From f270f72badd9c61d48f290c3396004802841b9df Mon Sep 17 00:00:00 2001 From: kinitrupti Date: Fri, 12 May 2017 18:53:46 +0530 Subject: Removed duplicates --- .../Chapter_3.ipynb | 240 +++++++++++++++++++++ 1 file changed, 240 insertions(+) create mode 100755 Mechanical_Engineering_Thermodynamics_by_Mooney_D_A/Chapter_3.ipynb (limited to 'Mechanical_Engineering_Thermodynamics_by_Mooney_D_A/Chapter_3.ipynb') diff --git a/Mechanical_Engineering_Thermodynamics_by_Mooney_D_A/Chapter_3.ipynb b/Mechanical_Engineering_Thermodynamics_by_Mooney_D_A/Chapter_3.ipynb new file mode 100755 index 00000000..f5793d78 --- /dev/null +++ b/Mechanical_Engineering_Thermodynamics_by_Mooney_D_A/Chapter_3.ipynb @@ -0,0 +1,240 @@ +{ + "metadata": { + "name": "" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter 3 - Temperature and Heat" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1 - Pg 33" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Calculate the mass of water required per pound of iron \n", + "import math\n", + "#Initialization of variables\n", + "T1=500 #F\n", + "T2=100 #F\n", + "Tf=75 #F\n", + "cpi=0.120 #B/lb F\n", + "cpw=1.0 #B/lb F\n", + "#calculations\n", + "Qw=1*cpw*(T2-Tf)\n", + "Qi=-1*cpi*(T2-T1)\n", + "mw=Qi/Qw\n", + "#results\n", + "print '%s %.2f %s' %(\"Mass of water = \",mw,\"lb water/lb iron\")\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Mass of water = 1.92 lb water/lb iron\n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 2 - Pg 34" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Calculate the net heat transferred\n", + "import math\n", + "import scipy\n", + "from scipy import integrate\n", + "#Initialization of variables\n", + "m=5 #lb\n", + "T1=1540+460 #R\n", + "T2=540+460 #R\n", + "#calculations\n", + "def q(T):\n", + "\tcp=m*(0.248+0.448*math.pow(10,-8) *T*T)\n", + "\treturn cp;\n", + "\n", + "Q=scipy.integrate.quad(q,T1,T2)\n", + "#results\n", + "print '%s %d %s' %(\"Heat transferred =\",Q[0],\"Btu\")\n", + "print '%s' %(\"The answer is a bit different due to rounding off error in textbook\")\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Heat transferred = -1292 Btu\n", + "The answer is a bit different due to rounding off error in textbook\n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3 - Pg 36" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Calculate the heat required for the process\n", + "#Initialization of variables\n", + "Tm=235 #F\n", + "Tb=832 #F\n", + "T=70 #F\n", + "cps=0.18 #B/lb F\n", + "cpl=0.235 #B/lb F\n", + "Lf=15.8 #B/lb\n", + "Lv=120 #B/lb\n", + "m=10 #lb\n", + "#calculations\n", + "Qa=m*cps*(Tm-T)\n", + "Qb=m*Lf\n", + "Qc=m*cpl*(Tb-Tm)\n", + "Qd=m*Lv\n", + "Q=Qa+Qb+Qc+Qd\n", + "#results\n", + "print '%s %.1f %s' %(\"Heat required =\",Q,\"Btu\")\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Heat required = 3057.9 Btu\n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 4 - Pg 36" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Calculate the mass of ice required for the process\n", + "#Initialization of variables\n", + "T1=22 #F\n", + "T2=32 #F\n", + "T3=40 #F\n", + "T4=70 #F\n", + "cps=0.501 #B/lb F\n", + "cpw=1 #B/lb F\n", + "Lf=143.3 #B/lb\n", + "m=40 #lb\n", + "#calculations\n", + "Qa=cps*(T2-T1)\n", + "Qb=Lf\n", + "Qc=cpw*(T3-T2)\n", + "Qd=m*cpw*(T3-T4)\n", + "mi=-Qd/(Qa+Qb+Qc)\n", + "#results\n", + "print '%s %.2f %s' %(\"Mass of ice required =\",mi,\"lb ice\")\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Mass of ice required = 7.68 lb ice\n" + ] + } + ], + "prompt_number": 5 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 5 - Pg 37" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Calculate the extra mass of ice required\n", + "#Initialization of variables\n", + "T1=22 #F\n", + "T2=32 #F\n", + "T3=40 #F\n", + "T4=70 #F\n", + "cps=0.501 #B/lb F\n", + "cpw=1 #B/lb F\n", + "Lf=143.3 #B/lb\n", + "m=40 #lb\n", + "cp=0.092\n", + "mc=10\n", + "#calculations\n", + "Qa=cps*(T2-T1)\n", + "Qb=Lf\n", + "Qc=cpw*(T3-T2)\n", + "Qe=mc*cp*(T3-T4)\n", + "mi=-Qe/(Qa+Qb+Qc)\n", + "#results\n", + "print '%s %.3f %s' %(\"Extra Mass of ice required =\",mi,\"lb ice\")" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Extra Mass of ice required = 0.177 lb ice\n" + ] + } + ], + "prompt_number": 6 + } + ], + "metadata": {} + } + ] +} \ No newline at end of file -- cgit