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+{

+ "metadata": {

+  "name": "CHAPTER2"

+ },

+ "nbformat": 3,

+ "nbformat_minor": 0,

+ "worksheets": [

+  {

+   "cells": [

+    {

+     "cell_type": "heading",

+     "level": 1,

+     "metadata": {},

+     "source": [

+      "chapter 2 :Steady state Conduction in one dimension"

+     ]

+    },

+    {

+     "cell_type": "heading",

+     "level": 3,

+     "metadata": {},

+     "source": [

+      "Example 2.1 Page No.42"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "#determination of the heat flow through a composite wall\n",

+      "\n",

+      "#Given\n",

+      "T3=-10.00\t\t# temperature of inside wall in degree Fahrenheit\n",

+      "T0=70.0 \t  \t# temperature of outside wall in degree Fahrenheit\n",

+      "dT=T0-T3          # overall temperature difference\n",

+      "# values of thermal conductivity in BTU/(hr.ft.degree Rankine) from appendix table B3\n",

+      "k1=0.38           # brick masonry\n",

+      "k2=0.02           # glass fibre\n",

+      "k3=0.063          # plywood\n",

+      "dx1=4/12.0          # thickness of brick layer in ft\n",

+      "dx2=3.5/12.0        # thickness of glass fibre layer in ft\n",

+      "dx3=0.5/12.0        # thickness of plywood layer in ft\n",

+      "A=1.0 # cross sectional area taken as 1 ft**2\n",

+      "\n",

+      "#Calculation\n",

+      "R1=dx1/(k1*A) # resistance of brick layer in (hr.degree Rankine)/BTU\n",

+      "R2=dx2/(k2*A) # resistance of glass fibre layer in (hr.degree Rankine)/BTU\n",

+      "R3=dx3/(k3*A) # resistance of plywood layer in (hr.degree Rankine)/BTU\n",

+      "qx=(T0-T3)/(R1+R2+R3) \n",

+      "\n",

+      "#Result\n",

+      "print\"Resistance of brick layer is  \",round(R1,3),\"(hr.degree Rankine)/BTU\"\n",

+      "print\"Resistance of glass fibre layer is \",round(R2,1),\"(hr.degree Rankine)/BTU\"\n",

+      "print\"Resistance of plywood layer is  \",round(R3,3),\"(hr.degree Rankine)/BTU\"\n",

+      "print\"Heat transfer through the composite wall is \",round(qx,2),\"(hr.degree Rankine)/BTU\"\n"

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "Resistance of brick layer is   0.877 (hr.degree Rankine)/BTU\n",

+        "Resistance of glass fibre layer is  14.6 (hr.degree Rankine)/BTU\n",

+        "Resistance of plywood layer is   0.661 (hr.degree Rankine)/BTU\n",

+        "Heat transfer through the composite wall is  4.96 (hr.degree Rankine)/BTU\n"

+       ]

+      }

+     ],

+     "prompt_number": 10

+    },

+    {

+     "cell_type": "heading",

+     "level": 3,

+     "metadata": {},

+     "source": [

+      "Example 2.2 Page No.45"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "# determination of heat transfer through composite wall for materials in parallel\n",

+      "\n",

+      "#Given\n",

+      "# values of thermal conductivities in W/(m.K) from appendix table B3\n",

+      "k1=0.45           # thermal conductivity of brick\n",

+      "k2a=0.15          # thermal conductivity of pine\n",

+      "k3=0.814         # thermal conductivity of plaster board\n",

+      "k2b=0.025       # thermal conductivity of air from appendix table D1\n",

+      "# Areas needed fpor evaluating heat transfer in sq.m\n",

+      "A1=0.41*3       # cross sectional area of brick layer \n",

+      "A2a=0.038*3     # cross sectional area of wall stud\n",

+      "A2b=(41-3.8)*0.01*3 # cross sectional area of air layer\n",

+      "A3=0.41*3     # cross sectional area of plastic layer \n",

+      "dx1=0.1        # thickness of brick layer in m\n",

+      "dx2=0.089      # thickness of wall stud and air layer in m\n",

+      "dx3=0.013      # thickness of plastic layer in m\n",

+      "\n",

+      "#Calculation\n",

+      "R1=dx1/(k1*A1) # Resistance of brick layer in K/W\n",

+      "R2=dx2/(k2a*A2a+k2b*A2b) # Resistance of wall stud and air layer in K/W\n",

