{ "metadata": { "name": "", "signature": "sha256:918b48292f00614bbba5ee6b8d2f7b2b929a3939332547fe2fb6585d1e0c7729" }, "nbformat": 3, "nbformat_minor": 0, "worksheets": [ { "cells": [ { "cell_type": "heading", "level": 1, "metadata": {}, "source": [ "Chapter 2:THERMODYNAMIC CONCEPTS AND THE ZEROTH LAW" ] }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 2.1, Page No:33" ] }, { "cell_type": "code", "collapsed": false, "input": [ "#Variable declaration\n", "patm = 14.5 ; # atmospheric pressure in psia\n", "pgauge = 2.5; # gauge pressure in psia\n", "A = 10; # Area of the piston in in^2\n", "g=9.80665; # Acceleration due to gravity in m/s^2\n", "#Calculation for (a)\n", "p = patm + pgauge; #total pressure of gas\n", "m=(p-patm)*A ; # mass of the piston\n", "#Result for (a)\n", "print \"In English units\"\n", "print \"Mass of the piston =\",m,\"lbm\"\n", "#Calculation for (b)\n", "p=(p*0.454*g)/(0.0254**2); # conversion of English unit to SI units\n", "patm=(patm*0.454*g)/(0.0254**2); # conversion of English unit to SI units\n", "m = ((p-patm)*(A*2.54**2*10**-4))/g; # Mass of the piston\n", "#Result for (b)\n", "print \"\\nIn SI units \\n\",\"Mass of the piston =\",m,\"kg\"" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "In English units\n", "Mass of the piston = 25.0 lbm\n", "\n", "In SI units \n", "Mass of the piston = 11.35 kg\n" ] } ], "prompt_number": 1 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 2.2, Page No:34" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math\n", "#Variable declaration\n", "d_r = 13600; # Density of manometric fluid (mercury) in kg/m^*3\n", "g = 8.92; # Gravitational acceleration in m/s^2\n", "#Variable declaration\n", "z1=0.589*math.sin(math.radians(60)); # vertical height of fluid at section 1\n", "z2=2*math.sin(math.radians(30)); # vertical height of fluid at section 2\n", "z=z2-z1; # Difference in vertical heights of fluid\n", "patm = 14.7; # Atmospheric pressure in lbf/in^2\n", "patm=(patm*4.44822*144/0.3048**2); # conversion of lbf/in^2 unit to N/m^2 unit\n", "p=patm + (d_r*g*(z2-z1)); # Balance of force at A\n", "#Result\n", "print \"Difference in vertical heights of fluid = \",round(z,2),\"m\"\n", "print \"\\nThe pressure of fluid in the vessel = \",round(p/1000,0),\"kPa\" \n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Difference in vertical heights of fluid = 0.49 m\n", "\n", "The pressure of fluid in the vessel = 161.0 kPa\n" ] } ], "prompt_number": 2 } ], "metadata": {} } ] }