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{
"cells": [
{
"cell_type": "heading",
"level": 1,
"metadata": {},
"source": [
"Chapter 6 : Introduction to Material Balances"
]
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
"Example 6.1 Page no. 142\n"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"from numpy import matrix\n",
"\n",
"# Variables\n",
"# Given\n",
"P_O = 89 ; # Premium octane -[octane/gal]\n",
"S_O = 93 ; # Supereme octane - [octane/gal]\n",
"R_O = 87 ; # Regular octane - [octane/gal]\n",
"CP = 1.269 ; # Cost of premium octane -[$/gal]\n",
"SP = 1.349 ; # Cost of supereme octane -[$/gal]\n",
"RP = 1.149 ; # Cost of regular octane -[$/gal]\n",
"\n",
"# Let x and y fraction of regular octane and supreme octane is blended respectively,therefore: x + y = 1 ...(a)\n",
"# and 89 = 87x + 93y ...(b)\n",
"# Solve equations (a) and (b) simultaneously\n",
"# Calculation\n",
"a = matrix([[1,1],[87,93]]) ; # Matrix of coefficients of unknown\n",
"b = matrix([[1.0],[89.0]]) ; # Matrix of constant\n",
"a = a.I\n",
"c = a * b\n",
"cost = c[0]*RP + c[1]*SP ; # Cost after blending - [$/gal]\n",
"sv = CP - cost ; # Save on blending - [$/gal]\n",
"\n",
"# Result\n",
"# Check whether there is loss or save\n",
"if (sv<0):\n",
" print 'We will not save money by blending.'\n",
"else:\n",
" print 'We will save money by blending, and save is %.3f $/gal.'%sv"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"We will save money by blending, and save is 0.053 $/gal.\n"
]
}
],
"prompt_number": 2
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
" Example 6.2 Page no. 147\n"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"\n",
"# Variables\n",
"fd= 1000.0 ; #feed rate-[L/hr]\n",
"cfd= 500.0; #Weight of cells per litre- [mg/L]\n",
"dn= 1.0 ; #Density of feed-[g/cm**3]\n",
"wp= 50.0 ; # Weight percent of cells in product stream\n",
"\n",
"# Calculation and Result\n",
"Pg=(fd*cfd*dn)/(1000*wp*.01) ; # Mass balance for cells \n",
"print ' Product flow(P) per hour is %.1f g'%Pg\n",
"Dg= (fd*dn*1000) - Pg*(wp*.01) ; # Mass balance for the fluid\n",
"print ' Discharge flow per hour is %.3e g'%Dg"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
" Product flow(P) per hour is 1000.0 g\n",
" Discharge flow per hour is 9.995e+05 g\n"
]
}
],
"prompt_number": 3
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
"Example 6.3 Page no. 154\n"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"dn = 0.80 ; #Density of motor oil-[g/cm**3]\n",
"\n",
"# Calculation and Result\n",
"in_ms = (10000*(0.1337)*62.4*dn) ; # Initial mass of motor oil in the tank -[lb]\n",
"print ' Initial mass of motor oil in the tank is %.1f lb'%in_ms\n",
"\n",
"m_fr = .0015 ; #Mass fractional loss\n",
"print ' Mass fractional loss is %.4f '%m_fr\n",
"\n",
"Dsg = m_fr*in_ms ; # Mass balance for the fluid\n",
"print ' Discharge of motor oil on flushing flow for 10000 gal motor oil is %.1f lb'%Dsg"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
" Initial mass of motor oil in the tank is 66743.0 lb\n",
" Mass fractional loss is 0.0015 \n",
" Discharge of motor oil on flushing flow for 10000 gal motor oil is 100.1 lb\n"
]
}
],
"prompt_number": 4
},
{
"cell_type": "code",
"collapsed": true,
"input": [],
"language": "python",
"metadata": {},
"outputs": []
}
],
"metadata": {}
}
]
}
|