+      "R3=dx3/(k3*A3) # Resistance of plastic layer in K/W\n",

+      "T1=25          # temperature of inside wall in degree celsius\n",

+      "T0=0           # temperature of outside wall in degree celsius\n",

+      "qx=(T1-T0)/(R1+R2+R3) # heat transfer through the composite wall in W\n",

+      "\n",

+      "#result\n",

+      "print\"Resistance of brick layer is \",round(R1,3),\"k/W\"\n",

+      "print\"Resistance of wall stud and air layer is  \",round(R2,2),\"k/W\"\n",

+      "print\"Resistance of plastic layer is \",round(R3,3),\"k/W\"\n",

+      "print\"Heat transfer through the composite wall is\",round(qx,1),\"W\"\n"

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "Resistance of brick layer is  0.181 k/W\n",

+        "Resistance of wall stud and air layer is   1.98 k/W\n",

+        "Resistance of plastic layer is  0.013 k/W\n",

+        "Heat transfer through the composite wall is 11.5 W\n"

+       ]

+      }

+     ],

+     "prompt_number": 15

+    },

+    {

+     "cell_type": "heading",

+     "level": 3,

+     "metadata": {},

+     "source": [

+      "Example 2.3 Page No. 50"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "# determination of heat transfer rate and overall heat transfer coefficient\n",

+      "\n",

+      "#Given\n",

+      "k1=24.8      # thermal conductivity of 1C steel in BTU/(hr.ft.degree Rankine)from appendix table B2 \n",

+      "k2=0.02      # thermal conductivity of styrofoam steel in BTU/(hr.ft.degree Rankine)\n",

+      "k3=0.09      # thermal conductivity of fibreglass in BTU/(hr.ft.degree Rankine)\n",

+      "hc1=0.79     # convection coefficient between the air and the vertical steel wall in BTU/(hr.ft**2.degree Rankine)\n",

+      "hc2=150.0      # the convection coefficient between the ice water and the fiberglass\n",

+      "A=1.0          # calculation based on per square foot\n",

+      "dx1=0.04/12.0  #  thickness of steel in ft\n",

+      "dx2=0.75/12.0  # thickness of styrofoam in ft\n",

+      "dx3=0.25/12.0  # thickness of fiberglass in ft\n",

+      "\n",

+      "#Calculation \n",

+      "\n",

+      "# Resistances in (degree Fahrenheit.hr)/BTU\n",

+      "Rc1=1/(hc1*A) # Resistance from air to sheet metal\n",

+      "Rk1=dx1/(k1*A) # Resistance of steel layer\n",

+      "Rk2=dx2/(k2*A) # Resistance of styrofoam layer\n",

+      "Rk3=dx3/(k3*A) # Resistance of fiberglass layer\n",

+      "Rc2=1/(hc2*A) # Resistance from ice water to fiberglass\n",

+      "U=1/(Rc1+Rk1+Rk2+Rk3+Rc2) # overall heat transfer coefficient \n",

+      "T_inf1=90     # temperature of air in degree F\n",

+      "T_inf2=32    # temperature of mixture of ice and water in degree F\n",

+      "q=U*A*(T_inf1-T_inf2)\n",

+      "\n",

+      "#result\n",

+      "print\"Resistance from air to sheet metal:  \",round(Rc1,3),\"degree F.hr/BTU\"\n",

+      "print\"Resistance of steel layer is \",round(Rk1,4),\"degree F.hr/BTU\"\n",

+      "print\"Resistance of styrofoam layer is \",round(Rk2,3),\"degree F.hr/BTU\"\n",

+      "print\"Resistance of fiberglass layer is \",round(Rk3,3),\"degree F.hr/BTU\"\n",

+      "print\"Resistance from ice water to fiberglass is \",round(Rc2,4),\"degree F.hr/BTU\"\n",

