path: root/Heat_Transfer:_Principles_And_Applications_by_B._K._Dutta/ch3_2.ipynb

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   "cells": [
    {
     "cell_type": "heading",
     "level": 1,
     "metadata": {},
     "source": [
      "Chapter 3 : Heat transfer coefficient"
     ]
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 3.1 Page No : 53"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "# Variables\n",
      "di = 0.06        \t\t\t#m,initial diameter of iceball\n",
      "T1 = 30.          \t\t\t#C, room temp.\n",
      "T2 = 0.           \t\t\t#ice ball temp.\n",
      "h = 11.4         \t\t\t#W/m**2 C, heat transfer coefficient\n",
      "x = 40.           \t\t\t#% for reduction\n",
      "rho = 929.        \t\t\t#kg/m**3, density of ice\n",
      "Lv = 3.35*10**5   \t\t\t#j/kg, latent heat of fusion\n",
      "\n",
      "# Calculations\n",
      "# m = 4/3*math.pi*r**3      \t\t\t#kg,mass of ice ball\n",
      "#rate of melting = -dm/dt\n",
      "#rate of heat adsorption  = -4*math.pi*r**2*rho*dr/dt*lamda\n",
      "#at initial time t = 0\n",
      "C1 = di/2        \t\t\t#consmath.tant of integration\n",
      "#if the volume of the ball is reduced by 40% of the original volume \n",
      "r = ((1-x/100)*(di/2)**3)**(1./3)\n",
      "#time required for melting umath.sing eq. 1\n",
      "t = (di/2-r)/(h*(T1-T2)/(rho*Lv))\n",
      "\n",
      "# Results\n",
      "print \"The time required for melting the ice is %.0f s\"%(t)\n",
      "\n",
      "# note : rounding off error."
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The time required for melting the ice is 4274 s\n"
       ]
      }
     ],
     "prompt_number": 1
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 3.2 Page No : 54"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "import math\n",
      "from scipy.integrate import quad \n",
      "#calculate the time required for the  heating coil.\n",
      "\n",
      "# Variables\n",
      "P = 1.*10**3           \t\t\t#W, electrical heating capacity\n",
      "V = 220.              \t\t\t#V, applied voltage\n",
      "d = 0.574*10**-3      \t\t\t#m, diameter of wire\n",
      "R = 4.167            \t\t\t#ohm, electrical resistance\n",
      "Tr = 21.              \t\t\t#C, room temp.\n",
      "h = 100.              \t\t\t#W/m**2 C, heat transfer coefficient\n",
      "rho = 8920.           \t\t\t#kg/m**3, density of wire\n",
      "cp = 384.             \t\t\t#j/kg C, specific heat of wire\n",
      "percent = 63.         \t\t\t#%, percent of the steady state\n",
      "\n",
      "#Calculation\n",
      "R_ = V**2/P           \t\t\t#ohm, total electrical resistance\n",
      "l = R_/R             \t\t\t#m, length of wire\n",
      "A = math.pi*d*l          \t\t\t#m**2, area of wire\n",
      "Tf = P/(h*A)+Tr      \t\t\t#final temp.\n",
      "dtf = Tf-Tr          \t\t\t#C. steady state temp. rise\n",
      "#temp. of wire after 63% rise\n",
      "T = Tr+(percent/100)*dtf   \n",
      "#rate of heat accumulation on the wire\n",
      "#d/dt(m*cp*T)                       (1)\n",
      "#rate of heat loss\n",
      "#h*A*(T-Tr).........................(2)\n",
      "#heat balance eq.       (1) = (2)\n",
      "m = math.pi*d**2*l*rho/4  \t\t\t#kg. mass of wire\n",
      "#integrating heat balance eq.\n",
      "\n",
      "def f6(T): \n",
      "    return 1/((P/(m*cp))-((h*A)/(m*cp))*(T-Tr))\n",
      "\n",
      "t =  quad(f6,21,322)[0]\n",
      "\n",
      "# Results\n",
      "print \"The time required for the heating coil is %.1f s\"%(t)\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The time required for the heating coil is 4.9 s\n"
       ]
      }
     ],
     "prompt_number": 2
