path: root/Elements_of_Thermodynamics_and_heat_transfer/Chapter_19_1.ipynb

{
 "metadata": {
  "name": "",
  "signature": "sha256:6dda6a8235cbc517967b1f5033dd3771eb08f1241054f5c218f55be7d9c59b8c"
 },
 "nbformat": 3,
 "nbformat_minor": 0,
 "worksheets": [
  {
   "cells": [
    {
     "cell_type": "heading",
     "level": 1,
     "metadata": {},
     "source": [
      "Chapter 19 - Fundamentals of heat transfer"
     ]
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 1 - Pg 381"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#calculate the experimental value of thermal conductivity and Required temperature\n",
      "#Initialization of variables\n",
      "import math\n",
      "r1=1.12 #in\n",
      "r2=3.06 #in\n",
      "t1=203 #F\n",
      "t2=184 #F\n",
      "r3=2.09 #in\n",
      "po=11.1 #watts\n",
      "#calculations\n",
      "km=po*3.413*(12/r1-12/r2)/(4*math.pi*(t1-t2))\n",
      "dt=po*3.413*(12/r1-12/r3)/(4*math.pi*km)\n",
      "t3d=t1-dt\n",
      "#results\n",
      "print '%s %.2f %s' %(\"The experimental value of thermal conductivity =\",km,\"Btu/hr ft F\")\n",
      "print '%s %.1f %s' %(\"\\n Required temperature =\",t3d,\" F\")"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The experimental value of thermal conductivity = 1.08 Btu/hr ft F\n",
        "\n",
        " Required temperature = 189.1  F\n"
       ]
      }
     ],
     "prompt_number": 1
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 2 - Pg 383"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#calculate the heat loss and Temperature required\n",
      "#Initialization of variables\n",
      "import math\n",
      "r1=4.035 #in\n",
      "r2=4.312 #in\n",
      "r3=5.312 #in\n",
      "r4=6.812 #in\n",
      "k12=25 #Btu/hr ft F\n",
      "k23=0.05 #Btu/hr ft F\n",
      "k34=0.04 #Btu/hr ft F\n",
      "t1=625. #F\n",
      "t4=125. #F\n",
      "l=100. #ft\n",
      "hr=1.7 #Btu/hr ft^2 F\n",
      "#calculations\n",
      "Rs=1/(2.*math.pi*l) *(math.log(r2/r1) /k12+math.log(r3/r2) /k23 +math.log(r4/r3) /k34)\n",
      "Qd=(t1-t4)/Rs\n",
      "dt=Qd*12/(hr*math.pi*2*l*6.812)\n",
      "t0=t4-dt\n",
      "#results\n",
      "print '%s %d %s' %(\"Heat loss =\",Qd,\"Btu/hr\")\n",
      "print '%s %d %s' %(\"\\n Temperature required =\",t0,\"F\")\n",
      "print '%s' %(\"The answers given in the textbook are a bit different due to rounding off error\")\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Heat loss = 30231 Btu/hr\n",
        "\n",
        " Temperature required = 75 F\n",
        "The answers given in the textbook are a bit different due to rounding off error\n"
       ]
      }
     ],
     "prompt_number": 2
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 3 - Pg 396"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#calculate the Coefficient of heat transfer\n",
      "#Initialization of variables\n",
      "import math\n",
      "dout=1 #in\n",
      "d1=0.049 #in\n",
      "t1=70. #F\n",
      "t2=80. #F\n",
      "rho=62.2 #lbm/ft^3\n",
      "mum=2.22 #lbm/ft hr\n",
      "k=0.352 #Btu/hr ft F\n",
      "cp=1 #Btu/lbm F\n",
      "vel=500000. #lbm/hr\n",
      "n=100. #tubes\n",
      "#calculations\n",
      "D=dout-2*d1\n",
      "t=(t1+t2)/2.\n",
      "V=vel/n *4*144/(math.pi*D**2 *rho)\n",
      "Re=rho*V*D/(mum*12)\n",
      "Pr=cp*mum/k\n",
      "Nu=0.023*Re**0.8 *Pr**0.4\n",
      "hc=Nu*k*12/D\n",
      "#results\n",
      "print '%s %d %s' %(\"Coefficient of heat transfer =\",hc,\"Btu/hr ft^2 F\")\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Coefficient of heat transfer = 1040 Btu/hr ft^2 F\n"
       ]
      }
     ],
     "prompt_number": 3
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 4 - Pg 397"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#calculate the Coefficient of heat transfer and Percentage change\n",
