From 64d949698432e05f2a372d9edc859c5b9df1f438 Mon Sep 17 00:00:00 2001 From: kinitrupti Date: Fri, 12 May 2017 18:40:35 +0530 Subject: Revised list of TBCs --- .../UmangAgarwal/Sample_Notebook.ipynb | 128 --------------------- 1 file changed, 128 deletions(-) delete mode 100755 sample_notebooks/UmangAgarwal/Sample_Notebook.ipynb (limited to 'sample_notebooks/UmangAgarwal/Sample_Notebook.ipynb') diff --git a/sample_notebooks/UmangAgarwal/Sample_Notebook.ipynb b/sample_notebooks/UmangAgarwal/Sample_Notebook.ipynb deleted file mode 100755 index 34fb4a40..00000000 --- a/sample_notebooks/UmangAgarwal/Sample_Notebook.ipynb +++ /dev/null @@ -1,128 +0,0 @@ -Sample Notebook - Heat and Mass Transfer by R.K. Rajput : Chapter 1 - Basic Concepts -author: Umang Agarwal - - -# Example 1.1 Page 16-17 - -L=.045; #[m] - Thickness of conducting wall -delT = 350 - 50; #[C] - Temperature Difference across the Wall -k=370; #[W/m.C] - Thermal Conductivity of Wall Material -#calculations -#Using Fourier's Law eq 1.1 -q = k*delT/(L*10**6); #[MW/m^2] - Heat Flux -#results -print '%s %.2f %s' %("\n \n Rate of Heat Transfer per unit area =",q," W"); -#END - -# Example 1.2 Page 17 - -L = .15; #[m] - Thickness of conducting wall -delT = 150 - 45; #[C] - Temperature Difference across the Wall -A = 4.5; #[m^2] - Wall Area -k=9.35; #[W/m.C] - Thermal Conductivity of Wall Material -#calculations -#Using Fourier's Law eq 1.1 -Q = k*A*delT/L; #[W] - Heat Transfer -#Temperature gradient using Fourier's Law -TG = - Q/(k*A); #[C/m] - Temperature Gradient -#results -print '%s %.2f %s' %("\n \n Rate of Heat Transfer per unit area =",Q," W"); -print '%s %.2f %s' %("\n \n The Temperature Gradient in the flow direction =",TG," C/m"); -#END - -# Example 1.3 Page 17-18 - -x = .0825; #[m] - Thickness of side wall of the conducting oven -delT = 175 - 75; #[C] - Temperature Difference across the Wall -k=0.044; #[W/m.C] - Thermal Conductivity of Wall Insulation -Q = 40.5; #[W] - Energy dissipitated by the electric coil withn the oven -#calculations -#Using Fourier's Law eq 1.1 -A = (Q*x)/(k*delT); #[m^2] - Area of wall -#results -print '%s %.2f %s' %("\n \n Area of the wall =",A," m^2"); -#END - -# Example 1.4 Page 18-19 - -delT = 300-20; #[C] - Temperature Difference across the Wall -h = 20; #[W/m^2.C] - Convective Heat Transfer Coefficient -A = 1*1.5; #[m^2] - Wall Area -#calculations -#Using Newton's Law of cooling eq 1.6 -Q = h*A*delT; #[W] - Heat Transfer -#results -print '%s %.2f %s' %("\n \n Rate of Heat Transfer =",Q," W"); -#END - -# Example 1.5 Page 19 - -L=.15; #[m] - Length of conducting wire -d = 0.0015; #[m] - Diameter of conducting wire -A = 22*d*L/7; #[m^2] - Surface Area exposed to Convection -delT = 120 - 100; #[C] - Temperature Difference across the Wire -h = 4500; #[W/m^2.C] - Convective Heat Transfer Coefficient -print 'Electric Power to be supplied = Convective Heat loss'; -#calculations -#Using Newton's Law of cooling eq 1.6 -Q = h*A*delT; #[W] - Heat Transfer -Q = round(Q,1); -#results -print '%s %.2f %s' %("\n \n Rate of Heat Transfer =",Q," W"); -#END - -# Example 1.6 Page 20-21 - -T1 = 300 + 273; #[K] - Temperature of 1st surface -T2 = 40 + 273; #[K] - Temperature of 2nd surface -A = 1.5; #[m^2] - Surface Area -F = 0.52; #[dimensionless] - The value of Factor due geometric location and emissivity -sigma = 5.67*(10**-8) #(W/(m^2 * K^4)) - Stephen - Boltzmann Constant -#calculations -#Using Stephen-Boltzmann Law eq 1.9 -Q = F*sigma*A*(T1**4 - T2**4) #[W] - Heat Transfer -#Equivalent Thermal Resistance using eq 1.10 -Rth = (T1-T2)/Q; #[C/W] - Equivalent Thermal Resistance -#Equivalent convectoin coefficient using h*A*(T1-T2) = Q -h = Q/(A*(T1-T2)); #[W/(m^2*C)] - Equivalent Convection Coefficient -#results -print '%s %.2f %s' %("\n \n Rate of Heat Transfer =",Q," W"); -print '%s %.2f %s' %("\n The equivalent thermal resistance =",Rth," C/W"); -print '%s %.2f %s' %("\n The equivalent convection coefficient =",h," W/(m^2 * C)"); -#END - -# Example 1.7 Page 21-22 - -L = 0.025; #[m] - Thickness of plate -A = 0.6*0.9; #[m^2] - Area of plate -Ts = 310; #[C] - Surface Temperature of plate -Tf = 15; #[C] - Temperature of fluid(air) -h = 22; #[W/m^2.C] - Convective Heat Transfer Coefficient -Qr = 250; #[W] - Heat lost from the plate due to radiation -k = 45; #[W/m.C] - Thermal Conductivity of Plate -#calculations -# In this problem, heat conducted by the plate is removed by a combination of convection and radiation -# Heat conducted through the plate = Convection Heat losses + Radiation Losses -# If Ti is the internal plate temperature, then heat conducted = k*A*(Ts-Ti)/L -Qc = h*A*(Ts-Tf); #[W] - Convection Heat Loss -Ti = Ts + L*(Qc + Qr)/(A*k); #[C] - Inside plate Temperature -#results -print '%s %.2f %s' %("\n \n Rate of Heat Transfer =",Ti," C"); -#END - -# Example 1.8 Page 22 - -Ts = 250; #[C] - Surface Temperature -Tsurr = 110; #[C] - Temperature of surroundings -h = 75; #[W/m^2.C] - Convective Heat Transfer Coefficient -F = 1; #[dimensionless] - The value of Factor due geometric location and emissivity -sigma = 5.67*(10**-8) #(W/(m^2 * K^4)) - Stephen - Boltzmann Constant -k = 10; #[W/m.C] - Thermal Conductivity of Solid -#calculations -# Heat conducted through the plate = Convection Heat losses + Radiation Losses -qr = F*sigma*((Ts+273)**4-(Tsurr+273)**4) #[W/m^2] - #[W] - Heat lost per unit area from the plate due to radiation -qc = h*(Ts-Tsurr); #[W/m^2] - Convection Heat Loss per unit area -TG = -(qc+qr)/k; #[C/m] - Temperature Gradient -#results -print '%s %.2f %s' %("\n \n The temperature Gradient =",TG," C/m"); -#END -- cgit