Revised list of TBCs

22 files changed, 1270 insertions, 7963 deletions

diff --git a/Introduction_to_Heat_Transfer_by_S._K._Som/Chapter1.ipynb b/Introduction_to_Heat_Transfer_by_S._K._Som/Chapter1.ipynb
index d3b728b1..d9d7f745 100644
--- a/Introduction_to_Heat_Transfer_by_S._K._Som/Chapter1.ipynb
+++ b/Introduction_to_Heat_Transfer_by_S._K._Som/Chapter1.ipynb
@@ -18,7 +18,7 @@
   },
   {
    "cell_type": "code",
-   "execution_count": 1,
+   "execution_count": 4,
    "metadata": {
     "collapsed": false
    },
@@ -34,6 +34,7 @@
     }
    ],
    "source": [
+    "import math\n",
     "print\"Introduction to heat transfer by S.K.Som, Chapter 1, Example 1\"\n",
     "#The temprature of two faces of the slabs are T1=40°C & T2=20°C \n",
     "#The thickness of the slab(L) is 80mm or .08m\n",
@@ -57,7 +58,7 @@
   },
   {
    "cell_type": "code",
-   "execution_count": 3,
+   "execution_count": 5,
    "metadata": {
     "collapsed": false
    },
@@ -73,7 +74,7 @@
     }
    ],
    "source": [
-    "\n",
+    "import math\n",
     "print\"Introduction to heat transfer by S.K.Som, Chapter 1, Example 2\"\n",
     "#The thermal conductivity(km)of masonry wall is .8 W/(mK)\n",
     "#The thermal conductivity(kc)of composite wall is .2 W/(mK)\n",
@@ -88,7 +89,7 @@
     "#The thickness of masonry wall is Lm.\n",
     "print\"The thickness of masonry wall is Lm in m\"\n",
     "Lm=(km/kc)*(Lc/(0.8))\n",
-    "print\"Lm=\",Lm"
+    "print\"Lm=\",Lm\n"
    ]
   },
   {
@@ -100,7 +101,7 @@
   },
   {
    "cell_type": "code",
-   "execution_count": 5,
+   "execution_count": 6,
    "metadata": {
     "collapsed": false
    },
@@ -116,7 +117,7 @@
     }
    ],
    "source": [
-    "\n",
+    "import math\n",
     "print\"Introduction to heat transfer by S.K.Som, Chapter 1, Example 4\"\n",
     "#The average forced convective heat transfer coefficient(hbr) is 200 W/( m**2 °C)\n",
     "#The fluid temprature(Tinf) upstream of the cold surface is 100°C\n",
@@ -139,7 +140,7 @@
   },
   {
    "cell_type": "code",
-   "execution_count": 6,
+   "execution_count": 7,
    "metadata": {
     "collapsed": false
    },
@@ -155,7 +156,7 @@
     }
    ],
    "source": [
-    "\n",
+    "import math\n",
     "print\"Introduction to heat transfer by S.K.Som, Chapter 1, Example 5\"\n",
     "#The average heat transfer coefficient(hbr) is 800 W/(m**2°C)\n",
     "#The surface temprature of heat exchanger is 75°C and air temprature is 25°C so deltaT=(75-25)\n",
@@ -178,7 +179,7 @@
   },
   {
    "cell_type": "code",
-   "execution_count": 14,
+   "execution_count": 3,
    "metadata": {
     "collapsed": false
    },
@@ -189,16 +190,16 @@
      "text": [
       "Introduction to heat transfer by S.K.Som, Chapter 1, Example 6\n",
       "The rate of heat transfer from the plate is given by Q=hbr*A*(Ts-Tinf)\n",
-      "Q= 224.0\n",
+      "Equating the above two equations we get hbr=(m*cp*|dT/dt|)/(A*(Ts-Tinf)) in W/(m**2°C)\n",
+      "hbr= 11.2\n",
       "The rate of heat transfer can also be written in the form of Q=m*cp*|dT/dt| from an energy balance.\n",
       "Q= 224.0\n",
-      "Equating the above two equations we get hbr=(m*cp*|dT/dt|)/(A*(Ts-Tinf)) in W/(m**2°C)\n",
-      "hbr= 11.2\n"
+      "Equating the above two equations we get hbr=(m*cp*|dT/dt|)/(A*(Ts-Tinf)) in W/(m**2°C)\n"
      ]
     }
    ],
    "source": [
-    "\n",
+    "import math\n",
     "print\"Introduction to heat transfer by S.K.Som, Chapter 1, Example 6\"\n",
     "#The temprature of the plate(Ts) is 225°C\n",
     "#The ambient temprature (Tinf) is 25°C\n",
@@ -214,13 +215,28 @@
     "m=4;\n",
     "cp=2.8;\n",
     "print\"The rate of heat transfer from the plate is given by Q=hbr*A*(Ts-Tinf)\"\n",
+    "print\"Equating the above two equations we get hbr=(m*cp*|dT/dt|)/(A*(Ts-Tinf)) in W/(m**2°C)\"\n",
+    "hbr=(m*cp*10**3*X)/(A*(Ts-Tinf))\n",
+    "print\"hbr=\",hbr\n",
     "Q=hbr*A*(Ts-Tinf)\n",
-    "print\"Q=\",Q\n",
     "print\"The rate of heat transfer can also be written in the form of Q=m*cp*|dT/dt| from an energy balance.\"\n",
     "print\"Q=\",Q\n",
     "print\"Equating the above two equations we get hbr=(m*cp*|dT/dt|)/(A*(Ts-Tinf)) in W/(m**2°C)\"\n",
-    "hbr=(m*cp*10**3*X)/(A*(Ts-Tinf))\n",
-    "print\"hbr=\",hbr"
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n"
    ]
   },
   {
@@ -232,7 +248,7 @@
   },
   {
    "cell_type": "code",
-   "execution_count": 2,
+   "execution_count": 8,
    "metadata": {
     "collapsed": false
    },
@@ -248,7 +264,7 @@
     }
    ],
    "source": [
-    "\n",
+    "import math\n",
     "print\"Introduction to heat transfer by S.K.Som, Chapter 1, Example 7\"\n",
     "#The temprature(T) of brick wall after sunset is 50°C\n",
     "#The emissity value(emi)=0.9\n",
@@ -272,7 +288,7 @@
   },
   {
    "cell_type": "code",
-   "execution_count": 4,
+   "execution_count": 9,
    "metadata": {
     "collapsed": false
    },
@@ -288,7 +304,7 @@
     }
    ],
    "source": [
-    "\n",
+    "import math\n",
     "print\"Introduction to heat transfer by S.K.Som, Chapter 1, Example 8\"\n",
     "#The temprature(T) of asphalt pavement = 50°C\n",
     "#The stefan-Boltzman constant(sigma)=5.6697*10**-8 W/(m**2*K**4).\n",
@@ -298,7 +314,7 @@
     "print\"The emitted radiant energy per unit surface area is given by Eb/A=sigma*T**4 in W/m**2\"\n",
     "#Let Eb/A=F\n",
     "F=sigma*(50+273.15)**4\n",
-    "print\"F=\",F"
+    "print\"F=\",F\n"
    ]
   },
   {
@@ -310,7 +326,7 @@
   },
   {
    "cell_type": "code",
-   "execution_count": 4,
+   "execution_count": 10,
    "metadata": {
     "collapsed": false
    },
@@ -330,7 +346,7 @@
     }
    ],
    "source": [
-    "\n",
+    "import math\n",
     "print\"Introduction to heat transfer by S.K.Som, Chapter 1, Example 9\"\n",
     "#The Thickness(L) of wall= 150 mm or 0.15 m.\n",
     "#The wall on one side is exposed to air at temprature(Ta)= 60°C and on the other side to air at temprature(Tb) = 20°C\n",
@@ -367,7 +383,7 @@
   },
   {
    "cell_type": "code",
-   "execution_count": 19,
+   "execution_count": 11,
    "metadata": {
     "collapsed": false
    },
@@ -378,9 +394,7 @@
      "text": [
       "Introduction to heat transfer by S.K.Som, Chapter 1, Example 10\n",
       "Heat transfer from the outer surface takes place only by radiation is given by Q/A=F1=emi*sigma*(T2**4-T0**4)in W/m**2 for different values of tempratures in K\n",
-      "F1= 332.029390022\n",
       "heat transfer from the outer surface can also be written as Q/A=F2=(Ti-To)/((1/hbri)+(L/k)+(1/hr)) in W/m**2 at different tempratures in K\n",
-      "F2= 332.132667923\n",
       "The values of temprature that are considered are <298 K\n",
       "Satisfactory solutions for Temprature in K is\n",
       "T2= 292.5\n",
@@ -391,7 +405,7 @@
     }
    ],
    "source": [
-    "\n",
+    "import math\n",
     "print\"Introduction to heat transfer by S.K.Som, Chapter 1, Example 10\"\n",
     "#The spacecraft panel has thickness(L)=.01 m\n",
     "#The spacecraft has inner temprature (Ti)=298 K\n",
@@ -413,9 +427,7 @@
     "#Radiation heat transfer coefficient(hr) is defined as Q/A=hr(T2-To)\n",
     "#so hr=4.536*10**-8*T2**3\n",
     "print\"Heat transfer from the outer surface takes place only by radiation is given by Q/A=F1=emi*sigma*(T2**4-T0**4)in W/m**2 for different values of tempratures in K\"\n",
-    "print\"F1=\",F1\n",
     "print\"heat transfer from the outer surface can also be written as Q/A=F2=(Ti-To)/((1/hbri)+(L/k)+(1/hr)) in W/m**2 at different tempratures in K\"\n",
-    "print\"F2=\",F2\n",
     "print\"The values of temprature that are considered are <298 K\"\n",
     "for i in range(285,292):\n",
     "    T2=i\n",
@@ -445,7 +457,7 @@
   },
   {
    "cell_type": "code",
-   "execution_count": 2,
+   "execution_count": 12,
    "metadata": {
     "collapsed": false
    },
@@ -488,7 +500,8 @@
     "print\"The total heat loss by The pipe per unit length is given by Q/L=hbr*A*(T1-T2)+sigma*emi*A*(T1**4-T2**4) in W/m\"\n",
     "#Let Q/L=F\n",
     "F=hbr*A*((T1+273.15)-(T2+273.15))+sigma*emi*A*((T1+273.15)**4-(T2+273.15)**4)\n",
-    "print\"F=\",F"
+    "print\"F=\",F\n",
+    "\n"
    ]
   }
  ],
diff --git a/Introduction_to_Heat_Transfer_by_S._K._Som/Chapter10.ipynb b/Introduction_to_Heat_Transfer_by_S._K._Som/Chapter10.ipynb
index 9eacc4ed..95f29d79 100644
--- a/Introduction_to_Heat_Transfer_by_S._K._Som/Chapter10.ipynb
+++ b/Introduction_to_Heat_Transfer_by_S._K._Som/Chapter10.ipynb
@@ -44,6 +44,10 @@
     }
    ],
    "source": [
+    "  \n",
+    " \n",
+    "  \n",
+    " \n",
     "import math\n",
     " \n",
     "print\"Introduction to heat transfer by S.K.Som, Chapter 10, Example 1\"\n",
@@ -90,7 +94,29 @@
     "print\"LMTD=\",LMTD\n",
     "print\"Area(A)=Q/(U*LMTD) in m**2\"\n",
     "A=Q/(U*LMTD)\n",
-    "print\"A=\",A"
+    "print\"A=\",A\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n"
    ]
   },
   {
@@ -124,6 +150,10 @@
     }
    ],
    "source": [
+    "  \n",
+    " \n",
+    "  \n",
+    " \n",
     "import math\n",
     " \n",
     "print\"Introduction to heat transfer by S.K.Som, Chapter 10, Example 2\"\n",
@@ -163,7 +193,7 @@
     "#Area(A)=Q/(U*LMTD) in m**2\n",
     "print\"Area(A)=Q/(U*LMTD) in m**2\"\n",
     "A=Q/(U*LMTD)\n",
-    "print\"A=\",A"
+    "print\"A=\",A\n"
    ]
   },
   {
@@ -203,6 +233,10 @@
     }
    ],
    "source": [
+    "  \n",
+    " \n",
+    "  \n",
+    " \n",
     "import math\n",
     " \n",
     "print\"Introduction to heat transfer by S.K.Som, Chapter 10, Example 3\"\n",
@@ -250,7 +284,13 @@
     "#overall heat transfer coefficient(U)=Q/(A*F*LMTD)\n",
     "print\"overall heat transfer coefficient(U)=Q/(A*F*LMTD)in W/(m**2*K)\"\n",
     "U=Q/(A*F*LMTD)\n",
-    "print\"U=\",U"
+    "print\"U=\",U\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n"
    ]
   },
   {
@@ -302,6 +342,10 @@
     }
    ],
    "source": [
+    "  \n",
+    " \n",
+    "  \n",
+    " \n",
     "import math\n",
     " \n",
     "print\"Introduction to heat transfer by S.K.Som, Chapter 10, Example 5\"\n",
@@ -380,7 +424,33 @@
     "print\"To provide this surface area ,The length(L) of the tube required is given by L=A/(pi*D) in m\"\n",
     "L=A/(math.pi*D)\n",
     "print\"Hence same result is obtained for both methods\"\n",
-    "print\"L=\",L"
+    "print\"L=\",L\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n"
    ]
   },
   {
@@ -412,6 +482,10 @@
     }
    ],
    "source": [
+    "  \n",
+    " \n",
+    "  \n",
+    " \n",
     "import math\n",
     " \n",
     "print\"Introduction to heat transfer by S.K.Som, Chapter 10, Example 6\"\n",
@@ -446,7 +520,13 @@
     "#Hence The total heat transfer rate (Q)=eff*Cmin*(Thi-Tci)in kW.\n",
     "print\"The total heat transfer rate (Q)=eff*Cmin*(Thi-Tci) in kW\"                           \n",
     "Q=eff*Cmin*(Thi-Tci)\n",
-    "print\"Q=\",Q"
+    "print\"Q=\",Q\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n"
    ]
   },
   {
@@ -480,6 +560,10 @@
     }
    ],
    "source": [
+    "  \n",
+    " \n",
+    "  \n",
+    " \n",
     "import math\n",
     " \n",
     "print\"Introduction to heat transfer by S.K.Som, Chapter 10, Example 7\"\n",
@@ -519,7 +603,22 @@
     "#The exit temprature(Tho) of air is given by Thi-(Q/(mdota*cpa))\n",
     "print\"The exit temprature of air in °C \"\n",
     "Tho=Thi-(Q/(mdota*1000*cpa))#NOTE:-The answer slightly varies from the answer in book(i.e Tho=26°C) because the value of Q taken in book is approximated to 1*10**6W.\n",
-    "print\"Tho=\",Tho"
+    "print\"Tho=\",Tho\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n"
    ]
   },
   {
@@ -557,6 +656,10 @@
     }
    ],
    "source": [
+    "  \n",
+    " \n",
+    "  \n",
+    " \n",
     "import math\n",
     " \n",
     "print\"Introduction to heat transfer by S.K.Som, Chapter 10, Example 8\"\n",
@@ -599,7 +702,28 @@
     "Tho=Tci;\n",
     "print\"Effectiveness of heat exchanger is \"\n",
     "eff=(mdoth*ch*(Thi-Tho))/(mdoth*ch*(Thi-Tci))\n",
-    "print\"eff=\",eff"
+    "print\"eff=\",eff\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n"
    ]
   }
  ],
diff --git a/Introduction_to_Heat_Transfer_by_S._K._Som/Chapter10_ljUjU8j.ipynb b/Introduction_to_Heat_Transfer_by_S._K._Som/Chapter10_ljUjU8j.ipynb
deleted file mode 100644
index 95f29d79..00000000
--- a/Introduction_to_Heat_Transfer_by_S._K._Som/Chapter10_ljUjU8j.ipynb
+++ /dev/null
@@ -1,751 +0,0 @@
-{
- "cells": [
-  {
-   "cell_type": "markdown",
-   "metadata": {
-    "collapsed": true
-   },
-   "source": [
-    "# Chapter 10:Principles of heat exchangers"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Ex10.1:pg-415"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 1,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Introduction to heat transfer by S.K.Som, Chapter 10, Example 1\n",
-      "The rate of heat transfer from water is given by Q=mdot*cp*(T3-T4) in W\n",
-      "Q= 11286.0\n",
-      "(a) For a parallel flow arrangement\n",
-      "LMTD is given by (deltaT2-deltaT1)/(ln(deltaT2/deltaT1)) in °C \n",
-      "LMTD= 35.4699026617\n",
-      "Area(A)=Q/(U*LMTD) in m**2\n",
-      "A= 0.397731567932\n",
-      "(b)For counterflow arrangement\n",
-      "LMTD=(deltaT2-deltaT1)/(ln(deltaT2/deltaT1))in°C \n",
-      "LMTD= 36.3143018164\n",
-      "Area(A)=Q/(U*LMTD) in m**2\n",
-      "A= 0.38848330532\n"
-     ]
-    }
-   ],
-   "source": [
-    "  \n",
-    " \n",
-    "  \n",
-    " \n",
-    "import math\n",
-    " \n",
-    "print\"Introduction to heat transfer by S.K.Som, Chapter 10, Example 1\"\n",
-    "#A brine solution is heated from temprature ,T1=8°C to temprature,T2=14°C in a double pipe heat exchanger.\n",
-    "T1=8.0;\n",
-    "T2=14.0;\n",
-    "#Water entering at temprature T3=55°C and leaving at temprature,T4=40°C at the mass flow rate of (mdot)=0.18kg/s\n",
-    "mdot=0.18;\n",
-    "T3=55.0;\n",
-    "T4=40.0;\n",
-    "#Specific heat (cp) of water =4.18kJ/(kg*K)\n",
-    "cp=4.18*10**3;\n",
-    "#overall heat transfer coefficient(U)=800 W/(m**2*K)\n",
-    "U=800.0;\n",
-    "#The rate of heat transfer from water is given by Q=mdot*cp*(T3-T4)\n",
-    "print\"The rate of heat transfer from water is given by Q=mdot*cp*(T3-T4) in W\"\n",
-    "Q=mdot*cp*(T3-T4)\n",
-    "print\"Q=\",Q\n",
-    "print\"(a) For a parallel flow arrangement\"\n",
-    "#For a parallel flow arrangement deltaT1=T3-T1 and deltaT2=T4-T2. \n",
-    "deltaT1=T3-T1;#deltaT1 is temprature difference \n",
-    "deltaT2=T4-T2;#deltaT2 is temprature difference \n",
-    "#LMTD(math.log mean temprature difference) is defined as (deltaT2-deltaT1)/(ln(deltaT2/deltaT1)) for both parallel and counter flow.\n",
-    "print\"LMTD is given by (deltaT2-deltaT1)/(ln(deltaT2/deltaT1)) in °C \"\n",
-    "#let X=math.log10((deltaT2/deltaT1)) and Y=math.log10(2.718281)\n",
-    "X=math.log10((deltaT2/deltaT1));\n",
-    "Y=math.log10(2.718281);\n",
-    "#ln=(ln(deltaT2/deltaT1))\n",
-    "ln=X/Y;\n",
-    "LMTD=(deltaT2-deltaT1)/ln\n",
-    "print\"LMTD=\",LMTD\n",
-    "#Area(A)=Q/(U*LMTD) in m**2\n",
-    "print\"Area(A)=Q/(U*LMTD) in m**2\"\n",
-    "A=Q/(U*LMTD)\n",
-    "print\"A=\",A\n",
-    "print\"(b)For counterflow arrangement\"\n",
-    "deltaT1=T3-T2;\n",
-    "deltaT2=T4-T1;\n",
-    "print\"LMTD=(deltaT2-deltaT1)/(ln(deltaT2/deltaT1))in°C \"\n",
-    "X=math.log10((deltaT2/deltaT1));\n",
-    "Y=math.log10(2.718281);\n",
-    "ln=X/Y;\n",
-    "LMTD=(deltaT2-deltaT1)/ln\n",
-    "print\"LMTD=\",LMTD\n",
-    "print\"Area(A)=Q/(U*LMTD) in m**2\"\n",
-    "A=Q/(U*LMTD)\n",
-    "print\"A=\",A\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Ex10.2:pg-416"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 3,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Introduction to heat transfer by S.K.Som, Chapter 10, Example 2\n",
-      "The inlet temprature(Tc1) of cold oil in °C \n",
-      "Tc1= 55.0948103792\n",
-      "The rate of heat transfer Q=mdoth*ch*(Th1-Th2) in W\n",
-      "Q= 190190.0\n",
-      "LMTD is given by (deltaT2-deltaT1)/(ln(deltaT2/deltaT1)) in °C \n",
-      "LMTD= 28.996452611\n",
-      "Area(A)=Q/(U*LMTD) in m**2\n",
-      "A= 10.0908895279\n"
-     ]
-    }
-   ],
-   "source": [
-    "  \n",
-    " \n",
-    "  \n",
-    " \n",
-    "import math\n",
-    " \n",
-    "print\"Introduction to heat transfer by S.K.Som, Chapter 10, Example 2\"\n",
-    "#Hot oil(specific heat,ch=2.09kJ/(kg*K)) flows through counter flow heat excahnger at the mass flow rate of mdoth=(0.7kg/s)\n",
-    "ch=2.09*10**3;\n",
-    "mdoth=0.7;\n",
-    "#overall heat transfer coefficient(U)=650 W/(m**2*K)\n",
-    "U=650;\n",
-    "#It enters at temprature,Th1=200°C and leaves at temprature,Th2=70°C \n",
-    "Th1=200;\n",
-    "Th2=70;\n",
-    "#Cold oil(specific heat,cc=1.67kJ/(kg*K) exits at temprature,Tc2=150°C at the mass flow rate of mdotc=(1.2kg/s)\n",
-    "mdotc=1.2;\n",
-    "cc=1.67*10**3;\n",
-    "Tc2=150;\n",
-    "#The unknown inlet temprature(Tc1) of cold oil may be found from energy balance mdotc*(Tc2-Tc1)=mdoth*(Th2-Th1)\n",
-    "print\"The inlet temprature(Tc1) of cold oil in °C \"\n",
-    "Tc1=Tc2-((mdoth*ch)/(mdotc*cc))*(Th1-Th2)\n",
-    "print\"Tc1=\",Tc1\n",
-    "#The rate of heat transfer can be calculate as Q=mdoth*ch*(Th1-Th2)\n",
-    "print\"The rate of heat transfer Q=mdoth*ch*(Th1-Th2) in W\"\n",
-    "Q=mdoth*ch*(Th1-Th2)\n",
-    "deltaT1=Th1-Tc2;\n",
-    "print\"Q=\",Q\n",
-    "#deltaT1 is temprature difference between hot oil inlet temprature and cold oil exit temprature\n",
-    "deltaT2=Th2-Tc1;\n",
-    "#deltaT2 is temprature difference between hot oil exit temprature and cold oil inlet temprature\n",
-    "#LMTD(math.log mean temprature difference) is defined as (deltaT2-deltaT1)/(ln(deltaT2/deltaT1)) for counter flow.\n",
-    "print\"LMTD is given by (deltaT2-deltaT1)/(ln(deltaT2/deltaT1)) in °C \"\n",
-    "#let X=log10((deltaT2/deltaT1)) and Y=log10(2.718281)\n",
-    "X=math.log10((deltaT2/deltaT1));\n",
-    "Y=math.log10(2.718281);\n",
-    "#ln=(ln(deltaT2/deltaT1))\n",
-    "ln=X/Y;\n",
-    "LMTD=(deltaT2-deltaT1)/ln\n",
-    "print\"LMTD=\",LMTD\n",
-    "#Area(A)=Q/(U*LMTD) in m**2\n",
-    "print\"Area(A)=Q/(U*LMTD) in m**2\"\n",
-    "A=Q/(U*LMTD)\n",
-    "print\"A=\",A\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Ex10.3:pg-417"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 5,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Introduction to heat transfer by S.K.Som, Chapter 10, Example 3\n",
-      "The outlet temprature(Tho) of oil in °C \n",
-      "Tho= 70.6666666667\n",
-      "For a counterflow heat exchanger\n",
-      "deltaT1= 10\n",
-      "deltaT2= 20.6666666667\n",
-      "LMTD is given by (deltaT2-deltaT1)/(ln(deltaT2/deltaT1)) in °C \n",
-      "LMTD= 14.6936488511\n",
-      "dimensionless parameters P and R are\n",
-      "P= 0\n",
-      "R= 0.733333333333\n",
-      "correction factor(F) for the cross flow arrangement as obtained from graph of F vs Single Pass flow with fluids unmixed\n",
-      "Q= 167200.0\n",
-      "overall heat transfer coefficient(U)=Q/(A*F*LMTD)in W/(m**2*K)\n",
-      "U= 758.604399737\n"
-     ]
-    }
-   ],
-   "source": [
-    "  \n",
-    " \n",
-    "  \n",
-    " \n",
-    "import math\n",
-    " \n",
-    "print\"Introduction to heat transfer by S.K.Som, Chapter 10, Example 3\"\n",
-    "#A cross flow heat exchanger with both fluids unmixed is used to heat water((specific heat,cc=4.18kJ/(kg*K)) from temprature Tci=50°C to Tco=90°C \n",
-    "#flowing at the mass flow rate of (mdotc)=1kg/s\n",
-    "Tci=50;\n",
-    "Tco=90;\n",
-    "cc=4.18*10**3;\n",
-    "mdotc=1;\n",
-    "#The hot engine oil has (specific heat,ch=1.9kJ/(kg*K)) flowing at the mass flow rate of mdoth=3kg/s enters at temprature Thi=100°C\n",
-    "mdoth=3;\n",
-    "Thi=100;\n",
-    "ch=1.9*10**3;\n",
-    "#The unknown outlet temprature(Tho) of  oil may be found from energy balance mdotc*(Tco-Tci)=mdoth*(Tho-Thi)\n",
-    "print\"The outlet temprature(Tho) of oil in °C \"\n",
-    "Tho=Thi-((mdotc*cc)/(mdoth*ch))*(Tco-Tci)\n",
-    "print\"Tho=\",Tho\n",
-    "print\"For a counterflow heat exchanger\"\n",
-    "deltaT1=Thi-Tco;#deltaT1 is temprature difference \n",
-    "deltaT2=Tho-Tci;#deltaT2 is temprature difference \n",
-    "print\"deltaT1=\",deltaT1\n",
-    "print\"deltaT2=\",deltaT2\n",
-    "#LMTD(math.log mean temprature difference) is defined as (deltaT2-deltaT1)/(ln(deltaT2/deltaT1)) for counter flow.\n",
-    "print\"LMTD is given by (deltaT2-deltaT1)/(ln(deltaT2/deltaT1)) in °C \"\n",
-    "#let X=log10((deltaT2/deltaT1)) and Y=log10(2.718281)\n",
-    "X=math.log10((deltaT2/deltaT1));\n",
-    "Y=math.log10(2.718281);\n",
-    "ln=X/Y;\n",
-    "LMTD=(deltaT2-deltaT1)/ln\n",
-    "print\"LMTD=\",LMTD\n",
-    "#Area(A)=20m**2\n",
-    "A=20;\n",
-    "#We have to employ correction factor(F) for the cross flow arrangement.\n",
-    "#We evaluate dimensionless parameters P=(Tco-Tci)/(Thi-Tco) and R=(Thi-Tho)/(Tco-Tci).\n",
-    "print\"dimensionless parameters P and R are\"\n",
-    "P=(Tco-Tci)/(Thi-Tci)\n",
-    "R=(Thi-Tho)/(Tco-Tci)\n",
-    "print\"P=\",P\n",
-    "print\"R=\",R\n",
-    "print\"correction factor(F) for the cross flow arrangement as obtained from graph of F vs Single Pass flow with fluids unmixed\"\n",
-    "F=0.75\n",
-    "#The rate of heat transfer can be calculate as Q=mdoth*ch*(Th1-Th2)\n",
-    "Q=mdotc*cc*(Tco-Tci);\n",
-    "print\"Q=\",Q\n",
-    "#overall heat transfer coefficient(U)=Q/(A*F*LMTD)\n",
-    "print\"overall heat transfer coefficient(U)=Q/(A*F*LMTD)in W/(m**2*K)\"\n",
-    "U=Q/(A*F*LMTD)\n",
-    "print\"U=\",U\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Ex10.5:pg-419"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 7,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Introduction to heat transfer by S.K.Som, Chapter 10, Example 5\n",
-      "(a)Applying LMTD method\n",
-      "The rate of heat transfer Q=mdotw*cpw*(Tout-Tin) in W\n",
-      "Q= 300960.0\n",
-      "The unknown outlet temprature(Thout) of geothermal fluid  in °C \n",
-      "Thout= 125.085846868\n",
-      "LMTD is given by (deltaT2-deltaT1)/(ln(deltaT2/deltaT1)) in °C \n",
-      "LMTD= 81.903612671\n",
-      "Area(A)=Q/(U*LMTD) in m**2\n",
-      "A= 6.12427197827\n",
-      "To provide this surface area ,The length(L) of the tube required is given by L=A/(pi*D) in m\n",
-      "L= 129.961087757\n",
-      "(b)Applying NTU method\n",
-      "The heat capacity rates are defined as Ch=mdoth*cph and Cc=mdotw*cpw in KW/°C\n",
-      "Ch= 8.62\n",
-      "Cc= 5.016\n",
-      "C=Cmin/Cmax\n",
-      "C= 0.581902552204\n",
-      "Heat transfer effectiveness is defined as eff=Q/(Cmin*(Thin-Tin))\n",
-      "eff= 0.461538461538\n",
-      "NTU is determined by NTU=(1/(C-1))*ln((eff-1)/(eff*C-1))\n",
-      "NTU= 0.732568418453\n",
-      "Area(A)=(NTU*Cmin)/U in m**2\n",
-      "A= 6.12427197827\n",
-      "To provide this surface area ,The length(L) of the tube required is given by L=A/(pi*D) in m\n",
-      "Hence same result is obtained for both methods\n",
-      "L= 129.961087757\n"
-     ]
-    }
-   ],
-   "source": [
-    "  \n",
-    " \n",
-    "  \n",
-    " \n",
-    "import math\n",
-    " \n",
-    "print\"Introduction to heat transfer by S.K.Som, Chapter 10, Example 5\"\n",
-    "#Water is heated from temprature ,Tin=30°C  to Tout=90°C  in a counter flow double pipe heat exchanger.\n",
-    "Tin=30;\n",
-    "Tout=90;\n",
-    "#Water flows at a mass flow rate of mdotw=1.2kg/s\n",
-    "mdotw=1.2;\n",
-    "#The heating is accomplished by a geothermal fluid which enters the heat exchanger at temprature ,Thin=160°C at the mass flow rate of mdoth=2kg/s\n",
-    "mdoth=2;\n",
-    "Thin=160;\n",
-    "#The inner tube is thin walled having diameter(D)=15mm or 0.015m\n",
-    "D=0.015;\n",
-    "#overall heat transfer coefficient(U)=600 W/(m**2*K)\n",
-    "U=600;\n",
-    "#The specific heat of water and geothermal fluid is (cpw=4.18kJ/(kg*K))and(cph=4.31kJ/(kg*K)) respectively\n",
-    "cpw=4.18*10**3;\n",
-    "cph=4.31*10**3;\n",
-    "#The rate of heat transfer in heat exchanger can be calculate as Q=mdotw*cpw*(Tout-Tin)\n",
-    "print\"(a)Applying LMTD method\"\n",
-    "print\"The rate of heat transfer Q=mdotw*cpw*(Tout-Tin) in W\"\n",
-    "Q=mdotw*cpw*(Tout-Tin)\n",
-    "print\"Q=\",Q\n",
-    "#The unknown outlet temprature(Thout) of geothermal fluid may be found from energy balance mdotw*cpw*(Tout-Tin)=mdoth*cph*(Thin-Thout)\n",
-    "print\"The unknown outlet temprature(Thout) of geothermal fluid  in °C \"\n",
-    "Thout=Thin-Q/(mdoth*cph)\n",
-    "print\"Thout=\",Thout\n",
-    "deltaT1=Thin-Tout;#Temprature difference between inlet temprature of hot fluid and outlet temprature of cold fluid\n",
-    "deltaT2=Thout-Tin;#Temprature difference between outlet temprature of hot fluid and inlet temprature of cold fluid\n",
-    "#LMTD is defined as (deltaT2-deltaT1)/(ln(deltaT2/deltaT1)) for counter flow.\n",
-    "print\"LMTD is given by (deltaT2-deltaT1)/(ln(deltaT2/deltaT1)) in °C \"\n",
-    "#let X=math.log10((deltaT2/deltaT1))and Y=math.log10(2.718281)\n",
-    "X=math.log10((deltaT2/deltaT1));\n",
-    "Y=math.log10(2.718281);\n",
-    "ln=X/Y;\n",
-    "LMTD=(deltaT2-deltaT1)/ln\n",
-    "print\"LMTD=\",LMTD\n",
-    "#Area(A)=Q/(U*LMTD) in m**2\n",
-    "print\"Area(A)=Q/(U*LMTD) in m**2\"\n",
-    "A=Q/(U*LMTD)\n",
-    "print\"A=\",A\n",
-    "print\"To provide this surface area ,The length(L) of the tube required is given by L=A/(pi*D) in m\"\n",
-    "L=A/(math.pi*D)\n",
-    "print\"L=\",L\n",
-    "print\"(b)Applying NTU method\"\n",
-    "#The heat capacity rates are defined as Ch=mdoth*cph and Cc=mdotw*cw in KW/°C\n",
-    "print\"The heat capacity rates are defined as Ch=mdoth*cph and Cc=mdotw*cpw in KW/°C\"\n",
-    "Ch=(mdoth*cph)/1000\n",
-    "Cc=(mdotw*cpw)/1000\n",
-    "print\"Ch=\",Ch\n",
-    "print\"Cc=\",Cc\n",
-    "#So Cmin=Cc and Cmax=Ch\n",
-    "Cmin=Cc;\n",
-    "Cmax=Ch;\n",
-    "#C is defined as Cmin/Cmax\n",
-    "print\"C=Cmin/Cmax\"\n",
-    "C=Cmin/Cmax\n",
-    "print\"C=\",C\n",
-    "#Heat transfer effectiveness is (eff)\n",
-    "print\"Heat transfer effectiveness is defined as eff=Q/(Cmin*(Thin-Tin))\"\n",
-    "eff=(Q/1000)/(Cmin*(Thin-Tin))\n",
-    "print\"eff=\",eff\n",
-    "print\"NTU is determined by NTU=(1/(C-1))*ln((eff-1)/(eff*C-1))\"\n",
-    "#let X=math.log10((eff-1)/(eff*C-1)) and Y=math.log10(2.718281)\n",
-    "X=math.log10((eff-1)/(eff*C-1));\n",
-    "Y=math.log10(2.718281);\n",
-    "#ln=ln((eff-1)/(eff*C-1))\n",
-    "ln=X/Y;\n",
-    "#NTU is Number of transfer units\n",
-    "NTU=(1/(C-1))*ln\n",
-    "print\"NTU=\",NTU\n",
-    "#NTU =U*A/Cmin\n",
-    "print\"Area(A)=(NTU*Cmin)/U in m**2\"\n",
-    "A=(NTU*Cmin*1000)/U\n",
-    "print\"A=\",A\n",
-    "print\"To provide this surface area ,The length(L) of the tube required is given by L=A/(pi*D) in m\"\n",
-    "L=A/(math.pi*D)\n",
-    "print\"Hence same result is obtained for both methods\"\n",
-    "print\"L=\",L\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Ex10.6:pg-422"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 8,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Introduction to heat transfer by S.K.Som, Chapter 10, Example 6\n",
-      "NTU is defined as (U*A)/Cmin \n",
-      "NTU= 1.86170212766\n",
-      "Heat transfer effectiveness(eff) is defined as (1-e**(-NTU*(1-C))/(1-C*e**(-NTU*(1-C))\n",
-      "eff= 0.683715054322\n",
-      "The total heat transfer rate (Q)=eff*Cmin*(Thi-Tci) in kW\n",
-      "Q= 163.886498521\n"
-     ]
-    }
-   ],
-   "source": [
-    "  \n",
-    " \n",
-    "  \n",
-    " \n",
-    "import math\n",
-    " \n",
-    "print\"Introduction to heat transfer by S.K.Som, Chapter 10, Example 6\"\n",
-    "#Water having specific heat,cw=4.18kJ/(kg*K) enters a counterflow double pipe heat exchanger at temprature,Tci=35°C flowing at the mass flow rate of mdotw=0.8 kg/s.\n",
-    "cw=4.18;\n",
-    "mdotw=0.8;\n",
-    "Tci=35;\n",
-    "#It is heated by oil having specific heat,co=1.88kJ/(kg*K) flowing at the mass flow rate of mdoto=1.5 kg/s from an inlet temprature(Thi) of 120°C.\n",
-    "co=1.88;\n",
-    "mdoto=1.5;\n",
-    "Thi=120;\n",
-    "#For an area(A) of 15m**2 and an overall heat transfer coefficient(U) of 350W/(m**2*K).\n",
-    "A=15;\n",
-    "U=350;\n",
-    "#Cwater and Co are heat capacities for water and oil respectively\n",
-    "#Cwater=mdotw*cw and Co=mdoto*co\n",
-    "Cwater=mdotw*cw;\n",
-    "Co=mdoto*co;\n",
-    "#C=Cmin/Cmax\n",
-    "Cmin=min(Cwater,Co);\n",
-    "Cmax=max(Cwater,Co);\n",
-    "C=Cmin/Cmax;\n",
-    "#NTU is number of transfer units\n",
-    "#NTU=(U*A)/Cmin\n",
-    "print\"NTU is defined as (U*A)/Cmin \"\n",
-    "NTU=(U*A)/(Cmin*1000)\n",
-    "print\"NTU=\",NTU\n",
-    "#Heat transfer effectiveness(eff) is defined as (1-e**(-NTU*(1-C))/(1-C*e**(-NTU*(1-C))\n",
-    "print\"Heat transfer effectiveness(eff) is defined as (1-e**(-NTU*(1-C))/(1-C*e**(-NTU*(1-C))\"\n",
-    "eff=(1-math.e**(-NTU*(1-C)))/(1-C*math.e**(-NTU*(1-C)))\n",
-    "print\"eff=\",eff\n",
-    "#Hence The total heat transfer rate (Q)=eff*Cmin*(Thi-Tci)in kW.\n",
-    "print\"The total heat transfer rate (Q)=eff*Cmin*(Thi-Tci) in kW\"                           \n",
-    "Q=eff*Cmin*(Thi-Tci)\n",
-    "print\"Q=\",Q\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Ex10.7:pg-424 "
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 9,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Introduction to heat transfer by S.K.Som, Chapter 10, Example 7\n",
-      "NTU is defined as (U*A)/Cmin \n",
-      "NTU= 5.44554455446\n",
-      "The effectiveness of heat exchanger is\n",
-      "eff= 9.98333889769\n",
-      "The total heat transfer rate(Q)=eff*Cmin*(Thi-Tci) in W\n",
-      "Q= 10587330.901\n",
-      "The exit temprature of air in °C \n",
-      "Tho= -923.250584257\n"
-     ]
-    }
-   ],
-   "source": [
-    "  \n",
-    " \n",
-    "  \n",
-    " \n",
-    "import math\n",
-    " \n",
-    "print\"Introduction to heat transfer by S.K.Som, Chapter 10, Example 7\"\n",
-    "#Water enters a cross flow heat exchanger (both fluids unmixed) at temprature(Tci)=20°C amd flows at a mass flow rate of mdotw=7kg/s\n",
-    "Tci=20;\n",
-    "mdotw=7;\n",
-    "#The air flows at a mass flow rate of mdota=10kg/s from Temprature(Thi)=125°C \n",
-    "mdota=10;\n",
-    "Thi=125;\n",
-    "#The overall heat transfer coefficient(U)=220W/(m**2*K)and Area(A)=250m**2.\n",
-    "U=220;\n",
-    "A=250;\n",
-    "#The specific heat of air (cpa=1.01kJ/(kg*K)) and water is (cpw=4.18kJ/(kg*K))\n",
-    "cpa=1.01;\n",
-    "cpw=4.18;\n",
-    "#Cair and Cwater are heat capacities of air and water respectively\n",
-    "Cair=mdota*cpa;\n",
-    "Cwater=mdotw*cpw;\n",
-    "#C=Cmin/Cmax\n",
-    "Cmin=min(Cwater,Cair);\n",
-    "Cmax=max(Cwater,Cair);\n",
-    "C=Cmin/Cmax;\n",
-    "#NTU is number of transfer units\n",
-    "#NTU=(U*A)/Cmin\n",
-    "print\"NTU is defined as (U*A)/Cmin \"\n",
-    "NTU=(U*A)/(Cmin*1000)\n",
-    "print\"NTU=\",NTU\n",
-    "#To determine the effectiveness of heat exchanger we have to find out the suitable expression \n",
-    "#For this type of heat exchanger The effectiveness(eff)is determined by (1-e**((NTU**.22*(e**-(C*NTU**0.78)-1)/C)\n",
-    "print\"The effectiveness of heat exchanger is\"\n",
-    "eff=(1-math.e**((NTU**0.22))*(math.e**(-C*NTU**0.78)-1)/C)\n",
-    "print\"eff=\",eff\n",
-    "#Hence The total heat transfer rate(Q)=eff*Cmin*(Thi-Tci)in W.\n",
-    "print\"The total heat transfer rate(Q)=eff*Cmin*(Thi-Tci) in W\"\n",
-    "Q=eff*Cmin*1000*(Thi-Tci)\n",
-    "print\"Q=\",Q\n",
-    "#The exit temprature(Tho) of air is given by Thi-(Q/(mdota*cpa))\n",
-    "print\"The exit temprature of air in °C \"\n",
-    "Tho=Thi-(Q/(mdota*1000*cpa))#NOTE:-The answer slightly varies from the answer in book(i.e Tho=26°C) because the value of Q taken in book is approximated to 1*10**6W.\n",
-    "print\"Tho=\",Tho\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Ex10.8:pg-437"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 10,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Introduction to heat transfer by S.K.Som, Chapter 10, Example 8\n",
-      "(a)Considering a parallel flow arrangement \n",
-      "Tho= 50\n",
-      "The minimum flow rate required for the oil in kg/s\n",
-      "mdoth= 2.22340425532\n",
-      "(b)Theoretical question\n",
-      "If LMTD--->0,Then for a finite value of heat transfer rate U*A--->infinity.For a given finite length this implies value of U which is not possible.\n",
-      "(c)Let us consider a counter flow arrangement\n",
-      "The minimum flow rate required for the oil in kg/s\n",
-      "Ch= 2.78666666667\n",
-      "Cc= 8.36\n",
-      "Effectiveness of heat exchanger is \n",
-      "eff= 1.0\n"
-     ]
-    }
-   ],
-   "source": [
-    "  \n",
-    " \n",
-    "  \n",
-    " \n",
-    "import math\n",
-    " \n",
-    "print\"Introduction to heat transfer by S.K.Som, Chapter 10, Example 8\"\n",
-    "#A double pipe heat exchanger of length(L)=0.30m is to be used to heat water(specific heat,cc=4.18kJ/(kg*K)) and mass flow rate(mdotw=2kg/s)\n",
-    "L=0.30;\n",
-    "cc=4.18;\n",
-    "mdotw=2;\n",
-    "#The water enters at temprature(Tci)=25°C  and leaves at temprature(Tco)=50°C\n",
-    "#The flow rate of oil is mdoth\n",
-    "Tci=25;\n",
-    "Tco=50; \n",
-    "#The oil used as hot fluid has(specific heat,ch=1.88kJ/(kg*K)) and has inlet temprature(Thi)=100°C \n",
-    "ch=1.88;\n",
-    "Thi=100;\n",
-    "print\"(a)Considering a parallel flow arrangement \"\n",
-    "#For minimum value of mdoth\n",
-    "#The theoretical minimum value of outlet temprature of hot fluid(Tho) under this situation is equal to Tco\n",
-    "Tho=Tco;\n",
-    "print\"Tho=\",Tho\n",
-    "#The mass flow rate of oil is given by energy balance as mdoth=(mdotw*cpw*(Tco-Tci))/(cph*(Thi-Tho))\n",
-    "print\"The minimum flow rate required for the oil in kg/s\"\n",
-    "mdoth=(mdotw*cc*(Tco-Tci))/(ch*(Thi-Tho))\n",
-    "print\"mdoth=\",mdoth\n",
-    "print\"(b)Theoretical question\"\n",
-    "print\"If LMTD--->0,Then for a finite value of heat transfer rate U*A--->infinity.For a given finite length this implies value of U which is not possible.\"\n",
-    "print\"(c)Let us consider a counter flow arrangement\"\n",
-    "#In this case value of Tho=Tci.\n",
-    "Tho=Tci;\n",
-    "#The mass flow rate of oil is given by energy balance as mdoth=(mdotw*cpw*(Tco-Tci))/(cph*(Thi-Tho))\n",
-    "print\"The minimum flow rate required for the oil in kg/s\"\n",
-    "mdoth=(mdotw*cc*(Tco-Tci))/(ch*(Thi-Tci))\n",
-    "#Now Heat capacities are Ch=mdoth*ch and Cc=mdotw*cc\n",
-    "Ch=mdoth*ch; \n",
-    "Cc=mdotw*cc;\n",
-    "print\"Ch=\",Ch\n",
-    "print\"Cc=\",Cc\n",
-    "Cmin=min(Ch,Cc);#minimum heat capacity in Ch and Cc \n",
-    "#Effectiveness of heat exchanger is eff.\n",
-    "#Tho=Tci for this kind of arrangement\n",
-    "Tho=Tci;\n",
-    "print\"Effectiveness of heat exchanger is \"\n",
-    "eff=(mdoth*ch*(Thi-Tho))/(mdoth*ch*(Thi-Tci))\n",
-    "print\"eff=\",eff\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\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": "2.7.11"
-  }
- },
- "nbformat": 4,
- "nbformat_minor": 0
-}
diff --git a/Introduction_to_Heat_Transfer_by_S._K._Som/Chapter11.ipynb b/Introduction_to_Heat_Transfer_by_S._K._Som/Chapter11.ipynb
index c44923f7..a46cced0 100644
--- a/Introduction_to_Heat_Transfer_by_S._K._Som/Chapter11.ipynb
+++ b/Introduction_to_Heat_Transfer_by_S._K._Som/Chapter11.ipynb
@@ -35,6 +35,10 @@
     }
    ],
    "source": [
+    " \n",
+    " \n",
+    " \n",
+    " \n",
     "import math\n",
     " \n",
