From c7fe425ef3c5e8804f2f5de3d8fffedf5e2f1131 Mon Sep 17 00:00:00 2001 From: hardythe1 Date: Tue, 7 Apr 2015 15:58:05 +0530 Subject: added books --- Thermodynamics_for_Engineers/Chapter_12.ipynb | 585 ++++++++++++++++++++++++++ 1 file changed, 585 insertions(+) create mode 100755 Thermodynamics_for_Engineers/Chapter_12.ipynb (limited to 'Thermodynamics_for_Engineers/Chapter_12.ipynb') diff --git a/Thermodynamics_for_Engineers/Chapter_12.ipynb b/Thermodynamics_for_Engineers/Chapter_12.ipynb new file mode 100755 index 00000000..adf56744 --- /dev/null +++ b/Thermodynamics_for_Engineers/Chapter_12.ipynb @@ -0,0 +1,585 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:aa78c4fecb9e6170fff80d7d4acf98ab991ec2677f1baef191ee2d45ad66490b" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter 12 - Heat Transfer" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1 - Pg 229" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#calculate the interface temperature\n", + "#Initialization of variables\n", + "km1=0.62\n", + "km2=0.16\n", + "km3=0.4\n", + "l1=8. #in\n", + "l2=4. #in\n", + "l3=4. #in\n", + "Tf=1600. #F\n", + "Tc=100. #F\n", + "#calculations\n", + "Rw=l1/12./km1 +l2/12./km2 +l3/12./km3\n", + "Rb=l1/12./km1\n", + "Ti=Tf-Rb/Rw *(Tf-Tc)\n", + "#results\n", + "print '%s %.1f %s' %(\"Interface temperature =\",Ti,\"F\")\n", + "print '%s' %(\"The answers might differ a bit from textbook due to rounding off error.\")\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Interface temperature = 1196.0 F\n", + "The answers might differ a bit from textbook due to rounding off error.\n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 2 - Pg 231" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#calculate the heat flow and the interface temperature\n", + "#Initialization of variables\n", + "import math\n", + "th=350. #F\n", + "tc=150. #F\n", + "od1=4.5\n", + "id1=4.026\n", + "od2=6.5\n", + "id2=4.5\n", + "k1=32.\n", + "k2=0.042\n", + "#calculations\n", + "Q=2*math.pi*(th-tc)/(math.log(od1/id1) /k1 + math.log(od2/id2) /k2)\n", + "r1=math.log(od1/id1) /k1\n", + "rt=math.log(od1/id1) /k1 + math.log(od2/id2) /k2\n", + "ti=th-r1/rt*(th-tc)\n", + "#results\n", + "print '%s %.1f %s' %(\"Heat flow =\",Q,\"Btu/hr\")\n", + "print '%s %.2f %s' %(\"\\n Interface temperature =\",ti,\" F\")\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Heat flow = 143.5 Btu/hr\n", + "\n", + " Interface temperature = 349.92 F\n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3 - Pg 239" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#calculate the net energy exchange in the process\n", + "#Initialization of variables\n", + "import math\n", + "Fa=0.045\n", + "l=4. #m\n", + "b=4. #m\n", + "Fe=1.\n", + "Ta=540.+460 #R\n", + "Tb=1540.+460 #R\n", + "#calculations\n", + "A=l*b\n", + "Q=0.173*A*Fa*Fe*(math.pow((Tb/100.),4) -math.pow((Ta/100.),4))\n", + "Q2=416000.\n", + "#results\n", + "print '%s %d %s' %(\"In case 1, Net energy exchange =\",Q,\"Btu/hr\")\n", + "print '%s %d %s' %(\"\\n In case 2, Net energy exchange =\",Q2,\"Btu/hr\")\n", + "print '%s' %('The answers are a bit different due to rounding off error in textbook')\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "In case 1, Net energy exchange = 18684 Btu/hr\n", + "\n", + " In case 2, Net energy exchange = 416000 Btu/hr\n", + "The answers are a bit different due to rounding off error in textbook\n" + ] + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 4 - Pg 239" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#calculate the net energy exchange\n", + "#Initialization of variables\n", + "import math\n", + "ea=0.8\n", + "eb=0.7\n", + "Fa=0.045\n", + "l=4. #m\n", + "b=4. #m\n", + "Fe=1.