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+{

+ "cells": [

+  {

+   "cell_type": "markdown",

+   "metadata": {},

+   "source": [

+    "# chapter 2:Zeroth Law of Thermodynamics"

+   ]

+  },

+  {

+   "cell_type": "markdown",

+   "metadata": {},

+   "source": [

+    "##example 2.1;pg no:46"

+   ]

+  },

+  {

+   "cell_type": "code",

+   "execution_count": 1,

+   "metadata": {

+    "collapsed": false

+   },

+   "outputs": [

+    {

+     "name": "stdout",

+     "output_type": "stream",

+     "text": [

+      "Example 2.1, Page:46  \n",

+      " \n",

+      "\n",

+      "Engineering Thermodynamics by Onkar Singh,Chapter 2,Example 1\n",

+      "degree celcius and farenheit are related as follows\n",

+      "Tc=(Tf-32)/1.8\n",

+      "so temperature of body in degree celcius 37.0\n"

+     ]

+    }

+   ],

+   "source": [

+    "#cal of temperature of body of human\n",

+    "#intiation of all variables\n",

+    "# Chapter 2\n",

+    "print\"Example 2.1, Page:46  \\n \\n\"\n",

+    "print(\"Engineering Thermodynamics by Onkar Singh,Chapter 2,Example 1\")\n",

+    "Tf=98.6;#temperature of body in farenheit\n",

+    "Tc=(Tf-32)/1.8\n",

+    "print(\"degree celcius and farenheit are related as follows\")\n",

+    "print(\"Tc=(Tf-32)/1.8\")\n",

+    "print(\"so temperature of body in degree celcius\"),round(Tc,2)\n"

+   ]

+  },

+  {

+   "cell_type": "markdown",

+   "metadata": {},

+   "source": [

+    "##example 2.2;pg no:47"

+   ]

+  },

+  {

+   "cell_type": "code",

+   "execution_count": 2,

+   "metadata": {

+    "collapsed": false

+   },

+   "outputs": [

+    {

+     "name": "stdout",

+     "output_type": "stream",

+     "text": [

+      "Example 2.2, Page:47  \n",

+      " \n",

+      "\n",

+      "Engineering Thermodynamics by Onkar Singh,Chapter 2,Example 2\n",

+      "using thermometric relation\n",

+      "t=a*log(p)+(b/2)\n",

+      "for ice point,b/a=\n",

+      "so b=2.1972*a\n",

+      "for steam point\n",

+      "a= 101.95\n",

+      "and b= 224.01\n",

+      "thus, t=in degree celcius\n",

+      "so for thermodynamic property of 6.5,t=302.83 degree celcius= 302.84\n"

+     ]

+    }

+   ],

+   "source": [

+    "#cal of celcius temperature\n",

+    "#intiation of all variables\n",

+    "# Chapter 2\n",

+    "import math\n",

+    "print\"Example 2.2, Page:47  \\n \\n\"\n",

+    "print(\"Engineering Thermodynamics by Onkar Singh,Chapter 2,Example 2\")\n",

+    "t1=0;#ice point temperature in degree celcius\n",

+    "p1=3;#thermometric property for ice point\n",

+    "t2=100;#steam point temperature in degree celcius\n",

+    "p2=8;#thermometric property for steam point\n",

+    "p3=6.5;#thermometric property for any temperature\n",

+    "print(\"using thermometric relation\")\n",

+    "print(\"t=a*log(p)+(b/2)\")\n",

+    "print(\"for ice point,b/a=\")\n",

+    "b=2*math.log(p1)\n",

+    "print(\"so b=2.1972*a\")\n",

+    "print(\"for steam point\")\n",

+    "a=t2/(math.log(p2)-(2.1972/2))\n",

+    "print(\"a=\"),round(a,2)\n",

+    "b=2.1972*a\n",

+    "print(\"and b=\"),round(b,2)\n",

+    "t=a*math.log(p3)+(b/2)\n",

+    "print(\"thus, t=in degree celcius\")\n",

+    "print(\"so for thermodynamic property of 6.5,t=302.83 degree celcius=\"),round(t,2)\n"

+   ]

+  },

+  {

+   "cell_type": "markdown",

+   "metadata": {},

+   "source": [

+    "##example 2.3;page no:47"

+   ]

