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+clear;

+clc;

+

+//Example3.6[Heat Loss through a Composite Wall]

+//Given:-

+//We consider a 1m deep and 0.25 m high portion of the wall since it is representative of the entire wall

+//Assuming any cross-section of the wall normal to the x-direction to be isothermal

+k_b=0.72;//thermal conductivity of bricks[W/m.K]

+k_p=0.22;//thermal conductivity of plaster layers[W/m.K]

+k_f=0.026;//thermal conductivity of foam layers[W/m.K]

+T_in=20;//Indoor Temperature[dgeree Celcius]

+T_out=-10;//Outdoor Temperature[dgeree Celcius]

+h_in=10;//Inner heat transfer coefficient[W/m^2.K]

+h_out=25;//Outer heat transfer coefficient[W/m^2.K]

+L_f=0.03;//Thickness of foam layer[m]

+L_p=0.02;//Thickness of plaster[m]

+L_b=0.16;//Thickness of brick wall[m]

+L_c=0.16;//Thickness of central plaster layer[m]

+A1=(0.25*1);//[m^2]

+A2=(0.015*1);//[m^2]

+A3=(0.22*1);//[m^2]

+//Resistances offered:-

+R_in=1/(h_in*A1);//Resistance to conevction heat transfer from inner surface[degree Celcius/W]

+R1=L_f/(k_f*A1);//Conduction Resistance offered by outer foam layer[degree Celcius/W]

+R2=L_p/(k_p*A1);//Conduction Resistance offered by Outer side Plaster Wall[degree Celcius/W]

+R6=R2;//Conduction Resistance offered by Inner side Plaster Wall[degree Celcius/W]

+R3=L_c/(k_p*A2);//Conduction Resistance offered by one side central Plaster wall[degree Celcius/W]

+R5=R3;//Conduction Resistance offered by other side central Plaster wall[degree Celcius/W]

+R4=L_b/(k_b*A3);//Conduction Resistance offered by Brick Wall[degree Celcius/W]

+R_out=1/(h_out*A1);//Convection Resistance from outer surface[degree Celcius/W]

+//R_in,R1,R2,R6,R_out are connected in series

+//R3,R4,R5 are connected in parallel

+R_mid=1/((1/R3)+(1/R4)+(1/R5));//Effective Parrallel Resistance

+R_total=(R_in+R1+R2+R_mid+R6+R_out);//[degree Celcius/W]

+disp("degree Celcius/W",R_total,"Net Resistance offered is")

+Q_=(T_in-T_out)/R_total;//[W]

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

+Q_p=Q_/A1;//[W/m^2]

+disp("W/m^2",Q_p,"Heat Transfer per unit area is")

+A_total=3*5;//Total Area of wall[m^2]

+Q_total=Q_p*A_total;//[W]

+disp("W",Q_total,"Thr rate of heat transfer through the entire wall")