+      "print\"The overall heat transfer coefficient is  \",round(U,3),\"BTU/hr ft**2\"\n",

+      "print\"The heat transfer rate is %.1f BTU/hr\",round(q,2),\"BTU/hr\"\n",

+      "\n"

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "Resistance from air to sheet metal:   1.266 degree F.hr/BTU\n",

+        "Resistance of steel layer is  0.0001 degree F.hr/BTU\n",

+        "Resistance of styrofoam layer is  3.125 degree F.hr/BTU\n",

+        "Resistance of fiberglass layer is  0.231 degree F.hr/BTU\n",

+        "Resistance from ice water to fiberglass is  0.0067 degree F.hr/BTU\n",

+        "The overall heat transfer coefficient is   0.216 BTU/hr ft**2\n",

+        "The heat transfer rate is %.1f BTU/hr 12.53 BTU/hr\n"

+       ]

+      }

+     ],

+     "prompt_number": 21

+    },

+    {

+     "cell_type": "heading",

+     "level": 3,

+     "metadata": {},

+     "source": [

+      "Example 2.4 Page No 55."

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "# determination of the heat transfer through the pipe wall per unit length of pipe.\n",

+      "\n",

+      "#given\n",

+      "k=14.4       # thermal conductivity of 304 stainless steel in W/(m.K) from appendix table B2\n",

+      "# dimensions of steel pipes in cm from appendix table F1\n",

+      "D2=32.39     #Diameter (cm)\n",

+      "D1=29.53\n",

+      "T1=40         #Temprature\n",

+      "T2=38\n",

+      "\n",

+      "#Calculation\n",

+      "import math\n",

+      "Qr_per_length=(2*3.14*k)*(T1-T2)/math.log(D2/D1)#format(6)\n",

+      "\n",

+      "#Result\n",

+      "print\"The heat transfer through the pipe wall per unit length of pipe is \",round(Qr_per_length/1000,2),\"kw/m\"\n"

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "The heat transfer through the pipe wall per unit length of pipe is  1.96 kw/m\n"

+       ]

+      }

+     ],

+     "prompt_number": 23

+    },

+    {

+     "cell_type": "heading",

+     "level": 3,

+     "metadata": {},

+     "source": [

+      "Example 2.5 Page NO. 58"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "# determination of the heat gain per unit length\n",

+      "\n",

+      "#Given\n",

+      "k1=231             # thermal conductivity of copper in BTU/(hr.ft.degree Rankine)from appendix table B1 \n",

+      "k2=0.02            # thermal conductivity of insuLtion in BTU/(hr.ft.degree Rankine)\n",

+      "# Specifications of 1 standard type M copper tubing from appendix table F2 are as follows\n",

+      "D2=1.125/12       # outer diameter in ft\n",

+      "D1=0.08792        # inner diameter in ft\n",

+      "R2=D2/2          # outer radius\n",

+      "R1=D1/2          # inner radius\n",

+      "t=0.5/12          # wall thickness of insulation in ft\n",

+      "R3=R2+t\n",

+      "LRk1=(log(R2/R1))/(2*3.14*k1) # product of length and copper layer resistance\n",

+      "LRk2=(log(R3/R2))/(2*3.14*k2) # product of length and insulation layer resistance\n",

+      "T1=40 # temperature of inside wall of tubing in degree fahrenheit\n",

+      "T3=70 # temperature of surface temperature of insulation degree fahrenheit\n",

+      "q_per_L=(T1-T3)/(LRk1+LRk2) # heat transferred per unit length in BTU/(hr.ft)\n",

+      "\n",

+      "#Result\n",

+      "print\"The heat transferred per unit length is \",round(q_per_L,2),\" BTU/(hr.ft\"\n"

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "The heat transferred per unit length is  -5.92  BTU/(hr.ft\n"

+       ]

+      }

+     ],

+     "prompt_number": 26

+    },

+    {

+     "cell_type": "heading",

+     "level": 3,

+     "metadata": {},

+     "source": [

+      "Example 2.6 Page No 63"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "# Determination of the overall heat transfer coefficient\n",