+  },

+  {

+   "cell_type": "code",

+   "execution_count": 3,

+   "metadata": {

+    "collapsed": false

+   },

+   "outputs": [

+    {

+     "name": "stdout",

+     "output_type": "stream",

+     "text": [

+      "Example 2.3, Page:47  \n",

+      " \n",

+      "\n",

+      "Engineering Thermodynamics by Onkar Singh,Chapter 2,Example 3\n",

+      "emf equation\n",

+      "E=(0.003*t)-((5*10^-7)*t^2))+(0.5*10^-3)\n",

+      "using emf equation at ice point,E_0 in volts\n",

+      "E_0= 0.0\n",

+      "using emf equation at steam point,E_100 in volts\n",

+      "E_100= 0.3\n",

+      "now emf at 30 degree celcius using emf equation(E_30)in volts\n",

+      "now the temperature(T) shown by this thermometer\n",

+      "T=in degree celcius 30.36\n",

+      "NOTE=>In this question,values of emf at 100 and 30 degree celcius is calculated wrong in book so it is corrected above so the answers may vary.\n"

+     ]

+    }

+   ],

+   "source": [

+    "#cal of temperature shown by this thermometer\n",

+    "#intiation of all variables\n",

+    "# Chapter 2\n",

+    "print\"Example 2.3, Page:47  \\n \\n\"\n",

+    "print(\"Engineering Thermodynamics by Onkar Singh,Chapter 2,Example 3\")\n",

+    "print(\"emf equation\")\n",

+    "print(\"E=(0.003*t)-((5*10^-7)*t^2))+(0.5*10^-3)\")\n",

+    "print(\"using emf equation at ice point,E_0 in volts\")\n",

+    "t=0.;#ice point temperature in degree celcius\n",

+    "E_0=(0.003*t)-((5.*10**-7)*t**2)+(0.5*10**-3)\n",

+    "print(\"E_0=\"),round(E_0,2)\n",

+    "print(\"using emf equation at steam point,E_100 in volts\")\n",

+    "t=100.;#steam point temperature in degree celcius\n",

+    "E_100=(0.003*t)-((5.*10**-7)*t**2)+(0.5*10**-3)\n",

+    "print(\"E_100=\"),round(E_100,2)\n",

+    "print(\"now emf at 30 degree celcius using emf equation(E_30)in volts\")\n",

+    "t=30.;#temperature of substance in degree celcius\n",

+    "E_30=(0.003*t)-((5.*10**-7)*t**2)+(0.5*10**-3)\n",

+    "T_100=100.;#steam point temperature in degree celcius\n",

+    "T_0=0.;#ice point temperature in degree celcius\n",

+    "T=((E_30-E_0)/(E_100-E_0))*(T_100-T_0)\n",

+    "print(\"now the temperature(T) shown by this thermometer\")\n",

+    "print(\"T=in degree celcius\"),round(T,2)\n",

+    "print(\"NOTE=>In this question,values of emf at 100 and 30 degree celcius is calculated wrong in book so it is corrected above so the answers may vary.\")\n"

+   ]

+  },

+  {

+   "cell_type": "markdown",

+   "metadata": {},

+   "source": [

+    "##example 2.4;pg no:48"

+   ]

+  },

+  {

+   "cell_type": "code",

+   "execution_count": 4,

+   "metadata": {

+    "collapsed": false

+   },

+   "outputs": [

+    {

+     "name": "stdout",

+     "output_type": "stream",

+     "text": [

+      "Example 2.4, Page:48  \n",

+      " \n",

+      "\n",

+      "Engineering Thermodynamics by Onkar Singh,Chapter 2,Example 4\n",

+      "emf equation,e=0.18*t-5.2*10^-4*t^2 in millivolts\n",

+      "as ice point and steam points are two reference points,so\n",

+      "at ice point,emf(e1)in mV\n",

+      "at steam point,emf(e2)in mV\n",

+      "at gas temperature,emf(e3)in mV\n",

+      "since emf variation is linear so,temperature(t)in degree celcius at emf of 7.7 mV 60.16\n",

+      "temperature of gas using thermocouple=60.16 degree celcius\n",

+      "percentage variation=((t-t3)/t3)*100variation in temperature reading with respect to gas thermometer reading of 50 degree celcius 20.31\n"

+     ]

