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diff --git a/Fundamentals_Of_Thermodynamics/Chapter11.ipynb b/Fundamentals_Of_Thermodynamics/Chapter11.ipynb
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-{

- "metadata": {

-  "name": "",

-  "signature": "sha256:099777bd4d735e894f7949924f63dca34258026e1381262643494860b41717f6"

- },

- "nbformat": 3,

- "nbformat_minor": 0,

- "worksheets": [

-  {

-   "cells": [

-    {

-     "cell_type": "heading",

-     "level": 1,

-     "metadata": {},

-     "source": [

-      "Chapter11:Power and Refrigeration Systems\u2014With Phase Change"

-     ]

-    },

-    {

-     "cell_type": "heading",

-     "level": 2,

-     "metadata": {},

-     "source": [

-      "Ex11.1:Pg-425"

-     ]

-    },

-    {

-     "cell_type": "code",

-     "collapsed": false,

-     "input": [

-      "#Ques 1\n",

-      "#To determine the efficiency of Rankine cycle\n",

-      "\n",

-      "#1-Inlet state of pump\n",

-      "#2-Exit state of pump\n",

-      "P2=2000;#Exit pressure in kPa\n",

-      "P1=10;#Inlet pressure in kPa\n",

-      "v=0.00101;#specific weight of water in m^3/kg\n",

-      "wp=v*(P2-P1);#work done in pipe in kJ/kg\n",

-      "h1=191.8;#Enthalpy in kJ/kg from table\n",

-      "h2=h1+wp;#enthalpy in kJ/kg\n",

-      "#2-Inlet state for boiler\n",

-      "#3-Exit state for boiler\n",

-      "h3=2799.5;#Enthalpy in kJ/kg\n",

-      "#3-Inlet state for turbine\n",

-      "#4-Exit state for turbine\n",

-      "#s3=s4(Entropy remain same)\n",

-      "s4=6.3409;#kJ/kg\n",

-      "sf=0.6493;#Entropy at liquid state in kJ/kg\n",

-      "sfg=7.5009;#Entropy difference for vapor and liquid state in kJ/kg\n",

-      "x4=(s4-sf)/sfg;#x-factor\n",

-      "hfg=2392.8;#Enthalpy difference in kJ/kg for turbine\n",

-      "h4=h1+x4*hfg;#Enthalpy in kJ/kg\n",

-      "\n",

-      "nth=((h3-h2)-(h4-h1))/(h3-h2);\n",

-      "print\" Percentage efficiency =\",round(nth*100,1)"

-     ],

-     "language": "python",

-     "metadata": {},

-     "outputs": [

-      {

-       "output_type": "stream",

-       "stream": "stdout",

-       "text": [

-        " Percentage efficiency = 30.3\n"

-       ]

-      }

-     ],

-     "prompt_number": 4

-    },

-    {

-     "cell_type": "heading",

-     "level": 2,

-     "metadata": {},

-     "source": [

-      "Ex11.2:Pg-429"

-     ]

-    },

-    {

-     "cell_type": "code",

-     "collapsed": false,

-     "input": [

-      "#Ques 2\n",

-      "#To determine the efficiency of Rankine cycle\n",

-      "\n",

-      "#1-Inlet state of pump\n",

-      "#2-Exit state of pump\n",

-      "P2=4000;#Exit pressure in kPa\n",

-      "P1=10;#Inlet pressure in kPa\n",

-      "v=0.00101;#specific weight of water in m^3/kg\n",

-      "wp=v*(P2-P1);#work done in pipe in kJ/kg\n",

-      "h1=191.8;#Enthalpy in kJ/kg from table\n",

-      "h2=h1+wp;#enthalpy in kJ/kg\n",

-      "#2-Inlet state for boiler\n",

-      "#3-Exit state for boiler\n",

-      "h3=3213.6;#Enthalpy in kJ/kg from table\n",

-      "#3-Inlet state for turbine\n",

-      "#4-Exit state for turbine\n",

-      "#s3=s4(Entropy remain same)\n",

-      "s4=6.7690;#Entropy in kJ/kg from table\n",

-      "sf=0.6493;#Entropy at liquid state in kJ/kg from table\n",

-      "sfg=7.5009;#Entropy difference for vapor and liquid state in kJ/kg from table\n",

