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-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

diff --git a/sample_notebooks/UmangAgarwal/Sample_Notebook_Umang.ipynb b/sample_notebooks/UmangAgarwal/UmangAgarwal_version_backup/Sample.ipynb
index 1eb49726..1eb49726 100755
--- a/sample_notebooks/UmangAgarwal/Sample_Notebook_Umang.ipynb
+++ b/sample_notebooks/UmangAgarwal/UmangAgarwal_version_backup/Sample.ipynb