{ "metadata": { "name": "" }, "nbformat": 3, "nbformat_minor": 0, "worksheets": [ { "cells": [ { "cell_type": "heading", "level": 1, "metadata": {}, "source": [ "Chapter 13 - Process Calculations for Stationary Systems" ] }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 1a - Pg 203" ] }, { "cell_type": "code", "collapsed": false, "input": [ "#Calculate the final temperature and heat transferred from the steam\n", "#initialization of varaibles\n", "P1=100. #psia\n", "T1=500+460. #R\n", "v=10. #cu ft\n", "P2=50. #psia\n", "cv=0.172\n", "R=53.34\n", "m=2.81 #lb\n", "#calculations\n", "T2=T1*P2/P1\n", "Q1=P1*144*v*cv*(T2-T1)/(R*T1)\n", "u1=165.26 #Btu/lb\n", "u2=81.77 #Btu/lb\n", "du=u2-u1\n", "Q2=m*du\n", "#results\n", "print '%s' %(\"Case 1,\")\n", "print '%s %d %s' %(\"\\n Final temperature of the steam = \",T2,\"R\")\n", "print '%s %.1f %s' %(\"\\n Heat transferred =\",Q1,\"Btu\")\n", "print '%s %.1f %s' %(\"\\n Heat transferred in case 2 =\",Q2,\"Btu\")\n", "print '%s' %(\"\\n The answer may be a bit different due to rounding off error in the textbook\")" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Case 1,\n", "\n", " Final temperature of the steam = 480 R\n", "\n", " Heat transferred = -232.2 Btu\n", "\n", " Heat transferred in case 2 = -234.6 Btu\n", "\n", " The answer may be a bit different due to rounding off error in the textbook\n" ] } ], "prompt_number": 1 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 1b - Pg 204" ] }, { "cell_type": "code", "collapsed": false, "input": [ "#Calculate the final temperature and heat transferred\n", "#initialization of varaibles\n", "P1=100. #psia\n", "T1=500.+460 #R\n", "V=10. #cu ft\n", "P2=50. #psia\n", "cv=0.172\n", "R=53.34\n", "v=5.589 #cu ft/lb\n", "#calculations\n", "m=V/v\n", "x2=(v-0.017)/8.498\n", "print '%s' %(\"From table 2,\")\n", "T2=281.01 #F\n", "h1=1279.1\n", "u1=h1-144*P1*v/778.\n", "uf=249.93\n", "ufg=845.4\n", "u2=uf+x2*ufg\n", "Q=m*(u2-u1)\n", "#results\n", "print '%s %.2f %s' %(\"Final temperature =\",T2,\"F\")\n", "print '%s %d %s' %(\"\\n Heat transferred =\",Q,\"Btu\")\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "From table 2,\n", "Final temperature = 281.01 F\n", "\n", " Heat transferred = -664 Btu\n" ] } ], "prompt_number": 2 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 2a - Pg 205" ] }, { "cell_type": "code", "collapsed": false, "input": [ "#Calculate the final temperature, Enthalpy and heat transferred in the process\n", "#initialization of varaibles\n", "T1=350+460 #R\n", "v1=6 #cu ft/lb\n", "m=1 #lb\n", "R=53.34\n", "v2=2*v1\n", "cp=0.24\n", "#calculations\n", "P=R*T1/(v1*144)\n", "W=P*144*(v2-v1)\n", "T2=T1*v2/v1\n", "Q=cp*(T2-T1)\n", "h1=194.25\n", "h2=401.09\n", "dh=h2-h1\n", "#results\n", "print '%s %d %s' %(\"Final temperature =\",T2-460,\"F\")\n", "print '%s %.2f %s' %(\"\\n Enthalpy =\",dh,\"B/lb\")\n", "print '%s %d %s' %(\"\\n Heat =\",Q,\"B/lb\")\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Final temperature = 1160 F\n", "\n", " Enthalpy = 206.84 B/lb\n", "\n", " Heat = 194 B/lb\n" ] } ], "prompt_number": 3 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 2b - Pg 206" ] }, { "cell_type": "code", "collapsed": false, "input": [ "#Calculate the final temperature, Enthalpy and heat transferred in the process\n", "#initialization of varaibles\n", "T1=350+460. #R\n", "v1=6 #cu ft/lb\n", "m=1 #lb\n", "R=53.34\n", "v2=2*v1\n", "cp=0.24\n", "#calculations\n", "print '%s' %(\"From steam tables,\")\n", "vg=3.342 #cu ft/lb\n", "P1=77.5 #psia\n", "P2=P1\n", "h1=1204.8 #B/lb\n", "v2=2*v1\n", "T2=1106 #F\n", "h2=1586.7 #B/lb\n", "Q=h2-h1\n", "W=P1*144*(v2-v1)\n", "#results\n", "print '%s %d %s' %(\"Final temperature =\",T2,\"F\")\n", "print '%s %d %s' %(\"\\n Work =\",W,\"ft