     "print\"Introduction to heat transfer by S.K.Som, Chapter 11, Example 3\"\n",
@@ -53,7 +57,7 @@
     "A1=2;\n",
     "A3=2.5;\n",
     "F31=(A1/A3)*F13\n",
-    "print\"F31=\",F31"
+    "print\"F31=\",F31\n"
    ]
   },
   {
@@ -83,6 +87,10 @@
     }
    ],
    "source": [
+    " \n",
+    " \n",
+    " \n",
+    " \n",
     "import math\n",
     " \n",
     "print\"Introduction to heat transfer by S.K.Som, Chapter 11, Example 4\"\n",
@@ -108,7 +116,11 @@
     "#This implies F14=((F1,2-4*(A1+A2)))-A2*F24)/A2\n",
     "print\"The view factor F14=((F1,2-4*(A1+A2)))-A2*F24)/A2\"\n",
     "F14=((F124*(A1+A2))-(A2*F24))/A2\n",
-    "print\"F14=\",F14"
+    "print\"F14=\",F14\n",
+    "\n",
+    "\n",
+    "\n",
+    " \n"
    ]
   },
   {
@@ -137,6 +149,10 @@
     }
    ],
    "source": [
+    " \n",
+    " \n",
+    " \n",
+    " \n",
     "import math\n",
     " \n",
     "print\"Introduction to heat transfer by S.K.Som, Chapter 11, Example 5\"\n",
@@ -155,7 +171,7 @@
     "#Let A1/A2=A\n",
     "A=1/4;\n",
     "F31=(A)*F13\n",
-    "print\"F31=\",F31"
+    "print\"F31=\",F31\n"
    ]
   },
   {
@@ -190,6 +206,10 @@
     }
    ],
    "source": [
+    " \n",
+    " \n",
+    " \n",
+    " \n",
     "import math\n",
     " \n",
     "print\"Introduction to heat transfer by S.K.Som, Chapter 11, Example 6\"\n",
@@ -232,7 +252,27 @@
     "#Therefore we can write Q1=A1*sigma*(T1**4-F12*T2**4-F1s*Ts**4)\n",
     "print\"The net rate of energy loss from the surface at 127°C if the surrounding other than the two surfaces act as black body at 300K in W \"\n",
     "Q1=A1*sigma*(T1**4-F12*T2**4-F1s*Ts**4)\n",
-    "print\"Q1=\",Q1"
+    "print\"Q1=\",Q1\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n"
    ]
   },
   {
@@ -262,6 +302,10 @@
     }
    ],
    "source": [
+    " \n",
+    " \n",
+    " \n",
+    " \n",
     "import math\n",
     " \n",
     "print\"Introduction to heat transfer by S.K.Som, Chapter 11, Example 7\"\n",
@@ -283,7 +327,17 @@
     "#So,The net rate of heat transfer when the two surfaces are black is Q/A=sigma*(T1**4-T2**4)\n",
     "print\"The net rate of heat transfer when the two surfaces are black is Q/A=sigma*(T1**4-T2**4) in W\"\n",
     "H=sigma*(T1**4-T2**4)\n",
-    "print\"H=\",H"
+    "print\"H=\",H\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n"
    ]
   },
   {
@@ -317,6 +371,10 @@
     }
    ],
    "source": [
+    " \n",
+    " \n",
+    " \n",
+    " \n",
     "import math\n",
     " \n",
     "print\"Introduction to heat transfer by S.K.Som, Chapter 11, Example 8\"\n",
@@ -353,7 +411,20 @@
     "print\"Q2=\",Q2\n",
     "print\"Error(E) is given By ((Q2-Q1)/Q1)*100 in percentage\"\n",
     "E=((Q2-Q1)/Q1)*100\n",
-    "print\"E=\",E"
+    "print\"E=\",E\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n"
    ]
   },
   {
@@ -384,6 +455,10 @@
     }
    ],
    "source": [
+    " \n",
+    " \n",
+    " \n",
+    " \n",
     "import math\n",
     " \n",
     "print\"Introduction to heat transfer by S.K.Som, Chapter 11, Example 10\"\n",
@@ -428,7 +503,19 @@
     "#So Q/A=(sigma*(T1**4-T2**4))/(R)\n",
     "#Let Q/A=H\n",
     "H=(sigma*(T1**4-T2**4))/(R)\n",
-    "print\"H=\",H"
+    "print\"H=\",H\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n"
    ]
   }
  ],
diff --git a/Introduction_to_Heat_Transfer_by_S._K._Som/Chapter11_5r7Matr.ipynb b/Introduction_to_Heat_Transfer_by_S._K._Som/Chapter11_5r7Matr.ipynb
deleted file mode 100644
index a46cced0..00000000
--- a/Introduction_to_Heat_Transfer_by_S._K._Som/Chapter11_5r7Matr.ipynb
+++ /dev/null
@@ -1,543 +0,0 @@
-{
- "cells": [
-  {
-   "cell_type": "markdown",
-   "metadata": {
-    "collapsed": true
-   },
-   "source": [
-    "# Chapter 11:Radiation heat transfer"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Ex11.3:pg-445"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 1,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Introduction to heat transfer by S.K.Som, Chapter 11, Example 3\n",
-      "The view factors F13 and F31 between the surfaces 1 and 3 are \n",
-      "F13= 0.04\n",
-      "F31= 0.032\n"
-     ]
-    }
-   ],
-   "source": [
-    " \n",
-    " \n",
-    " \n",
-    " \n",
-    "import math\n",
-    " \n",
-    "print\"Introduction to heat transfer by S.K.Som, Chapter 11, Example 3\"\n",
-    "print\"The view factors F13 and F31 between the surfaces 1 and 3 are \"\n",
-    "#Determine the view factors F13 and F31 between the surfaces 1 and 3.\n",
-    "#F1-2,3=F12+F13\n",
-    "#So F13=F1-2,3-F12\n",
-    "#Let F1-2,3=F123\n",
-    "#From Radiation Shape factor b/w two perpendicular rectangles with a commom edge table we get F12=.027,F1-2,3=0.31\n",
-    "F123=0.31;#View factor\n",
-    "F12=.27;#View factor\n",
-    "F13=F123-F12#View factor\n",
-    "print\"F13=\",F13\n",
-    "#A1,A2 and A3 are the emitting surface areas\n",
-    "#From reciprocity relation F31=(A1/A3)/F13\n",
-    "A1=2;\n",
-    "A3=2.5;\n",
-    "F31=(A1/A3)*F13\n",
-    "print\"F31=\",F31\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Ex11.4:pg-447"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 3,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Introduction to heat transfer by S.K.Som, Chapter 11, Example 4\n",
-      "F124= 0.04\n",
-      "F24= 0.06\n",
-      "The view factor F14=((F1,2-4*(A1+A2)))-A2*F24)/A2\n",
-      "F14= 0.02\n"
-     ]
-    }
-   ],
-   "source": [
-    " \n",
-    " \n",
-    " \n",
-    " \n",
-    "import math\n",
-    " \n",
-    "print\"Introduction to heat transfer by S.K.Som, Chapter 11, Example 4\"\n",
-    "#Determine the view factors F14 for the composite surface .\n",
-    "#From the table of radiation shape factor b/w two perpendicular surfaces F1,2-3,4=0.14 and F1,2-3=0.1\n",
-    "#By subdivision of the recieving surfaces we get F1,2-4=F1,2-3,4-F1,2-3\n",
-    "#Let F1,2-4=F124 , F1,2-3,4=F1234 , F1,2-3=F123\n",
-    "F1234=0.14;#View factor\n",
-    "F123=0.1;#View factor\n",
-    "F124=F1234-F123;#View factor\n",
-    "print\"F124=\",F124\n",
-    "#Again from the table of radiation shape factor b/w two perpendicular surfaces F2-3,4=0.24 , F23=0.18\n",
-    "#Let F2-3,4=F234\n",
-    "F234=0.24;#View factor\n",
-    "F23=0.18;#View factor\n",
-    "#By subdivision of the recieving surfaces we get F24=F2-3,4-F23\n",
-    "F24=F234-F23;#view factor\n",
-    "print\"F24=\",F24\n",
-    "#A1 and A2 are the emitting surface areas.\n",
-    "A1=12;\n",
-    "A2=12;\n",
-    "#Now by subdivision of emitting surfaces F1,2-4=(1/(A1+A2))*(A1*F14+A2*F24)\n",
-    "#This implies F14=((F1,2-4*(A1+A2)))-A2*F24)/A2\n",
-    "print\"The view factor F14=((F1,2-4*(A1+A2)))-A2*F24)/A2\"\n",
-    "F14=((F124*(A1+A2))-(A2*F24))/A2\n",
-    "print\"F14=\",F14\n",
-    "\n",
-    "\n",
-    "\n",
-    " \n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Ex11.5:pg-453"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 4,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Introduction to heat transfer by S.K.Som, Chapter 11, Example 5\n",
-      "The view factors of cylindrical surface with respect to the base are\n",
-      "F13= 0.84\n",
-      "F31= 0.0\n"
-     ]
-    }
-   ],
-   "source": [
-    " \n",
-    " \n",
-    " \n",
-    " \n",
-    "import math\n",
-    " \n",
-    "print\"Introduction to heat transfer by S.K.Som, Chapter 11, Example 5\"\n",
-    "#Consider a cylinder having length,L=2r determine the view factor of cylindrical surface with respect to the base.\n",
-    "#From the graph of radiation shape factor b/w parallel coaxial disks of equal diameter F12=0.16\n",
-    "F12=0.16;#View factor\n",
-    "#By the summation rule of an enclosure F11+F12+F13=1\n",
-    "#But F11=0(since the base surface is flat)\n",
-    "F11=0;#View factor\n",
-    "print\"The view factors of cylindrical surface with respect to the base are\"\n",
-    "F13=1-F12-F11#view factor\n",
-    "print\"F13=\",F13\n",
-    "#By making use of reciprocity theorem we have F31=(A1/A2)*F13\n",
-    "#A1 and A2 are emitting surface areas\n",
-    "#A1/A2=(pi*r**2)/(2*pi*r*2*r)=1/4\n",
-    "#Let A1/A2=A\n",
-    "A=1/4;\n",
-    "F31=(A)*F13\n",
-    "print\"F31=\",F31\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Ex11.6:pg-456"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 5,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Introduction to heat transfer by S.K.Som, Chapter 11, Example 6\n",
-      "The rate of heat transfer is given by Q=A1*F12*sigma*(T1**4-T2**4) in W\n",
-      "Q= -2918.916\n",
-      "Here minus sign indicates that the net heat transfer is from surface2 to surface1\n",
-      "The net rate of energy loss from the surface at 127°C if the surrounding other than the two surfaces act as black body at 0K in W\n",
-      "Q1= 2894.4216\n",
-      "The view factor of surface 1 with respect to surrounding is\n",
-      "F1s= 0.89\n",
-      "The net rate of energy loss from the surface at 127°C if the surrounding other than the two surfaces act as black body at 300K in W \n",
-      "Q1= 1055.04525\n"
-     ]
-    }
-   ],
-   "source": [
-    " \n",
-    " \n",
-    " \n",
-    " \n",
-    "import math\n",
-    " \n",
-    "print\"Introduction to heat transfer by S.K.Som, Chapter 11, Example 6\"\n",
-    "#Two rectangles length,L=1.5m by breadth,B=3.0m are parallel and directly opposed.\n",
-    "L=1.5;\n",
-    "B=3;\n",
-    "#They are 3m apart\n",
-    "#Temprature(T1) of surface 1 is 127°C or 400K and temprature(T2) of surface 2 is 327°C or 600K \n",
-    "T1=400;\n",
-    "T2=600;\n",
-    "#Area (A) is the product of L and B\n",
-    "A1=L*B;\n",
-    "#Stefan -Boltzman constant(sigma)=5.67*10**-8 W/(m**2*K**4)\n",
-    "sigma=5.67*10**-8;\n",
-    "#From the graph of radiation shape factor b/w parallel rectangles F12=0.11\n",
-    "F12=0.11;#View factor\n",
-    "#The rate of heat transfer is given by Q=A1*F12*sigma*(T1**4-T2**4)\n",
-    "print\"The rate of heat transfer is given by Q=A1*F12*sigma*(T1**4-T2**4) in W\"\n",
-    "Q=A1*F12*sigma*(T1**4-T2**4)\n",
-    "print\"Q=\",Q\n",
-    "print\"Here minus sign indicates that the net heat transfer is from surface2 to surface1\"\n",
-    "#Surface1 recieves energy only from surface 2,since the surrounding is at 0K.\n",
-    "#Therefore Q1=A1*Eb1-A2*F21*Eb2\n",
-    "#This implies Q1 can also be written as A1*sigma*(T1**4-F12*T2**4)\n",
-    "#From reciprocity theorem F21=F12 (since A1=A2)\n",
-    "F21=F12;#view factor\n",
-    "print\"The net rate of energy loss from the surface at 127°C if the surrounding other than the two surfaces act as black body at 0K in W\" \n",
-    "Q1=A1*sigma*(T1**4-F12*T2**4)\n",
-    "print\"Q1=\",Q1\n",
-    "#In the case when surrounding is at temprature, Ts=300K ,the energy recieved from the surrounding by the surface 1 has to be considered.\n",
-    "Ts=300;\n",
-    "#Applying summation rule of view factors F11+F12+F1s=1\n",
-    "F11=0;#view factor\n",
-    "print\"The view factor of surface 1 with respect to surrounding is\"\n",
-    "F1s=1-F11-F12\n",
-    "print\"F1s=\",F1s\n",
-    "#subscript s denotes the surroundings\n",
-    "#Q1=A1*Eb1-A2*F21*Eb2-As*Fs1*Ebs\n",
-    "#With the help of reciprocity theorem A2*F21=A1*F12 , As*Fs1=A1*F1s\n",
-    "#Therefore we can write Q1=A1*sigma*(T1**4-F12*T2**4-F1s*Ts**4)\n",
-    "print\"The net rate of energy loss from the surface at 127°C if the surrounding other than the two surfaces act as black body at 300K in W \"\n",
-    "Q1=A1*sigma*(T1**4-F12*T2**4-F1s*Ts**4)\n",
-    "print\"Q1=\",Q1\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Ex11.7:pg-470"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 10,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Introduction to heat transfer by S.K.Som, Chapter 11, Example 7\n",
-      "The net rate of heat transfer per unit area is given Q/A=(sigma*(T1**4-T2**4))/((1/Ɛ1)+(1/Ɛ2)-1) in W\n",
-      "H= 1764.51561476\n",
-      "The net rate of heat transfer when the two surfaces are black is Q/A=sigma*(T1**4-T2**4) in W\n",
-      "H= 3276.95757027\n"
-     ]
-    }
-   ],
-   "source": [
-    " \n",
-    " \n",
-    " \n",
-    " \n",
-    "import math\n",
-    " \n",
-    "print\"Introduction to heat transfer by S.K.Som, Chapter 11, Example 7\"\n",
-    "#Two parallel infinite surafces are maintained at tempratures T2=200°C or 473.15K and T1=300°C or 573.15K\n",
-    "T1=573.15;\n",
-    "T2=473.15;\n",
-    "#The emissivity(emi) is 0.7 for both the surfaces which are gray.\n",
-    "emi1=0.7;\n",
-    "emi2=0.7;\n",
-    "#stefan=boltzman constant(sigma)=5.67*10**-8W/(m**2*K**4)\n",
-    "sigma=5.67*10**-8;\n",
-    "#The net rate of heat transfer per unit area is given Q/A=(sigma*(T1**4-T2**4))/((1/Ɛ1)+(1/Ɛ2)-1)\n",
-    "#Let Q/A=H\n",
-    "print\"The net rate of heat transfer per unit area is given Q/A=(sigma*(T1**4-T2**4))/((1/Ɛ1)+(1/Ɛ2)-1) in W\"\n",
-    "H=(sigma*(T1**4-T2**4))/((1/emi1)+(1/emi2)-1)\n",
-    "print\"H=\",H\n",
-    "#When the two surfaces are black\n",
-    "#This implies emiisivity(emi)=1 for both surfaces\n",
-    "#So,The net rate of heat transfer when the two surfaces are black is Q/A=sigma*(T1**4-T2**4)\n",
-    "print\"The net rate of heat transfer when the two surfaces are black is Q/A=sigma*(T1**4-T2**4) in W\"\n",
-    "H=sigma*(T1**4-T2**4)\n",
-    "print\"H=\",H\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Ex11.8:pg-482"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 8,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Introduction to heat transfer by S.K.Som, Chapter 11, Example 8\n",
-      "The net rate of heat exchange between the spheres when the surfaces are black is Q=A1*sigma*(T1**4-T2**4) in W \n",
-      "Q= 779.311327631\n",
-      "The net rate of radiation exchange when one surface is gray and other is diffuse is given by Q1=(A1*sigma*(T1**4-T2**4))/((1/emi1)+(A1/A2)*(1/emi2-1)) in W\n",
-      "Q1= 346.360590058\n",
-      "The net rate of radiation exchange when outer surface is assumed to be black body i;e(emi2=1) in W\n",
-      "Q2= 389.655663816\n",
-      "Error(E) is given By ((Q2-Q1)/Q1)*100 in percentage\n",
-      "E= 12.5\n"
-     ]
-    }
-   ],
-   "source": [
-    " \n",
-    " \n",
-    " \n",
-    " \n",
-    "import math\n",
-    " \n",
-    "print\"Introduction to heat transfer by S.K.Som, Chapter 11, Example 8\"\n",
-    "#Two concentric spheres of diameters D1=0.5m and D2=1m are separated by an air space.\n",
-    "#The surface tempratures are T1=400K and T2=300K\n",
-    "T1=400;\n",
-    "T2=300;\n",
-    "D1=0.5;\n",
-    "D2=1;\n",
-    "#A1 and A2 are the areas in m**2 of surface 1 and surface 2 respectively\n",
-    "A1=(math.pi*D1**2);\n",
-    "A2=(math.pi*D2**2);\n",
-    "#Stefan-Boltzman constant(sigma)=5.67*10**-8 W/(m**2*K**4)\n",
-    "sigma=5.67*10**-8;\n",
-    "#The emissivity is represented by emi \n",
-    "#The radiation heat exchange in case of two concentric sphere is given by Q=(A1*sigma*(T1**4-T2**4))/((1/emi1)+(A1/A2)*(1/emi2-1)) \n",
-    "#When the spheres are black emi1=emi2=1\n",
-    "emi1=1;\n",
-    "emi2=1;\n",
-    "#Hence Q=A1*sigma*(T1**4-T2**4)\n",
-    "print\"The net rate of heat exchange between the spheres when the surfaces are black is Q=A1*sigma*(T1**4-T2**4) in W \"\n",
-    "Q=A1*sigma*(T1**4-T2**4)\n",
-    "print\"Q=\",Q\n",
-    "#The net rate of radiation exchange when one surface is gray and other is diffuse having emi1=0.5 and emi2=0.5  \n",
-    "emi1=0.5;\n",
-    "emi2=0.5;\n",
-    "print\"The net rate of radiation exchange when one surface is gray and other is diffuse is given by Q1=(A1*sigma*(T1**4-T2**4))/((1/emi1)+(A1/A2)*(1/emi2-1)) in W\"  \n",
-    "Q1=(A1*sigma*(T1**4-T2**4))/((1/emi1)+(A1/A2)*(1/emi2-1))\n",
-    "print\"Q1=\",Q1\n",
-    "#The net rate of radiation exchange when outer surface is assumed to be black body i;e(emi2=1)\n",
-    "emi2=1;#emissivity of outer surface\n",
-    "print\"The net rate of radiation exchange when outer surface is assumed to be black body i;e(emi2=1) in W\"\n",
-    "Q2=(A1*sigma*(T1**4-T2**4))/((1/emi1)+(A1/A2)*(1/emi2-1))\n",
-    "print\"Q2=\",Q2\n",
-    "print\"Error(E) is given By ((Q2-Q1)/Q1)*100 in percentage\"\n",
-    "E=((Q2-Q1)/Q1)*100\n",
-    "print\"E=\",E\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Ex11.10:pg-484"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 6,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Introduction to heat transfer by S.K.Som, Chapter 11, Example 10\n",
-      "F12= 0.5\n",
-      "Equivalent resistance of thermal network (R) is given by R1+((1/R12)+(1/(R13+R23)))**-1+R2\n",
-      "R= 2.09523809524\n",
-      "The radiation flux leaving the hot wall is Q/A=[sigma*(T1**4-T2**4)]/(A*R) in W/m**2\n",
-      "H= 26655.2740483\n"
-     ]
-    }
-   ],
-   "source": [
-    " \n",
-    " \n",
-    " \n",
-    " \n",
-    "import math\n",
-    " \n",
-    "print\"Introduction to heat transfer by S.K.Som, Chapter 11, Example 10\"\n",
-    "#Given a furnace which can be approximated as an equuilateral triangle duct\n",
-    "#The hot wall is maintained at temprature (T1)=1000K and has emmisivity(emi1)=0.75\n",
-    "#The cold wall is at temprature(T2)=350K and has emmisivity(emi2)=0.7\n",
-    "T1=1000;\n",
-    "T2=350;\n",
-    "emi1=0.75;\n",
-    "emi2=0.7;\n",
-    "#Stefan-Boltzman constant(sigma)=5.67*10**-8 W/(m**2*K**4)\n",
-    "sigma=5.67*10**-8;\n",
-    "#The third wall is reradiating zone having Q3=0\n",
-    "#The radiation flux leaving the hot wall is Q/A=(sigma*(T1**4-T2**4))/(A*R)\n",
-    "#By summation rule F33+F31+F32=1\n",
-    "#F33=0(in consideration of surface to be plane)\n",
-    "#From symmetry F31=F32\n",
-    "F31=0.5;#View factors\n",
-    "F32=F31;#View factors\n",
-    "F33=0;#View factors\n",
-    "#From reciprocity theorem F13=F31 and F23=F32=0.5 (since A1=A2=A3=A)\n",
-    "F13=F31;#View factors\n",
-    "F23=F32;#View factors\n",
-    "#Again F11+F12+F13=1 from summation rule\n",
-    "F11=0;#View factors\n",
-    "F12=1-F13-F11;#View factors\n",
-    "print\"F12=\",F12\n",
-    "#R1,R2,R12,R13,R23 are the resistances\n",
-    "#R is equivalent resistance of thermal network is given by R1+((1/R12)+(1/(R13+R23)))**-1+R2\n",
-    "R1=(1-emi1)/(emi1);\n",
-    "R2=(1-emi2)/(emi2);\n",
-    "R12=1/(F12);\n",
-    "R13=1/(F13);\n",
-    "R23=1/(F23);\n",
-    "#R is equivalent resistance of thermal network \n",
-    "print\"Equivalent resistance of thermal network (R) is given by R1+((1/R12)+(1/(R13+R23)))**-1+R2\"\n",
-    "R=R1+((1/R12)+(1/(R13+R23)))**-1+R2\n",
-    "print\"R=\",R\n",
-    "#The radiation flux leaving the hot wall is Q/A.\n",
-    "print\"The radiation flux leaving the hot wall is Q/A=[sigma*(T1**4-T2**4)]/(A*R) in W/m**2\"\n",
-    "#Since A gets cancelled in the factor (A*R)\n",
-    "#So Q/A=(sigma*(T1**4-T2**4))/(R)\n",
-    "#Let Q/A=H\n",
-    "H=(sigma*(T1**4-T2**4))/(R)\n",
-    "print\"H=\",H\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\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": "2.7.11"
-  }
- },
- "nbformat": 4,
- "nbformat_minor": 0
-}
diff --git a/Introduction_to_Heat_Transfer_by_S._K._Som/Chapter1_QhYeq33.ipynb b/Introduction_to_Heat_Transfer_by_S._K._Som/Chapter1_QhYeq33.ipynb
deleted file mode 100644
index d9d7f745..00000000
--- a/Introduction_to_Heat_Transfer_by_S._K._Som/Chapter1_QhYeq33.ipynb
+++ /dev/null
@@ -1,529 +0,0 @@
-{
- "cells": [
-  {
-   "cell_type": "markdown",
-   "metadata": {
-    "collapsed": true
-   },
-   "source": [
-    "# Chapter 01:Fundamental concepts"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Ex1.1:pg- 4"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 4,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Introduction to heat transfer by S.K.Som, Chapter 1, Example 1\n",
-      "The steady state heat transfer rate per unit area through the thick slab is given by q=k(T1-T2)/L in W/m**2 \n",
-      "q= 50.0\n"
-     ]
-    }
-   ],
-   "source": [
-    "import math\n",
-    "print\"Introduction to heat transfer by S.K.Som, Chapter 1, Example 1\"\n",
-    "#The temprature of two faces of the slabs are T1=40°C & T2=20°C \n",
-    "#The thickness of the slab(L) is 80mm or .08m\n",
-    "#The thermal conductivity(k)of the material is .20 W/(m*K)\n",
-    "T1=40;\n",
-    "T2=20;\n",
-    "L=.08;\n",
-    "k=.20;\n",
-    "#The steady state heat transfer rate per unit area through the thick slab is given by q=k(T1-T2)/L\n",
-    "print\"The steady state heat transfer rate per unit area through the thick slab is given by q=k(T1-T2)/L in W/m**2 \"\n",
-    "q=k*(T1-T2)/L\n",
-    "print\"q=\",q"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Ex1.2:pg- 4"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 5,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Introduction to heat transfer by S.K.Som, Chapter 1, Example 2\n",
-      "The thickness of masonry wall is Lm in m\n",
-      "Lm= 0.5\n"
-     ]
-    }
-   ],
-   "source": [
-    "import math\n",
-    "print\"Introduction to heat transfer by S.K.Som, Chapter 1, Example 2\"\n",
-    "#The thermal conductivity(km)of masonry wall is .8 W/(mK)\n",
-    "#The thermal conductivity(kc)of composite wall is .2 W/(mK)\n",
-    "#The thickness of composite wall(Lc) is 100 mm or .1 m\n",
-    "km=.8;\n",
-    "kc=.2;\n",
-    "Lc=.1;\n",
-    "#The thickness of masonry wall(Lm) is to be found. \n",
-    "#The steady state heat flow(qm)through masonry wall is km(T1-T2)/L\n",
-    "# The steady state heat flow(qc)through composite wall is kc(T1-T2)/L\n",
-    "#As the steady rate of heat flow through masonry wall is 80% that through composite wall and both the wall have same surface area and same temp. difference so qm/qc=0.8=(km/kc)*(Lc/Lm)\n",
-    "#The thickness of masonry wall is Lm.\n",
-    "print\"The thickness of masonry wall is Lm in m\"\n",
-    "Lm=(km/kc)*(Lc/(0.8))\n",
-    "print\"Lm=\",Lm\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Ex1.4:pg-8"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 6,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Introduction to heat transfer by S.K.Som, Chapter 1, Example 4\n",
-      "The rate of heat transfer per unit area q=hbr*(Tinf-Ts) in W/m**2\n",
-      "q= 16000\n"
-     ]
-    }
-   ],
-   "source": [
-    "import math\n",
-    "print\"Introduction to heat transfer by S.K.Som, Chapter 1, Example 4\"\n",
-    "#The average forced convective heat transfer coefficient(hbr) is 200 W/( m**2 °C)\n",
-    "#The fluid temprature(Tinf) upstream of the cold surface is 100°C\n",
-    "#The surface temprature(Ts) is 20°C\n",
-    "hbr=200;\n",
-    "Tinf=100;\n",
-    "Ts=20;\n",
-    "#The rate of heat transfer per unit area is q\n",
-    "print\"The rate of heat transfer per unit area q=hbr*(Tinf-Ts) in W/m**2\"\n",
-    "q=hbr*(Tinf-Ts)\n",
-    "print\"q=\",q"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Ex1.5:pg-9"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 7,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Introduction to heat transfer by S.K.Som, Chapter 1, Example 5\n",
-      "The heat exchanger surface area(A)in m**2 required for 20 MJ/h of heating is \n",
-      "A= 0\n"
-     ]
-    }
-   ],
-   "source": [
-    "import math\n",
-    "print\"Introduction to heat transfer by S.K.Som, Chapter 1, Example 5\"\n",
-    "#The average heat transfer coefficient(hbr) is 800 W/(m**2°C)\n",
-    "#The surface temprature of heat exchanger is 75°C and air temprature is 25°C so deltaT=(75-25)\n",
-    "#The amount of heat exchanged(Q) is 20 MJ/h\n",
-    "#The heat exchanger surface area(A) is given by A=Q/(hbr*∆T)\n",
-    "hbr=800;\n",
-    "deltaT=(75-25);\n",
-    "Q=20;\n",
-    "print\"The heat exchanger surface area(A)in m**2 required for 20 MJ/h of heating is \"\n",
-    "A = (Q*10**6)/(3600*hbr*deltaT)\n",
-    "print\"A=\",A"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Ex1.6:pg-9"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 3,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Introduction to heat transfer by S.K.Som, Chapter 1, Example 6\n",
-      "The rate of heat transfer from the plate is given by Q=hbr*A*(Ts-Tinf)\n",
-      "Equating the above two equations we get hbr=(m*cp*|dT/dt|)/(A*(Ts-Tinf)) in W/(m**2°C)\n",
-      "hbr= 11.2\n",
-      "The rate of heat transfer can also be written in the form of Q=m*cp*|dT/dt| from an energy balance.\n",
-      "Q= 224.0\n",
-      "Equating the above two equations we get hbr=(m*cp*|dT/dt|)/(A*(Ts-Tinf)) in W/(m**2°C)\n"
-     ]
-    }
-   ],
-   "source": [
-    "import math\n",
-    "print\"Introduction to heat transfer by S.K.Som, Chapter 1, Example 6\"\n",
-    "#The temprature of the plate(Ts) is 225°C\n",
-    "#The ambient temprature (Tinf) is 25°C\n",
-    "#The change in plate temprature with time is dT/dt=-.02K/s\n",
-    "#The plate area (A)=.1m**2 , mass(m)= 4Kg and specific heat(cp)=2.8KJ/(Kg*K)\n",
-    "#The average free convective heat coefficient(hbr) is to be found\n",
-    "Ts=225;\n",
-    "Tinf=25;\n",
-    "#|dT/dt|=0.2,because it is modulus function and it converts negative values to positive value.\n",
-    "#Let |dT/dt|=X\n",
-    "X=0.02;\n",
-    "A=.1;\n",
-    "m=4;\n",
-    "cp=2.8;\n",
-    "print\"The rate of heat transfer from the plate is given by Q=hbr*A*(Ts-Tinf)\"\n",
-    "print\"Equating the above two equations we get hbr=(m*cp*|dT/dt|)/(A*(Ts-Tinf)) in W/(m**2°C)\"\n",
-    "hbr=(m*cp*10**3*X)/(A*(Ts-Tinf))\n",
-    "print\"hbr=\",hbr\n",
-    "Q=hbr*A*(Ts-Tinf)\n",
-    "print\"The rate of heat transfer can also be written in the form of Q=m*cp*|dT/dt| from an energy balance.\"\n",
-    "print\"Q=\",Q\n",
-    "print\"Equating the above two equations we get hbr=(m*cp*|dT/dt|)/(A*(Ts-Tinf)) in W/(m**2°C)\"\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Ex1.7:pg-10"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 8,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Introduction to heat transfer by S.K.Som, Chapter 1, Example 7\n",
-      "The heat flux per square meter is given by E/A=emi*sigma*T**4 in W/m**2\n",
-      "F= 556.4411381\n"
-     ]
-    }
-   ],
-   "source": [
-    "import math\n",
-    "print\"Introduction to heat transfer by S.K.Som, Chapter 1, Example 7\"\n",
-    "#The temprature(T) of brick wall after sunset is 50°C\n",
-    "#The emissity value(emi)=0.9\n",
-    "#The radiant heat flux per square meter =E/A Where E is radiant heat energy and A is area of brick wall.\n",
-    "#The stefan-Boltzman constant(sigma)=5.6697*10**-8 W/(m**2*K**4).\n",
-    "T=50;\n",
-    "emi=.9;\n",
-    "sigma=5.6697*10**-8;\n",
-    "print\"The heat flux per square meter is given by E/A=emi*sigma*T**4 in W/m**2\"\n",
-    "#Let E/A=F\n",
-    "F=emi*sigma*(T+273.15)**4\n",
-    "print\"F=\",F"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Ex1.8:pg-11"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 9,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Introduction to heat transfer by S.K.Som, Chapter 1, Example 8\n",
-      "The emitted radiant energy per unit surface area is given by Eb/A=sigma*T**4 in W/m**2\n",
-      "F= 618.267931222\n"
-     ]
-    }
-   ],
-   "source": [
-    "import math\n",
-    "print\"Introduction to heat transfer by S.K.Som, Chapter 1, Example 8\"\n",
-    "#The temprature(T) of asphalt pavement = 50°C\n",
-    "#The stefan-Boltzman constant(sigma)=5.6697*10**-8 W/(m**2*K**4).\n",
-    "T=50;\n",
-    "sigma=5.6697*10**-8;\n",
-    "#The emitted radiant energy per unit surface area is given by (Eb/A)=sigma*T**4\n",
-    "print\"The emitted radiant energy per unit surface area is given by Eb/A=sigma*T**4 in W/m**2\"\n",
-    "#Let Eb/A=F\n",
-    "F=sigma*(50+273.15)**4\n",
-    "print\"F=\",F\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Ex1.9:pg-12"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 10,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Introduction to heat transfer by S.K.Som, Chapter 1, Example 9\n",
-      "The rate of heat transfer per unit surface area of wall is given by Q/A=(Ta-Tb)/((1/hbr1*A)+(L/(k*A))+(1/hbr2*A))in W/m**2\n",
-      "F= 213.333333333\n",
-      "The surface tempratures of wall on 60°C side is T1 =Ta-(Q/(A*hbr1)) in °C\n",
-      "T1= 54.6666666667\n",
-      "The surface tempratures of wall on 20°C side is T2 =Tb+(Q/(A*hbr2)) in °C\n",
-      "T2= 41.3333333333\n"
-     ]
-    }
-   ],
-   "source": [
-    "import math\n",
-    "print\"Introduction to heat transfer by S.K.Som, Chapter 1, Example 9\"\n",
-    "#The Thickness(L) of wall= 150 mm or 0.15 m.\n",
-    "#The wall on one side is exposed to air at temprature(Ta)= 60°C and on the other side to air at temprature(Tb) = 20°C\n",
-    "#The average convective heat transfer coefficients are hbr1=40 W/(m**2°C) on the 60°C and hbr2= 10 W/(m**2°C) on 20°C side.\n",
-    "#The thermal conductivity(k)=.8 W/(m°C)\n",
-    "L=0.15;\n",
-    "Ta=60;\n",
-    "Tb=20;\n",
-    "hbr1=40;\n",
-    "hbr2=10;\n",
-    "k=0.8;\n",
-    "#Area(A=1 m**2 )since unit surface area is required.\n",
-    "A=1;\n",
-    "#The rate of heat transfer per unit surface area of wall is given by (Q/A)=(Ta-Tb)/((1/hbr1*A)+(L/(k*A))+(1/hbr2*A))\n",
-    "print\"The rate of heat transfer per unit surface area of wall is given by Q/A=(Ta-Tb)/((1/hbr1*A)+(L/(k*A))+(1/hbr2*A))in W/m**2\"\n",
-    "#Let Q/A=F\n",
-    "F=(Ta-Tb)/((1/hbr1*A)+(L/(k*A))+(1/hbr2*A))\n",
-    "print\"F=\",F\n",
-    "#The surface tempratures of wall on 60°C side is T1 and on 20°C side is T2\n",
-    "print\"The surface tempratures of wall on 60°C side is T1 =Ta-(Q/(A*hbr1)) in °C\"\n",
-    "T1 =Ta-(F/hbr1)\n",
-    "print\"T1=\",T1\n",
-    "print\"The surface tempratures of wall on 20°C side is T2 =Tb+(Q/(A*hbr2)) in °C\"\n",
-    "T2 =Tb+(F/hbr2)\n",
-    "print\"T2=\",T2"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Ex1.10:pg-13"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 11,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Introduction to heat transfer by S.K.Som, Chapter 1, Example 10\n",
-      "Heat transfer from the outer surface takes place only by radiation is given by Q/A=F1=emi*sigma*(T2**4-T0**4)in W/m**2 for different values of tempratures in K\n",
-      "heat transfer from the outer surface can also be written as Q/A=F2=(Ti-To)/((1/hbri)+(L/k)+(1/hr)) in W/m**2 at different tempratures in K\n",
-      "The values of temprature that are considered are <298 K\n",
-      "Satisfactory solutions for Temprature in K is\n",
-      "T2= 292.5\n",
-      "Approximate Rate of Heat Transfer in W/m**2 is\n",
-      "F1= 332.029390022\n",
-      "F2= 332.132667923\n"
-     ]
-    }
-   ],
-   "source": [
-    "import math\n",
-    "print\"Introduction to heat transfer by S.K.Som, Chapter 1, Example 10\"\n",
-    "#The spacecraft panel has thickness(L)=.01 m\n",
-    "#The spacecraft has inner temprature (Ti)=298 K\n",
-    "#The spacecraft has outer temprature(T2)\n",
-    "#The panel is exposed to deep space where temprature(To)= 0K\n",
-    "#The material has Thermal conductivity(k)= 5.0 W/(m*K)\n",
-    "#The emissivity(emi)=0.8\n",
-    "#The inner surface of the panel is exposed to airflow resulting in an average heat transfer coefficient(hbri)=70 W/(m**2*K)\n",
-    "L=0.01;\n",
-    "Ti=298.0;\n",
-    "To=0.0;\n",
-    "k=5.0;\n",
-    "emi=0.8;\n",
-    "hbri=70.0;\n",
-    "#The stefan Boltzman constant(sigma)= 5.67*10**-8 W/(m**2/K**4)\n",
-    "sigma=5.67*10**(-8);\n",
-    "#Heat transfer from the outer surface takes place only by radiation is given by  Q/A=emi*sigma*(T2**4-T0**4)in W/m**2=F1\n",
-    "#heat transfer from the outer surface can also be written as Q/A=(Ti-To)/((1/hbri)+(L/k)+(1/hr))=F2\n",
-    "#Radiation heat transfer coefficient(hr) is defined as Q/A=hr(T2-To)\n",
-    "#so hr=4.536*10**-8*T2**3\n",
-    "print\"Heat transfer from the outer surface takes place only by radiation is given by Q/A=F1=emi*sigma*(T2**4-T0**4)in W/m**2 for different values of tempratures in K\"\n",
-    "print\"heat transfer from the outer surface can also be written as Q/A=F2=(Ti-To)/((1/hbri)+(L/k)+(1/hr)) in W/m**2 at different tempratures in K\"\n",
-    "print\"The values of temprature that are considered are <298 K\"\n",
-    "for i in range(285,292):\n",
-    "    T2=i\n",
-    "    hr=4.536*10**(-8)*i**3\n",
-    "    F1=emi*sigma*(T2**4-To**4)\n",
-    "    F2=(Ti-To)/((1/hbri)+(L/k)+(1/hr))\n",
-    "if F1==F2:\n",
-    "    T2=i\n",
-    "else: \n",
-    "     T2=292.5\n",
-    "     hr=4.536*10**(-8)*T2**3\n",
-    "     F1=emi*sigma*(T2**4-To**4)\n",
-    "     F2=(Ti-To)/((1/hbri)+(L/k)+(1/hr))\n",
-    "print\"Satisfactory solutions for Temprature in K is\"\n",
-    "print\"T2=\",T2\n",
-    "print\"Approximate Rate of Heat Transfer in W/m**2 is\"\n",
-    "print\"F1=\",F1\n",
-    "print\"F2=\",F2"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Ex1.11:pg-15"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 12,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Introduction to heat transfer by S.K.Som, Chapter 1, Example 11\n",
-      "L= 1\n",
-      "A= 0.251327412287\n",
-      "The total heat loss by The pipe per unit length is given by Q/L=hbr*A*(T1-T2)+sigma*emi*A*(T1**4-T2**4) in W/m\n",
-      "F= 121.586773684\n"
-     ]
-    }
-   ],
-   "source": [
-    "\n",
-    "import math \n",
-    "\n",
-    "print\"Introduction to heat transfer by S.K.Som, Chapter 1, Example 11\"\n",
-    "#The horizontal steel pipe has outer diameter(D)=80 mm or.08 m\n",
-    "#The pipe is maintained at a temprature(T1)=60°C where the air and wall temprature(T2)=20 °C \n",
-    "#The average free convective heat transfer coefficient(hbr)=6.5 W/(m**2/K) b/w the outer surface of the pipe and air\n",
-    "D=.08;\n",
-    "T1=60;\n",
-    "T2=20;\n",
-    "hbr=6.5;\n",
-    "#Length(L=1) since per unit length is considered\n",
-    "L=1;\n",
-    "#The surface area of pipe is given by A=(math.pi*D*L)\n",
-    "print\"L=\",L\n",
-    "A=(math.pi*D*L);\n",
-    "#The surface emissivity(emi) of steel = 0.8\n",
-    "#The stefan -Boltzman constant(sigma)= 5.7*10**-8 W/(m**2*K**4)\n",
-    "print\"A=\",A\n",
-    "sigma=5.67*10**-8;\n",
-    "emi=.8;\n",
-    "#The total heat loss by The pipe per unit length is given by Q/L=hbr*A*(T1-T2)+sigma*emi*A*(T1**4-T2**4)\n",
-    "print\"The total heat loss by The pipe per unit length is given by Q/L=hbr*A*(T1-T2)+sigma*emi*A*(T1**4-T2**4) in W/m\"\n",
-    "#Let Q/L=F\n",
-    "F=hbr*A*((T1+273.15)-(T2+273.15))+sigma*emi*A*((T1+273.15)**4-(T2+273.15)**4)\n",
-    "print\"F=\",F\n",
-    "\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": "2.7.11"
-  }
- },
- "nbformat": 4,
- "nbformat_minor": 0
-}
diff --git a/Introduction_to_Heat_Transfer_by_S._K._Som/Chapter2.ipynb b/Introduction_to_Heat_Transfer_by_S._K._Som/Chapter2.ipynb
index cda3a92c..ba64d857 100644
--- a/Introduction_to_Heat_Transfer_by_S._K._Som/Chapter2.ipynb
+++ b/Introduction_to_Heat_Transfer_by_S._K._Som/Chapter2.ipynb
@@ -18,7 +18,7 @@
   },
   {
    "cell_type": "code",
-   "execution_count": 1,
+   "execution_count": 2,
    "metadata": {
     "collapsed": false
    },
@@ -119,7 +119,31 @@
     "print\"Check for Ti(in °C)\"\n",
     "Ti=T4-(Q*Ri)\n",
     "print\"The value is same as given in the problem\"\n",
-    "print\"Ti=\",Ti"
+    "print\"Ti=\",Ti\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    " \n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n"
    ]
   },
   {
@@ -147,6 +171,10 @@
     }
    ],
    "source": [
+    "  \n",
+    "\n",
+    " \n",
+    " \n",
     "import math\n",
     " \n",
     "print\"Introduction to heat transfer by S.K.Som, Chapter 2, Example 2\"\n",
@@ -166,7 +194,7 @@
     "#Q=(Ti-To)/((Lb/kb)+(L/ki)) so L=ki*(((Ti-To)/Q)-(Lb/kb))\n",
     "print\"The thickness of insulating material L=ki*(((Ti-To)/Q)-(Lb/kb)) in m\"\n",
     "L=ki*(((Ti-To)/Q)-(Lb/kb))\n",
