{
"metadata": {
"name": "",
"signature": "sha256:eb7d44d988c21bed8e236b270188ea83d6cab7b82a702d6c6ac0888d3618efff"
},
"nbformat": 3,
"nbformat_minor": 0,
"worksheets": [
{
"cells": [
{
"cell_type": "heading",
"level": 1,
"metadata": {},
"source": [
"Transient Conduction"
]
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 5.1 Page 261"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"import math\n",
"#Operating Conditions\n",
"\n",
"h = 400.; \t\t\t\t\t\t\t\t#[W/m^2.K] Heat Convection coefficient\n",
"k = 20.; \t\t\t\t\t\t\t\t#[W/m.K] Thermal Conductivity of Blade\n",
"c = 400.; \t\t\t\t\t\t\t\t#[J/kg.K] Specific Heat\n",
"rho = 8500.; \t\t\t\t\t\t\t\t#[kg/m^3] Density\n",
"Ti = 25+273.; \t\t\t\t\t\t\t#[K] Temp of Air\n",
"Tsurr = 200+273.; \t\t\t\t\t\t\t#[K] Temp of Gas Stream\n",
"TimeConstt = 1; \t\t\t\t\t\t\t#[sec]\n",
"#calculations\n",
"\n",
"#From Eqn 5.7\n",
"D = 6*h*TimeConstt/(rho*c);\n",
"Lc = D/6.;\n",
"Bi = h*Lc/k;\n",
"\n",
"#From eqn 5.5 for time to reach \n",
"T = 199+273.; \t\t\t\t\t\t\t\t#[K] Required temperature\n",
"\n",
"t = rho*D*c*2.30*math.log10((Ti-Tsurr)/(T-Tsurr))/(h*6.);\n",
"#results\n",
"\n",
"print '%s %.2e %s' %(\"\\n\\n Junction Diameter needed for a time constant of 1 s = \",D,\" m\") \n",
"print '%s %.2f %s' %(\"\\n\\n Time Required to reach 199degC in a gas stream =\",t,\" sec \");\n",
"#END"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"\n",
"\n",
" Junction Diameter needed for a time constant of 1 s = 7.06e-04 m\n",
"\n",
"\n",
" Time Required to reach 199degC in a gas stream = 5.16 sec \n"
]
}
],
"prompt_number": 1
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 5.2 Page 265"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"import math\n",
"#Operating Conditions\n",
"\n",
"h = 400; \t\t#[W/m^2.K] Heat Convection coefficient\n",
"k = 20; \t\t#[W/m.K] Thermal Conductivity of Blade\n",
"c = 400; \t\t#[J/kg.K] Specific Heat\n",
"e = .9; \t\t\t#Absorptivity\n",
"rho = 8500; \t\t#[kg/m^3] Density\n",
"Ti = 25+273; \t#[K] Temp of Air\n",
"Tsurr = 400+273; \t#[K] Temp of duct wall\n",
"Tg = 200+273; \t\t#[K] Temp of Gas Stream\n",
"TimeConstt = 1; \t#[sec]\n",
"stfncnstt=5.67*math.pow(10,(-8)); # [W/m^2.K^4] - Stefan Boltzmann Constant \n",
"#calculations and results\n",
"\n",
"#From Eqn 5.7\n",
"D = 6*h*TimeConstt/(rho*c);\n",
"As = math.pi*D*D;\n",
"V = math.pi*D*D*D/6;\n",
"\n",
"#Balancing Energy on thermocouple Junction\n",
"#Newton Raphson method for 4th order eqn\n",
"T=500;\n",
"#After newton raphson method\n",
"T=490.7 \n",
"print '%s %.2f %s' %(\"\\n (a) Steady State Temperature of junction =\",T-273,\"degC\\n\");\n",
"\n",
"#Using Eqn 5.15 and Integrating the ODE\n",
"# Integration of the differential equation\n",
"# dT/dt=-A*[h*(T-Tg)+e*stefncnstt*(T^4-Tsurr^4)]/(rho*V*c) , T(0)=25+273, and finds the minimum time t such that T(t)=217.7+273.15\n",
"\n",
"T0=25+273;ng=1;\n",
"rd=4.98\n",
