// Implementation of example 3.5 // Basic and Applied Thermodynamics by P.K.Nag // page 57 clc clear rt=5000 // (rate of heat supply in kg/h) t1=15 // (in degree celsius) t2=1650 // (in degree celsius) mp=1535 // (melting point in degree celsius) lt=270 // (latent heat in kJ/kg*K) shs=0.502 // (specific heat in solid state in kJ/kg*K) shl=29.93 // (specific heat in liquid state in kJ/kg*K) e=0.7 // (efficiency) dn=6900 // (density in kg/m^3) wt=56 // (atomic wt of iron) ht=shs*(mp-t1)+lt+shl*(t2-mp)/wt; // ht is heat required to melt 1 kg of iron rm=(rt*ht); rate=(rm/e)/3600; disp("rating of furnace =") disp(rate) disp("kW") // since bath volume is 3 times the hourly melting rate V=(3*rt)/dn; // let d & l be the diameter & length and l=2d d=(V*2/%pi)^(1/3); l=(2*d); disp("diameter =") disp(d) disp("m") disp("length") disp(l) disp("m")