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+clear;
+clc;
+
+//Example3.15[Cost of Heat Loss through walls in winter]
+//Given:-
+R_va_insu=2.3;//thickness to thermal conductivity ratio[m^2.degreeCelcius/W]
+L1=12;//length of first wall of house[m]
+L2=12;//length of second wall of house[m]
+L3=9;//length of third wall of house[m]
+L4=9;//length of fourth wall of house[m]
+H=3;//height of all the walls[m]
+T_in=25;//Temperature inside house[degree Celcius]
+T_out=7;;//average temperature of outdoors on a certain day[degree Celcius]
+ucost=0.075;//Unit Cost of elctricity[$/kWh]
+h_in=8.29,h_out=34.0;//Heat transfer coefficients for inner and outer surface of the walls respectively[W/m^2.degree Celcius]
+v=24*(3600/1000);//velocity of wind[m/s]
+//Solution:-
+//Heat transfer Area of walls=(Perimeter*Height)
+A=(L1+L2+L3+L4)*H;//[m^2]
+//Individual Resistances
+R_conv_in=1/(h_in*A);//Convection Resistance on inner surface of wall[degree Celcius/W]
+R_conv_out=1/(h_out*A);//Convection Resistance on outer surface of wall[degree Celcius/W]
+R_wall=R_va_insu/A;//Conduction resistance to wall[degree Celcius/W]
+//All resistances are in series
+R_total=R_conv_in+R_wall+R_conv_out;//[degree Celcius/W]
+Q_=(T_in-T_out)/R_total;//[W]
+disp("W",Q_,"The steady rate of heat transfer through the walls of the house is")
+delta_t=24;//Time period[h]
+Q=(Q_/1000)*delta_t;//[kWh/day]
+disp("kWh/day",Q,"The total amount of heat lost through the walss during a 24 hour period ")
+cost=Q*ucost;//[$/day]
+disp("per day",cost,"Cost of heat consumption is $")