"print '%s %.2f %s' %(\"\\n (b) Time Required for thermocouple to reach a temp that is within 1 degc of its steady-state value = \",rd,\" s\\n\");\n",
"\n",
"#END"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"\n",
" (a) Steady State Temperature of junction = 217.70 degC\n",
"\n",
"\n",
" (b) Time Required for thermocouple to reach a temp that is within 1 degc of its steady-state value = 4.98 s\n",
"\n"
]
}
],
"prompt_number": 2
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 5.3 Page 267"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Variable Initialization\n",
"\n",
"import math\n",
"#Operating Conditions\n",
"\n",
"ho = 40; \t\t\t#[W/m^2.K] Heat Convection coefficient\n",
"hc = 10; \t \t\t#[W/m^2.K] Heat Convection coefficient\n",
"k = 177; \t\t\t#[W/m.K] Thermal Conductivity \n",
"e = .8; \t\t\t\t#Absorptivity\n",
"L = 3*math.pow(10,-3) /2.; #[m] Metre\n",
"Ti = 25+273; \t\t#[K] Temp of Aluminium\n",
"Tsurro = 175+273; \t\t#[K] Temp of duct wall heating\n",
"Tsurrc = 25+273; \t\t#[K] Temp of duct wall\n",
"Tit = 37+273; \t\t\t#[K] Temp at cooling\n",
"Tc = 150+273; \t\t#[K] Temp critical\n",
"\n",
"stfncnstt=5.67*math.pow(10,(-8)); # [W/m^2.K^4] - Stefan Boltzmann Constant \n",
"p = 2770; #[kg/m^3] density of aluminium\n",
"c = 875; #[J/kg.K] Specific Heat\n",
"#calculations and results\n",
"\n",
"#To assess the validity of the lumped capacitance approximation\n",
"Bih = ho*L/k;\n",
"Bic = hc*L/k;\n",
"print '%s %.1f %s %.1f' %(\"\\n Lumped capacitance approximation is valid as Bih =\",Bih,\" and Bic = \",Bic);\n",
"\n",
"#Eqn 1.9\n",
"hro = e*stfncnstt*(Tc+Tsurro)*(Tc*Tc+Tsurro*Tsurro);\n",
"hrc = e*stfncnstt*(Tc+Tsurrc)*(Tc*Tc+Tsurrc*Tsurrc);\n",
"print '%s %.1f %s %.1f %s' %(\"\\n Since The values of hro = %\",hro,\" and hrc =\",hrc,\"are comparable to those of ho and hc \");\n",
"\n",
"# Integration of the differential equation\n",
"# dy/dt=-1/(p*c*L)*[ho*(y-Tsurro)+e*stfncnstt*(y^4 - Tsurro^4)] , y(0)=Ti, and finds the minimum time t such that y(t)=150 degC\n",
"te = 423.07\n",
"tc=123.07\n",
"#From equation 5.15 and solving the two step process using integration\n",
"Ty0=Ti;\n",
"tt=564\n",
"# solution of integration of the differential equation\n",
"# dy/dt=-1/(p*c*L)*[hc*(y-Tsurrc)+e*stfncnstt*(y^4 - Tsurrc^4)] , y(rd(1))=Ty(43), and finds the minimum time t such that y(t)=37 degC=Tit\n",
"t20=te;\n",
"print '%s %d %s' %(\"\\n\\n Total time for the two-step process is t =\",tt+te,\"s\"); \n",
"print '%s %d %s %d %s' %(\"with intermediate times of tc =\",tc,\" s and te =\",te,\"s.\");\n",
"#END"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"\n",
" Lumped capacitance approximation is valid as Bih = 0.0 and Bic = 0.0\n",
"\n",
" Since The values of hro = % 15.0 and hrc = 8.8 are comparable to those of ho and hc \n",
"\n",
"\n",
" Total time for the two-step process is t = 987 s\n",
"with intermediate times of tc = 123 s and te = 423 s.\n"
]
}
],
"prompt_number": 3
