$(50$ points$)$ For flow inside tubes at constant volumetric flow rates $(Q),$ the mass transfer rate of a

component $A($either from the fluid stream to the tube inside surface or from the tube inside

surface to the fluid stream$)$ can be calculated using the following relations:

$Q(C_{A,\text{o}\text{u}\text{t}}-C_{A,in})=N_{A}S=S{k}_c$ubrace$(\frac{(C_{Ai}-C_{A,in})-(C_{Ai}-C_{A,\text{o}\text{u}\text{t}})}{ln(\frac{C_{At}-C_{A,in}}{C_{At}-C_{A,\text{o}\text{u}\text{t}}})}$ubrace${)}_{?}=(C_{Ai}-C_A{)}_{\text{m}\text{e}\text{a}\text{n}}$

where the final term is the log mean driving force at the inlet and outlet of the tube, $C_{AI}$ :

concentration of $A$ at the inside surface of the tube, $C_{A,\text{I}}$ : concentration of $A$ in the inlet fluid stream,

$C_{A,\text{o}\text{u}\text{t}\text{:}}$ concentration of $A$ in the outlet fluid stream, and $S$ : the inside surface area of the tube through

which component $A$ is transported.

a$)$ Derive this relation starting from a steady state component mass balance over a differential

element on the tube as given below.

$N_{A}dS=k_c(C_{Ai}-C_A)dS=Qd{C}_A$

$Q$ : volumetric flow rate $of$ the fluid,$\frac{m^3}{s}$

b$)$ Benzene$-$free air at $25C$ and $101\mathrm{.}3$kPa enters a $5-$cm$-$diameter tube at an average velocity of

$5\frac{m}{s}.$ The inner surface of the $6-m-$long tube is coated with a thin film of pure benzene at

$25C.$ The vapor pressure of benzene $(C6H6)$ at $25C$ is $13$kPa, and the solubility of air in

benzene is assumed to be negligible. Calculate the molar concentration of benzene in the

outlet air, and the evaporation rate of benzene in $k\frac{g}{h}$ given that the diffusivity of benzene

in air at $25C$ is ${0\mathrm{.}88}^{**}{10}^{-5}\frac{m^2}{s}.$