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 3.3 Page No : 56"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "# Variables\n",
      "t = 0.2           \t\t\t#m, thickness of wall\n",
      "k = 1.163         \t\t\t#W/m C, thermal conductivity of material\n",
      "Ta = 30           \t\t\t#C, ambient temp\n",
      "\n",
      "# Calculations and Results\n",
      "#(a) at x = 0.2   let T = T1 at x = x1\n",
      "x1 = 0.2\n",
      "T1 = 250-2750*x1**2\n",
      "#let     D = dT/dx\n",
      "D = -5500*0.2     \t\t\t#C/m, at x = 0.2\n",
      "h = -k*D/(T1-Ta)\n",
      "print \" the heat transfer coefficient is %.2f W/m**2 C \"%(h)\n",
      "\n",
      "#(b)at other surface of wall, x = 0 = x2 (say)\n",
      "x2 = 0\n",
      "a = -5500*0\n",
      "print \"So there is no heat flow at other surface of the wall \"\n",
      "\n",
      "#(c)\n",
      "A = 1            \t\t\t#m**2, area\n",
      "Vw = A*x1        \t\t\t#m**3, volume of wall\n",
      "HL = h*(T1-Ta)   \t\t\t#W, heat loss from unit area\n",
      "Vav = HL/x1\n",
      "print \"average volumetric rate of heat generation is %.0f W/m**3\"%(Vav)\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        " the heat transfer coefficient is 11.63 W/m**2 C \n",
        "So there is no heat flow at other surface of the wall \n",
        "average volumetric rate of heat generation is 6396 W/m**3\n"
       ]
      }
     ],
     "prompt_number": 3
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 3.4 Page No : 61"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "from scipy.optimize import fsolve \n",
      "import math\n",
      "# Variables\n",
      "id_ = 97.*10**-3         \t\t\t#m,internal diameter of steam pipe\n",
      "od = 114.*10**-3        \t\t\t#m,outer diameter of steam pipe\n",
      "pr = 30.               \t\t\t#bar, absolute pressure os saturated steam\n",
      "Ti = 234.                \t\t\t#C, temp. at 30 bar absolute pressure\n",
      "Ts = 55.               \t\t\t#C, skin temp.\n",
      "To = 30.               \t\t\t#C, ambient temp.\n",
      "kc = 0.1              \t\t\t#W/m C, thermal conductivity of wool\n",
      "kw = 43.               \t\t\t#W/m C, thermal conductivity of pipe\n",
      "h = 8.                 \t\t\t#W/m**2 C, external air film coefficient \n",
      "L = 1.                 \t\t\t#m, assume length\n",
      "\n",
      "#Calculation\n",
      "ri = id_/2             \t\t\t#m, \n",
      "r1 = (114.*10**-3)/2        \t\t\t#m,outer radius of steam pipe\n",
      "\n",
      "#thermal resistance of insulation\n",
      "#Ri = math.log(ro/r1)/(2*math.pi*L*kc)\n",
      "#Thermal resistance of pipe wall\n",
      "Rp = math.log(r1/ri)/(2*math.pi*L*kw)\n",
      "#RT = Ri+Rp\n",
      "DF = Ti-Ts            \t\t\t#C, driving force\n",
      "#At steady state the rate of heat flow through the insulation\n",
      "# and the outer air film are equ\n",
      "\n",
      "#by trial and error method :\n",
      "def f(ro): \n",
      "    return (Ti-Ts)/(math.log(ro/r1)/kc+math.log(r1/ri)/kw)-(h*ro*(Ts-To))\n",
      "ro = fsolve(f,0.1)\n",
      "th = ro-r1        \t\t\t#m, required thickness of insulation\n",
      "Q = 2*math.pi*ro*h*L*(Ts-To)\n",
      "\n",
      "# Results\n",
      "print \"The rate of heat loss is %.1f W\"%(Q)\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The rate of heat loss is 150.9 W\n"
       ]
      }
     ],
     "prompt_number": 4
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 3.5 Page No : 62"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "import math\n",
      "# Variables\n",
      "w1 = 8.            \t\t\t#%, solubility of alcohol\n",
      "w2 = 92.           \t\t\t#%, solubility of water\n",
      "k1 = 0.155        \t\t\t#W/m C, thermal conductivity of alcohol\n",
