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diff --git a/Introduction_to_Heat_Transfer_by_S._K._Som/Chapter4.ipynb b/Introduction_to_Heat_Transfer_by_S._K._Som/Chapter4.ipynb new file mode 100644 index 00000000..c6207f4b --- /dev/null +++ b/Introduction_to_Heat_Transfer_by_S._K._Som/Chapter4.ipynb @@ -0,0 +1,922 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": { + "collapsed": true + }, + "source": [ + "# Chapter 04:Unsteady conduction" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex4.1:pg-137" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Introduction to heat transfer by S.K.Som, Chapter 4, Example 1\n", + "Biot number is\n", + "Bi= 0.277777777778\n", + "Problem is not suitable for lumped parameter analysis\n" + ] + } + ], + "source": [ + " \n", + "\n", + " \n", + " \n", + "import math\n", + " \n", + "print\"Introduction to heat transfer by S.K.Som, Chapter 4, Example 1\"\n", + "#Diameter of apple in m\n", + "d = 100*(10**(-3));\n", + "#radius in m\n", + "r = d/2;\n", + "#Thermal conductivity of apple in W/(m*K)\n", + "k = 0.6;\n", + "#Heat transfer coefficient in W/(m**2*°C)\n", + "h = 10;\n", + "#Caculating characteristic dimension in m\n", + "Lc = (((((4*math.pi)*r)*r)*r)/3)/(((4*math.pi)*r)*r);\n", + "#Biot number\n", + "print\"Biot number is\"\n", + "Bi = (h*Lc)/k\n", + "print\"Bi=\",Bi\n", + "if Bi<0.1:\n", + " print\"Problem is suitable for lumped parameter analysis\"\n", + "else:\n", + " print\"Problem is not suitable for lumped parameter analysis\"\n", + "\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex4.2:pg-138 " + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Introduction to heat transfer by S.K.Som, Chapter 4, Example 2\n", + "Time constant in seconds is\n", + "tc= 8.0\n", + "Time required in seconds\n", + "t= 36.8413614879\n" + ] + } + ], + "source": [ + " \n", + " \n", + " \n", + " \n", + "import math\n", + " \n", + "print\"Introduction to heat transfer by S.K.Som, Chapter 4, Example 2\"\n", + "#Diameter of sphere in m\n", + "d = 1.5*(10**(-3));\n", + "#radius in m\n", + "r = d/2;\n", + "#Thermal conductivity of sphere in W/(m*°C)\n", + "k = 40.0;\n", + "#Density in kg/m**3\n", + "rho = 8000.0;\n", + "#Specific heat in J/(Kg*K)\n", + "c = 300.0;\n", + "#Heat transfer coefficient in W/(m**2*°C)\n", + "h = 75.0;\n", + "#Time constant in sec\n", + "tc = ((rho*c)*(((((4*math.pi)*r)*r)*r)/3))/((((h*4)*math.pi)*r)*r);\n", + "print\"Time constant in seconds is\"\n", + "print\"tc=\",tc\n", + "#Using eq. 4.4\n", + "#Given fraction is 0.01 (1 percent)\n", + "#Required time in sec\n", + "t = (-8)*math.log(0.01);\n", + "print\"Time required in seconds\"\n", + "print\"t=\",t\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex4.3:pg-138" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Introduction to heat transfer by S.K.Som, Chapter 4, Example 3\n", + "Maximum dimension in metre for lumped parameter analysis\n", + "a= 5.0\n" + ] + } + ], + "source": [ + " \n", + " \n", + " \n", + " \n", + "import math\n", + " \n", + "print\"Introduction to heat transfer by S.K.Som, Chapter 4, Example 3\"\n", + "#Heat transfer coefficient in W/(m**2*K)\n", + "h = 30;\n", + "#Thermal conductivity of sphere in W/(m*K)\n", + "k = 250;\n", + "#Biot number for lumped parameter analysis is 0.1\n", + "Bi = 