Example 18.5-1 Simultaneous Heat and Mass Transfer in Binary System

> restart; Starting with mass balances since they do not depend on Temperature

> eq1:=D(NAz)(z)=0; Eq. 18.5-1

> NAz:=z->-c*DAB*D(xA)(z)/(1-xA(z)); Eq. 18.5-3 (from 17.0-1 with NBz=0)

> s:=dsolve({eq1,xA(0)=xA0,xA(delta)=xAd},xA(z)):

> assign(s);xA:=unapply(xA(z),z);

> simplify(NAz(z),assume=positive); As stated in eq. 18.5-5

> dif4:=(1-xA(z))/(1-xA0)-((1-xAd)/(1-xA0))^(z/delta); Checking 18.5-4

> simplify(%,assume=positive); This does not get very far.

> expand(%); Proving eq. 18.5-4

> restart; Now moving to energy balances and restarting so NAz and xA will not be set.

> eq2:=D(ez)(z)=0; Eq. 18.5-2

> ez:=z->-k*D(T)(z)+NAz*CpAmol*(T(z)-T0); Eq. 18.5-7. Note that it is the molar specific heat of A that appears in this equation. It gives conduction plus convection in the film.

> s:=dsolve({eq2,T(0)=T0,T(delta)=Td},T(z)):

> assign(s);T:=unapply(T(z),z);

> dif8:=(T(z)-T0)/(Td-T0)-(1-exp(NAz*Cpmol*z/k))/(1-exp(NAz*Cpmol*delta/k)); Starting to check 18.5-8

> expand(%); This only gets us part way.

> simplify(%,assume=positive); This checks 18.5-8

> qzwall:=NAz->-k*D(T)(0); Treating the conductive term at the wall as a function of NAz

> qzwall(NAz)/limit(qzwall(NAz),NAz=0); so we can do this. This checks 18.5-9

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