{ "cells": [ { "cell_type": "markdown", "metadata": {}, "source": [ "# Chapter 5: Surface Tension" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "## Exa 5.1" ] }, { "cell_type": "code", "execution_count": 1, "metadata": { "collapsed": false }, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "Work (dyn cm or erg) = 5.03e+04\n", "press enter key to exit\n" ] }, { "data": { "text/plain": [ "''" ] }, "execution_count": 1, "metadata": {}, "output_type": "execute_result" } ], "source": [ "#If the surface tension of a soap bubble is 0.4 N/m, what work is required to\n", "#increase the diameter from 5 to 15 cm?\n", "import math\n", "#initialisation of variables\n", "St= 0.04 \t\t\t\t\t\t\t\t\t\t\t\t\t\t\t#N/m\n", "d1= 5. \t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t#cm\n", "d2= 15. \t\t\t\t\t\t\t\t\t\t\t\t\t\t\t#cm\n", "#CALCULATIONS\n", "W= St*1000.*2*4*math.pi*(math.pow((d2/2),2)-math.pow((d1/2),2))\t\t#Work\n", "#RESULTS\n", "print '%s %.2e' % ('Work (dyn cm or erg) = ',W)\n", "raw_input('press enter key to exit')" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "## Exa 5.2" ] }, { "cell_type": "code", "execution_count": 2, "metadata": { "collapsed": false }, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "Presuure difference (lbf/ft^2) = 14.12\n", "press enter key to exit\n" ] }, { "data": { "text/plain": [ "''" ] }, "execution_count": 2, "metadata": {}, "output_type": "execute_result" } ], "source": [ "#By much does the pressure inside a 0.017 in droplet of water at 20C\n", "#exceed the outside pressure.\n", "import math\n", "#initialisation of variables\n", "R= 0.017 \t\t\t\t\t\t\t\t\t\t\t\t#in\n", "sigma= 72.8 \t\t\t\t\t\t\t\t\t\t\t#m N/m\n", "#CALCULATIONS\n", "P= (2*sigma*0.005*0.017)/(72.8*R*7.08*math.pow(10,-4))\t#Presuure difference\n", "#RESULTS\n", "print '%s %.2f' % ('Presuure difference (lbf/ft^2) = ',P)\n", "raw_input('press enter key to exit')" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "## Exa 5.3" ] }, { "cell_type": "code", "execution_count": 3, "metadata": { "collapsed": false }, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ " Difference in mercury level (cm (depression)) = -2.48\n", "press enter key to exit\n" ] }, { "data": { "text/plain": [ "''" ] }, "execution_count": 3, "metadata": {}, "output_type": "execute_result" } ], "source": [ "#A 0.4 mm diameter glass tube stands vertically in a dish of mercury at 20C.\n", "#Determine the difference between the level of mercury in the dish and in\n", "#the tube. The specific gravity for mercury is 13.6 and the contact angle \n", "#with glass is 130C\n", "import math\n", "#initialisation of variables\n", "d= 13.6 \t\t\t\t\t\t\t#gm/cm^3\n", "g= 980 \t\t\t\t\t\t\t\t#cm/s^2\n", "D= 0.4 \t\t\t\t\t\t\t\t#mm\n", "angle= 130*math.pi/180. \t\t\t#radians\n", "s= 514. \t\t\t\t\t\t\t#dyn/cm\n", "#CALCULATIONS\n", "h= (4*s*10*math.cos(angle))/(d*g*D)\t#Depression in mercury level\n", "#RESULTS\n", "print '%s %.2f' % (' Difference in mercury level (cm (depression)) = ',h)\n", "raw_input('press enter key to exit')" ] } ], "metadata": { "kernelspec": { "display_name": "Python 2", "language": "python", "name": "python2" }, "language_info": { "codemirror_mode": { "name": "ipython", "version": 2 }, "file_extension": ".py", "mimetype": "text/x-python", "name": "python", "nbconvert_exporter": "python", "pygments_lexer": "ipython2", "version": "2.7.6" } }, "nbformat": 4, "nbformat_minor": 0 }