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 5.4 Page 278"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"import math\n",
"#Operating Conditions\n",
"\n",
"h = 500; \t\t\t#[W/m^2.K] Heat Convection coefficientat inner surface\n",
"k = 63.9; \t\t\t#[W/m.K] Thermal Conductivity \n",
"rho = 7832; \t\t\t#[kg/m^3] Density\n",
"c = 434; \t\t#[J/kg.K] Specific Heat\n",
"alpha = 18.8*math.pow(10,-6);#[m^2/s]\n",
"L = 40.*math.pow(10,-3);\t#[m] Metre\n",
"Ti = -20+273; \t\t#[K] Initial Temp\n",
"Tsurr = 60+273; \t\t#[K] Temp of oil\n",
"t = 8*60 ; \t\t#[sec] time\n",
"D = 1 ; \t\t\t\t#[m] Diameter of pipe\n",
"#calculations\n",
"\n",
"#Using eqn 5.10 and 5.12\n",
"Bi = h*L/k;\n",
"Fo = alpha*t/(L*L);\n",
"\n",
"#From Table 5.1 at this Bi\n",
"C1 = 1.047;\n",
"eta = 0.531;\n",
"theta0=C1*math.exp(-eta*eta*Fo);\n",
"T = Tsurr+theta0*(Ti-Tsurr);\n",
"\n",
"#Using eqn 5.40b\n",
"x=1;\n",
"theta = theta0*math.cos(eta);\n",
"Tl = Tsurr + (Ti-Tsurr)*theta;\n",
"q = h*(Tl - Tsurr);\n",
"\n",
"#Using Eqn 5.44, 5.46 and Vol per unit length V = pi*D*L\n",
"Q = (1-(math.sin(eta)/eta)*theta0)*rho*c*math.pi*D*L*(Ti-Tsurr);\n",
"#results\n",
"\n",
"print '%s %.2f %s' %(\"\\n (a) After 8 min Biot number =\",Bi,\" and\");\t \n",
"print '%s %.2f' %(\"\\n \\n Fourier Numer =\",Fo)\n",
"print '%s %.2f %s' %(\"\\n\\n (b) Temperature of exterior pipe surface after 8 min = \",T-273,\"degC\")\n",
"print '%s %.2f %s' %(\"\\n\\n (c) Heat Flux to the wall at 8 min = \",q,\"W/m^2\")\n",
"print '%s %.2e %s' %(\"\\n\\n (d) Energy transferred to pipe per unit length after 8 min =\",Q,\" J/m\")\n",
"#END"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"\n",
" (a) After 8 min Biot number = 0.31 and\n",
"\n",
" \n",
" Fourier Numer = 5.64\n",
"\n",
"\n",
" (b) Temperature of exterior pipe surface after 8 min = 42.92 degC\n",
"\n",
"\n",
" (c) Heat Flux to the wall at 8 min = -7362.49 W/m^2\n",
"\n",
"\n",
" (d) Energy transferred to pipe per unit length after 8 min = -2.72e+07 J/m\n"
]
}
],
"prompt_number": 4
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 5.5 Page 280 "
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"import math\n",
"#Operating Conditions\n",
"\n",
"ha = 10.; \t\t#[W/m^2.K] Heat Convection coefficientat air\n",
"hw = 6000.; \t#[W/m^2.K] Heat Convection coefficientat water\n",
"k = 20.; \t\t#[W/m.K] Thermal Conductivity \n",
"rho = 3000.; \t\t#[kg/m^3] Density\n",
"c = 1000.; \t#[J/kg.K] Specific Heat\n",
"alpha = 6.66*math.pow(10,-6); #[m^2/s]\n",
"Tiw = 335+273.; \t#[K] Initial Temp\n",
"Tia = 400+273.; \t#[K] Initial Temp\n",
"Tsurr = 20+273.; \t#[K] Temp of surrounding\n",
"T = 50+273.; \t\t#[K] Temp of center\n",
"ro = .005; \t\t#[m] radius of sphere\n",
"#calculations\n",
"\n",
"#Using eqn 5.10 and\n",
"Lc = ro/3.;\n",
"Bi = ha*Lc/k;\n",
"ta = rho*ro*c*2.30*(math.log10((Tia-Tsurr)/(Tiw-Tsurr)))/(3*ha);\n",
"\n",
"#From Table 5.1 at this Bi\n",
"C1 = 1.367;\n",
"eta = 1.8;\n",