      "k2 = 0.67         \t\t\t#W/m C thermal conductivity of water\n",
      "ka = 0.0263       \t\t\t#W/m C thermal conductivity of air\n",
      "kw = 45.           \t\t\t#W/m Cthermal conductivity of pipe wall\n",
      "ki = 0.068        \t\t\t#W/m C , thermal cond. of glass\n",
      "id_ = 53.*10**-3     \t\t\t#m, internal diameter of pipe\n",
      "od = 60.*10**-3     \t\t\t#m, outer  diameter of pipe\n",
      "t = 0.04          \t\t\t#m, thickness of insulation\n",
      "hi = 800.          \t\t\t#W/m**2 C, liquid film coefficient\n",
      "ho = 10.           \t\t\t#W/m**2 C, air film coefficient\n",
      "L = 1.             \t\t\t#m, length of pipe\n",
      "T1 = 75.           \t\t\t#C, initial temp.\n",
      "T2 = 28.           \t\t\t#C, ambient air temp.\n",
      "\n",
      "# Calculations and Results\n",
      "#(a)\n",
      "km = (w1/100)*k1+(w2/100)*k2-0.72*(w1/100)*(w2/100)*(-(k1-k2))\n",
      "deli = km/hi     \t\t\t#m, effective thickness of liquid film\n",
      "delo = ka/ho     \t\t\t#m, effective thickness of air film\n",
      "print \"effective thickness of air  is %.2f mm\"%(deli*10**3)\n",
      "print \"effective thickness of liquid films is %.1f mm.\"%(delo*10**3)\n",
      "\n",
      "#(b)\n",
      "Ai = 2*math.pi*id_/2*L      \t\t\t#m**2, inside area\n",
      "ri = id_/2                    \t\t\t#m,inside radius of pipe\n",
      "r_ = od/2                      \t\t\t#m, outside radius of pipe\n",
      "ro = r_+t              \t        \t\t#m, outer radius of insulation\n",
      "Ao = 2*math.pi*ro*L        \t\t    \t#m**2, outer area\n",
      "#from eq. 3.11, overall heat transfer coefficient\n",
      "Ui = 1/(1/hi+(Ai*math.log(r_/ri))/(2*math.pi*L*kw)+(Ai*math.log(ro/r_))/(2*math.pi*L*ki)+Ai/(Ao*ho))\n",
      "print \"the overall heat transfer coefficient based on i.d of pipe is %.3f W/m**2 C\"%(Ui)\n",
      "\n",
      "#(c)\n",
      "#frim eq. 3.14\n",
      "Uo = Ui*Ai/Ao  \n",
      "print \"the overall heat transfer coefficient based on od of pipe is %.3f W/m**2 C\"%(Uo)\n",
      "\n",
      "#(d)\n",
      "R = 1/(Ui*Ai)          \t\t\t#C/W, total heat transfer resistance\n",
      "Rair = 1/(Ao*ho)       \t\t\t#C/W, heat transfer resistance of air film\n",
      "p = Rair/R\n",
      "print \"the percentage of total resistance  offered by air film. is %.2f percent\"%(p*100)\n",
      "\n",
      "#(e)\n",
      "Q = Ui*Ai*(T1-T2)\n",
      "print \"Rate of heat loss is %.1f W\"%(Q)\n",
      "\n",
      "#(f)\n",
      "Ts = Uo*Ao*(T1-T2)/(ho*Ao)+T2\n",
      "print \"insulation skin temp.is %.1f C\"%(Ts)\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "effective thickness of air  is 0.75 mm\n",
        "effective thickness of liquid films is 2.6 mm.\n",
        "the overall heat transfer coefficient based on i.d of pipe is 2.707 W/m**2 C\n",
        "the overall heat transfer coefficient based on od of pipe is 1.025 W/m**2 C\n",
        "the percentage of total resistance  offered by air film. is 10.25 percent\n",
        "Rate of heat loss is 21.2 W\n",
        "insulation skin temp.is 32.8 C\n"
       ]
      }
     ],
     "prompt_number": 11
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 3.6 Page No : 64"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "import math\n",
      "# Variables\n",
      "id_ = 1.5                \t\t\t#m, internal diameter of math.tank\n",
      "h = 2.5                 \t\t\t#m, height of math.tank\n",
      "t1 = 0.006              \t\t\t#m, thickness of wall\n",
      "t2 = 0.04               \t\t\t#m, thickness of insulation\n",
      "Ta = 25.                 \t\t\t#C, ambient temp.\n",
      "T1 = 80.                 \t\t\t#C, outlet temp. of liquid\n",
