Revised list of TBCs

1 files changed, 221 insertions, 215 deletions

diff --git a/Basic_And_Applied_Thermodynamics_by_P._K._Nag/Chapter11.ipynb b/Basic_And_Applied_Thermodynamics_by_P._K._Nag/Chapter11.ipynb
index 75c1ae52..3e09ba67 100755..100644
--- a/Basic_And_Applied_Thermodynamics_by_P._K._Nag/Chapter11.ipynb
+++ b/Basic_And_Applied_Thermodynamics_by_P._K._Nag/Chapter11.ipynb
@@ -1,215 +1,221 @@
-{


- "metadata": {


-  "name": "",


-  "signature": "sha256:bac11063c240653dfd3c07e3907da1d648418ca108c3c127b610f8e4e00f83ef"


- },


- "nbformat": 3,


- "nbformat_minor": 0,


- "worksheets": [


-  {


-   "cells": [


-    {


-     "cell_type": "heading",


-     "level": 1,


-     "metadata": {},


-     "source": [


-      "Chapter 11:Thermodynamic relations Equilibrium and stability"


-     ]


-    },


-    {


-     "cell_type": "heading",


-     "level": 2,


-     "metadata": {},


-     "source": [


-      "Ex11.3:pg-436"


-     ]


-    },


-    {


-     "cell_type": "code",


-     "collapsed": false,


-     "input": [


-      "\n",


-      "Tb = 353.0 # boiling point of benzene in K\n",


-      "T = 303.0 # Operational temperature in K\n",


-      "R = 8.3143 #Gas constant\n",


-      "P = 101.325*math.exp((88/R)*(1.0-(Tb/T)))\n",


-      "\n",


-      "print \"\\n Example 11.3\"\n",


-      "print \"\\n Vapour pressure of benzene is \",P ,\" kPa\"\n",


-      "#The answers vary due to round off error\n",


-      "\n"


-     ],


-     "language": "python",


-     "metadata": {},


-     "outputs": [


-      {


-       "output_type": "stream",


-       "stream": "stdout",


-       "text": [


-        "\n",


-        " Example 11.3\n",


-        "\n",


-        " Vapour pressure of benzene is  17.6682592008  kPa\n"


-       ]


-      }


-     ],


-     "prompt_number": 2


-    },


-    {


-     "cell_type": "heading",


-     "level": 2,


-     "metadata": {},


-     "source": [


-      "Ex11.4:pg-436"


-     ]


-    },


-    {


-     "cell_type": "code",


-     "collapsed": false,


-     "input": [


-      "\n",


-      "T = (3754-3063)/(23.03-19.49) # Temperature at triple point in K\n",


-      "P = math.exp(23.03-(3754/195.2)) # Pressure at triple point\n",


-      "R = 8.3143 # Gas constant\n",


-      "Lsub = R*3754 # Latent heat of sublimation\n",


-      "Lvap = 3063*R # Latent heat of vaporisation\n",


-      "Lfu = Lsub-Lvap # Latent heat of fusion\n",


-      "\n",


-      "print \"\\n Example 11.4\"\n",


-      "print \"\\n Temperature at triple point is \",T ,\" K\"\n",


-      "print \"\\n Pressure at triple point is \",P ,\" mm Hg\"\n",


-      "print \"\\n\\n Latent heat of sublimation is \",Lsub ,\" kJ/kg mol\"\n",


-      "print \"\\n Latent heat of vapourization is is \",Lvap ,\" kJ/kg mol\"\n",


-      "print \"\\n Latent heat of fusion is \",Lfu ,\" kJ/kg mol\"\n",


-      "#The answers vary due to round off error\n"


-     ],


-     "language": "python",


-     "metadata": {},


-     "outputs": [


-      {


-       "output_type": "stream",


-       "stream": "stdout",


-       "text": [


-        "\n",


-        " Example 11.4\n",


-        "\n",


-        " Temperature at triple point is  195.197740113  K\n",


-        "\n",


-        " Pressure at triple point is  44.631622076  mm Hg\n",


-        "\n",


-        "\n",


-        " Latent heat of sublimation is  31211.8822  kJ/kg mol\n",


-        "\n",


-        " Latent heat of vapourization is is  25466.7009  kJ/kg mol\n",


-        "\n",


-        " Latent heat of fusion is  5745.1813  kJ/kg mol\n"


-       ]


-      }


-     ],


-     "prompt_number": 3


-    },


-    {


-     "cell_type": "heading",


-     "level": 2,


-     "metadata": {},


-     "source": [


-      "Ex11.6:pg-438"


-     ]