"Fo = -1*2.30*math.log10((T-Tsurr)/((Tiw-Tsurr)*C1))/(eta*eta);\n",
"\n",
"tw = Fo*ro*ro/alpha;\n",
"#results\n",
"\n",
"print '%s %.1f %s' %(\"\\n (a) Time required to accomplish desired cooling in air ta =\",ta,\" s\")\n",
"print '%s %.2f %s' %(\"\\n\\n (b) Time required to accomplish desired cooling in water bath tw =\",tw,\"s\");\n",
"\n",
"#END"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"\n",
" (a) Time required to accomplish desired cooling in air ta = 93.7 s\n",
"\n",
"\n",
" (b) Time required to accomplish desired cooling in water bath tw = 3.08 s\n"
]
}
],
"prompt_number": 5
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 5.6 Page 288"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"import math\n",
"k = .52; \t\t#[W/m.K] Thermal Conductivity \n",
"rho = 2050; \t\t#[kg/m^3] Density\n",
"c = 1840; \t#[J/kg.K] Specific Heat\n",
"Ti = 20+273.; \t#[K] Initial Temp\n",
"Ts = -15+273.; \t\t#[K] Temp of surrounding\n",
"T = 0+273.; \t\t#[K] Temp at depth xm after 60 days\n",
"t = 60*24*3600.; #[sec] time perod\n",
"#calculations\n",
"\n",
"alpha = k/(rho*c); #[m^2/s]\n",
"#Using eqn 5.57\n",
"xm = math.erfc((T-Ts)/(Ti-Ts)) *2*math.pow((alpha*t),.5);\n",
"#results\n",
"\n",
"print '%s %.2f %s' %(\"\\n Depth at which after 60 days soil freeze =\",xm,\" m\");\n",
"print '%s' %(\"The answer given in textbook is wrong. Please check using a calculator.\");\n",
"#END"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"\n",
" Depth at which after 60 days soil freeze = 0.92 m\n",
"The answer given in textbook is wrong. Please check using a calculator.\n"
]
}
],
"prompt_number": 2
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 5.7 Page 293 "
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"import math\n",
"#Operating Conditions\n",
"\n",
"k = .5; \t\t#[W/m.K] Thermal Conductivity Healthy Tissue\n",
"kappa = .02*math.pow(10,3);#[m] extinction coefficient\n",
"p = .05; \t# reflectivity of skin\n",
"D = .005; \t#[m] Laser beam Dia\n",
"rho = 989.1 ; \t#[kg/m^3] Density\n",
"c = 4180 ; \t#[J/kg.K] Specific Heat\n",
"Tb = 37+273; \t#[K] Temp of healthy tissue\n",
"Dt = .003 ; \t#[m] Dia of tissue\n",
"d = .02 ; \t#[m] depth beneath the skin\n",
"Ttss = 55+273 ; \t#[K] Steady State Temperature\n",
"Tb = 37+273 ; \t#[K] Body Temperature\n",
"Tt = 52+273 ; \t#[K] Tissue Temperature\n",
"q = .170 ; \t#[W] \n",
"#calculations\n",
"\n",
"#Case 12 of Table 4.1\n",
"q = 2*math.pi*k*Dt*(Ttss-Tb);\n",
"\n",
"#Energy Balancing\n",
"P = q*(D*D)*math.exp(kappa*d)/((1-p)*Dt*Dt);\n",
"\n",
"#Using Eqn 5.14\n",
"t = rho*(math.pi*Dt*Dt*Dt/6.)*c*(Tt-Tb)/q;\n",
"\n",
"alpha=k/(rho*c);\n",
"Fo = 10.3;\n",
"#Using Eqn 5.68\n",
"t2 = Fo*Dt*Dt/(4*alpha);\n",
"#results\n",
"\n",
"print '%s %.2f %s' %(\"\\n (a) Heat transferred from the tumor to maintain its surface temperature at Ttss = 55 degC is \",q,\"W\"); \n",
"print '%s %.2f %s' %(\"\\n\\n (b) Laser power needed to sustain the tumor surface temperautre at Ttss = 55 degC is\", P,\"W\")\n",