+      "\n",

+      "#Given data\n",

+      "k12=24.8      # thermal conductivity of 1C steel in BTU/(hr.ft.degree Rankine)from appendix table B2 \n",

+      "k23=.023      # thermal conductivity of glass wool insulation in BTU/(hr.ft.degree Rankine)from appendix table B3 \n",

+      "# Specifications of 6 nominal, schedule 40 pipe (no schedule was specified, so the standard is assumed) from appendix table F1 are as follows\n",

+      "D2=6.625/12.0  # outer diameter in ft\n",

+      "D1=0.5054    # inner diameter in ft\n",

+      "t=2/12.0        # wall thickness of insulation in ft\n",

+      "D3=D2+t\n",

+      "hc1=12        # convection coefficient between the air and the pipe wall in BTU/(hr. sq.ft.degree Rankine).\n",

+      "hc2=1.5       # convection coefficient between the glass wool and the ambient air in  BTU/(hr. sq.ft.degree Rankine).\n",

+      "\n",

+      "#calculation\n",

+      "U=1/((1/hc1)+(D1*log(D2/D1)/k12)+(D1*log(D3/D2)/k23)+(D1/(hc2*D3)))\n",

+      "\n",

+      "#Result\n",

+      "print\"Overall heat transfer coefficient is  \",round(U,4),\" BTU/(hr.sq.ft. Fahrenheit)\"\n"

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "Overall heat transfer coefficient is   0.1596  BTU/(hr.sq.ft. Fahrenheit)\n"

+       ]

+      }

+     ],

+     "prompt_number": 31

+    },

+    {

+     "cell_type": "heading",

+     "level": 3,

+     "metadata": {},

+     "source": [

+      "Example 2.7 Page NO.72"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "# Determination of the thermal contact resistance\n",

+      "\n",

+      "#Given\n",

+      "k=14.4       # thermal conductivity of 304 stainless steel in  W/(m.K)from appendix table B2 \n",

+      "T1=543.0     # temperature in K at point 1\n",

+      "T2=460.0     # temperature in K at point 2\n",

+      "dT=T1-T2     # temperature difference between point 1 and 2\n",

+      "dz12=0.035   # distance between thermocouple 1 and 2 in cm\n",

+      "\n",

+      "dz56=4.45    # distance between thermocouple 5 and 6 in cm\n",

+      "dz6i=3.81    # distance between thermocouple 6 and interface in cm\n",

+      "dz5i=dz56+dz6i # distance between thermocouple 5 and interface in cm\n",

+      "T5=374       # temperature in K at point 5\n",

+      "T6=366       # temperature in K at point 6\n",

+      "dzi7=2.45       # distance between thermocouple 7 and interface in cm\n",

+      "dz78=4.45       # distance between thermocouple 7 and 8 in cm\n",

+      "dzi8=dzi7+dz78  # distance between thermocouple 8 and interface in cm\n",

+      "T7=349          #  temperature in K at point 7\n",

+      "T8=337          # temperature in K at point 8\n",

+      "\n",

+      "#Calculation\n",

+      "qz_per_A=k*dT/dz12 # heat flow calculated in W/m**2 calculated using Fourier's law\n",

+      "T_ial=T5-(dz5i*(T5-T6)/dz56) # temperature of aluminium interface in K\n",

+      "T_img=dzi8*(T7-T8)/dz78+T8   # temperature of magnesium interface in K\n",

+      "T_img_=355.8                 #Approx value in the book\n",

+      "Rtc=(T_ial-T_img_)/(qz_per_A)\n",

+      "\n",

+      "#Result\n",

+      "\n",

+      "print\"The required thermal contact resistance is\",round(Rtc,7),\"K sq.m/W\"\n"

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "The required thermal contact resistance is 9.81e-05 K sq.m/W\n"

+       ]

+      }

+     ],

+     "prompt_number": 30

+    },

+    {

+     "cell_type": "heading",

+     "level": 3,

+     "metadata": {},

+     "source": [

+      "Example 2.8 Page No. 85"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "%pylab inline"

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": []

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "# determination of temperature profile, heat transferred, efficiency, effectiveness.\n",