-      "x4=(s4-sf)/sfg;#x-factor\n",

-      "hfg=2392.8;#Enthalpy difference in kJ/kg for turbine\n",

-      "h4=h1+x4*hfg;#Enthalpy in kJ/kg\n",

-      "\n",

-      "nth=((h3-h2)-(h4-h1))/(h3-h2);\n",

-      "print\"Percentage efficiency =\",round(nth*100,1)"

-     ],

-     "language": "python",

-     "metadata": {},

-     "outputs": [

-      {

-       "output_type": "stream",

-       "stream": "stdout",

-       "text": [

-        "Percentage efficiency = 35.3\n"

-       ]

-      }

-     ],

-     "prompt_number": 5

-    },

-    {

-     "cell_type": "heading",

-     "level": 2,

-     "metadata": {},

-     "source": [

-      "Ex11.3:Pg-433"

-     ]

-    },

-    {

-     "cell_type": "code",

-     "collapsed": false,

-     "input": [

-      "#Ques 3\n",

-      "#To determine the efficiency of a cycle\n",

-      "\n",

-      "#1-Inlet state of pump\n",

-      "#2-Exit state of pump\n",

-      "P2=4000;#Exit pressure in kPa\n",

-      "P1=10;#Inlet pressure in kPa\n",

-      "v=0.00101;#specific weight of water in m^3/kg\n",

-      "wp=v*(P2-P1);#work done in pipe in kJ/kg\n",

-      "h1=191.8;#Enthalpy in kJ/kg from table\n",

-      "h2=h1+wp;#enthalpy in kJ/kg\n",

-      "#2-Inlet state for boiler\n",

-      "#3-Exit state for Boiler\n",

-      "h3=3213.6;#Enthalpy in kJ/kg from table\n",

-      "#3-Inlet state for high pressure turbine\n",

-      "#4-Exit state for high pressure turbine\n",

-      "#s3=s4(Entropy remain same)\n",

-      "s4=6.7690;#Entropy in kJ/kg from table\n",

-      "sf=1.7766;#Entropy at liquid state in kJ/kg from table\n",

-      "sfg=5.1193;#Entropy difference for vapor and liquid state in kJ/kg from table\n",

-      "x4=(s4-sf)/sfg;#x-factor\n",

-      "hf=604.7#Enthalpy of liquid state in kJ/kg\n",

-      "hfg=2133.8;#Enthalpy difference in kJ/kg for turbine\n",

-      "h4=hf+x4*hfg;#Enthalpy in kJ/kg\n",

-      "#5-Inlet state for low pressure turbine\n",

-      "#6-Exit pressure for low pressure turbine\n",

-      "sf=0.6493;#Entropy in liquid state in kJ/kg for turbine\n",

-      "h5=3273.4;#enthalpy in kJ/kg \n",

-      "s5=7.8985;#Entropy in kJ/kg\n",

-      "sfg=7.5009;#entropy diff in kJ/kg \n",

-      "x6=(s5-sf)/sfg;#x-factor\n",

-      "hfg=2392.8;#enthalpy difference for low pressure turbine in kj/kg\n",

-      "h6=h1+x6*hfg;#entropy in kg/kg\n",

-      "wt=(h3-h4)+(h5-h6);#work output in kJ/kg\n",

-      "qh=(h3-h2)+(h5-h4);\n",

-      "\n",

-      "nth=(wt-wp)/qh;\n",

-      "print\" Percentage efficiency =\",round(nth*100,1)"

-     ],

-     "language": "python",

-     "metadata": {},

-     "outputs": [

-      {

-       "output_type": "stream",

-       "stream": "stdout",

-       "text": [

-        " Percentage efficiency = 35.9\n"

-       ]

-      }

-     ],

-     "prompt_number": 6

-    },

-    {

-     "cell_type": "heading",

-     "level": 2,

-     "metadata": {},

-     "source": [

-      "Ex11.4:Pg-438"

-     ]

