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diff --git a/Fundamentals_Of_Thermodynamics/Chapter3.ipynb b/Fundamentals_Of_Thermodynamics/Chapter3.ipynb
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+{

+ "metadata": {

+  "name": "",

+  "signature": "sha256:6916c81bc8c1ac256991650a35c94801d5197d3ce593f7738220895278cf5f8b"

+ },

+ "nbformat": 3,

+ "nbformat_minor": 0,

+ "worksheets": [

+  {

+   "cells": [

+    {

+     "cell_type": "heading",

+     "level": 1,

+     "metadata": {},

+     "source": [

+      "Chapter3:PROPERTIES OF A PURE SUBSTANCE"

+     ]

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Ex3.3:pg-59"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "#example 3\n",

+      "#determining the quality and specific volume\n",

+      "\n",

+      "v1=0.5 #given specific volume in m^3/kg\n",

+      "vf=0.001073 #specific volume when only liquid phase is present in m^3/kg\n",

+      "vfg=0.60475 #in m^3/kg\n",

+      "x=(v1-vf)/vfg #quality\n",

+      "print\"For water at a pressure of 300 kPa,the state at which v1 is 0.5 m^3/kg is seen to be in the liquid-vapor two-phase region,at which T=133.6 C and the quality is\",round(x,3)\n",

+      "\n",

+      "v2=1 #given specific volume in m^3/kg\n",

+      "\n",

+      " # using the method of interplotation\n",

+      "T=((400-300)*(1.0-0.8753))/(1.0315-0.8753)+300 #temperature of the water\n",

+      "print\"For water at a pressure of 300 kPa,the state at which v2 is 1 m^3/kg is seen to be in the liquid-vapor two-phase region,the temperature is\",round(T,1)"

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "For water at a pressure of 300 kPa,the state at which v1 is 0.5 m^3/kg is seen to be in the liquid-vapor two-phase region,at which T=133.6 C and the quality is 0.825\n",

+        "For water at a pressure of 300 kPa,the state at which v2 is 1 m^3/kg is seen to be in the liquid-vapor two-phase region,the temperature is 379.8\n"

+       ]

+      }

+     ],

+     "prompt_number": 13

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Ex3.4:pg-60"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      " #example 4\n",

+      "#percentage of vapor \n",

+      "\n",

+      "vliq=0.1 #volume of saturated liquid in m^3\n",

+      "vf=0.000843 #in m^3/kg\n",

+      "vvap=0.9 #volume of saturated vapor R-134a in equilbrium\n",

+      "vg=0.02671 #in m^3/kg\n",

+      "mliq=vliq/vf #mass of liquid in kg \n",

+      "mvap=vvap/vg #mass of vapor in kg\n",

+      "m=mliq+mvap #total mass in kg\n",

+      "x=mvap*100/m #percentage of vapor on mass basis\n",

+      "print\"hence,% vapor on mass basis is\",round(x,1),\"%\""

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "hence,% vapor on mass basis is 22.1 %\n"

+       ]

+      }

+     ],

+     "prompt_number": 7

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Ex3.6:pg-61"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "#example 6\n",

+      "#Determinig the missing property\n",

+      "\n",

+      "T1=273-53.2 #given temperature in K\n",

+      "P1=600 #given pressure in kPa\n",

+      "print\"This temperature is higher than the critical temperature (critical temp. at P=600 kPa) is 96.37 K.Hence,v=0.10788 m^3/kg\"\n",

+      "T2=100 #given temp. in K\n",

+      "v2=0.008 #given specific volume in m^3/kg\n",

+      "vf=0.001452 #in m^3/kg\n",

+      "vg=0.0312 #in m^3/kg\n",

+      "Psat=779.2 #saturation pressure in kPa\n",

+      "vfg=vg-vf #in m^3/kg\n",

+      "x=(v2-vf)/vfg #quality\n",

+      "print\"\\n hence, the pressure is\",round(Psat,1),\"kPa\"\n",

+      "print\"\\n and quality is\",round(x,4),\"%\""

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "This temperature is higher than the critical temperature (critical temp. at P=600 kPa) is 96.37 K.Hence,v=0.10788 m^3/kg\n",

+        "\n",

+        " hence, the pressure is 779.2 kPa\n",

+        "\n",

+        " and quality is 0.2201 %\n"

+       ]

+      }

+     ],

+     "prompt_number": 19

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Ex3.7:pg-62"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "#example 7\n",

+      "#determining the pressure of water\n",

+      "\n",

+      "vg=0.12736 #specific volume in m^3/kg for water at 200C\n",

+      "v=0.4 #specific volume in m^3/kg\n",

+      "P1=500 #in kPa\n",

+      "v1=0.42492 #specific volume at P1 in m^3/kg\n",

+      "P2=600 #in kPa\n",

+      "v2=0.35202 #specific volume at P2 in m^3/kg\n",

+      "P=P1+(P2-P1)*(v-v1)/(v2-v1) #calculating pressure by interplotation\n",

+      "print \"hence,the pressure of water is\",round(P,1),\" kPa\""

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        " hence,the pressure of water is 534.2  kPa\n"

+       ]

+      }

+     ],

+     "prompt_number": 21

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Ex3.8:pg-66"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "#example 8\n",

+      "#calculating mass of air\n",

+      "\n",

+      "P=100 #pressure in kPa\n",

+      "V=6*10*4 #volume of room in m^3\n",

+      "R=0.287 #in kN-m/kg-K\n",

+      "T=25 #temperature in Celsius\n",

+      "m=P*V/(R*(T+273.1)) #mass of air contained in room\n",

+      "print\"\\n hence, mass of air contained in room is\",round(m,1),\"kg\""