\n", + "Ta=540.+460 #R\n", + "Tb=1540.+460 #R\n", + "#calculations\n", + "A=l*b\n", + "ef=ea*eb\n", + "Q=0.173*A*Fa*Fe*ef*(math.pow((Tb/100),4) -math.pow((Ta/100),4))\n", + "#results\n", + "print '%s %d %s' %(\"Net energy exchange =\",\tQ,\"Btu/hr\")\n", + "print '%s' %('The answers are a bit different due to rounding off error in textbook')\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Net energy exchange = 10463 Btu/hr\n", + "The answers are a bit different due to rounding off error in textbook\n" + ] + } + ], + "prompt_number": 4 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 5 - Pg 246" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#calculate the inside film coefficient\n", + "#Initialization of variables\n", + "import math\n", + "den=61.995 #lb/cu ft\n", + "vel=6 #ft/s\n", + "t1=100. #F\n", + "t2=160. #F\n", + "de=2.067 #in\n", + "mu=1.238\n", + "pr=3.3\n", + "#calculations\n", + "G=den*vel*3600.\n", + "tm=(t1+t2)/2\n", + "hc=0.023*0.377/(de/12.) *math.pow(de/12 *G/mu,0.8) *math.pow(pr,0.4)\n", + "#results\n", + "print '%s %d %s' %(\"Inside film coefficient =\",hc,\"Btu/sq ft hr F\")\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Inside film coefficient = 1335 Btu/sq ft hr F\n" + ] + } + ], + "prompt_number": 5 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6 - Pg 247" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#calculate the inside film coefficient\n", + "#Initialization of variables\n", + "import math\n", + "d=0.5 #in\n", + "tm=1000. #F\n", + "v=5#ft/s\n", + "k=38.2\n", + "den=51.2\n", + "mu=0.3\n", + "#calculations\n", + "Nu=7+ 0.025*math.pow((d/12 *v*den*mu/k*3600),0.8)\n", + "h=Nu*k/(d/12.)\n", + "#results\n", + "print '%s %d %s' %(\"Inside film coefficient =\",h,\"Btu/sq ft hr F\")\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Inside film coefficient = 8624 Btu/sq ft hr F\n" + ] + } + ], + "prompt_number": 6 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 7 - Pg 249" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#calculate the convective film coefficient\n", + "#Initialization of variables\n", + "import math\n", + "do=2 #in\n", + "tf=120. #F\n", + "ti=80. #F\n", + "rho=0.0709\n", + "g=32.17\n", + "bet=1/560.\n", + "cp=0.24\n", + "mu=0.0461\n", + "k=0.0157\n", + "d=2. #in\n", + "Cd=0.45\n", + "#calculations\n", + "GrPr=math.pow(d/12.,3) *rho*rho *g*3600*3600. *bet*(tf-ti)*cp/(mu*k)\n", + "hc=Cd*k/math.pow(d/12.,(1./4.)) *math.pow(GrPr,(1./4.))\n", + "#results\n", + "print '%s %.3f %s' %(\"Convective film coefficient =\",hc,\"Btu/sq ft hr F\")\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Convective film coefficient = 0.242 Btu/sq ft hr F\n" + ] + } + ], + "prompt_number": 7 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 8 - Pg 251" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#calculate the outer film coefficient\n", + "#Initialization of variables\n", + "import math\n", + "tf=220. #F\n", + "ti=200. #F\n", + "d=2. #in\n", + "C=103.7\n", + "k=0.394\n", + "rho=59.37\n", + "hfg=965.2\n", + "mu=0.70\n", + "#calculations\n", + "h=C*math.pow(k*k*k *rho*rho *hfg/((d/12.) *mu*(tf-ti)),(1./4.))\n", + "#results\n", + "print '%s %d %s' %(\"Outer film coefficient =\",h,\"Btu/sq ft hr F\")\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Outer film coefficient = 1792 Btu/sq ft hr F\n" + ] + } + ], + "prompt_number": 8 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 9 - Pg 252" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#calculate the boiling film coefficient\n", + "#Initialization of variables\n", + "tf=225. #F\n", + "a=190.