-    "print\"L=\",L"
+    "print\"L=\",L\n"
    ]
   },
   {
@@ -178,7 +206,7 @@
   },
   {
    "cell_type": "code",
-   "execution_count": 3,
+   "execution_count": 1,
    "metadata": {
     "collapsed": false
    },
@@ -207,6 +235,7 @@
     }
    ],
    "source": [
+    " \n",
     "import math\n",
     " \n",
     "print\"Introduction to heat transfer by S.K.Som, Chapter 2, Example 4\"\n",
@@ -252,7 +281,17 @@
     "print\"q2=\",q2\n",
     "print\"Heat flux qo=q1+q2 in W/m**2 \"\n",
     "qo=q1+q2\n",
-    "print\"qo=\",qo"
+    "print\"qo=\",qo\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n"
    ]
   },
   {
@@ -290,6 +329,7 @@
     }
    ],
    "source": [
+    "  \n",
     "import math\n",
     " \n",
     "print\"Introduction to heat transfer by S.K.Som, Chapter 2, Example 6\"\n",
@@ -332,7 +372,9 @@
     "X=(2*10**3)-(4*10**5*x);\n",
     "Q=-k*X/10**6\n",
     "#A check for the above results can be made from an energy balance of the plate as |(q/A)|@x=0+|(q/A)|@x=0.02=qG*0.02\n",
-    "print\"Q=\",Q"
+    "print\"Q=\",Q\n",
+    "\n",
+    "\n"
    ]
   },
   {
@@ -344,7 +386,7 @@
   },
   {
    "cell_type": "code",
-   "execution_count": 16,
+   "execution_count": 2,
    "metadata": {
     "collapsed": false
    },
@@ -368,6 +410,7 @@
     }
    ],
    "source": [
+    " \n",
     "import math\n",
     " \n",
     "print\"Introduction to heat transfer by S.K.Som, Chapter 2, Example 9\"\n",
@@ -415,7 +458,16 @@
     "Q=(T1-Tinf)/(((X)/(2*math.pi*L*k))+(1/(h*2*math.pi*r2*L)))\n",
     "#It is important to note that Q increases by 5.2% when the insulation thickness increases from 0.002m to critical thickness. \n",
     "#Addition of insulation beyond the critical thickness decreases the value of Q (The heat loss).\n",
-    "print\"Q=\",Q"
+    "print\"Q=\",Q\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n"
    ]
   },
   {
@@ -427,7 +479,7 @@
   },
   {
    "cell_type": "code",
-   "execution_count": 8,
+   "execution_count": 3,
    "metadata": {
     "collapsed": false
    },
@@ -443,6 +495,7 @@
     }
    ],
    "source": [
+    " \n",
     "import math\n",
     " \n",
     "print\"Introduction to heat transfer by S.K.Som, Chapter 2, Example 10\"\n",
@@ -472,7 +525,22 @@
     "#Therefore the thickness of insulation is given by t=r3-Do\n",
     "print\"the thickness of insulation in metre is\"\n",
     "t=r3-Do\n",
-    "print\"t=\",t"
+    "print\"t=\",t\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n"
    ]
   },
   {
@@ -484,7 +552,7 @@
   },
   {
    "cell_type": "code",
-   "execution_count": 9,
+   "execution_count": 4,
    "metadata": {
     "collapsed": false
    },
@@ -502,6 +570,7 @@
     }
    ],
    "source": [
+    " \n",
     "import math\n",
     " \n",
     "print\"Introduction to heat transfer by S.K.Som, Chapter 2, Example 11\"\n",
@@ -522,7 +591,21 @@
     "print\"The temprature of wire at the centre in K is \"\n",
     "To=Tw+((qG*ro**2)/(4*k))\n",
     "#Note:The answer in the book is incorrect(value of D has been put instead of ro)\n",
-    "print\"To=\",To"
+    "print\"To=\",To\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n"
    ]
   },
   {
@@ -572,6 +655,10 @@
     }
    ],
    "source": [
+    "\n",
+    " \n",
+    " \n",
+    " \n",
     "import math\n",
     " \n",
     "print\"Introduction to heat transfer by S.K.Som, Chapter 2, Example 12\"\n",
@@ -643,7 +730,12 @@
     "#the amount of ice in kG which melts during a 24 hour period is (mice)\n",
     "print\"Therefore,the amount of ice(mice)in kG which melts during a 24 hour period is\"\n",
     "mice=Qt/deltahf\n",
-    "print\"mice=\",mice"
+    "print\"mice=\",mice\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n"
    ]
   },
   {
@@ -655,7 +747,7 @@
   },
   {
    "cell_type": "code",
-   "execution_count": 13,
+   "execution_count": 5,
    "metadata": {
     "collapsed": false
    },
@@ -690,6 +782,7 @@
     }
    ],
    "source": [
+    " \n",
     "import math\n",
     " \n",
     "print\"Introduction to heat transfer by S.K.Som, Chapter 2, Example 13\"\n",
@@ -753,7 +846,15 @@
     "Qinf=(h*P*k*A)**0.5*thetab\n",
     "print\"We see that since k is large there is significant difference  between the finite length and the infinte length cases\"\n",
     "print\"However when the length of the rod approaches 1m,the result become almost same.\" \n",
-    "print\"Qinf=\",Qinf"
+    "print\"Qinf=\",Qinf\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n"
    ]
   },
   {
@@ -765,7 +866,7 @@
   },
   {
    "cell_type": "code",
-   "execution_count": 14,
+   "execution_count": 6,
    "metadata": {
     "collapsed": false
    },
@@ -781,6 +882,7 @@
     }
    ],
    "source": [
+    " \n",
     "import math\n",
     " \n",
     "print\"Introduction to heat transfer by S.K.Som, Chapter 2, Example 14\"\n",
@@ -797,7 +899,41 @@
     "#The thermal conductivity of Rod B iskB\n",
     "print\"The thermal conductivity of Rod B kB in W/(m*K) is \"\n",
     "kB=kA*(xB/xA)**2\n",
-    "print\"kB=\",kB"
+    "print\"kB=\",kB\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n"
    ]
   },
   {
@@ -809,7 +945,7 @@
   },
   {
    "cell_type": "code",
-   "execution_count": 15,
+   "execution_count": 7,
    "metadata": {
     "collapsed": false
    },
@@ -835,6 +971,7 @@
     }
    ],
    "source": [
+    "  \n",
     "import math\n",
     " \n",
     "print\"Introduction to heat transfer by S.K.Som, Chapter 2, Example 15\"\n",
@@ -876,7 +1013,17 @@
     "#Heat loss from the plate is Qb\n",
     "print\"Heat loss from the plate at 400K in W is\"\n",
     "Qb=(N*(h*P*kal*A)**0.5*thetab*((math.cosh(m*L)-(thetaL/thetab))/(math.sinh(m*L))))+(((l*b)-(N*A))*h*thetab)+(l*b*h*thetab)\n",
-    "print\"Qb=\",Qb"
+    "print\"Qb=\",Qb\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n"
    ]
   }
  ],
diff --git a/Introduction_to_Heat_Transfer_by_S._K._Som/Chapter2_GK3uH9r.ipynb b/Introduction_to_Heat_Transfer_by_S._K._Som/Chapter2_GK3uH9r.ipynb
deleted file mode 100644
index ba64d857..00000000
--- a/Introduction_to_Heat_Transfer_by_S._K._Som/Chapter2_GK3uH9r.ipynb
+++ /dev/null
@@ -1,1051 +0,0 @@
-{
- "cells": [
-  {
-   "cell_type": "markdown",
-   "metadata": {
-    "collapsed": true
-   },
-   "source": [
-    "# Chapter 02:One dimensional steady-state heat conduction"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Ex2.1:pg-33"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 2,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Introduction to heat transfer by S.K.Som, Chapter 2, Example 1\n",
-      "The convective resistance Ro= 1/(ho*A) at the outer surface in KW**-1 is\n",
-      "Ro= 0.0666666666667\n",
-      "The conduction resistance Rs= Ls/(ks*A) of steel sheet in KW**-1 is\n",
-      "Rs= 8e-05\n",
-      "The conduction resistance Rg= Lg/(kg*A) of glass wool in KW**-1 is\n",
-      "Rg= 2.66666666667\n",
-      "The conduction resistance Rp= Lp/(kp*A) of plywood in KW**-1 is\n",
-      "Rp= 0.266666666667\n",
-      "The convective resistance Ri= 1/(hi*A) at the outer surface in KW**-1 is\n",
-      "Ri= 0.111111111111\n",
-      "The rate of heat flow Q=(To-Ti)/(Ro+Rs+Rg+Rp+Ri) in W is\n",
-      "Q= 12.5353919471\n",
-      "The temprature at the outer surface of wall is T1=To-(Q*Ro) in °C\n",
-      "T1= 23.1643072035\n",
-      "The temprature at the interface b/w steel sheet and glass wool is T2=T1-(Q*Rs) in°C\n",
-      "T2= 23.1633043722\n",
-      "The temprature at the interface b/w glass wool and plywood is T3=T2-(Q*Rg) in°C\n",
-      "T3= -10.2644074867\n",
-      "The temprature at the inner surface of wall is T4=T3-(Q*Rp)in °C\n",
-      "T4= -13.6071786725\n",
-      "Check for Ti(in °C)\n",
-      "The value is same as given in the problem\n",
-      "Ti= -15.0\n"
-     ]
-    }
-   ],
-   "source": [
-    "import math\n",
-    " \n",
-    "print\"Introduction to heat transfer by S.K.Som, Chapter 2, Example 1\"\n",
-    "#The length of steel sheet (Ls)=1.5 mm and thermal conductivity (ks)=25 W/(mK) at the outer surface.\n",
-    "Ls=1.5;\n",
-    "ks=25;\n",
-    "#The length of plywood (Lp)=10 mm and thermal conductivity (kp)= .05 W/(mK) at the inner surface.\n",
-    "Lp=10;\n",
-    "kp=.05;\n",
-    "#The length of glass wool (Lg)=20 mm and thermal conductivity (kg)= .01W/(mK) in between steel sheet and plywood.\n",
-    "Lg=20;\n",
-    "kg=.01;\n",
-    "#The temprature of van inside cold Enviroment is (Ti)= -15°C  while the outside surface is exposed to a surrounding ambient temprature (To)=24°C \n",
-    "To=24;\n",
-    "Ti=-15;\n",
-    "#The average value of heat transfer coefficients at the inner and outside surfaces of the wall are hi=12 W/(m**2*K) and ho= 20 W/(m**2*K) \n",
-    "hi=12;\n",
-    "ho=20;\n",
-    "#The surface area of wall (A)= .75 m**2 \n",
-    "A=.75;\n",
-    "#The convective resistance is Ro= 1/(ho*A) at the outer surface\n",
-    "print\"The convective resistance Ro= 1/(ho*A) at the outer surface in KW**-1 is\"\n",
-    "Ro=1/(ho*A)\n",
-    "print\"Ro=\",Ro\n",
-    "#The conduction resistance is Rs= Ls/(ks*A) of steel sheet\n",
-    "print\"The conduction resistance Rs= Ls/(ks*A) of steel sheet in KW**-1 is\"\n",
-    "Rs=Ls*10**-3/(ks*A)\n",
-    "print\"Rs=\",Rs\n",
-    "#The conduction resistance is Rg= Lg/(kg*A) of glass wool\n",
-    "print\"The conduction resistance Rg= Lg/(kg*A) of glass wool in KW**-1 is\"\n",
-    "Rg= Lg*10**-3/(kg*A)\n",
-    "print\"Rg=\",Rg\n",
-    "#The conduction resistance is Rp= Lp/(kp*A) of plywood\n",
-    "print\"The conduction resistance Rp= Lp/(kp*A) of plywood in KW**-1 is\"\n",
-    "Rp= Lp*10**-3/(kp*A)\n",
-    "print\"Rp=\",Rp\n",
-    "#The convective resistance is Ri= 1/(hi*A) at the outer surface\n",
-    "print\"The convective resistance Ri= 1/(hi*A) at the outer surface in KW**-1 is\"\n",
-    "Ri= 1/(hi*A)\n",
-    "print\"Ri=\",Ri\n",
-    "#The rate of heat flow is Q=(To-Ti)/(Ro+Rs+Rg+Rp+Ri)\n",
-    "print\"The rate of heat flow Q=(To-Ti)/(Ro+Rs+Rg+Rp+Ri) in W is\"\n",
-    "Q=(To-Ti)/(Ro+Rs+Rg+Rp+Ri)\n",
-    "print\"Q=\",Q\n",
-    "#The tempraure at the outer surface of wall is T1.\n",
-    "#The temprature at the interface b/w steel sheet and glass wool is T2.\n",
-    "#The temprature at the interface b/w glass wool and plywood is T3.\n",
-    "#The tempraure at the inner surface of wall is T4.\n",
-    "print\"The temprature at the outer surface of wall is T1=To-(Q*Ro) in °C\"\n",
-    "T1=To-(Q*Ro)\n",
-    "print\"T1=\",T1\n",
-    "print\"The temprature at the interface b/w steel sheet and glass wool is T2=T1-(Q*Rs) in°C\"\n",
-    "T2=T1-(Q*Rs)\n",
-    "print\"T2=\",T2\n",
-    "print\"The temprature at the interface b/w glass wool and plywood is T3=T2-(Q*Rg) in°C\"\n",
-    "T3=T2-(Q*Rg)\n",
-    "print\"T3=\",T3\n",
-    "print\"The temprature at the inner surface of wall is T4=T3-(Q*Rp)in °C\"\n",
-    "T4=T3-(Q*Rp)\n",
-    "print\"T4=\",T4\n",
-    "#Check for Ti(Temprature inside the van)\n",
-    "print\"Check for Ti(in °C)\"\n",
-    "Ti=T4-(Q*Ri)\n",
-    "print\"The value is same as given in the problem\"\n",
-    "print\"Ti=\",Ti\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    " \n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Ex2.2:pg-36"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 1,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Introduction to heat transfer by S.K.Som, Chapter 2, Example 2\n",
-      "The thickness of insulating material L=ki*(((Ti-To)/Q)-(Lb/kb)) in m\n",
-      "L= 0.056\n"
-     ]
-    }
-   ],
-   "source": [
-    "  \n",
-    "\n",
-    " \n",
-    " \n",
-    "import math\n",
-    " \n",
-    "print\"Introduction to heat transfer by S.K.Som, Chapter 2, Example 2\"\n",
-    "#The Thickness of fire clay bricks (Lb)=.2 m\n",
-    "Lb=0.2;\n",
-    "#The thermal conductivity of fire clay bricks(kb)=1.0 W/(m*K)\n",
-    "kb=1;\n",
-    "#the Thicknes of insulating material is L\n",
-    "#The thermal conductivity of insulating material(ki)=.07 W/(m*K)\n",
-    "ki=.07;\n",
-    "#The furnace inner brick surface is at temprature Ti=1250 K\n",
-    "Ti=1250;\n",
-    "#The furnace outer brick surface is at temprature To=310 K\n",
-    "To=310;\n",
-    "#The maximum allowable heat transfer rate(Q) from wall = 900 W/m**2\n",
-    "Q=900;\n",
-    "#Q=(Ti-To)/((Lb/kb)+(L/ki)) so L=ki*(((Ti-To)/Q)-(Lb/kb))\n",
-    "print\"The thickness of insulating material L=ki*(((Ti-To)/Q)-(Lb/kb)) in m\"\n",
-    "L=ki*(((Ti-To)/Q)-(Lb/kb))\n",
-    "print\"L=\",L\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Ex2.4:pg-36"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 1,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Introduction to heat transfer by S.K.Som, Chapter 2, Example 4\n",
-      "Heat transfer per unit surface area through film qf=(To-Tinf)/((1/h)+(Lf/kf))in W/(m**2)\n",
-      "qf= 4000.0\n",
-      "Heat transfer per unit surface area through substrate qs=(To-T1/(Ls/ks)in W/(m**2)\n",
-      "qs= 1000.0\n",
-      "Heat flux qo=qs+qf in W/(m**2)\n",
-      "qo= 5000.0\n",
-      "If the film is not transparent and all of the radiant heat flux is absorbed at its upper surface then\n",
-      "Rate of heat conduction through the film and substrate q1=(To-T1)/(Ls/ks) in W/m**2\n",
-      "q1= 1000.0\n",
-      "The temprature of the top surface of the film T2=(q1*(Lf*10**-3/kf))+To in °C\n",
-      "T2= 70.0\n",
-      "Rate of convective heat transfer from the upper surface of film to air q2=h*(T2-Tinf)in W/m**2\n",
-      "q2= 1250.0\n",
-      "Heat flux qo=q1+q2 in W/m**2 \n",
-      "qo= 2250.0\n"
-     ]
-    }
-   ],
-   "source": [
-    " \n",
-    "import math\n",
-    " \n",
-    "print\"Introduction to heat transfer by S.K.Som, Chapter 2, Example 4\"\n",
-    "#To cure the bond at a temprature(To),a radiant source used to provide a heat flux qo W/m**2 \n",
-    "#The back of the substrate is maintained at a temprature T1\n",
-    "#The film is exposed to air at a temprature (Tinf)\n",
-    "#the convective heat transfer coefficient is h.\n",
-    "#Given To=60°C,Tinf=20°C,h=25 W/(m**K)and T1=30°C\n",
-    "To=60;\n",
-    "Tinf=20;\n",
-    "h=25;\n",
-    "T1=30;\n",
-    "#Ls is the thickness of substrate and Lf is thickness of film in mm.\n",
-    "Ls=1.5;\n",
-    "Lf=.25;\n",
-    "#kf and ks are thermal conductivity of film and substrate respectively in W/(m*K)\n",
-    "kf=.025;\n",
-    "ks=.05;\n",
-    "#qo is Heat flux.\n",
-    "#qo=qf+qs where qf and qs are rate of heat transfer per unit surface area through the film and the substrate respectively.\n",
-    "print\"Heat transfer per unit surface area through film qf=(To-Tinf)/((1/h)+(Lf/kf))in W/(m**2)\"\n",
-    "qf=(To-Tinf)/((1/h)+(Lf*10**-3/kf))\n",
-    "print\"qf=\",qf\n",
-    "print\"Heat transfer per unit surface area through substrate qs=(To-T1/(Ls/ks)in W/(m**2)\"\n",
-    "qs=(To-T1)/(Ls*10**-3/ks)\n",
-    "print\"qs=\",qs\n",
-    "print\"Heat flux qo=qs+qf in W/(m**2)\"\n",
-    "qo=qs+qf\n",
-    "print\"qo=\",qo\n",
-    "#If the film is not transparent and all of the radiant heat flux is absorbed at its upper surface then qo=q1+q2\n",
-    "#q1 is rate of heat conduction through the film and substrate\n",
-    "#q2 is rate of convective heat transfer from the upper surface of film to air\n",
-    "print\"If the film is not transparent and all of the radiant heat flux is absorbed at its upper surface then\"\n",
-    "print\"Rate of heat conduction through the film and substrate q1=(To-T1)/(Ls/ks) in W/m**2\"\n",
-    "q1=(To-T1)/(Ls*10**-3/ks)\n",
-    "print\"q1=\",q1\n",
-    "#To determine q2 we need to find the temprature(T2) of the top surface \n",
-    "print\"The temprature of the top surface of the film T2=(q1*(Lf*10**-3/kf))+To in °C\"\n",
-    "T2=(q1*(Lf*10**-3/kf))+To\n",
-    "print\"T2=\",T2\n",
-    "print\"Rate of convective heat transfer from the upper surface of film to air q2=h*(T2-Tinf)in W/m**2\"\n",
-    "q2=h*(T2-Tinf)\n",
-    "print\"q2=\",q2\n",
-    "print\"Heat flux qo=q1+q2 in W/m**2 \"\n",
-    "qo=q1+q2\n",
-    "print\"qo=\",qo\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Ex2.6:pg-38"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 6,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      " Introduction to heat transfer by S.K.Som, Chapter 2, Example 6\n",
-      "(a)The expression for the temprature distribution in the plate T=160+2*10**3*(x-100*x**2)\n",
-      "(b)The maximum temprature occurs at x=5mm or 0.005m away from the left surface towards the right\n",
-      "The maximum temprature(Tmax) in °C is\n",
-      "Tmax= 165.0\n",
-      "(c(i))The rate of heat transfer at the left face in MW/m**2 is\n",
-      "Q= -1\n",
-      "The minus sign indicates that the heat flow in the negative direction\n",
-      "(c(ii))The rate of heat transfer at the right face in MW/m**2 is\n",
-      "Q= 1.2\n",
-      "The minus sign for (q/A)@x=0 and the plus sign for (q/A)@x=0.02 indicates that heat is lost from both the surfaces to surroundings\n",
-      "(c(iii))The rate of heat transfer at the centre in MW/m**2 is\n",
-      "Q= 0.4\n"
-     ]
-    }
-   ],
-   "source": [
-    "  \n",
-    "import math\n",
-    " \n",
-    "print\"Introduction to heat transfer by S.K.Som, Chapter 2, Example 6\"\n",
-    "#The thickness of plate 20mm or .02m with uniform heat generation qg=80MW/m**3\n",
-    "#The thermal conductivity of wall(k)is 200W/m*K.\n",
-    "k=200;\n",
-    "#The left and right faces are kept at tempratures T=160°C and T=120°C respectively\n",
-    "#We start with the equation (d**2T/dx**2)+(qg/k)=0 or T=-(qg/2k)*x**2 +(c1*x)+c2 \n",
-    "#With qg=80*10**6W/m**3 and k=200W/(m*K)\n",
-    "#Applying boundary condition at x=0,T=160°C and x=0.02,T=120°C we get constant c2=160°C and c2=2000m**-1\n",
-    "#Hence T=160+2*10**3*(x-100*x**2)----->eq.1\n",
-    "print\"(a)The expression for the temprature distribution in the plate T=160+2*10**3*(x-100*x**2)\" \n",
-    "#For maximum temprature differentiating eq.1 with respect to x and equating it to zero...we get dT/dx=(2*10**3)-(4*10**5*x)=0,which gives x=0.005m=5mm\n",
-    "print\"(b)The maximum temprature occurs at x=5mm or 0.005m away from the left surface towards the right\"\n",
-    "x=0.005;\n",
-    "#The maximum temprature is Tmax\n",
-    "print\"The maximum temprature(Tmax) in °C is\"\n",
-    "Tmax=160+2*10**3*(x-100*x**2)\n",
-    "print\"Tmax=\",Tmax\n",
-    "#The rate of heat transfer(q/A) is given by -k*(dT/dx)\n",
-    "#Let dT/dx=X and (q/A)=Q\n",
-    "print\"(c(i))The rate of heat transfer at the left face in MW/m**2 is\"\n",
-    "#For left face x=0\n",
-    "x=0;\n",
-    "X=(2*10**3)-(4*10**5*x);\n",
-    "Q=-k*X/10**6\n",
-    "print\"Q=\",Q\n",
-    "print\"The minus sign indicates that the heat flow in the negative direction\"\n",
-    "print\"(c(ii))The rate of heat transfer at the right face in MW/m**2 is\"\n",
-    "#For right face x=0.02\n",
-    "x=0.02;\n",
-    "X=(2*10**3)-(4*10**5*x);\n",
-    "Q=-k*X/10**6\n",
-    "print\"Q=\",Q\n",
-    "#(q/A)@x=0 implies rate of heat transfer at the position where x=0. \n",
-    "print\"The minus sign for (q/A)@x=0 and the plus sign for (q/A)@x=0.02 indicates that heat is lost from both the surfaces to surroundings\"\n",
-    "print\"(c(iii))The rate of heat transfer at the centre in MW/m**2 is\"\n",
-    "#For centre x=0.01\n",
-    "x=0.01;\n",
-    "X=(2*10**3)-(4*10**5*x);\n",
-    "Q=-k*X/10**6\n",
-    "#A check for the above results can be made from an energy balance of the plate as |(q/A)|@x=0+|(q/A)|@x=0.02=qG*0.02\n",
-    "print\"Q=\",Q\n",
-    "\n",
-    "\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Ex2.9:pg- 48"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 2,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Introduction to heat transfer by S.K.Som, Chapter 2, Example 9\n",
-      "The critical thickness of insulation in metre is\n",
-      "X= 0.182340462632\n",
-      "The heat transfer rate Q per metre of tube length in W/m is \n",
-      "Q= 26.807459995\n",
-      "X= 0.559673817307\n",
-      "The heat transfer rate per metre of tube length Q in W/m is \n",
-      "Q= 28.1996765363\n",
-      "X= 1.79194526577\n",
-      "The heat transfer rate per metre of tube length Q in W/m is \n",
-      "Q= 21.1086798197\n"
-     ]
-    }
-   ],
-   "source": [
-    " \n",
-    "import math\n",
-    " \n",
-    "print\"Introduction to heat transfer by S.K.Som, Chapter 2, Example 9\"\n",
-    "#A thin walled copper tube of outside metal radius r=0.01m carries steam at temprature, T1=400K.It is inside a room where the surrounding air temprature is Tinf=300K.\n",
-    "T1=400;\n",
-    "Tinf=300;\n",
-    "r=0.01;\n",
-    "#The tube is insulated with magnesia insulation of an approximate thermal conductivity of k=0.07W/(m*K)\n",
-    "k=0.07;\n",
-    "#External convective Coefficient h=4W/(m**2*K)\n",
-    "h=4;\n",
-    "#Critical thickness(rc) is given by k/h\n",
-    "print\"The critical thickness of insulation in metre is\"\n",
-    "rc=k/h\n",
-    "#We use the rate of heat transfer per metre of tube length as Q=(Ti-Tinf)/((ln(r2/r1)/(2*math.pi*L*k))+(1/(h*2*math.pi*r2*L))) where length,L=1m\n",
-    "L=1;\n",
-    "#When 0.002m thick layer of insulation r1=0.01m,r2=0.01+0.002=0.012m\n",
-    "r1=0.01;#inner radius\n",
-    "r2=0.012;#outer radius\n",
-    "#Let ln(r2/r1)=X\n",
-    "X=math.log(r2/r1)/math.log(2.718);\n",
-    "print\"X=\",X\n",
-    "#The heat transfer rate per metre of tube length is Q\n",
-    "print\"The heat transfer rate Q per metre of tube length in W/m is \"\n",
-    "Q=(T1-Tinf)/(((X)/(2*math.pi*L*k))+(1/(h*2*math.pi*r2*L)))\n",
-    "print\"Q=\",Q\n",
-    "#When critical thickness of insulation r1=0.01m,r2=0.0175m\n",
-    "r2=0.0175;#outer radius\n",
-    "r1=0.01;#inner radius\n",
-    "#Let ln(r2/r1)=X\n",
-    "X=math.log(r2/r1)/math.log(2.718);\n",
-    "print\"X=\",X\n",
-    "#The heat transfer rate per metre of tube length is Q \n",
-    "print\"The heat transfer rate per metre of tube length Q in W/m is \"\n",
-    "Q=(T1-Tinf)/(((X)/(2*math.pi*L*k))+(1/(h*2*math.pi*r2*L)))\n",
-    "print\"Q=\",Q\n",
-    "#When there is a 0.05 m thick layer of insulation r1=0.01m,r2=.01+0.05=0.06m\n",
-    "r1=0.01;#inner radius\n",
-    "r2=0.06;#outer radius\n",
-    "#Let ln(r2/r1)=X\n",
-    "X=math.log(r2/r1)/math.log(2.718);\n",
-    "print\"X=\",X\n",
-    "#The heat transfer rate per metre of tube length is Q \n",
-    "print\"The heat transfer rate per metre of tube length Q in W/m is \"\n",
-    "Q=(T1-Tinf)/(((X)/(2*math.pi*L*k))+(1/(h*2*math.pi*r2*L)))\n",
-    "#It is important to note that Q increases by 5.2% when the insulation thickness increases from 0.002m to critical thickness. \n",
-    "#Addition of insulation beyond the critical thickness decreases the value of Q (The heat loss).\n",
-    "print\"Q=\",Q\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Ex2.10:pg-49"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 3,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Introduction to heat transfer by S.K.Som, Chapter 2, Example 10\n",
-      "the thickness of insulation in metre is\n",
-      "t= 0.019\n"
-     ]
-    }
-   ],
-   "source": [
-    " \n",
-    "import math\n",
-    " \n",
-    "print\"Introduction to heat transfer by S.K.Som, Chapter 2, Example 10\"\n",
-    "#A copper pipe having 35mm outer diameter(Do) and 30mm inner diameter(Di) carries liquid oxygen to the storage site of a space shuttle at temprature,T1=-182°C\n",
-    "#mass flow rate is ,mdot=0.06m**3/min. \n",
-    "Di=0.03;#in metre\n",
-    "Do=0.035;# in metre\n",
-    "T1=-182;\n",
-    "mdot=0.06;\n",
-    "#The ambient air is at temprature(Ta)=20°C and has a dew point(T3)=10°C.\n",
-    "Ta=20;\n",
-    "T3=10;\n",
-    "#The thermal conductivity(k) of insulating material is 0.02W/(m*k)\n",
-    "k=0.02;\n",
-    "#The convective heat transfer coefficient on the outside is h=17W/(m**2*K)\n",
-    "h=17;\n",
-    "#The thermal conductivity of copper kcu=400W/(m*K)\n",
-    "kcu=400;\n",
-    "#We can write Q=((Ta-T1)/(R1+R2+R3))=((Ta-T3)/(R3)),Rearranging we get ((R1+R2+R3)/(R3))=((Ta-T1)/(Ta-T3))--------eq.1\n",
-    "#The conduction Resistance of copper pipe(R1)=ln(0.035/0.03)/(2*math.pi*L*kcu)=3.85*10**-4/(2*math.pi*L)K/W\n",
-    "#The conduction resistance of insulating material (R2)=ln(r3/0.035)/(2*math.pi*L*k)=(1/(2*math.pi*L))((50*ln(r3/0.035)))K/W where r3 is the outer radius of insulation in metres.\n",
-    "#The convective resistance at the outer surface(R3)=1/(2*math.pi*L*h*r3)=(1/2*math.pi*L)*(mdot/r3)K/W\n",
-    "#Substituting the values in eq.1 we have 1+((50*ln(r3/0.035)+(3.85*10**-4))/(mdot/r3))=20-(-182)/(20-10)\n",
-    "#A rearrangement of the above equation gives r3*ln(r3)+3.35*r3=0.023\n",
-    "#The equation is solved by trial and error method which finally gives r3=0.054m\n",
-    "r3=0.054;#outer radius of insulation\n",
-    "#Therefore the thickness of insulation is given by t=r3-Do\n",
-    "print\"the thickness of insulation in metre is\"\n",
-    "t=r3-Do\n",
-    "print\"t=\",t\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Ex2.11:pg-52"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 4,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Introduction to heat transfer by S.K.Som, Chapter 2, Example 11\n",
-      "The rate of heat generation of thermal energy in W/m**3 is\n",
-      "qG= 407436654.315\n",
-      "The temprature of wire at the centre in K is \n",
-      "To= 656.631455962\n"
-     ]
-    }
-   ],
-   "source": [
-    " \n",
-    "import math\n",
-    " \n",
-    "print\"Introduction to heat transfer by S.K.Som, Chapter 2, Example 11\"\n",
-    "#An electrical resistance wire 2.5mm or 2.5*10**-3m in diameter(D) and L=0.5m long has a measured voltage drop of E=25V for a current flow of I=40A.\n",
-    "D=2.5*10**-3;\n",
-    "I=40;\n",
-    "E=25;\n",
-    "L=0.5;\n",
-    "ro=D/2;#ro is radius of wire\n",
-    "#The thermal conductivity(k) of wire material is 24W/(m*K) \n",
-    "k=24;\n",
-    "#The rate of generation of thermal energy per unit volume is given by qG=(E*I)/(L*math.pi*D**2/4)\n",
-    "print\"The rate of heat generation of thermal energy in W/m**3 is\"\n",
-    "qG=(E*I)/(L*math.pi*D**2/4)\n",
-    "print\"qG=\",qG\n",
-    "#The temprature at the centre is given by To=Tw+((qG*ro**2)/(4*k)) where Tw=650K is surface temprature \n",
-    "Tw=650;\n",
-    "print\"The temprature of wire at the centre in K is \"\n",
-    "To=Tw+((qG*ro**2)/(4*k))\n",
-    "#Note:The answer in the book is incorrect(value of D has been put instead of ro)\n",
-    "print\"To=\",To\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Ex2.12:pg-56"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 10,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Introduction to heat transfer by S.K.Som, Chapter 2, Example 12\n",
-      "The inner(A1) and outer surfaces(A2) areas of the tank in m**2 are\n",
-      "The convective resistance(Ri) at the inner surface in K/W is \n",
-      "Ri= 0.000159154943092\n",
-      "The conduction resistance(Rs)of the tank in K/W is\n",
-      "Rs= 0.0\n",
-      "The convective resistance(Roc) at the outer surface in K/W is\n",
-      "Roc= 0.00127323954474\n",
-      "The radiative heat transfer coefficient hr in W/(m**2*K) is\n",
-      "hr= 5.26265474661\n",
-      "Therefore the radiative resistance(Ror) at the outer surface in K/W is\n",
-      "Ror= 0.00241938642385\n",
-      "The equivalent resistance in K/W is\n",
-      "Ro= 0.000834218925785\n",
-      "The total resistance in K/W is\n",
-      "Rtotal= 0.000993373868877\n",
-      "The rate of heat transfer,Q in W is\n",
-      "Q= 20133.406592\n",
-      "The outer surface temprature in °C is\n",
-      "T2= 3.2043311804\n",
-      "which is sufficiently close to the assumption.So there is no need of further iteration\n",
-      "The total heat transfer(Qt) during a 24-hour period in KJ is\n",
-      "Qt= 1739526.32955\n",
-      "Therefore,the amount of ice(mice)in kG which melts during a 24 hour period is\n",
-      "mice= 5208.16266333\n"
-     ]
-    }
-   ],
-   "source": [
-    "\n",
-    " \n",
-    " \n",
-    " \n",
-    "import math\n",
-    " \n",
-    "print\"Introduction to heat transfer by S.K.Som, Chapter 2, Example 12\"\n",
-    "#A spherical tank of internal diameter, Di=5m ,made of t=25mm thick stainless steel(thermal conductivity,k=15W/(m*K)),is used to store water at temprature(Ti)=0°C.Do is outer diameter\n",
-    "Di=5;\n",
-    "t=25;\n",
-    "Do=5+2*(t/1000);#in metre\n",
-    "k=15;\n",
-    "Ti=0;\n",
-    "#The tank is located in a room whose temprature is (To)=20°C.\n",
-    "To=20;\n",
-    "#Emmisivity is 1.\n",
-    "#The convection heat transfer coefficients at the inner and outer surfaces of the tank are hi=80W/(m**2*K) and ho=10W/(m**2*K)\n",
-    "hi=80;\n",
-    "ho=10;\n",
-    "#The heat of fusion of ice at atmospheric pressure is deltahf=334kJ/kg.The stefan-boltzman constant(sigma) is 5.67*10**-8W/m**2.\n",
-    "sigma=5.67*10**-8;\n",
-    "deltahf=334;\n",
-    "#The inner surface area is (A1) and outer surface area is (A2)of the tank\n",
-    "print\"The inner(A1) and outer surfaces(A2) areas of the tank in m**2 are\"\n",
-    "A1=math.pi*Di**2\n",
-    "A2=math.pi*Do**2\n",
-    "#The individual thermal resistances can be determined as\n",
-    "#The convective resistance is (Ri)\n",
-    "print\"The convective resistance(Ri) at the inner surface in K/W is \"\n",
-    "Ri=1/(hi*A1)\n",
-    "print\"Ri=\",Ri\n",
-    "#The conduction resistance is(Rs)\n",
-    "print\"The conduction resistance(Rs)of the tank in K/W is\"\n",
-    "Rs=(Do-Di)/(2*k*math.pi*Di*Do)\n",
-    "print\"Rs=\",Rs\n",
-    "#The convective resistance is(Roc)\n",
-    "print\"The convective resistance(Roc) at the outer surface in K/W is\"\n",
-    "Roc=1/(ho*A2)\n",
-    "print\"Roc=\",Roc\n",
-    "#The radiative resistance(Ror) at the outer surface is Ror=1/(A2*hr) \n",
-    "#The radiative heat transfer coefficient hr is determined by hr=sigma*(T2**2+293.15**2)*(T2+293.15)\n",
-    "#But we do not know the outer surface temprature T2 of the tank and hence we cannot determine the value of hr.\n",
-    "#Therfore we adopt an iterative procedure.We assume T2=4°C=277.15K.Putting the value in hr=sigma*(T2**2+293.15**2)*(T2+293.15) we get\n",
-    "T2=277.15;\n",
-    "print\"The radiative heat transfer coefficient hr in W/(m**2*K) is\"\n",
-    "hr=sigma*(T2**2+293.15**2)*(T2+293.15)\n",
-    "print\"hr=\",hr\n",
-    "print\"Therefore the radiative resistance(Ror) at the outer surface in K/W is\"\n",
-    "Ror=1/(A2*hr)\n",
-    "print\"Ror=\",Ror\n",
-    "#The two parallel resistances Roc and Ror can be replaced by an equivalent resistance Ro as X=(1/Ro)=(1/Roc)+(1/Ror)\n",
-    "print\"The equivalent resistance in K/W is\"\n",
-    "X=(1/Roc)+(1/Ror);\n",
-    "Ro=1/X\n",
-    "print\"Ro=\",Ro\n",
-    "#Now the resistance Ri,Rs and Ro are in series an hence the total resistance becomes Rtotal=Ri+Rs+Ro \n",
-    "print\"The total resistance in K/W is\"\n",
-    "Rtotal=Ri+Rs+Ro \n",
-    "print\"Rtotal=\",Rtotal\n",
-    "#The rate of heat transfer is given by Q=(To-Ti)/Rtotal\n",
-    "print\"The rate of heat transfer,Q in W is\"\n",
-    "Q=(To-Ti)/Rtotal\n",
-    "print\"Q=\",Q\n",
-    "#The outer surface(T2) is calculated as T2=To-Q*Ro\n",
-    "print\"The outer surface temprature in °C is\"\n",
-    "T2=To-Q*Ro\n",
-    "print\"T2=\",T2\n",
-    "print\"which is sufficiently close to the assumption.So there is no need of further iteration\"\n",
-    "#The total heat transfer is (Qt),during a 24-hour period\n",
-    "print\"The total heat transfer(Qt) during a 24-hour period in KJ is\"\n",
-    "Qt=Q*24*3600/1000\n",
-    "print\"Qt=\",Qt\n",
-    "#the amount of ice in kG which melts during a 24 hour period is (mice)\n",
-    "print\"Therefore,the amount of ice(mice)in kG which melts during a 24 hour period is\"\n",
-    "mice=Qt/deltahf\n",
-    "print\"mice=\",mice\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Ex2.13:pg-68"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 5,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Introduction to heat transfer by S.K.Som, Chapter 2, Example 13\n",
-      "P/A in m**-1 is\n",
-      "X= 400.0\n",
-      "m in m**-1 is\n",
-      "m= 3.28797974611\n",
-      "M in W/K is\n",
-      "M= 0.0955478103936\n",
-      "thetab in °C is \n",
-      "thetab= 100\n",
-      "Heat loss from rod in Watt, for different value of length(in m) is \n",
-      "Q= 0.705573384326\n",
-      "Q= 1.32655528327\n",
-      "Q= 2.53006298798\n",
-      "Q= 5.56299390901\n",
-      "Q= 8.29001416681\n",
-      "Q= 9.45769063154\n",
-      "Q= 9.52862252977\n",
-      "Q= 9.55478103936\n",
-      "For an infintely long rod heat loss in W is\n",
-      "We see that since k is large there is significant difference  between the finite length and the infinte length cases\n",
-      "However when the length of the rod approaches 1m,the result become almost same.\n",
-      "Qinf= 9.55478103936\n"
-     ]
-    }
-   ],
-   "source": [
-    " \n",
-    "import math\n",
-    " \n",
-    "print\"Introduction to heat transfer by S.K.Som, Chapter 2, Example 13\"\n",
-    "#A very long,10mm diameter(D) copper rod(thermal conductivity,k=370W/(m*K))is exposed to an enviroment at temprature,Tinf=20°C.\n",
-    "D=0.01;\n",
-    "k=370;\n",
-    "Tinf=20;\n",
-    "#The base temprature of the radius maintained at Tb=120°C.\n",
-    "Tb=120;\n",
-    "#The heat transfer coefficient between the rod and the surrounding air is h=10W/(m*K**2)\n",
-    "h=10;\n",
-    "#The rate of heat transfer for all finite lengths will be given by P/A=(4*pi*D)/(pi*D**2)\n",
-    "#Let P/A=X\n",
-    "print\"P/A in m**-1 is\"\n",
-    "X=(4*math.pi*D)/(math.pi*D**2)\n",
-    "print\"X=\",X\n",
-    "#m is defined as ((h*p)/(k*A)]**0.5 \n",
-    "print\"m in m**-1 is\"\n",
-    "m=(h*X/k)**0.5\n",
-    "print\"m=\",m\n",
-    "#Let Y=h/(m*k)\n",
-    "Y=h/(m*k)\n",
-    "#Let M=(h*P*k*A)**0.5\n",
-    "P=(math.pi*D);#perimeter of the rod\n",
-    "A=(math.pi*D**2)/4;#Area of the rod\n",
-    "print\"M in W/K is\"\n",
-    "M=(h*P*k*A)**0.5\n",
-    "print\"M=\",M\n",
-    "#thetab is the parameter that defines the base temprature\n",
-    "print\"thetab in °C is \"\n",
-    "thetab=Tb-Tinf\n",
-    "print\"thetab=\",thetab\n",
-    "#Heat loss from the rod is defined as Q=(h*P*k*A)*thetab*(((h/m*k)+math.tanh(m*L)]/(1+(h/m*k)*math.tanh(m*L)]}\n",
-    "print\"Heat loss from rod in Watt, for different value of length(in m) is \"\n",
-    "L=0.02#Length of rod\n",
-    "Q=M*thetab*(((Y)+math.tanh(m*L))/(1+(Y)*math.tanh(m*L)))\n",
-    "print\"Q=\",Q\n",
-    "L=0.04#length of rod\n",
-    "Q=M*thetab*(((Y)+math.tanh(m*L))/(1+(Y)*math.tanh(m*L)))\n",
-    "print\"Q=\",Q\n",
-    "L=0.08#length of rod\n",
-    "Q=M*thetab*(((Y)+math.tanh(m*L))/(1+(Y)*math.tanh(m*L)))\n",
-    "print\"Q=\",Q\n",
-    "L=0.20#length of rod\n",
-    "Q=M*thetab*(((Y)+math.tanh(m*L))/(1+(Y)*math.tanh(m*L)))\n",
-    "print\"Q=\",Q\n",
-    "L=0.40#length of rod\n",
-    "Q=M*thetab*(((Y)+math.tanh(m*L))/(1+(Y)*math.tanh(m*L)))\n",
-    "print\"Q=\",Q\n",
-    "L=0.80#length of rod\n",
-    "Q=M*thetab*(((Y)+math.tanh(m*L))/(1+(Y)*math.tanh(m*L)))\n",
-    "print\"Q=\",Q\n",
-    "L=1.00#length of rod\n",
-    "Q=M*thetab*(((Y)+math.tanh(m*L))/(1+(Y)*math.tanh(m*L)))\n",
-    "print\"Q=\",Q\n",
-    "L=10.00#length of rod\n",
-    "Q=M*thetab*(((Y)+math.tanh(m*L))/(1+(Y)*math.tanh(m*L)))\n",
-    "print\"Q=\",Q\n",
-    "#For an infinitely long rod we use heat loss as ,Qinf=(h*P*k*A)**0.5*thetab\n",
-    "print\"For an infintely long rod heat loss in W is\"\n",
-    "Qinf=(h*P*k*A)**0.5*thetab\n",