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "\n",

+        " hence, mass of air contained in room is 280.5 kg\n"

+       ]

+      }

+     ],

+     "prompt_number": 27

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Ex3.9:pg-67"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "#example 9\n",

+      "#calculating pressure inside tank\n",

+      "\n",

+      "V=0.5 #volumr of tank in m^3\n",

+      "m=10 #mass of ideal gas in kg\n",

+      "T=25 #temperature of tank in Celsius\n",

+      "M=24 #molecular mass of gas in kg/kmol\n",

+      "Ru=8.3145 #universal gas constant in kN-m/kmol-K\n",

+      "R=Ru/M #gas constant for given ideal gas in kN-m/kg-K\n",

+      "P=m*R*(T+273.2)/V #pressure inside tank\n",

+      "print\"\\n hence,pressure inside tank is\",round(P),\"kpa\" "

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "\n",

+        " hence,pressure inside tank is 2066.0 kpa\n"

+       ]

+      }

+     ],

+     "prompt_number": 30

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Ex3.10:pg-67"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "#example 10\n",

+      "#mass  flow rate\n",

+      "\n",

+      "dt=185 #time period in seconds over which there is incrrease in volume \n",

+      "dV=0.75 #increase in volume in 0.75 in m^3\n",

+      "V=dV/dt #volume flow rate in m^3/s\n",

+      "P=105 #pressure inside gas bell kPa\n",

+      "T=21 #temperature in celsius\n",

+      "R=0.1889 #ideal gas constant in kJ/kg-K\n",

+      "m=P*V/(R*(T+273.15)) #mass flow rate of the flow in kg/s\n",

+      "print\"\\n hence,mass flow rate is\",round(m,5),\"kg/s\"\n",

+      "print\"\\n and volume flow rate is\",round(V,5),\"m^3/s\"\n",

+      "#The answer of volume flow rate in the book is wrong."

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "\n",

+        " hence,mass flow rate is 0.00766 kg/s\n",

+        "\n",

+        " and volume flow rate is 0.00405 m^3/s\n"

+       ]

+      }

+     ],

+     "prompt_number": 40

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Ex3.12:pg-71"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "#example 12\n",

+      "#determining specific using diffenet laws\n",

+      "\n",

+      "T=100.0 #given temp.in 100 celsius\n",

+      "P=3.0 #given pressure in MPa\n",

+      "v1=0.0065 #specific volume in m^3/kg using table\n",

+      "print\"\\n hence,the specific volume for R-134a using R-134a tables is\",round(v1,3),\"m^3/kg\"\n",

+      "M=102.3 #molecular mass in kg\n",

+      "R=8.3145 #in kJ/K\n",

+      "Ru=R/M #in kJ/K-kg\n",

+      "v2=Ru*(T+273)/(P*1000) #specific volume assuming R-134a to be ideal gas in m^3/kg\n",

+      "print\"\\n hence,the specific volume for R-134a using R-134a the ideal gas laws is\",round(v2,5),\"m^3/kg\"\n",

+      "Tr=373.2/374.2 #reduced temperature using generalized chart\n",

+      "Pr=3.0/4.06 #reduced pressure using generalized chart\n",

+      "Z=0.67 #compressibility factor \n",

+      "v3=Z*v2 # specific volume using generalized chart in m^3/kg\n",

+      "print\"\\n hence,the specific volume for R-134a using the generalized chart is\",round(v3,5),\"m^3/kg\"\n"

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "\n",

+        " hence,the specific volume for R-134a using R-134a tables is 0.006 m^3/kg\n",

+        "\n",

+        " hence,the specific volume for R-134a using R-134a the ideal gas laws is 0.01011 m^3/kg\n",

+        "\n",

+        " hence,the specific volume for R-134a using the generalized chart is 0.00677 m^3/kg\n"

+       ]

+      }

+     ],

+     "prompt_number": 51

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Ex3.13:pg-71"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "#example 13\n",

+      "#calculating mass of gas\n",

+      "\n",

+      "Pc=4250 #critical pressure of propane in kPa\n",

+      "Tc=369.8 #critical temperature in K\n",

+      "T=15 #temperature of propane in celsius\n",

+      "Tr=T/Tc #reduced temperature\n",

+      "Prsat=0.2 # reduced pressure \n",

+      "P=Prsat*Pc #pressure in kPa\n",

+      "x=0.1 #given quality\n",

+      "Zf=0.035 #from graph\n",

+      "Zg=0.83 #from graph\n",

+      "Z=(1-x)*Zf+x*Zg #overall compressibility factor\n",

+      "V=0.1 #volume of steel bottle in m^3\n",

+      "R=0.1887 #in kPa-m^3/kg-K\n",

+      "m=P*V/(Z*R*(T+273)) #total propane mass in kg\n",

+      "print\"\\n hence,the total propane mass is\",round(m,2),\"kg\"\n",

+      "print\"\\n and pressure is\",round(P,2),\"kPa\""

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "\n",

+        " hence,the total propane mass is 13.66 kg\n",

+        "\n",

+        " and pressure is 850.0 kPa\n"

+       ]

+      }

+     ],

+     "prompt_number": 44

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [],

+     "language": "python",

+     "metadata": {},

+     "outputs": []

+    }

+   ],

+   "metadata": {}

+  }

+ ]

+}
\ No newline at end of file