\n", + "b=0.043\n", + "ti=212. #F\n", + "#calculations\n", + "hc=a/(1-b*(tf-ti))\n", + "hcti=hc*1.25\n", + "#results\n", + "print '%s %.1f %s' %(\"For a flat copper plate, boiling film coefficient =\",hc,\" Btu/sq ft hr F\")\n", + "print '%s %d %s' %(\"\\n For an inclined copper plate, boiling film coefficient =\",hcti,\"Btu/sq ft hr F\")\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "For a flat copper plate, boiling film coefficient = 430.8 Btu/sq ft hr F\n", + "\n", + " For an inclined copper plate, boiling film coefficient = 538 Btu/sq ft hr F\n" + ] + } + ], + "prompt_number": 9 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 10 - Pg 255" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#calculate the heat transferred per foot length of pipe\n", + "#Initialization of variables\n", + "import math\n", + "Do=2.375 #in\n", + "hi=1200.\n", + "Di=2.067 #in\n", + "km=29.2\n", + "h0=1500.\n", + "L=2.375 #in\n", + "t1=220. #F\n", + "t4=140. #F\n", + "#calculations\n", + "U0= 1/(Do/(Di*hi) + (Do/12. *math.log(Do/Di) /(2*km)) + 1./h0)\n", + "Q=U0*L*math.pi*(t1-t4)/12.\n", + "#results\n", + "print '%s %d %s' %(\"Heat transferred per foot length of pipe =\",Q,\"btu/hr\")\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Heat transferred per foot length of pipe = 23744 btu/hr\n" + ] + } + ], + "prompt_number": 10 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 11 - Pg 255" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#calculate the temperature of inner and outer surfaces of pipe\n", + "#Initialization of variables\n", + "import math\n", + "Do=2.375 #in\n", + "hi=1200.\n", + "Di=2.067 #in\n", + "km=29.2\n", + "h0=1500.\n", + "L=2.375 #in\n", + "t1=220. #F\n", + "t4=140. #F\n", + "#calculations\n", + "Re=Do/(Di*hi)\n", + "R0=Do/(Di*hi) + (Do/12. *math.log(Do/Di) /(2*km)) + 1./h0\n", + "td=Re/R0 *(t1-t4)\n", + "ti=t4+td\n", + "Req=1./h0\n", + "td2=Req/R0 *(t1-t4)\n", + "to=t1-td2\n", + "#results\n", + "print '%s %.1f %s' %(\"The temperature of the inner surface of pipe =\",ti,\" F\")\n", + "print '%s %.1f %s' %(\"\\n The temperature of the outer surface of pipe =\",to,\" F\")\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The temperature of the inner surface of pipe = 176.6 F\n", + "\n", + " The temperature of the outer surface of pipe = 194.5 F\n" + ] + } + ], + "prompt_number": 11 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 12 - Pg 259" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#calculate the Logarithmic Mean temperature difference \n", + "#Initialization of variables\n", + "import math\n", + "th1=800. #F\n", + "th2=300. #F\n", + "tc1=100. #F\n", + "tc2=400. #F\n", + "#calculations\n", + "lmtd= ((th1-tc2) - (th2-tc1) )/(math.log((th1-tc2)/(th2-tc1)))\n", + "#results\n", + "print '%s %d %s' %(\"Logarithmic Mean temperature difference =\",lmtd,\"F\")\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Logarithmic Mean temperature difference = 288 F\n" + ] + } + ], + "prompt_number": 12 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 13 - Pg 262" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#calculate the True Mean temperature difference\n", + "#Initialization of variables\n", + "import math\n", + "th1=200. #F\n", + "th2=100. #F\n", + "tc1=80. #F\n", + "tc2=110. #F\n", + "#calculations\n", + "print '%s' %(\"From the lmtd graph,\")\n", + "R=(tc1-tc2)/(th2-th1)\n", + "P=(th2-th1)/(tc1-th1)\n", + "F=0.62\n", + "lmtd= F* ((th1-tc2) - (th2-tc1) )/(math.log((th1-tc2)/(th2-tc1)))\n", + "#results\n", + "print '%s %.1f %s' %(\"True Mean temperature difference =\",lmtd,\" F\")\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "From the lmtd graph,\n", + "True Mean temperature difference = 28.9 F\n" + ] + } + ], + "prompt_number": 13 + } + ], + "metadata": {} + } + ] +} \ No newline at end of file -- cgit