      "#Initialization of variables\n",
      "import math\n",
      "d1=0.5 #ft\n",
      "t1=200. #F\n",
      "t2=80. #F\n",
      "ta=400. #F\n",
      "rho=0.0662 #lbm/ft**3\n",
      "mum=0.0483 #lbm/ft hr\n",
      "k=0.0167 #Btu/hr ft F\n",
      "cp=0.2408 #Btu/lbm F\n",
      "rho2=0.0567 #lbm/ft**3\n",
      "mum2=0.0542 #lbm/ft hr\n",
      "k2=0.0190 #Btu/hr ft F\n",
      "cp2=0.2419 #Btu/lbm F\n",
      "g=32.17\n",
      "#calculations\n",
      "ti=(t1+t2)/2.\n",
      "bet=1/(460.+ti)\n",
      "Pr1=cp*mum/k\n",
      "Gr1=d1**3 *rho**2 *3600**2 *g*bet*(t1-t2)/mum**2\n",
      "Gr1pr1=Gr1*Pr1\n",
      "hc1=k/d1 *0.53*(Gr1pr1)**0.25\n",
      "Q1=hc1*(t1-t2)\n",
      "tf=(ta+t2)/2.\n",
      "bet2=1/(460.+tf)\n",
      "Pr2=cp2*mum2/k2\n",
      "Gr2=d1**3 *rho2**2 *3600**2 *g*bet2*(ta-t2)/mum2**2\n",
      "Gr2pr2=Gr2*Pr2\n",
      "hc2=k2/d1 *0.53*(Gr2pr2)**0.25\n",
      "Q2=hc2*(ta-t2)\n",
      "per=100*(Q2-Q1)/Q1\n",
      "#results\n",
      "print '%s %.3f %s' %(\"Coefficient of heat transfer in case 1=\",hc1,\" Btu/hr ft^2 F\")\n",
      "print '%s %.3f %s' %(\"\\n Coefficient of heat transfer in case 2 =\",hc2,\"Btu/hr ft^2 F\")\n",
      "print '%s %d %s' %(\"\\n Percentage change =\",per,\"percent\")\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Coefficient of heat transfer in case 1= 1.076  Btu/hr ft^2 F\n",
        "\n",
        " Coefficient of heat transfer in case 2 = 1.312 Btu/hr ft^2 F\n",
        "\n",
        " Percentage change = 225 percent\n"
       ]
      }
     ],
     "prompt_number": 4
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 5 - Pg 398"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#calculate the Temperature o wing surface and Heat transfer convective\n",
      "#Initialization of variables\n",
      "chord=40. #ft\n",
      "v=1200. #mph\n",
      "t1=80. #F\n",
      "t2=200. #F\n",
      "mu=0.0447 #lbm/ft hr\n",
      "rho=5280. #lbm/ft**3\n",
      "cp=0.2404 #Btu/lbm F\n",
      "k=0.0152 #Btu/hr ft F\n",
      "J=778.\n",
      "gc=32.17 #ft/s**2\n",
      "mu2=0.0514 #lbm/ft hr\n",
      "k2=0.0179 #Btu/hr ft F\n",
      "cp2=0.2414 #Btu/lbm F\n",
      "#calculations\n",
      "Re=rho*v*chord*0.0735/mu\n",
      "r=(mu*cp/k)**(1./3.)\n",
      "tav=t1+ r*v**2 *rho**2 /(2*gc*J*cp*3600**2)\n",
      "ts=t1+ 0.5*(t2-t1)+ 0.22*(tav-t1)\n",
      "Re2=v*rho*chord*0.0610/mu2\n",
      "Pr2=cp2*mu2/k2\n",
      "hc=cp2*v*rho*0.0610 *0.037*Re2**(-0.2) *Pr2**(-0.667)\n",
      "Q2=hc*(t2-tav)\n",
      "#results\n",
      "print '%s %.1f %s' %(\"Temperature of wing surface =\",tav,\" F\")\n",
      "print '%s %d %s' %(\"\\n Heat transfer convective =\",Q2,\"Btu/hr ft^2\")\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": [
        "Temperature of wing surface = 309.3  F\n",
        "\n",
        " Heat transfer convective = -9711 Btu/hr ft^2\n",
        "The answers are a bit different due to rounding off error in textbook\n"
       ]
      }
     ],
     "prompt_number": 6
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 6 - Pg 413"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#calculate the percent of radiation emitted by surface and percent absorbed\n",
      "#Initialization of variables\n",
      "import math\n",
      "r1=1. #in\n",
      "r2=5. #in\n",
      "F12=1.\n",
      "#calculations\n",
      "F21=4*math.pi*r1**2 *F12/(4*math.pi*r2**2)*100\n",
      "F22=(1-F21/100.)*100.\n",
      "#results\n",
      "print '%s %d %s' %(\"Percent of radiation emitted by surface 2 on small sphere =\",F21,\" percent\")\n",
      "print '%s %d %s' %(\"\\n Remaining\",F22, \"percent is absorbed by inner surface of larger sphere\")\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Percent of radiation emitted by surface 2 on small sphere = 4  percent\n",