      "cp = 2000.               \t\t\t#j/kg C, specific heat of liquid\n",
      "FR = 700./3600           \t\t\t#KG/s, Liquid flow rate\n",
      "\n",
      "# Calculations and Results\n",
      "ri = id_/2+t1            \t\t\t#m, inner radius of insulation\n",
      "ro = ri+t2              \t\t\t#m, outer radius of insulation\n",
      "ki = 0.05               \t\t\t#W/m C, thermal conductivity of insulation\n",
      "hc = 4                  \t\t\t#W/m**2 C, heat transfer coefficient at cylindrical surface\n",
      "ht = 5.5                \t\t\t#W/m**2 C, heat transfer coefficient at flat surface\n",
      "l = h+t1+t2             \t\t\t#m, height of the top of insulation\n",
      "#fromm eq. 3.10\n",
      "#heat transfer resistance of cylindrical wall\n",
      "Rc = math.log(ro/ri)/(2*math.pi*l*ki)+1/(2*math.pi*ro*l*hc)\n",
      "#heat transfer resistance of flat insulated top surface\n",
      "Ri = (1/(math.pi*ro**2))*((ro-ri)/ki+1/ht)\n",
      "tdf = T1-Ta             \t\t\t#C, temp. driving force\n",
      "Q = tdf/Rc + tdf/Ri       \t\t\t#W, total rate of heat loss\n",
      "Tt = Q/(FR*cp)+T1        \t\t\t#C, inlet temp. of liquid\n",
      "print \"Inlet liquid temp. should be %.0f C \"%(Tt)\n",
      "Q1 = tdf/Ri   \t\t\t#W, rate of heat loss from flat surface\n",
      "T1 = Q1/(math.pi*ro**2*ht)+Ta    \n",
      "print \" the insulation skin temp. at the flat top surface is %.0f C \"%(T1)\n",
      "#similarly\n",
      "T2 = 38\n",
      "print \"similarly  the insulation skin temp at cylindrical surface is %.0f C\"%(T2)\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Inlet liquid temp. should be 82 C \n",
        " the insulation skin temp. at the flat top surface is 35 C \n",
        "similarly  the insulation skin temp at cylindrical surface is 38 C\n"
       ]
      }
     ],
     "prompt_number": 6
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 3.7 Page No : 66"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "import math\n",
      "# Variables\n",
      "id_ = 2.5*10**-2              \t\t\t#m, internal diameter of glass tube\n",
      "t = 0.3*10**-2               \t\t\t#m, thickness of wall\n",
      "l = 2.5                     \t\t\t#m, length of nichrome wire\n",
      "L = 0.12                    \t\t\t#m, length of steel covered with heating coil\n",
      "Re = 16.7                   \t\t\t#ohm, electrical resistance\n",
      "ti = 2.5*10**-2              \t\t\t#m, thickness of layer of insulation\n",
      "kg = 1.4                    \t\t\t#W/m C, thermal conductivity of glass\n",
      "ki = 0.041                  \t\t\t#W/m C, thermal conductivity of insulation\n",
      "T1 = 91.                     \t\t\t#C, boiling temp. of liquid\n",
      "T2 = 27.                     \t\t\t#C, ambient temp.\n",
      "ho = 5.8                    \t\t\t#W/m **2 C outside air film coefficient\n",
      "V = 90.                      \t\t\t#V,  voltage\n",
      "\n",
      "#Calculation\n",
      "Rc = Re*l                   \t\t\t#ohm, resistance of heating coil\n",
      "Q = V**2/Rc                  \t\t\t#W, rate of heat generation\n",
      "ri = id_/2                   \t\t\t#m, inner radius of glass tube\n",
      "r_ = ri+t                   \t\t\t#m, outer radius of glass tube\n",
      "ro = r_+ti                   \t\t\t#m,outer radius of insulation\n",
      "#heat transfer resistance of glass wall\n",
      "Rg = math.log(r_/ri)/(2*math.pi*L*kg)\n",
      "#combined resistance of insulation and outer air film\n",
      "Rt = math.log(ro/r_)/(2*math.pi*L*ki)+1/(2*math.pi*ro*L*ho)\n",
      "#Rate of heat input to the boiling liquid in steel = Q1 = (Ts-T1)/Rg\n",
      "#Rate of heat loss through insulation ,Q2 = (Ts-To)/(Rt)\n",