-    "print\"We see that since k is large there is significant difference  between the finite length and the infinte length cases\"\n",
-    "print\"However when the length of the rod approaches 1m,the result become almost same.\" \n",
-    "print\"Qinf=\",Qinf\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Ex2.14:pg-81"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 6,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Introduction to heat transfer by S.K.Som, Chapter 2, Example 14\n",
-      "The thermal conductivity of Rod B kB in W/(m*K) is \n",
-      "kB= 18.0\n"
-     ]
-    }
-   ],
-   "source": [
-    " \n",
-    "import math\n",
-    " \n",
-    "print\"Introduction to heat transfer by S.K.Som, Chapter 2, Example 14\"\n",
-    "#Considering two very long slender rods of the same diameter but of different materials.\n",
-    "#The base of each rod is maintained at 100°C while the surfaces of the rod are exposed to 20°C\n",
-    "#By traversing length of each rod with a thermocouple it was observed that tempratures of rod were equal at the position xA=0.15m and xB=0.075 from base.\n",
-    "xA=0.15;\n",
-    "xB=0.075;\n",
-    "#Thermal conductivity of rod A is known to be kA=72 W/(m*K)\n",
-    "kA=72;\n",
-    "#In case of a very long slender rod we use the tip boundary condition thetaL=0 as L--->infinity\n",
-    "#Therfore we can write for the locations where the tempratures are equal thetab*e**(-mA*xA)=thetab*e**(-mB*xB) or xA/xB=mB/mA,Again mB/mA=(kA/kB)**0.5\n",
-    "#So kB=kA*(xB/xA)**2\n",
-    "#The thermal conductivity of Rod B iskB\n",
-    "print\"The thermal conductivity of Rod B kB in W/(m*K) is \"\n",
-    "kB=kA*(xB/xA)**2\n",
-    "print\"kB=\",kB\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Ex2.15:pg-83"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 7,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Introduction to heat transfer by S.K.Som, Chapter 2, Example 15\n",
-      " perimeter of each fin in m is\n",
-      "P= 0.202\n",
-      "Cross sectional area of fin in m**2 is\n",
-      "A= 0.0001\n",
-      "M= 0.834805366538\n",
-      "m= 36.2958855016\n",
-      "Temprature parameter at fin base in K is\n",
-      "thetab= 100\n",
-      "Fixed temprature at fin tip in K is\n",
-      "thetaL= 50\n",
-      "Heat loss from the plate at 400K in W is\n",
-      "Qb= 13028.1662009\n"
-     ]
-    }
-   ],
-   "source": [
-    "  \n",
-    "import math\n",
-    " \n",
-    "print\"Introduction to heat transfer by S.K.Som, Chapter 2, Example 15\"\n",
-    "#A stack that is b=300mm wide and l=100mm deep contains N=60 fins each of length L=12mm.\n",
-    "L=0.012;#in metre\n",
-    "b=0.3;#in metre\n",
-    "l=0.1;#in metre\n",
-    "N=60;\n",
-    "#The entire stack is made of aluminum which is everywhere t=1.0 mm thick.\n",
-    "t=0.001;#in metre\n",
-    "#The temprature limitations associated with electrical components are  Tb=400K and TL=350K.\n",
-    "#Tb is base temprature and TL is end temprature\n",
-    "Tb=400;\n",
-    "TL=350; \n",
-    "#Given convection heat transfer coefficient(h=150W/(m**2*K)),Surrounding Temprature(Tinf=300K),thermal conductivity of aluminium(kaluminium=230W/(m*K))\n",
-    "h=150;\n",
-    "Tinf=300;\n",
-    "kal=230;\n",
-    "#Here both the ends of the fins are at fixed tempratures .Therefore we use M=(h*P*k*A)**0.5 and m=((h*P)/(k*A))**0.5,thetab=Tb-Tinf,thetaL=TL-Tinf\n",
-    "#from the given data perimeter of each fin is given by P= 2*(l+t)in m and area of each fin is A=t*l\n",
-    "print\" perimeter of each fin in m is\"\n",
-    "P= 2*(l+t)\n",
-    "print\"P=\",P\n",
-    "print\"Cross sectional area of fin in m**2 is\"\n",
-    "A=t*l\n",
-    "print\"A=\",A\n",
-    "#M is defined as (h*P*kal*A)**0.5 and m is defined as ((h*P)/(kal*A))**0.5\n",
-    "M=(h*P*kal*A)**0.5\n",
-    "m=((h*P)/(kal*A))**0.5\n",
-    "print\"M=\",M\n",
-    "print\"m=\",m\n",
-    "#thetab and thetaL are the parameters that define the fin tempratures at base and tip respectively.\n",
-    "print\"Temprature parameter at fin base in K is\"\n",
-    "thetab=Tb-Tinf\n",
-    "print\"thetab=\",thetab\n",
-    "print\"Fixed temprature at fin tip in K is\"\n",
-    "thetaL=TL-Tinf\n",
-    "print\"thetaL=\",thetaL\n",
-    "#Heat loss from the plate is Qb\n",
-    "print\"Heat loss from the plate at 400K in W is\"\n",
-    "Qb=(N*(h*P*kal*A)**0.5*thetab*((math.cosh(m*L)-(thetaL/thetab))/(math.sinh(m*L))))+(((l*b)-(N*A))*h*thetab)+(l*b*h*thetab)\n",
-    "print\"Qb=\",Qb\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\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": "2.7.11"
-  }
- },
- "nbformat": 4,
- "nbformat_minor": 0
-}
diff --git a/Introduction_to_Heat_Transfer_by_S._K._Som/Chapter3.ipynb b/Introduction_to_Heat_Transfer_by_S._K._Som/Chapter3.ipynb
index 6b51de29..55a11dc9 100644
--- a/Introduction_to_Heat_Transfer_by_S._K._Som/Chapter3.ipynb
+++ b/Introduction_to_Heat_Transfer_by_S._K._Som/Chapter3.ipynb
@@ -34,6 +34,9 @@
     }
    ],
    "source": [
+    "\n",
+    "\n",
+    " \n",
     "import math\n",
     " \n",
     "print\"Introduction to heat transfer by S.K.Som, Chapter 3, Example 1\"\n",
@@ -56,7 +59,7 @@
     "#Temperature in degree celcius\n",
     "print\"Temperature at the centre in Degree C is\"\n",
     "T = theta*100+100\n",
-    "print\"T=\",T"
+    "print\"T=\",T\n"
    ]
   },
   {
@@ -87,6 +90,9 @@
     }
    ],
    "source": [
+    " \n",
+    " \n",
+    " \n",
     "import math\n",
     " \n",
     "print\"Introduction to heat transfer by S.K.Som, Chapter 3, Example 2\"\n",
@@ -111,7 +117,7 @@
     "print\"theta=\",theta\n",
     "print\"Temperature in K at centre point\"\n",
     "T = theta*100+300\n",
-    "print\"T=\",T"
+    "print\"T=\",T\n"
    ]
   },
   {
@@ -160,6 +166,7 @@
     }
    ],
    "source": [
+    " \n",
     "import math\n",
     "import numpy\n",
     " \n",
@@ -206,7 +213,7 @@
     "print T7\n",
     "print\"T8 in degree K\"\n",
     "T8 = T[7]\n",
-    "print T8"
+    "print T8\n"
    ]
   },
   {
@@ -248,6 +255,7 @@
     }
    ],
    "source": [
+    " \n",
     "import math\n",
     "import numpy\n",
     " \n",
@@ -292,7 +300,7 @@
     "print T5\n",
     "print\"T6 in degree C\"\n",
     "T6 = T[4]\n",
-    "print T6"
+    "print T6\n"
    ]
   },
   {
@@ -346,6 +354,7 @@
     }
    ],
    "source": [
+    " \n",
     "import math\n",
     " \n",
     "print\"Introduction to heat transfer by S.K.Som, Chapter 3, Example 6\"\n",
@@ -403,7 +412,7 @@
     "print T8\n",
     "print\"T9 in degree C\"\n",
     "T9 = T[8]\n",
-    "print T9"
+    "print T9\n"
    ]
   }
  ],
diff --git a/Introduction_to_Heat_Transfer_by_S._K._Som/Chapter3_18OmDFC.ipynb b/Introduction_to_Heat_Transfer_by_S._K._Som/Chapter3_18OmDFC.ipynb
deleted file mode 100644
index 55a11dc9..00000000
--- a/Introduction_to_Heat_Transfer_by_S._K._Som/Chapter3_18OmDFC.ipynb
+++ /dev/null
@@ -1,440 +0,0 @@
-{
- "cells": [
-  {
-   "cell_type": "markdown",
-   "metadata": {
-    "collapsed": true
-   },
-   "source": [
-    "# Chapter 03:Multidimensional steady-state heat conduction"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Ex3.1:pg-92"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 1,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Introduction to heat transfer by S.K.Som, Chapter 3, Example 1\n",
-      "Temperature at the centre in Degree C is\n",
-      "T= 125.371641666\n"
-     ]
-    }
-   ],
-   "source": [
-    "\n",
-    "\n",
-    " \n",
-    "import math\n",
-    " \n",
-    "print\"Introduction to heat transfer by S.K.Som, Chapter 3, Example 1\"\n",
-    "#Length and breadth is given as 1 unit (Gemoetry is Square)\n",
-    "L = 1;#length\n",
-    "#Problem can be divided into two modules\n",
-    "#Solution to module 1 is given by Eq. 3.21, considering the first three terms\n",
-    "#n is the looping parameter\n",
-    "#theta is the non dimensional temperature defined as ((T-100)/100) where T is actual temperature in degree Celcius.\n",
-    "#Initialising theta as zero\n",
-    "theta = 0;\n",
-    "for n in range(1,3):\n",
-    "  theta = theta+((2/math.pi)*((math.sin((n*math.pi)/2)*math.sinh((n*math.pi)/2))*((-1)**(n+1)+1)))/(n*math.sinh(n*math.pi));\n",
-    " \n",
-    "#Solution to module 2 is given by Eq. 3.24, considering the first three terms\n",
-    "for n in range(1,3):\n",
-    "  theta2 = theta+(((3*2)/math.pi)*((math.sin((n*math.pi)/2)*math.sinh((n*math.pi)/2))*((-1)**(n+1)+1)))/(n*math.sinh(n*math.pi));\n",
-    " \n",
-    "#Calculating value of temperature from the value of theta\n",
-    "#Temperature in degree celcius\n",
-    "print\"Temperature at the centre in Degree C is\"\n",
-    "T = theta*100+100\n",
-    "print\"T=\",T\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Ex3.2:pg-94"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 2,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      " Introduction to heat transfer by S.K.Som, Chapter 3, Example 2\n",
-      "Steady state non dimensional temperature is\n",
-      "theta=2*math.sinh(pi*y/a)*math.sin(pi*x/a)/(math.sinh(pi)) + math.sinh(pi*x/a)*math.sin(pi*y/a)/(math.sinh(pi))\n",
-      "theta= 0.597805223008\n",
-      "Temperature in K at centre point\n",
-      "T= 359.780522301\n"
-     ]
-    }
-   ],
-   "source": [
-    " \n",
-    " \n",
-    " \n",
-    "import math\n",
-    " \n",
-    "print\"Introduction to heat transfer by S.K.Som, Chapter 3, Example 2\"\n",
-    "#Temperature in K at four edges are given\n",
-    "#Theta is non dimensional temperature defined as ((T-300)/100) where T is actual temperature in K.\n",
-    "#Given length as well as the breadth of square plate is ''a''\n",
-    "#Problem can be divided into two modules\n",
-    "#Solution to module 1 is given by Eq. 3.23\n",
-    "#Solution of first module is non dimensional temperature theta1\n",
-    "#theta1=2*math.sinh(pi*y/a)*math.sin(pi*x/a)/(math.sinh(pi))\n",
-    "#Solution to module 2 is given by Eq. 3.24\n",
-    "#Solution of second module is non dimensional temperature theta2\n",
-    "#theta2=math.sinh(pi*x/a)*math.sin(pi*y/a)/(math.sinh(pi))\n",
-    "#Therefore\n",
-    "print\"Steady state non dimensional temperature is\"\n",
-    "print\"theta=2*math.sinh(pi*y/a)*math.sin(pi*x/a)/(math.sinh(pi)) + math.sinh(pi*x/a)*math.sin(pi*y/a)/(math.sinh(pi))\"\n",
-    "#At the centre, x coordinate and y coordinate in unit are\n",
-    "#x=a/2, y=a/2\n",
-    "#Non dimensional temperature at centre point\n",
-    "theta = (2*math.sinh(math.pi/2))/math.sinh(math.pi)+math.sinh(math.pi/2)/math.sinh(math.pi);\n",
-    "#Temperature in K at centre point\n",
-    "print\"theta=\",theta\n",
-    "print\"Temperature in K at centre point\"\n",
-    "T = theta*100+300\n",
-    "print\"T=\",T\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Ex3.3:pg-96"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 14,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Introduction to heat transfer by S.K.Som, Chapter 3, Example 3\n",
-      "Temperatures at nodal points in degree K\n",
-      "T1 in degree K\n",
-      "[ 398.67699539  155.83601706   66.53320567  119.43598224   79.06693359\n",
-      "   40.99573505   14.15266777    9.19140047]\n",
-      "T2 in degree K\n",
-      "[  77.91800853  232.60510053   92.0706763    39.53346679   80.21585865\n",
-      "   48.72486726   19.11393507   11.30646706]\n",
-      "T3 in degree K\n",
-      "[  33.26660284   92.0706763   237.56636783   20.49786753   48.72486726\n",
-      "   82.33092523   46.76647228   21.51623292]\n",
-      "T4 in degree K\n",
-      "[  14.15266777   38.22787014   93.53294456    9.19140047   22.61293411\n",
-      "   43.03246584  124.91948821   27.99199234]\n",
-      "T5 in degree K\n",
-      "[-0. -0. -0. -0. -0. -0. -0. -0.]\n",
-      "T6 in degree K\n",
-      "[-0. -0. -0. -0. -0. -0. -0. -0.]\n",
-      "T7 in degree K\n",
-      "[  95.65671512  227.3827139   384.21098442   77.62207329  214.83157803\n",
-      "  554.32152494  100.40908695  109.12176865]\n",
-      "T8 in degree K\n",
-      "[  24.51040125   60.30115763  114.75324223   18.87022369   50.97049352\n",
-      "  124.71059274   74.64531291  166.55931761]\n"
-     ]
-    }
-   ],
-   "source": [
-    " \n",
-    "import math\n",
-    "import numpy\n",
-    " \n",
-    "print\"Introduction to heat transfer by S.K.Som, Chapter 3, Example 3\"\n",
-    "#internodal distance in x direction in m\n",
-    "deltax = 1.0/4;\n",
-    "#internodal distance in y direction in m\n",
-    "deltay = 1.0/4;\n",
-    "#Air temperature in degree K\n",
-    "Tinfinity = 400;\n",
-    "#Heat transfer coefficient in W/(m**2*K)\n",
-    "h = 10;\n",
-    "#T1, T2, T3, T4, T5, T6, T7, T8 are nodal temperatures in degree K.\n",
-    "#T is the temperature matrix and is transpose of [T1 T2 T3 T4 T5 T6 T7 T8]\n",
-    "#using Nodal Equations, we have Coefficeint Matrix A as\n",
-    "A = [[-4,1,0,0,1,0,0,0],[1,-4,1,0,0,1,0,0],[0,1,-4,1,0,0,1,0],[2,0,0,0,-4,1,0,0],[0,2,0,0,1,-4,1,0],[0,0,2,0,0,1,-4,1],[0,0,2,-6,0,0,0,1],[0,0,0,2,0,0,2,-6]]#Coefficient matrix B\n",
-    "B = [[-1200],[-600],[-600],[-600],[0],[0],[-1400],[-800]]\n",
-    "\n",
-    "\n",
-    "#Therefore the temperature matrix is\n",
-    "T = numpy.linalg.inv(A)*B;\n",
-    "#Temperature at nodal points in degree K\n",
-    "print\"Temperatures at nodal points in degree K\"\n",
-    "print\"T1 in degree K\"\n",
-    "T1 = T[0]\n",
-    "print T1\n",
-    "print\"T2 in degree K\"\n",
-    "T2 = T[1]\n",
-    "print T2\n",
-    "print\"T3 in degree K\"\n",
-    "T3 = T[2]\n",
-    "print T3\n",
-    "print\"T4 in degree K\"\n",
-    "T4 = T[3]\n",
-    "print T4\n",
-    "print\"T5 in degree K\"\n",
-    "T5 = T[4]\n",
-    "print T5\n",
-    "print\"T6 in degree K\"\n",
-    "T6 = T[5]\n",
-    "print T6\n",
-    "print\"T7 in degree K\"\n",
-    "T7 = T[6]\n",
-    "print T7\n",
-    "print\"T8 in degree K\"\n",
-    "T8 = T[7]\n",
-    "print T8\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Ex3.5:pg-98"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 33,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Introduction to heat transfer by S.K.Som, Chapter 3, Example 5\n",
-      "Temperatures at nodal points in degree C\n",
-      "T2 in degree C\n",
-      "[  1.83976243e-36   4.79441040e-01   3.66134997e-01   3.07581515e-01\n",
-      "   5.38080937e-01]\n",
-      "T3 in degree C\n",
-      "[  1.46972670e-67   1.92742646e+00   1.47191880e+00   1.23652483e+00\n",
-      "   1.07886949e+00]\n",
-      "T4 in degree C\n",
-      "[  4.52446173e-92   1.47191880e+00   3.54032873e+00   2.97414801e+00\n",
-      "   1.67320356e+00]\n",
-      "T5 in degree C\n",
-      "[  6.50142301e-108   1.23652483e+000   2.97414801e+000   5.91733919e+000\n",
-      "   2.36010173e+000]\n",
-      "T6 in degree C\n",
-      "[  2.06473580e-113   1.16395938e+000   2.79961015e+000   5.57008016e+000\n",
-      "   3.18199172e+000]\n"
-     ]
-    }
-   ],
-   "source": [
-    " \n",
-    "import math\n",
-    "import numpy\n",
-    " \n",
-    "print\"Introduction to heat transfer by S.K.Som, Chapter 3, Example 5\"\n",
-    "#Thermal conductivity of aluminium in W/(m*K)\n",
-    "k = 200.0\n",
-    "#Diameter in m\n",
-    "d = 20*(10**(-3));\n",
-    "#Length of fin in m\n",
-    "L = 0.2;\n",
-    "#Wall temperature in degree C\n",
-    "Tw = 400.0;\n",
-    "#Air temperature in degree C\n",
-    "Tinfinity = 30;\n",
-    "#Heat transfer coefficient in W/(m**2*K)\n",
-    "h = 40.0;\n",
-    "#internodal distance in x direction in m\n",
-    "deltax = L/5;\n",
-    "#Node 1 temperature is equal to  wall temperature in degree C\n",
-    "T1 = Tw;\n",
-    "#using Nodal Equations, we have Coefficeint Matrix A as\n",
-    "A = [[2.064,-1,0,0,0],[-1,2.064,-1,0,0],[0,-1,2.064,-1,0],[0,0,-1,2.064,-1],[0,0,0,-1,1.032]]\n",
-    "#Coefficient matrix B\n",
-    "B = [401.92,1.92,1.92,1.92,0.96]\n",
-    "#T2, T3, T4, T5, T6 are nodal temperature in degree C\n",
-    "#T is the temperature matrix and is transpose of [T2 T3 T4 T5 T6]\n",
-    "#Therefore the temperature matrix is\n",
-    "T = numpy.linalg.inv(A)**B;\n",
-    "#Temperature at nodal points in degree C\n",
-    "print\"Temperatures at nodal points in degree C\"\n",
-    "print\"T2 in degree C\"\n",
-    "T2 = T[0]\n",
-    "print T2\n",
-    "print\"T3 in degree C\"\n",
-    "T3 = T[1]\n",
-    "print T3\n",
-    "print\"T4 in degree C\"\n",
-    "T4 = T[2]\n",
-    "print T4\n",
-    "print\"T5 in degree C\"\n",
-    "T5 = T[3]\n",
-    "print T5\n",
-    "print\"T6 in degree C\"\n",
-    "T6 = T[4]\n",
-    "print T6\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Ex3.6:pg-104"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 40,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Introduction to heat transfer by S.K.Som, Chapter 3, Example 6\n",
-      "Temperatures at nodal points in degree C\n",
-      "T1 in degree C\n",
-      "[  -0.           -0.           -0.          186.04651163    1.86046512\n",
-      "    2.79069767    1.86046512    0.46511628   74.41860465]\n",
-      "T2 in degree C\n",
-      "[  -0.           -0.           -0.           74.41860465    1.69263965\n",
-      "    3.56988732    2.51738192    0.62934548  157.39630784]\n",
-      "T3 in degree C\n",
-      "[ -0.          -0.          -0.          83.72093023   3.88875569\n",
-      "   4.80220571   3.06401343   0.76600336  65.85950611]\n",
-      "T4 in degree C\n",
-      "[ -0.          -0.          -0.          55.81395349   2.45504675\n",
-      "   5.7444258    4.10453129   1.02613282  77.58331335]\n",
-      "T5 in degree C\n",
-      "[ -0.          -0.          -0.          37.20930233   1.77415488\n",
-      "   4.6199952    7.34116519   1.8352913   51.37856629]\n",
-      "T6 in degree C\n",
-      "[ -0.          -0.          -0.          37.20930233   8.78446416\n",
-      "   4.92927356   2.18652601   0.5466315   33.8527931 ]\n",
-      "T7 in degree C\n",
-      "[ -0.          -0.          -0.          27.90697674   2.46463678\n",
-      "   9.98561496   3.49556461   0.87389115  35.69887317]\n",
-      "T8 in degree C\n",
-      "[ -0.          -0.          -0.          18.60465116   1.09326301\n",
-      "   3.49556461  10.57779909   2.64444977  25.17381923]\n",
-      "T9 in degree C\n",
-      "[ -0.          -0.          -0.           9.30232558   0.5466315\n",
-      "   1.74778231   5.28889954  11.32222489  12.58690961]\n"
-     ]
-    }
-   ],
-   "source": [
-    " \n",
-    "import math\n",
-    " \n",
-    "print\"Introduction to heat transfer by S.K.Som, Chapter 3, Example 6\"\n",
-    "#Thermal conductivity of concrete in W/mK\n",
-    "k = 2;\n",
-    "#Length in m\n",
-    "L = 0.2;\n",
-    "#Breadth in m\n",
-    "b = 0.2;\n",
-    "#Depth in m\n",
-    "d = 0.2;\n",
-    "#Temperature of hot gas in chimney in degree C\n",
-    "Tg = 400;\n",
-    "#Air temperature in degree C\n",
-    "Tinfinity = 20;\n",
-    "#internodal distance in x direction in m\n",
-    "deltax = 0.1;\n",
-    "#internodal distance in y direction in m\n",
-    "deltay = 0.1;\n",
-    "#Heat transfer coefficient in W/(m**2*K)\n",
-    "h = 20;\n",
-    "#T1, T2, T3, T4, T5, T6, T7, T8, T9 are nodal temperatures in degree K.\n",
-    "#T is the temperature matrix and is transpose of [T1 T2 T3 T4 T5 T6 T7 T8 T9]\n",
-    "#using Nodal Equations, we have Coefficeint Matrix A as\n",
-    "A = numpy.array([[1,0,-4,2,0,1,0,0,0],[0,1,1,-4,1,0,1,0,0],[0,0,0,2,-4,0,0,2,0],[-3,1,1,0,0,0,0,0,0],[0,0,1,0,0,-3,1,0,0],[0,0,0,2,0,1,-6,1,0],[0,0,0,0,2,0,1,-6,1],[0,0,0,0,0,0,0,1,-2],[1,-4,0,2,0,0,0,0,0]]);\n",
-    "#Coefficient matrix B\n",
-    "B = numpy.array([0,0,0,-400,-20,-40,-40,-20,-400]);\n",
-    "#Therefore the temperature matrix is\n",
-    "T = numpy.linalg.inv(A)*B;\n",
-    "#Temperature at nodal points in degree C\n",
-    "print\"Temperatures at nodal points in degree C\"\n",
-    "print\"T1 in degree C\"\n",
-    "T1 = T[0]\n",
-    "print T1\n",
-    "print\"T2 in degree C\"\n",
-    "T2 = T[1]\n",
-    "print T2\n",
-    "print\"T3 in degree C\"\n",
-    "T3 = T[2]\n",
-    "print T3\n",
-    "print\"T4 in degree C\"\n",
-    "T4 = T[3]\n",
-    "print T4\n",
-    "print\"T5 in degree C\"\n",
-    "T5 = T[4]\n",
-    "print T5\n",
-    "print\"T6 in degree C\"\n",
-    "T6 = T[5]\n",
-    "print T6\n",
-    "print\"T7 in degree C\"\n",
-    "T7 = T[6]\n",
-    "print T7\n",
-    "print\"T8 in degree C\"\n",
-    "T8 = T[7]\n",
-    "print T8\n",
-    "print\"T9 in degree C\"\n",
-    "T9 = T[8]\n",
-    "print T9\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": "2.7.11"
-  }
- },
- "nbformat": 4,
- "nbformat_minor": 0
-}
diff --git a/Introduction_to_Heat_Transfer_by_S._K._Som/Chapter4.ipynb b/Introduction_to_Heat_Transfer_by_S._K._Som/Chapter4.ipynb
index 6c85f1a7..54a5875b 100644
--- a/Introduction_to_Heat_Transfer_by_S._K._Som/Chapter4.ipynb
+++ b/Introduction_to_Heat_Transfer_by_S._K._Som/Chapter4.ipynb
@@ -35,6 +35,10 @@
     }
    ],
    "source": [
+    " \n",
+    "\n",
+    " \n",
+    " \n",
     "import math\n",
     " \n",
     "print\"Introduction to heat transfer by S.K.Som, Chapter 4, Example 1\"\n",
@@ -55,7 +59,8 @@
     "if Bi<0.1:\n",
     "  print\"Problem is suitable for lumped parameter analysis\"\n",
     "else:\n",
-    "  print\"Problem is not suitable for lumped parameter analysis\""
+    "  print\"Problem is not suitable for lumped parameter analysis\"\n",
+    "\n"
    ]
   },
   {
@@ -67,7 +72,7 @@
   },
   {
    "cell_type": "code",
-   "execution_count": 3,
+   "execution_count": 1,
    "metadata": {
     "collapsed": false
    },
@@ -85,6 +90,7 @@
     }
    ],
    "source": [
+    " \n",
     "import math\n",
     " \n",
     "print\"Introduction to heat transfer by S.K.Som, Chapter 4, Example 2\"\n",
@@ -109,7 +115,7 @@
     "#Required time in sec\n",
     "t = (-8)*math.log(0.01);\n",
     "print\"Time required in seconds\"\n",
-    "print\"t=\",t"
+    "print\"t=\",t\n"
    ]
   },
   {
@@ -121,7 +127,7 @@
   },
   {
    "cell_type": "code",
-   "execution_count": 4,
+   "execution_count": 2,
    "metadata": {
     "collapsed": false
    },
@@ -137,6 +143,7 @@
     }
    ],
    "source": [
+    " \n",
     "import math\n",
     " \n",
     "print\"Introduction to heat transfer by S.K.Som, Chapter 4, Example 3\"\n",
@@ -150,7 +157,7 @@
     "#Maximum dimension in metre\n",
     "a = ((6*k)*Bi)/h;\n",
     "print\"Maximum dimension in metre for lumped parameter analysis\"\n",
-    "print\"a=\",a"
+    "print\"a=\",a\n"
    ]
   },
   {
@@ -179,6 +186,7 @@
     }
    ],
    "source": [
+    " \n",
     "from scipy.integrate  import quad\n",
     "import math\n",
     "print\"Introduction to heat transfer by S.K.Som, Chapter 4, Example 4\"\n",
@@ -214,7 +222,7 @@
     "E = (((h*math.pi)*d)*H)*quad(lambda t:(80.0-25.0)*math.e*(-t/472.5),0,60.0*t)[0];\n",
     "print\"Energy required for cooling in KJ\"\n",
     "E = E/1000.0\n",
-    "print \"E=\",E"
+    "print \"E=\",E\n"
    ]
   },
   {
@@ -226,7 +234,7 @@
   },
   {
    "cell_type": "code",
-   "execution_count": 6,
+   "execution_count": 3,
    "metadata": {
     "collapsed": false
    },
@@ -244,6 +252,7 @@
     }
    ],
    "source": [
+    " \n",
     "import math\n",
     " \n",
     "print\"Introduction to heat transfer by S.K.Som, Chapter 4, Example 5\"\n",
@@ -279,7 +288,7 @@
     "Q = ((((0.69*k)*2)*L)*(Tinfinity-Ti))/alpha;\n",
     "print\"Heat transfer rate in MJ\"\n",
     "Q = Q/(10**6)\n",
-    "print\"Q=\",Q"
+    "print\"Q=\",Q\n"
    ]
   },
   {
@@ -291,7 +300,7 @@
   },
   {
    "cell_type": "code",
-   "execution_count": 7,
+   "execution_count": 4,
    "metadata": {
     "collapsed": false
    },
@@ -309,6 +318,7 @@
     }
    ],
    "source": [
+    " \n",
     "import math\n",
     " \n",
     "print\"Introduction to heat transfer by S.K.Som, Chapter 4, Example 6\"\n",
@@ -346,7 +356,7 @@
     "Q = (((0.4*k)*L)*(Ti-Tinfinity))/alpha;\n",
     "print\"Heat transfer rate in MJ\"\n",
     "Q = Q/(10**6)\n",
-    "print\"Q=\",Q"
+    "print\"Q=\",Q\n"
    ]
   },
   {
@@ -358,7 +368,7 @@
   },
   {
    "cell_type": "code",
-   "execution_count": 8,
+   "execution_count": 5,
    "metadata": {
     "collapsed": false
    },
@@ -376,6 +386,7 @@
     }
    ],
    "source": [
+    " \n",
     "import math\n",
     " \n",
     "print\"Introduction to heat transfer by S.K.Som, Chapter 4, Example 7\"\n",
@@ -413,7 +424,7 @@
     "Q = (((((0.4*k)*math.pi)*ro)*ro)*(Ti-Tinfinity))/alpha;\n",
     "print\"Heat transfer rate per unit length in MJ/m\"\n",
     "Q = Q/(10**6)\n",
-    "print\"Q=\",Q"
+    "print\"Q=\",Q\n"
    ]
   },
   {
@@ -425,7 +436,7 @@
   },
   {
    "cell_type": "code",
-   "execution_count": 9,
+   "execution_count": 6,
    "metadata": {
     "collapsed": false
    },
@@ -441,6 +452,7 @@
     }
    ],
    "source": [
+    " \n",
     "import math\n",
     "\n",
     "print\"Introduction to heat transfer by S.K.Som, Chapter 4, Example 8\"\n",
@@ -469,7 +481,7 @@
     "t = ((Fo*ro)*ro)/alpha;\n",
     "print\"Time required in minutes\"\n",
     "t = t/60\n",
-    "print\"t=\",t"
+    "print\"t=\",t\n"
    ]
   },
   {
@@ -481,7 +493,7 @@
   },
   {
    "cell_type": "code",
-   "execution_count": 10,
+   "execution_count": 7,
    "metadata": {
     "collapsed": false
    },
@@ -497,6 +509,7 @@
     }
    ],
    "source": [
+    " \n",
     "import math\n",
     " \n",
     "print\"Introduction to heat transfer by S.K.Som, Chapter 4, Example 9\"\n",
@@ -539,7 +552,7 @@
     "#Temperature in °C\n",
     "T = Tinfinity+z*(Ti-Tinfinity);\n",
     "print\"Tempearture of bar in °C\"\n",
-    "print\"T=\",T"
+    "print\"T=\",T\n"
    ]
   },
   {
@@ -551,7 +564,7 @@
   },
   {
    "cell_type": "code",
-   "execution_count": 13,
+   "execution_count": 8,
    "metadata": {
     "collapsed": false
    },
@@ -576,6 +589,7 @@
     }
    ],
    "source": [
+    " \n",
     "import math\n",
     " \n",
     "print\"Introduction to heat transfer by S.K.Som, Chapter 4, Example 10\"\n",
@@ -645,7 +659,17 @@
     "#Therefore ((To-Tinf)/(Ti-Tinf))plate1*((To-Tinf)/(Ti-Tinf))plate2=A*B\n",
     "T=A*B\n",
     "print\"The calculated value is very close to the required value of 0.6.Hence the time required for the centre of the beam to reach 310°C is nearly 1200s or 20 minutes.\" \n",
-    "print\"T=\",T"
+    "print\"T=\",T\n",
+    " \n",
+    " \n",
+    " \n",
+    " \n",
+    " \n",
+    " \n",
+    " \n",
+    " \n",
+    " \n",
+    " \n"
    ]
   },
   {
@@ -657,7 +681,7 @@
   },
   {
    "cell_type": "code",
-   "execution_count": 14,
+   "execution_count": 9,
    "metadata": {
     "collapsed": false
    },
@@ -673,6 +697,7 @@
     }
    ],
    "source": [
+    " \n",
     "import math\n",
     " \n",
     "print\"Introduction to heat transfer by S.K.Som, Chapter 4, Example 11\"\n",
@@ -694,7 +719,22 @@
     "#Therefore 10/t**0.5=0.38...this implies t=(10/0.38)**2\n",
     "print\"The time required for the temprature to reach 255°C at a depth of 80mm, in minutes is\"\n",
     "t=(10/0.38)**2/60\n",
-    "print\"T=\",T"
+    "print\"T=\",T\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n"
    ]
   },
   {
@@ -724,6 +764,7 @@
     }
    ],
    "source": [
+    " \n",
     "import math\n",
     "import scipy \n",
     "print\"Introduction to heat transfer by S.K.Som, Chapter 4, Example 12\"\n",
@@ -747,7 +788,18 @@
     "print\"The temprature at a depth(x) of 100mm after a time(t) of 100 seconds,in °C is\"\n",
     "T=Ti+((2*qo*(alpha*t/math.pi)**0.5)/(k))*math.e**((-x**2.0)/(4*alpha*t))-((qo*x)/(k))*scipy.special.erf(x/(2*(alpha*t)**0.5))\n",
     "print\"T=\",T\n",
-    "#NOTE:The answer in the book is incorrect(Calculation mistake)"
+    "#NOTE:The answer in the book is incorrect(Calculation mistake)\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n"
    ]
   },
   {
@@ -759,7 +811,7 @@
   },
   {
    "cell_type": "code",
-   "execution_count": 2,
+   "execution_count": 17,
    "metadata": {
     "collapsed": false
    },
@@ -770,7 +822,7 @@
      "text": [
       "Introduction to heat transfer by S.K.Som, Chapter 4, Example 14\n",
       "Temperature distribution after 25 mins in °C\n",
-      "T= [[  2.29192547e+02   2.91925466e+00   1.11801242e+00   4.34782609e-01\n",
+      "[[  2.29192547e+02   2.91925466e+00   1.11801242e+00   4.34782609e-01\n",
       "    1.86335404e-01   6.21118012e-02]\n",
       " [  8.75776398e+01   8.75776398e+00   3.35403727e+00   1.30434783e+00\n",
       "    5.59006211e-01   1.86335404e-01]\n",
@@ -786,6 +838,7 @@
     }
    ],
    "source": [
+    " \n",
     "import math\n",
     "import numpy\n",
     " \n",
@@ -811,8 +864,10 @@
     "#From Eq. 4.126\n",
     "#Temperature distribution after one time step\n",
     "T = numpy.linalg.inv(A)*B;\n",
+    "\n",
+    " \n",
     "print\"Temperature distribution after 25 mins in °C\"\n",
-    "print\"T=\",T"
+    "print T\n"
    ]
   }
  ],
diff --git a/Introduction_to_Heat_Transfer_by_S._K._Som/Chapter4_fMX8RWT.ipynb b/Introduction_to_Heat_Transfer_by_S._K._Som/Chapter4_fMX8RWT.ipynb
deleted file mode 100644
index 54a5875b..00000000
--- a/Introduction_to_Heat_Transfer_by_S._K._Som/Chapter4_fMX8RWT.ipynb
+++ /dev/null
@@ -1,895 +0,0 @@
-{
- "cells": [
-  {
-   "cell_type": "markdown",
-   "metadata": {
-    "collapsed": true
-   },
-   "source": [
-    "# Chapter 04:Unsteady conduction"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Ex4.1:pg-137"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 1,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Introduction to heat transfer by S.K.Som, Chapter 4, Example 1\n",
-      "Biot number is\n",
-      "Bi= 0.277777777778\n",
-      "Problem is not suitable for lumped parameter analysis\n"
-     ]
-    }
-   ],
-   "source": [
-    " \n",
-    "\n",
-    " \n",
-    " \n",
-    "import math\n",
-    " \n",
-    "print\"Introduction to heat transfer by S.K.Som, Chapter 4, Example 1\"\n",
-    "#Diameter of apple in m\n",
-    "d = 100*(10**(-3));\n",
-    "#radius in m\n",
-    "r = d/2;\n",
-    "#Thermal conductivity of apple in W/(m*K)\n",
-    "k = 0.6;\n",
-    "#Heat transfer coefficient in W/(m**2*°C)\n",
-    "h = 10;\n",
-    "#Caculating characteristic dimension in m\n",
-    "Lc = (((((4*math.pi)*r)*r)*r)/3)/(((4*math.pi)*r)*r);\n",
-    "#Biot number\n",
-    "print\"Biot number is\"\n",
-    "Bi = (h*Lc)/k\n",
-    "print\"Bi=\",Bi\n",
-    "if Bi<0.1:\n",
-    "  print\"Problem is suitable for lumped parameter analysis\"\n",
-    "else:\n",
-    "  print\"Problem is not suitable for lumped parameter analysis\"\n",
-    "\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Ex4.2:pg-138 "
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 1,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Introduction to heat transfer by S.K.Som, Chapter 4, Example 2\n",
-      "Time constant in seconds is\n",
-      "tc= 8.0\n",
-      "Time required in seconds\n",
-      "t= 36.8413614879\n"
-     ]
-    }
-   ],
-   "source": [
-    " \n",
-    "import math\n",
-    " \n",
-    "print\"Introduction to heat transfer by S.K.Som, Chapter 4, Example 2\"\n",
-    "#Diameter of sphere in m\n",
-    "d = 1.5*(10**(-3));\n",
-    "#radius in m\n",
-    "r = d/2;\n",
-    "#Thermal conductivity of sphere in W/(m*°C)\n",
-    "k = 40.0;\n",
-    "#Density in kg/m**3\n",
-    "rho = 8000.0;\n",
-    "#Specific heat in J/(Kg*K)\n",
-    "c = 300.0;\n",
-    "#Heat transfer coefficient in W/(m**2*°C)\n",
-    "h = 75.0;\n",
-    "#Time constant in sec\n",
-    "tc = ((rho*c)*(((((4*math.pi)*r)*r)*r)/3))/((((h*4)*math.pi)*r)*r);\n",
-    "print\"Time constant in seconds is\"\n",
-    "print\"tc=\",tc\n",
-    "#Using eq. 4.4\n",
-    "#Given fraction is 0.01 (1 percent)\n",
-    "#Required time in sec\n",
-    "t = (-8)*math.log(0.01);\n",
-    "print\"Time required in seconds\"\n",
-    "print\"t=\",t\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Ex4.3:pg-138"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 2,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Introduction to heat transfer by S.K.Som, Chapter 4, Example 3\n",
-      "Maximum dimension in metre for lumped parameter analysis\n",
-      "a= 5.0\n"
-     ]
-    }
-   ],
-   "source": [
-    " \n",
-    "import math\n",
-    " \n",
-    "print\"Introduction to heat transfer by S.K.Som, Chapter 4, Example 3\"\n",
-    "#Heat transfer coefficient in W/(m**2*K)\n",
-    "h = 30;\n",
-    "#Thermal conductivity of sphere in W/(m*K)\n",
-    "k = 250;\n",
-    "#Biot number for lumped parameter analysis is 0.1\n",
-    "Bi = 0.1;\n",
-    "#Characteristic dimension of a cube is (a/6) where a is the side of cube in metre\n",
-    "#Maximum dimension in metre\n",
-    "a = ((6*k)*Bi)/h;\n",
-    "print\"Maximum dimension in metre for lumped parameter analysis\"\n",
-    "print\"a=\",a\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Ex4.4:pg-146"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 5,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Introduction to heat transfer by S.K.Som, Chapter 4, Example 4\n",
-      "Time required to cool milk in minutes\n",
-      "Energy required for cooling in KJ\n",
-      "E= -319.013666564\n"
-     ]
-    }
-   ],
-   "source": [
-    " \n",
-    "from scipy.integrate  import quad\n",
-    "import math\n",
-    "print\"Introduction to heat transfer by S.K.Som, Chapter 4, Example 4\"\n",
-    "#Diameter of glass in m\n",
-    "d = 50*(10**(-3));\n",
-    "#radius in m\n",
-    "r = d/2;\n",
-    "#Height of milk in glass in m\n",
-    "H = 0.1;\n",
-    "#Initial temperature of milk in °C\n",
-    "T = 80.0;\n",
-    "#Cold water temperature in °C\n",
-    "Tf = 25.0;\n",
-    "#Heat transfer coefficient in W/(m**2*°C)\n",
-    "h = 100.0;\n",
-    "#Thermal conductivity of milk in W/(m*K)\n",
-    "k = 0.6;\n",
-    "#Density of milk in kg/m**3\n",
-    "rho = 900.0;\n",
-    "#Specific heat in J/(Kg*K)\n",
-    "c = 4.2*(10**3);\n",
-    "#Since the milk temperature is always maintained as constant.\n",
-    "#Therefore it can be assumed as lumped paramteter analysis.\n",
-    "#Time constant n seconds\n",
-    "tcs = (((((rho*c)*math.pi)*r)*r)*H)/(((h*math.pi)*d)*H);\n",
-    "#Time constant in minutes\n",
-    "tc = tcs/60;\n",
-    "#Calculating from eq. 4.3 time taken to cool milk from 80°C to 30°C\n",
-    "t = -tc*math.log((30.0-Tf)/(T-Tf));\n",
-    "print\"Time required to cool milk in minutes\"\n",
-    "t\n",
-    "#Energy transferred during cooling\n",
-    "E = (((h*math.pi)*d)*H)*quad(lambda t:(80.0-25.0)*math.e*(-t/472.5),0,60.0*t)[0];\n",
-    "print\"Energy required for cooling in KJ\"\n",
-    "E = E/1000.0\n",
-    "print \"E=\",E\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Ex4.5:pg-159"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 3,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Introduction to heat transfer by S.K.Som, Chapter 4, Example 5\n",
-      "Time required in hours\n",
-      "t= 7.5\n",
-      "Heat transfer rate in MJ\n",
-      "Q= 186.3\n"
-     ]
-    }
-   ],
-   "source": [
-    " \n",
-    "import math\n",
-    " \n",
-    "print\"Introduction to heat transfer by S.K.Som, Chapter 4, Example 5\"\n",
-    "#Thermal conductivity of wall in W/(m*K)\n",
-    "k = 0.6;\n",
-    "#Thermal diffusivity in m**2/s\n",
-    "alpha = 5*(10**(-7));\n",
-    "#Thickness in m\n",
-    "L = 0.15;\n",
-    "#Initial temperature in °C\n",
-    "Ti = 30;\n",
-    "#Temperature of hot gas in °C\n",
-    "Tinfinity = 780;\n",
-    "#Heat transfer coefficient in W/(m**2*K)\n",
-    "h = 20;\n",
-    "#Surface temperaute to be achieved in °C\n",
-    "To = 480;\n",
-    "#Dimensionless temperature ratio\n",
-    "z = (To-Tinfinity)/(Ti-Tinfinity);\n",
