Added(A)/Deleted(D) following books

 A  Heat_Transfer_Principles_And_Applications_by_Dutta/README.txt
A  Heat_Transfer_Principles_And_Applications_by_Dutta/ch10.ipynb
A  Heat_Transfer_Principles_And_Applications_by_Dutta/ch11.ipynb
A  Heat_Transfer_Principles_And_Applications_by_Dutta/ch2.ipynb
A  Heat_Transfer_Principles_And_Applications_by_Dutta/ch3.ipynb
A  Heat_Transfer_Principles_And_Applications_by_Dutta/ch4.ipynb
A  Heat_Transfer_Principles_And_Applications_by_Dutta/ch5.ipynb
A  Heat_Transfer_Principles_And_Applications_by_Dutta/ch6.ipynb
A  Heat_Transfer_Principles_And_Applications_by_Dutta/ch7.ipynb
A  Heat_Transfer_Principles_And_Applications_by_Dutta/ch8.ipynb
A  Heat_Transfer_Principles_And_Applications_by_Dutta/ch9.ipynb
A  Heat_Transfer_Principles_And_Applications_by_Dutta/screenshots/10.png
A  Heat_Transfer_Principles_And_Applications_by_Dutta/screenshots/5.png
A  Heat_Transfer_Principles_And_Applications_by_Dutta/screenshots/51.png
A  Heat_Transfer_in_SI_units_by_Holman/Chapter1.ipynb
A  Heat_Transfer_in_SI_units_by_Holman/Chapter10.ipynb
A  Heat_Transfer_in_SI_units_by_Holman/Chapter11.ipynb
A  Heat_Transfer_in_SI_units_by_Holman/Chapter2.ipynb
A  Heat_Transfer_in_SI_units_by_Holman/Chapter3.ipynb
A  Heat_Transfer_in_SI_units_by_Holman/Chapter4.ipynb
A  Heat_Transfer_in_SI_units_by_Holman/Chapter5.ipynb
A  Heat_Transfer_in_SI_units_by_Holman/Chapter6.ipynb
A  Heat_Transfer_in_SI_units_by_Holman/Chapter7.ipynb
A  Heat_Transfer_in_SI_units_by_Holman/Chapter8.ipynb
A  Heat_Transfer_in_SI_units_by_Holman/Chapter9.ipynb
A  Heat_Transfer_in_SI_units_by_Holman/README.txt
A  Heat_Transfer_in_SI_units_by_Holman/screenshots/9.1.png
A  Heat_Transfer_in_SI_units_by_Holman/screenshots/9.2.png
A  Heat_Transfer_in_SI_units_by_Holman/screenshots/9.4.png
A  Power_Electronics_Principles_and_Applications_by_Jacob/Chapter1.ipynb
A  Power_Electronics_Principles_and_Applications_by_Jacob/Chapter2.ipynb
A  Power_Electronics_Principles_and_Applications_by_Jacob/Chapter3.ipynb
A  Power_Electronics_Principles_and_Applications_by_Jacob/Chapter4.ipynb
A  Power_Electronics_Principles_and_Applications_by_Jacob/Chapter5.ipynb
A  Power_Electronics_Principles_and_Applications_by_Jacob/Chapter6.ipynb
A  Power_Electronics_Principles_and_Applications_by_Jacob/Chapter7.ipynb
A  Power_Electronics_Principles_and_Applications_by_Jacob/Chapter8.ipynb
A  Power_Electronics_Principles_and_Applications_by_Jacob/Chapter9.ipynb
A  Power_Electronics_Principles_and_Applications_by_Jacob/README.txt
A  Power_Electronics_Principles_and_Applications_by_Jacob/screenshots/4.png
A  Power_Electronics_Principles_and_Applications_by_Jacob/screenshots/5.png
A  Power_Electronics_Principles_and_Applications_by_Jacob/screenshots/6.png
A  sample_notebooks/AviralYadav/Chapter5.ipynb

1 files changed, 580 insertions, 0 deletions

diff --git a/Heat_Transfer_Principles_And_Applications_by_Dutta/ch2.ipynb b/Heat_Transfer_Principles_And_Applications_by_Dutta/ch2.ipynb
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+{
+ "cells": [
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "# Chapter 2 :Steady State conduction In one  dimension"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 2.1 Page No : 14"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 1,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "the rate of heat gain is 16.27 W\n",
+      "interface temp. between brick and cork is 24.2 C\n",
