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diff --git a/Heat_Transfer_Principles_And_Applications_by_Dutta/ch2.ipynb b/Heat_Transfer_Principles_And_Applications_by_Dutta/ch2.ipynb new file mode 100644 index 00000000..5fcf0a36 --- /dev/null +++ b/Heat_Transfer_Principles_And_Applications_by_Dutta/ch2.ipynb @@ -0,0 +1,580 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 2 :Steady State conduction In one dimension" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 2.1 Page No : 14" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "the rate of heat gain is 16.27 W\n", + "interface temp. between brick and cork is 24.2 C\n", + "interface temp. between cement and cork is -13.6 C\n", + "thermal resistance offered by brick layer is 12.9 percent\n", + "thermal resistance offered by cork layer is 84.1 percent\n", + "thermal resistance offered by cement layer is 3.0 percent\n", + "Additional thickness of cork to be provided = 5.1 cm\n" + ] + } + ], + "source": [ + "# Variables\n", + "A = 1. \t\t\t#m**2, area\n", + "#for inner layer (cement)\n", + "ti = 0.06 \t\t\t#m, thickness\n", + "ki = 0.72 \t\t\t#W/m C, thermal conductivity\n", + "Ti = -15. \t\t\t#C, temprature\n", + "#for middle layer (cork)\n", + "tm = 0.1 \t\t\t#m, thickness\n", + "km = 0.043 \t\t\t#W/m C, thermal conductivity\n", + "#for outer layer(brick)\n", + "to = 0.25 \t\t\t#m, thickness\n", + "ko = 0.7 \t\t\t#W/m C, thermal conductivity\n", + "To = 30. \t\t\t#C, temprature\n", + "\n", + "# Calculation and Results\n", + "#Thermal resistance of outer layer \t\t\t#C/W\n", + "Ro = to/(ko*A) \n", + "#Thermal resistance of middle layer \t\t\t#C/W\n", + "Rm = tm/(km*A) \n", + "#Thermal resistance of inner layer \t\t\t#C/W\n", + "Ri = ti/(ki*A)\n", + "Rt = Ro+Rm+Ri\n", + "tdf = To-Ti \t\t\t#temp driving force\n", + "#(a)\n", + "Q = tdf/Rt \t\t\t#rate of heat gain\n", + "print \"the rate of heat gain is %.2f W\"%(Q)\n", + "\n", + "#(b)\n", + "#from fig. 2.4\n", + "td1 = Q*to/(ko*A) \t\t\t#C temp. drop across the brick layer\n", + "T1 = To-td1 \t\t\t#interface temp. between brick and cork\n", + "#similarly\n", + "td2 = Q*tm/(km*A) \t\t\t#C temp. drop across the cork layer\n", + "T2 = T1-td2 \t\t\t#C, interface temp. between cement and cork\n", + "print \"interface temp. between brick and cork is %.1f C\"%(T1)\n", + "print \"interface temp. between cement and cork is %.1f C\"%(T2)\n", + "\n", + "\n", + "#(c)\n", + "Rpo = Ro/Rt \t\t\t#thermal resistance offered by brick layer\n", + "Rpm = Rm/Rt \t\t\t#thermal resistance offered by cork layer\n", + "Rpi = Ri/Rt \t\t\t#thermal resistance offered by cement layer\n", + "print \"thermal resistance offered by brick layer is %.1f percent\"%(Rpo*100)\n", + "print \"thermal resistance offered by cork layer is %.1f percent\"%(Rpm*100)\n", + "print \"thermal resistance offered by cement layer is %.1f percent\"%(Rpi*100)\n", + "\n", + "#second part\n", + "x = 30. \t\t\t#percentage dec in heat transfer \n", + "Q1 = Q*(1-x/100) \t\t\t#W, desired rate of heat flow\n", + "Rth = tdf/Q1 \t\t\t#C/W, required thermal resistance\n", + "Rad = Rth-Rt \t\t\t#additional thermal resistance\n", + "Tad = Rad*km*A\n", + "print \"Additional thickness of cork to be provided = %.1f cm\"%(Tad*100)\n", + "\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 2.2 Page No : 15" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Rate of heat loss is 50.7 W\n", + "thermal conductivities of insulating layer is 0.1633 W/m C\n" + ] + } + ], + "source": [ + "# Variables\n", + "#outer thickness of brickwork (to) & inner thickness (ti)\n", + "to = 0.15 \t\t\t#m thickness\n", + "ti = 0.012 \t\t\t#m thickness\n", + "#thickness of intermediate layer(til)\n", + "til = 0.07 \t\t\t#m thick\n", + "#thermal conductivities of brick and wood\n", + "kb = 0.70 \t\t\t#W/m celcius\n", + "kw = 0.18 \t\t\t#W/m celcius\n", + "#temp. of outside and inside wall\n", + "To = -15 \t\t\t#celcius\n", + "Ti = 21 \t\t\t#celcius\n", + "#area\n", + "A = 1 \t\t\t#m**2\n", + "\n", + "\n", + "# Calculations and Results\n", + "#(a)\n", + "#Thermal resistance of brick , wood and insulating layer\n", + "TRb = to/(kb*A) \t\t\t#C/W\n", + "TRw = ti/(kw*A) \t\t\t#C/W\n", + "TRi = 2*TRb \t\t\t#C/W\n", + "#Total thermal resistance\n", + "TR = TRb+TRw+TRi \t\t\t#C/W\n", + "#Temp. driving force\n", + "T = Ti-To \t\t\t#C\n", + "#Rate of heat loss\n", + "Q = T/TR\n", + "print \"Rate of heat loss is %.1f W\"%(Q)\n", + "#(b)thermal conductivities of insulating layer\n", + "k = til/(A*TRi)\n", + "print \"thermal conductivities of insulating layer is %.4f W/m C\"%(k)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 2.3 Page No : 19" + ] + }, + { + "cell_type": "code", + "execution_count": 5, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Rate of heat loss is 4095 W\n", + "interface temp.is 183 C\n", + "Fractional resistance offered by the special brick layer is 0.353 \n" + ] + } + ], + "source": [ + "\n", + "import math \n", + "\n", + "# Variables\n", + "#Length & Inside rdius of gas duct\n", + "L = 1. \t\t\t#m\n", + "ri = 0.5 \t\t\t#m radius\n", + "#Properties of inner and outer layer\n", + "ki = 1.3 \t\t\t#W/m C, thermal conductivity of inner bricks\n", + "ti = 0.27 \t\t\t#m, inner layer thickness \n", + "ko = 0.92 \t\t\t#W/m C, thermal conductivity of special bricks \n", + "to = 0.14 \t\t\t#m, outer layer thickness\n", + "Ti = 400. \t\t\t#C, inner layer temp.\n", + "To = 65. \t\t\t#C, outer layer temp.\n", + "\n", + "#calculation and Results\n", + "r_ = ri+ti \t\t\t#m, outer radius of fireclay brick layer\n", + "ro = r_+to \t\t\t#m, outer radius of special brick layer\n", + "#Heat transfer resistance\n", + "#Heat transfer resistance of fireclay brick\n", + "R1 = (math.log(r_/ri))/(2*math.pi*L*ki)\n", + "#Heat transfer resistance of special brick\n", + "R2 = (math.log(ro/r_))/(2*math.pi*L*ko)\n", + "#Total resistance\n", + "R = round(R1+R2,4)\n", + "#Driving force\n", + "T = Ti-To\n", + "#Rate of heat loss\n", + "Q = T/(R)\n", + "print \"Rate of heat loss is %d W\"%(Q)\n", + "#interface temp.