{ "metadata": { "name": "", "signature": "sha256:efb3cb597d2918b83a867fe2c379b3952acba398aec14a9edf762b5c40043583" }, "nbformat": 3, "nbformat_minor": 0, "worksheets": [ { "cells": [ { "cell_type": "heading", "level": 1, "metadata": {}, "source": [ "Chapter 8 - Second and third law topics" ] }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 1 - Pg 125" ] }, { "cell_type": "code", "collapsed": false, "input": [ "#calculate the value of dp/ds at constant volume\n", "#Initialization of variables\n", "P=500 #psia\n", "T=700. #F\n", "J=778.\n", "#calculations\n", "dpds=1490 *144/J\n", "#results\n", "print '%s %d %s' %(\"dp by ds at constant volume =\",dpds,\" F/ft^3/lbm\")\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": [ "dp by ds at constant volume = 275 F/ft^3/lbm\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 2 - Pg 131" ] }, { "cell_type": "code", "collapsed": false, "input": [ "#calculate the thermal efficiency\n", "#Initialization of variables\n", "import math\n", "cp=0.25 #Btu/lbm R\n", "T0=520. #R\n", "T1=3460. #R\n", "#calculations\n", "dq=cp*(T0-T1)\n", "ds=cp*math.log(T0/T1)\n", "dE=dq-T0*ds\n", "eta=dE/dq*100.\n", "#results\n", "print '%s %.1f %s' %(\"Thermal efficiency =\",eta,\"percent\")\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Thermal efficiency = 66.5 percent\n" ] } ], "prompt_number": 2 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 3 - Pg 134" ] }, { "cell_type": "code", "collapsed": false, "input": [ "#calculate the Loss of available energy\n", "#Initialization of variables\n", "import math\n", "cp=0.25 #Btu/lbm R\n", "T0=520. #R\n", "T1=3460. #R\n", "dG=21069. #Btu/lbm\n", "dH=21502. #Btu/lbm\n", "#calculations\n", "dq=cp*(T0-T1)\n", "ds=cp*math.log(T0/T1)\n", "dE=dq-T0*ds\n", "eta=dE/dq\n", "dw=eta*dH\n", "de=-dG+dw\n", "#results\n", "print '%s %d %s' %(\"Loss of available energy =\",de,\"Btu/lbm\")\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Loss of available energy = -6774 Btu/lbm\n" ] } ], "prompt_number": 3 } ], "metadata": {} } ] }