kcal/h m**2 C\n",
    "\n",
    "#shell side(organic)\n",
    "B = bf                       \t\t\t#m, baffel spacing\n",
    "p = 0.0254                   \t\t\t#m,radius of 1 tube\n",
    "Ds = 0.254                   \t\t\t#m, inside diameter of shell\n",
    "c = 0.0064  \n",
    "#from eq. 8.32\n",
    "As = c*B*Ds/p                \t\t\t#m**2, flow area\n",
    "Gs = Cp/As                   \t\t\t#kg/m**2 h, mass flow rate of shell fluid\n",
    "do = od/10                   \t\t\t#cm, outside diameter of shell\n",
    "#from eq. 8.31\n",
    "Dh = 4*((0.5*p*100)*(0.86*p*100)-((math.pi*(do)**2)/8))/((math.pi*do)/2)\n",
    "Dh_ = Dh*10**-2              \t\t\t#m, hydrolic diameter\n",
    "Re1 = (Dh_*Gs)/(3600*(mu1*10**-3))   \t\t\t#Reynold no.\n",
    "#from fig 8.11(b)\n",
    "jh1 = 32                           \t\t\t#colburn factor\n",
    "ho = jh1*(k1_/Dh_)*((6)**(1./3))\n",
    "#from eq. 8.28\n",
    "ratio = od/id_                      \t\t\t#ratio = Ao/Ai\n",
    "Rdo = 0.21*10**-3                   \t\t\t#outside dirt factor\n",
    "Rdi = 0.35*10**-3                   \t\t\t#inside dirt factor\n",
    "Udo = 1/((1/ho)+Rdo+(ratio)*Rdi+(ratio)*(1/hi))\n",
    "\n",
    "#SECOND TRIAL\n",
    "#estimation of heat transfer coefficient\n",
    "#Tube side (water)\n",
    "np1 = 12               \t\t\t#\n",
    "fa2 = (math.pi/4)*(id_*10**-3)**2*np1   \t\t\t#m**2, flow area\n",
    "v2 = (wfr/1000)/(3600*fa2)      \t\t\t#m/s, velocity\n",
    "Re2 = (id_*10**-3)*v2*rho2/mu2     \t\t\t#Reynold no.\n",
    "#from fig . 8.11(a)\n",
    "jht = 83.                         \t\t\t#colburn factor\n",
    "#jh = (hi*di)/k*(cp*mu/k)**-1/3     \n",
    "#assume,   (cp*mu/k) = x\n",
    "hit = jht*(k2_/(id_*10**-3))*(cp2*1000*mu2/k2)**(1./3)  \t\t\t#kcal/h m**2 C\n",
    "\n",
    "#shell side\n",
    "B2 = 0.1                        \t\t\t#m, baffel spacing\n",
    "p2 = 0.0254                     \t\t\t#m,radius of 1 tube\n",
    "Ds2 = 0.254                     \t\t\t#m, inside diameter of shell\n",
    "c2 = .0064\n",
    "#from eq. 8.32\n",
    "As2 = c2*B2*Ds2/p2              \t\t\t#m**2, flow area\n",
    "Gs2 = Cp/As2                    \t\t\t#kg/m**2 h, mass flow rate of shell fluid\n",
    "do2 = od/10                     \t\t\t#cm, outside diameter of shell\n",
    "#from eq. 8.30\n",
    "Dh2 = 4*((p2*100)**2-((math.pi*(do2)**2)/4))/((math.pi*do2))\n",
    "Dh2_ = Dh2*10**-2                \t\t\t#m, hydrolic diameter\n",