-    "#Biot number\n",
-    "Bi = (h*L)/k;\n",
-    "#For this value of (1/Bi) and dimensionless temp. ratio\n",
-    "#From Fig. 4.11 Fourier number is\n",
-    "Fo = 0.6;\n",
-    "#Time required in seconds\n",
-    "t = ((Fo*L)*L)/alpha;\n",
-    "print\"Time required in hours\"\n",
-    "t = t/3600\n",
-    "print\"t=\",t\n",
-    "#From fig. 4.13, for this Bi and Fo*Bi*Bi, we have ratio of heats as\n",
-    "#Q/Qi=0.69\n",
-    "#Heat transfer in J\n",
-    "Q = ((((0.69*k)*2)*L)*(Tinfinity-Ti))/alpha;\n",
-    "print\"Heat transfer rate in MJ\"\n",
-    "Q = Q/(10**6)\n",
-    "print\"Q=\",Q\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Ex4.6:pg-166"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 4,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Introduction to heat transfer by S.K.Som, Chapter 4, Example 6\n",
-      "Temperature at this distance in °C\n",
-      "T= 355.5\n",
-      "Heat transfer rate in MJ\n",
-      "Q= 100.0\n"
-     ]
-    }
-   ],
-   "source": [
-    " \n",
-    "import math\n",
-    " \n",
-    "print\"Introduction to heat transfer by S.K.Som, Chapter 4, Example 6\"\n",
-    "#Thickness of plate in m\n",
-    "L = 0.2;\n",
-    "#Initial temperature in °C\n",
-    "Ti = 530;\n",
-    "#Heat transfer coefficient in W/(m**2*K)\n",
-    "h = 500;\n",
-    "#Given distance in m\n",
-    "x = L-20*(10**(-3));\n",
-    "#Temperature of surrounding in °C\n",
-    "Tinfinity = 30;\n",
-    "#Given time in seconds\n",
-    "t = 225;\n",
-    "#Thermal conductivity of aluminium in W/(m*K)\n",
-    "k = 200;\n",
-    "#Thermal diffusivity in m**2/s\n",
-    "alpha = 8*(10**(-5));\n",
-    "#Biot number\n",
-    "Bi = (h*L)/k;\n",
-    "#Fourier number\n",
-    "Fo = (alpha*t)/(L*L);\n",
-    "#From fig. 4.11, at this Fo and (1/Bi), we have dimensionless temperature\n",
-    "#ratio to be 0.7\n",
-    "#From fig. 4.12 for this (1/Bi) and (x/L), we have another dimensionless\n",
-    "#temperature to be 0.93\n",
-    "#Temperature in °C\n",
-    "T = Tinfinity+(0.93*0.7)*(Ti-Tinfinity);\n",
-    "print\"Temperature at this distance in °C\"\n",
-    "print\"T=\",T\n",
-    "#From fig. 4.13, for this Bi and Fo*Bi*Bi, we have ratio of heats as\n",
-    "#Q/Qi=0.4\n",
-    "#Heat transfer in J\n",
-    "Q = (((0.4*k)*L)*(Ti-Tinfinity))/alpha;\n",
-    "print\"Heat transfer rate in MJ\"\n",
-    "Q = Q/(10**6)\n",
-    "print\"Q=\",Q\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Ex4.7:pg-171"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 5,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Introduction to heat transfer by S.K.Som, Chapter 4, Example 7\n",
-      "Temperature at this radius in °C\n",
-      "T= 300.0\n",
-      "Heat transfer rate per unit length in MJ/m\n",
-      "Q= 33.2639222145\n"
-     ]
-    }
-   ],
-   "source": [
-    " \n",
-    "import math\n",
-    " \n",
-    "print\"Introduction to heat transfer by S.K.Som, Chapter 4, Example 7\"\n",
-    "#Radius in m\n",
-    "ro = 0.15;\n",
-    "#Initial temperature in °C\n",
-    "Ti = 530;\n",
-    "#Temperature of surrounding in °C\n",
-    "Tinfinity = 30;\n",
-    "#Heat transfer coefficient in W/(m**2*K)\n",
-    "h = 380;\n",
-    "#Thermal conductivity of aluminium in W/(m*K)\n",
-    "k = 200;\n",
-    "#Thermal diffusivity in m**2/s\n",
-    "alpha = 8.5*(10**(-5));\n",
-    "#Given radius at which temperature has to be find out in m\n",
-    "r = 0.12;\n",
-    "#Given time in seconds\n",
-    "t = 265;\n",
-    "#Fourier number\n",
-    "Fo = (alpha*t)/(ro**2);\n",
-    "#Biot number\n",
-    "Bi = (h*ro)/k;\n",
-    "#From fig. 4.15, at this fourier number,Fo and (1/Bi), we have dimensionless temperature\n",
-    "#ratio to be 0.6\n",
-    "#From fig. 4.16 for this (1/Bi) and (r/ro), we have another dimensionless\n",
-    "#temperature to be 0.9\n",
-    "#Temperature in °C\n",
-    "T = Tinfinity+(0.9*0.6)*(Ti-Tinfinity);\n",
-    "print\"Temperature at this radius in °C\"\n",
-    "print\"T=\",T\n",
-    "#From fig. 4.17, for this Bi and Fo*Bi*Bi, we have ratio of heats as\n",
-    "#Q/Qi=0.4\n",
-    "#Heat transfer per metre in J/m\n",
-    "Q = (((((0.4*k)*math.pi)*ro)*ro)*(Ti-Tinfinity))/alpha;\n",
-    "print\"Heat transfer rate per unit length in MJ/m\"\n",
-    "Q = Q/(10**6)\n",
-    "print\"Q=\",Q\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Ex4.8:pg-174"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 6,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Introduction to heat transfer by S.K.Som, Chapter 4, Example 8\n",
-      "Time required in minutes\n",
-      "t= 4.16666666667\n"
-     ]
-    }
-   ],
-   "source": [
-    " \n",
-    "import math\n",
-    "\n",
-    "print\"Introduction to heat transfer by S.K.Som, Chapter 4, Example 8\"\n",
-    "#Radius in m\n",
-    "ro = 0.05;\n",
-    "#Initial temperature in °C\n",
-    "Ti = 530;\n",
-    "#Temperature of surrounding in °C\n",
-    "Tinfinity = 30;\n",
-    "#Heat transfer coefficient in W/(m**2*K)\n",
-    "h = 500;\n",
-    "#Thermal conductivity of aluminium in W/(m*K)\n",
-    "k = 50;\n",
-    "#Thermal diffusivity in m**2/s\n",
-    "alpha = 1.5*(10**(-5));\n",
-    "#Required centre temperature to achieve in °C\n",
-    "To = 105;\n",
-    "#Dimensionless temperature\n",
-    "z = (To-Tinfinity)/(Ti-Tinfinity);\n",
-    "#Biot number\n",
-    "Bi = (h*ro)/k;\n",
-    "#For this value of (1/Bi) and dimensionless temp. ratio\n",
-    "#From Fig. 4.19 Fourier number is\n",
-    "Fo = 1.5;\n",
-    "#Time required in seconds\n",
-    "t = ((Fo*ro)*ro)/alpha;\n",
-    "print\"Time required in minutes\"\n",
-    "t = t/60\n",
-    "print\"t=\",t\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Ex4.9:pg-177"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 7,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Introduction to heat transfer by S.K.Som, Chapter 4, Example 9\n",
-      "Tempearture of bar in °C\n",
-      "T= 260.3\n"
-     ]
-    }
-   ],
-   "source": [
-    " \n",
-    "import math\n",
-    " \n",
-    "print\"Introduction to heat transfer by S.K.Som, Chapter 4, Example 9\"\n",
-    "#Thermal conductivity of aluminium in W/(m*K)\n",
-    "k = 198;\n",
-    "#Length in m\n",
-    "L = 0.18;\n",
-    "#Breadth in m\n",
-    "b = 0.104;\n",
-    "#Initial temperature in °C\n",
-    "Ti = 730;\n",
-    "#Temperature of surrounding in °C\n",
-    "Tinfinity = 30;\n",
-    "#Heat transfer coefficient in W/(m**2*K)\n",
-    "h = 1100;\n",
-    "#Thermal diffusivity in m**2/s\n",
-    "alpha = 8.1*(10**(-5));\n",
-    "#Given time in seconds\n",
-    "t = 100;\n",
-    "#Bar can be considered to be an intersection of two infinite plates of\n",
-    "#thickness L1 and L2 in m\n",
-    "L1 = L/2;\n",
-    "L2 = b/2;\n",
-    "#For plate 1\n",
-    "#Fourier number\n",
-    "Fo1 = (alpha*t)/(L1**2);\n",
-    "#Biot number\n",
-    "Bi1 = (h*L1)/k;\n",
-    "#From fig. 4.11, at this Fo and (1/Bi), we have dimensionless temperature\n",
-    "#ratio to be 0.7\n",
-    "#For plate 2\n",
-    "#Fourier number\n",
-    "Fo2 = (alpha*t)/(L2**2);\n",
-    "#Biot number\n",
-    "Bi2 = (h*L2)/k;\n",
-    "#From fig. 4.11, at this Fo and (1/Bi), we have dimensionless temperature\n",
-    "#ratio to be 0.47\n",
-    "#Therefore combined dimensionless temperature ratio is multiply of two\n",
-    "z = 0.47*0.7;\n",
-    "#Temperature in °C\n",
-    "T = Tinfinity+z*(Ti-Tinfinity);\n",
-    "print\"Tempearture of bar in °C\"\n",
-    "print\"T=\",T\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Ex4.10:pg-180"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 8,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Introduction to heat transfer by S.K.Som, Chapter 4, Example 10\n",
-      "The factor((To-Tinf)/(Ti-Tinf)) is \n",
-      "For plate 1\n",
-      "A= 0.85\n",
-      "For plate 2\n",
-      "B= 0.8\n",
-      "For plate 1\n",
-      "A= 0.83\n",
-      "For plate 2\n",
-      "B= 0.72\n",
-      "The calculated value is very close to the required value of 0.6.Hence the time required for the centre of the beam to reach 310°C is nearly 1200s or 20 minutes.\n",
-      "T= 0.5976\n"
-     ]
-    }
-   ],
-   "source": [
-    " \n",
-    "import math\n",
-    " \n",
-    "print\"Introduction to heat transfer by S.K.Som, Chapter 4, Example 10\"\n",
-    "#An iron beam of rectangular cross section of size length,L=300mm by breadth,B=200 mm is used in the construction of a building\n",
-    "#Initially the beam is at a uniform temprature(Ti) of 30°C.\n",
-    "#Due to an accidental fire,the beam is suddenly exposed to hot gases at temprature,Tinf=730°C,with a convective heat transfer coefficient(h) of 100 W/(m**2*K)\n",
-    "#To determine the time required for the centre plane of the beam to reach a temprature(To) of 310°C.\n",
-    "To=310;\n",
-    "Tinf=730;\n",
-    "Ti=30;\n",
-    "#Take thermal conductivity k=73W/(m*K) and thermal diffusivity of the beam alpha=2.034*10**-5m**2/s \n",
-    "alpha=2.034*10**-5; \n",
-    "k=73; \n",
-    "h=100; \n",
-    "#The rectangular iron beam can be considered as an intersection of an infinite plate 1 having thickness 2*L1=300mm and a second infinite plate 2 of thickness 2*L2=200mm \n",
-    "L1=0.15;#in metre\n",
-    "L2=0.10;#in metre\n",
-    "#Here the faactor X=((To-Tinf)/(Ti-Tinf))\n",
-    "print\"The factor((To-Tinf)/(Ti-Tinf)) is \"\n",
-    "X=((To-Tinf)/(Ti-Tinf))\n",
-    "#Therefore we can write 0.6=((To-Tinf)/(Ti-Tinf))plate 1 *((To-Tinf)/(Ti-Tinf))plate2\n",
-    "#A straight forward solution is not possible.We have to adopt an iterative method of solution \n",
-    "#At first ,a value of time(t) is assumed to determine the centre-line temprature of the beam.The value of t at which((To-Tinf)/(Ti-Tinf))beam =0.6 is satisfied\n",
-    "#Let us first assume time, t=900s\n",
-    "t=900;\n",
-    "print\"For plate 1\"\n",
-    "#For plate1 Biot number Bi1=h*L1/k \n",
-    "Bi1=h*L1/k  \n",
-    "Y=1/Bi1\n",
-    "#Fourier number(Fo1) is\n",
-    "Fo1=alpha*t/L1**2\n",
-    "#At Fo=0.814 and (1/Bi)=4.87...We read from graphs  A=((To-Tinf)/(Ti-Tinf))plate1= 0.85\n",
-    "A=0.85;\n",
-    "print\"A=\",A\n",
-    "print\"For plate 2\"\n",
-    "#For plate1 Biot number Bi2=h*L2/k \n",
-    "Bi2=h*L2/k  \n",
-    "Y=1/Bi2\n",
-    "#Fourier number(Fo2) is\n",
-    "Fo2=alpha*t/L2**2\n",
-    "#At Fo=1.83 and (1/Bi)=7.3...We read from graphs  B=((To-Tinf)/(Ti-Tinf))plate2= 0.8\n",
-    "B=0.8;\n",
-    "print\"B=\",B\n",
-    "#Therefore ((To-Tinf)/(Ti-Tinf))plate1*((To-Tinf)/(Ti-Tinf))plate2=A*B\n",
-    "T=A*B\n",
-    "#Since the calculated value of 0.68 is greater than the required value of 0.60 and Tinf>To>Ti,The assume dvalue of t is less.\n",
-    "#So let us take time,t=1200s for the second iteration\n",
-    "t=1200;\n",
-    "print\"For plate 1\"\n",
-    "#For plate1 Biot number Bi1=h*L1/k \n",
-    "Bi1=h*L1/k  \n",
-    "Y=1/Bi1\n",
-    "#Fourier number (Fo1)\n",
-    "Fo1=alpha*t/L1**2\n",
-    "#At Fo=1.08 and (1/Bi)=4.87...We read from graphs  A=((To-Tinf)/(Ti-Tinf))plate1= 0.83\n",
-    "A=0.83;\n",
-    "print\"A=\",A\n",
-    "print\"For plate 2\"\n",
-    "#For plate1 Biot number Bi2=h*L2/k \n",
-    "Bi2=h*L2/k  \n",
-    "Y=1/Bi2\n",
-    "#Fourier number (Fo2)\n",
-    "Fo2=alpha*t/L2**2\n",
-    "#At Fo=2.44 and (1/Bi)=7.3...We read from graphs  B=((To-Tinf)/(Ti-Tinf))plate2= 0.72\n",
-    "B=0.72;\n",
-    "print\"B=\",B\n",
-    "#Therefore ((To-Tinf)/(Ti-Tinf))plate1*((To-Tinf)/(Ti-Tinf))plate2=A*B\n",
-    "T=A*B\n",
-    "print\"The calculated value is very close to the required value of 0.6.Hence the time required for the centre of the beam to reach 310°C is nearly 1200s or 20 minutes.\" \n",
-    "print\"T=\",T\n",
-    " \n",
-    " \n",
-    " \n",
-    " \n",
-    " \n",
-    " \n",
-    " \n",
-    " \n",
-    " \n",
-    " \n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Ex4.11:pg-182"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 9,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Introduction to heat transfer by S.K.Som, Chapter 4, Example 11\n",
-      "The time required for the temprature to reach 255°C at a depth of 80mm, in minutes is\n",
-      "T= 255\n"
-     ]
-    }
-   ],
-   "source": [
-    " \n",
-    "import math\n",
-    " \n",
-    "print\"Introduction to heat transfer by S.K.Som, Chapter 4, Example 11\"\n",
-    "#A large slab wrought-iron is at a uniform temprature of Ti=550°C.\n",
-    "#The temprature of one surface is suddenly changed to Tinf=50°C\n",
-    "Tinf=50;\n",
-    "Ti=550; \n",
-    "#For slab conductivity(k=60W/(m*K)),Thermal diffusivity(alpha=1.6*10**-5m**2/s)\n",
-    "#To calculate the time(t) required for the temprature to reach T=255°C at a depth of 80mm\n",
-    "k=60;\n",
-    "T=255;\n",
-    "alpha=1.6**10-5;\n",
-    "#Similarity parameter,eta=x/(2*(alpha*t)**0.5)=(10/t**0.5)\n",
-    "#((T-Tinf)/(Ti-Tinf))=erf(10/t**0.5)...where erf is the error function.\n",
-    "#Let ((T-Tinf)/(Ti-Tinf))=X\n",
-    "X=((T-Tinf)/(Ti-Tinf));\n",
-    "#This implies erf(10/t**0.5)=0.41\n",
-    "#We read from the table the value of eta(=10/t**0.5)=0.38....corresponding to erf(eta)=0.41\n",
-    "#Therefore 10/t**0.5=0.38...this implies t=(10/0.38)**2\n",
-    "print\"The time required for the temprature to reach 255°C at a depth of 80mm, in minutes is\"\n",
-    "t=(10/0.38)**2/60\n",
-    "print\"T=\",T\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Ex4.12:pg-186"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 15,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Introduction to heat transfer by S.K.Som, Chapter 4, Example 12\n",
-      "gaussian error function is \n",
-      "E= 0.998109069322\n",
-      "The temprature at a depth(x) of 100mm after a time(t) of 100 seconds,in °C is\n",
-      "T= -29851.5095103\n"
-     ]
-    }
-   ],
-   "source": [
-    " \n",
-    "import math\n",
-    "import scipy \n",
-    "print\"Introduction to heat transfer by S.K.Som, Chapter 4, Example 12\"\n",
-    "#A large block of nickel steel conductivity(k=20W/(m*K)),thermal diffusivity(alpha=0.518*10-5 m**2/s) is at uniform temprature(Ti) of 30°C.\n",
-    "Ti=30.0;\n",
-    "k=20.0;\n",
-    "alpha=0.518*10.0**-5.0;\n",
-    "#One surface of the block is suddenly exposed to a constant surface heat flux(qo) of 6MW/m**2.\n",
-    "qo=6*10**6;#in W/m**2\n",
-    "#To determine the temprature at a depth(x) of 100mm after a time(t) of 100 seconds.\n",
-    "t=100.0;\n",
-    "x=0.1;#in metre\n",
-    "#Similarity parameter,eta=x/(4*alpha*t)\n",
-    "eta=x/((4.0*alpha*t)**0.5)\n",
-    "#E is gaussian error function\n",
-    "print\"gaussian error function is \"\n",
-    "E=scipy.special.erf(eta)\n",
-    "print\"E=\",E\n",
-    "#The equation to determine temprature is T-Ti=((2*qo(alpha*t/math.pi)**0.5)/(k))*e**((-x**2)/(4*alpha*t))-((qo*x)/(k))*erf(x/(2.0*(alpha*t)**0.5))\n",
-    "#Above equation can also be written as T=Ti+((2*qo(alpha*t/math.pi)**0.5)/(k))*e**((-x**2)/(4*alpha*t))-((qo*x)/(k))*erf(x/(2.0*(alpha*t)**0.5))\n",
-    "print\"The temprature at a depth(x) of 100mm after a time(t) of 100 seconds,in °C is\"\n",
-    "T=Ti+((2*qo*(alpha*t/math.pi)**0.5)/(k))*math.e**((-x**2.0)/(4*alpha*t))-((qo*x)/(k))*scipy.special.erf(x/(2*(alpha*t)**0.5))\n",
-    "print\"T=\",T\n",
-    "#NOTE:The answer in the book is incorrect(Calculation mistake)\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Ex4.14:pg-187 "
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 17,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Introduction to heat transfer by S.K.Som, Chapter 4, Example 14\n",
-      "Temperature distribution after 25 mins in °C\n",
-      "[[  2.29192547e+02   2.91925466e+00   1.11801242e+00   4.34782609e-01\n",
-      "    1.86335404e-01   6.21118012e-02]\n",
-      " [  8.75776398e+01   8.75776398e+00   3.35403727e+00   1.30434783e+00\n",
-      "    5.59006211e-01   1.86335404e-01]\n",
-      " [  3.35403727e+01   3.35403727e+00   8.94409938e+00   3.47826087e+00\n",
-      "    1.49068323e+00   4.96894410e-01]\n",
-      " [  1.30434783e+01   1.30434783e+00   3.47826087e+00   9.13043478e+00\n",
-      "    3.91304348e+00   1.30434783e+00]\n",
-      " [  5.59006211e+00   5.59006211e-01   1.49068323e+00   3.91304348e+00\n",
-      "    1.02484472e+01   3.41614907e+00]\n",
-      " [  3.72670807e+00   3.72670807e-01   9.93788820e-01   2.60869565e+00\n",
-      "    6.83229814e+00   8.94409938e+00]]\n"
-     ]
-    }
-   ],
-   "source": [
-    " \n",
-    "import math\n",
-    "import numpy\n",
-    " \n",
-    "print\"Introduction to heat transfer by S.K.Som, Chapter 4, Example 14\"\n",
-    "#Nodal distance Deltax in m\n",
-    "deltax = 0.1;\n",
-    "#Time in seconds\n",
-    "t = 25*60;\n",
-    "#timestep deltaT in seconds\n",
-    "deltaT = 500;\n",
-    "#Number of increment\n",
-    "n = t/deltaT;\n",
-    "#Temperature raised in °C\n",
-    "To = 580.0;\n",
-    "#Using Eq. 4.114 for interior grid points, table 4.8 for exterior node\n",
-    "#Using Eq. 4.125a to 4.125f are written in matrix form\n",
-    "#Coefficient matrix A is\n",
-    "A = [[-3,1,0,0,0,0],[1,-3,1,0,0,0],[0,1,-3,1,0,0],[0,0,1,-3,1,0],[0,0,0,1,-3,1],[0,0,0,0,2,-3]]\n",
-    "#Coefficient matrix B is\n",
-    "B = [-600,-20,-20,-20,-20,-20];\n",
-    "#Temperature matrix is transpose of [T2 T3 T4 T5 T6 T7] where\n",
-    "#T2 to T7 are temperature in °C\n",
-    "#From Eq. 4.126\n",
-    "#Temperature distribution after one time step\n",
-    "T = numpy.linalg.inv(A)*B;\n",
-    "\n",
-    " \n",
-    "print\"Temperature distribution after 25 mins in °C\"\n",
-    "print T\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": "2.7.11"
-  }
- },
- "nbformat": 4,
- "nbformat_minor": 0
-}
diff --git a/Introduction_to_Heat_Transfer_by_S._K._Som/Chapter5.ipynb b/Introduction_to_Heat_Transfer_by_S._K._Som/Chapter5.ipynb
index 5b3d46e7..1751c7ce 100644
--- a/Introduction_to_Heat_Transfer_by_S._K._Som/Chapter5.ipynb
+++ b/Introduction_to_Heat_Transfer_by_S._K._Som/Chapter5.ipynb
@@ -38,6 +38,7 @@
     }
    ],
    "source": [
+    " \n",
     "import math \n",
     "from scipy.integrate import quad\n",
     " \n",
@@ -73,7 +74,23 @@
     "#Q is the rate of heat transfer\n",
     "print\"The rate of heat transfer in W/m of width is\"\n",
     "Q=hbarL*L*(T2-T1)\n",
-    "print\"Q=\",Q"
+    "print\"Q=\",Q\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n"
    ]
   },
   {
@@ -85,7 +102,7 @@
   },
   {
    "cell_type": "code",
-   "execution_count": 27,
+   "execution_count": 2,
    "metadata": {
     "collapsed": false
    },
@@ -102,6 +119,7 @@
     }
    ],
    "source": [
+    " \n",
     "import math\n",
     " \n",
     "print\"Introduction to heat transfer by S.K.Som, Chapter 5, Example 4\"\n",
@@ -126,7 +144,18 @@
     "#from an enrgy balance we can write as E=27.063*U**0.85*L*B*(Ts-Tinf)\n",
     "print\"The minimum flow velocity in m/s is\"\n",
     "U=(E/(27.063*L*B*(Ts-Tinf)))**(1/0.85)\n",
-    "print\"U=\",U"
+    "print\"U=\",U\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n"
    ]
   },
   {
@@ -138,7 +167,7 @@
   },
   {
    "cell_type": "code",
-   "execution_count": 1,
+   "execution_count": 3,
    "metadata": {
     "collapsed": false
    },
@@ -161,8 +190,10 @@
     }
    ],
    "source": [
+    " \n",
     "import math\n",
     " \n",
+    " \n",
     "print\"Introduction to heat transfer by S.K.Som, Chapter 5, Example 6\"\n",
     "#Air at 1atm pressure and temprature(Tin)=30°C enters a tube of 25mm diameter(D) with a velocity(U) of 10m/s\n",
     "D=0.025;#in metre\n",
@@ -200,7 +231,25 @@
     "k=0.0285;\n",
     "print\"Overall Nusselt number is \"\n",
     "NuL=hx*D/k\n",
-    "print\"NuL=\",NuL"
+    "print\"NuL=\",NuL\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n"
    ]
   },
   {
@@ -212,7 +261,7 @@
   },
   {
    "cell_type": "code",
-   "execution_count": 3,
+   "execution_count": 4,
    "metadata": {
     "collapsed": false
    },
@@ -236,6 +285,7 @@
     }
    ],
    "source": [
+    " \n",
     "import math\n",
     " \n",
     "print\"Introduction to heat transfer by S.K.Som, Chapter 5, Example 7\"\n",
@@ -274,7 +324,21 @@
     "#Q is the heat loss from the plate\n",
     "print\"The heat loss from the plate in W is\"\n",
     "Q=hbar*A*(Ts-Tinf)\n",
-    "print\"Q=\",Q"
+    "print\"Q=\",Q\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n"
    ]
   },
   {
@@ -306,6 +370,7 @@
     }
    ],
    "source": [
+    " \n",
     "import math\n",
     " \n",
     "print\"Introduction to heat transfer by S.K.Som, Chapter 5, Example 8\"\n",
@@ -333,7 +398,28 @@
     "#I is the current flow.\n",
     "print\"The current in Ampere is\"\n",
     "I=(Q/(R*L))**0.5\n",
-    "print\"I=\",I"
+    "print\"I=\",I\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n"
    ]
   }
  ],
diff --git a/Introduction_to_Heat_Transfer_by_S._K._Som/Chapter5_2AAnLS8.ipynb b/Introduction_to_Heat_Transfer_by_S._K._Som/Chapter5_2AAnLS8.ipynb
deleted file mode 100644
index 1751c7ce..00000000
--- a/Introduction_to_Heat_Transfer_by_S._K._Som/Chapter5_2AAnLS8.ipynb
+++ /dev/null
@@ -1,447 +0,0 @@
-{
- "cells": [
-  {
-   "cell_type": "markdown",
-   "metadata": {
-    "collapsed": true
-   },
-   "source": [
-    "# Chapter 5:Convection"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Ex5.3:pg-206"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 1,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Introduction to heat transfer by S.K.Som, Chapter 5, Example 3\n",
-      "Reynolds number is\n",
-      "ReL= 55263.1578947\n",
-      "The average heat transfer coefficient over a length(L)= 1m ,in W/m**2 is\n",
-      "hbarL= 4.16\n",
-      "The rate of heat transfer in W/m of width is\n",
-      "Q= 332.8\n"
-     ]
-    }
-   ],
-   "source": [
-    " \n",
-    "import math \n",
-    "from scipy.integrate import quad\n",
-    " \n",
-    "print\"Introduction to heat transfer by S.K.Som, Chapter 5, Example 3\"\n",
-    "#Air at temprature (T1=20°C) and 1 atmospheric pressure flows over a flat plate with a free stream velocity(Uinf) of 1m/s.\n",
-    "Uinf=1;\n",
-    "T1=20;\n",
-    "#The length of plate is 1m and is heated over its entire length to a constant temprature of T2=100°C.\n",
-    "T2=100;\n",
-    "#For air at 20°C(The mean temprature of 100°C and 20°C),viscosity(mu=1.9*10**-5kg/(m*s)),density(rho=1.05kg/m**3),conductivity(k=0.03W/(m*K)),specific heat(cp=1.007kJ/(kg*K))\n",
-    "#Prandtl number is Pr=0.7\n",
-    "mu=1.9*10**-5;\n",
-    "rho=1.05;\n",
-    "k=0.03;\n",
-    "cp=1.007;\n",
-    "Pr=0.7;\n",
-    "#For laminar flow over a plate Nusselt number is Nux=0.332*Rex**0.5*Pr**(1/3)\n",
-    "#The boundary layer flow over a flat plate will be laminar if Reynolds number is Rex=(rho*Uinf*x)/mu<5*10**5\n",
-    "#First of all we have to check whether the flow is laminar or not.\n",
-    "#Let us check at x=1m\n",
-    "x=1.0;\n",
-    "print\"Reynolds number is\"\n",
-    "ReL=(rho*Uinf*x)/mu\n",
-    "print\"ReL=\",ReL\n",
-    "#There fore the flow is laminar and we can use the relationships of Nux,\n",
-    "#Thus Rex=(1.05*1*x)/(1.9*10**-5)=0.5526*10**5*x\n",
-    "#Therefore we can write Nux=(hx*x/k)=0.332*(0.5526*10**5*x)**0.5*Pr**(1/3)....or hx=2.08*x**(-1/2) W/(m**2*°C)\n",
-    "#hbarL is the average heat transfer coefficient over a length(L)\n",
-    "print\"The average heat transfer coefficient over a length(L)= 1m ,in W/m**2 is\"\n",
-    "L=1;\n",
-    "hbarL=(1.0/L)*quad(lambda x:2.08*x**(-1/2.0),0,L)[0]\n",
-    "print\"hbarL=\",hbarL\n",
-    "#Q is the rate of heat transfer\n",
-    "print\"The rate of heat transfer in W/m of width is\"\n",
-    "Q=hbarL*L*(T2-T1)\n",
-    "print\"Q=\",Q\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Ex5.4:pg-207"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 2,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Introduction to heat transfer by S.K.Som, Chapter 5, Example 4\n",
-      "The local heat transfer coefficient hx is hx=27.063*U**0.85\n",
-      "The minimum flow velocity in m/s is\n",
-      "U= 15.4806813943\n"
-     ]
-    }
-   ],
-   "source": [
-    " \n",
-    "import math\n",
-    " \n",
-    "print\"Introduction to heat transfer by S.K.Som, Chapter 5, Example 4\"\n",
-    "#Air at atmospheric pressure is required to flow over a circuit board to cool the electronics element mounted on it.\n",
-    "#Chip has length (L)=3mm and width(B)=3mm located x=0.1m from the leading edge\n",
-    "L=0.003;#in metre\n",
-    "B=0.003;#in metre\n",
-    "x=0.1;\n",
-    "#The Nusselt no. is given by Nux=0.06*Rex**0.85*Pr**0.33\n",
-    "#The chip has to dissipate E=50mW of energy while its surface temprature has to be kept below temprature,Ts=45°C and free strem Temptrature of air is Tinf=25°C\n",
-    "Ts=45;\n",
-    "Tinf=25;\n",
-    "E=50*10**-3;#in watt\n",
-    "#For air ,density(rho=1.2kg/m**3),viscosity(mu=1.8*10**5kg/(m*s)),conductivity(k=0.03W/(m*K)) and specific heat(cp=1000J/(kg*K))\n",
-    "rho=1.2;\n",
-    "mu=1.8*10**5;\n",
-    "k=0.03;\n",
-    "cp=1000;\n",
-    "#Let the minimum flow velocity be U.\n",
-    "#The local heat transfer coefficient hx where the chip is mounted is determined as hx=(k/x)*0.06*(rho*U*x/mu)**0.85*(mu*cp/k)**0.33\n",
-    "print\"The local heat transfer coefficient hx is hx=27.063*U**0.85\"\n",
-    "#from an enrgy balance we can write as E=27.063*U**0.85*L*B*(Ts-Tinf)\n",
-    "print\"The minimum flow velocity in m/s is\"\n",
-    "U=(E/(27.063*L*B*(Ts-Tinf)))**(1/0.85)\n",
-    "print\"U=\",U\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Ex5.6:pg-208"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 3,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Introduction to heat transfer by S.K.Som, Chapter 5, Example 6\n",
-      "mdot= 0.0235619449019\n",
-      "(dTb/dx)in °C/m is\n",
-      "Y= 26.6666666667\n",
-      "Therefore Exit bulk mean temprature Tb2 in °C is\n",
-      "Tb2= 83.3333333333\n",
-      "Heat flux(hx) in W/(m**2*°C) is \n",
-      "hx= 100\n",
-      "Overall Nusselt number is \n",
-      "NuL= 87.7192982456\n"
-     ]
-    }
-   ],
-   "source": [
-    " \n",
-    "import math\n",
-    " \n",
-    " \n",
-    "print\"Introduction to heat transfer by S.K.Som, Chapter 5, Example 6\"\n",
-    "#Air at 1atm pressure and temprature(Tin)=30°C enters a tube of 25mm diameter(D) with a velocity(U) of 10m/s\n",
-    "D=0.025;#in metre\n",
-    "U=10;\n",
-    "Tin=30;\n",
-    "#Tube is heated so that a constant heat flux(q) of 2kW/m**2 is maintained at the wall whose temprature is deltaT=20°C above the bulk mean air temprature through the length of tube \n",
-    "#Let Tw-Tb=T\n",
-    "q=2000;\n",
-    "deltaT=20;\n",
-    "#The length(L)= 2m\n",
-    "L=2;\n",
-    "#For air density(rho=1.2kg/m**3),specific heat(cp=1000J/(kg*K))\n",
-    "rho=1.2;\n",
-    "cp=1000;\n",
-    "#From an energy balance of a control volume of air we get mdot*cp*(Tb+(dTb/dx)*deltax-Tb)=q*pi*D*deltax or (dTb/dx)=(q*pi*D)/(mdot*cp)\n",
-    "#mass flow rate=mdot\n",
-    "mdot=rho*math.pi*D**2*U;\n",
-    "print\"mdot=\",mdot\n",
-    "#let (dTb/dx)=Y\n",
-    "print\"(dTb/dx)in °C/m is\"\n",
-    "Y=(4*q*math.pi*D)/(mdot*cp)\n",
-    "print\"Y=\",Y\n",
-    "#Tb2 is Exit bulk mean temprature\n",
-    "print\"Therefore Exit bulk mean temprature Tb2 in °C is\"\n",
-    "Tb2=Tin+Y*2\n",
-    "print\"Tb2=\",Tb2\n",
-    "#Again we can write at any section of the tube hx*(Tw-Tb)=q or hx=q/(Tw-Tb)\n",
-    "#hx is heat flux\n",
-    "print\"Heat flux(hx) in W/(m**2*°C) is \"\n",
-    "hx=q/(deltaT)\n",
-    "print\"hx=\",hx\n",
-    "#Since Tw-Tb remains the same,The heat transfer coefficient at all sections are the same\n",
-    "#Now Overall Nusselt number,NuL=hx*D/k\n",
-    "#The thermal conductivity of air at mean temprature of (30+83.4)/2=56.7°C is k=0.0285 W/(m*K)\n",
-    "k=0.0285;\n",
-    "print\"Overall Nusselt number is \"\n",
-    "NuL=hx*D/k\n",
-    "print\"NuL=\",NuL\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Ex5.7:pg-210"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 4,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Introduction to heat transfer by S.K.Som, Chapter 5, Example 7\n",
-      "Beta(The volumetric coefficient of expansion in K**-1 is\n",
-      "Beta= 0\n",
-      "Grashoff number is\n",
-      "Gr= 0.0\n",
-      "The average nusselt number is\n",
-      "NuHbar= 0.13\n",
-      "Heat flux hbar in W/(m**2*°C)\n",
-      "hbar= 0.00169\n",
-      "The heat loss from the plate in W is\n",
-      "Q= 0.4394\n"
-     ]
-    }
-   ],
-   "source": [
-    " \n",
-    "import math\n",
-    " \n",
-    "print\"Introduction to heat transfer by S.K.Som, Chapter 5, Example 7\"\n",
-    "#A wall is exposed to nitrogen at one atmospheric pressure and temprature,Tinf=4°C.\n",
-    "Tinf=4;\n",
-    "#The wall is H=2.0m high and B=2.5m wide and is maintained at temprature,Ts=56°C\n",
-    "Ts=56;\n",
-    "H=2;\n",
-    "B=2.5;\n",
-    "A=H*B;#area is(A)\n",
-    "#The average nusselt number NuHbar over the height of the plate is given by NuHbar=0.13*(Gr*Pr)**(1/3)\n",
-    "#The properties of nitrogen at mean film temprature(Tf) is (56+4)/2=30°C are given as density(rho=1.142kg/m**3) ,conductivity(k=0.026W/(m*K)),\n",
-    "#kinematic viscosity(nu=15.630*10**-6 m**2/s) ,Prandtl number(Pr=0.713)\n",
-    "rho=1.142;\n",
-    "k=0.026;\n",
-    "nu=15.630*10**-6;\n",
-    "Pr=0.713;\n",
-    "Tf=30;\n",
-    "#We first have to detrmine the value of Grashoff number,Gr.In consideration of nitrogen as an ideal gas,we can write\n",
-    "#Beta(The volumetric coefficient of expansion)=1/T\n",
-    "print\"Beta(The volumetric coefficient of expansion in K**-1 is\"\n",
-    "Beta=1/(273+Tf)\n",
-    "print\"Beta=\",Beta\n",
-    "#Now Gr=(g*Beta*(Ts-Tinf)*H**3)/nu**2\n",
-    "g=9.81;#acceleration due to gravity\n",
-    "print\"Grashoff number is\"\n",
-    "Gr=(g*Beta*(Ts-Tinf)*H**3)/nu**2\n",
-    "print\"Gr=\",Gr\n",
-    "print\"The average nusselt number is\"\n",
-    "NuHbar=0.13*(Gr*Pr)**(1/3)\n",
-    "print\"NuHbar=\",NuHbar\n",
-    "#hbar is the heat flux\n",
-    "print\"Heat flux hbar in W/(m**2*°C)\"\n",
-    "hbar=NuHbar*k/H\n",
-    "print\"hbar=\",hbar\n",
-    "#Q is the heat loss from the plate\n",
-    "print\"The heat loss from the plate in W is\"\n",
-    "Q=hbar*A*(Ts-Tinf)\n",
-    "print\"Q=\",Q\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Ex5.8:pg-211"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 1,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Introduction to heat transfer by S.K.Som, Chapter 5, Example 8\n",
-      "The heat flux in W/(m**2*K) is\n",
-      "hbar 120.0\n",
-      "The heat loss from the wire is Q=hbar*pi*D*L*(Twire-Tair) in Watt\n",
-      "Q= 1.88495559215\n",
-      "The current in Ampere is\n",
-      "I= 17.7245385091\n"
-     ]
-    }
-   ],
-   "source": [
-    " \n",
-    "import math\n",
-    " \n",
-    "print\"Introduction to heat transfer by S.K.Som, Chapter 5, Example 8\"\n",
-    "#Eletric current passes through a L=0.5m long horizontal wire of D=0.1mm diameter.\n",
-    "L=0.5;\n",
-    "D=0.1*10**-3;\n",
-    "#The wire is to be maintained at temprature,Twire=400K and the air is at temprature,Tair=300K.\n",
-    "Twire=400;\n",
-    "Tair=300;\n",
-    "#The resistance of the wire(R) is 0.012ohm per meter.Nusselt number(NuL) over the length of wire to be 0.4.\n",
-    "NuL=0.4;\n",
-    "R=0.012;\n",
-    "#At mean temprature of Tf=350K, The thermal conductivity of air is k=0.03W/(m*K)\n",
-    "k=0.03;\n",
-    "#Nusselt number is NuL=hbar*D/k\n",
-    "#hbar is the heat flux\n",
-    "print\"The heat flux in W/(m**2*K) is\"\n",
-    "hbar=NuL*k/D\n",
-    "print\"hbar\",hbar\n",
-    "#Q is the heat loss from the wire\n",
-    "print\"The heat loss from the wire is Q=hbar*pi*D*L*(Twire-Tair) in Watt\"\n",
-    "Q=hbar*math.pi*D*L*(Twire-Tair)\n",
-    "print\"Q=\",Q\n",
-    "#At steady state the ohmic loss in the wire equals the heat loss from its surface Therfore I**2*R=Q\n",
-    "#I is the current flow.\n",
-    "print\"The current in Ampere is\"\n",
-    "I=(Q/(R*L))**0.5\n",
-    "print\"I=\",I\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\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": "2.7.11"
-  }
- },
- "nbformat": 4,
- "nbformat_minor": 0
-}
diff --git a/Introduction_to_Heat_Transfer_by_S._K._Som/Chapter6.ipynb b/Introduction_to_Heat_Transfer_by_S._K._Som/Chapter6.ipynb
index 5a31da63..1c386970 100644
--- a/Introduction_to_Heat_Transfer_by_S._K._Som/Chapter6.ipynb
+++ b/Introduction_to_Heat_Transfer_by_S._K._Som/Chapter6.ipynb
@@ -18,7 +18,7 @@
   },
   {
    "cell_type": "code",
-   "execution_count": 10,
+   "execution_count": 3,
    "metadata": {
     "collapsed": false
    },
@@ -45,6 +45,7 @@
     }
    ],
    "source": [
+    " \n",
     "import math\n",
     " \n",
     "print\"Introduction to heat transfer by S.K.Som, Chapter 6, Example 1\"\n",
@@ -54,7 +55,7 @@
     "mu=0.1;\n",
     "b=0.005;  #in metre\n",
     " #Umax is maximum velocity\n",
-    " Umax=(3/2)*Uav\n",
+    " Umax=(3.0/2)*Uav\n",
     "print\"Umax in m/s is\"\n",
     "Umax=(3/2)*Uav\n",
     "print\"Umax=\",Umax\n",
@@ -81,7 +82,31 @@
     " #Since pressure drop is considered at a distance of 2m so L=2m\n",
     "L=2;\n",
     "deltaP=(-X)*L\n",
-    "print\"deltaP=\",deltaP"
+    "print\"deltaP=\",deltaP\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n"
    ]
   },
   {
@@ -93,7 +118,7 @@
   },
   {
    "cell_type": "code",
-   "execution_count": 14,
+   "execution_count": 2,
    "metadata": {
     "collapsed": false
    },
@@ -113,6 +138,7 @@
     }
    ],
    "source": [
+    " \n",
     "import math\n",
     " \n",
     "print\"Introduction to heat transfer by S.K.Som, Chapter 6, Example 3\"\n",
@@ -133,7 +159,28 @@
     " #The viscosity of oil is mu=(pi*D**4*X)/(128*Q*dz)\n",
     "print\"The viscosity of oil(mu)in kg/(m*s)\"\n",
     "mu=(math.pi*D**4*X)/(128*Q)\n",
-    "print\"mu=\",mu"
+    "print\"mu=\",mu\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n"
    ]
   },
   {
@@ -145,7 +192,7 @@
   },
   {
    "cell_type": "code",
-   "execution_count": 16,
+   "execution_count": 1,
    "metadata": {
     "collapsed": false
    },
@@ -154,7 +201,7 @@
      "name": "stdout",
      "output_type": "stream",
      "text": [
-      " Introduction to heat transfer by S.K.Som, Chapter 6, Example 7\n",
+      "Introduction to heat transfer by S.K.Som, Chapter 6, Example 7\n",
       "The maximum length of plate in m is \n",
       "L= 2.5\n",
       "The average skin friction coefficient is\n",
@@ -165,6 +212,7 @@
     }
    ],
    "source": [
+    " \n",
     "import math\n",
     " \n",
     "print\"Introduction to heat transfer by S.K.Som, Chapter 6, Example 7\"\n",
@@ -188,7 +236,26 @@
     " #Fd is drag force\n",
     "print\"Drag force on one side of plate in N is\"\n",
     "Fd=cfL*(rhoair*Uinf**2/2)*B*L\n",
-    "print\"Fd=\",Fd"
+    "print\"Fd=\",Fd\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n"
    ]
   },
   {
@@ -200,7 +267,7 @@
   },
   {
    "cell_type": "code",
-   "execution_count": 17,
+   "execution_count": 2,
    "metadata": {
     "collapsed": false
    },
@@ -224,6 +291,7 @@
     }
    ],
    "source": [
+    " \n",
     "import math\n",
     " \n",
     "print\"Introduction to heat transfer by S.K.Som, Chapter 6, Example 10\"\n",
@@ -252,7 +320,21 @@
     " #The turbulent boundary layer thickness at the trailing edge is given by delta=L*(0.379/ReL**(1/5))\n",
     "print\"The turbulent boundary layer thickness at the trailing edge in metre is \"\n",
     "delta=L*(0.379/ReL**(1/5))\n",
-    "print\"delta=\",delta"
+    "print\"delta=\",delta\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n"
    ]
   }
  ],
