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{
"metadata": {
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"worksheets": [
{
"cells": [
{
"cell_type": "heading",
"level": 1,
"metadata": {},
"source": [
"Chapter 28 : Heats of Solution and Mixing"
]
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
" Example 28.1 page no. 869\n"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"# Application of Heats of Solution data\n",
"\n",
"# Solution \n",
"# Variables\n",
"Ref_T = 77. ;\t\t\t#Reference temperature-[degree F]\n",
"\n",
"#(a)\n",
"mol_NH3 = 1. ;\t\t\t# Moles of NH3 - [lb mol]\n",
"mw_NH3 = 17. ;\t\t\t#Molecular t. of NH3 -[lb]\n",
"mw_H2O = 18. ;\t\t\t#Molecular t. of H2O -[lb]\n",
"\n",
"# Calculations and Results\n",
"f1_NH3 = 3./100 ;\t\t\t# Fraction of NH3 in solution \n",
"m_H2O = (mw_NH3/f1_NH3) - mw_NH3 ;\t\t\t# Mass of water in solution -[lb]\n",
"mol_H2O = m_H2O/mw_H2O ;\t\t\t# Moles of H2O in solution -[lb mol]\n",
"\n",
"print '(a) Moles of H2O in solution is %.1f lb mol . '%mol_H2O\n",
"print ' As we can see that moles of water is 30 lb mol(approx), hence we will see H_soln from table corresponding to 30 lb mol water . '\n",
"H_soln = -14800. ;\t\t\t# From table given in question in book -[Btu/lb mol NH3]\n",
"print ' The amount of cooling needed is, %.0f Btu heat removed. '%(abs(H_soln))\n",
"\n",
"#(b)\n",
"V = 100. ;\t\t\t# Volume of solution produced -[gal]\n",
"f2_NH3 = 32./100 ;\t\t\t# Fraction of NH3 in solution \n",
"sg_NH3 = .889 ;\t\t\t# Specific gravity of NH3 \n",
"sg_H2O = 1.003 ;\t\t\t# Specific gravity of H2O\n",
"d_soln = sg_NH3*62.4*sg_H2O*100/7.48 ;\t\t\t# Density of solution - [lb / 100 gal]\n",
"NH3 = d_soln*f2_NH3/mw_NH3 ;\t\t\t# Mass of NH3 - [ lb mol/ 100 gal]\n",
"m1_H2O = (mw_NH3/f2_NH3) - mw_NH3 ;\t\t\t# Mass of water in solution -[lb]\n",
"mol1_H2O = m1_H2O/mw_H2O ;\t\t\t# Moles of H2O in solution -[lb mol]\n",
"\n",
"print ' (b) Moles of H2O in solution is %.1f lb mol . '%mol1_H2O\n",
"print ' As we can see that moles of water is 2 lb mol , hence we will see H_soln from table corresponding to 2 lb mol water . '\n",
"H_soln = -13700 ;\t\t\t# From table given in question in book -[Btu/lb mol NH3]\n",
"total_H = abs(NH3*H_soln) ;\t\t\t# Total heat removed from solution -[Btu]\n",
"print ' The amount of cooling needed is, %.0f Btu heat removed. '%total_H\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"(a) Moles of H2O in solution is 30.5 lb mol . \n",
" As we can see that moles of water is 30 lb mol(approx), hence we will see H_soln from table corresponding to 30 lb mol water . \n",
" The amount of cooling needed is, 14800 Btu heat removed. \n",
" (b) Moles of H2O in solution is 2.0 lb mol . \n",
" As we can see that moles of water is 2 lb mol , hence we will see H_soln from table corresponding to 2 lb mol water . \n",
" The amount of cooling needed is, 191826 Btu heat removed. \n"
]
}
],
"prompt_number": 1
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
"Example 28.2 page no. 872"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"# Application of the heat of solution data\n",
"\n",
"from scipy.integrate import quad\n",
"# Variables\n",
"p = 100. ;\t\t\t# Mass of product - [kg]\n",
"f_HCl = 25./100 ;\t\t\t#Fraction of HCl in product \n",
"HCl = f_HCl*p ;\t\t\t# Mass of HCl in product - [kg]\n",
"H2O = (1.-f_HCl)*p ;\t\t\t# Mass of H2O in product -[kg]\n",
"mw_HCl = 36.37 ;\t\t\t# Molecular weight of HCl -[kg]\n",
"mw_H2O = 18.02 ;\t\t\t# Molecular weight of H2O -[kg]\n",
"\n",
"# Calculations\n",
"mol_HCl = HCl /mw_HCl ;\t\t\t# Moles of HCl - [kg mol]\n",
"mol_H2O = H2O /mw_H2O; \t\t\t# Moles of H2O - [kg mol]\n",
"total_mol = mol_HCl + mol_H2O ;\t\t\t# Total no. of moles -[kg mol]\n",
"mf_HCl = mol_HCl / total_mol ;\t\t\t# mole fraction of HCl \n",