+      "\n",

+      "#Given\n",

+      "import math\n",

+      "k=24.8                  # thermal conductivity of 1C steel in BTU/(hr.ft.degree Rankine)from appendix table B2\n",

+      "D=(5.0/16.0)/12.0           # diameter of the rod in ft\n",

+      "P=(math.pi*D)           # Circumference of the rod in ft\n",

+      "#print\"The perimeter is %.4f ft\",P)\n",

+      "A=(math.pi/4)*D**2      # Cross sectional area of the rod in sq.ft\n",

+      "#print\"The Cross sectional area is %.6f sq.ft\",A)\n",

+      "hc=1.0 # assuming the convective heat transfer coefficient as 1 BTU/(hr. sq.ft. degree Rankine)\n",

+      "\n",

+      "#Calculation\n",

+      "m=math.sqrt(hc*P/(k*A))\n",

+      "#print\"The value of parameter m is: /ft\",m)\n",

+      "L=(9/2.0)/12.0 # length of rod in ft\n",

+      "# using the equation (T-T_inf)/(T_w-T_inf)=(cosh[m(L-z)])/(cosh(mL)) for temperature profile\n",

+      "T_inf=70.0\n",

+      "T_w=200.0\n",

+      "dT=T_w-T_inf\n",

+      "const=dT/math.cosh(m*L)\n",

+      "\n",

+      "#(b) the heat transferred can be calculated using the equation qz=k*A*m*(T_w-T_inf)*tanh(m*L)\n",

+      "qz=k*A*m*dT*tanh(m*L)\n",

+      "\n",

+      "#(c)\n",

+      "mL=m*L\n",

+      "efficiency=0.78       # from fig. 2.30\n",

+      "\n",

+      "#(d)\n",

+      "effectiveness=math.sqrt(k*P/(hc*A))*tanh(mL)\n",

+      "\n",

+      "\n",

+      "print\"(b)The heat transferred is \",round(qz,2),\"BTU/hr\"\n",

+      "print\"(c)The efficiency found from the graph in figure 2.30 is\",efficiency\n",

+      "print\"(d)The effectiveness is found to be\",round(effectiveness,1)\n",

+      "\n",

+      "\n",

+      "#Plot\n",

+      "import matplotlib.pyplot as plt\n",

+      "fig = plt.figure()\n",

+      "ax = fig.add_subplot(111)\n",

+      "\n",

+      "mL1=[0,1,1.2,5]\n",

+      "n1=[0,0.2,0.25,0.25]\n",

+      "\n",

+      "mL2=[0,1,1.8,5]\n",

+      "n2=[0,0.35,0.5,0.5]\n",

+      "\n",

+      "mL3=[0,1,2,5]\n",

+      "n3=[0,0.75,1,1]\n",

+      "mL4=[0,1,2,5]\n",

+      "n4=[0,1.5,2,2]\n",

+      "\n",

+      "xlabel(\"mL\") \n",

+      "ylabel(\"n  \") \n",

+      "plt.xlim((0,6))\n",

+      "plt.ylim((0,2.5))\n",

+      "\n",

+      "ax.annotate('(0.25)', xy=(5,0.25))\n",

+      "ax.annotate('(0.5)', xy=(5,0.5))\n",

+      "ax.annotate('(1)', xy=(5,1))\n",

+      "ax.annotate('(2)', xy=(5,2))\n",

+      "a1=plot(mL1,n1)\n",

+      "a2=plot(mL2,n2)\n",

+      "a3=plot(mL3,n3)\n",

+      "a4=plot(mL4,n4)\n",

+      "\n",

+      "import matplotlib.pyplot as plt\n",

+      "fig = plt.figure()\n",

+      "ax = fig.add_subplot(111)\n",

+      "\n",

+      "z1=[0,3,4.8]\n",

+      "T1=[200,165,158]\n",

+      "z2=[0,4.8]\n",

+      "T2=[140,140]\n",

+      "z3=[0,4.8]\n",

+      "T3=[130,130]\n",

+      "z4=[4.8,4.8]\n",

+      "T4=[140,130]\n",

+      "\n",

+      "xlabel(\"z (m)\") \n",

+      "ylabel(\"T  (K)\") \n",

+      "plt.xlim((0,5.5))\n",

+      "plt.ylim((120,200))\n",

+      "\n",

+      "ax.annotate('(5/16 in)', xy=(4.85,135))\n",

+      "a1=plot(z1,T1)\n",

+      "a2=plot(z2,T2)\n",

+      "a3=plot(z3,T3)\n",

+      "a4=plot(z4,T4)\n",

+      "\n"

