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diff --git a/Problems_In_Hydraulics/ch2.ipynb b/Problems_In_Hydraulics/ch2.ipynb new file mode 100755 index 00000000..d68673ef --- /dev/null +++ b/Problems_In_Hydraulics/ch2.ipynb @@ -0,0 +1,222 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:2c442a1cbc28b933d555165b6cb09fa7f45de31e28c837593e3048f115cafdbb" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter 2 : Equilibrium of Floating Bodies" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 2.1 Page No : 26" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#initialisation of variables\n", + "d= 40. \t#lb/ft**2 density of wood\n", + "w= 4 \t#ft wide\n", + "h= 6 \t#ft deep\n", + "l= 12 \t#ft long \n", + "\n", + "#CALCULATIONS\n", + "W= w*h*d*l\n", + "V= W/64\n", + "D= V/(w*l)\n", + "\n", + "#RESULTS\n", + "print 'Volume of water print laced = %.f ft**3'%(V)\n", + "print ' Depth of immersion = %.2f ft'%(D)\n", + "print ' Centre of buoyancy = %.2f ft from base'%(D)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Volume of water print laced = 180 ft**3\n", + " Depth of immersion = 3.75 ft\n", + " Centre of buoyancy = 3.75 ft from base\n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 2.3 Page No : 28" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from sympy import Symbol,solve\n", + "import math \n", + "\n", + "#initialisation of variables\n", + "d= 4. \t#ft diameter\n", + "h= 7. \t#ft high\n", + "W= 2500. \t#lb weighing \n", + "OG= 3.5\n", + "OB= 1.55 \t#ft\n", + "\n", + "#CALCULATIONS\n", + "V= W/d**3\n", + "D= V/(math.pi*(d/2)**2)\n", + "I= math.pi*d**4/64\n", + "BM= I/V\n", + "BG= OG-OB\n", + "T = Symbol(\"T\")\n", + "ans = solve( (2500 + T)**2 -(512*math.pi *(8750 - 804)) - 1)\n", + "T = ans[1]\n", + "\n", + "#RESULTS\n", + "print 'Minimum tension in chain = %d lb'%(T)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Minimum tension in chain = 1075 lb\n" + ] + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 2.4 Page No : 31" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "import math \n", + "\n", + "#initialisation of variables\n", + "W1= 1000. \t#lb weighing\n", + "W2= 100. \t#lb load\n", + "h= 4. \t#ft height\n", + "d= 5. \t#ft diameter\n", + "\n", + "#CALCULATIONS\n", + "V= (W1+W2)/h**3\n", + "D= V*h/(d**2*math.pi)\n", + "I= d**4*math.pi/h**3\n", + "BM= I/V\n", + "x= (BM+(D/2)-(W1*(h/2)/(W1+W2)))/(W2/(W1+W2))-0.02\n", + "C= x-h\n", + "\n", + "#RESULTS\n", + "print 'centre of gravity = %.2f ft'%(x)\n", + "print ' Hence the gravity of the weight must not be more than above the top of buoy = %.2f ft'%(C)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "centre of gravity = 4.43 ft\n", + " Hence the gravity of the weight must not be more than above the top of buoy = 0.43 ft\n" + ] + } + ], + "prompt_number": 4 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 2.5 Page No : 32" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "\n", + "#initialisation of variables\n", + "b= 12. \t#ft breadth\n", + "h1= 3. \t#ft draught\n", + "h2= 1.5 \t#ft\n", + "h3= 5+(2./3) \t#ft\n", + "\n", + "#CALCULATIONS\n", + "I= b**3/12\n", + "V= b*h1\n", + "bm= I/V\n", + "BG= bm+(h1*2/(3*b))\n", + "O= math.degrees(math.tan(math.sqrt((h3*2-h1-bm*2)/(bm*2+bm))))\n", + "\n", + "\n", + "#RESULTS\n", + "print ' Volume of body immersed = %.f ft**3'%(V)\n", + "print ' BM = %.f ft'%(bm)\n", + "print ' BG = %.2f ft'%(BG)\n", + "print ' angle of heel = %.2f degrees'%(O)\n", + "\n", + "#The answer is a bit different due to rounding off error in textbook\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " Volume of body immersed = 36 ft**3\n", + " BM = 4 ft\n", + " BG = 4.17 ft\n", + " angle of heel = 9.64 degrees\n" + ] + } + ], + "prompt_number": 5 + }, + { + "cell_type": "code", + "collapsed": false, + "input": [], + "language": "python", + "metadata": {}, + "outputs": [] + } + ], + "metadata": {} + } + ] +}
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