      "#Q1+Q2 = Q\n",
      "Ts = (Q+ T1/Rg +T2/Rt)/(1/Rg +1/Rt)\n",
      "Q1 = (Ts-T1)/Rg\n",
      "Q2 = Q-Q1\n",
      "\n",
      "# Results\n",
      "print \"the heat imput to the boiling.is %.1f W\"%(Q1)\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "the heat imput to the boiling.is 191.2 W\n"
       ]
      }
     ],
     "prompt_number": 7
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 3.8 Page No : 68"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "import math\n",
      "\n",
      "# Variables\n",
      "ri = 1.3*10**-3            \t\t\t#m, radius of 10 gauge wire\n",
      "t = 1.3*10**-3             \t\t\t#m, thickness of rubber insulation\n",
      "Ti = 90.                   \t\t\t#C, temp. 0f insulation\n",
      "To = 30.                   \t\t\t#C, ambient temp.\n",
      "h = 15.                    \t\t\t#W/m**2 C, air film coefficient\n",
      "km = 380.                  \t\t\t#W/m C, thermal cond. of copper\n",
      "kc = 0.14                 \t\t\t#W/m C, thermal cond. of rubber(insulation)\n",
      "Rc = 0.422/100            \t\t\t#ohm/m, eletrical resistance of copper wire\n",
      "\n",
      "# Calculations and Results\n",
      "Tcmax = 90.                \t\t\t#X, the maximum temp. in insulation\n",
      "ro = ri+t                 \t\t\t#m, outside radius of 10 gauge wire\n",
      "Sv = ((Tcmax-To)*(2*kc/ri**2))/(math.log(ro/ri)+kc/(h*ro))\n",
      "I = (math.pi*ri**2*Sv/Rc)**0.5      \t\t\t#A, Current strength\n",
      "print \"maximum allowable current is %.2f A\"%(I)\n",
      "\n",
      "#(b) at r = 0\n",
      "Tm = To+(ri**2*Sv/2)*(1/km+(math.log(ro/ri))/kc+1/(h*ro))\n",
      "print \"remp. at the centre of wire is %.3f C\"%(Tm)\n",
      "\n",
      "#at r = ro\n",
      "Tc = 30+(ri**2*Sv/(2*kc))*(kc/(h*ro))\n",
      "print \"The temprature at the outer surface of insulation is %.1f C\"%(Tc)\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "maximum allowable current is 54.04 A\n",
        "remp. at the centre of wire is 90.005 C\n",
        "The temprature at the outer surface of insulation is 80.3 C\n"
       ]
      }
     ],
     "prompt_number": 2
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 3.9 Page No : 72"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "# Variables\n",
      "tA = 0.25          \t\t\t#m, thickness of slab A\n",
      "tB = 0.1           \t\t\t#m, thickness of slab B\n",
      "tC = 0.15          \t\t\t#m, thickness of slab C\n",
      "kA = 15.            \t\t\t#W/m C, thermal comductivity of slab A\n",
      "kB = 10.            \t\t\t#W/m C, thermal comductivity of slab B\n",
      "kC = 30.            \t\t\t#W/m C, thermal comductivity of slab C\n",
      "#Temprature distribution in slab A\n",
      "T1 = 40.            \t\t\t#C, fluid temp.\n",
      "T2 = 35.            \t\t\t#C, medium temp.\n",
      "\n",
      "# Calculations and Results\n",
      "#(a)\n",
      "x1 = tB           \n",
      "TA1 = 90.+4500*x1-11000*x1**2\n",
      "#similarly at the right surface\n",
      "x2 = tA+tB\n",
      "TA2 = 90.+4500*x2-11000*x2**2\n",
      "#let dTA/dx = D\n",
      "D = 0              \t\t\t#for maximum temp.\n",
      "x3 = 4500./22000\n",
      "TAmax = 90.+4500*x3-11000*x3**2\n",
      "print \"At x = 0.1 the temp. at the surface of slab A  is %.0f C\"%(TA1)\n",
      "print \"At x = 0.35 the temp. at the surface of slab A  is %.0f C\"%(TA2)\n",
      "print \" the maximum Temp. in A occurs at  %.4f m\"%(x3)\n",
      "print \" the maximum Temp. in A is %.1f TAmax \"%(TAmax)\n",
      "\n",
      "#(b)\n",
      "#At the interface 2\n",
      "D1 = 4500-2.*11000*x1       \t\t\t#C/W, D1 = dTA/dx, at x = 0.1\n",
      "#At the interface 3\n",