+      "interface temp. between cement and cork is -13.6 C\n",
+      "thermal resistance offered by brick layer is 12.9 percent\n",
+      "thermal resistance offered by cork layer is 84.1 percent\n",
+      "thermal resistance offered by cement layer is 3.0 percent\n",
+      "Additional thickness of cork to be provided  = 5.1 cm\n"
+     ]
+    }
+   ],
+   "source": [
+    "# Variables\n",
+    "A = 1.           \t\t\t#m**2, area\n",
+    "#for inner layer (cement)\n",
+    "ti = 0.06       \t\t\t#m, thickness\n",
+    "ki = 0.72       \t\t\t#W/m C,  thermal conductivity\n",
+    "Ti = -15.        \t\t\t#C, temprature\n",
+    "#for middle layer (cork)\n",
+    "tm = 0.1        \t\t\t#m, thickness\n",
+    "km = 0.043      \t\t\t#W/m C,  thermal conductivity\n",
+    "#for outer layer(brick)\n",
+    "to = 0.25       \t\t\t#m, thickness\n",
+    "ko = 0.7        \t\t\t#W/m C,  thermal conductivity\n",
+    "To = 30.         \t\t\t#C, temprature\n",
+    "\n",
+    "# Calculation and Results\n",
+    "#Thermal resistance of outer layer  \t\t\t#C/W\n",
+    "Ro = to/(ko*A) \n",
+    "#Thermal resistance of middle layer   \t\t\t#C/W\n",
+    "Rm = tm/(km*A) \n",
+    "#Thermal resistance of inner layer    \t\t\t#C/W\n",
+    "Ri = ti/(ki*A)\n",
+    "Rt = Ro+Rm+Ri\n",
+    "tdf = To-Ti        \t\t\t#temp driving force\n",
+    "#(a)\n",
+    "Q = tdf/Rt      \t\t\t#rate of heat gain\n",
+    "print \"the rate of heat gain is %.2f W\"%(Q)\n",
+    "\n",
+    "#(b)\n",
+    "#from fig. 2.4\n",
+    "td1 = Q*to/(ko*A)   \t\t\t#C temp. drop across the brick layer\n",
+    "T1 = To-td1         \t\t\t#interface temp. between brick and cork\n",
+    "#similarly\n",
+    "td2 = Q*tm/(km*A)   \t\t\t#C temp. drop across the cork layer\n",
+    "T2 = T1-td2         \t\t\t#C, interface temp. between cement and cork\n",
+    "print \"interface temp. between brick and cork is %.1f C\"%(T1)\n",
+    "print \"interface temp. between cement and cork is %.1f C\"%(T2)\n",
+    "\n",
+    "\n",
+    "#(c)\n",
+    "Rpo = Ro/Rt         \t\t\t#thermal resistance offered by brick layer\n",
+    "Rpm = Rm/Rt         \t\t\t#thermal resistance offered by cork layer\n",
+    "Rpi = Ri/Rt         \t\t\t#thermal resistance offered by cement layer\n",
+    "print \"thermal resistance offered by brick layer is %.1f percent\"%(Rpo*100)\n",
+    "print \"thermal resistance offered by cork layer is %.1f percent\"%(Rpm*100)\n",
+    "print \"thermal resistance offered by cement layer is %.1f percent\"%(Rpi*100)\n",
+    "\n",
+    "#second part\n",
+    "x = 30.              \t\t\t#percentage dec in heat transfer \n",
+    "Q1 = Q*(1-x/100)    \t\t\t#W, desired rate of heat flow\n",
+    "Rth = tdf/Q1        \t\t\t#C/W, required thermal resistance\n",
+    "Rad = Rth-Rt        \t\t\t#additional thermal resistance\n",
+    "Tad = Rad*km*A\n",
+    "print \"Additional thickness of cork to be provided  = %.1f cm\"%(Tad*100)\n",
+    "\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 2.2 Page No : 15"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 2,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Rate of heat loss is 50.7 W\n",
+      "thermal conductivities of insulating layer is 0.1633 W/m C\n"
+     ]
+    }
+   ],
+   "source": [
+    "# Variables\n",
+    "#outer thickness of brickwork (to) & inner thickness (ti)\n",
+    "to = 0.15    \t\t\t#m thickness\n",
+    "ti = 0.012   \t\t\t#m thickness\n",