\n", + "Tif = Ti-(Q*R1)\n", + "print \"interface temp.is %d C\"%(Tif)\n", + "#Fractional resistance offered by the special brick layer\n", + "FR = R2/(R1+R2)\n", + "print \"Fractional resistance offered by the special brick layer is %.3f \"%(FR)\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 2.4 Page No : 20" + ] + }, + { + "cell_type": "code", + "execution_count": 6, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The hot end temp. is 148 C\n", + "The temprature gradient at hot end is -294.7 C/m\n", + "The temprature gradient at cold end is -1179 C/m\n", + "the temprature at 0.15m away from the cold end is 131 C\n" + ] + } + ], + "source": [ + "import math\n", + "\n", + "# Variables\n", + "d1 = 0.06 \t\t\t#m, one end diameter of steel rod\n", + "d2 = 0.12 \t\t\t#m,other end diameter of steel rod\n", + "l = 0.2 \t\t\t#m length of rod\n", + "T2 = 30. \t\t\t#C, temp. at end 2\n", + "Q = 50. \t\t\t#W, heat loss\n", + "k = 15. \t\t\t#W/m c, thermal conductivity of rod\n", + "\n", + "# Calculation and Results\n", + "#T = 265.8-(7.07/(0.06-0.15*x))........(a)\n", + "#(a)\n", + "x1 = 0\n", + "#from eq. (a)\n", + "T1 = 265.8-(7.07/(0.06-0.15*x1))\n", + "print \"The hot end temp. is %.0f C\"%(T1)\n", + "#(b) from eq. (i)\n", + "C = 50 \t\t\t#integration consmath.tant\n", + "#from eq. (i)\n", + "D1 = -C/(math.pi*d1**2*k) \t\t\t#D = dT/dx, temprature gradient\n", + "print \"The temprature gradient at hot end is %.1f C/m\"%(D1)\n", + "#similarly\n", + "D2 = -1179 \t\t\t#at x = 0.2m\n", + "print \"The temprature gradient at cold end is %.0f C/m\"%(D2)\n", + "\n", + "#(c)\n", + "x2 = 0.15 \t\t\t#m, given,\n", + "x3 = l-x2 \t\t\t#m, section away from the cold end\n", + "#from eq. (a)\n", + "T2 = 265.8-(7.07/(0.06-0.15*x3))\n", + "print \"the temprature at 0.15m away from the cold end is %.0f C\"%(T2)\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 2.5 Page No : 24" + ] + }, + { + "cell_type": "code", + "execution_count": 7, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "the rate of heat transfer is -3746 W\n", + "Refrigeration capacity is 1.07 tons\n" + ] + } + ], + "source": [ + "import math\n", + "\n", + "# Variables\n", + "#inside and outside diameter and Temp. of sphorical vessel\n", + "do = 16. \t\t\t#m, diameter \n", + "t = 0.1 \t\t\t#m, thick \n", + "Ri = do/2 \t\t\t#m, inside radius \n", + "Ro = Ri+t \t\t\t#m. outside radius\n", + "To = 27. \t\t\t#C, temperature\n", + "Ti = 4. \t\t\t#C ammonia\n", + "k = 0.02 \t\t\t#W/m C, thermal conductivity of foam layer \n", + "\n", + "# Calculations and Results\n", + "#from eq. 2.23 the rate of heat transfer\n", + "Q = (Ti-To)*(4*math.pi*k*Ro*Ri)/(Ro-Ri)\n", + "print \"the rate of heat transfer is %.0f W\"%(Q)\n", + "#Refrigeration capacity(RC)\n", + "#3516 Watt = 1 ton\n", + "RC = -Q/3516\n", + "print \"Refrigeration capacity is %.2f tons\"%(RC)\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 2.6 Page No : 28" + ] + }, + { + "cell_type": "code", + "execution_count": 8, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "the temprature midway in the rod at steady state is 167.3 C\n", + "Temprature gradient at one end of the rod is 559 C/W\n", + "Temprature gradient at other end of the rod is 521.8 C/W\n" + ] + } + ], + "source": [ + "import math \n", + "# Variables\n", + "d = 0.05 \t\t\t#m, diameter of rod\n", + "l = 0.5 \t\t\t#m, length of rod\n", + "T1 = 30. \t\t\t#CTemp. at one end (1)\n", + "T2 = 300. \t\t\t#C, temp at other end (2)\n", + "\n", + "# Calculations and Results\n", + "x1 = l/2 \t\t\t#m, at mid plane\n", + "#temprature distribution ,\n", + "#comparing with quadratic eq. ax**2+bx+c \n", + "#and its solution as x = (-b+math.sqrt(b**2-4*a*c))/2*a\n", + "a = 1.35*10**-4\n", + "b = 1\n", + "c = -(564*x1+30.1)\n", + "T = (-b+math.sqrt(b**2-4*a*c))/(2*a)\n", + "print \"the temprature midway in the rod at steady state is %.1f C\"%(T)\n", + "\n", + "#Temprature gradient at the ends of the rod\n", + "x2 = 0 \t\t\t#m, at one end\n", + "a1 = 1.35*10**-4\n", + "b1 = 1\n", + "c1 = -(564*x2+30.1)\n", + "T1 = (-b1+math.sqrt(b1**2-4*a1*c1))/(2*a1)\n", + "k1 = 202+0.0545*T1 \n", + "C1 = 113930 \t\t\t#integration consmath.tant from eq. (1)\n", + "TG1 = C1/k1 \t\t\t#C/W, temprature gradient, dT/dx\n", + "#similarly\n", + "x3 = 0.5\n", + "a2 = 1.35*10**-4\n", + "b2 = 1\n", + "c2 = -(564*x3+30.1)\n", + "T2 = (-b2+math.sqrt(b2**2-4*a2*c2))/(2*a2)\n", + "k2 = 202+0.0545*T2\n", + "TG2 = C1/k2\n", + "print \"Temprature gradient at one end of the rod is %.0f C/W\"%(TG1)\n", + "print \"Temprature gradient at other end of the rod is %.1f C/W\"%(TG2)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 2.7 Page No : 29" + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "At the surface x = 0, the temp. is 600 C\n", + "At the surface x = 0.3m, the temp. is 270 C\n", + "Rhe average temprature of the wall is 615 C\n", + "The maximum temprature occurs at 0.104 m\n", + "The maximum temp. is 730 C\n", + "heat flux at left surface is -58750 W/m**2\n", + "heat flux at right surface is 110450 W/m**2\n", + "The average volumetric rate if heat genaration is 564000 W/m**3 \n" + ] + } + ], + "source": [ + "import math \n", + "from scipy.integrate import quad \n", + "# Variables\n", + "#temprature distribution in wall\n", + "\n", + "t = 0.3 \t\t\t#m, thickness of wall\n", + "k = 23.5 \t\t\t#W/m c thermal conductivity of wall\n", + "\n", + "# Calculations and Results\n", + "x1 = 0\n", + "T1 = 600+2500*x1-12000*x1**2 \t\t\t#C, at surface\n", + "x2 = 0.3\n", + "T2 = 600+2500*x2-12000*x2**2 \t\t\t#C, at x = 0.3\n", + "\n", + "def f3(x): \n", + " return 600+2500*x-12000*x**2\n", + "\n", + "Tav = 1/t* quad(f3,0,0.3)[0]\n", + "\n", + "print \"At the surface x = 0, the temp. is %.0f C\"%(T1)\n", + "print \"At the surface x = 0.3m, the temp. is %.0f C\"%(T2)\n", + "print \"Rhe average temprature of the wall is %.0f C\"%(Tav)\n", + "\n", + "#(b)\n", + "\n", + "#for maximum temprature D = 0\n", + "x3 = 2500/24000.\n", + "print \"The maximum temprature occurs at %.3f m\"%(x3)\n", + "Tmax = 600+2500*x3-12000*x3**2\n", + "print \"The maximum temp. is %.0f C\"%(Tmax)\n", + "\n", + "#(c)\n", + "D1 = 2500-24000*x1 \t\t\t#at x = 0, temprature gradient\n", + "Hf1 = -k*D1 \t\t\t#W/m**2, heat flux at left surface(x = 0)\n", + "D2 = 2500-24000*x2 \t\t\t#at x = 0.3, temprature gradient\n", + "Hf2 = -k*D2 \t\t\t#W/m**2, heat flux at right surface(x = 0.3)\n", + "print \"heat flux at left surface is %.0f W/m**2\"%(Hf1)\n", + "print \"heat flux at right surface is %.0f W/m**2\"%(Hf2)\n", + "\n", + "#(d)\n", + "Qt = Hf2-Hf1 \t\t\t#W/m**2, total rate of heat loss\n", + "Vw = 0.3 \t\t\t#m**3/m**2, volume of wall per unit surface area\n", + "Hav = Qt/Vw \t\t\t#W/m**3, average volumetric rate\n", + "print \"The average volumetric rate if heat genaration is %.0f W/m**3 \"%(Hav) \n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 2.8 Page No : 30" + ] + }, + { + "cell_type": "code", + "execution_count": 10, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The maximum temp. will occur at a position 0.209 m\n", + "The maximum temprature is 152.6 C\n" + ] + } + ], + "source": [ + "import math \n", + "# Variables\n", + "ka = 24 \t\t\t#W/mC thermal conductivitiy of material A\n", + "tA = 0.1 \t\t\t#m, thickness of A material\n", + "kB = 230 \t\t\t#W/mC thermal conductivity of metl B\n", + "kC = 200 \t\t\t#W/mC thermal conductivity of metal C\n", + "tB = 0.1 \t\t\t#m, thickness of B metal\n", + "tC = 0.1 \t\t\t#m, thickness of C metal\n", + "TBo = 100 \t\t\t#C, outer surface temp. of B wall\n", + "TCo = 100 \t\t\t#C, outer surface temp. of C wall\n", + "Q = 2.5*10**5 \t\t\t#W/m**3, heat generated\n", + "\n", + "# Calculations and Results\n", + "#At D = 0\n", + "x = 2175./10416\n", + "print \"The maximum temp. will occur at a position %.3f m\"%(x)\n", + "x1 = x\n", + "TA = -5208*x1**2+2175*x1-74.5\n", + "print \"The maximum temprature is %.1f C\"%(TA)\n", + "\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 2.9 Page No : 36" + ] + }, + { + "cell_type": "code", + "execution_count": 11, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "This radial position does not fall within layer 1. Therefore no temprature maximum occurs in this layer.\n", + " Similarly no temprature maximum occurs in layer 2.\n", + "The maximum temprature at the outer boundary is 200 C\n" + ] + } + ], + "source": [ + "import math\n", + "\n", + "# Variables\n", + "di = 0.15 \t\t\t#m, inner diameter\n", + "do = 0.3 \t\t\t#m, outer diameter\n", + "Q1 = 100.*10**3 \t\t\t#W/,m**3,inner rate of heat generation\n", + "Q2 = 40.*10**3 \t\t\t#W/m**3, outer rate of heat generation\n", + "Ti = 100. \t\t\t#C, temp.at inside surface\n", + "To = 200. \t\t\t#C, temp. at outside surface\n", + "k1 = 30. \t\t\t#W/m C, thermal conductivity of material for inner layer\n", + "k2 = 10. \t\t\t#W/m C, thermal conductivity of material for outer layer\n", + "\n", + "# Calculations and Results\n", + "#T1 = 364+100*math.log(r)-833.3*r**2 (1)\n", + "#T2 = 718+216*math.log(r)-1000*r**2 (2)\n", + "#(b)from eq. 1\n", + "r = math.sqrt(100./2*833.3)\n", + "print \"This radial position does not fall within layer 1. Therefore no temprature maximum occurs in this layer.\"\n", + "#similarly\n", + "print \" Similarly no temprature maximum occurs in layer 2.\"\n", + "ro = di \t\t # m, outer boundary\n", + "Tmax = To\n", + "print \"The maximum temprature at the outer boundary is %.0f C\"%(Tmax)\n" + ] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 2", + "language": "python", + "name": "python2" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 2 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython2", + "version": "2.7.6" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} |