-    },


-    {


-     "cell_type": "code",


-     "collapsed": false,


-     "input": [


-      "\n",


-      "R = 8.3143 # Gas constant in kJ/kg-mol-K\n",


-      "N1 = 0.5 # Mole no. of first system\n",


-      "N2 = 0.75 # Mole no. of second system\n",


-      "T1 = 200 # Initial temperature of first system in K\n",


-      "T2 = 300 # Initial temperature of second system in K\n",


-      "v = 0.02 # Total volume in m**3\n",


-      "print \"\\n Example 11.6\\n\"\n",


-      "Tf = (T2*N2+T1*N1)/(N1+N2)\n",


-      "Uf_1 = (3.0/2.0)*(R*N1*Tf)*(10**-3)\n",


-      "Uf_2 = (3.0/2.0)*(R*N2*Tf)*(10**-3)\n",


-      "pf = (R*Tf*(N1+N2)*(10**-3))/v\n",


-      "Vf_1 = R*N1*(10**-3)*Tf/pf\n",


-      "Vf_2 = v-Vf_1\n",


-      "print \"\\n Energy of first system is \",Uf_1 ,\" kJ,\\n Energy of second system is \",Uf_2 ,\" kJ,\\n Volume of first system is \",Vf_1 ,\" m**3,\\n Volume of second system is \",Vf_2 ,\" m**3,\\n Pressure is \",pf ,\" kN/m**2,\\n Temperature is \",Tf ,\" K.\"\n",


-      "#The answers vary due to round off error\n"


-     ],


-     "language": "python",


-     "metadata": {},


-     "outputs": [


-      {


-       "output_type": "stream",


-       "stream": "stdout",


-       "text": [


-        "\n",


-        " Example 11.6\n",


-        "\n",


-        "\n",


-        " Energy of first system is  1.6212885  kJ,\n",


-        " Energy of second system is  2.43193275  kJ,\n",


-        " Volume of first system is  0.008  m**3,\n",


-        " Volume of second system is  0.012  m**3,\n",


-        " Pressure is  135.107375  kN/m**2,\n",


-        " Temperature is  260.0  K.\n"


-       ]


-      }


-     ],


-     "prompt_number": 4


-    },


-    {


-     "cell_type": "heading",


-     "level": 2,


-     "metadata": {},


-     "source": [


-      "Ex11.10:pg-446"


-     ]


-    },


-    {


-     "cell_type": "code",


-     "collapsed": false,


-     "input": [


-      "R = 0.082 # Gas constant in litre-atm/gmol-K\n",


-      "m = 1.5 # Mass flow rate in kg/s\n",


-      "p1 = 1.0 # Pressure in atm\n",


-      "t2 = 300.0 # Temperature after compression in K\n",


-      "p2 = 400.0 # Pressure after compression in atm\n",


-      "Tc = 151.0 # For Argon in K\n",


-      "pc = 48.0 # For Argon in atm\n",


-      "print \"\\n Example 11.10 \"\n",


-      "a = 0.42748*((R*1000)**2)*((Tc)**2)/pc\n",


-      "b = 0.08664*(R*1000)*(Tc)/pc\n",


-      "# By solving equation v2**2 - 49.24*v2**2 + 335.6*v2 - 43440 = 0\n",


-      "v2 = 56.8 # In cm**3/g mol\n",


-      "v1 = (R*1000)*(t2)/p1\n",


-      "delta_h = -1790 # In J/g mol\n",


-      "delta_s = -57 # In J/g mol\n",


-      "Q = (t2*delta_s*(10**5)/39.8)/(3600*1000)\n",


-      "W = Q - (delta_h*(10**5)/39.8)/(3600*1000)\n",


-      "print \"\\n Power required to run the compressor = \",W ,\" kW, \\n The rate at which heat must be removed from the compressor = \",Q ,\" kW\"\n",


-      "# Answers vary due to round off error.\n"


-     ],


-     "language": "python",


-     "metadata": {},


-     "outputs": [


-      {


-       "output_type": "stream",


-       "stream": "stdout",


-       "text": [


-        "\n",


-        " Example 11.10 \n",


-        "\n",


-        " Power required to run the compressor =  -10.6853713009  kW, \n",


-        " The rate at which heat must be removed from the compressor =  -11.9346733668  kW\n"


-       ]


-      }


-     ],


-     "prompt_number": 5


-    }


-   ],


-   "metadata": {}


-  }


- ]