+    "#thickness of intermediate layer(til)\n",
+    "til = 0.07   \t\t\t#m  thick\n",
+    "#thermal conductivities of brick  and wood\n",
+    "kb = 0.70    \t\t\t#W/m celcius\n",
+    "kw = 0.18    \t\t\t#W/m celcius\n",
+    "#temp. of outside and inside wall\n",
+    "To = -15    \t\t\t#celcius\n",
+    "Ti = 21     \t\t\t#celcius\n",
+    "#area\n",
+    "A = 1       \t\t\t#m**2\n",
+    "\n",
+    "\n",
+    "# Calculations and Results\n",
+    "#(a)\n",
+    "#Thermal resistance of brick  , wood and insulating layer\n",
+    "TRb = to/(kb*A)   \t\t\t#C/W\n",
+    "TRw = ti/(kw*A)   \t\t\t#C/W\n",
+    "TRi = 2*TRb       \t\t\t#C/W\n",
+    "#Total thermal resistance\n",
+    "TR = TRb+TRw+TRi  \t\t\t#C/W\n",
+    "#Temp. driving force\n",
+    "T = Ti-To         \t\t\t#C\n",
+    "#Rate of heat loss\n",
+    "Q = T/TR\n",
+    "print \"Rate of heat loss is %.1f W\"%(Q)\n",
+    "#(b)thermal conductivities of insulating layer\n",
+    "k = til/(A*TRi)\n",
+    "print \"thermal conductivities of insulating layer is %.4f W/m C\"%(k)"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 2.3 Page No : 19"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 5,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Rate of heat loss is 4095 W\n",
+      "interface temp.is  183 C\n",
+      "Fractional resistance offered  by the special brick layer is 0.353 \n"
+     ]
+    }
+   ],
+   "source": [
+    "\n",
+    "import math \n",
+    "\n",
+    "# Variables\n",
+    "#Length & Inside rdius of gas duct\n",
+    "L = 1.       \t\t\t#m\n",
+    "ri = 0.5    \t\t\t#m radius\n",
+    "#Properties of inner and outer layer\n",
+    "ki = 1.3    \t\t\t#W/m C, thermal conductivity of inner bricks\n",
+    "ti = 0.27   \t\t\t#m, inner layer thickness \n",
+    "ko = 0.92   \t\t\t#W/m C, thermal conductivity of special bricks \n",
+    "to = 0.14   \t\t\t#m, outer layer thickness\n",
+    "Ti = 400.    \t\t\t#C, inner layer temp.\n",
+    "To = 65.     \t\t\t#C, outer layer temp.\n",
+    "\n",
+    "#calculation and Results\n",
+    "r_ = ri+ti   \t\t\t#m, outer radius of fireclay  brick layer\n",
+    "ro = r_+to   \t\t\t#m, outer radius of special brick layer\n",
+    "#Heat transfer resistance\n",
+    "#Heat transfer resistance of fireclay brick\n",
+    "R1 = (math.log(r_/ri))/(2*math.pi*L*ki)\n",
+    "#Heat transfer resistance of special brick\n",
+    "R2 = (math.log(ro/r_))/(2*math.pi*L*ko)\n",
+    "#Total resistance\n",
+    "R = round(R1+R2,4)\n",
+    "#Driving force\n",
+    "T = Ti-To\n",
+    "#Rate of heat loss\n",
+    "Q = T/(R)\n",
+    "print \"Rate of heat loss is %d W\"%(Q)\n",
+    "#interface temp.\n",
+    "Tif = Ti-(Q*R1)\n",
+    "print \"interface temp.is  %d C\"%(Tif)\n",
+    "#Fractional resistance offered  by the special brick layer\n",
+    "FR = R2/(R1+R2)\n",
+    "print \"Fractional resistance offered  by the special brick layer is %.3f \"%(FR)\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 2.4 Page No : 20"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 6,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "The hot  end temp. is 148 C\n",
+      "The temprature gradient at hot end is -294.7 C/m\n",
+      "The temprature gradient at cold end is -1179 C/m\n",
+      "the temprature at 0.15m away from the cold end is 131 C\n"
+     ]
+    }
+   ],
+   "source": [
+    "import math\n",
+    "\n",
+    "# Variables\n",
