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diff --git a/sample_notebooks/UmangAgarwal/Sample_Notebook.ipynb b/sample_notebooks/UmangAgarwal/Sample_Notebook.ipynb new file mode 100644 index 00000000..34fb4a40 --- /dev/null +++ b/sample_notebooks/UmangAgarwal/Sample_Notebook.ipynb @@ -0,0 +1,128 @@ +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
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