0.1;\n", + "#Characteristic dimension of a cube is (a/6) where a is the side of cube in metre\n", + "#Maximum dimension in metre\n", + "a = ((6*k)*Bi)/h;\n", + "print\"Maximum dimension in metre for lumped parameter analysis\"\n", + "print\"a=\",a\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex4.4:pg-146" + ] + }, + { + "cell_type": "code", + "execution_count": 5, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Introduction to heat transfer by S.K.Som, Chapter 4, Example 4\n", + "Time required to cool milk in minutes\n", + "Energy required for cooling in KJ\n", + "E= -319.013666564\n" + ] + } + ], + "source": [ + " \n", + "from scipy.integrate import quad\n", + "import math\n", + "print\"Introduction to heat transfer by S.K.Som, Chapter 4, Example 4\"\n", + "#Diameter of glass in m\n", + "d = 50*(10**(-3));\n", + "#radius in m\n", + "r = d/2;\n", + "#Height of milk in glass in m\n", + "H = 0.1;\n", + "#Initial temperature of milk in °C\n", + "T = 80.0;\n", + "#Cold water temperature in °C\n", + "Tf = 25.0;\n", + "#Heat transfer coefficient in W/(m**2*°C)\n", + "h = 100.0;\n", + "#Thermal conductivity of milk in W/(m*K)\n", + "k = 0.6;\n", + "#Density of milk in kg/m**3\n", + "rho = 900.0;\n", + "#Specific heat in J/(Kg*K)\n", + "c = 4.2*(10**3);\n", + "#Since the milk temperature is always maintained as constant.\n", + "#Therefore it can be assumed as lumped paramteter analysis.\n", + "#Time constant n seconds\n", + "tcs = (((((rho*c)*math.pi)*r)*r)*H)/(((h*math.pi)*d)*H);\n", + "#Time constant in minutes\n", + "tc = tcs/60;\n", + "#Calculating from eq. 4.3 time taken to cool milk from 80°C to 30°C\n", + "t = -tc*math.log((30.0-Tf)/(T-Tf));\n", + "print\"Time required to cool milk in minutes\"\n", + "t\n", + "#Energy transferred during cooling\n", + "E = (((h*math.pi)*d)*H)*quad(lambda t:(80.0-25.0)*math.e*(-t/472.5),0,60.0*t)[0];\n", + "print\"Energy required for cooling in KJ\"\n", + "E = E/1000.0\n", + "print \"E=\",E\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex4.5:pg-159" + ] + }, + { + "cell_type": "code", + "execution_count": 6, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Introduction to heat transfer by S.K.Som, Chapter 4, Example 5\n", + "Time required in hours\n", + "t= 7.5\n", + "Heat transfer rate in MJ\n", + "Q= 186.3\n" + ] + } + ], + "source": [ + " \n", + " \n", + " \n", + " \n", + "import math\n", + " \n", + "print\"Introduction to heat transfer by S.K.Som, Chapter 4, Example 5\"\n", + "#Thermal conductivity of wall in W/(m*K)\n", + "k = 0.6;\n", + "#Thermal diffusivity in m**2/s\n", + "alpha = 5*(10**(-7));\n", + "#Thickness in m\n", + "L = 0.15;\n", + "#Initial temperature in °C\n", + "Ti = 30;\n", + "#Temperature of hot gas in °C\n", + "Tinfinity = 780;\n", + "#Heat transfer coefficient in W/(m**2*K)\n", + "h = 20;\n", + "#Surface temperaute to be achieved in °C\n", + "To = 480;\n", + "#Dimensionless temperature ratio\n", + "z = (To-Tinfinity)/(Ti-Tinfinity);\n", + "#Biot number\n", + "Bi = (h*L)/k;\n", + "#For this value of (1/Bi) and dimensionless temp. ratio\n", + "#From Fig. 4.11 Fourier number is\n", + "Fo = 0.6;\n", + "#Time required in seconds\n", + "t = ((Fo*L)*L)/alpha;\n", + "print\"Time required in hours\"\n", + "t = t/3600\n", + "print\"t=\",t\n", + "#From fig. 4.13, for this Bi