        "\n",
        " Remaining 96 percent is absorbed by inner surface of larger sphere\n"
       ]
      }
     ],
     "prompt_number": 7
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 7 - Pg 413"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#calculate the Net exchange of radiation\n",
      "#Initialization of variables\n",
      "short=2. #ft\n",
      "apart=3. #ft\n",
      "lon=4. #ft\n",
      "T1=2260. #R\n",
      "T2=530. #R\n",
      "sigma=0.1714\n",
      "#calculations\n",
      "A1=short*lon\n",
      "ratio=short/apart\n",
      "print '%s' %(\"from curve 3\")\n",
      "F=0.165\n",
      "Q12=A1*F*sigma*((T1/100)**4 -(T2/100)**4)\n",
      "#results\n",
      "print '%s %d %s' %(\"Net exchange of radiation =\",Q12,\"Btu/hr\")\n",
      "print '%s' %(\"The answer in the textbook is a bit different due to rounding off error in textbook.\")\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "from curve 3\n",
        "Net exchange of radiation = 58844 Btu/hr\n",
        "The answer in the textbook is a bit different due to rounding off error in textbook.\n"
       ]
      }
     ],
     "prompt_number": 9
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 8 - Pg 416"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#calculate the Net exchange of radiation\n",
      "#Initialization of variables\n",
      "F=0.51\n",
      "A1=8 #ft^2\n",
      "sigma=0.1714\n",
      "T1=2260. #R\n",
      "T2=530. #R\n",
      "#calculations\n",
      "Q12=A1*F*sigma*((T1/100)**4 -(T2/100)**4)\n",
      "#results\n",
      "print '%s %d %s' %(\"Net exchange of radiation =\",Q12,\"Btu/hr\")\n",
      "print '%s' %(\"The answer in the textbook is a bit different due to rounding off error in textbook.\")\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Net exchange of radiation = 181881 Btu/hr\n",
        "The answer in the textbook is a bit different due to rounding off error in textbook.\n"
       ]
      }
     ],
     "prompt_number": 10
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 9 - Pg 417"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#calculate the Net exchange of radiation\n",
      "#Initialization of variables\n",
      "F=0.51\n",
      "A1=8. #f^2\n",
      "sigma=0.1714\n",
      "T1=2260. #R\n",
      "T2=530. #R\n",
      "#calculations\n",
      "F12=1/(1/0.51 +(1/0.9 -1) +(1/0.6 -1))\n",
      "Q12=A1*F12*sigma*((T1/100)**4 -(T2/100)**4)\n",
      "#results\n",
      "print '%s %d %s' %(\"Net exchange of radiation =\",Q12,\"Btu/hr\")\n",
      "print '%s' %(\"The answer in the textbook is a bit different due to rounding off error in textbook.\")\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Net exchange of radiation = 130225 Btu/hr\n",
        "The answer in the textbook is a bit different due to rounding off error in textbook.\n"
       ]
      }
     ],
     "prompt_number": 11
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 10 - Pg 418"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#calculate the Percentage change in total heat transfer\n",
      "#Initialization of variables\n",
      "em=0.79\n",
      "sigma=0.1714\n",
      "T1=660. #R\n",
      "T2=540. #R\n",
      "T3=860. #R\n",
      "#calculations\n",
      "Q1=em*sigma*((T1/100)**4 -(T2/100)**4)\n",
      "Q2=em*sigma*((T3/100)**4 -(T2/100)**4)\n",
      "Qh1=129+Q1\n",
      "Qh2=419+Q2\n",
      "per=100*(Qh2-Qh1)/Qh1\n",
      "#results\n",
      "print '%s %.1f %s' %(\"Percentage change in total heat transfer =\",per,\"percent\")\n",
      "print '%s' %(\"The answer in the textbook is a bit different due to rounding off error in textbook.\")\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Percentage change in total heat transfer = 285.7 percent\n",