      "D2 = 4500-2.*11000*x2       \t\t\t#D12 = dTA/dx, at x = 0.35\n",
      "#Temprature gradient in slab B and C\n",
      "#by umath.sing the continuity of heat flux at interface (2)\n",
      "D3 = -kA*D1/(-kB)          \t\t\t#D3 = dTB/dx,  at x = 0.1\n",
      "#at interface (1)\n",
      "D4 = D3                    \t\t\t#D4 = dTB/dx  at x = 0\n",
      "#similarly \n",
      "D5 = -1600.                 \t\t\t#C/W, dTB/dx, x = 0.35\n",
      "D6 = D5                    \t\t\t#at interface 4\n",
      "print \"temp. gradient at interface 2 of the slabs A is %.0f C/W\"%(D1)\n",
      "print \"temp. gradient at interface 3 of the slabs A is %.0f C/W\"%(D2)\n",
      "print \"temp. gradient at interface 2 of the slabs B is %.0f C/W\"%(D3)\n",
      "print \"temp. gradient at interface 1 of the slabs B is %.0f C/W\"%(D4)\n",
      "print \"temp. gradient at interface 3 of the slabs C is %.0f C/W\"%(D5)\n",
      "print \"temp. gradient at interface 4 of the slabs C is %.0f C/W\"%(D6)\n",
      "\n",
      "#(c)\n",
      "#from D3 = 3450  and TB = beeta1*x+beeta2\n",
      "beeta1 = 3450.\n",
      "beeta2 = 85.\n",
      "x = 0.\n",
      "TB = beeta1*x+beeta2\n",
      "#similary\n",
      "TC = 877.5-1600*x\n",
      "h1 = -kB*D4/(T1-TB)\n",
      "#similarly\n",
      "h2 = 1129.\n",
      "print \"The  heat transfer coefficient at one surface of solid fluid interface is %.1f W/m**2 C\"%(h1)\n",
      "print \"The  heat transfer coefficient at other surface of solid fluid interface is %.0f W/m**2 C\"%(h2)\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "At x = 0.1 the temp. at the surface of slab A  is 430 C\n",
        "At x = 0.35 the temp. at the surface of slab A  is 318 C\n",
        " the maximum Temp. in A occurs at  0.2045 m\n",
        " the maximum Temp. in A is 550.2 TAmax \n",
        "temp. gradient at interface 2 of the slabs A is 2300 C/W\n",
        "temp. gradient at interface 3 of the slabs A is -3200 C/W\n",
        "temp. gradient at interface 2 of the slabs B is 3450 C/W\n",
        "temp. gradient at interface 1 of the slabs B is 3450 C/W\n",
        "temp. gradient at interface 3 of the slabs C is -1600 C/W\n",
        "temp. gradient at interface 4 of the slabs C is -1600 C/W\n",
        "The  heat transfer coefficient at one surface of solid fluid interface is 766.7 W/m**2 C\n",
        "The  heat transfer coefficient at other surface of solid fluid interface is 1129 W/m**2 C\n"
       ]
      }
     ],
     "prompt_number": 12
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 3.10 Page No : 79"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "import math\n",
      "\n",
      "# Variables\n",
      "id_ = 78.*10**-3       \t\t\t#m, actual internal dia of pipe\n",
      "tw = 5.5*10**-3      \t\t\t#m, wall thickness\n",
      "nl = 8.              \t\t\t#no. of longitudinal fins\n",
      "tf = 1.5*10**-3      \t\t\t#m, thickness of fin\n",
      "w = 30.*10**-3        \t\t\t#m,breadth of fin\n",
      "kf = 45.             \t\t\t#W/m C, thermal conductivity of fin \n",
      "Tw = 150.            \t\t\t#C, wall temp.\n",
      "To = 28.             \t\t\t#C, ambient temp.\n",
      "h = 75.              \t\t\t#W/m**2C, surface heat transfer coefficient\n",
      "\n",
      "#Calculation\n",
      "#from eq. 3.27\n",
      "e = math.sqrt(2*h/(kf*tf))    \n",
      "n = (1./(e*w))*math.tanh(e*w)  \t\t\t#efficiency of fin\n",
      "L = 1.              \t\t\t#m, length of fin\n",
      "Af = 2.*L*w         \t\t\t#m**2, area of  math.single fin\n",
      "Atf = nl*Af          \t\t\t#m**2 total area of fin\n",
      "Qmax = h*Atf*(Tw-To)   \t\t\t#W, maximum rate of heat transfer\n",
      "Qa = n*Qmax           \t\t\t#W, actual rate of heat transfer\n",
