{
"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": [
"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*Â°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\""
]
},
{
"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": [
"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*Â°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*Â°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"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Ex4.3:pg-138"
]
},
{
"cell_type": "code",
"execution_count": 4,
"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": [
"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"
]
},
{
"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": [
"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 Â°C\n",
"T = 80.0;\n",
"#Cold water temperature in Â°C\n",
"Tf = 25.0;\n",
"#Heat transfer coefficient in W/(m**2*Â°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 80Â°C to 30Â°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"
]
},
{
"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": [
"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 Â°C\n",
"Ti = 30;\n",
"#Temperature of hot gas in Â°C\n",
"Tinfinity = 780;\n",
"#Heat transfer coefficient in W/(m**2*K)\n",
"h = 20;\n",
"#Surface temperaute to be achieved in Â°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"
]
},
{
"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 Â°C\n",
"T= 355.5\n",
"Heat transfer rate in MJ\n",
"Q= 100.0\n"
]
}
],
"source": [
"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 Â°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 Â°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 Â°C\n",
"T = Tinfinity+(0.93*0.7)*(Ti-Tinfinity);\n",
"print\"Temperature at this distance in Â°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"
]
},
{
"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 Â°C\n",
"T= 300.0\n",
"Heat transfer rate per unit length in MJ/m\n",
"Q= 33.2639222145\n"
]
}
],
"source": [
"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 Â°C\n",
"Ti = 530;\n",
"#Temperature of surrounding in Â°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 Â°C\n",
"T = Tinfinity+(0.9*0.6)*(Ti-Tinfinity);\n",
"print\"Temperature at this radius in Â°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"
]
},
{
"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": [
"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 Â°C\n",
"Ti = 530;\n",
"#Temperature of surrounding in Â°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 Â°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"
]
},
{
"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 Â°C\n",
"T= 260.3\n"
]
}
],
"source": [
"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 Â°C\n",
"Ti = 730;\n",
"#Temperature of surrounding in Â°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 Â°C\n",
"T = Tinfinity+z*(Ti-Tinfinity);\n",
"print\"Tempearture of bar in Â°C\"\n",
"print\"T=\",T"
]
},
{
"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 310Â°C is nearly 1200s or 20 minutes.\n",
"T= 0.5976\n"
]
}
],
"source": [
"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 30Â°C.\n",
"#Due to an accidental fire,the beam is suddenly exposed to hot gases at temprature,Tinf=730Â°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 310Â°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 310Â°C is nearly 1200s or 20 minutes.\" \n",
"print\"T=\",T"
]
},
{
"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 255Â°C at a depth of 80mm, in minutes is\n",
"T= 255\n"
]
}
],
"source": [
"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=550Â°C.\n",
"#The temprature of one surface is suddenly changed to Tinf=50Â°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=255Â°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 255Â°C at a depth of 80mm, in minutes is\"\n",
"t=(10/0.38)**2/60\n",
"print\"T=\",T"
]
},
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"metadata": {},
"source": [
"## Ex4.12:pg-186"
]
},
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{
"name": "stdout",
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"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 Â°C is\n",
"T= -29851.5095103\n"
]
}
],
"source": [
"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 30Â°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 Â°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)"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Ex4.14:pg-187 "
]
},
{
"cell_type": "code",
"execution_count": 2,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
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"text": [
"Introduction to heat transfer by S.K.Som, Chapter 4, Example 14\n",
"Temperature distribution after 25 mins in Â°C\n",
"T= [[ 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": [
"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 Â°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 Â°C\n",
"#From Eq. 4.126\n",
"#Temperature distribution after one time step\n",
"T = numpy.linalg.inv(A)*B;\n",
"print\"Temperature distribution after 25 mins in Â°C\"\n",
"print\"T=\",T"
]
}
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