        "The answer in the textbook is a bit different due to rounding off error in textbook.\n"
       ]
      }
     ],
     "prompt_number": 12
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 11 - Pg 419"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#calculate the Error in probe reading\n",
      "#Initialization of variables\n",
      "Tp=12.57\n",
      "Tw=10.73\n",
      "ep=0.8\n",
      "sig=0.1714\n",
      "hc=7\n",
      "#calculations\n",
      "dt=ep*sig*(Tp**4-Tw**4)/hc\n",
      "#results\n",
      "print '%s %d %s' %(\"Error in probe reading =\",dt,\"F\")\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Error in probe reading = 229 F\n"
       ]
      }
     ],
     "prompt_number": 13
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 12 - Pg 420"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#calculate the Heat transfer in both cases\n",
      "#Initialization of variables\n",
      "import math\n",
      "l=6 #ft\n",
      "d1=0.55 #in\n",
      "d2=0.75 #in\n",
      "h1=280. #Btu/hr ft^2 F\n",
      "h2=2000. #Btu/fr ft^2 F\n",
      "k=220. #Btu/hr ft F\n",
      "t2=212. #F\n",
      "t1=60. #F\n",
      "f=500. #Btu/hr ft^2 F\n",
      "#calculations\n",
      "A2=math.pi*d1*l/12\n",
      "A3=math.pi*d2*l/12\n",
      "Rt=1/(h1*A2) + 1/(h2*A3) +math.log(d2/d1) /(2*math.pi*k*l)\n",
      "Q=(t2-t1)/Rt\n",
      "Rt2=Rt+ 1/(f*A2)\n",
      "Q2=(t2-t1)/Rt2\n",
      "#results\n",
      "print '%s %d %s' %(\"Heat transfer =\",Q,\"Btu/hr\")\n",
      "print '%s %d %s' %(\"\\n Heat transfer in case 2=\",Q2,\" Btu/hr\")\n",
      "print '%s' %(\"The answer in the textbook is a bit different due to rounding off error in textbook.\")\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Heat transfer = 33074 Btu/hr\n",
        "\n",
        " Heat transfer in case 2= 21994  Btu/hr\n",
        "The answer in the textbook is a bit different due to rounding off error in textbook.\n"
       ]
      }
     ],
     "prompt_number": 14
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 13 - Pg 422"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#calculate the Overall Heat transfer coefficient\n",
      "#Initialization of variables\n",
      "import math\n",
      "l=6 #ft\n",
      "d1=0.55 #in\n",
      "d2=0.75 #in\n",
      "h1=280. #Btu/hr ft^2 F\n",
      "h2=2000. #Btu/fr ft^2 F\n",
      "k=220. #Btu/hr ft F\n",
      "t2=212. #F\n",
      "t1=60. #F\n",
      "#calculations\n",
      "A2=math.pi*d1*l/12\n",
      "A3=math.pi*d2*l/12\n",
      "Rt=1/(h1*A2) + 1/(h2*A3) +math.log(d2/d1) /(2*math.pi*k*l)\n",
      "U3=1/(A3*Rt)\n",
      "#results\n",
      "print '%s %.1f %s' %(\"Overall Heat transfer coefficient =\",U3,\"Btu/hr ft^2 F\")\n",
      "print '%s' %(\"The answer in the textbook is a bit different due to rounding off error in textbook.\")\n",
      "\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Overall Heat transfer coefficient = 184.7 Btu/hr ft^2 F\n",
        "The answer in the textbook is a bit different due to rounding off error in textbook.\n"
       ]
      }
     ],
     "prompt_number": 15
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 14 - Pg 427"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#calculate the Parameters X and Z\n",
      "#Initialization of variables\n",
      "t1=300. #F\n",
      "t2=260. #F\n",
      "t3=200. #F\n",
      "t4=160. #F\n",
      "#calculations\n",
      "X=(t2-t4)/(t1-t4)\n",
      "Z=(t1-t3)/(t2-t4)\n",
      "#results\n",
      "print '%s %.3f %s %.1f %s' %(\"Parameters X and Z are\",X, \"and\",Z,\"respectively\")\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Parameters X and Z are 0.714 and 1.0 respectively\n"
       ]
      }
     ],
     "prompt_number": 16
    }
   ],
   "metadata": {}
  }
 ]
}