{ "metadata": { "name": "", "signature": "sha256:c8eaf5948eadfe6d1b6a7dc0019961e94ca5ac36c2e0b4b3d3661a7c7823315b" }, "nbformat": 3, "nbformat_minor": 0, "worksheets": [ { "cells": [ { "cell_type": "heading", "level": 1, "metadata": {}, "source": [ "Chapter 1 : Introduction and Basic Concepts" ] }, { "cell_type": "heading", "level": 3, "metadata": {}, "source": [ "Example 1.1, Page no:5 " ] }, { "cell_type": "code", "collapsed": false, "input": [ "\n", "\n", "# Variables\n", "F = 300; \t\t\t #[N]\n", "g_local = 4.5; \t\t\t#local gravitational acceleration[m/s**2]\n", "g_earth = 9.81; \t\t#earth's gravitational acceleration[m/s**2]\n", "\n", "# Calculations\n", "#To find man's mass and weight on earth\n", "m = F/g_local;\t\t\t#mass of man[kg]\n", "w = m*g_earth; \t\t\t# weight of man on earth[N]\n", "\n", "# Results\n", "print 'Mass of man is %f kg'%m\n", "print '\\nWeight of man on earth is %f N'%w\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Mass of man is 66.666667 kg\n", "\n", "Weight of man on earth is 654.000000 N\n" ] } ], "prompt_number": 2 }, { "cell_type": "heading", "level": 3, "metadata": {}, "source": [ "Example 1.2, Page no:6 " ] }, { "cell_type": "code", "collapsed": false, "input": [ "\n", "\n", "# Variables\n", "p1 = 1.15*10**5; \t\t\t#measured pressure[N/m**2]\n", "p2 = 1.01325*10**5; \t\t#atmospheric pressure[N/m**2]\n", "rho = 2.95*10**3; \t\t\t#specific gravity of fluid\n", "g = 9.8067\n", "\n", "# Calculations\n", "#To find height of manometer fluid\n", "p = p1-p2; \t \t\t#difference in pressure\n", "h = p/(rho*g); \t\t\t#height of manometer fluid[m]\n", "\n", "# Results\n", "print 'Height of manometer fluid is %f m'%h\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Height of manometer fluid is 0.472697 m\n" ] } ], "prompt_number": 2 }, { "cell_type": "heading", "level": 3, "metadata": {}, "source": [ "Example 1.3, Page no:8" ] }, { "cell_type": "code", "collapsed": false, "input": [ "\n", "# Variables\n", "PE = 1.5*10**3; \t#potential energy[J]\n", "m = 10; \t\t\t#mass in kg\n", "u = 50; \t\t\t# velocity in m/s\n", "g = 9.8067\n", "\n", "# Calculations\n", "h = PE/(m*g);\t\t\t# height from ground in m\n", "#Using equation 1.9 (Page no. 8)\n", "KE = 0.5*m*(u**2);\t\t\t# Kinetic energy in J\n", "\n", "# Results\n", "print 'Height from ground is %f m'%h\n", "print '\\nKinetic Energy of body is %3.2e J'%KE\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Height from ground is 15.295665 m\n", "\n", "Kinetic Energy of body is 1.25e+04 J\n" ] } ], "prompt_number": 3 }, { "cell_type": "heading", "level": 3, "metadata": {}, "source": [ "Example 1.4, Page no:9 " ] }, { "cell_type": "code", "collapsed": false, "input": [ "\n", "# Variables\n", "#Given\n", "F = 600.; \t\t\t#weight in N\n", "t = 120.; \t\t\t#time in sec\n", "h = 0.18; \t\t\t#height of stairs in m\n", "\n", "# Calculations\n", "#To determine the power developed in man\n", "S = 20*h; \t\t\t#total vertical displacement in m\n", "W = F*S; \t\t\t#work done in J\n", "P = W/t; \t\t\t#power developed\n", "\n", "# Results\n", "print 'Power developed is %i W'%P\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Power developed is 18 W\n" ] } ], "prompt_number": 10 }, { "cell_type": "heading", "level": 3, "metadata": {}, "source": [ "Example 1.5, Page no:9 " ] }, { "cell_type": "code", "collapsed": false, "input": [ "\n", "\n", "\n", "import math\n", "\n", "# Variables\n", "#Given:\n", "A = (math.pi/4)*(0.1**2); \t\t\t#area