+    "d1 = 0.06       \t\t\t#m, one end diameter of steel rod\n",
+    "d2 = 0.12       \t\t\t#m,other end diameter of steel rod\n",
+    "l = 0.2         \t\t\t#m length of rod\n",
+    "T2 = 30.         \t\t\t#C, temp. at end 2\n",
+    "Q = 50.          \t\t\t#W, heat loss\n",
+    "k = 15.           \t\t\t#W/m c, thermal conductivity of rod\n",
+    "\n",
+    "# Calculation and Results\n",
+    "#T = 265.8-(7.07/(0.06-0.15*x))........(a)\n",
+    "#(a)\n",
+    "x1 = 0\n",
+    "#from eq. (a)\n",
+    "T1 = 265.8-(7.07/(0.06-0.15*x1))\n",
+    "print \"The hot  end temp. is %.0f C\"%(T1)\n",
+    "#(b)  from eq. (i)\n",
+    "C = 50                 \t\t\t#integration consmath.tant\n",
+    "#from eq. (i)\n",
+    "D1 = -C/(math.pi*d1**2*k)   \t\t\t#D = dT/dx, temprature gradient\n",
+    "print \"The temprature gradient at hot end is %.1f C/m\"%(D1)\n",
+    "#similarly\n",
+    "D2 = -1179             \t\t\t#at x =  0.2m\n",
+    "print \"The temprature gradient at cold end is %.0f C/m\"%(D2)\n",
+    "\n",
+    "#(c)\n",
+    "x2 = 0.15              \t\t\t#m, given,\n",
+    "x3 = l-x2              \t\t\t#m, section away from the cold end\n",
+    "#from eq. (a)\n",
+    "T2 = 265.8-(7.07/(0.06-0.15*x3))\n",
+    "print \"the temprature at 0.15m away from the cold end is %.0f C\"%(T2)\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 2.5 Page No : 24"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 7,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "the rate of heat transfer is -3746 W\n",
+      "Refrigeration capacity is 1.07 tons\n"
+     ]
+    }
+   ],
+   "source": [
+    "import math\n",
+    "\n",
+    "# Variables\n",
+    "#inside and outside diameter and Temp.  of sphorical vessel\n",
+    "do = 16.         \t\t\t#m, diameter  \n",
+    "t = 0.1          \t\t\t#m, thick \n",
+    "Ri = do/2        \t\t\t#m, inside radius \n",
+    "Ro = Ri+t        \t\t\t#m. outside radius\n",
+    "To = 27.          \t\t\t#C, temperature\n",
+    "Ti = 4.           \t\t\t#C  ammonia\n",
+    "k = 0.02         \t\t\t#W/m C, thermal conductivity of foam layer \n",
+    "\n",
+    "# Calculations and Results\n",
+    "#from eq. 2.23 the rate of heat transfer\n",
+    "Q = (Ti-To)*(4*math.pi*k*Ro*Ri)/(Ro-Ri)\n",
+    "print \"the rate of heat transfer is %.0f W\"%(Q)\n",
+    "#Refrigeration capacity(RC)\n",
+    "#3516 Watt =  1 ton\n",
+    "RC = -Q/3516\n",
+    "print \"Refrigeration capacity is %.2f tons\"%(RC)\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 2.6 Page No : 28"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 8,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "the temprature midway in the rod at steady state is 167.3 C\n",
+      "Temprature gradient at one end of the rod is 559 C/W\n",
+      "Temprature gradient at other end of the rod is 521.8 C/W\n"
+     ]
+    }
+   ],
+   "source": [
+    "import math \n",
+    "# Variables\n",
+    "d = 0.05        \t\t\t#m, diameter of rod\n",
+    "l = 0.5         \t\t\t#m, length of rod\n",
+    "T1 = 30.         \t\t\t#CTemp. at  one end (1)\n",
+    "T2 = 300.        \t\t\t#C, temp  at other end (2)\n",
+    "\n",
+    "# Calculations and Results\n",
+    "x1 = l/2         \t\t\t#m, at mid plane\n",
+    "#temprature distribution ,\n",
+    "#comparing with quadratic eq. ax**2+bx+c \n",
+    "#and its solution as x = (-b+math.sqrt(b**2-4*a*c))/2*a\n",
+    "a = 1.35*10**-4\n",
+    "b = 1\n",
+    "c = -(564*x1+30.1)\n",
