Added(A)/Deleted(D) following books

 A  Microwave_engineering__by_D.M.Pozar_/Chapter_10_ACTIVE_MICROWAVE_CIRCUITS.ipynb
A  Microwave_engineering__by_D.M.Pozar_/Chapter_12_INTRODUCTION_TO_MICROWAVE_SYSTEMS.ipynb
A  Microwave_engineering__by_D.M.Pozar_/Chapter_1_ELECTROMAGNETIC_THEORY.ipynb
A  Microwave_engineering__by_D.M.Pozar_/Chapter_2_TRANSMISSION_LINE_THEORY.ipynb
A  Microwave_engineering__by_D.M.Pozar_/Chapter_3_TRANSMISSION_LINE_AND_WAVEGUIDES.ipynb
A  Microwave_engineering__by_D.M.Pozar_/Chapter_4_MICROWAVE_NETWORK_ANALYSIS.ipynb
A  Microwave_engineering__by_D.M.Pozar_/Chapter_5_IMPEDENCE_MATCHING_AND_TUNNING.ipynb
A  Microwave_engineering__by_D.M.Pozar_/Chapter_6_MICROWAVE_RESONATORS.ipynb
A  Microwave_engineering__by_D.M.Pozar_/Chapter_7_POWER_DIVIDERS_DIRECTIONAL_COUPLERS_AND_HYBRIDS.ipynb
A  Microwave_engineering__by_D.M.Pozar_/Chapter_8_MICROWAVE_FILTERS.ipynb
A  Microwave_engineering__by_D.M.Pozar_/Chapter_9_THEORY_AND_DESIGN_OF_FERRIMAGNETIC_COMPONENTS.ipynb
A  Microwave_engineering__by_D.M.Pozar_/screenshots/Screen_Shot_2015-12-21_at_10.35.42_pm.png
A  Microwave_engineering__by_D.M.Pozar_/screenshots/Screen_Shot_2015-12-21_at_10.37.21_pm.png
A  Microwave_engineering__by_D.M.Pozar_/screenshots/Screen_Shot_2015-12-21_at_10.37.54_pm.png
A  sample_notebooks/PrashantSahu/Chapter-2-Molecular_Diffusion_-_Principles_of_Mass_Transfer_and_Separation_Process_by_Binay_K_Dutta.ipynb
A  sample_notebooks/UmangAgarwal/Sample_Notebook.ipynb
A  sample_notebooks/UmangAgarwal/Sample_Notebook_1.ipynb

2 files changed, 256 insertions, 0 deletions

diff --git a/sample_notebooks/UmangAgarwal/Sample_Notebook.ipynb b/sample_notebooks/UmangAgarwal/Sample_Notebook.ipynb
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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_1.ipynb b/sample_notebooks/UmangAgarwal/Sample_Notebook_1.ipynb
new file mode 100644
index 00000000..34fb4a40
--- /dev/null
+++ b/sample_notebooks/UmangAgarwal/Sample_Notebook_1.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