QUESTION $2$

Acetone is being absorbed by water in a packed tower having a cross$-$sectional area of $0\mathrm{.}186m2$ at $293K$ and atm$).$ The inlet air contains $2\mathrm{.}6$mol$\%$ acetone and the outlet contains $0\mathrm{.}5$mol$\%.$ The gas flow is $13\mathrm{.}65$kgmolinertai$\frac{r}{h}.$ The pure$-$water inlet flow is $45\mathrm{.}36$kgmol water $\frac{?}{h}.$ The following are the mass transfer film coefficients and the corresponding values at the interface. The equilibrium data is given by $y=1\mathrm{.}186x.$

Film mass transfer coefficient Value Unit

$3\mathrm{.}78{10}^{-2}$Kgmo$\frac{l}{s}.m^3.$mol frac

$6\mathrm{.}16{10}^{-2}$kgmo$\frac{l}{s}.m^3{?}^3$mol frac

$($a$)$ Determine the flowrate and composition of each stream: $L_1,x_1,V_1,y_1,L_2,x_2,V_2,y_2($some are already given but some you may have to calculate from the mass balance$).$

$($b$)$ Use $H_6$ and $N_6$ and also $H_L$ and $N_L$ to determine the tower height. The formula is given below. Compare the results.

$$ The interface composition is

$H_C=\frac{V}{k_{}^{\text{'}}aS}=\frac{V}{k_{y}a(1-y{)}_{MS}S}$

$H_L=\frac{L}{K_x^{\text{'}}aS}=\frac{L}{K_{x}a(1-x{)}_{M,A}S}$

$H_{OC}=\frac{V}{K_y^{\text{'}}aS}=\frac{V}{K_{y}a(1-y{)}_{**MN}S}$

$H_{OL}=\frac{L}{K_x^{\text{'}}aS}=\frac{L}{K_{x}a(1-x{)}_{**M}S}$

$N_o=\frac{y_1-y_0}{(y-y_0{)}_1}$

$(22\mathrm{.}5-36)$

$,y_{i1}=0\mathrm{.}0154$

$y_{i2}=0\mathrm{.}0020$

Where

$(y-y_i{)}_M=\frac{(y_1-y_{i1})-(y_2-y_{i2})}{ln[\frac{y_1-y_{i1}}{y_2-y_{i2}}]}$