lb/lb\")\n", "print '%s %.1f %s' %(\"\\n Heat =\",Q,\"B/lb\")\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "From steam tables,\n", "Final temperature = 1106 F\n", "\n", " Work = 66960 ft lb/lb\n", "\n", " Heat = 381.9 B/lb\n" ] } ], "prompt_number": 4 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 3a - Pg 206" ] }, { "cell_type": "code", "collapsed": false, "input": [ "#calculate the final pressure, work done and change in internal energy of the process\n", "#initialization of varaibles\n", "import math\n", "T1=400+460. #R\n", "P1=50. #psia\n", "ratio=1/10.\n", "R=53.34\n", "#calculations\n", "P2=P1/ratio\n", "W=R*T1*math.log(ratio)\n", "du=0\n", "#results\n", "print '%s %d %s' %(\"Final pressure =\",P2,\"psia\")\n", "print '%s %.1f %s' %(\"\\n Work done =\",W,\"B/lb\")\n", "print '%s %d %s' %(\"\\n Change in Internal energy \",du,\"Btu/lb\")\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Final pressure = 500 psia\n", "\n", " Work done = -105625.1 B/lb\n", "\n", " Change in Internal energy 0 Btu/lb\n" ] } ], "prompt_number": 5 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 3b - Pg 207" ] }, { "cell_type": "code", "collapsed": false, "input": [ "#calculate the final pressure, work done and change in internal energy of the process\n", "#initialization of varaibles\n", "T1=400+460. #R\n", "P1=50. #psia\n", "ratio=1/10.\n", "R=53.34\n", "v1=10.065 #cu ft/lb\n", "vfg=1.8447 #cu ft/lb\n", "vg=1.8633 #cu ft/lb\n", "#calculations\n", "v2=v1*ratio\n", "dx=(v2-vg)/vfg\n", "P2=247.3 #psia\n", "print '%s' %(\"From steam tables,\")\n", "u2=773. #B/lb\n", "u1=1141.6 #B/lb\n", "du=u2-u1\n", "s1=1.7349 #B/lb R\n", "s2=1.082 #B/lb R\n", "W=T1*(s2-s1) - du\n", "#results\n", "print '%s %.1f %s' %(\"Final pressure = \",P2,\"psia\")\n", "print '%s %d %s' %(\"\\n Work done =\",W,\"B/lb\")\n", "print '%s %d %s' %(\"\\n Change in Internal energy = \",du,\"B/lb \")\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "From steam tables,\n", "Final pressure = 247.3 psia\n", "\n", " Work done = -192 B/lb\n", "\n", " Change in Internal energy = -368 B/lb \n" ] } ], "prompt_number": 6 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 4a - Pg 208" ] }, { "cell_type": "code", "collapsed": false, "input": [ "#Calculate final specific volume and work per pound of fluid in the problem\n", "#initialization of varaibles\n", "import math\n", "P1=150. #psia\n", "T1=400.+460 #R\n", "P2=15. #psia\n", "g=1.4\n", "R=53.34\n", "#calculations\n", "Tratio=math.pow((P2/P1),((g-1)/g))\n", "W=53.34*T1*(Tratio-1)/(1-g)\n", "T2=T1*Tratio\n", "v2=R*T2/(P2*144)\n", "u1=147.50\n", "Pr1=7.149\n", "Pr2=Pr1*P2/P1\n", "print '%s' %(\"From tables,\")\n", "Pr=0.7149\n", "T2=447. #R\n", "u2=76.13 #B/lb\n", "W=-(u2-u1)\n", "v2=R*T2/(P2*144)\n", "#results\n", "print '%s %.1f %s' %(\"Final specific volume =\",v2,\"cu ft/lb\")\n", "print '%s %.1f %s' %(\"\\n Work per pound of fluid =\",W,\"B/lb\")\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "From tables,\n", "Final specific volume = 11.0 cu ft/lb\n", "\n", " Work per pound of fluid = 71.4 B/lb\n" ] } ], "prompt_number": 7 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 4b - Pg 209" ] }, { "cell_type": "code", "collapsed": false, "input": [ "#Calculate final specific volume and work per pound of fluid in the problem\n", "#initialization of varaibles\n", "print '%s' %(\"From Steam tables,\")\n", "h1=1219.4\n", "P1=150 #psia\n", "v1=0.59733 #cu ft/lb\n", "s1=1.5995 #B/lb R\n", "#calculations\n", "u1=h1-P1*v1\n", "sg=1.7549\n", "sfg=1.4415\n", "s2=s1\n", "dx=(sg-s2)/sfg\n", "u2=981.3\n", "W=u1-u2\n", "v2=23.48\n", "#results\n", "print '%s %.2f %s' %(\"Final specific volume =\",v2,\"cu ft/lb\")\n", "print '%s %.1f %s' %(\"\\n Work per pound of fluid = \",W,\"B/lb\")\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "From Steam tables,\n", "Final specific volume = 23.48 cu ft/lb\n", "\n", " Work per pound of fluid = 148.5 B/lb\n" ] } ], "prompt_number": 8 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 5a - Pg 210" ] }, { "cell_type": "code", "collapsed": false, "input": [ "#calculate the final specific volume, temperature and the net heat transferred in the process\n", "#initialization of varaibles\n", "import math\n", "P1=150. #psia\n", "T1=400.