//Example 6.4 //Calculate the physical properties of the liquid. //Given W1=200 //kg/h, rate of entering toluene muv=10^-5 //kg/m s, viscosity of toluene vapour mul=2.31*10^-4 //kg/m s, viscosity of benzene rhol=753 //kg/m^3, density of benzene rhov=3.7 //kg/m^3, density of toluene vapour Cpl=1968 //j/kg C, specific heat of benzene kl=0.112 //W/m C, thermal conductivity of benzene T1=160 //C tube wall temp. T2=120 //C , saturated temp. Te=T1-T2 //C, excess temp. Lv=3.63*10^5 //j/kg, enthalpy of vaporization s=1.66*10^-2 //N/m, surface tension //Calculation of hc & hb w=0.125 //m, mean step size d=0.0211 //, internal diameter of tube G=W1/(3600*%pi/4*(d^2)) //kg/m^2 s, mass flow rate Re1=G*(1-w)*d/mul //Reynold no. Prl=Cpl*mul/kl //Prandtl no. //from eq. 6.23 x=(w/(1-w))^(0.9)*(rhol/rhov)^(0.5)*(muv/mul)^0.1 //let x=1/succepsibility //from eq. 6.22 F=2.35*(x+0.231)^0.736 //factor signifies 'liquid only reynold no.' to a two phase reynold no. //from eq. 7.21 Re2=10^-4*Re1*F^1.25 //Reynold no. //from eq. 6.18 S=(1+0.12*Re2^1.14)^-1 //boiling supression factor //from eq. 6.15 hc=0.023*Re1^(0.8)*Prl^(0.4)*(kl/d)*F //W/m^2 C, forced convection boiling part //from eq. 6.16 mulv=(1/rhov)-(1/rhol) //m^3/kg, kinetic viscosity of liquid vpaour dpsat=Te*Lv/((T2+273)*mulv) //N/m^2, change in saturated presssure //nucleate boiling part hb hb=1.218*10^-3*(kl^0.79*Cpl^0.45*rhol^0.49*Te^0.24*dpsat^0.75*S/(s^0.5*mul^0.29*Lv^0.24*rhov^0.24)) h=hc+hb //W/m^2 C, total heat transfer coefficient //calculation of required heat transfer area a=5 //%, persentage change in rate of vaporization W2=W1*a/100 //kg/h, rate of vaporization W2_=W2/3600 //kg/s Q=W2_*Lv //W,heat load A=Q/(h*Te) //m^2, area of heat transfer l=A/(%pi*d) //m, required length of tube //from table 6.2 Tl=0.393 printf("The total tube length is %f m",Tl)