    "Re2 = (Dh2_*Gs2)/(3600*(mu1*10**-3))\n",
    "#from fig 8.11(b)\n",
    "jh2 = 48                       \t\t\t#colburn factor\n",
    "ho2 = jh2*(k1_/Dh2_)*((6)**(1./3))\n",
    "#from eq. 8.28\n",
    "ratio = od/id_                   \t\t\t#ratio = Ao/Ai\n",
    "Rdo2 = 0.21*10**-3               \t\t\t#outside dirt factor\n",
    "Rdi2 = 0.35*10**-3               \t\t\t#inside dirt factor\n",
    "Udo2 = 1/((1/ho2)+Rdo+(ratio)*Rdi+(ratio)*(1/hit))\n",
    "\n",
    "#from eq. 8.10(a)\n",
    "tauc = (T2-T3)/(T1-T3)         \t\t\t#Temprature ratio\n",
    "R = (T1-T2)/(T2-T3)            \t\t\t#Temprature ratio\n",
    "Ft = 0.8                       \t\t\t#LMTD correction ftor\n",
    "Areq = HD_/(Udo2*Ft*LMTD)      \t\t\t#area required\n",
    "tubes = 48.                     \t\t\t#no. of tubes\n",
    "lnt = 4.5                      \t\t\t#length of 1 tube\n",
    "Aavl = (math.pi*od*10**-3)*tubes*lnt      \t\t\t#available area\n",
    "excA = ((Aavl-Areq)/Areq)*100         \t\t\t#% excess area\n",
    "\n",
    "#Pressure drop calculation\n",
    "#Tube side\n",
    "#from eq. 8.33\n",
    "Gt = wfr/(3600*fa2)          \t\t\t#kg/m**2 s, mass flow rate of tube fluid\n",
    "n2 = 4.                       \t\t\t#tube passes\n",
    "fit = 1.                      \t\t\t#dimensionless vismath.cosity ratio\n",
    "g = 9.8                      \t\t\t#gravitational consmath.tant\n",
    "f = 0.0037                   \t\t\t#friction factor\n",
    "dpt = f*Gt**2*lnt*n2/(2*g*rho2*id_*10**-3*fit)      \t\t\t#kg/m**2, tube side pressure drop\n",
    "\n",
    "#eq.8.35\n",
    "dpr = 4*n2*v2**2*rho2/(2*g)            \t\t\t#kg/m**2, return tube pressure loss\n",
    "dpr_ = dpr*9.801                      \t\t\t#N/m**2\n",
    "tpr = dpt+dpr                         \t\t\t#kg/m**2, total pressure drop\n",
    "#shell side\n",
    "fs = 0.052                            \t\t\t#friction factor for shell\n",
    "bf1 = 0.1                             \t\t\t#m, baffel spacing\n",
    "Nb = lnt/bf1-1                        \t\t\t#no. of baffles\n",
    "dps = fs*(Gs2/3600)**2*Ds*(Nb+1)/(2*g*rho1*Dh2_*fit)   \t\t\t#kg/m**2, shell side pressure drop\n",
    "dps_ = dps*9.81                       \t\t\t#N/m**2, shell side pressure drop\n",
    "print \"Tube side Pressure drop is %1.3e N/m**2 \"%(dpr_)\n",
    "print \"Shell side Pressure drop is %.0f N/m**2 \"%(round(dps_,-1))\n",
    "\n",
    "# note : rounding off error."