in m**2\n", "P = 1.01325*10**5; \t\t\t#pressure in N/m**2\n", "m = 50; \t\t\t#mass of piston and weight in kg\n", "g = 9.81; \t\t\t#acceleration due to gravity (N/m**2)\n", "\n", "\n", "# Calculations and Results\n", "#To determine the force exerted pressure work done and change in potential energy\n", "#(a)\n", "Fa = P*A; \t\t\t#force exerted by atmosphere in N\n", "Fp = m*g; \t\t\t#force exerted by piston and weight in N\n", "F = Fp+Fa; \t\t\t#total force exerted in N\n", "print 'Total force exerted by the atmosphere, the piston and the weight is %f N'%F\n", "\n", "#(b)\n", "Pg = F/A; \t\t\t#pressure of gas in N/m**2\n", "print 'Pressure of gas is %5.4e N/m^2'%Pg\n", "\n", "#(c)\n", "S = 0.4; \t\t\t#displacement of gas in m\n", "W = F*S; \t\t\t#work done by gas in J\n", "print 'Work done by gas is %f J'%W\n", "\n", "#(d)\n", "PE = m*g*S; \t\t\t#change in potential energy in J\n", "print 'Change in potential energy is %f J'%PE\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Total force exerted by the atmosphere, the piston and the weight is 1286.304689 N\n", "Pressure of gas is 1.6378e+05 N/m^2\n", "Work done by gas is 514.521876 J\n", "Change in potential energy is 196.200000 J\n" ] } ], "prompt_number": 4 }, { "cell_type": "heading", "level": 3, "metadata": {}, "source": [ "Example 1.6, Page no:10" ] }, { "cell_type": "code", "collapsed": false, "input": [ "\n", "\n", "import math\n", "\n", "# Variables\n", "#P =(2*10**5)*D\n", "Df = 2.5; \t\t\t#final diameter (m)\n", "Di = 0.5; \t\t\t#initial diameter(m)\n", "\n", "# Calculations\n", "#To determine work done by gas\n", "W = (math.pi/4)*10**5*((Df**4)-Di**4);\n", "\n", "# Results\n", "print 'Work done by gas is %6.4e J'%W\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Work done by gas is 3.0631e+06 J\n" ] } ], "prompt_number": 12 }, { "cell_type": "heading", "level": 3, "metadata": {}, "source": [ "Example 1.7, Page no:19" ] }, { "cell_type": "code", "collapsed": false, "input": [ "\n", "\n", "# Variables\n", "T = 300.; \t\t\t #temperature in K\n", "P = 6.5*10**5; \t\t\t#pressure in N/m**2\n", "Pa = 1.01325*10**5; \t#atmospheric pressure in N/m**2\n", "R = 8.314; \t\t\t #ideal gas constant\n", "m = 2.; \t\t\t #mass of gas (kg)\n", "M = 44.; \t\t\t #molecular weihgt of gas\n", "\n", "# Calculations\n", "#To find the work done on surrounding\n", "n = m/M; \t\t\t # n is number of kmoles\n", "Vi = (n*R*10**3*T)/P; \t\t\t# initial volume in m**3\n", "Vf = 2*Vi; \t\t \t#final volume in m**3\n", "V = Vf-Vi; \t\t\t #change in volume\n", "Ps = Pa+(5000*9.8067); \t\t\t#pressure on surroundings\n", "W = Ps*V; \t\t\t #work done on the surroundings\n", "\n", "# Results\n", "print 'Work done on surroundings is %5.2e J'%W\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Work done on surroundings is 2.62e+04 J\n" ] } ], "prompt_number": 1 }, { "cell_type": "code", "collapsed": false, "input": [ "import math\n", "from scipy import integrate\n", "\n", "#Taking 3rd and 2nd order derivative respectively, we get the following expression \n", "#x = ((2*10**5*math.pi)/2)*\n", "D = lambda x: x**3\n", "integ,err = integrate.quad(D,0.5,2.5)\n", "print integ\n", "#integ,err = integrate.quad(D**3,0.5,2.5)\n", "#print integ" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "9.75\n" ] } ], "prompt_number": 12 }, { "cell_type": "code", "collapsed": false, "input": [], "language": "python", "metadata": {}, "outputs": [] } ], "metadata": {} } ] }