{ "metadata": { "name": "", "signature": "sha256:63ec806cc789e6d4e52abc481b4c69dbae196d3a1063d71df1549638fb7b9752" }, "nbformat": 3, "nbformat_minor": 0, "worksheets": [ { "cells": [ { "cell_type": "heading", "level": 1, "metadata": {}, "source": [ "Chapter 2:Properties of Pure Substances" ] }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Ex2.1, PG-28" ] }, { "cell_type": "code", "collapsed": false, "input": [ "# initialization of variables\n", "\n", "m=10; #mass of saturated water in kg\n", "\n", " # All the necessary values are taken from table C.2\n", " \n", "# part (a)\n", " \n", "P=0.001; # Pressure in MPa\n", "vf=0.001; #specific volume of saturated liquid at 0.001 Mpa in Kg/m^3\n", "vg=129.2; # specific volume of saturated vapour at 0.001 Mpa in Kg/m^3\n", "deltaV=m*(vg-vf) # by properties of pure substance \n", "# result\n", "print \"The Volume change at pressure \",(P),\" MPa is\",round(deltaV),\" m^3 \\n\"\n", "\n", "# part (b) \n", "\n", "P=0.26; # Pressure in MPa\n", "vf=0.0011; # specific volume of saturated liquid at 0.26 MPa( it is same from at 0.2 and 0.3 MPa upto 4 decimals)\n", "vg=(P-0.2)*(0.6058-0.8857)/(0.3-0.2)+0.8857; # specific volume of saturated vapour by interpolation of Values at 0.2 MPa and 0.3 MPa\n", "deltaV=m*(vg-vf) # by properties of pure substance \n", "# result\n", "print \"The Volume change at pressure \",(P),\" MPa is\",round(deltaV,2),\" m^3 \\n\"\n", "\n", "# part (c) \n", "P=10; # Pressure in MPa\n", "vf=0.00145; # specific volume of saturated liquid at 10 MPa\n", "vg=0.01803; # specific volume of saturated vapour at 10 MPa\n", "deltaV=m*(vg-vf) # by properties of pure substance \n", "# result\n", "print \"The Volume change at pressure \",(P),\" MPa is\",round(deltaV,4),\" m^3 \\n\"" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "The Volume change at pressure 0.001 MPa is 1292.0 m^3 \n", "\n", "The Volume change at pressure 0.26 MPa is 7.17 m^3 \n", "\n", "The Volume change at pressure 10 MPa is 0.1658 m^3 \n", "\n" ] } ], "prompt_number": 4 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Ex2.2, PG-29" ] }, { "cell_type": "code", "collapsed": false, "input": [ "# initialization of variables\n", "m=4.0 # mass of water in kg\n", "V=1.0 # volume in m^3\n", "T=150 # temperature of water in degree centigrade\n", "\n", "# TABLE C.1 is used for values in wet region\n", "# Part (a)\n", "P=475.8 # pressure in KPa in wet region at temperature of 150 *C\n", "print \" The pressure is\",round(P,2),\" kPa \\n\"\n", "\n", "# Part (b)\n", "#first we determine the dryness fraction\n", "v=V/m # specific volume of water\n", "vg=0.3928 # specific volume of saturated vapour @150 degree celsius\n", "vf=0.00109 # specific volume of saturated liquid @150 degree celsius\n", "x=(v-vf)/(vg-vf); # dryness fraction\n", "mg=m*x; # mass of vapour\n", "print \" The mass of vapour present is\",round(mg,3),\" kg \\n\"\n", "\n", "# Part(c) \n", "Vg=mg*vg; # volume of vapour\n", "print \" The volume of vapour is\",round(Vg,3),\" m^3\"" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ " The pressure is 475.8 kPa \n", "\n", " The mass of vapour present is 2.542 kg \n", "\n", " The volume of vapour is 0.998 m^3\n" ] } ], "prompt_number": 8 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Ex2.3, PG-29" ] }, { "cell_type": "code", "collapsed": false, "input": [ "# initialization of variables\n", "m=2 # mass of water in kg\n", "P=220 # pressure in KPa\n", "x=0.8 # quality of steam\n", "\n", "# Table C.2 is used for values\n", "\n", "vg=(P-200)*(0.6058-0.8857)/(300-200)+0.8857 # specific volume of saturated vapour @ given pressure by interpolating\n", "vf=0.0011 # specific volume of saturated liquid @ 220 KPa\n", "v=vf+x*(vg-vf)# property of pure substance\n", "V=m*v # total volume\n", "#result\n", "print \"The Total volume of the mixture is \",round(V,3),\" m^3\"\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "The Total volume of the mixture is 1.328 m^3\n" ] } ], "prompt_number": 10 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Ex2.4, PG-30" ] }, { "cell_type": "code", "collapsed": false, "input": [ "# initialization of variables\n", "m=2 # mass of water in kg\n", "P=2.2 # pressure in Mpa\n", "T=800 # temperature in degree centigrade\n", " # Table C.3 is used for values\n", "v=0.2467+(P-2)*(0.1972-0.2467)/(2.5-2) # specific volue by interpolatin between 2 and 2.5 MPa\n", "V=m*v # final volume\n", "print \"The Final Volume is\",round(V,3),\" m^3\"\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "The Final Volume is 0.454 m^3\n" ] } ], "prompt_number": 12 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Ex2.5, PG-32" ] }, { "cell_type": "code", "collapsed": false, "input": [ "# initialization of variables\n", "V=0.6 # volume of tyre in m^3\n", "Pgauge=200 # gauge pressure in KPa\n", "T=20+273 # temperature converted to kelvin\n", "Patm=100 # atmospheric pressure in KPa\n", "R=287 # gas constant in Nm/kg.K\n", "Pabs=(Pgauge+Patm)*1000 # calculating absolute pressue in Pa \n", "\n", "m=Pabs*V/(R*T)# mass from ideal gas equation\n", "print \"The Mass of air is\",round(m,2),\" Kg\"\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "The Mass of air is 2.14 Kg\n" ] } ], "prompt_number": 14 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Ex2.6, PG-33" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math\n", "# initialization of variables\n", "T=500+273 # temperature of steam in kelvin\n", "rho=24.0 # density in Kg/m^3\n", "R=0.462 # gas constant from Table B.2\n", "v=1/rho # specific volume and density relation\n", "# PART (a)\n", "P=rho*R*T # from Ideal gas equation\n", "print \" PART (a) The Pressure is \",int(P),\" KPa \\n\"\n", "# answer is approximated in textbook\n", "\n", "# PART (b)\n", "a=1.703 # van der Waal's constant a value from Table B.7\n", "b=0.00169 # van der Waal's constant b value from Table B.7\n", "P=(R*T/(v-b))-(a/v**2) # Pressure from van der Waal's equation\n", "print \" PART (b) The Pressure is \",int(P),\" KPa \\n\"\n", "# answer is approximated in textbook\n", "\n", "# PART (c)\n", "a=43.9 # van der Waal's constant a value from Table B.7\n", "b=0.00117 # van der Waal's constant b value from Table B.7\n", "\n", "P=(R*T/(v-b))-(a/(v*(v+b)*math.sqrt(T))) # Redlich-Kwong equation\n", "print \" PART (c) The Pressure is \",int(P),\" KPa \\n\"\n", "# answer is approximated in textbook\n", "\n", "# PART (d)\n", "Tcr=947.4 # compressibilty temperature from table B.3\n", "Pcr=22100 # compressibility pressure from table B.3\n", "\n", "TR=T/Tcr # reduced temperature\n", "PR=P/Pcr # reduced pressure\n", "Z=0.93 # from compressiblility chart\n", "P=Z*R*T/v # Pressure in KPa\n", "print \" PART (d) The Pressure is \",int(P),\" KPa \\n\"\n", "# answer is approximated in textbook\n", "\n", "# PART (e)\n", "P=8000 # pressure from steam table @ 500*c and v= 0.0417 m^3\n", "print \" PART (e) The Pressure is \",int(P),\" KPa \\n\"\n", "# answer is approximated in textbook\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ " PART (a) The Pressure is 8571 KPa \n", "\n", " PART (b) The Pressure is 7952 KPa \n", "\n", " PART (c) The Pressure is 7934 KPa \n", "\n", " PART (d) The Pressure is 7971 KPa \n", "\n", " PART (e) The Pressure is 8000 KPa \n", "\n" ] } ], "prompt_number": 23 } ], "metadata": {} } ] }