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Example 8.3 Page No : 320"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 3,
   "metadata": {
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   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "Th2 = 49.5 C\n",
      "The new rate of heat transfer : 161003 kcal/h\n",
      "the heat teansfer rate will be affected by 1.3 percent \n"
     ]
    }
   ],
   "source": [
    "import math\n",
    "# Variables\n",
    "#for hot stream\n",
    "Wh = 10000.                \t\t\t#kg/h, Rate of leaving a hydrolic system by the oil\n",
    "Cph = 0.454               \t\t\t#Kcal/Kg C, specific heat of oil\n",
    "Th1 = 85.                  \t\t\t#C initial temp. of oil\n",
    "Th2 = 50.                  \t\t\t#C final temp. of oil \n",
    "\n",
    "#For  cold stream\n",
    "Cpc = 1.                   \t\t\t#Kcal/Kg C, specific heat of water\n",
    "Tc2 = 30.                   \t\t\t#C final temp. of water\n",
    "Tc1 = 38.                   \t\t\t#C initial temp. of water\n",
    "\n",
    "# Calculations\n",
    "#from heat balance eq.\n",
    "#kg/h, Rate of leaving a hydrolic system by the water\n",
    "Wc = Wh*Cph*(Th1-Th2)/(Cpc*(Tc1-Tc2))\n",
    "#For the hot stream\n",
    "Cmin = Wh*Cph            \t\t\t#Kcal/h C.Taking hot stream as min. stream\n",
    "#For cold stream\n",
    "Cmax = Wc*Cpc             \t\t\t#Kcal/h C.Taking cold  stream as max. stream\n",
    "Cr = Cmin/Cmax             \t\t\t#Capacity ratio\n",
    "n = (Th1-Th2)/(Th1-Tc2)     \t\t\t#effectiveness factor\n",
    "#From eq. 8.57\n",
    "#No. of transfer units\n",
    "NTU = -(1+(Cr)**2)**-(1./2)*math.log(((2/n)-(1+Cr)-(1+(Cr)**2)**(1./2))/((2./n)-(1+Cr)+(1+(Cr)**2)**(1./2)))\n",
    "Ud = 400.                  \t\t\t#kcal/h m**2C , overall dirty heat transfer coefficient\n",
    "#from eq. 8.53\n",
    "A = (NTU*Cmin)/Ud         \t\t\t#Area required\n",
    "#if the water rate is increased by 20 %,\n",
    "a = 20.\n",
    "Wc_ = Wc+(Wc*(a/100))\n",
    "Cmax_ = Wc_*Cpc\n",
    "Cr_ = Cmin/Cmax_\n",
    "#From eq. 8.56\n",
    "n_ = 2*((1+Cr_)+(1+(Cr_)**2)**(1./2)*(1+math.exp(-(1+(Cr_)**2)**(1./2)*NTU))/(1-math.exp(-(1+(Cr_)**2)**(1./2)*NTU)))**(-1)\n",
    "Th2_ = Th1-(n_*(Th1-Tc2))\n",
    "q1 = Wh*Cph*(Th1-Th2)   \t\t\t#kcal/h previous rate of heat transfer\n",
    "q2 = Wh*Cph*(Th1-Th2_)   \t\t\t#kcal/h new rate of heat transfer\n",
    "#increase in rate of heat transfer\n",
    "dq = (q2-q1)/q1  \n",
    "\n",
    "# Results\n",
    "print \"Th2 = %.1f C\"%Th2_\n",
    "print \"The new rate of heat transfer : %d kcal/h\"%q2\n",
    "print \"the heat teansfer rate will be affected by %.1f percent \"%(dq*100 )\n",
    "\n",
    "# note : rounding off error would be there."