+    }

+   ],

+   "source": [

+    "#cal of percentage variation in temperature\n",

+    "#intiation of all variables\n",

+    "# Chapter 2\n",

+    "import math\n",

+    "print\"Example 2.4, Page:48  \\n \\n\"\n",

+    "print(\"Engineering Thermodynamics by Onkar Singh,Chapter 2,Example 4\")\n",

+    "t1=0;#temperature at ice point\n",

+    "t2=100;#temperature at steam point\n",

+    "t3=50;#temperature of gas\n",

+    "print(\"emf equation,e=0.18*t-5.2*10^-4*t^2 in millivolts\")\n",

+    "print(\"as ice point and steam points are two reference points,so\")\n",

+    "print(\"at ice point,emf(e1)in mV\")\n",

+    "e1=0.18*t1-5.2*10**-4*t1**2\n",

+    "print(\"at steam point,emf(e2)in mV\")\n",

+    "e2=0.18*t2-5.2*10**-4*t2**2\n",

+    "print(\"at gas temperature,emf(e3)in mV\")\n",

+    "e3=0.18*t3-5.2*10**-4*t3**2\n",

+    "t=((t2-t1)/(e2-e1))*e3\n",

+    "variation=((t-t3)/t3)*100\n",

+    "print(\"since emf variation is linear so,temperature(t)in degree celcius at emf of 7.7 mV\"),round(t,2)\n",

+    "print(\"temperature of gas using thermocouple=60.16 degree celcius\")\n",

+    "print(\"percentage variation=((t-t3)/t3)*100variation in temperature reading with respect to gas thermometer reading of 50 degree celcius\"),round(variation,2)\n"

+   ]

+  },

+  {

+   "cell_type": "markdown",

+   "metadata": {},

+   "source": [

+    "##example 2.5;pg no:48"

+   ]

+  },

+  {

+   "cell_type": "code",

+   "execution_count": 5,

+   "metadata": {

+    "collapsed": false

+   },

+   "outputs": [

+    {

+     "name": "stdout",

+     "output_type": "stream",

+     "text": [

+      "Example 2.5, Page:48  \n",

+      " \n",

+      "\n",

+      "Engineering Thermodynamics by Onkar Singh,Chapter 2,Example 5\n",

+      "let the conversion relation be X=aC+b\n",

+      "where C is temperature in degree celcius,a&b are constants and X is temperature in X degree \n",

+      "at freezing point,temperature=0 degree celcius,0 degree X\n",

+      "so by equation X=aC+b\n",

+      "we get b=0\n",

+      "at boiling point,temperature=100 degree celcius,1000 degree X\n",

+      "conversion relation\n",

+      "X=10*C\n",

+      "absolute zero temperature in degree celcius=-273.15\n",

+      "absolute zero temperature in degree X= -2731.5\n"

+     ]

+    }

+   ],

+   "source": [

+    "#cal of absolute zero temperature\n",

+    "#intiation of all variables\n",

+    "# Chapter 2\n",

+    "print\"Example 2.5, Page:48  \\n \\n\"\n",

+    "print(\"Engineering Thermodynamics by Onkar Singh,Chapter 2,Example 5\")\n",

+    "print(\"let the conversion relation be X=aC+b\")\n",

+    "print(\"where C is temperature in degree celcius,a&b are constants and X is temperature in X degree \")\n",

+    "print(\"at freezing point,temperature=0 degree celcius,0 degree X\")\n",

+    "print(\"so by equation X=aC+b\")\n",

+    "X=0;#temperature in degree X\n",

+    "C=0;#temperature in degree celcius\n",

+    "print(\"we get b=0\")\n",

+    "b=0;\n",

+    "print(\"at boiling point,temperature=100 degree celcius,1000 degree X\")\n",

+    "X=1000;#temperature in degree X\n",

+    "C=100;#temperature in degree celcius\n",

+    "a=(X-b)/C\n",

+    "print(\"conversion relation\")\n",

+    "print(\"X=10*C\")\n",

+    "print(\"absolute zero temperature in degree celcius=-273.15\")\n",

+    "X=10*-273.15\n",

+    "print(\"absolute zero temperature in degree X=\"),round(X,2)\n"

+   ]

+  }

+ ],

+ "metadata": {

+  "kernelspec": {

+   "display_name": "Python 2",

+   "language": "python",

+   "name": "python2"

+  },

+  "language_info": {

+   "codemirror_mode": {

+    "name": "ipython",

+    "version": 2

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+   "file_extension": ".py",

+   "mimetype": "text/x-python",

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