and Fo*Bi*Bi, we have ratio of heats as\n", + "#Q/Qi=0.69\n", + "#Heat transfer in J\n", + "Q = ((((0.69*k)*2)*L)*(Tinfinity-Ti))/alpha;\n", + "print\"Heat transfer rate in MJ\"\n", + "Q = Q/(10**6)\n", + "print\"Q=\",Q\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex4.6:pg-166" + ] + }, + { + "cell_type": "code", + "execution_count": 7, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Introduction to heat transfer by S.K.Som, Chapter 4, Example 6\n", + "Temperature at this distance in °C\n", + "T= 355.5\n", + "Heat transfer rate in MJ\n", + "Q= 100.0\n" + ] + } + ], + "source": [ + " \n", + " \n", + " \n", + " \n", + "import math\n", + " \n", + "print\"Introduction to heat transfer by S.K.Som, Chapter 4, Example 6\"\n", + "#Thickness of plate in m\n", + "L = 0.2;\n", + "#Initial temperature in °C\n", + "Ti = 530;\n", + "#Heat transfer coefficient in W/(m**2*K)\n", + "h = 500;\n", + "#Given distance in m\n", + "x = L-20*(10**(-3));\n", + "#Temperature of surrounding in °C\n", + "Tinfinity = 30;\n", + "#Given time in seconds\n", + "t = 225;\n", + "#Thermal conductivity of aluminium in W/(m*K)\n", + "k = 200;\n", + "#Thermal diffusivity in m**2/s\n", + "alpha = 8*(10**(-5));\n", + "#Biot number\n", + "Bi = (h*L)/k;\n", + "#Fourier number\n", + "Fo = (alpha*t)/(L*L);\n", + "#From fig. 4.11, at this Fo and (1/Bi), we have dimensionless temperature\n", + "#ratio to be 0.7\n", + "#From fig. 4.12 for this (1/Bi) and (x/L), we have another dimensionless\n", + "#temperature to be 0.93\n", + "#Temperature in °C\n", + "T = Tinfinity+(0.93*0.7)*(Ti-Tinfinity);\n", + "print\"Temperature at this distance in °C\"\n", + "print\"T=\",T\n", + "#From fig. 4.13, for this Bi and Fo*Bi*Bi, we have ratio of heats as\n", + "#Q/Qi=0.4\n", + "#Heat transfer in J\n", + "Q = (((0.4*k)*L)*(Ti-Tinfinity))/alpha;\n", + "print\"Heat transfer rate in MJ\"\n", + "Q = Q/(10**6)\n", + "print\"Q=\",Q\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex4.7:pg-171" + ] + }, + { + "cell_type": "code", + "execution_count": 8, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Introduction to heat transfer by S.K.Som, Chapter 4, Example 7\n", + "Temperature at this radius in °C\n", + "T= 300.0\n", + "Heat transfer rate per unit length in MJ/m\n", + "Q= 33.2639222145\n" + ] + } + ], + "source": [ + " \n", + " \n", + " \n", + " \n", + "import math\n", + " \n", + "print\"Introduction to heat transfer by S.K.Som, Chapter 4, Example 7\"\n", + "#Radius in m\n", + "ro = 0.15;\n", + "#Initial temperature in °C\n", + "Ti = 530;\n", + "#Temperature of surrounding in °C\n", + "Tinfinity = 30;\n", + "#Heat transfer coefficient in W/(m**2*K)\n", + "h = 380;\n", + "#Thermal conductivity of aluminium in W/(m*K)\n", + "k = 200;\n", + "#Thermal diffusivity in m**2/s\n", + "alpha = 8.5*(10**(-5));\n", + "#Given radius at which temperature has to be find out in m\n", + "r = 0.12;\n", + "#Given time in seconds\n", + "t = 265;\n", + "#Fourier number\n", + "Fo = (alpha*t)/(ro**2);\n", + "#Biot number\n", + "Bi = (h*ro)/k;\n", + "#From fig. 4.15, at this fourier number,Fo and (1/Bi), we have dimensionless temperature\n", + "#ratio to be 0.6\n", + "#From fig. 4.16 for this (1/Bi) and (r/ro), we have another dimensionless\n", + "#temperature to be 0.9\n", + "#Temperature in °C\n", + "T = Tinfinity+(0.9*0.6)*(Ti-Tinfinity);\n", + "print\"Temperature at this radius in °C\"\n", + "print\"T=\",T\n", + "#From fig. 4.17, for this Bi and Fo*Bi*Bi, we have ratio of heats as\n", + "#Q/Qi=0.4\n", + "#Heat transfer per metre in J/m\n", + "Q = (((((0.4*k)*math.pi)*ro)*ro)*(Ti-Tinfinity))/alpha;\n", + "print\"Heat transfer rate per unit length in MJ/m\"\n", + "Q = Q/(10**6)\n", + "print\"Q=\",Q\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex4.8:pg-174" + ] + }, + { + "cell_type": "code", + "execution_count": 9, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Introduction to heat transfer by S.K.Som, Chapter 4, Example 8\n", + "Time required in minutes\n", + "t= 4.16666666667\n" + ] + } + ], + "source": [ + " \n", + " \n", + " \n", + " \n", + "import math\n", + "\n", + "print\"Introduction to heat transfer by S.K.Som, Chapter 4, Example 8\"\n", + "#Radius in m\n", + "ro = 0.05;\n", + "#Initial temperature in °C\n", + "Ti = 530;\n", + "#Temperature of surrounding in °C\n", + "Tinfinity = 30;\n", + "#Heat transfer coefficient in W/(m**2*K)\n", + "h = 500;\n", + "#Thermal conductivity of aluminium in W/(m*K)\n", + "k = 50;\n", + "#Thermal diffusivity in m**2/s\n", + "alpha = 1.5*(10**(-5));\n", + "#Required centre temperature to achieve in °C\n", + "To = 105;\n", + "#Dimensionless temperature\n", + "z = (To-Tinfinity)/(Ti-Tinfinity);\n", + "#Biot number\n", + "Bi = (h*ro)/k;\n", + "#For this value of (1/Bi) and dimensionless temp. ratio\n", + "#From Fig. 4.19 Fourier number is\n", + "Fo = 1.5;\n", + "#Time required in seconds\n", + "t = ((Fo*ro)*ro)/alpha;\n", + "print\"Time required in minutes\"\n", + "t = t/60\n", + "print\"t=\",t\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex4.9:pg-177" + ] + }, + { + "cell_type": "code", + "execution_count": 10, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Introduction to heat transfer by S.K.Som, Chapter 4, Example 9\n", + "Tempearture of bar in °C\n", + "T= 260.3\n" + ] + } + ], + "source": [ + " \n", + " \n", + " \n", + " \n", + "import math\n", + " \n", + "print\"Introduction to heat transfer by S.K.Som, Chapter 4, Example 9\"\n", + "#Thermal conductivity of aluminium in W/(m*K)\n", + "k = 198;\n", + "#Length in m\n", + "L = 0.18;\n", + "#Breadth in m\n", + "b = 0.104;\n", + "#Initial temperature in °C\n", + "Ti = 730;\n", + "#Temperature of surrounding in °C\n", + "Tinfinity = 30;\n", + "#Heat transfer coefficient in W/(m**2*K)\n", + "h = 1100;\n", + "#Thermal diffusivity in m**2/s\n", + "alpha = 8.1*(10**(-5));\n", + "#Given time in seconds\n", + "t = 100;\n", + "#Bar can be considered to be an intersection of two infinite plates of\n", + "#thickness L1 and L2 in m\n", + "L1 = L/2;\n", + "L2 = b/2;\n", + "#For plate 1\n", + "#Fourier number\n", + "Fo1 = (alpha*t)/(L1**2);\n", + "#Biot number\n", + "Bi1 = (h*L1)/k;\n", + "#From fig. 4.11, at this Fo and (1/Bi), we have dimensionless temperature\n", + "#ratio to be 0.7\n", + "#For plate 2\n", + "#Fourier number\n", + "Fo2 = (alpha*t)/(L2**2);\n", + "#Biot number\n", + "Bi2 = (h*L2)/k;\n", + "#From fig. 4.11, at this Fo and (1/Bi), we have dimensionless temperature\n", + "#ratio to be 0.47\n", + "#Therefore combined dimensionless temperature ratio is multiply of two\n", + "z = 0.47*0.7;\n", + "#Temperature in °C\n", + "T = Tinfinity+z*(Ti-Tinfinity);\n", + "print\"Tempearture of bar in °C\"\n", + "print\"T=\",T\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex4.10:pg-180" + ] + }, + { + "cell_type": "code", + "execution_count": 13, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Introduction to heat transfer by S.K.Som, Chapter 4, Example 10\n", + "The factor((To-Tinf)/(Ti-Tinf)) is \n", + "For plate 1\n", + "A= 0.85\n", + "For plate 2\n", + "B= 0.8\n", + "For plate 1\n", + "A= 0.83\n", + "For plate 2\n", + "B= 0.72\n", + "The calculated value is very close to the required value of 0.6.Hence the time required for the centre of the beam to reach 310°C is nearly 1200s or 20 minutes.\n", + "T= 0.5976\n" + ] + } + ], + "source": [ + " \n", + " \n", + " \n", + " \n", + "import math\n", + " \n", + "print\"Introduction to heat transfer by S.K.Som, Chapter 4, Example 10\"\n", + "#An iron beam of rectangular cross section of size length,L=300mm by breadth,B=200 mm is used in the construction of a building\n", + "#Initially the beam is at a uniform temprature(Ti) of 30°C.\n", + "#Due to an accidental fire,the beam is suddenly exposed to hot gases at temprature,Tinf=730°C,with a convective heat transfer coefficient(h) of 100 W/(m**2*K)\n", + "#To determine the time required for the centre plane of the beam to reach a temprature(To) of 310°C.\n", + "To=310;\n", + "Tinf=730;\n", + "Ti=30;\n", + "#Take thermal conductivity k=73W/(m*K) and thermal diffusivity of the beam alpha=2.034*10**-5m**2/s \n", + "alpha=2.034*10**-5; \n", + "k=73; \n", + "h=100; \n", + "#The rectangular iron beam can be considered as an intersection of an infinite plate 1 having thickness 2*L1=300mm and a second infinite plate 2 of thickness 2*L2=200mm \n", + "L1=0.15;#in metre\n", + "L2=0.10;#in metre\n", + "#Here the faactor X=((To-Tinf)/(Ti-Tinf))\n", + "print\"The factor((To-Tinf)/(Ti-Tinf)) is \"\n", + "X=((To-Tinf)/(Ti-Tinf))\n", + "#Therefore we can write 0.6=((To-Tinf)/(Ti-Tinf))plate 1 *((To-Tinf)/(Ti-Tinf))plate2\n", + "#A straight forward solution is not possible.We have to adopt an iterative method of solution \n", + "#At first ,a value of time(t) is assumed to determine the centre-line temprature of the beam.The value of t at which((To-Tinf)/(Ti-Tinf))beam =0.6 is satisfied\n", + "#Let us first assume time, t=900s\n", + "t=900;\n", + "print\"For plate 1\"\n", + "#For plate1 Biot number Bi1=h*L1/k \n", + "Bi1=h*L1/k \n", + "Y=1/Bi1\n", + "#Fourier number(Fo1) is\n", + "Fo1=alpha*t/L1**2\n", + "#At Fo=0.814 and (1/Bi)=4.87...We read from graphs A=((To-Tinf)/(Ti-Tinf))plate1= 0.85\n", + "A=0.85;\n", + "print\"A=\",A\n", + "print\"For plate 2\"\n", + "#For plate1 Biot number Bi2=h*L2/k \n", + "Bi2=h*L2/k \n", + "Y=1/Bi2\n", + "#Fourier number(Fo2) is\n", + "Fo2=alpha*t/L2**2\n", + "#At Fo=1.83 and (1/Bi)=7.3...We read from graphs B=((To-Tinf)/(Ti-Tinf))plate2= 0.8\n", + "B=0.8;\n", + "print\"B=\",B\n", + "#Therefore ((To-Tinf)/(Ti-Tinf))plate1*((To-Tinf)/(Ti-Tinf))plate2=A*B\n", + "T=A*B\n", + "#Since the calculated value of 0.68 is greater than the required value of 0.60 and Tinf>To>Ti,The assume dvalue of t is less.