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Example 8.4 Page No : 337"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 8,
   "metadata": {
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   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "the time required to heat the charge 22 min\n"
     ]
    }
   ],
   "source": [
    "import math\n",
    "# Variables\n",
    "p = 0.0795           \t\t\t#m. pitch of the coil\n",
    "d1 = 0.0525         \t\t\t#m,coil diameter\n",
    "h = 1.464             \t\t\t#m,height of the limpetted section\n",
    "d2 = 1.5              \t\t\t#m,diameter of batch polymerization reactor\n",
    "d3 = 0.5              \t\t\t#m,diameter of agitator\n",
    "rpm = 150.           \t\t\t#speed of agitator\n",
    "rho = 850.            \t\t\t#kg/m3,density of monomer\n",
    "rho1 = 900.          \t\t\t#kg/m3,density of fluid\n",
    "mu = 0.7*10**-3   \t\t\t#poise, vismath.cosity of monomer\n",
    "mu1 = 4*10.**-3    \t\t\t#poise, vismath.cosity of fluid\n",
    "cp = 0.45            \t\t\t#kcal/kg C, specific heat of monomer\n",
    "cp1 = 0.5            \t\t\t#kcal/kg C, specific heat of fluid\n",
    "k = 0.15              \t\t\t#kcal/h mC, thermal conductivity of monomer\n",
    "k1 = 0.28             \t\t\t#kcal/h mC, thermal conductivity of fluid\n",
    "Rdi = 0.0002         \t\t\t#h m2 C/kcal, fouling factor for vessel\n",
    "Rdc = 0.0002        \t\t\t#h m2 C/kcal, fouling factor for coil\n",
    "Tci = 120.             \t\t\t#C, initial temp. of coil liquid\n",
    "Tvi = 25.               \t\t\t#C, initial temp. of vessel liquid\n",
    "Tvf = 80.               \t\t\t#C, final temp. of vessel liquid\n",
    "\n",
    "#calculation\n",
    "a = math.pi*d2*h       \t\t\t#outside area of the vessel\n",
    "x = 60.                  \t\t\t#%.  added of the unwetted area to the wetted area\n",
    "ao = ((d1+(x/100)*(p-d1))/p)*a  \t\t\t#m**2,effective outside heat transfer area of vessel\n",
    "ai = 6.9                        \t\t\t#m**2,inside heat transfer area of vessel\n",
    "#same as outside area , if thickness is very small\n",
    "#vessel side heat transfer coefficient\n",
    "Re = (d3**2*(rpm/60)*rho)/mu      \t\t\t#reynold no.\n",
    "Pr = ((cp*3600)*(mu))/k\n",
    "#from eq. 8.66\n",
    "y = 1                            \t\t\t#x = mu/muw = 1\n",
    "Nu = 0.74*(Re**(0.67))*(Pr**(0.33))*(y**(0.14))        \t\t\t#Nusslet no\n",
    "hi = Nu*(k/d2)                                      \t\t\t#heat transfer coefficient\n",
    "\n",
    "#coil side heat transfer coefficient\n",
    "v = 1.5                \t\t\t#m/s, linear velocity of fluid\n",
    "fa = ((math.pi/4)*d1**2)     \t\t\t#m2, flow area of coil\n",
    "fr = v*fa*3600           \t\t\t#m3/h , flow rate of the fluid\n",
    "Wc = fr*rho             \t\t\t#kg/h , flow rate\n",
    "dh = (4*(math.pi/8)*d1**2)/(d1+(math.pi/2)*d1)     \t\t\t#m,hydrolic diameter of limpet coil\n",
    "Re1 = v*rho1*dh/mu1                        \t\t\t#coil reynold no.\n",
    "Pr1 = cp1*mu1*3600/k1                      \t\t\t#prandtl no. of the coil fluid\n",
    "#from eq. 8.68\n",
    "d4 = 0.0321                               \t\t\t#m, inside diameter of the tube\n",
    "Nu1 = 0.021*(Re1**(0.85)*Pr1**(0.4)*(d4/d2)**(0.1)*y**0.14)        \n",
    "hc = Nu1*(k1/dh)                       \t\t\t#coil side coefficient\n",
    "\n",
    "U = 1/((1/hi)+(ai/(hc*ao))+Rdi+Rdc)     \t\t\t#overall heat transfer corfficient\n",
    "#from eq. 8.63\n",
    "beeta = math.exp(U*ai/(Wc*cp1))\n",
    "Wv = 2200.                              \t\t\t#kg, mass of fluid vessel\n",
    "t = (beeta/(beeta-1))*((Wv*cp)/(Wc*cp1))*math.log((Tci-Tvi)/(Tci-Tvf))      \n",
    "\n",
    "# Results\n",
    "print \"the time required to heat the charge %.0f min\"%(t*60)\n"
   ]
  }
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