diff --git a/Introduction_to_Heat_Transfer_by_S._K._Som/Chapter6_lGPDUWp.ipynb b/Introduction_to_Heat_Transfer_by_S._K._Som/Chapter6_lGPDUWp.ipynb
deleted file mode 100644
index 1c386970..00000000
--- a/Introduction_to_Heat_Transfer_by_S._K._Som/Chapter6_lGPDUWp.ipynb
+++ /dev/null
@@ -1,362 +0,0 @@
-{
- "cells": [
-  {
-   "cell_type": "markdown",
-   "metadata": {
-    "collapsed": true
-   },
-   "source": [
-    "# Chapter 06:Incompressible viscous flow: A brief review"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Ex6.1:pg-226"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 3,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Introduction to heat transfer by S.K.Som, Chapter 6, Example 1\n",
-      "Umax in m/s is\n",
-      "Umax= 1.6\n",
-      "The shear stress T in N/m**2\n",
-      "T= 64.0\n",
-      "(dp/dx) in N/m**3 is\n",
-      "X= -19200.0\n",
-      "The Shear stress at a distance of 0.002m from the lower plate in N/m**2\n",
-      "t= -57.6\n",
-      "The shear stress at a distance of 0.002m from the upper plate in N/m**2\n",
-      "t= 57.6\n",
-      "The opposite signs in t represents the opposite directions.The plus sign is in the direction of flow and the minus sign is in the direction opposite to the flow \n",
-      "The pressure drop over a distance of 2m in N/m**2 is\n",
-      "deltaP= 38400.0\n"
-     ]
-    }
-   ],
-   "source": [
-    " \n",
-    "import math\n",
-    " \n",
-    "print\"Introduction to heat transfer by S.K.Som, Chapter 6, Example 1\"\n",
-    " #Oil of specific gravity 0.90 and dynamic viscosity (mu=0.1Pa*s) flows between two fixed plates kept 2*b=10mm apart,So b=5mm.\n",
-    "#The average velocity is Uav=1.60m/s\n",
-    "Uav=1.60;\n",
-    "mu=0.1;\n",
-    "b=0.005;  #in metre\n",
-    " #Umax is maximum velocity\n",
-    " Umax=(3.0/2)*Uav\n",
-    "print\"Umax in m/s is\"\n",
-    "Umax=(3/2)*Uav\n",
-    "print\"Umax=\",Umax\n",
-    " #The shear stress at the plate is given by T=2*µ*(Umax/b)\n",
-    "print\"The shear stress T in N/m**2\"\n",
-    "T=2*mu*(Umax/b) \n",
-    " #The shear sress at a distance from plate is given by t=y*(dp/dx)\n",
-    "#(dp/dx)=X=-3*mu*(Uav/b**2)\n",
-    "print\"T=\",T\n",
-    "print\"(dp/dx) in N/m**3 is\"\n",
-    "X=-3*mu*(Uav/b**2)\n",
-    " #Taking modulus of X by multipying it with negative sign.\n",
-    "print\"X=\",X\n",
-    "print\"The Shear stress at a distance of 0.002m from the lower plate in N/m**2\"\n",
-    "y=b-0.002;\n",
-    "t=y*(X) #NOTE:Answer given in the book is incorrect (Calculation mistake)\n",
-    "print\"t=\",t\n",
-    "print\"The shear stress at a distance of 0.002m from the upper plate in N/m**2\"\n",
-    "t=-y*(X) #NOTE:Answer given in the book is incorrect (Calculation mistake)\n",
-    "print\"t=\",t\n",
-    "print\"The opposite signs in t represents the opposite directions.The plus sign is in the direction of flow and the minus sign is in the direction opposite to the flow \"\n",
-    " #deltaP is the pressure drop\n",
-    "print\"The pressure drop over a distance of 2m in N/m**2 is\"\n",
-    " #Since pressure drop is considered at a distance of 2m so L=2m\n",
-    "L=2;\n",
-    "deltaP=(-X)*L\n",
-    "print\"deltaP=\",deltaP\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Ex6.3:pg-229"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 2,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Introduction to heat transfer by S.K.Som, Chapter 6, Example 3\n",
-      "The rate of change of pressure with respect to length in N/m**3\n",
-      "X= 2000\n",
-      "Flow rate(Q) in m**3/s is)\n",
-      "Q= 0.00333333333333\n",
-      "The viscosity of oil(mu)in kg/(m*s)\n",
-      "mu= 0.0920388472731\n"
-     ]
-    }
-   ],
-   "source": [
-    " \n",
-    "import math\n",
-    " \n",
-    "print\"Introduction to heat transfer by S.K.Som, Chapter 6, Example 3\"\n",
-    " #Oil of specific gravity (sg)=0.90 is discharged at a rate(mdot)=3kg/s under a pressure difference dp=10KN/m**2 over a length dz=5m of a pipe having a diameter(D) of 50mm.\n",
-    "dp=10*10**3; #in N/m**2\n",
-    "dz=5;\n",
-    "D=0.05; #in metre\n",
-    "mdot=3;\n",
-    "sg=0.90;\n",
-    " #X=dp/dz is the rate of change of pressure\n",
-    "print\"The rate of change of pressure with respect to length in N/m**3\"\n",
-    "X=dp/dz\n",
-    "print\"X=\",X\n",
-    " #Flow rate is Q\n",
-    "print\"Flow rate(Q) in m**3/s is)\"\n",
-    "Q=mdot/(sg*10**3)\n",
-    "print\"Q=\",Q\n",
-    " #The viscosity of oil is mu=(pi*D**4*X)/(128*Q*dz)\n",
-    "print\"The viscosity of oil(mu)in kg/(m*s)\"\n",
-    "mu=(math.pi*D**4*X)/(128*Q)\n",
-    "print\"mu=\",mu\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Ex6.7:pg-250   "
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 1,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Introduction to heat transfer by S.K.Som, Chapter 6, Example 7\n",
-      "The maximum length of plate in m is \n",
-      "L= 2.5\n",
-      "The average skin friction coefficient is\n",
-      "cfL= 1.328\n",
-      "Drag force on one side of plate in N is\n",
-      "Fd= 21.5136\n"
-     ]
-    }
-   ],
-   "source": [
-    " \n",
-    "import math\n",
-    " \n",
-    "print\"Introduction to heat transfer by S.K.Som, Chapter 6, Example 7\"\n",
-    " #A flat plate B=1.2m wide and of length L is kept parallel to a uniform stream of air of velocity Uinf=3m/s in a wind tunnel.\n",
-    "Uinf=3;\n",
-    "B=1.2;\n",
-    " #If it is desired to have a laminar boundary layer only on the plate \n",
-    "#Assume that the laminar flow exists up to a reynold number(ReL)=5*10**5.Take density of air as rhoair=1.2kg/m**3 and viscosity of air as nuair=1.5*10**-5 m**2/s.\n",
-    "nuair=1.5*10**-5;\n",
-    "rhoair=1.2;\n",
-    "ReL=5*10**5;\n",
-    " #For maximum length of the plate reynolds number is ReL=Uinf*L/nuair\n",
-    "#so L=ReL*nuair/Uinf\n",
-    "print\"The maximum length of plate in m is \"\n",
-    "L=ReL*nuair/Uinf\n",
-    "print\"L=\",L\n",
-    " #The average skin friction coefficient is cfL=1.328/(ReL)**(1/2)\n",
-    "print\"The average skin friction coefficient is\"\n",
-    "cfL=1.328/(ReL)**(1/2)\n",
-    "print\"cfL=\",cfL\n",
-    " #Fd is drag force\n",
-    "print\"Drag force on one side of plate in N is\"\n",
-    "Fd=cfL*(rhoair*Uinf**2/2)*B*L\n",
-    "print\"Fd=\",Fd\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Ex6.10:pg-268   "
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 2,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Introduction to heat transfer by S.K.Som, Chapter 6, Example 10\n",
-      "Wind velocity(Uinf)in m/s is\n",
-      "Uinf= 10\n",
-      "Reynolds number is\n",
-      "ReL= 4000000.0\n",
-      "Friction coefficient is\n",
-      "CbarfL= 0.0735645\n",
-      "Drag force on one side of the plate per unit metre width in Newton is \n",
-      "FD= 26.48322\n",
-      "The turbulent boundary layer thickness at the trailing edge in metre is \n",
-      "delta= 2.274\n"
-     ]
-    }
-   ],
-   "source": [
-    " \n",
-    "import math\n",
-    " \n",
-    "print\"Introduction to heat transfer by S.K.Som, Chapter 6, Example 10\"\n",
-    " #Wind at a speed of U=36km/hr blows over a flat plate of length,L=6m .If the density and kinematic viscosity of air are rho=1.2kg/m**3 and mu=1.5*10**-5m**2/s respectively.\n",
-    "U=36;\n",
-    "L=6;\n",
-    "rho=1.2;\n",
-    "mu=1.5*10**-5;\n",
-    " #Wind velocity in m/s is Uinf\n",
-    "print\"Wind velocity(Uinf)in m/s is\"\n",
-    "Uinf=U*1000/3600\n",
-    "print\"Uinf=\",Uinf\n",
-    " #Reynolds number is given by ReL=L*Uinf/mu\n",
-    "print\"Reynolds number is\"\n",
-    "ReL=L*Uinf/mu\n",
-    "print\"ReL=\",ReL\n",
-    " #We consider that transition of boundary layer takes place from laminar to turbulent takes place at ReL=5*10**5.\n",
-    "#Therfore the corresponding friction coefficient is given by  CbarfL=(0.074-ReL**(1/5))-(1742/ReL)\n",
-    "print\"Friction coefficient is\"\n",
-    "CbarfL=(0.074/ReL**(1/5))-(1742/ReL)\n",
-    "print\"CbarfL=\",CbarfL\n",
-    " #Drag force on one side of the plate per unit metre width is given by FD=CbarfL*rho*Uinf**2*L/2\n",
-    "print\"Drag force on one side of the plate per unit metre width in Newton is \"\n",
-    "FD=CbarfL*rho*Uinf**2*L/2\n",
-    "print\"FD=\",FD\n",
-    " #The turbulent boundary layer thickness at the trailing edge is given by delta=L*(0.379/ReL**(1/5))\n",
-    "print\"The turbulent boundary layer thickness at the trailing edge in metre is \"\n",
-    "delta=L*(0.379/ReL**(1/5))\n",
-    "print\"delta=\",delta\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\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": "2.7.11"
-  }
- },
- "nbformat": 4,
- "nbformat_minor": 0
-}
diff --git a/Introduction_to_Heat_Transfer_by_S._K._Som/Chapter7.ipynb b/Introduction_to_Heat_Transfer_by_S._K._Som/Chapter7.ipynb
index bffd25f6..85b7eec5 100644
--- a/Introduction_to_Heat_Transfer_by_S._K._Som/Chapter7.ipynb
+++ b/Introduction_to_Heat_Transfer_by_S._K._Som/Chapter7.ipynb
@@ -42,6 +42,7 @@
     }
    ],
    "source": [
+    " \n",
     "import math \n",
     " \n",
     "print\"Introduction to heat transfer by S.K.Som, Chapter 7, Example 1\"\n",
@@ -114,7 +115,11 @@
     }
    ],
    "source": [
-    "import math\n",
+    " \n",
+    " \n",
+    " \n",
+    " \n",
+    " import math\n",
     " \n",
     "print\"Introduction to heat transfer by S.K.Som, Chapter 7, Example 2\"\n",
     "#Atmospheric air at temprature,Tinf=300K and with a free stream Velocity Uinf=30m/s flows over a flat plate parallel to a side of length(L)=2m.\n",
@@ -156,7 +161,24 @@
     "A=L*B;\n",
     "print\"The rate of heat transfer per unit width in W is\"\n",
     "Q=hbarL*A*(Tw-Tinf)\n",
-    "print\"Q=\",Q"
+    "print\"Q=\",Q\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n"
    ]
   },
   {
@@ -196,7 +218,11 @@
     }
    ],
    "source": [
-    "import math\n",
+    " \n",
+    " \n",
+    " \n",
+    " \n",
+    " import math\n",
     " \n",
     "print\"Introduction to heat transfer by S.K.Som, Chapter 7, Example 3\"\n",
     "#Air at a pressure of 101kPa and temprature,Tinf=20°C flows with a velocity(Uinf) of 5m/s over a flat plate whose temprature is kept constant at Tw=140°C.\n",
@@ -246,7 +272,33 @@
     "#Q is the rate of heat transfer\n",
     "print\"The rate of heat transfer per unit width in W is\"\n",
     "Q=h*A*(Tw-Tinf)\n",
-    "print\"Q=\",Q"
+    "print\"Q=\",Q\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n"
    ]
   },
   {
@@ -288,7 +340,11 @@
     }
    ],
    "source": [
-    "import math\n",
+    " \n",
+    " \n",
+    " \n",
+    " \n",
+    " import math\n",
     " \n",
     "print\"Introduction to heat transfer by S.K.Som, Chapter 7, Example 4\"\n",
     "#Castor oil at temprature,Tinf=36°C flows over a heated plate of length,L=6m and breadth,B=1m at velocity,Uinf=0.06m/s\n",
@@ -336,7 +392,29 @@
     "A=L*B;\n",
     "print\"(c)The rate of heat transfer in W is\"\n",
     "Q=hbarL*A*(Tw-Tinf)\n",
-    "print\"Q=\",Q"
+    "print\"Q=\",Q\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n"
    ]
   },
   {
@@ -369,7 +447,11 @@
     }
    ],
    "source": [
-    "import math\n",
+    " \n",
+    " \n",
+    " \n",
+    " \n",
+    " import math\n",
     " \n",
     "print\"Introduction to heat transfer by S.K.Som, Chapter 7, Example 5\"\n",
     "#A flat plate of width B=1m is maintained at a uniform surface temprtaure(Tw)=225°C\n",
@@ -405,7 +487,24 @@
     "#If qm be the power generation in W/m**2 within the module ,we can write from energy balance qm*(t/0.1000)*(l/0.1000)*(B)=hbarL*(t/0.1000)*(B)*(Tw-Tinf)\n",
     "print\"The required power generation in W/m**3 is\"\n",
     "qm=(hL*(l/0.1000)*(B)*(Tw-Tinf))/((t/0.1000)*(l/0.1000)*(B))\n",
-    "print\"qm=\",qm"
+    "print\"qm=\",qm\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n"
    ]
   },
   {
@@ -441,7 +540,11 @@
     }
    ],
    "source": [
-    "import math\n",
+    " \n",
+    " \n",
+    " \n",
+    " \n",
+    " import math\n",
     "\n",
     "print\"Introduction to heat transfer by S.K.Som, Chapter 7, Example 6\"\n",
     "#An aircraft is moving at a velocity of Uinf=150m/s in air at an altitude where the pressure is 0.7bar and the temprature is Tinf=-5°C.\n",
@@ -476,7 +579,21 @@
     "#Therefore we can write Surface temprature of wing, Tw=Tinf+(Qr/(2*hbarL))\n",
     "print\"Surface temprature of wing in kelvin is\"\n",
     "Tw=(273+Tinf)+(Qr/(2*hbarL))\n",
-    "print\"Tw=\",Tw"
+    "print\"Tw=\",Tw\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n"
    ]
   },
   {
@@ -516,7 +633,11 @@
     }
    ],
    "source": [
-    "import math\n",
+    " \n",
+    " \n",
+    " \n",
+    " \n",
+    " import math\n",
     " \n",
     "print\"Introduction to heat transfer by S.K.Som, Chapter 7, Example 7\"\n",
     "#A fine wire having a diameter(D)=0.04mm is placed in an air stream at temprature,Tinf=25°C having a flow velocity of Uinf=60m/s perpendicular to wire.\n",
@@ -558,7 +679,32 @@
     "#Heat transfer per unit length(qL) is given by pi*D*hbar*(Tw-Tinf)\n",
     "print\"Heat transfer per unit length in W/m is\"\n",
     "qL=math.pi*(D*10**-3)*hbar*(Tw-Tinf)\n",
-    "print\"qL=\",qL"
+    "print\"qL=\",qL\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n"
    ]
   },
   {
@@ -596,7 +742,11 @@
     }
    ],
    "source": [
-    "import math\n",
+    " \n",
+    " \n",
+    " \n",
+    " \n",
+    " import math\n",
     "\n",
     "print\"Introduction to heat transfer by S.K.Som, Chapter 7, Example 8\"\n",
     "#Mercury and a light oil flowing at Uinf=4mm/s in a smooth tube having diameter(D)=25mm at a bulk temprature of 80°C.\n",
@@ -633,7 +783,24 @@
     "#Ltoil is the thermal entry length for oil\n",
     "print\"The thermal entry length for oil in m is\"\n",
     "Ltoil=0.05*Reoil*Proil*D\n",
-    "print\"Ltoil=\",Ltoil"
+    "print\"Ltoil=\",Ltoil\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n"
    ]
   },
   {
@@ -673,7 +840,11 @@
     }
    ],
    "source": [
-    "import math\n",
+    " \n",
+    " \n",
+    " \n",
+    " \n",
+    " import math\n",
     " \n",
     "print\"Introduction to heat transfer by S.K.Som, Chapter 7, Example 9\"\n",
     "#Air at one atmospheric pressure and temprature(Tbi=75°C) enters a tube of internal diameter(D)=4.0mm with average velocity(U)=2m/s\n",
@@ -724,7 +895,21 @@
     "#Let Twe be the surface temprature at the exit plane.Then we can write hL*(Twe-Tbo)=qw\n",
     "print\"The tube surface temprature at the exit plane in °C is \"\n",
     "Twe=Tbo+(qw/hL)\n",
-    "print\"Twe=\",Twe"
+    "print\"Twe=\",Twe\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n"
    ]
   },
   {
@@ -767,7 +952,11 @@
     }
    ],
    "source": [
-    "import math\n",
+    " \n",
+    " \n",
+    " \n",
+    " \n",
+    " import math\n",
     " \n",
     "print\"Introduction to heat transfer by S.K.Som, Chapter 7, Example 10\"\n",
     "#Air at one atmospheric pressure and temprature(Tbi=75°C) enters a tube of internal diameter(D)=4.0mm with average velocity(U)=2m/s\n",
@@ -822,7 +1011,18 @@
     "#Let Twe be the surface temprature at the exit plane.Then we can write hL*(Twe-Tbo)=qw\n",
     "print\"The tube surface temprature at the exit plane in °C is \"\n",
     "Twe=Tbo+(qw/hL)\n",
-    "print\"Twe=\",Twe"
+    "print\"Twe=\",Twe\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n"
    ]
   },
   {
@@ -867,7 +1067,11 @@
     }
    ],
    "source": [
-    "import math\n",
+    " \n",
+    " \n",
+    " \n",
+    " \n",
+    " import math\n",
     " \n",
     "print\"Introduction to heat transfer by S.K.Som, Chapter 7, Example 11\"\n",
     "#Liquid sulphur di oxide in a saturated state flows inside a L=5m long tube and D=25mm internal diameter with a mass flow rate(mdot) of 0.15 kg/s.\n",
diff --git a/Introduction_to_Heat_Transfer_by_S._K._Som/Chapter7_Ie2FcUI.ipynb b/Introduction_to_Heat_Transfer_by_S._K._Som/Chapter7_Ie2FcUI.ipynb
deleted file mode 100644
index 85b7eec5..00000000
--- a/Introduction_to_Heat_Transfer_by_S._K._Som/Chapter7_Ie2FcUI.ipynb
+++ /dev/null
@@ -1,1176 +0,0 @@
-{
- "cells": [
-  {
-   "cell_type": "markdown",
-   "metadata": {
-    "collapsed": true
-   },
-   "source": [
-    "# Chapter 07:Principles of forced convection"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Ex7.1:pg-296"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 1,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Introduction to heat transfer by S.K.Som, Chapter 7, Example 1\n",
-      "First we check from reynolds no. that the flow is laminar or tubulent\n",
-      "Reynold number is\n",
-      "Re= 20000.0\n",
-      "which is less than critical reynolds number,So the flow is laminar.\n",
-      "The average nusselt number over the entire length under the situation is given by NuL=0.664*Re**0.5*Pr**(1/3)\n",
-      "NuL= 93.9037805416\n",
-      "Heat flux in W/(m**2*K) is\n",
-      "h= 2.72320963571\n",
-      "The rate of heat transfer per unit width in W is\n",
-      "Q= 408.481445356\n"
-     ]
-    }
-   ],
-   "source": [
-    " \n",
-    "import math \n",
-    " \n",
-    "print\"Introduction to heat transfer by S.K.Som, Chapter 7, Example 1\"\n",
-    "#Engine oil at temprature,Tinf=60°C with a velocity of Uinf=1m/s flows over plate of length(L)=5m whose temprature(Tw)=30°C\n",
-    "Tw=30;\n",
-    "L=5;\n",
-    "Uinf=1;\n",
-    "Tinf=60;\n",
-    "#The properties at a film temprature of 45°C are as follows density(rho=870kg/m**3),Prandtl number(Pr=2850),conductivity(k=0.145W/(m*°C)),kinematic viscosity(nu=250*10**-6m**2/s).\n",
-    "rho=870;\n",
-    "Pr=2850;\n",
-    "k=0.145;\n",
-    "nu=250*10**-6;\n",
-    "print\"First we check from reynolds no. that the flow is laminar or tubulent\"\n",
-    "#Reynolds number is given by Re=(Uinf*L)/nu\n",
-    "print\"Reynold number is\"\n",
-    "Re=(Uinf*L)/nu\n",
-    "print\"Re=\",Re\n",
-    "print\"which is less than critical reynolds number,So the flow is laminar.\"\n",
-    "#NuL is the average nusselt number\n",
-    "print\"The average nusselt number over the entire length under the situation is given by NuL=0.664*Re**0.5*Pr**(1/3)\"\n",
-    "NuL=0.664*Re**0.5*Pr**(1/3)\n",
-    "print\"NuL=\",NuL\n",
-    "#Heat flux is given by h=(k/L)*NuL\n",
-    "print\"Heat flux in W/(m**2*K) is\"\n",
-    "h=(k/L)*NuL\n",
-    "print\"h=\",h\n",
-    "#The rate of heat transfer per unit width is Q=h*A*(Tinf-Tw)\n",
-    "#Since unit width is considerd so B=1\n",
-    "#Area(A)=L*B\n",
-    "B=1;\n",
-    "A=L*B;\n",
-    "print\"The rate of heat transfer per unit width in W is\"\n",
-    "Q=h*A*(Tinf-Tw)\n",
-    "print\"Q=\",Q"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Ex7.2:pg-298"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 2,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Introduction to heat transfer by S.K.Som, Chapter 7, Example 2\n",
-      "The location x in m where the transition occurs\n",
-      "x= 0.275\n",
-      "The average Nusselt number for the laminar zone is\n",
-      "Nux= 469.518902708\n",
-      "Heat flux in W/(m**2*K) is\n",
-      "h= 44.3908780742\n",
-      "The reynolds number at L=2m is\n",
-      "ReL= 3636363.63636\n",
-      "The average heat transfer coefficient over L=2m in W/(m**2*K)\n",
-      "hbarL= -11.322519\n",
-      "The rate of heat transfer per unit width in W is\n",
-      "Q= -2264.5038\n"
-     ]
-    }
-   ],
-   "source": [
-    " \n",
-    " \n",
-    " \n",
-    " \n",
-    " import math\n",
-    " \n",
-    "print\"Introduction to heat transfer by S.K.Som, Chapter 7, Example 2\"\n",
-    "#Atmospheric air at temprature,Tinf=300K and with a free stream Velocity Uinf=30m/s flows over a flat plate parallel to a side of length(L)=2m.\n",
-    "Tinf=300;\n",
-    "Uinf=30;\n",
-    "L=2;\n",
-    "#It is maintained at a uniform temprature of Tw=400K.\n",
-    "Tw=400;\n",
-    "#The properties of air at the film temprature of 350K are Prandtl number(Pr=0.705),conductivity(k=0.026W/(m*°C)),kinematic viscosity(nu=16.5*10**-6m**2/s)\n",
-    "Pr=0.705; \n",
-    "k=0.026;\n",
-    "nu=16.5*10**-6;\n",
-    "#We first find the location x(for reynolds number,Re=5*10**5) where the transition occurs\n",
-    "#Rex is reynolds number\n",
-    "print\"The location x in m where the transition occurs\"\n",
-    "Rex=5*10**5;\n",
-    "x=(nu*Rex)/Uinf\n",
-    "print\"x=\",x\n",
-    "#The average Nusselt number for the laminar zone is given by Nux=0.664*Re**0.5*Pr**(1/3)\n",
-    "print\"The average Nusselt number for the laminar zone is\"\n",
-    "Nux=0.664*Rex**0.5*Pr**(1/3)\n",
-    "print\"Nux=\",Nux\n",
-    "#Heat flux is given by h=(k/x)*Nux\n",
-    "print\"Heat flux in W/(m**2*K) is\"\n",
-    "h=(k/x)*Nux\n",
-    "print\"h=\",h\n",
-    "#Reynolds number is given by ReL=(Uinf*L)/nu\n",
-    "print\"The reynolds number at L=2m is\"\n",
-    "ReL=(Uinf*L)/nu\n",
-    "print\"ReL=\",ReL\n",
-    "#The average heat transfer coefficient over L=2m is determined from hbarL=(k/L)*(0.037*(ReL)**(4/5)-871)*Pr**(1/3)\n",
-    "print\"The average heat transfer coefficient over L=2m in W/(m**2*K)\"\n",
-    "hbarL=(k/L)*(0.037*(ReL)**(4/5)-871)*Pr**(1/3)\n",
-    "print\"hbarL=\",hbarL\n",
-    "#The rate of heat transfer per unit width is Q=h*A*(Tinf-Tw)\n",
-    "#Since unit width is considerd so B=1\n",
-    "#Area(A)=L*B\n",
-    "B=1;\n",
-    "A=L*B;\n",
-    "print\"The rate of heat transfer per unit width in W is\"\n",
-    "Q=hbarL*A*(Tw-Tinf)\n",
-    "print\"Q=\",Q\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Ex7.3:pg-314"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 4,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Introduction to heat transfer by S.K.Som, Chapter 7, Example 3\n",
-      "(a)When the air flows parallel to the long side we have L=5 and the Reynolds no. becomes\n",
-      "ReL= 1250000.0\n",
-      "which is greater than critical Reynolds number.\n",
-      "The average heat transfer coefficient over L=5m in W/(m**2*K)\n",
-      "hbarL= -5.225778\n",
-      "The rate of heat transfer per unit width in W is\n",
-      "Q= -3135.4668\n",
-      "(b)When the air flow is parallel to the 1m side we have L=1 an the Reynolds no. becomes \n",
-      "which is less than critical Reynolds number.\n",
-      "ReL= 250000.0\n",
-      "Heat flux in W/(m**2*K) is\n",
-      "h= 9.96\n",
-      "The rate of heat transfer per unit width in W is\n",
-      "Q= 5976.0\n"
-     ]
-    }
-   ],
-   "source": [
-    " \n",
-    " \n",
-    " \n",
-    " \n",
-    " import math\n",
-    " \n",
-    "print\"Introduction to heat transfer by S.K.Som, Chapter 7, Example 3\"\n",
-    "#Air at a pressure of 101kPa and temprature,Tinf=20°C flows with a velocity(Uinf) of 5m/s over a flat plate whose temprature is kept constant at Tw=140°C.\n",
-    "Tw=140;\n",
-    "Tinf=20;\n",
-    "Uinf=5;\n",
-    "#The properties at the film temprature of 80°C are Prandtl number(Pr=0.706),Conductivity(k=0.03W/(m*°C)),kinematic viscosity(nu=2*10**-5m**2/s)\n",
-    "Pr=0.706;\n",
-    "k=0.03;\n",
-    "nu=2*10**-5;\n",
-    "#ReL is reynolds number and L is length of flat plate\n",
-    "print\"(a)When the air flows parallel to the long side we have L=5 and the Reynolds no. becomes\"\n",
-    "L=5;\n",
-    "ReL=(Uinf*L)/nu\n",
-    "print\"ReL=\",ReL\n",
-    "print\"which is greater than critical Reynolds number.\"\n",
-    "#Thus we have combined laminar and tubulent flow.\n",
-    "# So The average heat transfer coefficient over L=5m is determined from hbarL=(k/L)*(0.037*(ReL)**(4/5)-871)*Pr**(1/3)\n",
-    "print\"The average heat transfer coefficient over L=5m in W/(m**2*K)\"\n",
-    "hbarL=(k/L)*(0.037*(ReL)**(4/5)-871)*Pr**(1/3)\n",
-    "print\"hbarL=\",hbarL\n",
-    "#The rate of heat transfer per unit width is Q=h*A*(Tinf-Tw)\n",
-    "#Since width is 1m so B=1\n",
-    "#Area(A)=L*B\n",
-    "B=1;\n",
-    "A=L*B;\n",
-    "#Q is the rate of heat transfer\n",
-    "print\"The rate of heat transfer per unit width in W is\"\n",
-    "Q=hbarL*A*(Tw-Tinf)\n",
-    "print\"Q=\",Q\n",
-    "#When the air flow is parallel to the 1m side we have L=1\n",
-    "print\"(b)When the air flow is parallel to the 1m side we have L=1 an the Reynolds no. becomes \"\n",
-    "L=1;\n",
-    "ReL=(Uinf*L)/nu\n",
-    "print\"which is less than critical Reynolds number.\"\n",
-    "print\"ReL=\",ReL\n",
-    "#Thus we have laminar flow\n",
-    "#Heat flux is given by h=(k/L)*0.664*ReL**0.5*Pr**(1/3)\n",
-    "print\"Heat flux in W/(m**2*K) is\"\n",
-    "h=(k/L)*0.664*ReL**0.5*Pr**(1/3)\n",
-    "print\"h=\",h\n",
-    "#The rate of heat transfer per unit width is Q=h*A*(Tinf-Tw)\n",
-    "#Now width is 5m so B=5\n",
-    "#Area(A)=L*B\n",
-    "B=5;\n",
-    "A=L*B;\n",
-    "#Q is the rate of heat transfer\n",
-    "print\"The rate of heat transfer per unit width in W is\"\n",
-    "Q=h*A*(Tw-Tinf)\n",
-    "print\"Q=\",Q\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Ex7.4:pg-322"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 5,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Introduction to heat transfer by S.K.Som, Chapter 7, Example 4\n",
-      "(a)Reynolds number is\n",
-      "ReL= 6000.0\n",
-      "The boundary layer thickness in m is\n",
-      "delta= 0.387298334621\n",
-      "Prandtl no. is\n",
-      "Pr= 831.024930748\n",
-      "The thermal boundary layer thickness in m is\n",
-      "deltaT= 0.387298334621\n",
-      "(b)Since the prandtl number is high So Nusselt no. is\n",
-      "NuL= 26.2588270873\n",
-      "Heat flux in W/(m**2*K) is\n",
-      "hL= 0.919058948055\n",
-      "hbarL in W/(m**2*K) is\n",
-      "hbarL= 1.83811789611\n",
-      "(c)The rate of heat transfer in W is\n",
-      "Q= 661.7224426\n"
-     ]
-    }
-   ],
-   "source": [
-    " \n",
-    " \n",
-    " \n",
-    " \n",
-    " import math\n",
-    " \n",
-    "print\"Introduction to heat transfer by S.K.Som, Chapter 7, Example 4\"\n",
-    "#Castor oil at temprature,Tinf=36°C flows over a heated plate of length,L=6m and breadth,B=1m at velocity,Uinf=0.06m/s\n",
-    "Tinf=36;\n",
-    "L=6;\n",
-    "B=1;\n",
-    "Uinf=0.06;\n",
-    "#For a surface temprature at Tw=96°C\n",
-    "Tw=96;\n",
-    "#The properties at film temprature 66°C conductivity(k=0.21W/(m*K)),kinematic viscosity(nu=6*10**-5m**2/s),Thermal diffusivity(alpha=7.22*10**-8 m**2/s)\n",
-    "nu=6*10**-5;\n",
-    "k=0.21;\n",
-    "alpha=7.22*10**-8;\n",
-    "#ReL is reynolds number\n",
-    "print\"(a)Reynolds number is\"\n",
-    "ReL=(Uinf*L)/nu\n",
-    "print\"ReL=\",ReL\n",
-    "#Therefore the boundary layer is laminar over the entire plate.\n",
-    "#delta is the boundary layer thickness\n",
-    "print\"The boundary layer thickness in m is\"\n",
-    "delta=(5*L)/(ReL)**0.5\n",
-    "print\"delta=\",delta\n",
-    "#Pr is prandtl number.\n",
-    "print\"Prandtl no. is\"\n",
-    "Pr=nu/alpha\n",
-    "print\"Pr=\",Pr\n",
-    "#deltaT is thermal boundary layer thickness\n",
-    "print\"The thermal boundary layer thickness in m is\"\n",
-    "deltaT=delta/(Pr**(1/3))#NOTE:Answer in the book is incorrect(calculation mistake)\n",
-    "print\"deltaT=\",deltaT\n",
-    "#NuL is the nusselt number\n",
-    "print\"(b)Since the prandtl number is high So Nusselt no. is\"\n",
-    "NuL=0.339*(ReL)**0.5*Pr**(1/3)\n",
-    "print\"NuL=\",NuL\n",
-    "#Heat flux is given by hL=(k/L)*NuL\n",
-    "print\"Heat flux in W/(m**2*K) is\"\n",
-    "hL=(k/L)*NuL\n",
-    "print\"hL=\",hL\n",
-    "#hbarL is the average heat flux over length L\n",
-    "print\"hbarL in W/(m**2*K) is\"\n",
-    "hbarL=2*hL\n",
-    "print\"hbarL=\",hbarL\n",
-    "#The rate of heat transfer is Q=h*A*(Tinf-Tw)\n",
-    "#Area(A)=L*B\n",
-    "A=L*B;\n",
-    "print\"(c)The rate of heat transfer in W is\"\n",
-    "Q=hbarL*A*(Tw-Tinf)\n",
-    "print\"Q=\",Q\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Ex7.5:pg-322"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 2,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Introduction to heat transfer by S.K.Som, Chapter 7, Example 5\n",
-      "Reynolds number is\n",
-      "ReL= 11181.8181818\n",
-      "Therefore the flow is turbulent over the module \n",
-      "The local heat transfer coefficient at L in W/(m**2*K)is\n",
-      "hL= 8.32911955901e+30\n",
-      "The required power generation in W/m**3 is\n",
-      "qm= 1.6658239118e+31\n"
-     ]
-    }
-   ],
-   "source": [
-    " \n",
-    " \n",
-    " \n",
-    " \n",
-    " import math\n",
-    " \n",
-    "print\"Introduction to heat transfer by S.K.Som, Chapter 7, Example 5\"\n",
-    "#A flat plate of width B=1m is maintained at a uniform surface temprtaure(Tw)=225°C\n",
-    "Tw=225;\n",
-    "B=1;\n",
-    "#Heating is done by rectangular modules of thickness t=10mm and length l=40mm.\n",
-    "t=10;\n",
-    "l=40;\n",
-    "#atmospheric air at temprature,Tinf=25°C flows over the plate at velocity(Uinf)=30m/s.\n",
-    "Tinf=25;\n",
-    "Uinf=30;\n",
-    "#The thermophysical properties of module are conductivity(km=5.2W/(m*K)),specific heat(cp=320J/(kg/K)),density(rho=2300kg/m**3).\n",
-    "km=5.2;\n",
-    "cp=320;\n",
-    "rho=2300;\n",
-    "#Assume the air properties at the film temprature of 125°C conductivity(ka=0.031W/(m*K)),kinematic viscosity(nu=22*10**-6m**2/s),Prandtl number(Pr=0.7)\n",
-    "ka=0.031;\n",
-    "nu=22*10**-6;\n",
-    "Pr=0.7;\n",
-    "#Module is placed at a distance of 800mm from the leading edge\n",
-    "#The distance from leading edge to the centre-line of the module,L=800+20=820mm.\n",
-    "L=0.0082;#in metre\n",
-    "#ReL is the reynolds number \n",
-    "print\"Reynolds number is\"\n",
-    "ReL=(Uinf*L)/nu\n",
-    "print\"ReL=\",ReL\n",
-    "print\"Therefore the flow is turbulent over the module \"\n",
-    "#The local heat transfer coefficient at L is calculated using hL=(k/L)*0.0296*(ReL)**(4/5)*(Pr)**(1/3)\n",
-    "print\"The local heat transfer coefficient at L in W/(m**2*K)is\"\n",
-    "hL=(ka/L)*0.0296*(ReL)**(4/0.5)*(Pr)**(1/0.3)\n",
-    "print\"hL=\",hL\n",
-    "#We consider that the local heat transfer coefficient at L=0.82m remains the same over the module which extends from L=0.80m to 0.84m \n",
-    "#If qm be the power generation in W/m**2 within the module ,we can write from energy balance qm*(t/0.1000)*(l/0.1000)*(B)=hbarL*(t/0.1000)*(B)*(Tw-Tinf)\n",
-    "print\"The required power generation in W/m**3 is\"\n",
-    "qm=(hL*(l/0.1000)*(B)*(Tw-Tinf))/((t/0.1000)*(l/0.1000)*(B))\n",
-    "print\"qm=\",qm\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Ex7.6:pg-327"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 6,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Introduction to heat transfer by S.K.Som, Chapter 7, Example 6\n",
-      "Reynolds number is\n",
-      "ReL= 15000000.0\n",
-      "Since ReL>Rec(=5*10**5) the flow is approximated as turbulent over the entire surface of the wing \n",
-      "Nux= 0.0308\n",
-      "Nusselt number is \n",
-      "NubarL= 0.0308\n",
-      "Average heat transfer coefficient in W/(m**2*K) is\n",
-      "hbarL= 0.0003696\n",
-      "Surface temprature of wing in kelvin is\n",
-      "Tw= 1217800.46753\n"
-     ]
-    }
-   ],
-   "source": [
-    " \n",
-    " \n",
-    " \n",
-    " \n",
-    " import math\n",
-    "\n",
-    "print\"Introduction to heat transfer by S.K.Som, Chapter 7, Example 6\"\n",
-    "#An aircraft is moving at a velocity of Uinf=150m/s in air at an altitude where the pressure is 0.7bar and the temprature is Tinf=-5°C.\n",
-    "Tinf=-5;\n",
-    "Uinf=150;\n",
-    "#The top surface of the wing absorbs solar radiation at a rate of Qr=900W/m**2.\n",
-    "Qr=900;\n",
-    "#Considering the wing as a flat plate of length(L)=2m and to be of solid construction with a single uniform surface temprature .\n",
-    "L=2;\n",
-    "#The properties of air at 268K and 0.7 bar are conductivity(k=0.024W/(m*K)),kinematic viscosity(nu=2*10**-5m**2/s),Prandtl number(Pr=0.72)\n",
-    "k=0.024;\n",
-    "nu=2*10**-5;\n",
-    "Pr=0.72;\n",
-    "#ReL is reynolds number\n",
-    "print\"Reynolds number is\"\n",
-    "ReL=Uinf*L/nu\n",
-    "print\"ReL=\",ReL\n",
-    "#Rec is critical reynolds number\n",
-    "print\"Since ReL>Rec(=5*10**5) the flow is approximated as turbulent over the entire surface of the wing \"\n",
-    "#Nusselt number is given by Nux=0.0308*ReL**(4/5)*Pr**(1/3)\n",
-    "Nux=0.0308*ReL**(4/5)*Pr**(1/3);\n",
-    "print\"Nux=\",Nux\n",
-    "#NubarL is average nusselt number over length L\n",
-    "print\"Nusselt number is \"\n",
-    "NubarL=(5/4)*Nux\n",
-    "print\"NubarL=\",NubarL\n",
-    "#Average heat transfer coefficient is given by hbarL=(k/L)*NubarL\n",
-    "print\"Average heat transfer coefficient in W/(m**2*K) is\"\n",
-    "hbarL=(k/L)*NubarL\n",
-    "print\"hbarL=\",hbarL\n",
-    "#From an energy balance the airfoil at steady state,Qr*As=2*hbarL*As*(Tw-Tinf) where Qr=radiation flux,As=upper or lower surface area.\n",
-    "#Therefore we can write Surface temprature of wing, Tw=Tinf+(Qr/(2*hbarL))\n",
-    "print\"Surface temprature of wing in kelvin is\"\n",
-    "Tw=(273+Tinf)+(Qr/(2*hbarL))\n",
-    "print\"Tw=\",Tw\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Ex7.7:pg-331"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 7,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Introduction to heat transfer by S.K.Som, Chapter 7, Example 7\n",
-      "Reynolds number is\n",
-      "Re= 141.176470588\n",
-      "Nusselt number is\n",
-      "NuD= 6.85819682626\n",
-      "The average Heat transfer coefficient in W/(m**2*K) is\n",
-      "hbar= 4629.28285773\n",
-      "Heat transfer per unit length in W/m is\n",
-      "qL= 14.5433210172\n",
-      "If we use eq NuD=0.3+((0.62*Re**0.5*Pr**(1/3))/(1+(0.4/Pr**(2/3))**(1/4))*(1+(Re/282000)**(5/8))**(4/5)\n",
-      "NuD= 7.66669771975\n",
-      "The average Heat transfer coefficient in W/(m**2*K) is\n",
-      "hbar= 5175.02096083\n",
-      "Heat transfer per unit length in W/m is\n",
-      "qL= 16.2578078327\n"
-     ]
-    }
-   ],
-   "source": [
-    " \n",
-    " \n",
-    " \n",
-    " \n",
-    " import math\n",
-    " \n",
-    "print\"Introduction to heat transfer by S.K.Som, Chapter 7, Example 7\"\n",
-    "#A fine wire having a diameter(D)=0.04mm is placed in an air stream at temprature,Tinf=25°C having a flow velocity of Uinf=60m/s perpendicular to wire.\n",
-    "D=0.04;\n",
-    "Tinf=25;\n",
-    "Uinf=60;\n",
-    "#An electric current is passed through the wire ,raising its surface temprature to Tw=50°C\n",
-    "Tw=50;\n",
-    "#For air at the film temprature of 37.5°C,conductivity(k=0.027 W/(m*K)),kinematic viscosity(nu=17*10**-6m**2/s) and Prandtl number(Pr=0.71)\n",
-    "k=0.027;\n",
-    "nu=17*10**-6;\n",
-    "Pr=0.71;\n",
-    "#Re is reynolds number\n",
-    "print\"Reynolds number is\"\n",
-    "Re=Uinf*(D*10**-3)/nu\n",
-    "print\"Re=\",Re\n",
-    "#C and n are constants\n",
-    "#The values of C and n are found  for Re=141 are C=0.683 and n=0.466\n",
-    "#NuD is nusselt number\n",
-    "print\"Nusselt number is\"\n",
-    "NuD=(0.683)*Re**0.466*Pr**(1/3)\n",
-    "print\"NuD=\",NuD\n",
-    "#hbar is the average Heat transfer coefficient\n",
-    "print\"The average Heat transfer coefficient in W/(m**2*K) is\"\n",
-    "hbar=(k/(D*10**-3))*NuD\n",
-    "print\"hbar=\",hbar\n",
-    "#Heat transfer per unit length(qL) is given by pi*D*hbar*(Tw-Tinf)\n",
-    "print\"Heat transfer per unit length in W/m is\"\n",
-    "qL=math.pi*(D*10**-3)*hbar*(Tw-Tinf)\n",
-    "print\"qL=\",qL\n",
-    "#NuD is nusselt number\n",
-    "print\"If we use eq NuD=0.3+((0.62*Re**0.5*Pr**(1/3))/(1+(0.4/Pr**(2/3))**(1/4))*(1+(Re/282000)**(5/8))**(4/5)\"\n",
