Example 21.2-1 Evaporation of a Freely Falling Drop

> restart;

> eq:=WmA=kxm*A*(xA0-xAinf)+xA0*(WmA+WmB); eq. 21.1.-12

> WmB:=0; The air is assumed to be stationary.

> A:=Pi*D1^2; Area of the surface of the drop.

> WmA:=solve(eq,WmA); eq. 21.2-26

> T0:=70*F;

> Tinf:=140*F;

> Tf:=(T0+Tinf)/2;

> xA0:=.0247*atm/P; As given in problem and confirmed with steam. Steam gives 0.363 psia as the vapor pressure at 70F.

> xAinf:=0*atm/P;

> xAf:=(xA0+xAinf)/2;

> P:=1*atm;

> c:=T->P/(Rgas*(T+460*F)*R/F);

> cf:=c(Tf);

> Rgas:=82.05*cm^3*atm/(mol*K);

> cf;

> cf:=cf*1.8*R/K;

> rhof:=cf*MW;

> MW:=28.8*g/mol;

> rhof;

> TfC:=(Tf-32*F)*C/(1.8*F);

> muf:=.0191*.01*g/(cm*s); Table 1.1-1 give .0191 cp. Multiply by 0.01 to get Poise which is g/(cm*s)

> DABf:=2.295e-5*m^2/s*(100*cm/m)^2; From dcalc in matlab. note that it differs quite a bit from the value given by BS&L

> Scf:=muf/(rhof*DABf); The difference is from DABf.

> Ren:=D1*vinf*rhof/muf;

> D1:=.05*cm;

> vinf:=215*cm/s;

> Ren;

> kxm:=cf*DABf/D1*(2+0.6*Ren^.5*Scf^(1/3)); eq. 21.2-25

> evalf(WmA);

>

The observed value for Dwater-air at 313K is 0.292. so it looks like

BS&L are closer than the Matlab dcalc fn.

About 17% lower than the value found by the text