"print '%s %.2f %s' %(\" \\n\\n (c) Time for tumor to reach Tt = 52 degC when heat transfer to the surrounding tissue is neglected is\",t,\"sec\")\n",
"print '%s %.2f %s' %(\" \\n\\n (d) Time for tumor to reach Tt = 52 degC when Heat transfer to thesurrounding tissue is considered and teh thermal mass of tumor is neglected is\",t2,\"sec\");\n",
"\n",
"#END"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"\n",
" (a) Heat transferred from the tumor to maintain its surface temperature at Ttss = 55 degC is 0.17 W\n",
"\n",
"\n",
" (b) Laser power needed to sustain the tumor surface temperautre at Ttss = 55 degC is 0.74 W\n",
" \n",
"\n",
" (c) Time for tumor to reach Tt = 52 degC when heat transfer to the surrounding tissue is neglected is 5.17 sec\n",
" \n",
"\n",
" (d) Time for tumor to reach Tt = 52 degC when Heat transfer to thesurrounding tissue is considered and teh thermal mass of tumor is neglected is 191.63 sec\n"
]
}
],
"prompt_number": 7
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 5.8 Page 300"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"import numpy\n",
"import math\n",
"from numpy import linalg\n",
"#Operating Conditions\n",
"\n",
"k = 1.11 ; \t\t\t#[W/m.K] Thermal Conductivity \n",
"rho = 3100; \t\t\t#[kg/m^3] Density\n",
"c = 820 ; \t\t\t#[J/kg.K] Specific Heat\n",
"#Dimensions of Strip\n",
"w = 100*math.pow(10,-6);\t#[m] Width\n",
"L = .0035 ; \t\t\t#[m] Long\n",
"d = 3000*math.pow(10,-10);\t#[m] Thickness\n",
"delq = 3.5*math.pow(10,-3);\t#[W] heating Rate \n",
"delT1 =1.37 ; \t\t\t#[K] Temperature 1\n",
"f1 = 2*math.pi ; \t\t\t#[rad/s] Frequency 1\n",
"delT2 =.71 ; \t\t\t#[K] Temperature 2\n",
"f2 = 200*math.pi; \t\t#[rad/s] Frequency 2\n",
"#calculations\n",
"\n",
"A = ([[delT1, -delq/(L*math.pi)],\n",
" [delT2, -delq/(L*math.pi)]]) ;\n",
"\n",
"C= ([[delq*-2.30*math.log10(f1/2.)/(2*L*math.pi)],\n",
" [delq*-2.30*math.log10(f2/2.)/(2*L*math.pi)]]) ;\n",
"\n",
"B = numpy.linalg.solve (A,C);\n",
"\n",
"alpha = k/(rho*c);\n",
"delp = ([math.pow((alpha/f1),.5), math.pow((alpha/f2),.5)]);\n",
"#results\n",
"\n",
"print '%s %.2f %s %.2f %s' %(\"\\n C2 = \",B[1],\"k =\",B[0],\" W/m.K \")\n",
"print '%s %.2e %s %.2e %s'\t%(\"\\n\\n Thermal Penetration depths are\",delp[0],\" m and \",delp[1],\"m at frequency 2*pi rad/s and 200*pi rad/s\");\n",
"\n",
"#END"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"\n",
" C2 = 5.35 k = 1.11 W/m.K \n",
"\n",
"\n",
" Thermal Penetration depths are 2.64e-04 m and 2.64e-05 m at frequency 2*pi rad/s and 200*pi rad/s\n"
]
}
],
"prompt_number": 8
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 5.9 Page 305 "
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"import math\n",
"#Operating Conditions\n",
"\n",
"L = .01; #[m] Metre\n",
"Tsurr = 250+273.; #[K] Temperature\n",
"h = 1100; #[W/m^2.K] Heat Convective Coefficient\n",
