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+clear;

+clc;

+

+//Example3.3[:Heat Loss through double pane windows]

+//Given:-

+k_g=0.78;//Thermal conductitvity of glass [W/m.K]

+k_a=0.026;//Thermal conductivity of air space[W/m.K]

+L_g=.004;//Thickness of glass layer[m]

+L_a=0.01;//Thickness of air space[m]

+h_in=10;//ConvectionHeat transfer coefficient on the inner surface of the window[W/m^2]

+h_out=40;//ConvectionHeat transfer coefficient on the outer surface of the window[W/m^2]

+T_1=20;//Outer wall Temperature [degree Celcius]

+T_2=-10;//Inner wall Temperature [degree Celcius]

+//Solution:-

+A=(0.8*1.5);//Area of glass window[m^2]

+//Convection Resistances

+R_conv1=1/(h_in*A);//Due to convection heat transfer between inner atmosphere and glass[degree Celcius/W]

+R_conv2=1/(h_out*A);//Due to convection heat transfer between outer atmosphere and glass[degree Celcius/W]

+//Conduction Resistances

+R_cond1=L_g/(k_g*A);//Due to conduction heat transfer through the glass[degree Celcius/W]

+R_cond2=R_cond1;//Glass Medium is seperated by air spac hence two glass mediums are created[degree Celcius/W]

+R_cond3=L_a/(k_a*A);//Due to conduction heat transfer through the air space[degree Celcius/W]

+//Net Resistance offered by window is the sum of all the individual resistances written in the oreder of their occurence

+R_total=R_conv1+R_cond1+R_cond2+R_cond3+R_conv2;//[degree Celcius/W]

+disp("degree Celcius/W",R_total,"The net resistance offered is")

+Q_=(T_1-T_2)/R_total;//[W]

+disp("W",Q_,"The steady rate of Heat transfer through the window is")

+//Inner surface temperature of the window is given by

+T1=T_1-(Q_*R_conv1);//[degree Celcius]

+disp("degree Celcius",T1,"Inner Surface Temperature of the window is")
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