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "(b)The heat transferred is  3.13 BTU/hr\n",

+        "(c)The efficiency found from the graph in figure 2.30 is 0.78\n",

+        "(d)The effectiveness is found to be 45.2\n"

+       ]

+      },

+      {

+       "output_type": "display_data",

+       "png": 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+      },

+      {

+       "output_type": "display_data",

+       "png": 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FdOjQgRtuuIELLriAG2+8kSNHjtTYzmKxeI5ILBYLZWVlfPbZZzz99NM1mtFW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\n"

+      }

+     ],

+     "prompt_number": 1

+    },

+    {

+     "cell_type": "heading",

+     "level": 3,

+     "metadata": {},

+     "source": [

+      "Example 2.9 Page No. 90"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "# determination of heat transferred\n",

+      "\n",

+      "#Given\n",

+      "k=136.0             # thermal conductivity of aluminium in BTU/(hr.ft.degree Rankine)from appendix table B1\n",

+      "L=9/(8*12.0)        #length in ft\n",

+      "W=9/(4*12.0)          #width in ft\n",

+      "delta=0.002604       #ft\n",

+      "hc=0.8             # the convective heat transfer coefficient estimated as 1 BTU/(hr.ft**2. degree Rankine)\n",

+      "T_w=1000.0             # the root temperature in degree fahrenheit\n",

+      "T_inf=90.0             # the ambient temperature in degree fahrenheit\n",

+      "\n",

+      "#Calculation\n",

+      "import math\n",

+      "m=math.sqrt(hc/(k*delta))\n",

+      "P=2*W\n",

+      "A=2*delta*W\n",

+      "qz1=math.sqrt(hc*P*k*A)*(T_w-T_inf)*(sinh(m*L)+(hc/(m*k)*cosh(m*L)))/(cosh(m*L)+(hc/(m*k)*sinh(m*L)))\n",

+      "qz2=math.sqrt(k*A*hc*P)*(T_w-T_inf)*math.tanh(m*L)\n",

+      "Lc=L+delta\n",

+      "qz3=k*A*m*(T_w-T_inf)*math.tanh(m*L*(1+delta/Lc))\n",

+      "\n",

+      "#Result\n",

+      "print\"(a)The heat transferred is  \",round(qz1,2),\"BTU/hr\"\n",

+      "print\"(b)The heat transferred is \",round(qz2,2),\"BTU/hr      In the book the answer is incorrect\"\n",

+      "print\"(c)The heat transferred is \",round(qz3,2),\" BTU/hr\"\n"

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "(a)The heat transferred is   26.12 BTU/hr\n",

+        "(b)The heat transferred is  25.43 BTU/hr      In the book the answer is incorrect\n",

+        "(c)The heat transferred is  26.1  BTU/hr\n"

+       ]

+      }

+     ],

+     "prompt_number": 75

+    },

+    {

+     "cell_type": "heading",

+     "level": 3,

+     "metadata": {},

+     "source": [

+      "Example 2.10 Page No 94"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "# determination of optimum fin length and heat transferred by fin\n",

+      "\n",

+      "#Given\n",

+      "k=8.32           # thermal conductivity  BTU/(hr.ft.degree Rankine)\n",

+      "hc=400.0           # the convective heat transfer coefficient given in BTU/(hr.ft**2. degree Rankine)\n",

+      "\n",

+      "#Calculation\n",

+      "import math\n",

+      "delta_opt=0.55/(12*2)\n",

+      "# determination of dimension of one fin using the equation delta_opt=0.583*hc*Lc**2/k\n",

