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+clear;

+clc;

+

+//Example3.9[Heat Loss from an Insulated Electric Wire]

+//Given:-

+k_insu=0.15;//[W/m.degree Celcius]

+V=8;//Voltage drop across wire[Volts]

+I=10;//Current flowimg through the wire[Amperes]

+T_atm=30;//Temperature of atmosphere to which wire is exposed[degree Celcius]

+h=12;//heat transfer coefficient[W/m^2.degree Celcius]

+L=5;//length of wire[m]

+D=0.003;//diameter of wire[m]

+t=0.002;//thickness of insulation[m]

+r=(D/2)+t;//Effective radius[m]

+//Solution:-

+//Rate of heat generated in the wire becomes equal to the rate of heat transfer

+Q_=V*I;//[W]

+disp("W",Q_,"Heat generated in the wire is")

+A2=2*%pi*r*L;//Outer surface area[m^2]

+//Resistances offered

+R_conv=1/(h*A2);//Convection resistance for the outer sueface of insulation[degree Celcius/W]

+R_insu=(log(r/(D/2)))/(2*%pi*k_insu*L);//Conduction resitance for the plastic insulation[degree Celcius/W]

+//Effective Resistance

+R_total=R_conv+R_insu;//[degree Celcius/W]

+//Interface Temperature can be determined from

+T1=T_atm+(Q_*R_total);//[degree Celcius]

+disp("degree Celcius",T1,"The interface temperature is")

+//Critical radius 

+r_cr=k_insu/h;//[m]

+disp("mm",r_cr*1000,"The critical radius of insulation of the plastic cover is")

+//Larger value of critical radius ensures that increasing the thickness of insulation upto critical radius will increase the rate of heat transfer
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