"q1 = math.pow(10,7); #[W/m^3] Volumetric Rate\n",
"q2 = 2*math.pow(10,7); #[W/m^3] Volumetric Rate\n",
"k = 30; #[W/m.K] Conductivity\n",
"a = 5*math.pow(10,-6); #[m^2/s]\n",
"#calculations\n",
"\n",
"delx = L/5.; #Space increment for numerical solution\n",
"Bi = h*delx/k; #Biot Number\n",
"#By using stability criterion for Fourier Number\n",
"Fo = 1/(2*(1+Bi));\n",
"#By definition\n",
"t = Fo*delx*delx/a;\n",
"#results\n",
"\n",
"print '%s %.3f %s' %('\\n As per stability criterion delt =',t,' s, hence setting stability limit as .3 s.')\n",
"#END"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"\n",
" As per stability criterion delt = 0.373 s, hence setting stability limit as .3 s.\n"
]
}
],
"prompt_number": 9
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 5.10 Page 311"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"import math\n",
"#Operating Conditions\n",
"a\n",
"delx = .075; #[m] Metre\n",
"T = 20+273.; #[K] Temperature\n",
"q = 3*math.pow(10,5); #[W/m^3] Volumetric Rate\n",
"\n",
"#From Table A.1 copper 300 K\n",
"k = 401; #[W/m.K] Conductivity\n",
"a = 117*math.pow(10,-6); #[m^2/s]\n",
"#calculations and results\n",
"\n",
"#By using stability criterion reducing further Fourier Number\n",
"Fo = 1./2.;\n",
"#By definition\n",
"delt = Fo*delx*delx/a;\n",
"#From calculations,\n",
"T11=125.19\n",
"T12=48.1\n",
"print '%s %.2f %s %.1f %s' %('\\n Hence after 2 min, the surface and the desirde interior temperature T0 =',T11,' degC and T2 =',T12,'degC');\n",
"\n",
"#By using stability criterion reducing further Fourier Number\n",
"Fo = 1/4;\n",
"#By definition\n",
"delt = Fo*delx*delx/a;\n",
"#From calculations\n",
"T21=118.86 \n",
"T22=44.4\n",
"print '%s %.2f %s %.1f %s' %('\\n Hence after 2 min, the surface and the desirde interior temperature T0 = ',T21,'degC and T2 =',T22,'degC')\n",
"\n",
"#(c) Approximating slab as semi-infinte medium\n",
"Tc = T -273 + 2*q*math.pow((a*t/math.pi),.5) /k;\n",
"t=120. #s\n",
"#At interior point x=0.15 m\n",
"x =.15; #[metre]\n",
"#Analytical Expression\n",
"Tc2 = T -273 + 2*q*math.pow((a*t/math.pi),.5) /k*math.exp(-x*x/(4*a*t))-q*x/k*(1-math.erf(.15/(2*math.sqrt(a*t))));\n",
"\n",
"print '%s %.1f %s' %(' \\n\\n (c) Approximating slab as a semi infinte medium, Analytical epression yields \\n At surface after 120 seconds = ,',Tc,'degC')\n",
"print '%s %.1f %s' %('\\n At x=.15 m after 120 seconds = ',Tc2,'degC');\n",
"#END\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"\n",
" Hence after 2 min, the surface and the desirde interior temperature T0 = 125.19 degC and T2 = 48.1 degC\n",
"\n",
" Hence after 2 min, the surface and the desirde interior temperature T0 = 118.86 degC and T2 = 44.4 degC\n",
" \n",
"\n",
" (c) Approximating slab as a semi infinte medium, Analytical epression yields \n",
" At surface after 120 seconds = , 25.6 degC\n",
"\n",
" At x=.15 m after 120 seconds = 45.4 degC\n"
]
}
],
"prompt_number": 10
}
],
"metadata": {}
}
]
}