-}
\ No newline at end of file
+{
+ "cells": [
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "# Chapter 11:Thermodynamic relations Equilibrium and stability"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Ex11.3:pg-436"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 1,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "\n",
+      " Example 11.3\n",
+      "\n",
+      " Vapour pressure of benzene is  17.6682592008  kPa\n"
+     ]
+    }
+   ],
+   "source": [
+    "import math\n",
+    "Tb = 353.0 # boiling point of benzene in K\n",
+    "T = 303.0 # Operational temperature in K\n",
+    "R = 8.3143 #Gas constant\n",
+    "P = 101.325*math.exp((88/R)*(1.0-(Tb/T)))\n",
+    "\n",
+    "print \"\\n Example 11.3\"\n",
+    "print \"\\n Vapour pressure of benzene is \",P ,\" kPa\"\n",
+    "#The answers vary due to round off error\n",
+    "\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Ex11.4:pg-436"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 2,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "\n",
+      " Example 11.4\n",
+      "\n",
+      " Temperature at triple point is  195.197740113  K\n",
+      "\n",
+      " Pressure at triple point is  44.631622076  mm Hg\n",
+      "\n",
+      "\n",
+      " Latent heat of sublimation is  31211.8822  kJ/kg mol\n",
+      "\n",
+      " Latent heat of vapourization is is  25466.7009  kJ/kg mol\n",
+      "\n",
+      " Latent heat of fusion is  5745.1813  kJ/kg mol\n"
+     ]
+    }
+   ],
+   "source": [
+    "import math\n",
+    "T = (3754-3063)/(23.03-19.49) # Temperature at triple point in K\n",
+    "P = math.exp(23.03-(3754/195.2)) # Pressure at triple point\n",
+    "R = 8.3143 # Gas constant\n",
+    "Lsub = R*3754 # Latent heat of sublimation\n",
+    "Lvap = 3063*R # Latent heat of vaporisation\n",
+    "Lfu = Lsub-Lvap # Latent heat of fusion\n",
+    "\n",
+    "print \"\\n Example 11.4\"\n",
+    "print \"\\n Temperature at triple point is \",T ,\" K\"\n",
+    "print \"\\n Pressure at triple point is \",P ,\" mm Hg\"\n",
+    "print \"\\n\\n Latent heat of sublimation is \",Lsub ,\" kJ/kg mol\"\n",
+    "print \"\\n Latent heat of vapourization is is \",Lvap ,\" kJ/kg mol\"\n",
+    "print \"\\n Latent heat of fusion is \",Lfu ,\" kJ/kg mol\"\n",
+    "#The answers vary due to round off error\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Ex11.6:pg-438"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 4,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "\n",
+      " Example 11.6\n",
+      "\n",
+      "\n",
+      " Energy of first system is  1.6212885  kJ,\n",
+      " Energy of second system is  2.43193275  kJ,\n",
+      " Volume of first system is  0.008  m**3,\n",
+      " Volume of second system is  0.012  m**3,\n",
+      " Pressure is  135.107375  kN/m**2,\n",
+      " Temperature is  260.0  K.\n"
+     ]
+    }
+   ],
+   "source": [
+    "\n",
+    "R = 8.3143 # Gas constant in kJ/kg-mol-K\n",
+    "N1 = 0.5 # Mole no. of first system\n",
+    "N2 = 0.75 # Mole no. of second system\n",
+    "T1 = 200 # Initial temperature of first system in K\n",
+    "T2 = 300 # Initial temperature of second system in K\n",
+    "v = 0.02 # Total volume in m**3\n",
+    "print \"\\n Example 11.6\\n\"\n",
+    "Tf = (T2*N2+T1*N1)/(N1+N2)\n",
+    "Uf_1 = (3.0/2.0)*(R*N1*Tf)*(10**-3)\n",
+    "Uf_2 = (3.0/2.0)*(R*N2*Tf)*(10**-3)\n",
+    "pf = (R*Tf*(N1+N2)*(10**-3))/v\n",
+    "Vf_1 = R*N1*(10**-3)*Tf/pf\n",
+    "Vf_2 = v-Vf_1\n",
+    "print \"\\n Energy of first system is \",Uf_1 ,\" kJ,\\n Energy of second system is \",Uf_2 ,\" kJ,\\n Volume of first system is \",Vf_1 ,\" m**3,\\n Volume of second system is \",Vf_2 ,\" m**3,\\n Pressure is \",pf ,\" kN/m**2,\\n Temperature is \",Tf ,\" K.\"\n",
+    "#The answers vary due to round off error\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Ex11.10:pg-446"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 3,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "\n",
+      " Example 11.10 \n",
+      "\n",
+      " Power required to run the compressor =  -10.6853713009  kW, \n",
+      " The rate at which heat must be removed from the compressor =  -11.9346733668  kW\n"
+     ]
+    }
+   ],
+   "source": [
+    "import math\n",
+    "R = 0.082 # Gas constant in litre-atm/gmol-K\n",
+    "m = 1.5 # Mass flow rate in kg/s\n",
+    "p1 = 1.0 # Pressure in atm\n",
+    "t2 = 300.0 # Temperature after compression in K\n",
+    "p2 = 400.0 # Pressure after compression in atm\n",
+    "Tc = 151.0 # For Argon in K\n",
+    "pc = 48.0 # For Argon in atm\n",
+    "print \"\\n Example 11.10 \"\n",
+    "a = 0.42748*((R*1000)**2)*((Tc)**2)/pc\n",
+    "b = 0.08664*(R*1000)*(Tc)/pc\n",
+    "# By solving equation v2**2 - 49.24*v2**2 + 335.6*v2 - 43440 = 0\n",
+    "v2 = 56.8 # In cm**3/g mol\n",
+    "v1 = (R*1000)*(t2)/p1\n",
+    "delta_h = -1790 # In J/g mol\n",
+    "delta_s = -57 # In J/g mol\n",
+    "Q = (t2*delta_s*(10**5)/39.8)/(3600*1000)\n",
+    "W = Q - (delta_h*(10**5)/39.8)/(3600*1000)\n",
+    "print \"\\n Power required to run the compressor = \",W ,\" kW, \\n The rate at which heat must be removed from the compressor = \",Q ,\" kW\"\n",
+    "# Answers vary due to round off error.\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
+}