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{
"metadata": {
"name": ""
},
"nbformat": 3,
"nbformat_minor": 0,
"worksheets": [
{
"cells": [
{
"cell_type": "heading",
"level": 1,
"metadata": {},
"source": [
"Chapter 9 : Agitation"
]
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
"Example 9.3 - Page No :389\n"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"# Calculate the power per unit volume and the torque per unit volume\n",
"\n",
"import math\n",
"\n",
"# Variables\n",
"Nblades = 4.; \t\t\t # no. of blades\n",
"d = 9/12.; \t\t\t #[ft] - diameter of the impeller\n",
"dt = 30/12.; \t\t\t #[ft] - diameter of the math.tank\n",
"Nbaffles = 4. \t\t\t # no. of baffles\n",
"h = 30.; \t \t\t # [inch] - height of unit\n",
"mu = 10.; \t\t\t #[cP] - vismath.cosity of fluid_\n",
"sg = 1.1; \t\t \t # specific gravity of fluid_\n",
"s = 300. \t\t\t #[rpm] - speed of agitator\n",
"CbyT = 0.3; \n",
"\n",
"# Calculations\n",
"V = (math.pi*dt**3)/4; \t #volume of math.tank in ft**3\n",
"V1 = V*7.48; \t\t\t #[gal] - volume of math.tank in gallons\n",
"mu = mu*(6.72*10**-4); #[lb/ft*sec]\n",
"p = sg*62.4; \t\t\t #[lb/ft**3] - density of fluid_\n",
"N = s/60.; \t\t\t #[rps] - impeller speed in revolutions per second\n",
"Nre = ((d**2)*N*p)/mu;\n",
"\n",
"# Results\n",
"print \"Nre = %.2e\"%Nre\n",
"print \" Therefore the agitator operates in the turbulent region\"\n",
"Npo = 1.62;\n",
"gc = 32.174;\n",
"P = (Npo*(p*(N**3)*(d**5)))/(gc*550);\n",
"Cf = 63025.;\n",
"Tq = (P/s)*Cf;\n",
"PbyV = P/V;\n",
"PbyV1 = P/V1;\n",
"TqbyV = Tq/V;\n",
"TqbyV1 = Tq/V1;\n",
"print \" The power per unit volume and the torque per unit volume is \\nP/V = %.2ef hp/ft**3 = %.2e \\\n",
"hp/gal \\nTq/V = %.2f in*lb/ft**3 = %.3f in*lb/gal\"%(PbyV,PbyV1,TqbyV,TqbyV1);"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Nre = 2.87e+04\n",
" Therefore the agitator operates in the turbulent region\n",
" The power per unit volume and the torque per unit volume is \n",
"P/V = 1.52e-02f hp/ft**3 = 2.03e-03 hp/gal \n",
"Tq/V = 3.19 in*lb/ft**3 = 0.427 in*lb/gal\n"
]
}
],
"prompt_number": 6
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
"Example 9.4 - Page No :391\n"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"# designing a reactor of capacity\n",
"\n",
"# Variables\n",
"Tpilot = 30.;\n",
"Tlab = 10.;\n",
"N1 = 690.;\n",
"N2 = 271.;\n",
"D2 = 3.;\n",
"D1 = 1.;\n",
"\n",
"# Calculations\n",
"n = (math.log(N1/N2))/(math.log(D2/D1));\n",
"V = 12000/7.48; \t\t\t #[ft**3]\n",
"T = ((4.*V)/math.pi)**(1./3); \t\t\t #[ft]\n",
"R = 12.69/(30/12.);\n",
"N3 = N2*(1./R)**n; \t\t\t #[rpm] - impeller speed in the reactor\n",
"\n",
"# Results\n",
"print \"impeller speed in rpm = %f\"%round(N3,4)\n",
"D3 = 0.75*R; \t\t\t #[ft] - reactor impeller diameter\n",
"print \"reactor impeller diameter in ft = %.3f\"%D3\n",
"P = 0.1374*((N3/N2)**3)*(R**5);\n",
"print \"power in hp = %.3f\"%P\n",
"Cf = 63025.;\n",
"Tq = (P/N3)*Cf; \t\t\t #[inch*lb]\n",
"print \"torque in inch*lb = %.0f\"%Tq\n",
"print \"At this point, the design is complete. \\nA sarc ard size impeller would be chosen as \\\n",
" well as a tan ard size motor7.5 hp or 10 hp\"\n",
"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"impeller speed in rpm = 68.044500\n",
"reactor impeller diameter in ft = 3.807\n",
"power in hp = 7.329\n",
"torque in inch*lb = 6789\n",
"At this point, the design is complete. \n",
"A sarc ard size impeller would be chosen as well as a tan ard size motor7.5 hp or 10 hp\n"
]
}
],
"prompt_number": 19
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
"Example 9.5 - Page No : 393\n"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"'''\n",
"Suppose the pilot plant data in Table 9.8 never existed. Use the\n",
"laboratory data only from Example 9.4 to design the 12 000-gal reactor.\n",
"'''\n",
"\n",
"from numpy import *\n",
"\n",
"\n",
"# Variables\n",
"# given\n",
"n = array([0.5, 0.6, 0.7, 0.8, 0.9, 1.0]);\n",
"D2 = 3.806;\n",
"D1 = 0.25;\n",
"R = D2/D1;\n",
"N1 = 690.;\n",
"\n",
"# Calculations\n",
"N2 = N1*((D1/D2)**n);\n",
"P1 = 9.33*10**-3; \t\t\t #[hp]\n",
"P2 = P1*R**(5.-3*n);\n",
"\n",
"# Results\n",
"print \" n N,rpm P,hp\"\n",
"for i in range(6):\n",
" print \" %f %4.0f %4.0f\"%(n[i],N2[i],P2[i]);\n",
"\n",
"\n",
"# Answers may be differ because of rounding error."
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
" n N,rpm P,hp\n",
" 0.500000 177 128\n",
" 0.600000 135 57\n",
" 0.700000 103 25\n",
" 0.800000 78 11\n",
" 0.900000 60 5\n",
" 1.000000 45 2\n"
]
}
],
"prompt_number": 29
}
],
"metadata": {}
}
]
}
|
