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diff --git a/Thermodynamics_for_Engineers/Chapter_12_3.ipynb b/Thermodynamics_for_Engineers/Chapter_12_3.ipynb deleted file mode 100755 index adf56744..00000000 --- a/Thermodynamics_for_Engineers/Chapter_12_3.ipynb +++ /dev/null @@ -1,585 +0,0 @@ -{
- "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 |