-    },

-    {

-     "cell_type": "code",

-     "collapsed": false,

-     "input": [

-      "#ques4\n",

-      "#Efficiency of Refrigeration cycle\n",

-      "\n",

-      "#from previous examples\n",

-      "h1=191.8;#kJ/kg\n",

-      "h5=3213.6;#kg/kg\n",

-      "h6=2685.7;#kJ/kg\n",

-      "h7=2144.1;#kJ/kg\n",

-      "h3=604.7;#kJ/kg\n",

-      "#1-Inlet state of pump\n",

-      "#2-Exit state of pump\n",

-      "P2=400;#Exit pressure in kPa\n",

-      "P1=10;  #Inlet pressure in kPa\n",

-      "v=0.00101;#specific weight of water in m^3/kg\n",

-      "wp1=v*(P2-P1);#work done for low pressure pump in kJ/kg\n",

-      "h1=191.8;#Enthalpy in kJ/kg from table\n",

-      "h2=h1+wp1;#enthalpy in kJ/kg\n",

-      "#5-Inlet state for turbine\n",

-      "#6,7-Exit state for turbine\n",

-      "y=(h3-h2)/(h6-h2);#extraction fraction\n",

-      "wt=(h5-h6)+(1-y)*(h6-h7);#turbine work in kJ/kg\n",

-      "#3-Inlet for high pressure pump\n",

-      "#4-Exit for high pressure pump\n",

-      "P3=400;#kPa\n",

-      "P4=4000;#kPa\n",

-      "v=0.001084;#specific heat for 3-4 process in m^3/kg\n",

-      "wp2=v*(P4-P3);#work done for high pressure pump\n",

-      "h4=h3+wp2;#Enthalpy in kJ/kg\n",

-      "wnet=wt-(1-y)*wp1-wp2;\n",

-      "qh=h5-h4;#Heat output in kJ/kg\n",

-      "nth=wnet/qh;\n",

-      "print\" Refrigerator Efficiency =\",round(nth*100,1)\n"

-     ],

-     "language": "python",

-     "metadata": {},

-     "outputs": [

-      {

-       "output_type": "stream",

-       "stream": "stdout",

-       "text": [

-        " Refrigerator Efficiency = 37.5\n"

-       ]

-      }

-     ],

-     "prompt_number": 7

-    },

-    {

-     "cell_type": "heading",

-     "level": 2,

-     "metadata": {},

-     "source": [

-      "Ex11.5:Pg-443"

-     ]

-    },

-    {

-     "cell_type": "code",

-     "collapsed": false,

-     "input": [

-      "#ques5\n",

-      "#To determine thermal efficiency of cycle\n",

-      "\n",

-      "#5-Inlet state for turbine\n",

-      "#6-Exit state for turbine\n",

-      "#h-Enthalpy at a state \n",

-      "#s-Entropy at a state\n",

-      "#from steam table\n",

-      "h5=3169.1;#kJ/kg\n",

-      "s5=6.7235;#kJ/kg\n",

-      "s6s=s5;\n",

-      "sf=0.6493;#Entropy for liquid state in kJ/kg\n",

-      "sfg=7.5009;#Entropy difference in kJ/kg\n",

-      "hf=191.8;#kJ/kg\n",

-      "hfg=2392.8;#Enthalpy difference in kJ/kg\n",

-      "x6s=(s6s-sf)/sfg;#x-factor\n",

-      "h6s=hf+x6s*hfg;#kJ/Kg at state 6s\n",

-      "nt=0.86;#turbine efficiency given\n",

-      "wt=nt*(h5-h6s);\n",

-      "#1-Inlet state for pump\n",

-      "#2-Exit state for pump\n",

-      "np=0.80;#pump efficiency given\n",

-      "v=0.001009;#specific heat in m^3/kg\n",

-      "P2=5000;#kPa\n",

-      "P1=10;#kPa\n",

-      "wp=v*(P2-P1)/np;#Work done in pump in kJ/kg\n",

-      "wnet=wt-wp;#net work in kJ/kg\n",

-      "#3-Inlet state for boiler\n",

-      "#4-Exit state for boiler\n",

-      "h3=171.8;#in kJ/kg from table\n",

-      "h4=3213.6;#kJ/kg from table\n",

-      "qh=h4-h3;\n",

-      "nth=wnet/qh;\n",

-      "print \"Cycle Efficiency =\",round(nth*100,1)"

-     ],

-     "language": "python",

-     "metadata": {},

-     "outputs": [

-      {

-       "output_type": "stream",

-       "stream": "stdout",

-       "text": [

-        "Cycle Efficiency = 29.2\n"

-       ]

-      }

-     ],

-     "prompt_number": 8

-    },

-    {

-     "cell_type": "heading",

-     "level": 2,

-     "metadata": {},

-     "source": [

-      "Ex11.6:Pg-451"

-     ]