\n", + "#So let us take time,t=1200s for the second iteration\n", + "t=1200;\n", + "print\"For plate 1\"\n", + "#For plate1 Biot number Bi1=h*L1/k \n", + "Bi1=h*L1/k \n", + "Y=1/Bi1\n", + "#Fourier number (Fo1)\n", + "Fo1=alpha*t/L1**2\n", + "#At Fo=1.08 and (1/Bi)=4.87...We read from graphs A=((To-Tinf)/(Ti-Tinf))plate1= 0.83\n", + "A=0.83;\n", + "print\"A=\",A\n", + "print\"For plate 2\"\n", + "#For plate1 Biot number Bi2=h*L2/k \n", + "Bi2=h*L2/k \n", + "Y=1/Bi2\n", + "#Fourier number (Fo2)\n", + "Fo2=alpha*t/L2**2\n", + "#At Fo=2.44 and (1/Bi)=7.3...We read from graphs B=((To-Tinf)/(Ti-Tinf))plate2= 0.72\n", + "B=0.72;\n", + "print\"B=\",B\n", + "#Therefore ((To-Tinf)/(Ti-Tinf))plate1*((To-Tinf)/(Ti-Tinf))plate2=A*B\n", + "T=A*B\n", + "print\"The calculated value is very close to the required value of 0.6.Hence the time required for the centre of the beam to reach 310°C is nearly 1200s or 20 minutes.\" \n", + "print\"T=\",T\n", + " \n", + " \n", + " \n", + " \n", + " \n", + " \n", + " \n", + " \n", + " \n", + " \n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex4.11:pg-182" + ] + }, + { + "cell_type": "code", + "execution_count": 14, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Introduction to heat transfer by S.K.Som, Chapter 4, Example 11\n", + "The time required for the temprature to reach 255°C at a depth of 80mm, in minutes is\n", + "T= 255\n" + ] + } + ], + "source": [ + " \n", + " \n", + " \n", + " \n", + "import math\n", + " \n", + "print\"Introduction to heat transfer by S.K.Som, Chapter 4, Example 11\"\n", + "#A large slab wrought-iron is at a uniform temprature of Ti=550°C.\n", + "#The temprature of one surface is suddenly changed to Tinf=50°C\n", + "Tinf=50;\n", + "Ti=550; \n", + "#For slab conductivity(k=60W/(m*K)),Thermal diffusivity(alpha=1.6*10**-5m**2/s)\n", + "#To calculate the time(t) required for the temprature to reach T=255°C at a depth of 80mm\n", + "k=60;\n", + "T=255;\n", + "alpha=1.6**10-5;\n", + "#Similarity parameter,eta=x/(2*(alpha*t)**0.5)=(10/t**0.5)\n", + "#((T-Tinf)/(Ti-Tinf))=erf(10/t**0.5)...where erf is the error function.\n", + "#Let ((T-Tinf)/(Ti-Tinf))=X\n", + "X=((T-Tinf)/(Ti-Tinf));\n", + "#This implies erf(10/t**0.5)=0.41\n", + "#We read from the table the value of eta(=10/t**0.5)=0.38....corresponding to erf(eta)=0.41\n", + "#Therefore 10/t**0.5=0.38...this implies t=(10/0.38)**2\n", + "print\"The time required for the temprature to reach 255°C at a depth of 80mm, in minutes is\"\n", + "t=(10/0.38)**2/60\n", + "print\"T=\",T\n", + "\n", + "\n", + "\n", + "\n", + "\n", + "\n", + "\n", + "\n", + "\n", + "\n", + "\n", + "\n", + "\n", + "\n", + "\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex4.12:pg-186" + ] + }, + { + "cell_type": "code", + "execution_count": 15, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Introduction to heat transfer by S.K.Som, Chapter 4, Example 12\n", + "gaussian error function is \n", + "E= 0.998109069322\n", + "The temprature at a depth(x) of 100mm after a time(t) of 100 seconds,in °C is\n", + "T= -29851.5095103\n" + ] + } + ], + "source": [ + " \n", + "import math\n", + "import scipy \n", + "print\"Introduction to heat transfer by S.K.Som, Chapter 4, Example 12\"\n", + "#A large block of nickel steel conductivity(k=20W/(m*K)),thermal diffusivity(alpha=0.518*10-5 m**2/s) is at uniform temprature(Ti) of 30°C.\n", + "Ti=30.0;\n", + "k=20.0;\n", + "alpha=0.518*10.0**-5.0;\n", + "#One surface of the block is suddenly exposed to a constant surface heat flux(qo) of 6MW/m**2.