      "Afw = L*tf            \t\t\t#m**2, area of contact of fin with pipe wall\n",
      "Atfw = Afw*nl         \t\t\t#m**2 , area of contact of all fin with pipe wall\n",
      "ro = id_/2+tw          \t\t\t#m, outer  pipe radius\n",
      "A = 2*math.pi*L*ro        \t\t\t#m**2  area per meter\n",
      "Afree = A-Atfw        \t\t\t#m**2, free outside area of finned pipe\n",
      "#Rate of heat transfer from free area of pipe wall\n",
      "Q1 = h*Afree*(Tw-To)  \t\t\t#W, \n",
      "#total rate of hewat gtransfer from finned pipe\n",
      "Qtotal = Qa+Q1        \t\t\t#W\n",
      "#Rate of heat transfer fromm unfinned pipe\n",
      "Q2 = h*A*(Tw-To)\n",
      "per = (Qtotal-Q2)/Q2\n",
      "\n",
      "# Results\n",
      "print \"the percentage increase in the rate of heat transfer is %.1f percent \"%(per*100)\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "the percentage increase in the rate of heat transfer is 103.6 percent \n"
       ]
      }
     ],
     "prompt_number": 13
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 3.11 Page No : 80"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "import math\n",
      "\n",
      "# Variables\n",
      "id_ = 90.*10**-2       \t\t\t#m, internal diameter of steel\n",
      "od = 110.*10**-2      \t\t\t#m, outer diameter of steel\n",
      "Ti = 180.            \t\t\t#C, inside temp. of steel\n",
      "To = 170.            \t\t\t#C, outside temp. of steel\n",
      "k = 37.             \t\t\t#W/m C, thermal conductivity of alloy\n",
      "Q = 5.18*10**3       \t\t\t#W, Rate of heat loss\n",
      "\n",
      "# Calculations and Results\n",
      "ri = id_/2           \t\t\t#m, inside radius of shell\n",
      "ro = od/2           \t\t\t#m, outside radius of shell\n",
      "r_ = 0.5            \t\t\t#m, boundary between the layers\n",
      "L = 1               \t\t\t#m, length of shell\n",
      "#Rate of heat transfer in the absence of contact resistance\n",
      "Q1 = 2*math.pi*L*k*(Ti-To)/(math.log(ro/ri))             \n",
      "print \"Rate of heat transfer in the absence of contact resistance is %.3f KW\"%(Q1/1000)\n",
      "print \"The actual rate of heat loss is 5.18kW is much less than this value\\\n",
      ". So there is a thermal contact resistance at the interface between the layers \"\n",
      "\n",
      "#(b)\n",
      "Ri = (math.log(r_/ri)/(2*math.pi*L*k))  \t\t\t#C/W, resistance of inner layer\n",
      "Ro = (math.log(ro/r_)/(2*math.pi*L*k))  \t\t\t#C/W, resistance of outer layer\n",
      "Rc = ((Ti-To)/(Q))-(Ri+Ro)     \t\t\t#C/W, contact resistance\n",
      "print \"The contact resistance is %f C/W \"%(Rc)\n",
      "Ac = 2*math.pi*L*r_                \t\t\t#m**2, area of contact surface of shell\n",
      "hc = 1/(Ac*Rc)                 \t\t\t    #W/m**2 c, contact heat transfer coefficient\n",
      "print \"contact heat transfer coefficient is %.1f W/m**2 C \"%(hc)\n",
      "\n",
      "#(c)\n",
      "dt = Q/(hc*Ac)\n",
      "print \"The temprature jump is %.1f C\"%(dt)\n",
      "\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Rate of heat transfer in the absence of contact resistance is 11.585 KW\n",
        "The actual rate of heat loss is 5.18kW is much less than this value. So there is a thermal contact resistance at the interface between the layers \n",
        "The contact resistance is 0.001067 C/W \n",
        "contact heat transfer coefficient is 298.2 W/m**2 C \n",
        "The temprature jump is 5.5 C\n"
       ]
      }
     ],
     "prompt_number": 14
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 3.12 Page No : 84"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "# Variables\n",
      "d = 5.2*10**-3       \t\t\t#m, diameter of copper wire\n",
