{ "metadata": { "name": "" }, "nbformat": 3, "nbformat_minor": 0, "worksheets": [ { "cells": [ { "cell_type": "heading", "level": 1, "metadata": {}, "source": [ "Chapter 15 - Basic Flow equations" ] }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 1 - Pg 407" ] }, { "cell_type": "code", "collapsed": false, "input": [ "#calcualte the final temperature and pressure of the gas\n", "#Initialization of variables\n", "print '%s' %(\"From Table B-4,\")\n", "h=1187.2 #Btu/lbm\n", "t=328. #F\n", "#calculations\n", "p2=100 #psia\n", "u2=1187.2 #Btu/lbm\n", "t2=540. #F\n", "dt=t2-t\n", "#results\n", "print '%s %d %s' %(\"Final temperature of steam =\",t2,\"F\")\n", "print '%s %d %s' %(\"\\n Final pressure =\",p2,\"psia\")\n", "print '%s %d %s' %(\"\\n Change in temperature =\",dt,\"F\")\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "From Table B-4,\n", "Final temperature of steam = 540 F\n", "\n", " Final pressure = 100 psia\n", "\n", " Change in temperature = 212 F\n" ] } ], "prompt_number": 1 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 3 - Pg 409" ] }, { "cell_type": "code", "collapsed": false, "input": [ "#calculate the work done in the process\n", "#Initialization of variables\n", "import math\n", "p1=100. #psia\n", "p2=14.7 #psia\n", "k=1.4\n", "T1=700. #R\n", "R=10.73/29\n", "V=50.\n", "cv=0.171\n", "cp=0.24\n", "R2=1.986/29.\n", "#calculations\n", "T2=T1/ math.pow((p1/p2),((k-1)/k))\n", "m1=p1*V/(R*T1)\n", "m2=p2*V/(R*T2)\n", "Wrev= cv*(m1*T1 - m2*T2) - (m1-m2)*(T2)*cp\n", "#results\n", "print '%s %d %s' %(\"Work done in case 1 =\",Wrev,\" Btu\")\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Work done in case 1 = 572 Btu\n" ] } ], "prompt_number": 2 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 4 - Pg 420" ] }, { "cell_type": "code", "collapsed": false, "input": [ "#calculate the friction of the process\n", "#Initialization of variables\n", "import math\n", "p1=100. #psia\n", "p2=10. #psia\n", "n=1.3\n", "T1=800. #R\n", "cv=0.172\n", "R=1.986/29.\n", "T0=537. #R\n", "cp=0.24\n", "#calculations\n", "T2=T1*math.pow((p2/p1),((n-1)/n))\n", "dwir=cv*(T1-T2)\n", "dwr=R*(T2-T1)/(1-n)\n", "dq=dwr-dwir\n", "#results\n", "print '%s %.1f %s' %(\"The friction of the process per pound of air =\",dq,\"Btu/lbm\")\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "The friction of the process per pound of air = 18.6 Btu/lbm\n" ] } ], "prompt_number": 3 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 5 - Pg 421" ] }, { "cell_type": "code", "collapsed": false, "input": [ "#calculate the friction in the process\n", "#Initialization of variables\n", "ms=10 #lbm\n", "den=62.3 #lbm/ft^3\n", "A1=0.0218 #ft^2\n", "A2=0.00545 #ft^2\n", "p2=50. #psia\n", "p1=100. #psia\n", "gc=32.2 #ft/s^2\n", "dz=30. #ft\n", "T0=537. #R\n", "T1=620. #R\n", "T2=420. #R\n", "#calculations\n", "V1=ms/(A1*den)\n", "V2=ms/(A2*den)\n", "df=-144/den*(p2-p1) - (V2*V2 -V1*V1)/(2*gc) - dz\n", "#results\n", "print '%s %.1f %s' %(\"Friction =\",df,\"ft-lbf/lbm\")\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Friction = 72.9 ft-lbf/lbm\n" ] } ], "prompt_number": 4 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 6 - Pg 432" ] }, { "cell_type": "code", "collapsed": false, "input": [ "#calculate the efficiency of the cycle and the loss of available energy\n", "#Initialization of variables\n", "cp1=0.25\n", "T=3460 #R\n", "T0=946.2 #R\n", "T00=520 #R\n", "dG=1228 #Btu/lbm\n", "cp=0.45\n", "#calculations\n", "dqa=cp1*(T-T0)\n", "w=cp*dqa\n", "dg=489.\n", "eff=w/dg*100\n", "dI=-dg+w\n", "#results\n", "print '%s %.1f %s' %(\"\\n Efficiency of cycle =\",eff,\" percent\")\n", "print '%s %.1f %s' %(\"\\n Loss of available energy =\",dI,\"Btu/lbm\")\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "\n", " Efficiency of cycle = 57.8 percent\n", "\n", " Loss of available energy = -206.2 Btu/lbm\n" ] } ], "prompt_number": 5 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 7 - Pg 434" ] }, { "cell_type": "code", "collapsed": false, "input": [ "#calculate the engine efficiency and effectiveness. Also, calculate the loss of available energy\n", "#Initialization of variables\n", "p1=400. #psia\n", "t1=600. #F\n", "h1=1306.9 #Btu/lbm\n", "b1=480.9 #Btu/lbm\n", "p2=50 #psia\n", "h2=1122 #Btu/lbm\n", "h3=1169.5 #Btu/lbm\n", "b3=310.9 #Btu/lbm\n", "#calculations\n", "print '%s' %(\"All the values are obtained from Mollier chart,\")\n", "dw13=h1-h3\n", "dw12=h1-h2\n", "dasf=b3-b1\n", "etae=dw13/dw12*100\n", "eta=abs(dw13/dasf)*100\n", "dq=dw13+dasf\n", "#results\n", "print '%s %.1f %s' % (\"Engine efficiency =\",etae,\"percent\")\n", "print '%s %.1f %s' %(\"\\n Effectiveness =\",eta,\"percent\")\n", "print '%s %.1f %s' %(\"\\n Loss of available energy =\",dq,\"Btu/lbm\")\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "All the values are obtained from Mollier chart,\n", "Engine efficiency = 74.3 percent\n", "\n", " Effectiveness = 80.8 percent\n", "\n", " Loss of available energy = -32.6 Btu/lbm\n" ] } ], "prompt_number": 6 } ], "metadata": {} } ] }