-    "NuD=0.3+((0.62*Re**0.5*Pr**(1/3))/(1+(0.4/Pr)**(2/3))**(1/4))*(1+(Re/282000)**(5/8))**(4/5)\n",
-    "print\"NuD=\",NuD\n",
-    "#hbar is the average Heat transfer coefficient\n",
-    "print\"The average Heat transfer coefficient in W/(m**2*K) is\"\n",
-    "hbar=(k/(D*10**-3))*NuD\n",
-    "print\"hbar=\",hbar\n",
-    "#Heat transfer per unit length(qL) is given by pi*D*hbar*(Tw-Tinf)\n",
-    "print\"Heat transfer per unit length in W/m is\"\n",
-    "qL=math.pi*(D*10**-3)*hbar*(Tw-Tinf)\n",
-    "print\"qL=\",qL\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Ex7.8:pg-334"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 8,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Introduction to heat transfer by S.K.Som, Chapter 7, Example 8\n",
-      "Reynolds number for mercury is\n",
-      "ReHg= 1000.0\n",
-      "Reynolds number for oil is\n",
-      "Reoil= 15.3846153846\n",
-      "The hydrodynamic entry length for mercury in m is\n",
-      "LeHg= 1.25\n",
-      "The hydrodynamic entry length for oil in m is\n",
-      "Leoil= 0.0192307692308\n",
-      "The thermal entry length for mercury in m is \n",
-      "LtHg= 0.02375\n",
-      "The thermal entry length for oil in m is\n",
-      "Ltoil= 1.63461538462\n"
-     ]
-    }
-   ],
-   "source": [
-    " \n",
-    " \n",
-    " \n",
-    " \n",
-    " import math\n",
-    "\n",
-    "print\"Introduction to heat transfer by S.K.Som, Chapter 7, Example 8\"\n",
-    "#Mercury and a light oil flowing at Uinf=4mm/s in a smooth tube having diameter(D)=25mm at a bulk temprature of 80°C.\n",
-    "Uinf=4*10**-3;#in metre\n",
-    "D=25*10**-3;#in metre\n",
-    "#The pertinent properties of the fluid at that temprature are kinematic viscosity of mercury(nuHg=1*10**-7m**2/s),kinematic viscosity of oil(nuoil=6.5*10**-6m**2/s)\n",
-    "#Prandtl number of mercury(PrHg=0.019),Prandtl number of oil(Proil=85).\n",
-    "nuHg=1*10**-7;\n",
-    "nuoil=6.5*10**-6;\n",
-    "PrHg=0.019;\n",
-    "Proil=85;\n",
-    "#ReHg is Reynolds number for mercury\n",
-    "print\"Reynolds number for mercury is\"\n",
-    "ReHg=Uinf*D/nuHg\n",
-    "print\"ReHg=\",ReHg\n",
-    "#Reoil is Reynolds number for oil\n",
-    "print\"Reynolds number for oil is\"\n",
-    "Reoil=Uinf*D/nuoil\n",
-    "print\"Reoil=\",Reoil\n",
-    "#The hydrodynamic length are given by L=0.05*Re*D\n",
-    "#LeHg is the hydrodynamic entry length for mercury\n",
-    "print\"The hydrodynamic entry length for mercury in m is\"\n",
-    "LeHg=0.05*ReHg*D\n",
-    "print\"LeHg=\",LeHg\n",
-    "#Leoil the hydrodynamic entry length for oil\n",
-    "print\"The hydrodynamic entry length for oil in m is\"\n",
-    "Leoil=0.05*Reoil*D\n",
-    "print\"Leoil=\",Leoil\n",
-    "#The thermal entry length are given by L=0.05*Re*Pr*D\n",
-    "#LtHg is the thermal entry length for mercury\n",
-    "print\"The thermal entry length for mercury in m is \"\n",
-    "LtHg=0.05*ReHg*PrHg*D\n",
-    "print\"LtHg=\",LtHg\n",
-    "#Ltoil is the thermal entry length for oil\n",
-    "print\"The thermal entry length for oil in m is\"\n",
-    "Ltoil=0.05*Reoil*Proil*D\n",
-    "print\"Ltoil=\",Ltoil\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Ex7.9:pg-336"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 9,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Introduction to heat transfer by S.K.Som, Chapter 7, Example 9\n",
-      "Reynold number is\n",
-      "Re= 348.623853211\n",
-      "Therefore the flow is laminar.The hydrodynamic entrance length in m is\n",
-      "Leh= 0.0697247706422\n",
-      "The thermal entrance length in m is\n",
-      "Let= 0.0488073394495\n",
-      "The heat transfer coefficient in W/(m**2*K) is \n",
-      "h= 32.7\n",
-      "The mass flow rate of air in kg/s is\n",
-      "mdot= 2.38761041673e-05\n",
-      "Therefore the constant surface heat flux qw in W/m**2 is\n",
-      "qw= 95.95\n",
-      "The tube surface temprature at the exit plane in °C is \n",
-      "Twe= 127.934250765\n"
-     ]
-    }
-   ],
-   "source": [
-    " \n",
-    " \n",
-    " \n",
-    " \n",
-    " import math\n",
-    " \n",
-    "print\"Introduction to heat transfer by S.K.Som, Chapter 7, Example 9\"\n",
-    "#Air at one atmospheric pressure and temprature(Tbi=75°C) enters a tube of internal diameter(D)=4.0mm with average velocity(U)=2m/s\n",
-    "Tbi=75;\n",
-    "D=4*10**-3;#in metre\n",
-    "U=2;\n",
-    "#The tube length is L=1.0m and a constant heat flux is imposed by the tube surface on the air over the entire length.\n",
-    "L=1;\n",
-    "#An exit bulk mean temprature(Tbo)=125°C is required.\n",
-    "Tbo=125;\n",
-    "#The properties of air 100°C are density(rho=0.95kg/m**3),Prandtl number(Pr=0.70),conductivity(k=0.03W/(m*K)),viscosity(mu=2.18*10**-5kg/(m*s)),specific heat(cp=1.01kJ/(kg/K))\n",
-    "rho=0.95;\n",
-    "Pr=0.70;\n",
-    "k=0.03;\n",
-    "mu=2.18*10**-5;\n",
-    "cp=1.01*10**3;\n",
-    "#Re is reynolds number\n",
-    "print\"Reynold number is\"\n",
-    "Re=rho*U*D/mu\n",
-    "print\"Re=\",Re\n",
-    "#Leh is the hydrodynamic entrance length\n",
-    "print\"Therefore the flow is laminar.The hydrodynamic entrance length in m is\"\n",
-    "Leh=0.05*Re*D\n",
-    "print\"Leh=\",Leh\n",
-    "#Let is the thermal entrance length\n",
-    "print\"The thermal entrance length in m is\"\n",
-    "Let=0.05*Re*Pr*D\n",
-    "print\"Let=\",Let\n",
-    "#The length of tube is given as 1m.A reasonable approach is to consider the flow to be fully developed for both velocity and tempratures over the entire profile lengths.\n",
-    "#For a fully developed flow with constant surface heat flux,Nusselt number is Nu=4.36\n",
-    "Nu=4.36;\n",
-    "#h is the heat transfer coefficient\n",
-    "print\"The heat transfer coefficient in W/(m**2*K) is \"\n",
-    "h=Nu*(k/D)\n",
-    "print\"h=\",h\n",
-    "#Here h=hL Since the heat transfer coefficient is constant over the entire length of tube.\n",
-    "#hL is the local heat transfer coefficient\n",
-    "hL=h;\n",
-    "#from an energy balance qw*pi*D*L=mdot*cp*(Tbo-Tbi)\n",
-    "#mdot is mass flow rate\n",
-    "print\"The mass flow rate of air in kg/s is\"\n",
-    "mdot=rho*(math.pi/4)*D**2*U\n",
-    "print\"mdot=\",mdot\n",
-    "#qw is the constant surface heat flux\n",
-    "print\"Therefore the constant surface heat flux qw in W/m**2 is\"\n",
-    "qw=(mdot*cp*(Tbo-Tbi))/(math.pi*D*L)\n",
-    "print\"qw=\",qw\n",
-    "#Let Twe be the surface temprature at the exit plane.Then we can write hL*(Twe-Tbo)=qw\n",
-    "print\"The tube surface temprature at the exit plane in °C is \"\n",
-    "Twe=Tbo+(qw/hL)\n",
-    "print\"Twe=\",Twe\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Ex7.10:pg-338"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 10,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Introduction to heat transfer by S.K.Som, Chapter 7, Example 10\n",
-      "Reynold number is\n",
-      "Re= 348.623853211\n",
-      "Therefore the flow is laminar.The hydrodynamic entrance length in m is\n",
-      "Leh= 0.0697247706422\n",
-      "The thermal entrance length in m is\n",
-      "Let= 0.0488073394495\n",
-      "The thermal entrance length is greater than the tube length Therefore the flow is hydrodynamically developed but not thermally developed\n",
-      "The inverse of graetz number Gr_1 is\n",
-      "Gr_1= 0.040977443609\n",
-      "Therefore the local heat transfer coefficient in W/(m**2*K) is\n",
-      "hL= 35.25\n",
-      "The mass flow rate of air in kg/s is\n",
-      "mdot= 2.38761041673e-05\n",
-      "Therefore surafce heat flux qw in W/m**2 is\n",
-      "qw= 2398.75\n",
-      "The tube surface temprature at the exit plane in °C is \n",
-      "Twe= 193.04964539\n"
-     ]
-    }
-   ],
-   "source": [
-    " \n",
-    " \n",
-    " \n",
-    " \n",
-    " import math\n",
-    " \n",
-    "print\"Introduction to heat transfer by S.K.Som, Chapter 7, Example 10\"\n",
-    "#Air at one atmospheric pressure and temprature(Tbi=75°C) enters a tube of internal diameter(D)=4.0mm with average velocity(U)=2m/s\n",
-    "Tbi=75;\n",
-    "D=4*10**-3;\n",
-    "U=2;\n",
-    "#The heated tube length is L=0.04m and a constant heat flux is imposed by the tube surface on the air over the entire length.\n",
-    "L=0.04;\n",
-    "#An exit bulk mean temprature(Tbo)=125°C is required.\n",
-    "Tbo=125;\n",
-    "#The properties of air 100°C are density(rho=0.95kg/m**3),Prandtl number(Pr=0.70),conductivity(k=0.03W/(m*K)),viscosity(mu=2.18*10**-5kg/(m*s)),specific heat(cp=1.01kJ/(kg/K))\n",
-    "rho=0.95;\n",
-    "Pr=0.70;\n",
-    "k=0.03;\n",
-    "mu=2.18*10**-5;\n",
-    "cp=1.01*10**3;\n",
-    "#Re is the reynolds number \n",
-    "print\"Reynold number is\"\n",
-    "Re=rho*U*D/mu\n",
-    "print\"Re=\",Re\n",
-    "#Leh is the hydrodynamic entrance length\n",
-    "print\"Therefore the flow is laminar.The hydrodynamic entrance length in m is\"\n",
-    "Leh=0.05*Re*D\n",
-    "print\"Leh=\",Leh\n",
-    "#Let is thermal entrance length\n",
-    "print\"The thermal entrance length in m is\"\n",
-    "Let=0.05*Re*Pr*D\n",
-    "print\"Let=\",Let\n",
-    "print\"The thermal entrance length is greater than the tube length Therefore the flow is hydrodynamically developed but not thermally developed\" \n",
-    "#We calculate the inverse graetz number at x=L=0.04m\n",
-    "x=0.04;\n",
-    "#Gr_1 is inverse of graetz number\n",
-    "print\"The inverse of graetz number Gr_1 is\"\n",
-    "Gr_1=(x/D)*(1/(Re*Pr))\n",
-    "print\"Gr_1=\",Gr_1\n",
-    "#For constant surface heat flux nusselt number is Nu=4.7 and Graetz number is Gr=4.1*10**-2\n",
-    "Nu=4.7;\n",
-    "Gr=4.1*10**-2;\n",
-    "#hL is the local heat transfer coefficient\n",
-    "print\"Therefore the local heat transfer coefficient in W/(m**2*K) is\"\n",
-    "hL=Nu*(k/D)\n",
-    "print\"hL=\",hL\n",
-    "#from an energy balance qw*pi*D*L=mdot*cp*(Tbo-Tbi)\n",
-    "#mdot is the mass flow rate\n",
-    "print\"The mass flow rate of air in kg/s is\"\n",
-    "mdot=rho*(math.pi/4)*D**2*U\n",
-    "print\"mdot=\",mdot\n",
-    "#qw is the surface heat flux\n",
-    "print\"Therefore surafce heat flux qw in W/m**2 is\"\n",
-    "qw=(mdot*cp*(Tbo-Tbi))/(math.pi*D*L)\n",
-    "print\"qw=\",qw\n",
-    "#Let Twe be the surface temprature at the exit plane.Then we can write hL*(Twe-Tbo)=qw\n",
-    "print\"The tube surface temprature at the exit plane in °C is \"\n",
-    "Twe=Tbo+(qw/hL)\n",
-    "print\"Twe=\",Twe\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Ex7.11:pg-339"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 6,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      " Introduction to heat transfer by S.K.Som, Chapter 7, Example 11\n",
-      "Reynold number is\n",
-      "Re1= 13541.3214942\n",
-      "Nusselt number is\n",
-      "Nubar1= 56.808608087\n",
-      "The heat transfer transfer coefficient in W/(m**2*°C) \n",
-      "hbar1= 522.6391944\n",
-      "Outlet fluid temprature in first iteration is Tbo2 in °C is\n",
-      "Tb2 in °C is\n",
-      "Tb2= -30.4912413164\n",
-      "Reynold number is\n",
-      "Re2= 13938.8493187\n",
-      "Nusselt number is\n",
-      "The heat transfer transfer coefficient in W/(m**2*°C) \n",
-      "hbar2= 784.03829067\n",
-      "Outlet fluid temprature in second iteration is Tbo3 in °C is\n",
-      "Tbo3= -16.646852652\n",
-      "Tb3 in °C is\n",
-      "The Exit fluid temprature after second iteration is obtained as Tbo=-16.67°C\n",
-      "Tb3= -28.323426326\n"
-     ]
-    }
-   ],
-   "source": [
-    " \n",
-    " \n",
-    " \n",
-    " \n",
-    " import math\n",
-    " \n",
-    "print\"Introduction to heat transfer by S.K.Som, Chapter 7, Example 11\"\n",
-    "#Liquid sulphur di oxide in a saturated state flows inside a L=5m long tube and D=25mm internal diameter with a mass flow rate(mdot) of 0.15 kg/s.\n",
-    "#The tube is heated at a constant surface temprature(Tw) of -10°C and the inlet fluid temprature is Tbi=-40°C\n",
-    "Tw=-10;\n",
-    "Tbi=-40;\n",
-    "mdot=0.15;\n",
-    "D=0.025;#in metre\n",
-    "L=5;\n",
-    "#The properties to be used shoud be estimated at a temprature which is arithmetic mean of Tbi and Tbo.\n",
-    "#Since (outlet fluid temprature Tbo) is not known a priori,the solution has to be based on an iterative method starting with a guess value of Tb1=(Tbi+Tbo)/2\n",
-    "#Here we denote bulk mean temprature as Tb.The superscript refers to the no. of trials\n",
-    "#For first trial,guess Tbo1=-20°C;so Tb1=-30°C\n",
-    "#We have the property values as follows at a temprature of -30°C.\n",
-    "rhob1=1520.64;#density in kg/m**3\n",
-    "nub1=0.371*10**-6;#kinematic viscosity in m**2/s\n",
-    "kb1=0.23;#conductivity in W/(m*°C)\n",
-    "Prb1=3.31;#Prandtl number\n",
-    "mub1=nub1*rhob1;#viscosity in kg/(m*s)\n",
-    "cpb1=1361.6;#specific heat in J/(kg*K)\n",
-    "#muw=nuw*rhow at Tw=10°C\n",
-    "nuw=0.288*10**-6;#kinematic viscosity at Tw in m**2/s\n",
-    "rhow=1463.61;#density at Tw in kg/m**3\n",
-    "muw=nuw*rhow;#viscosity at Tw in kg/(m*s)\n",
-    "#The reynolds number is found as Re1=(4*mdot)/(math.pi*D*mub1)\n",
-    "print\"Reynold number is\"\n",
-    "Re1=(4*mdot)/(math.pi*D*mub1)\n",
-    "print\"Re1=\",Re1\n",
-    "#Hence the flow is turbulent\n",
-    "#Now using equation, nusselt number is,Nubar1=0.027*(Re1)**0.8*Prb1**(1/3)*(mub1/muw)**0.14\n",
-    "print\"Nusselt number is\"\n",
-    "Nubar1=0.027*(Re1)**0.8*Prb1**(1/3)*(mub1/muw)**0.14\n",
-    "print\"Nubar1=\",Nubar1\n",
-    "#The heat transfer transfer coefficient hbar1=(kb1/D)*Nubar1\n",
-    "print\"The heat transfer transfer coefficient in W/(m**2*°C) \"\n",
-    "hbar1=(kb1/D)*Nubar1\n",
-    "print\"hbar1=\",hbar1\n",
-    "#The outlet fluid temprature can be found by making use of eqn Tbo2=Tw-(Tw-Tbi)*math.e((-math.pi*D*L*hbar1)/(mdot*cpb1))\n",
-    "print\"Outlet fluid temprature in first iteration is Tbo2 in °C is\"\n",
-    "Tbo2=Tw-(Tw-Tbi)*math.e**((-math.pi*D*L*hbar1)/(mdot*cpb1))\n",
-    "#Tb2 is the bulk mean temprature.\n",
-    "print\"Tb2 in °C is\"\n",
-    "Tb2=(Tbi+Tbo2)/2\n",
-    "print\"Tb2=\",Tb2\n",
-    "#Since the value differs from the assumed value of Tb1=-30°C,WE require furtheriteration,Therfore we start second trial with Tb2=-28.36°C\n",
-    "#We have the property value at a temprature of -28.36°C as follows\n",
-    "rhob2=1514;#density in kg/m**3\n",
-    "nub2=0.362*10**-6;#kinematic viscosity in m**2/s\n",
-    "kb2=0.229;#conductivity in W/(m*°C)\n",
-    "Prb2=3.23;#Prandtl number\n",
-    "mub2=nub2*rhob2;#viscosity in kg/(m*s)\n",
-    "cpb2=1362;#specific heat in J/(kg*K)\n",
-    "#muw=nuw*rhow at Tw=10°C\n",
-    "nuw=0.288*10**-6;#viscosity at Tw in m**2/s\n",
-    "rhow=1463.61;#density at Tw in kg/m**3\n",
-    "muw=nuw*rhow;#kinematic viscosity at Tw in kg/(m*s)\n",
-    "#The reynolds number is found as Re2=(4*mdot)/(math.pi*D*mub2)\n",
-    "print\"Reynold number is\"\n",
-    "Re2=(4*mdot)/(math.pi*D*mub2)\n",
-    "print\"Re2=\",Re2\n",
-    "#Now using equation, nusselt number is,Nubar2=0.027*(Re2)**0.8*Prb2**(1/3.0)*(mub2/muw)**0.14\n",
-    "print\"Nusselt number is\"\n",
-    "Nubar2=0.027*(Re2)**0.8*Prb2**(1/3.0)*(mub2/muw)**0.14\n",
-    "#The heat transfer transfer coefficient hbar2=(kb2/D)*Nubar2\n",
-    "print\"The heat transfer transfer coefficient in W/(m**2*°C) \"\n",
-    "hbar2=(kb2/D)*Nubar2\n",
-    "print\"hbar2=\",hbar2\n",
-    "#The outlet fluid temprature can be found by making use of eqn Tbo3=Tw-(Tw-Tbi)*math.e((-math.pi*D*L*hbar2)/(mdot*cpb2))\n",
-    "print\"Outlet fluid temprature in second iteration is Tbo3 in °C is\"\n",
-    "Tbo3=Tw-(Tw-Tbi)*math.e**((-math.pi*D*L*hbar2)/(mdot*cpb2))\n",
-    "print\"Tbo3=\",Tbo3\n",
-    "#Tb3 is the bulk mean temprature.\n",
-    "print\"Tb3 in °C is\"\n",
-    "Tb3=(Tbi+Tbo3)/2\n",
-    "#We see that difference between Tbo2 and Tbo3 and that between Tb2 and Tb3 is marginal.Therfore we can stop iteration and present the result as Tbo=-16.67°C\n",
-    "print\"The Exit fluid temprature after second iteration is obtained as Tbo=-16.67°C\"\n",
-    "print\"Tb3=\",Tb3"
-   ]
-  }
- ],
- "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": "2.7.11"
-  }
- },
- "nbformat": 4,
- "nbformat_minor": 0
-}
diff --git a/Introduction_to_Heat_Transfer_by_S._K._Som/Chapter8.ipynb b/Introduction_to_Heat_Transfer_by_S._K._Som/Chapter8.ipynb
index 882c8bf9..a6c237e4 100644
--- a/Introduction_to_Heat_Transfer_by_S._K._Som/Chapter8.ipynb
+++ b/Introduction_to_Heat_Transfer_by_S._K._Som/Chapter8.ipynb
@@ -79,7 +79,20 @@
     "#The rate of heat transfer is given by q=hbarL*A*(Tw-Tinf)\n",
     "print\"The rate of heat transfer in W is\"\n",
     "q=hbarL*A*(Tw-Tinf)\n",
-    "print\"q=\",q"
+    "print\"q=\",q\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n"
    ]
   },
   {
@@ -115,6 +128,7 @@
     }
    ],
    "source": [
+    " \n",
     "import math \n",
     " \n",
     "print\"Introduction to heat transfer by S.K.Som, Chapter 8, Example 2\"\n",
@@ -151,7 +165,24 @@
     "#spac is the minimum spacing \n",
     "print\"The minimum spacing in metre is\"\n",
     "spac=2*delta\n",
-    "print\"spac=\",spac"
+    "print\"spac=\",spac\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n"
    ]
   },
   {
@@ -188,6 +219,7 @@
     }
    ],
    "source": [
+    "\n",
     "from scipy.integrate import quad\n",
     "print \"Introduction to heat transfer by S.K.Som, Chapter 8, Example 3\"\n",
     "#Considering question 5.7\n",
@@ -243,7 +275,14 @@
     "print \"Mass flow rate at x=0.8m,in kG is\"\n",
     "I=quad(lambda y:465.9*(y-116*y*2+3341*y*3),0,delta)\n",
     "mdot=rho*B*I[0]\n",
-    "print\"mdot=\",mdot"
+    "print\"mdot=\",mdot\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n"
    ]
   },
   {
@@ -275,6 +314,7 @@
     }
    ],
    "source": [
+    " \n",
     "import math \n",
     " \n",
     "print\"Introduction to heat transfer by S.K.Som, Chapter 8, Example 4\"\n",
@@ -312,7 +352,33 @@
     "hL=(2*k)/delta;\n",
     "hbarL=(4.0/3)*(hL)#NOTE:The answer in the book is incorrect(calculation mistake)\n",
     "print\"hL=\",hL\n",
-    "print\"hbarL=\",hbarL"
+    "print\"hbarL=\",hbarL\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n"
    ]
   },
   {
@@ -366,6 +432,7 @@
     }
    ],
    "source": [
+    " \n",
     "import math \n",
     " \n",
     "print\"Introduction to heat transfer by S.K.Som, Chapter 8, Example 5\"\n",
@@ -442,7 +509,51 @@
     "print hbarL\n",
     "print\"The rate of heat transfer in W is \"\n",
     "Q=hbarL*A*(Tw-Tinf)\n",
-    "print Q"
+    "print Q\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n"
    ]
   },
   {
@@ -488,6 +599,10 @@
     }
    ],
    "source": [
+    " \n",
+    " \n",
+    " \n",
+    " \n",
     "import math \n",
     " \n",
     "print\"Introduction to heat transfer by S.K.Som, Chapter 8, Example 6\"\n",
@@ -548,7 +663,16 @@
     "#The current flowing in the wire I=(q/(R*L)**(1/2.0)\n",
     "print\"The current flowing in the wire in Ampere is\"\n",
     "I=(q/(R*L))**(1/2.0)\n",
-    "print\"I=\",I"
+    "print\"I=\",I\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n"
    ]
   },
   {
@@ -585,6 +709,7 @@
     }
    ],
    "source": [
+    " \n",
     "import math \n",
     " \n",
     "print\"Introduction to heat transfer by S.K.Som, Chapter 8, Example 7\"\n",
@@ -625,7 +750,26 @@
     "#The heat loss per meter length is given by q=hbar*A*(Tw-Tinf)\n",
     "print\"The heat loss per meter length in W is\"\n",
     "q=hbar*A*(Tw-Tinf)\n",
-    "print\"q=\",q"
+    "print\"q=\",q\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n"
    ]
   },
   {
@@ -664,6 +808,7 @@
     }
    ],
    "source": [
+    " \n",
     "import math \n",
     " \n",
     "print\"Introduction to heat transfer by S.K.Som, Chapter 8, Example 8\"\n",
@@ -709,7 +854,18 @@
     "print\"Hence,steady state Surface temprature in °C is\"\n",
     "Tw=Tinf+(P/(hbarD*math.pi*D*L))\n",
     "print\"Hence we see that our guess is in excellent agreement with the calculated value\"\n",
-    "print\"Tw=\",Tw"
+    "print\"Tw=\",Tw\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n"
    ]
   }
  ],
diff --git a/Introduction_to_Heat_Transfer_by_S._K._Som/Chapter8_vyeGLD8.ipynb b/Introduction_to_Heat_Transfer_by_S._K._Som/Chapter8_vyeGLD8.ipynb
deleted file mode 100644
index a6c237e4..00000000
--- a/Introduction_to_Heat_Transfer_by_S._K._Som/Chapter8_vyeGLD8.ipynb
+++ /dev/null
@@ -1,893 +0,0 @@
-{
- "cells": [
-  {
-   "cell_type": "markdown",
-   "metadata": {
-    "collapsed": true
-   },
-   "source": [
-    "# Chapter 08:Principles of free convection"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Ex8.1:pg-355"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 1,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Introduction to heat transfer by S.K.Som, Chapter 8, Example 1\n",
-      "Grashoff number is\n",
-      "GrL= 2175146201.53\n",
-      "Rayleigh number is\n",
-      "RaL= 9440134514.65\n",
-      "Therefore the flow is turbulent\n",
-      "Now we use [(hbarL*L)/k]=0.10*(GrL*Pr)**(1/3)\n",
-      "The average heat transfer coefficient in W/(m**2*K) is\n",
-      "hbarL= 0.314\n",
-      "The rate of heat transfer in W is\n",
-      "q= 0.5024\n"
-     ]
-    }
-   ],
-   "source": [
-    "import math \n",
-    " \n",
-    "print\"Introduction to heat transfer by S.K.Som, Chapter 8, Example 1\"\n",
-    "#Water is heated by a vertical flat plate  length(L=200mm or .2m )by breadth(B=200mm) which is maintained at temprature,Tw=60°C\n",
-    "Tw=60;\n",
-    "L=.2;\n",
-    "B=.2;# in metre\n",
-    "#Area(A) is L*B \n",
-    "A=L*B;\n",
-    "#Water is at temprature,Tinf=20°C\n",
-    "Tinf=20;\n",
-    "#At mean film temprature 40°C The physical properties parameters can be taken as \n",
-    "#conductivity(k=0.0628W/(m*K)),Prandtl number(Pr=4.34),density(rho=994.59kg/m**3),kinematic viscosity(nu=0.658*10**-6m**2/s),volume expnasion coefficient(Beta=3*10**-4K**-1))\n",
-    "k=0.628;\n",
-    "Pr=4.34;\n",
-    "rho=994.59;\n",
-    "nu=0.658*10**-6;\n",
-    "Beta=3*10**-4;\n",
-    "#g is acceleration due to gravity =9.81m/s**2\n",
-    "g=9.81;\n",
-    "#Grashoff number is given by GrL=(g*beta*(Tw-Tinf)*L**3)/(nu)**2\n",
-    "print\"Grashoff number is\"\n",
-    "GrL=(g*Beta*(Tw-Tinf)*L**3)/(nu)**2\n",
-    "print\"GrL=\",GrL\n",
-    "#Rayleigh number is defined as RaL=GrL*Pr\n",
-    "print\"Rayleigh number is\"\n",
-    "RaL=GrL*Pr\n",
-    "print\"RaL=\",RaL\n",
-    "print\"Therefore the flow is turbulent\"\n",
-    "print\"Now we use [(hbarL*L)/k]=0.10*(GrL*Pr)**(1/3)\"\n",
-    "#hbarL is the average heat transfer coefficient\n",
-    "print\"The average heat transfer coefficient in W/(m**2*K) is\"\n",
-    "hbarL=(0.10*(GrL*Pr)**(1/3)*k)/L\n",
-    "print\"hbarL=\",hbarL\n",
-    "#The rate of heat transfer is given by q=hbarL*A*(Tw-Tinf)\n",
-    "print\"The rate of heat transfer in W is\"\n",
-    "q=hbarL*A*(Tw-Tinf)\n",
-    "print\"q=\",q\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Ex8.2:pg-357"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 2,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Introduction to heat transfer by S.K.Som, Chapter 8, Example 2\n",
-      "The minimum spacing between the plates will be twice the thickness of the boundary layer at the trailing edge where x=0.09\n",
-      "Grashoff number is\n",
-      "GrL= 198210197.615\n",
-      "Rayleigh number is\n",
-      "RaL= 860232257.647\n",
-      "Since Ra<10**9,Therefore the flow is laminar\n",
-      "The thickness of the boundary layer in metre is\n",
-      "delta= 4.11168026839e-10\n",
-      "The minimum spacing in metre is\n",
-      "spac= 8.22336053678e-10\n"
-     ]
-    }
-   ],
-   "source": [
-    " \n",
-    "import math \n",
-    " \n",
-    "print\"Introduction to heat transfer by S.K.Som, Chapter 8, Example 2\"\n",
-    "#The thin plates are kept at temprature(Tw)=60°C while the temprature of water bath(Tinf)=20°C\n",
-    "Tw=60;\n",
-    "Tinf=20;\n",
-    "#The plates have length(L)=90mm or .09m\n",
-    "L=.09;\n",
-    "#The minimum spacing between the plates will be twice the thickness of the boundary layer at the trailing edge where x=0.09.\n",
-    "print\"The minimum spacing between the plates will be twice the thickness of the boundary layer at the trailing edge where x=0.09\"\n",
-    "x=.09;\n",
-    "#At mean film temprature 40°C The physical properties parameters can be taken as\n",
-    "# conducivity(k=0.0628W/(m*K)),Prandtl number(Pr=4.34),Density(rho=994.59kg/m**3),kinematic viscosity(nu=0.658*10**-6m**2/s),Volume expansion coefficient(Beta=3*10**-4K**-1)\n",
-    "k=0.628;\n",
-    "Pr=4.34;\n",
-    "rho=994.59;\n",
-    "nu=0.658*10**-6;\n",
-    "Beta=3*10**-4;\n",
-    "#g is acceleration due to gravity =9.81m/s**2\n",
-    "g=9.81;\n",
-    "#Grashoff number is given by GrL=(g*beta*(Tw-Tinf)*L**3)/(nu)**2\n",
-    "print\"Grashoff number is\"\n",
-    "GrL=(g*Beta*(Tw-Tinf)*L**3)/(nu)**2\n",
-    "print\"GrL=\",GrL\n",
-    "#Rayleigh number is defined as RaL=GrL*Pr\n",
-    "print\"Rayleigh number is\"\n",
-    "RaL=GrL*Pr\n",
-    "print\"RaL=\",RaL\n",
-    "print\"Since Ra<10**9,Therefore the flow is laminar\"\n",
-    "#delta is the thickness of the boundary layer\n",
-    "print\"The thickness of the boundary layer in metre is\"\n",
-    "delta=x*3.93*Pr**(-1/2)*(0.952+Pr)**(1/4)*GrL**(-1/4)\n",
-    "print\"delta=\",delta\n",
-    "#spac is the minimum spacing \n",
-    "print\"The minimum spacing in metre is\"\n",
-    "spac=2*delta\n",
-    "print\"spac=\",spac\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Ex8.3:pg-366"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 3,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Introduction to heat transfer by S.K.Som, Chapter 8, Example 3\n",
-      "Grashoff number is\n",
-      "Grx= 517025.52213\n",
-      "The boundary layer thickness in metre is\n",
-      "delta= 60304.3038858\n",
-      "The velocity at point x is ux in m/s is\n",
-      "ux= 3247798354.51\n",
-      "For maximum value of velocity,u\n",
-      "Maximum velocity in m/s is\n",
-      "Umax= 3.57089835848\n",
-      "Mass flow rate at x=0.8m,in kG is\n",
-      "mdot= 2.36830073295e+16\n"
-     ]
-    }
-   ],
-   "source": [
-    "\n",
-    "from scipy.integrate import quad\n",
-    "print \"Introduction to heat transfer by S.K.Som, Chapter 8, Example 3\"\n",
-    "#Considering question 5.7\n",
-    "#A wall is exposed to nitrogen at one atmospheric pressure and temprature,Tinf=4°C.\n",
-    "Tinf=4.0;\n",
-    "#The wall is H=2.0m high and 2.5m wide and is maintained at temprature,Tw=56°C\n",
-    "Tw=56.0;\n",
-    "H=2.0;\n",
-    "B=2.5;\n",
-    "A=H*B;#Area of wall in m**2\n",
-    "#The properties of nitrogen at mean film temprature (56+4)/2=30°C are given as \n",
-    "#density(rho=1.142kg/m*3) ,conductivity(k=0.026W/(m*K)),kinematic viscosity(nu=15.630*10-6 m*2/s) ,prandtl number(Pr=0.713)\n",
-    "rho=1.142;\n",
-    "k=0.026;\n",
-    "nu=15.630*10**-6;\n",
-    "Pr=0.713;\n",
-    "Tf=30.0;#mean film temprature\n",
-    "Beta=1/(273.0+Tf);#volume expansion coefficient:unit K**-1\n",
-    "#Now Grashoff number is Grx=(g*Beta*(Tw-Tinf)*x*3)/nu*2\n",
-    "g=9.81;#acceleration due to gravity\n",
-    "print \"Grashoff number is\"\n",
-    "x=0.8;#distance from the bottom of wall\n",
-    "Grx=(g*Beta*(Tw-Tinf)*x*3)/nu*2\n",
-    "print\"Grx=\",Grx\n",
-    "#Using equation delta=x*Pr*(-0.5)(0.952+Pr)*(0.25)*Grx*(-0.25)\n",
-    "#delta is the boundary layer thickness\n",
-    "print \"The boundary layer thickness in metre is\"\n",
-    "delta=x*3.93*Pr*(-0.5)*(0.952+Pr)*(0.25)*Grx*(-0.25)\n",
-    "print\"delta=\",delta\n",
-    "#Now using equation ux=(g*Beta*delta*2(Tw-Tinf))/(4*nu)\n",
-    "#ux is the velocity at point x\n",
-    "print \"The velocity at point x is ux in m/s is\"\n",
-    "ux=(g*Beta*delta*2*(Tw-Tinf))/(4*nu)\n",
-    "print\"ux=\",ux\n",
-    "# (u/ux)=(y/delta)*(1-y/delta)**2\n",
-    "#Putting value of ux we get velovity function,u=465.9*(y-116*y*2+3341*y*3)\n",
-    "#For maximum value of u,du/dy=465.9*(1-232*y+10023*y**2)=0...this is a quadratic equation in which coefficients a=10023,b=232,c=1\n",
-    "a=10023;\n",
-    "b=232;\n",
-    "c=1;\n",
-    "#Solution for quadratic equation is given by y=(-b+-(b*2-4ac)*0.5)/2*a\n",
-    "print \"For maximum value of velocity,u\"\n",
-    "y=(b+(b*2-4*a*c)*0.5)/(2*a)#root of the quadratic equation\n",
-    "y=(b-(b*2-4*a*c)*0.5)/(2*a)#root of the quadratic equation\n",
-    "#The value of 0.0173 is at the edge of boundary layer,where u=0\n",
-    "#Therefore the maximum value occurs at y=0.00573m i.e Umax=465.9*y*(1-57.8*y)**2\n",
-    "y=0.00573;\n",
-    "#Umax is maximum velocity\n",
-    "print \"Maximum velocity in m/s is\"\n",
-    "Umax=465.9*y*(1-57.8*y)*2#NOTE:The answer given in the book is incorrect,in this expresssion they considered square on y only,however it is on whole expression (1-57.8*y)*2\n",
-    "#mdot is mass flow rate\n",
-    "print\"Umax=\",Umax\n",
-    "print \"Mass flow rate at x=0.8m,in kG is\"\n",
-    "I=quad(lambda y:465.9*(y-116*y*2+3341*y*3),0,delta)\n",
-    "mdot=rho*B*I[0]\n",
-    "print\"mdot=\",mdot\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Ex8.4:pg-369"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 11,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Introduction to heat transfer by S.K.Som, Chapter 8, Example 4\n",
-      "Grashoff number is\n",
-      "Rayleigh number is\n",
-      "Hence,the flow is laminar\n",
-      "The thickness of the boundary layer in metre is\n",
-      "The average heat transfer coeficient in W/(m**2*K) is\n",
-      "0.101781170483\n"
-     ]
-    }
-   ],
-   "source": [
-    " \n",
-    "import math \n",
-    " \n",
-    "print\"Introduction to heat transfer by S.K.Som, Chapter 8, Example 4\"\n",
-    "#A square plate length,L=0.2m by breadth,B=0.2m is suspended vertically in a quiescent atmospheric air at a temprature(Tinf)=300K\n",
-    "L=0.2;\n",
-    "B=0.2;\n",
-    "Tinf=300;\n",
-    "#The Temprature of plate(Tw) is maintained at 400K\n",
-    "Tw=400;\n",
-    "#The required property value of air at a film temprature(Tf)=350K,kinematic viscosity (nu=20.75*10**-6),Prandtl number(Pr=0.69),conductivity(k=0.03W/(m*K))\n",
-    "Tf=350;\n",
-    "nu=20.75*10**-6;\n",
-    "Pr=0.69;\n",
-    "k=0.03;\n",
-    "#volume expansion coefficient is Beta\n",
-    "Beta=(1/Tf);\n",
-    "#g is acceleration due to gravity =9.81m/s**2\n",
-    "g=9.81;\n",
-    "#Grashoff number is given by GrL=(g*beta*(Tw-Tinf)*L**3)/(nu)**2\n",
-    "print\"Grashoff number is\"\n",
-    "GrL=(g*Beta*(Tw-Tinf)*L**3)/(nu)**2 \n",
-    "print\"GrL=\",GrL\n",
-    "#Rayleigh number is defined as RaL=GrL*Pr\n",
-    "print\"Rayleigh number is\"\n",
-    "RaL=GrL*Pr\n",
-    "print\"Hence,the flow is laminar\"\n",
-    "print\"RaL=\",RaL\n",
-    "#delta is the thickness of the boundary layer\n",
-    "print\"The thickness of the boundary layer in metre is\"\n",
-    "x=0.2;#location of trailing edge of plate\n",
-    "delta=(x*3.93*(0.952+Pr)**(1/4))/(Pr**(1/2)*(GrL)**(1/4))#NOTE:The answer in the book is incorrect(calculation mistake)\n",
-    "print\"delta=\",delta\n",
-    "#hL and hbarL are local and average heat transfer coefficient respectively\n",
-    "print\"The average heat transfer coeficient in W/(m**2*K) is\"\n",
-    "hL=(2*k)/delta;\n",
-    "hbarL=(4.0/3)*(hL)#NOTE:The answer in the book is incorrect(calculation mistake)\n",
-    "print\"hL=\",hL\n",
-    "print\"hbarL=\",hbarL\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Ex8.5:pg-373"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 4,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Introduction to heat transfer by S.K.Som, Chapter 8, Example 5\n",
-      "Grashoff number is\n",
-      "503958851.066\n",
-      "Rayleigh number is\n",
-      "357810784.257\n",
-      "Therefore the flow is laminar\n",
-      "Nusselt number is\n",
-      "75.3134665126\n",
-      "Average heat transfer coefficient(hbarL)in W/(m**2*°C)\n",
-      "The rate of heat transfer in W is \n",
-      "Now if we use NuL2=0.59*RaL**(1/4) with the value of C=0.59,n=(1/4)\n",
-      "Nusselt number is\n",
-      "Average heat transfer coefficient(hbarL)in W/(m**2*°C)\n",
-      "The rate of heat transfer in W is \n",
-      "49.9857347505\n",
-      "(b)For the horizontal plate facing up\n",
-      "Now RaL2=Gr*Pr*(Lc/L)**3\n",
-      "Rayleigh number is\n",
-      "Nusselt number is given by NuL3=C*(GrL*Pr)**n\n",
-      "Average heat transfer coefficient(hbarL)in W/(m**2*°C)\n",
-      "The rate of heat transfer in W is \n",
-      "64.6997833306\n",
-      "(c)When the hot surface faces is down\n",
-      "Nusselt number is given by NuL4=0.27*RaL2**(1/4)\n",
-      "13.1290144745\n",
-      "Average heat transfer coefficient(hbarL) in W/(m**2)\n",
-      "2.9408992423\n",
-      "The rate of heat transfer in W is \n",
-      "32.3498916653\n"
-     ]
-    }
-   ],
-   "source": [
-    " \n",
-    "import math \n",
-    " \n",
-    "print\"Introduction to heat transfer by S.K.Som, Chapter 8, Example 5\"\n",
-    "#A square plate of length(L)=0.5m by breadth,B=0.5m in a room at temprature,Tinf=30°C\n",
-    "#One side of plate is kept a uniform temprature(Tw)=74°C\n",
-    "Tw=74;\n",
-    "L=0.5;\n",
-    "B=0.5;\n",
-    "Tinf=30.0;\n",
-    "#The required properties at the film temprature(Tf)=52°C are kinematic viscosity(nu=1.815*10**-5),Prandtl number(Pr=0.71),conductivity(k=0.028W/(m*°C))\n",
-    "Tf=52.0;\n",
-    "Pr=0.71;\n",
-    "nu=1.815*10**-5;\n",
-    "k=0.028;\n",
-    "#Area(A)=L*B m**2\n",
-    "A=L*B;\n",
-    "#Volume expansion coefficient is Beta\n",
-    "Beta=1/(273+Tf);\n",
-    "#g is acceleration due to gravity =9.81m/s**2\n",
-    "g=9.81;\n",
-    "#Grashoff number is given by GrL=(g*beta*(Tw-Tinf)*L**3)/(nu)**2\n",
-    "print\"Grashoff number is\"\n",
-    "GrL=(g*Beta*(Tw-Tinf)*L**3)/(nu)**2 \n",
-    "print GrL\n",
-    "#Rayleigh number is defined as RaL1=GrL*Pr\n",
-    "print\"Rayleigh number is\"\n",
-    "RaL1=GrL*Pr\n",
-    "print RaL1\n",
-    "print\"Therefore the flow is laminar\"\n",
-    "#We make use of following equation to find Nusselt number,NuL1=(4/3)*(0.508*Pr**(-1/2)*(0.952+Pr)**(-1/4)*Gr**(1/4))\n",
-    "print\"Nusselt number is\"\n",
-    "NuL1=(4.0/3)*(0.508*Pr**(1.0/2)*(0.952+Pr)**(-1.0/4)*GrL**(1.0/4))\n",
-    "#Average heat transfer coefficient(hbarL) is given by (NuL*k)/L\n",
-    "print NuL1\n",
-    "print\"Average heat transfer coefficient(hbarL)in W/(m**2*°C)\"\n",
-    "hbarL=(NuL1*k)/L\n",
-    "#The rate of heat transfer(Q) from the plate by free convection is given by Q=hbarL*A*(Tw-Tinf)\n",
-    "print\"The rate of heat transfer in W is \"\n",
-    "Q=hbarL*A*(Tw-Tinf)\n",
-    "print\"Now if we use NuL2=0.59*RaL**(1/4) with the value of C=0.59,n=(1/4)\"\n",
-    "print\"Nusselt number is\"\n",
-    "NuL2=0.59*RaL1**(1.0/4)\n",
-    "#Average heat transfer coefficient(hbarL) is given by (NuL*k)/L\n",
-    "print\"Average heat transfer coefficient(hbarL)in W/(m**2*°C)\"\n",
-    "hbarL=(NuL2*k)/L\n",
-    "#The rate of heat transfer(Q) from the plate by free convection is given by Q=hbarL*A*(Tw-Tinf)\n",
-    "print\"The rate of heat transfer in W is \"\n",
-    "Q=hbarL*A*(Tw-Tinf)\n",
-    "print Q\n",
-    "print\"(b)For the horizontal plate facing up\"\n",
-    "#Perimeter(P) for a square plate is P=4*L\n",
-    "P=4*L;\n",
-    "#Characterstic length(Lc)=A/P\n",
-    "Lc=A/P\n",
-    "print\"Now RaL2=Gr*Pr*(Lc/L)**3\"\n",
-    "print\"Rayleigh number is\"\n",
-    "RaL2=GrL*Pr*(Lc/L)**3\n",
-    "#The values of constants,C=0.54 and n=(1/4)\n",
-    "C=0.54;\n",