"mf_H2O = mol_H2O / total_mol ; \t\t\t# mole fraction of H2O\n",
"mr = mol_H2O/mol_HCl ;\t\t\t# Mole ratio of H2O to HCl \n",
"MW = mf_HCl*mw_HCl + mf_H2O*mw_H2O ;\t\t\t# Molecular t. of solution-[kg]\n",
"Ref_T = 25. ;\t\t\t#Reference temperature-[degree C]\n",
"\n",
"mol1_HCl = total_mol ;\t\t\t# Moles of HCl \t\t\t# Moles of HCl output -[g mol]\n",
"Hf1_HCl = -157753. ;\t\t\t# Heat of formation of HCl output-[J/ g mol HCl ]\n",
"Hf_HCl = -92311. ;\t\t\t# Heat of formation of HCl input-[J/ g mol HCl ]\n",
"Hf_H2O = 0 ;\t\t\t# Heat of formation of H2O input-[J/ g mol HCl ]\n",
"H1_HCl = 556. ;\t\t\t# Change in enthalpy during temperature change from 25 C to 35 C of HCl - [J/g mol] \n",
"\n",
"def f(T):\n",
" return (29.13 - 0.134*.01*T)\n",
"\n",
"H_HCl = quad(f,298,393)[0]\t# Change in enthalpy during temperature change from 25 C to 120 C of HCl - [J/g mol] \n",
"\n",
"H_H2O = 0 ;\t\t\t# Change in enthalpy during temperature change from 25 C to 25 C of H2O - [J/g mol] \n",
"\n",
"H_in = (Hf_HCl + H_HCl)*mol_HCl + (Hf_H2O + H_H2O)*mol_H2O ;\t\t\t# Enthalpy change of input -[J]\n",
"H_out = Hf1_HCl*mol_HCl +H1_HCl*mol1_HCl ;\t\t\t# Enthalpy change of output -[J]\n",
"\n",
"del_H = H_out - H_in ;\t\t\t# Net enthalpy change n process - [J]\n",
"Q = del_H; \t\t\t# By energy balance - [J]\n",
"\n",
"# Results\n",
"print 'The amount of heat removed from the absorber by cooling water is, %.0f J. '%Q\n",
"print 'It Seems answer is wrong in book'\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The amount of heat removed from the absorber by cooling water is, -44159 J. \n",
"It Seems answer is wrong in book\n"
]
}
],
"prompt_number": 7
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
" Example 28.3 page no. 875\n"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"# Application of an Enthalpy Concentration Chart\n",
"\n",
"# Solution fig. 28.3\n",
"\n",
"# Variables\n",
"soln1 = 600. ; \t\t\t# Mass flow rate of entering solution 1 -[lb/hr]\n",
"c1_NaOH = 10./100 ;\t\t\t# Fraction of NaOH in entering solution 1\n",
"T1 = 200. ;\t\t\t# Temperature at entry \n",
"soln2 = 400. ;\t\t\t# Mass flow rate of another solution 2 entering -[lb/hr]\n",
"c2_NaOH = 50./100 ;\t\t\t# Fraction of NaOH in another entering solution 2\n",
"\n",
"# Calculations\n",
"F = soln1 + soln2; \t\t\t# Mass flow rate of final solution - [lb/hr]\n",
"F_NaOH = c1_NaOH * soln1 + c2_NaOH * soln2 ;\t\t\t# Mass of NaOH in final solution-[lb]\n",
"F_H2O = F - F_NaOH ;\t\t\t# Mass of H2O in final solution-[lb]\n",
"H_soln1 = 152. ;\t\t\t# Specific enthalpy change for solution 1-[Btu/lb]\n",
"H_soln2 = 290. ;\t\t\t# Specific enthalpy change for solution 2-[Btu/lb]\n",
"H_F = (soln1*H_soln1 + soln2*H_soln2)/F ;\t\t\t# Specific enthalpy change for final solution -[Btu/lb]\n",
"\n",
"# Results\n",
"print ' (a) The final temperature of the exit solution from figure E28.3 using the obtained condition of final solution is 232 degree F '\n",
"\n",
"cF = F_NaOH*100/F; \t\t\t# Concentration of final solution -[wt % NaOH ]\n",
"print ' (b) The concentration of final solution is %.0f wt.%% NaOH . '%cF\n",
"\n",
"x = (F*H_F - F*175)/(1158.0 - 175) ;\t\t\t# H2O evaporated per hour -[lb]\n",
"print ' (c) H2O evaporated per hour is %.1f lb . '%x\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
" (a) The final temperature of the exit solution from figure E28.3 using the obtained condition of final solution is 232 degree F \n",
" (b) The concentration of final solution is 26 wt.% NaOH . \n",
" (c) H2O evaporated per hour is 32.8 lb . \n"
]
}
],
"prompt_number": 3
},
{
"cell_type": "code",
"collapsed": true,
"input": [],
"language": "python",
"metadata": {},
"outputs": [],
"prompt_number": 3
}
],
"metadata": {}
}
]
}
|
