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{
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"worksheets": [
{
"cells": [
{
"cell_type": "heading",
"level": 1,
"metadata": {},
"source": [
"Chapter 11 : Liquefaction of Gases"
]
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 11.1 Page No : 195"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"#Given\n",
"P1 = 8.74;#Initial pressure in Kgf/sq cm\n",
"P2 = 2.41;#Final pressure in Kgf/sq cm\n",
"H1 = 327.13;#Enthalpy of inlet stream in Kcal/Kg\n",
"Hl = 26.8;#Enthalpy of liquid at the final condition in Kcal/Kg\n",
"H2 = H1#Enthalpy of exit stream in Kcal/Kg ,math.since throttling is isenthalpic\n",
"Hg = 340.3;#Enthalpy of gas at the final condition in Kcal/Kg\n",
"vl = 152*10**-5;#Specific volume of liquid at the final condition in cubic meter/Kg\n",
"vg = 0.509;#Specific volume of gas at the final condition in cubic meter/Kg\n",
"v1 = 0.1494;#Initial specific volume in cubic meter/Kg\n",
"\n",
"#To Calculate the dryness fraction of exit stream and the ratio of upstream to downstream diameters\n",
"#(i)Calculation of the dryness fraction of exit stream\n",
"#From equation 3.13(a) (page no 82)\n",
"x = (H2- Hl)/(Hg-Hl);\n",
"print \"i)The dryness fraction of the exit stream is %f\"%(x);\n",
"\n",
"#(ii)Calculation of the ratio of upstream to downstream pipe diameters\n",
"#From equation 3.13(b) (page no 82)\n",
"v2 = (vl*(1-x))+(x*vg);#Total specific volume at the final condition in cubic meter/Kg\n",
"#u1 = u2; math.since KE changes are negligible\n",
"#From continuity equation: A2/A1 = D2**2/D1**2 = v2/v1 ; let required ratio,r = D2/D1;\n",
"r = (v2/v1)**(1/2);\n",
"print \" ii)The ratio of upstream to downstream diameters is %f\"%(r);\n",
"#end\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"i)The dryness fraction of the exit stream is 0.957990\n",
" ii)The ratio of upstream to downstream diameters is 1.000000\n"
]
}
],
"prompt_number": 1
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 11.2 Page No : 199"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Given\n",
"P1 = 1000*1.033*10**4;#Initial pressure in Kgf/sq m\n",
"P2 = 1*1.033*10**4;#Final pressure in Kgf/sq m\n",
"T1 = 300.0;#Inital temperature in K\n",
"Cp = 7.0;#Specific heat of the gas in Kcal/Kgmole K\n",
"#Gas obeys the relation: v = (R*T)/P+(b*(T**2))\n",
"b = 5.4392*10**-8;#in cubic meter/Kgmole K**2\n",
"\n",
"#To Calculate the temperature of the throttled gas\n",
"#From equation (a) (page no 212);which we got after integration \n",
"T2 = 1/((1/T1)-((b/Cp)*((P2-P1)/427)));\n",
"print \"The throttled gas is cooled to %f K\"%(T2);\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The throttled gas is cooled to 284.000191 K\n"
]
}
],
"prompt_number": 2
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 11.3 Page No : 203"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Given\n",
"#From the figure 11.8 (page no 216) & from figure A.2.7\n",
"H3 = 0.0;\n",
"H7 = -47.0;#in Kcal/Kg\n",
"H6 = -93.0;#in Kcal/Kg\n",
"H8 = 7.0;#in Kcal/Kg\n",
"\n",
"#To Calculate the fraction of air liquified at steady state and temperature of air before throttling\n",
"#(i)Calculation of fraction of air liquified\n",
"#From equation 11.3 (page no 215)\n",
"x = (H8-H3)/(H8-H6);\n",
"print \"The fraction of air liquified is %f\"%(x);\n",
"\n",
"#(ii)Calculation of temperature \n",
"H4 = H3+(H7*(1-x))-(H8*(1-x));#enthalpy of the gas before throttling\n",
"#From figure A.2.7 temperature corresponds to pressure 160 atm and the enthalpy H4 is\n",
"T = -112;\n",
"print \" The temperature of air before throttling is %d deg celsius\"%(T);\n",
"#end\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The fraction of air liquified is 0.070000\n",
" The temperature of air before throttling is -112 deg celsius\n"
]
}
],
"prompt_number": 3
}
],
"metadata": {}
}
]
}
|