+    "T = (-b+math.sqrt(b**2-4*a*c))/(2*a)\n",
+    "print \"the temprature midway in the rod at steady state is %.1f C\"%(T)\n",
+    "\n",
+    "#Temprature gradient at the ends of the rod\n",
+    "x2 = 0               \t\t\t#m, at one end\n",
+    "a1 = 1.35*10**-4\n",
+    "b1 = 1\n",
+    "c1 = -(564*x2+30.1)\n",
+    "T1 = (-b1+math.sqrt(b1**2-4*a1*c1))/(2*a1)\n",
+    "k1 = 202+0.0545*T1   \n",
+    "C1 = 113930           \t\t\t#integration consmath.tant from eq. (1)\n",
+    "TG1 = C1/k1           \t\t\t#C/W, temprature gradient, dT/dx\n",
+    "#similarly\n",
+    "x3 = 0.5\n",
+    "a2 = 1.35*10**-4\n",
+    "b2 = 1\n",
+    "c2 = -(564*x3+30.1)\n",
+    "T2 = (-b2+math.sqrt(b2**2-4*a2*c2))/(2*a2)\n",
+    "k2 = 202+0.0545*T2\n",
+    "TG2 = C1/k2\n",
+    "print \"Temprature gradient at one end of the rod is %.0f C/W\"%(TG1)\n",
+    "print \"Temprature gradient at other end of the rod is %.1f C/W\"%(TG2)"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 2.7 Page No : 29"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 3,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "At the surface x = 0, the temp. is 600 C\n",
+      "At the surface x = 0.3m, the temp. is 270 C\n",
+      "Rhe average temprature of the wall is 615 C\n",
+      "The maximum temprature occurs at 0.104 m\n",
+      "The maximum temp. is 730 C\n",
+      "heat flux at left surface is -58750 W/m**2\n",
+      "heat flux at right surface is 110450 W/m**2\n",
+      "The average volumetric rate if heat genaration is 564000 W/m**3 \n"
+     ]
+    }
+   ],
+   "source": [
+    "import math \n",
+    "from scipy.integrate import quad \n",
+    "# Variables\n",
+    "#temprature distribution in wall\n",
+    "\n",
+    "t = 0.3             \t\t\t#m, thickness of wall\n",
+    "k = 23.5            \t\t\t#W/m c  thermal conductivity of wall\n",
+    "\n",
+    "# Calculations and Results\n",
+    "x1 = 0\n",
+    "T1 = 600+2500*x1-12000*x1**2    \t\t\t#C, at surface\n",
+    "x2 = 0.3\n",
+    "T2 = 600+2500*x2-12000*x2**2    \t\t\t#C, at x = 0.3\n",
+    "\n",
+    "def f3(x): \n",
+    "    return 600+2500*x-12000*x**2\n",
+    "\n",
+    "Tav = 1/t* quad(f3,0,0.3)[0]\n",
+    "\n",
+    "print \"At the surface x = 0, the temp. is %.0f C\"%(T1)\n",
+    "print \"At the surface x = 0.3m, the temp. is %.0f C\"%(T2)\n",
+    "print \"Rhe average temprature of the wall is %.0f C\"%(Tav)\n",
+    "\n",
+    "#(b)\n",
+    "\n",
+    "#for maximum temprature D = 0\n",
+    "x3 = 2500/24000.\n",
+    "print \"The maximum temprature occurs at %.3f m\"%(x3)\n",
+    "Tmax = 600+2500*x3-12000*x3**2\n",
+    "print \"The maximum temp. is %.0f C\"%(Tmax)\n",
+    "\n",
+    "#(c)\n",
+    "D1 = 2500-24000*x1         \t\t\t#at x = 0, temprature gradient\n",
+    "Hf1 = -k*D1                \t\t\t#W/m**2, heat flux at left surface(x = 0)\n",
+    "D2 = 2500-24000*x2         \t\t\t#at x = 0.3, temprature gradient\n",
+    "Hf2 = -k*D2                \t\t\t#W/m**2, heat flux at right surface(x = 0.3)\n",
+    "print \"heat flux at left surface is %.0f W/m**2\"%(Hf1)\n",
+    "print \"heat flux at right surface is %.0f W/m**2\"%(Hf2)\n",
+    "\n",
+    "#(d)\n",
+    "Qt = Hf2-Hf1              \t\t\t#W/m**2, total rate of heat loss\n",
+    "Vw = 0.3                  \t\t\t#m**3/m**2, volume of wall per unit surface area\n",
+    "Hav = Qt/Vw               \t\t\t#W/m**3, average volumetric rate\n",