+460 #R\n", "P2=15. #psia\n", "n=1.15\n", "cv=0.172\n", "R=53.34\n", "#calculations\n", "v2=R*T1*math.pow((P1/P2),(1/n)) /(P1*144.)\n", "v1=R*T1/(P1*144.)\n", "T2=T1*P2*v2/(P1*v1)\n", "Q=(cv - 0.458)*(T2-T1)\n", "#results\n", "print '%s %.1f %s' %(\"Final specific volume = \",v2,\"cu ft/lb\")\n", "print '%s %d %s' %(\"\\n Final temperature =\",T2,\"R\")\n", "print '%s %.1f %s' %(\"\\n Heat transferred =\",Q,\"B/lb\")\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Final specific volume = 15.7 cu ft/lb\n", "\n", " Final temperature = 636 R\n", "\n", " Heat transferred = 63.8 B/lb\n" ] } ], "prompt_number": 9 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 5b - Pg 211" ] }, { "cell_type": "code", "collapsed": false, "input": [ "#calculate the final specific volume, temperature and the net heat transferred in the process\n", "#initialization of varaibles\n", "import math\n", "print '%s' %(\"From table 3,\")\n", "v1=3.223 #cu ft/lb\n", "P1=150. #psia\n", "T1=400.+460 #R\n", "P2=15. #psia\n", "n=1.15\n", "#calculations\n", "v2=v1*math.pow((P1/P2),(1/n))\n", "T2=213 #F\n", "W=144*(P2*v2-P1*v1)*0.00129/(1-n)\n", "u1=1129.8 #B/lb\n", "v2=23.9\n", "vg=26.29\n", "vfg=26.27\n", "dx=(vg-v2)/vfg\n", "u2=996.1\n", "Q=(u2-u1)+W\n", "#results\n", "print '%s %.1f %s' %(\"Final specific volume =\",v2,\"cu ft/lb\")\n", "print '%s %d %s' %(\"\\n Final temperature =\",T2,\" F\")\n", "print '%s %.1f %s' %(\"\\n Heat transferred =\",Q,\"B/lb\")\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "From table 3,\n", "Final specific volume = 23.9 cu ft/lb\n", "\n", " Final temperature = 213 F\n", "\n", " Heat transferred = 21.6 B/lb\n" ] } ], "prompt_number": 10 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 6a - Pg 212" ] }, { "cell_type": "code", "collapsed": false, "input": [ "#calculate the final specific volume, temperature and the net work done in the process\n", "#initialization of varaibles\n", "v2=15.7 #cu ft/lb\n", "T2=640 #R\n", "cv=0.172\n", "T1=400+460. #R\n", "#calculations\n", "du=cv*(T2-T1)\n", "W=-du\n", "#results\n", "print '%s %.1f %s' %(\"Final specific volume =\",v2,\"cu ft/lb\")\n", "print '%s %d %s' %(\"\\n Final temperature =\",T2,\"R \")\n", "print '%s %.1f %s' %(\"\\n Work done =\",W,\"B/lb\")\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Final specific volume = 15.7 cu ft/lb\n", "\n", " Final temperature = 640 R \n", "\n", " Work done = 37.8 B/lb\n" ] } ], "prompt_number": 11 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 6b - Pg 213" ] }, { "cell_type": "code", "collapsed": false, "input": [ "#calculate the final specific volume, temperature and the net work done in the process\n", "#initialization of varaibles\n", "print '%s' %(\"From steam tables,\")\n", "T2=213 #F\n", "v2=23.9 #cu ft/lb\n", "W=133.7 #B/lb\n", "#results\n", "print '%s %.1f %s' %(\"Final specific volume =\",v2,\"cu ft/lb\")\n", "print '%s %d %s' %(\"\\n Final temperature =\",T2,\"F\")\n", "print '%s %.1f %s' %(\"\\n Work done = \",W,\" B/lb\")\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "From steam tables,\n", "Final specific volume = 23.9 cu ft/lb\n", "\n", " Final temperature = 213 F\n", "\n", " Work done = 133.7 B/lb\n" ] } ], "prompt_number": 12 } ], "metadata": {} } ] }