      "ri = d/2            \t\t\t#inner radius of insulation\n",
      "kc = 0.43           \t\t\t#W/m C, thermal conductivity of PVC\n",
      "Tw = 60.             \t\t\t#C, temp. 0f wire\n",
      "h = 11.35           \t\t\t#W/m**2 C, film coefficient\n",
      "To = 21.             \t\t\t#C, ambient temp.\n",
      "\n",
      "#calculation\n",
      "Ro = kc/h           \t\t\t#m,critical outer radius of insulation\n",
      "t = Ro-ri\n",
      "\n",
      "# Results\n",
      "print \"the critical thickness is %.2f mm\"%(t*10**3)\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "the critical thickness is 35.29 mm\n"
       ]
      }
     ],
     "prompt_number": 6
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 3.13 Page No : 85"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "# calculate the critical  insulation thickness.\n",
      "\n",
      "# Variables\n",
      "d = 15.*10**-2        \t\t\t#m, length of steam main\n",
      "t = 10.*10**-2        \t\t\t#m, thickness  of insulation\n",
      "ki = 0.035          \t\t\t#W/m C, thermal conductivity of insulation\n",
      "h = 10.              \t\t\t#W/m**2 C, heat transfer coefficient\n",
      "\n",
      "#calculation\n",
      "#from eq. 3.29\n",
      "ro = ki/h\n",
      "\n",
      "# Results\n",
      "print \"ro =  %.1f cm \"%(ro*10**3)\n",
      "print \"Radius of bare pipe is larger than outer radius of insulation  So critical \\\n",
      "  insulation thickness does not exist \"\n",
      "\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "ro =  3.5 cm \n",
        "Radius of bare pipe is larger than outer radius of insulation  So critical   insulation thickness does not exist \n"
       ]
      }
     ],
     "prompt_number": 7
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 3.14 Page No : 87"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "from scipy.optimize import fsolve\n",
      "import math\n",
      "\n",
      "# Variables\n",
      "Ti = 172.           \t\t\t#C, saturation temp.\n",
      "To = 20.            \t\t\t#C, ambient temp.\n",
      "Cs = 700.           \t\t\t#per ton, math.cost of steam\n",
      "Lv = 487.           \t\t\t#kcal/kg, latent heat of steam\n",
      "ho = 10.32           \t\t\t#kcal/h m**2 C, outer heat transfer coefficient\n",
      "kc = 0.031             \t\t\t#W/m C, thermal conductivity of insulation\n",
      "n = 5.              \t\t\t#yr, service life of insulation\n",
      "i = 0.18            \t\t\t#Re/(yr)(Re), interest rate\n",
      "\n",
      "#Calculation\n",
      "di = 0.168           \t\t\t#m, inner diameter of insulation\n",
      "#Cost of insulation\n",
      "Ci = 17360.-(1.91*10**4)*di         \t\t\t#Rs/m**3\n",
      "Ch = Cs/(1000*Lv)                 \t\t\t#Rs/cal, math.cost of heat energy in steam\n",
      "sm = 1./(1+i)+1/(1+i)**2+1/(1+i)**3+1/(1+i)**4+1/(1+i)**n\n",
      "#from eq. 3.33\n",
      "ri = di/2         \t\t\t#m  inner radius of insulation\n",
      "L = 1             \t\t\t#m, length of pipe\n",
      "#Pt = Ch*sm*2*math.pi*ri*L*( 1/(((ri/kc)*('math.log(ro/ri)'))+ri/(ho*ro)))*7.2*10**3*(Ti-To)+math.pi*(ro**2-ri**2)*L*Ci\n",
      "#On differentiating , dpt/dro = -957.7*((1/ro)-(0.003/ro**2))/(math.log(ro)+(0.003/ro)+2.477)**2\n",
      "def f(ro): \n",
      "    return -957.7*((1/ro)-(0.003/ro**2))/(math.log(ro)+(0.003/ro)+2.477)**2+98960*ro\n",
      "ro = fsolve(f,0.1)\n",
      "t = ro-ri\n",
      "\n",
      "# Results\n",
      "print \"The optimum insulation thickness is %.0f mm\"%(t*1000)\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The optimum insulation thickness is 71 mm\n"
       ]
      }
     ],
     "prompt_number": 1
    }
   ],
   "metadata": {}
  }
 ]
}