-    },

-    {

-     "cell_type": "code",

-     "collapsed": false,

-     "input": [

-      "#ques6\n",

-      "#to determine the rate of refrigeration\n",

-      "\n",

-      "# refer to fig 11.21 in book\n",

-      "mdot=0.03 # mass flow rate in Kg/s\n",

-      "T1=-20 # temperature in evaporator in celsius\n",

-      "T3=40 #temperature in evaporator in Celsius\n",

-      "P2=1017 # saturation pressure in KPa\n",

-      "\n",

-      "# from table of R-134a refrigerant\n",

-      "h1=386.1 # enthalpy at state 1 in kJ/kg,\n",

-      "S1=1.7395 # entropy at state 1 in kJ/kg.K\n",

-      "S2=S1 # isentropic process\n",

-      "T2=47.7# corresponding value to S2 in table of R-134a in degree celsius\n",

-      "h2=428.4 # corresponding value to S2 in table of R-134a in kJ/kg\n",

-      "wc=h2-h1 # work done in compressor in kJ/kg\n",

-      "h4=h3=256.5 #enthalpy at state 4 and 3 in kJ/kg\n",

-      "qL=h1-h4 #Heat rejected in kJ/kg\n",

-      "\n",

-      "B=qL/wc # COP\n",

-      "\n",

-      "print\" the COP of the plant is\",round(B,2)\n",

-      "print\" the refrigeration rate is\",round(mdot*qL,2)"

-     ],

-     "language": "python",

-     "metadata": {},

-     "outputs": [

-      {

-       "output_type": "stream",

-       "stream": "stdout",

-       "text": [

-        " the COP of the plant is 3.06\n",

-        " the refrigeration rate is 3.89\n"

-       ]

-      }

-     ],

-     "prompt_number": 16

-    },

-    {

-     "cell_type": "heading",

-     "level": 2,

-     "metadata": {},

-     "source": [

-      "Ex11.7:Pg-454"

-     ]

-    },

-    {

-     "cell_type": "code",

-     "collapsed": false,

-     "input": [

-      "#ques7\n",

-      "#to determine the COP of cycle\n",

-      "\n",

-      "P1=125 # pressure at state 1 in kPa\n",

-      "P2=1.2 # pressure at state 2 in MPa\n",

-      "P3=1.19 # pressure at state 3 in MPa,\n",

-      "P4=1.16 # pressure at state 4 in MPa,\n",

-      "P5=1.15 # pressure at state 5 in MPa,\n",

-      "P6=P7=140 # pressure at state 6 and 7 in kPa,\n",

-      "P8=130 # pressure at state 8 in kPa,\n",

-      "T1=-10 #temperaure at state  1 in \u25e6C\n",

-      "T2=100 #temperaure at state  2 in \u25e6C\n",

-      "T3=80 #temperaure at state  3 in \u25e6C\n",

-      "T4=45 #temperaure at state  4 in \u25e6C\n",

-      "T5=40 #temperaure at state  5 in \u25e6C\n",

-      "T8=-20 #temperaure at state  8 in \u25e6C\n",

-      "q=-4 # heat transfer in kJ/Kg\n",

-      "\n",

-      "#x6=x7 quality condition given in question\n",

-      "\n",

-      "\n",

-      "# the following values are taken from table for refrigerant R-134a\n",

-      "h1=394.9 # enthalpy at state 1 in kJ/kg\n",

-      "h2=480.9 # enthalpy at state 2 in kJ/kg\n",

-      "h8=386.6 # enthalpy at state 8 in kJ/kg\n",

-      "wc=h2-h1-q # from first law\n",

-      "h5=h6=h7=256.4 # as x6=x7 and from table at state 5 in Kj/Kg\n",

-      "qL=h8-h7 # from first law \n",

-      "B=qL/wc # COP\n",

-      "print\" the COP of the plant is\",round(B,3)\n",

-      "\n"

-     ],

-     "language": "python",

-     "metadata": {},

-     "outputs": [

-      {

-       "output_type": "stream",

-       "stream": "stdout",

-       "text": [

-        " the COP of the plant is 1.447\n"

-       ]

-      }

-     ],

-     "prompt_number": 20

-    },

-    {

-     "cell_type": "code",

-     "collapsed": false,

-     "input": [],

-     "language": "python",

-     "metadata": {},

-     "outputs": []

-    }

-   ],

-   "metadata": {}

-  }

- ]

-}
\ No newline at end of file