-    "n=(1.0/4);\n",
-    "print\"Nusselt number is given by NuL3=C*(GrL*Pr)**n\"\n",
-    "NuL3=C*(RaL2)**n\n",
-    "print\"Average heat transfer coefficient(hbarL)in W/(m**2*°C)\"\n",
-    "hbarL=(NuL3*k)/Lc\n",
-    "print\"The rate of heat transfer in W is \"\n",
-    "Q=hbarL*A*(Tw-Tinf)\n",
-    "print Q\n",
-    "print\"(c)When the hot surface faces is down\"\n",
-    "print\"Nusselt number is given by NuL4=0.27*RaL2**(1/4)\"\n",
-    "NuL4=0.27*RaL2**(1.0/4)\n",
-    "print NuL4\n",
-    "print\"Average heat transfer coefficient(hbarL) in W/(m**2)\"\n",
-    "hbarL=(NuL4*k)/Lc\n",
-    "print hbarL\n",
-    "print\"The rate of heat transfer in W is \"\n",
-    "Q=hbarL*A*(Tw-Tinf)\n",
-    "print Q\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Ex8.6:pg-375"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 5,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Introduction to heat transfer by S.K.Som, Chapter 8, Example 6\n",
-      "Grashoff number is\n",
-      "GrL= 813719594.384\n",
-      "Rayleigh number is\n",
-      "RaL= 569603716.069\n",
-      "Therefore the flow is laminar\n",
-      "Now we use NuL=0.59*RaL**(1/4.0) with the value of constants C=0.59,n=(1/4.0)\n",
-      "Nusselt number is\n",
-      "NuL= 91.1475952489\n",
-      "Average heat transfer coefficient in W/(m**2*K)\n",
-      "hbarL1= 5.46885571493\n",
-      "Grashoff number GrD=GrL*(D/L)**3\n",
-      "GrD= 0.00650975675508\n",
-      "The correction factor is\n",
-      "F= 39.485281111\n",
-      "The correct value of Average heat transfer coefficient(hbarL2)=hbarL1*F in W/(m**2*K) is\n",
-      "hbarL2= 215.939305259\n",
-      "The ohmic loss in W is \n",
-      "q= 3.39196667512\n",
-      "The current flowing in the wire in Ampere is\n",
-      "I= 7.51882822777\n"
-     ]
-    }
-   ],
-   "source": [
-    " \n",
-    " \n",
-    " \n",
-    " \n",
-    "import math \n",
-    " \n",
-    "print\"Introduction to heat transfer by S.K.Som, Chapter 8, Example 6\"\n",
-    "#A vertical wire of length(L)=0.5m and  Dimeter(D)=0.1mm is maintained at temprature, Tw=400K\n",
-    "#The temprature of quicsent air is Tinf=300K\n",
-    "#Resistance(R) per  meter length is 0.12ohm\n",
-    "R=0.12;\n",
-    "Tw=400.0;\n",
-    "L=0.5;\n",
-    "D=0.1*10**-3;#in metre\n",
-    "Tinf=300;\n",
-    "#The required properties at the film temprature(Tf)=350K are kinematic viscosity(nu=20.75*10**-6m**2/s),Prandtl number(Pr=0.70),conductivity(k=0.03W/(m*°C))\n",
-    "Tf=350.0;\n",
-    "Pr=0.70;\n",
-    "nu=20.75*10**-6;\n",
-    "k=0.03;\n",
-    "#Area(A)=L*B m**2\n",
-    "A=math.pi*D*L;\n",
-    "#Volume expansion Coefficient is Beta\n",
-    "Beta=1/(Tf);\n",
-    "#g is acceleration due to gravity =9.81m/s**2\n",
-    "g=9.81;\n",
-    "#Grashoff number is given by GrL=(g*beta*(Tw-Tinf)*L**3)/(nu)**2\n",
-    "print\"Grashoff number is\"\n",
-    "GrL=(g*Beta*(Tw-Tinf)*L**3)/(nu)**2 \n",
-    "print\"GrL=\",GrL\n",
-    "#Rayleigh number is defined as RaL=GrL*Pr\n",
-    "print\"Rayleigh number is\"\n",
-    "RaL=GrL*Pr\n",
-    "print\"RaL=\",RaL\n",
-    "print\"Therefore the flow is laminar\"\n",
-    "#NuL is nusselt number\n",
-    "#C and n are constants\n",
-    "print\"Now we use NuL=0.59*RaL**(1/4.0) with the value of constants C=0.59,n=(1/4.0)\"\n",
-    "print\"Nusselt number is\"\n",
-    "NuL=0.59*RaL**(1/4.0)\n",
-    "print\"NuL=\",NuL\n",
-    "#hbarL1 is the Average heat transfer coefficient\n",
-    "print\"Average heat transfer coefficient in W/(m**2*K)\"\n",
-    "hbarL1=(NuL*k)/L\n",
-    "print\"hbarL1=\",hbarL1\n",
-    "#Grashoff number GrD=GrL*(D/L)**3\n",
-    "print\"Grashoff number GrD=GrL*(D/L)**3\"\n",
-    "GrD=GrL*(D/L)**3\n",
-    "print\"GrD=\",GrD\n",
-    "#The correction factor is given By F=1.3*((L/D)/GrD)**(1/4.0)+1.0\n",
-    "print\"The correction factor is\"\n",
-    "F=1.3*((L/D)/GrD)**(1/4.0)+1.0\n",
-    "print\"F=\",F\n",
-    "print\"The correct value of Average heat transfer coefficient(hbarL2)=hbarL1*F in W/(m**2*K) is\"\n",
-    "hbarL2=hbarL1*F\n",
-    "print\"hbarL2=\",hbarL2\n",
-    "#The ohmic power loss is given by energy balance I**2*R=q=hbar2*A*(Tw-Tinf)\n",
-    "#q is the ohmic power loss\n",
-    "print\"The ohmic loss in W is \"\n",
-    "q=hbarL2*A*(Tw-Tinf)\n",
-    "print\"q=\",q\n",
-    "#The current flowing in the wire I=(q/(R*L)**(1/2.0)\n",
-    "print\"The current flowing in the wire in Ampere is\"\n",
-    "I=(q/(R*L))**(1/2.0)\n",
-    "print\"I=\",I\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Ex8.7:pg-378"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 6,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Introduction to heat transfer by S.K.Som, Chapter 8, Example 7\n",
-      "Grashoff number is\n",
-      "GrD= 53311595.6796\n",
-      "Rayleigh number is\n",
-      "RaD= 37318116.9757\n",
-      "The flow is laminar over the entire cylinder\n",
-      "we use following equation to find Nusselt number NuD=(0.60+((0.387*RaD**(1/6))/(1+(0.559/Pr**(9/16)))**(8/27)))**2\n",
-      "NuD= 0.974169\n",
-      "Average heat transfer coefficient in W/(m**2*K)\n",
-      "hbar= 0.14612535\n",
-      "The heat loss per meter length in W is\n",
-      "q= 9.64039284733\n"
-     ]
-    }
-   ],
-   "source": [
-    " \n",
-    "import math \n",
-    " \n",
-    "print\"Introduction to heat transfer by S.K.Som, Chapter 8, Example 7\"\n",
-    "#A long horizontal pressurized hot water of diameter(D)=200mm passes through a room where the air temprature is Tinf=25°C\n",
-    "D=.2;\n",
-    "Tinf=25;\n",
-    "#Length(L)=1m ,since the unit length is considered\n",
-    "L=1;\n",
-    "#Area(A)=pi*L*D\n",
-    "A=math.pi*L*D;\n",
-    "#The pipe surface temprature is Tw=130°C\n",
-    "Tw=130;\n",
-    "#The properties of air at the film temprature Tf=77.5°C are kinematic viscosity(nu=21*10**-6m**2/s),Prandtl number(Pr=0.70),Conductivity(k=0.03W/(m*K))\n",
-    "Tf=77.5;\n",
-    "nu=21*10**-6;\n",
-    "k=0.03;\n",
-    "Beta=(1/(273+Tf));#Volume expansion coefficient in k**-1)\n",
-    "Pr=0.70;\n",
-    "#g is acceleration due to gravity =9.81m/s**2\n",
-    "g=9.81;\n",
-    "#Grashoff number is given by GrD=(g*beta*(Tw-Tinf)*L**3)/(nu)**2\n",
-    "print\"Grashoff number is\"\n",
-    "GrD=(g*Beta*(Tw-Tinf)*D**3)/(nu)**2 \n",
-    "print\"GrD=\",GrD\n",
-    "#Rayleigh number is defined as RaD=GrD*Pr\n",
-    "print\"Rayleigh number is\"\n",
-    "RaD=GrD*Pr\n",
-    "print\"RaD=\",RaD\n",
-    "print\"The flow is laminar over the entire cylinder\"\n",
-    "#NuD is the nusselt number\n",
-    "print\"we use following equation to find Nusselt number NuD=(0.60+((0.387*RaD**(1/6))/(1+(0.559/Pr**(9/16)))**(8/27)))**2\"\n",
-    "NuD=(0.60+((0.387*RaD**(1/6))/(1+(0.559/Pr**(9/16)))**(8/27)))**2\n",
-    "print\"NuD=\",NuD\n",
-    "#hbar is the avearge heat transfer coefficient\n",
-    "print\"Average heat transfer coefficient in W/(m**2*K)\"\n",
-    "hbar=(NuD*k)/D\n",
-    "print\"hbar=\",hbar\n",
-    "#The heat loss per meter length is given by q=hbar*A*(Tw-Tinf)\n",
-    "print\"The heat loss per meter length in W is\"\n",
-    "q=hbar*A*(Tw-Tinf)\n",
-    "print\"q=\",q\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Ex8.8:pg-381"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 7,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Introduction to heat transfer by S.K.Som, Chapter 8, Example 8\n",
-      "Let us take first trial Tw=64°C\n",
-      "Grashoff number is\n",
-      "GrD= 226303.67232\n",
-      "Rayleigh number is\n",
-      "The flow is laminar \n",
-      "RaD= 941423.276851\n",
-      "we use following equation to find Nusselt number NuD=(0.60+((0.387*RaD**(1/6))/(1+(0.559/Pr**(9/16)))**(8/27)))**2\n",
-      "NuD= 0.974169\n",
-      "Average heat transfer coefficient in W/(m**2*K)\n",
-      "hbarD= 77.20289325\n",
-      "Hence,steady state Surface temprature in °C is\n",
-      "Hence we see that our guess is in excellent agreement with the calculated value\n",
-      "Tw= 793.068225127\n"
-     ]
-    }
-   ],
-   "source": [
-    " \n",
-    "import math \n",
-    " \n",
-    "print\"Introduction to heat transfer by S.K.Som, Chapter 8, Example 8\"\n",
-    "#An electric immersion heater diameter(D)=8mm and length(L)=300mm is rated at power input,P=450W\n",
-    "P=450;\n",
-    "L=0.3;#in metre\n",
-    "D=0.008;#in metre\n",
-    "#If the heater is horizontally positioned in a large tank of stationery water at temprature,Tinf=20°C\n",
-    "Tinf=20;\n",
-    "#At steady state ,The electrical power input(P)=(Q)Heat loss from the heater\n",
-    "#P=Q\n",
-    "#Q=hbarD*(pi*D)*L*(Tw-Tinf)\n",
-    "#This gives Tw(surface temprature)=Tinf+(P/(hbarD*pi*D*L))\n",
-    "#So we need to find Average heat transfer coefficient,hbarD.\n",
-    "#In this problem we need to take guess of steady state surface temprature(Tw) and iterate the solution for Tw till a desired convergence is achieved.\n",
-    "print\"Let us take first trial Tw=64°C\"\n",
-    "Tw=64;\n",
-    "Tf=(Tw+Tinf)/2;#mean film temprature\n",
-    "#At this temprature of 42°C,The required properties of water kinematic viscosity(nu=6.25*10**-7m**2/s),Prandtl number(Pr=4.16),Conductivity(k=0.634W/(m*K)),Beta=4*10**-4K**-1\n",
-    "Beta=4*10**-4;#Volume expansion coefficient\n",
-    "nu=6.25*10**-7;\n",
-    "Pr=4.16;\n",
-    "k=0.634;\n",
-    "#g is acceleration due to gravity =9.81m/s**2\n",
-    "g=9.81;\n",
-    "#Grashoff number is given by GrD=(g*beta*(Tw-Tinf)*L**3)/(nu)**2\n",
-    "print\"Grashoff number is\"\n",
-    "GrD=(g*Beta*(Tw-Tinf)*D**3)/(nu)**2 \n",
-    "print\"GrD=\",GrD\n",
-    "#Rayleigh number is defined as RaD=GrD*Pr\n",
-    "print\"Rayleigh number is\"\n",
-    "RaD=GrD*Pr\n",
-    "print\"The flow is laminar \"\n",
-    "print\"RaD=\",RaD\n",
-    "#/NuD is nusselt number\n",
-    "#hbarD is Average heat transfer coefficient\n",
-    "print\"we use following equation to find Nusselt number NuD=(0.60+((0.387*RaD**(1/6))/(1+(0.559/Pr**(9/16)))**(8/27)))**2\"\n",
-    "NuD=(0.60+((0.387*RaD**(1/6))/(1+(0.559/Pr**(9/16)))**(8/27)))**2\n",
-    "print\"NuD=\",NuD\n",
-    "print\"Average heat transfer coefficient in W/(m**2*K)\"\n",
-    "hbarD=(NuD*k)/D\n",
-    "print\"hbarD=\",hbarD\n",
-    "print\"Hence,steady state Surface temprature in °C is\"\n",
-    "Tw=Tinf+(P/(hbarD*math.pi*D*L))\n",
-    "print\"Hence we see that our guess is in excellent agreement with the calculated value\"\n",
-    "print\"Tw=\",Tw\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\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": "2.7.11"
-  }
- },
- "nbformat": 4,
- "nbformat_minor": 0
-}
diff --git a/Introduction_to_Heat_Transfer_by_S._K._Som/Chapter9.ipynb b/Introduction_to_Heat_Transfer_by_S._K._Som/Chapter9.ipynb
index 2944362a..7d76afa3 100644
--- a/Introduction_to_Heat_Transfer_by_S._K._Som/Chapter9.ipynb
+++ b/Introduction_to_Heat_Transfer_by_S._K._Som/Chapter9.ipynb
@@ -46,6 +46,10 @@
     }
    ],
    "source": [
+    " \n",
+    " \n",
+    " \n",
+    " \n",
     "import math\n",
     " \n",
     "print\"Introduction to heat transfer by S.K.Som, Chapter 9, Example 1\"\n",
@@ -86,7 +90,18 @@
     "print\"Reynolds no. is\"\n",
     "ReL=(4*mdotc)/(mu)\n",
     "print\"Therefore the flow is laminar and hence the use of the equation is justified\"\n",
-    "print\"ReL=\",ReL"
+    "print\"ReL=\",ReL\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n"
    ]
   },
   {
@@ -122,6 +137,10 @@
     }
    ],
    "source": [
+    " \n",
+    " \n",
+    " \n",
+    " \n",
     "import math\n",
     " \n",
     "print\"Introduction to heat transfer by S.K.Som, Chapter 9, Example 2\"\n",
@@ -159,7 +178,19 @@
     "#Re is reynolds number\n",
     "print\"Reynolds number is\"\n",
     "Re=(4*mdotc)/(mu*P)\n",
-    "print\"Re=\",Re"
+    "print\"Re=\",Re\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n"
    ]
   },
   {
@@ -197,6 +228,10 @@
     }
    ],
    "source": [
+    " \n",
+    " \n",
+    " \n",
+    " \n",
     "import math\n",
     " \n",
     "print\"Introduction to heat transfer by S.K.Som, Chapter 9, Example 3\"\n",
@@ -243,7 +278,29 @@
     "#v is the average flow velocity\n",
     "print\"Hence the average flow velocity at the trailing edge in m/s is\"\n",
     "v=(mdotc)/(rho*delta*B)\n",
-    "print\"v=\",v"
+    "print\"v=\",v\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n"
    ]
   },
   {
@@ -277,6 +334,10 @@
     }
    ],
    "source": [
+    " \n",
+    " \n",
+    " \n",
+    " \n",
     "import math\n",
     " \n",
     "print\"Introduction to heat transfer by S.K.Som, Chapter 9, Example 4\"\n",
@@ -312,7 +373,29 @@
     "#The rate of condensation is given by mdotc=(hbar*(pi*D*L)*(Tg-Tw))/hfg\n",
     "print\"The total rate of condensation in kg/hr\"\n",
     "mdotc=((hbar*(math.pi*D*L)*(Tg-Tw))/hfg)*3600\n",
-    "print\"mdotc=\",mdotc"
+    "print\"mdotc=\",mdotc\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n"
    ]
   },
   {
@@ -340,6 +423,10 @@
     }
    ],
    "source": [
+    " \n",
+    " \n",
+    " \n",
+    " \n",
     "import math\n",
     " \n",
     "print\"Introduction to heat transfer by S.K.Som, Chapter 9, Example 5\"\n",
@@ -358,7 +445,15 @@
     "#h is heat transfer coefficient\n",
     "print\"Heat transfer coefficient in W/m**2 is\"\n",
     "h=(E*I)/(A*(T1-T2))\n",
-    "print\"h=\",h"
+    "print\"h=\",h\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n"
    ]
   },
   {
@@ -388,6 +483,10 @@
     }
    ],
    "source": [
+    " \n",
+    " \n",
+    " \n",
+    " \n",
     "import math\n",
     " \n",
     "print\"Introduction to heat transfer by S.K.Som, Chapter 9, Example 6\"\n",
@@ -414,7 +513,23 @@
     "#E is the burn out voltage\n",
     "print\"The burn out voltage in Volts is  \"\n",
     "E=(qc*A)/I\n",
-    "print\"E=\",E"
+    "print\"E=\",E\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n"
    ]
   },
   {
@@ -443,6 +558,10 @@
     }
    ],
    "source": [
+    " \n",
+    " \n",
+    " \n",
+    " \n",
     "import math\n",
     " \n",
     "print\"Introduction to heat transfer by S.K.Som, Chapter 9, Example 7\"\n",
@@ -468,7 +587,20 @@
     "print\"Heat flux q in W/m**2 is\"\n",
     "q=(mul*hfg)*(((rhol-rhov)*g)/sigma)**(1/2)*((cpl*(T1-T2))/(csf*hfg*Prl**n))**3 \n",
     "print\"The peak heat flux for water at one atmospheric pressure is qc=1.24*10**6(found in example 9.6).Since q<qc,The regime of boiling is nucleate.\"\n",
-    "print\"q=\",q"
+    "print\"q=\",q\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n"
    ]
   },
   {
@@ -480,7 +612,7 @@
   },
   {
    "cell_type": "code",
-   "execution_count": 21,
+   "execution_count": 1,
    "metadata": {
     "collapsed": false
    },
@@ -495,11 +627,15 @@
       "The surface temprature in °C is\n",
       "Tw= 120.0\n",
       "The value of the coefficient csf is \n",
-      "csf= 0.0214423761571\n"
+      "csf= 0.0151329179422\n"
      ]
     }
    ],
    "source": [
+    " \n",
+    " \n",
+    " \n",
+    " \n",
     "import math\n",
     " \n",
     "print\"Introduction to heat transfer by S.K.Som, Chapter 9, Example 8\"\n",
@@ -540,8 +676,30 @@
     "#Now we use following equation to determine csf,q=(mul*hfg)*(((rhol-rhov)*g)/sigma1)**(1/2)*((cpl*(Tw-T))/(csf*hfg*Prl**n))**3 \n",
     "#Manipulating above equation to find csf we get csf=((cpl*(Tw-T))/(((q/((mul*hfg)*(((rhol-rhov)*g)/sigma1)**(1/2))**(1/3))*hfg*Prl**n))\n",
     "print\"The value of the coefficient csf is \"\n",
-    "csf=((cpl*(Tw-T))/(((q/((mul*hfg)*(((rhol-rhov)*g)/sigma1)**(1/2)))**(1/3))*hfg*Prl**n))#[NOTE:The answer in the book is incorrect.(Calcultion mistake)]\n",
-    "print\"csf=\",csf"
+    "csf=((cpl*(Tw-T))/(((q/((mul*hfg)*(((rhol-rhov)*g)/sigma1)**(1.0/2)))**(1.0/3))*hfg*Prl**n))#[NOTE:The answer in the book is incorrect.(Calcultion mistake)]\n",
+    "print\"csf=\",csf\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n"
    ]
   }
  ],
diff --git a/Introduction_to_Heat_Transfer_by_S._K._Som/Chapter9_4YOTRPU.ipynb b/Introduction_to_Heat_Transfer_by_S._K._Som/Chapter9_4YOTRPU.ipynb
deleted file mode 100644
index 7d76afa3..00000000
--- a/Introduction_to_Heat_Transfer_by_S._K._Som/Chapter9_4YOTRPU.ipynb
+++ /dev/null
@@ -1,727 +0,0 @@
-{
- "cells": [
-  {
-   "cell_type": "markdown",
-   "metadata": {
-    "collapsed": true
-   },
-   "source": [
-    "# Chapter 09:Heat transfer in condensation and boiling"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Ex9.1:pg-392"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 17,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Introduction to heat transfer by S.K.Som, Chapter 9, Example 1\n",
-      "The properties of condensate(liquid water) are evaluated at the mean film temprature \n",
-      "The mean film temprature in°C is\n",
-      "tf= 95\n",
-      "hfg= 2270000.0\n",
-      "The average heat transfer coefficient over length L in W/(m**2*K)\n",
-      "hbar= 0.745\n",
-      "The rate of heat transfer per unit width in W/m \n",
-      "Q= 3.772\n",
-      "The total rate of condensation in kg/(s*m)\n",
-      "mdotc= 1.66167400881e-06\n",
-      "We have to check whether the flow is laminar or not \n",
-      "Reynolds no. is\n",
-      "Therefore the flow is laminar and hence the use of the equation is justified\n",
-      "ReL= 0.0221556534508\n"
-     ]
-    }
-   ],
-   "source": [
-    " \n",
-    " \n",
-    " \n",
-    " \n",
-    "import math\n",
-    " \n",
-    "print\"Introduction to heat transfer by S.K.Som, Chapter 9, Example 1\"\n",
-    "#A vertical cooling fin, Approximately a flat plate length,(L)=0.4m high is exposed to saturated steam(temprature,Tg=100°C) at atmospheric pressure.\n",
-    "L=0.4;\n",
-    "Tg=100;\n",
-    "#The fin is maintained at temprature,Tw=90°C by cooling water.\n",
-    "Tw=90;\n",
-    "print\"The properties of condensate(liquid water) are evaluated at the mean film temprature \"\n",
-    "#tf is mean film temprature\n",
-    "print\"The mean film temprature in°C is\"\n",
-    "tf=(Tg+Tw)/2\n",
-    "print\"tf=\",tf\n",
-    "#The properties of condensate are density(rho=962kg/m**3),conductivity(k=0.677W/(m*K)),viscosity(mu=3*10**-4 kg/(m*s))\n",
-    "rho=962;\n",
-    "k=0.677;\n",
-    "mu=3*10**-4;\n",
-    "#The value rhov=0.598kg/m**3 and hfg=2.27*10**6J/kg at 100°C are found from steam table\n",
-    "#g is acceleration due to gravity =9.81m/s**2\n",
-    "g=9.81;\n",
-    "rhov=0.598;#rhov is vapour density\n",
-    "hfg=2.27*10**6;#hfg is enthalpy of vaporisation\n",
-    "print\"hfg=\",hfg\n",
-    "#The average heat transfer coefficient over length L is hbarL=0.943*((rho*(rho-rhov)*g*h*L**3)/(mu*k*(Tg-Tw)))**(1/4)\n",
-    "print\"The average heat transfer coefficient over length L in W/(m**2*K)\"\n",
-    "hbarL=0.943*((rho*(rho-rhov)*g*hfg*k**3)/(mu*L*(Tg-Tw)))**(1/4)\n",
-    "print\"hbar=\",hbar\n",
-    "#The rate of heat transfer per unit width is Q=hbarL*L*(Tg-Tw)\n",
-    "print\"The rate of heat transfer per unit width in W/m \"\n",
-    "Q=hbarL*L*(Tg-Tw)\n",
-    "print\"Q=\",Q\n",
-    "#The rate of condensation is given by mdotc=(Q/hfg)\n",
-    "print\"The total rate of condensation in kg/(s*m)\"\n",
-    "mdotc=(Q/hfg)\n",
-    "print\"mdotc=\",mdotc\n",
-    "print\"We have to check whether the flow is laminar or not \"\n",
-    "#Reynolds no is given by ReL=(4*mdotc)/(mu)\n",
-    "print\"Reynolds no. is\"\n",
-    "ReL=(4*mdotc)/(mu)\n",
-    "print\"Therefore the flow is laminar and hence the use of the equation is justified\"\n",
-    "print\"ReL=\",ReL\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Ex9.2:pg-393"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 16,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Introduction to heat transfer by S.K.Som, Chapter 9, Example 2\n",
-      "The mean film temprature in°C is\n",
-      "tf= 95\n",
-      "hfg= 2270000.0\n",
-      "The average heat transfer coefficient in W/(m**2*K)\n",
-      "hbar= 0.745\n",
-      "The total rate of condensation in kg/s\n",
-      "Check for reynolds no.\n",
-      "mdotc= 1.54657700017e-06\n",
-      "Reynolds number is\n",
-      "Re= 0.00343683777816\n"
-     ]
-    }
-   ],
-   "source": [
-    " \n",
-    " \n",
-    " \n",
-    " \n",
-    "import math\n",
-    " \n",
-    "print\"Introduction to heat transfer by S.K.Som, Chapter 9, Example 2\"\n",
-    "#Steam is condensed at temprature(Tg=100°C) on the outer surafce of a horizontal tube of length(L=3m) and diameter(d)=50mm or .05m\n",
-    "Tg=100;\n",
-    "L=3;\n",
-    "D=0.05;\n",
-    "#The Tube surface is maintained at temprature,Tw=90°C \n",
-    "Tw=90;\n",
-    "#tf is mean film temprature\n",
-    "print\"The mean film temprature in°C is\"\n",
-    "tf=(Tg+Tw)/2\n",
-    "print\"tf=\",tf\n",
-    "#The properties of condensate are density(rho=962kg/m**3),conductivity(k=0.677W/(m*K)),viscosity(mu=3*10**-4 kg/(m*s))\n",
-    "rho=962;\n",
-    "k=0.677;\n",
-    "mu=3*10**-4;\n",
-    "#The value rhov=0.598kg/m**3 and hfg=2.27*10**6J/kg at 100°C are found from steam table\n",
-    "#g is acceleration due to gravity =9.81m/s**2\n",
-    "g=9.81;\n",
-    "rhov=0.598;#vapour density\n",
-    "hfg=2.27*10**6;#enthalpy of vaporisation\n",
-    "print\"hfg=\",hfg\n",
-    "#The average heat transfer coefficient hbar=0.745*((rho*(rho-rhov)*g*hfg*k**3)/(mu*D*(Tg-Tw)))**(1/4)\n",
-    "print\"The average heat transfer coefficient in W/(m**2*K)\"\n",
-    "hbar=0.745*((rho*(rho-rhov)*g*hfg*k**3)/(mu*D*(Tg-Tw)))**(1/4)\n",
-    "print\"hbar=\",hbar\n",
-    "#The rate of condensation is given by mdotc=(hbar*(pi*D*L)*(Tg-Tw))/hfg\n",
-    "print\"The total rate of condensation in kg/s\"\n",
-    "mdotc=(hbar*(math.pi*D*L)*(Tg-Tw))/hfg\n",
-    "print\"Check for reynolds no.\"\n",
-    "print\"mdotc=\",mdotc\n",
-    "#For a horizontal tube having length,L,perimeter is P=2L\n",
-    "P=2*L;\n",
-    "#Re is reynolds number\n",
-    "print\"Reynolds number is\"\n",
-    "Re=(4*mdotc)/(mu*P)\n",
-    "print\"Re=\",Re\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Ex9.3:pg-394"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 13,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Introduction to heat transfer by S.K.Som, Chapter 9, Example 3\n",
-      "The mean film temprature in°C is\n",
-      "tf= 80\n",
-      "The average heat transfer coefficient over length L in W/(m**2*K)\n",
-      "hbar= 0.943\n",
-      "The rate of heat transfer  in kW \n",
-      "Q= 0.016974\n",
-      "(b)The film thickness at the trailing edges in m is\n",
-      "delta= 1.0\n",
-      "The total rate of condensation in kg/s\n",
-      "mdotc= 7.47753303965e-06\n",
-      "Hence the average flow velocity at the trailing edge in m/s is\n",
-      "v= 2.56431174199e-08\n"
-     ]
-    }
-   ],
-   "source": [
-    " \n",
-    " \n",
-    " \n",
-    " \n",
-    "import math\n",
-    " \n",
-    "print\"Introduction to heat transfer by S.K.Som, Chapter 9, Example 3\"\n",
-    "#A vertical plate having length,(L)=1.5m is maintained at temprature(Tw) of 60°C in the presence of saturated steam(temprature,Tg=100°C) at atmospheric pressure.\n",
-    "L=1.5;\n",
-    "Tg=100;\n",
-    "Tw=60;\n",
-    "#Consider the width of plate to be (B)=0.3m\n",
-    "B=0.3;\n",
-    "#tf is the mean film temprature\n",
-    "print\"The mean film temprature in°C is\"\n",
-    "tf=(Tg+Tw)/2\n",
-    "print\"tf=\",tf\n",
-    "#The relevant properties are desity(rho=972kg/m**3),conductivity(k=0.670W/(m*K)),viscosity(mu=3.54*10**-4 kg/(m*s))\n",
-    "#specific heat(cp=4.2J/(kg*K)),vapur density(rhov(100°C)=0.598k/m**3),Enthalpy of vaporisation(hfg(100°C)=2.27*10**6J/kg)\n",
-    "#g is acceleration due to gravity =9.81m/s**2\n",
-    "g=9.81;\n",
-    "rho=972;\n",
-    "k=0.670;\n",
-    "mu=3.54*10**-4;\n",
-    "cp=4.2;\n",
-    "rhov=0.598;\n",
-    "hfg=2.27*10**6;\n",
-    "#The average heat transfer coefficient over length L is hbar=0.943*((rho*(rho-rhov)*g*h*L**3)/(mu*k*(Tg-Tw)))**(1/4)\n",
-    "print\"The average heat transfer coefficient over length L in W/(m**2*K)\"\n",
-    "hbar=0.943*((rho*(rho-rhov)*g*hfg*k**3)/(mu*L*(Tg-Tw)))**(1/4)\n",
-    "print\"hbar=\",hbar\n",
-    "#The rate of heat transfer Q=hbarL*A*(Tg-Tw)\n",
-    "#Area(A)=L*B\n",
-    "A=L*B;\n",
-    "print\"The rate of heat transfer  in kW \"\n",
-    "Q=(hbar*A*(Tg-Tw))/1000\n",
-    "print\"Q=\",Q\n",
-    "#The film thickness at the trailing edges is found out by delta=((4*mu*k*x*(Tg-Tw))/(g*hfg*rho*(rho-rhov)))**(1/4)\n",
-    "print\"(b)The film thickness at the trailing edges in m is\"\n",
-    "#at trailing edges x=1.5m\n",
-    "x=1.5;\n",
-    "delta=((4*mu*k*x*(Tg-Tw))/(g*hfg*rho*(rho-rhov)))**(1/4)\n",
-    "print\"delta=\",delta\n",
-    "#The rate of condensation is given by mdotc=(Q/hfg)\n",
-    "print\"The total rate of condensation in kg/s\"\n",
-    "mdotc=((Q*1000)/hfg)\n",
-    "print\"mdotc=\",mdotc\n",
-    "#v is the average flow velocity\n",
-    "print\"Hence the average flow velocity at the trailing edge in m/s is\"\n",
-    "v=(mdotc)/(rho*delta*B)\n",
-    "print\"v=\",v\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Ex9.4:pg-396"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 11,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Introduction to heat transfer by S.K.Som, Chapter 9, Example 4\n",
-      "The mean film temprature in°C is\n",
-      "tf= 30\n",
-      " Modified enthalpy in J/kg is\n",
-      "hfgdash= 131330.0\n",
-      "The average heat transfer coefficient over length L in W/(m**2*K)\n",
-      "hbar= 0.555\n",
-      "The total rate of condensation in kg/hr\n",
-      "mdotc= 0.00716923260703\n"
-     ]
-    }
-   ],
-   "source": [
-    " \n",
-    " \n",
-    " \n",
-    " \n",
-    "import math\n",
-    " \n",
-    "print\"Introduction to heat transfer by S.K.Som, Chapter 9, Example 4\"\n",
-    "#Saturated freon-012 at Temprature(Tg)=35°C is condensed horizontal tube of diameter(D)=15mm or.015m at a lower vapour velocity.\n",
-    "#length,L=1m,Since per meter of tube is considered.\n",
-    "L=1;\n",
-    "Tg=35;\n",
-    "D=0.015;\n",
-    "#The tube wall is maintained at temprature(Tw)=25°C\n",
-    "Tw=25;\n",
-    "#For freon-12 at 35°C,enthalpy of vaporisation(hfg=131.33kJ/kg) and vapour density(rhov=42.68kg/m**3)\n",
-    "hfg=131.33*10**3;\n",
-    "rhov=42.68;\n",
-    "#tf is mean film temprature\n",
-    "print\"The mean film temprature in°C is\"\n",
-    "tf=(Tg+Tw)/2\n",
-    "print\"tf=\",tf\n",
-    "#The relevant properties at 30°C are density(rho=1.29*10**3kg/m**3),conductivity(k=0.071W/(mK)),viscosity(mu=2.50*10**-4kg/(m*s)),specific heat(cp=983J/(kg*°C))\n",
-    "rho=1.29*10**3;\n",
-    "k=0.071;\n",
-    "mu=2.50*10**-4;\n",
-    "cp=983;\n",
-    "#g is acceleration due to gravity =9.81m/s**2\n",
-    "g=9.81;\n",
-    "#we found the modified enthalpy by using following equation hfgdash=hfg+(3/8)*cp*(Tg-Tw)\n",
-    "print\" Modified enthalpy in J/kg is\"\n",
-    "hfgdash=hfg+((3/8)*cp*(Tg-Tw))\n",
-    "print\"hfgdash=\",hfgdash\n",
-    "#The average heat transfer coefficient over length L is hbar=0.555*((rho*(rho-rhov)*g*hfgdash*k**3)/(mu*D*(Tg-Tw)))**(1/4)\n",
-    "print\"The average heat transfer coefficient over length L in W/(m**2*K)\"\n",
-    "hbar=0.555*((rho*(rho-rhov)*g*hfgdash*k**3)/(mu*D*(Tg-Tw)))**(1/4)\n",
-    "print\"hbar=\",hbar\n",
-    "#The rate of condensation is given by mdotc=(hbar*(pi*D*L)*(Tg-Tw))/hfg\n",
-    "print\"The total rate of condensation in kg/hr\"\n",
-    "mdotc=((hbar*(math.pi*D*L)*(Tg-Tw))/hfg)*3600\n",
-    "print\"mdotc=\",mdotc\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Ex9.5:pg-397"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 7,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Introduction to heat transfer by S.K.Som, Chapter 9, Example 5\n",
-      "Heat transfer coefficient in W/m**2 is\n",
-      "h= 105042.262441\n"
-     ]
-    }
-   ],
-   "source": [
-    " \n",
-    " \n",
-    " \n",
-    " \n",
-    "import math\n",
-    " \n",
-    "print\"Introduction to heat transfer by S.K.Som, Chapter 9, Example 5\"\n",
-    "#A nickel wire of length(L)=0.1m,Diameter(D)=1mm or .001m \n",
-    "#Submerged horizontally in water at pressure=1 atm(101kPa) requires current,I=150A at voltage ,E=2.2V to maintain wire at temprature(T1)=110°C\n",
-    "L=0.1;\n",
-    "T1=110;\n",
-    "D=0.001;\n",
-    "I=150;\n",
-    "E=2.2;\n",
-    "#Area(A)=(math.pi*D*L)\n",
-    "A=math.pi*D*L;\n",
-    "#The saturation temprature of water at one atmospheric pressure(101kPa) is T2=100°C.\n",
-    "T2=100;\n",
-    "#We can write from energy balance E*I=h*A*(T1-T2),we can find heat transfer coefficient from it.\n",
-    "#h is heat transfer coefficient\n",
-    "print\"Heat transfer coefficient in W/m**2 is\"\n",
-    "h=(E*I)/(A*(T1-T2))\n",
-    "print\"h=\",h\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Ex9.6:pg-398"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 18,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Introduction to heat transfer by S.K.Som, Chapter 9, Example 6\n",
-      "Critical Heat flux in W/m**2 is\n",
-      "qc= 202044.0\n",
-      "The burn out voltage in Volts is  \n",
-      "E= 1.90421983831\n"
-     ]
-    }
-   ],
-   "source": [
-    " \n",
-    " \n",
-    " \n",
-    " \n",
-    "import math\n",
-    " \n",
-    "print\"Introduction to heat transfer by S.K.Som, Chapter 9, Example 6\"\n",
-    "#In a laboratory experiment,A current(I)=100A burns out a nickel wire having Diameter(D)=1mm or 0.001mm,length(L)=0.3m\n",
-    "I=100;\n",
-    "D=.001;\n",
-    "L=0.3;\n",
-    "#It is submerged horizontally in water at one atmospheric pressure.\n",
-    "#For saturated water at one atmospheric pressure,density(rhol=960kg/m**3),vapour density(rhov=0.60kg/m**3),enthalpy of vaporisation(hfg=2.26*10**6J/kg),surface tension(sigma=0.055N/m).\n",
-    "rhol=960;\n",
-    "rhov=0.60;\n",
-    "hfg=2.26*10**6;\n",
-    "sigma=0.055;\n",
-    "#Area(A)=(pi*D*L)\n",
-    "A=math.pi*D*L;\n",
-    "#g is acceleration due to gravity =9.81m/s**2\n",
-    "g=9.81;\n",
-    "#The wire is burnt out when heat reaches its peak\n",
-    "#We use following expression to determine critical heat flux qc=0.149*hfg*rhov*((sigma*g*(rhol-rhov))/rhov**2)**(1/4)*((rhol+rhov)/rhol)**(1/2) \n",
-    "print\"Critical Heat flux in W/m**2 is\"\n",
-    "qc=0.149*hfg*rhov*((sigma*g*(rhol-rhov))/rhov**2)**(1/4)*((rhol+rhov)/rhol)**(1/2) \n",
-    "print\"qc=\",qc\n",
-    "#From the energy balance E*I=qc*A\n",
-    "#E is the burn out voltage\n",
-    "print\"The burn out voltage in Volts is  \"\n",
-    "E=(qc*A)/I\n",
-    "print\"E=\",E\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Ex9.7:pg-399"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 19,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Introduction to heat transfer by S.K.Som, Chapter 9, Example 7\n",
-      "Heat flux q in W/m**2 is\n",
-      "The peak heat flux for water at one atmospheric pressure is qc=1.24*10**6(found in example 9.6).Since q<qc,The regime of boiling is nucleate.\n",
-      "q= 3636.07255495\n"
-     ]
-    }
-   ],
-   "source": [
-    " \n",
-    " \n",
-    " \n",
-    " \n",
-    "import math\n",
-    " \n",
-    "print\"Introduction to heat transfer by S.K.Som, Chapter 9, Example 7\"\n",
-    "#A heated nickel plate at temprature (T1)=110°C is submereged in water at one atmospheric pressure.\n",
-    "T1=110;\n",
-    "#For nucleate boiling coefficient(csf=0.006) and n=1\n",
-    "csf=0.006;\n",
-    "n=1;\n",
-    "#For saturated water at one atmospheric pressure,density of liquid(rhol=960kg/m**3),vapour density(rhov=0.60kg/m**3)\n",
-    "#enthalpy of vaporisation(hfg=2.26*10**6J/kg),surface tension(sigma=0.055N/m),saturation temprature(T2)=100°C\n",
-    "T2=100;\n",
-    "rhol=960;\n",
-    "rhov=0.60;\n",
-    "hfg=2.26*10**6;\n",
-    "sigma=0.055;\n",
-    "#g is acceleration due to gravity =9.81m/s**2\n",
-    "g=9.81;\n",
-    "#We take specific heat of liquid(cpl=4.216kJ/(kg*K)),prandtl number of liquid(Prl=1.74),viscosity of liquid(mul=2.82*10**-4kg/(m*s))\n",
-    "cpl=4.216*10**3;\n",
-    "Prl=1.74;\n",
-    "mul=2.82*10**-4;\n",
-    "#The heat flux q is given by expression q=(mul*hfg)*(((rhol-rhov)*g)/sigma)**(1/2)*((cpl*(T1-T2))*(csf*hfg*prl**n))**3 \n",
-    "print\"Heat flux q in W/m**2 is\"\n",
-    "q=(mul*hfg)*(((rhol-rhov)*g)/sigma)**(1/2)*((cpl*(T1-T2))/(csf*hfg*Prl**n))**3 \n",
-    "print\"The peak heat flux for water at one atmospheric pressure is qc=1.24*10**6(found in example 9.6).Since q<qc,The regime of boiling is nucleate.\"\n",
-    "print\"q=\",q\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Ex9.8:pg-401"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 1,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Introduction to heat transfer by S.K.Som, Chapter 9, Example 8\n",
-      "The heat flux in W/m**2 is\n",
-      "q= 750000.0\n",
-      "The surface temprature in °C is\n",
-      "Tw= 120.0\n",
-      "The value of the coefficient csf is \n",
-      "csf= 0.0151329179422\n"
-     ]
-    }
-   ],
-   "source": [
-    " \n",
-    " \n",
-    " \n",
-    " \n",
-    "import math\n",
-    " \n",
-    "print\"Introduction to heat transfer by S.K.Som, Chapter 9, Example 8\"\n",
-    "#A Copper bar whose one end is exposed to boiling water while the other end is encapsulated by an electric heater.\n",
-    "#Thermocouples are inserted in the bar to measure the tempratures at two locations A and b at distances xA=10mm and xB=30mm from the surface.\n",
-    "xA=.010;\n",
-    "xB=.030;\n",
-    "#Under steady condition nucleate boiling is maintained in saturated water at atmospheric pressure and the tempratures are TA=140°C and TB=180°C,n=1\n",
-    "TA=140;\n",
-    "TB=180;\n",
-    "n=1;\n",
-    "#The values of relevant properties of water and other parameters are \n",
-    "#density of liquid(rhol=960kg/m**3),vapour density(rhov=0.60kg/m**3),specific heat of liquid(cpl=4.216 kJ/(kg*K))\n",
-    "#enthalpy of vaporisation(hfg=2.26*106J/kg),prandtl number of liquiid(Prl=1.74),viscosity of liquid(mul=2.82*10**-4kg/(m*s)),surface tension(sigma1=0.055N/m).\n",
-    "rhol=960;\n",
-    "rhov=0.60;\n",
-    "cpl=4.216*10**3;\n",
-    "hfg=2.26*10**6;\n",
-    "Prl=1.74;\n",
-    "mul=2.82*10**-4;\n",
-    "sigma1=0.055;\n",
-    "#We have to know the value of heat flux(q) and the surface temprature(Tw).\n",
-    "#Since we know the tempratures at location A and B,The heat flux q is determined by fourier law of heat conduction in the bar at steady-state as\n",
-    "#q=k*((TB-TA)/(xB-xA))\n",
-    "#We take for copper conductivity,k=375W/(m*K)\n",
-    "k=375;\n",
-    "print\"The heat flux in W/m**2 is\"\n",
-    "q=k*((TB-TA)/(xB-xA))\n",
-    "print\"q=\",q\n",
-    "#g is acceleration due to gravity =9.81m/s**2\n",
-    "g=9.81;\n",
-    "#The surface temprature is given by Tw=TA-((TB-TA)/(xB-xA))*xA\n",
-    "print\"The surface temprature in °C is\"\n",
-    "Tw=TA-((TB-TA)/(xB-xA))*xA\n",
-    "print\"Tw=\",Tw\n",
-    "#Temprature,T=100°C,since copper bar is exposed to boiling water. \n",
-    "T=100;\n",
-    "#Now we use following equation to determine csf,q=(mul*hfg)*(((rhol-rhov)*g)/sigma1)**(1/2)*((cpl*(Tw-T))/(csf*hfg*Prl**n))**3 \n",
-    "#Manipulating above equation to find csf we get csf=((cpl*(Tw-T))/(((q/((mul*hfg)*(((rhol-rhov)*g)/sigma1)**(1/2))**(1/3))*hfg*Prl**n))\n",
-    "print\"The value of the coefficient csf is \"\n",
-    "csf=((cpl*(Tw-T))/(((q/((mul*hfg)*(((rhol-rhov)*g)/sigma1)**(1.0/2)))**(1.0/3))*hfg*Prl**n))#[NOTE:The answer in the book is incorrect.(Calcultion mistake)]\n",
-    "print\"csf=\",csf\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n"
-   ]
-  }
- ],
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