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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": [
+ "\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": [
+ "\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",
+ "\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": {}
+ }
+ ]
+} \ No newline at end of file