+    "print \"The average volumetric rate if heat genaration is %.0f W/m**3 \"%(Hav)   \n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 2.8 Page No : 30"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 10,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "The maximum temp. will occur at a position 0.209 m\n",
+      "The maximum temprature is 152.6 C\n"
+     ]
+    }
+   ],
+   "source": [
+    "import math \n",
+    "# Variables\n",
+    "ka = 24         \t\t\t#W/mC thermal conductivitiy of material A\n",
+    "tA = 0.1        \t\t\t#m, thickness of A material\n",
+    "kB = 230        \t\t\t#W/mC thermal conductivity of metl B\n",
+    "kC = 200        \t\t\t#W/mC thermal conductivity of metal C\n",
+    "tB = 0.1        \t\t\t#m, thickness of B metal\n",
+    "tC = 0.1        \t\t\t#m, thickness of C metal\n",
+    "TBo = 100       \t\t\t#C, outer surface temp. of B wall\n",
+    "TCo = 100       \t\t\t#C, outer surface temp. of C wall\n",
+    "Q = 2.5*10**5    \t\t\t#W/m**3, heat generated\n",
+    "\n",
+    "# Calculations and Results\n",
+    "#At D = 0\n",
+    "x = 2175./10416\n",
+    "print \"The maximum temp. will occur at a position %.3f m\"%(x)\n",
+    "x1 = x\n",
+    "TA = -5208*x1**2+2175*x1-74.5\n",
+    "print \"The maximum temprature is %.1f C\"%(TA)\n",
+    "\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 2.9 Page No : 36"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 11,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "This radial position does not fall within layer 1. Therefore no temprature maximum occurs in this layer.\n",
+      " Similarly no temprature maximum occurs in  layer 2.\n",
+      "The maximum temprature at the outer boundary is 200 C\n"
+     ]
+    }
+   ],
+   "source": [
+    "import math\n",
+    "\n",
+    "# Variables\n",
+    "di = 0.15         \t\t\t#m, inner diameter\n",
+    "do = 0.3          \t\t\t#m, outer diameter\n",
+    "Q1 = 100.*10**3  \t\t\t#W/,m**3,inner  rate of heat generation\n",
+    "Q2 = 40.*10**3   \t\t\t#W/m**3, outer rate of heat generation\n",
+    "Ti = 100.       \t\t\t#C, temp.at inside surface\n",
+    "To = 200.       \t\t\t#C, temp. at outside surface\n",
+    "k1 = 30.        \t\t\t#W/m C, thermal conductivity of material for inner layer\n",
+    "k2 = 10.        \t\t\t#W/m C, thermal conductivity of material for outer layer\n",
+    "\n",
+    "# Calculations and Results\n",
+    "#T1 = 364+100*math.log(r)-833.3*r**2          (1)\n",
+    "#T2 = 718+216*math.log(r)-1000*r**2           (2)\n",
+    "#(b)from eq. 1\n",
+    "r = math.sqrt(100./2*833.3)\n",
+    "print \"This radial position does not fall within layer 1. Therefore no temprature maximum occurs in this layer.\"\n",
+    "#similarly\n",
+    "print \" Similarly no temprature maximum occurs in  layer 2.\"\n",
+    "ro = di        \t\t        # m, outer boundary\n",
+    "Tmax = To\n",
+    "print \"The maximum temprature at the outer boundary is %.0f C\"%(Tmax)\n"
+   ]
+  }
+ ],
+ "metadata": {
+  "kernelspec": {
+   "display_name": "Python 2",
+   "language": "python",
+   "name": "python2"
+  },
+  "language_info": {
+   "codemirror_mode": {
+    "name": "ipython",
+    "version": 2
+   },
+   "file_extension": ".py",
+   "mimetype": "text/x-python",
+   "name": "python",
+   "nbconvert_exporter": "python",
+   "pygments_lexer": "ipython2",
+   "version": "2.7.6"
+  }
+ },
+ "nbformat": 4,
+ "nbformat_minor": 0
+}