+      "Lc=math.sqrt(delta_opt*k/(0.583*hc))\n",

+      "\n",

+      "A=Lc*delta_opt\n",

+      "# determination of parameter for finding out efficiency from graph\n",

+      "parameter=Lc**1.5*math.sqrt(hc/(k*A))\n",

+      "efficiency=0.6\n",

+      "W=1/(2.0*12.0)   # width in ft\n",

+      "T_w=190.0       # wall temperature in degree fahrenheit\n",

+      "T_inf=58.0      # ambient temperature in degree fahrenheit\n",

+      "L=1.0           # length in ft\n",

+      "delta=W/2.0 \n",

+      "q_ac=efficiency*hc*2*W*math.sqrt(L**2+delta**2)*(T_w-T_inf)\n",

+      "\n",

+      "#Result\n",

+      "print\"(a)The optimum length is  \",round(Lc*12,2),\"inch\"\n",

+      "print\"(b)The actual heat transferred is \",round(q_ac,2),\"BTU/hr.  NOTE: In the book answer is incorrect\"\n"

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "(a)The optimum length is   0.34 inch\n",

+        "(b)The actual heat transferred is  2640.57 BTU/hr.  NOTE: In the book answer is incorrect\n"

+       ]

+      }

+     ],

+     "prompt_number": 35

+    },

+    {

+     "cell_type": "heading",

+     "level": 3,

+     "metadata": {},

+     "source": [

+      "Example 2.11 Page No 95"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "# determination of heat transferred and fin effectiveness\n",

+      "\n",

+      "#Given\n",

+      "#parameters of the problem are\n",

+      "N=9            # number of fins\n",

+      "delta=0.003/2.0 \n",

+      "L=0.025\n",

+      "Lc=L+delta\n",

+      "R=0.219/2\n",

+      "R2c=R+delta\n",

+      "R1=R-L\n",

+      "T_w=260       # root wall temperature in degree celsius\n",

+      "T_inf=27      # ambient temperature in degree celsius\n",

+      "hc=15 \n",

+      "k=52          # thermal conductivity of cast iron in W/(m.K)from appendix table B2\n",

+      "\n",

+      "#calculation\n",

+      "import math\n",

+      "#(a)\n",

+      "Ap=2*delta*Lc\n",

+      "As=2*math.pi*(R2c**2-R1**2)\n",

+      "radius_ratio=R2c/R1    # for finding efficiency from figure 2.38\n",

+      "variable=Lc**1.5*math.sqrt(hc/(k*Ap))\n",

+      "efficiency=0.93        # efficiency from figure 2.38\n",

+      "qf=N*efficiency*As*hc*(T_w-T_inf)\n",

+      "Sp=0.0127              # fin spacing\n",

+      "Asw=2*math.pi*R1*Sp*N     # exposed surface area\n",

+      "qw=hc*Asw*(T_w-T_inf)\n",

+      "q=qf+qw\n",

+      "\n",

+      "#(b)H=N*(Sp+2*delta)   # height of cylinder\n",

+      "H=N*(Sp+2*delta)\n",

+      "Aso=2*math.pi*R1*H       # surface area without fins\n",

+      "qo=hc*Aso*(T_w-T_inf)\n",

+      "\n",

+      "#(c)\n",

+      "effectiveness=q/qo     # effectiveness defined as ratio of heat transferred with fins to heat transferred without fins\n",

+      "\n",

+      "#Result\n",

+      "print\"(a)The total heat transferred from the cylinder is  \",round(q,0),\"W\"\n",

+      "print\"(b)The Heat transferred without fins is W\",round(qo,0),\"W\"\n",

+      "print\"(c)The fin effectiveness is \",round(effectiveness,2)\n"

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "(a)The total heat transferred from the cylinder is   1164.0 W\n",

+        "(b)The Heat transferred without fins is W 262.0 W\n",

+        "(c)The fin effectiveness is  4.44\n"

+       ]

+      }

+     ],

+     "prompt_number": 9

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [],

+     "language": "python",

+     "metadata": {},

+     "outputs": []

+    }

+   ],

+   "metadata": {}

+  }

+ ]

+}
\ No newline at end of file