\n", + "qo=6*10**6;#in W/m**2\n", + "#To determine the temprature at a depth(x) of 100mm after a time(t) of 100 seconds.\n", + "t=100.0;\n", + "x=0.1;#in metre\n", + "#Similarity parameter,eta=x/(4*alpha*t)\n", + "eta=x/((4.0*alpha*t)**0.5)\n", + "#E is gaussian error function\n", + "print\"gaussian error function is \"\n", + "E=scipy.special.erf(eta)\n", + "print\"E=\",E\n", + "#The equation to determine temprature is T-Ti=((2*qo(alpha*t/math.pi)**0.5)/(k))*e**((-x**2)/(4*alpha*t))-((qo*x)/(k))*erf(x/(2.0*(alpha*t)**0.5))\n", + "#Above equation can also be written as T=Ti+((2*qo(alpha*t/math.pi)**0.5)/(k))*e**((-x**2)/(4*alpha*t))-((qo*x)/(k))*erf(x/(2.0*(alpha*t)**0.5))\n", + "print\"The temprature at a depth(x) of 100mm after a time(t) of 100 seconds,in °C is\"\n", + "T=Ti+((2*qo*(alpha*t/math.pi)**0.5)/(k))*math.e**((-x**2.0)/(4*alpha*t))-((qo*x)/(k))*scipy.special.erf(x/(2*(alpha*t)**0.5))\n", + "print\"T=\",T\n", + "#NOTE:The answer in the book is incorrect(Calculation mistake)\n", + "\n", + "\n", + "\n", + "\n", + "\n", + "\n", + "\n", + "\n", + "\n", + "\n", + "\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex4.14:pg-187 " + ] + }, + { + "cell_type": "code", + "execution_count": 17, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Introduction to heat transfer by S.K.Som, Chapter 4, Example 14\n", + "Temperature distribution after 25 mins in °C\n", + "[[ 2.29192547e+02 2.91925466e+00 1.11801242e+00 4.34782609e-01\n", + " 1.86335404e-01 6.21118012e-02]\n", + " [ 8.75776398e+01 8.75776398e+00 3.35403727e+00 1.30434783e+00\n", + " 5.59006211e-01 1.86335404e-01]\n", + " [ 3.35403727e+01 3.35403727e+00 8.94409938e+00 3.47826087e+00\n", + " 1.49068323e+00 4.96894410e-01]\n", + " [ 1.30434783e+01 1.30434783e+00 3.47826087e+00 9.13043478e+00\n", + " 3.91304348e+00 1.30434783e+00]\n", + " [ 5.59006211e+00 5.59006211e-01 1.49068323e+00 3.91304348e+00\n", + " 1.02484472e+01 3.41614907e+00]\n", + " [ 3.72670807e+00 3.72670807e-01 9.93788820e-01 2.60869565e+00\n", + " 6.83229814e+00 8.94409938e+00]]\n" + ] + } + ], + "source": [ + " \n", + "import math\n", + "import numpy\n", + " \n", + "print\"Introduction to heat transfer by S.K.Som, Chapter 4, Example 14\"\n", + "#Nodal distance Deltax in m\n", + "deltax = 0.1;\n", + "#Time in seconds\n", + "t = 25*60;\n", + "#timestep deltaT in seconds\n", + "deltaT = 500;\n", + "#Number of increment\n", + "n = t/deltaT;\n", + "#Temperature raised in °C\n", + "To = 580.0;\n", + "#Using Eq. 4.114 for interior grid points, table 4.8 for exterior node\n", + "#Using Eq. 4.125a to 4.125f are written in matrix form\n", + "#Coefficient matrix A is\n", + "A = [[-3,1,0,0,0,0],[1,-3,1,0,0,0],[0,1,-3,1,0,0],[0,0,1,-3,1,0],[0,0,0,1,-3,1],[0,0,0,0,2,-3]]\n", + "#Coefficient matrix B is\n", + "B = [-600,-20,-20,-20,-20,-20];\n", + "#Temperature matrix is transpose of [T2 T3 T4 T5 T6 T7] where\n", + "#T2 to T7 are temperature in °C\n", + "#From Eq. 4.126\n", + "#Temperature distribution after one time step\n", + "T = numpy.linalg.inv(A)*B;\n", + "\n", + " \n", + "print\"Temperature distribution after 25 mins in °C\"\n", + "print T\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": "2.7.11" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} |