{
"metadata": {
"name": "",
"signature": "sha256:d3b8a40a5268a38ad3fc311ebed2078460c9192ee65f2f5e3922c8f65a52bf49"
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"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-22"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"# initialization of variables\n",
"\n",
"m=1; #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,1),\" m^3/kg \\n\"\n",
"\n",
"# part (b) \n",
"\n",
"P=0.10; # Pressure in MPa\n",
"vf=0.001; # 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=1.694; # specific volume of saturated vapour at 0.1 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,3),\" m^3/kg \\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,5),\" m^3/kg \\n\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The Volume change at pressure 0.001 MPa is 129.2 m^3/kg \n",
"\n",
"The Volume change at pressure 0.1 MPa is 1.693 m^3/kg \n",
"\n",
"The Volume change at pressure 10 MPa is 0.01658 m^3/kg \n",
"\n"
]
}
],
"prompt_number": 8
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Ex2.2:pg-23"
]
},
{
"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,1),\" 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=vg*mg; # volume of vapour\n",
"print \" The volume of vapour is\",round(Vg,4),\" 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 0.9984 m^3\n"
]
}
],
"prompt_number": 12
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Ex2.3:pg-23"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"# initialization of variables\n",
"m=4 # 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 at 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\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The Total volume of the mixture is 2.656 m^3\n"
]
}
],
"prompt_number": 5
},
{
"cell_type": "heading",
"level": 1,
"metadata": {},
"source": [
"Ex2.4:pg-23"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"# initialization of variables\n",
"m=2 # mass of water in lb\n",
"P=540 # pressure in psi\n",
"T=700 # temperature in degree fahrenheit\n",
" # Table C-3E is used for values\n",
"v=1.3040+(P-500)*(1.0727-1.3030)/(600-500) # specific volue by interpolatin between 500 and 600 psi\n",
"V=m*v # final volume\n",
"print \"The Final Volume is\",round(V,3),\" ft^3\"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The Final Volume is 2.424 ft^3\n"
]
}
],
"prompt_number": 9
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Ex2.5:pg-25"
]
},
{
"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": 10
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Ex2.6:pg-26"
]
},
{
"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-8\n",
"b=0.00169 # van der Waal's constant b value from Table B-8\n",
"P=(R*T/(v-b))-(a/v**2) # Pressure from van der Waals 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-8\n",
"b=0.00117 # van der Waal's constant b value from Table B-8\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"
],
"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": 13
}
],
"metadata": {}
}
]
}