Height of a Cyanogen Absorption Column: 23-Chem-A3 Heat and Mass Transfer, May 2014, Q5

Question 5: Height of a Cyanogen Absorption Column (equal value)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. A dilute packed absorber with individual gas- and liquid-film coefficients and a curved equilibrium line, so the local slope (and hence the overall coefficient) changes up the column.

Quantity	Value
$k^{\prime}_ya$ / $k^{\prime}_xa$	312 / 1800 kmol/m³·h
Equilibrium	$y^*=5.3\,x^{1.07}$
$y_1$ (bottom) / $y_2$ (top)	0.0125 / $75\times10^{-6}$
Inlet water $x_2$	0
Gas $G$ / water $L$	85 / 350 kmol/m²·h

Find. The packed height $z$.

Figure 5 — The operating line (blue) lies above the convex equilibrium curve $y^*=5.3x^{1.07}$ (green). For each bulk point the interface composition is found along a tie line of slope $-k_xa/k_ya$ (red); the gas-film driving force is $y-y_i$.

Approach. Set the dilute operating line from the solute balance, then at each gas composition find the interface along a tie line of slope $-k^{\prime}_xa/k^{\prime}_ya$ (interface on the equilibrium curve), and integrate the gas-film transfer units $\int dy/(y-y_i)$; multiply by $H_G=G/k^{\prime}_ya$.

Operating line. A dilute solute balance with $x_2=0$ gives $x = \dfrac{G}{L}(y-y_2) = 0.2429(y-y_2)$; at the bottom $x_1 = 0.2429(0.0125-7.5\times10^{-5}) = 0.00302.$
Height of a gas-film transfer unit. $H_G = \dfrac{G}{k^{\prime}_ya} = \dfrac{85}{312} = \boxed{0.272\ \text{m}}$ (constant, since $k^{\prime}_ya$ is given).
Interface compositions (curved equilibrium). At each $y$ the interface $(x_i,y_i)$ satisfies flux continuity $k^{\prime}_ya(y-y_i)=k^{\prime}_xa(x_i-x)$ with $y_i=5.3x_i^{1.07}$ — i.e. a tie line of slope $-k^{\prime}_xa/k^{\prime}_ya=-5.77$ from the operating point to the equilibrium curve. This is solved point-by-point.
Number of gas-film transfer units. Integrating the gas-side driving force up the column, $$N_G = \int_{y_2}^{y_1}\frac{dy}{y-y_i} = 16.2,$$ the integrand peaking at the dilute (top) end where the 75 ppm target makes the driving force small.
Packed height. $$z = H_G\,N_G = 0.272\times16.2 = \boxed{4.4\ \text{m}}.$$

Quantity	Result
Bottom liquid $x_1$	0.00302
$H_G=G/k^{\prime}_ya$	0.272 m
Gas-film transfer units $N_G$	16.2
Packed height $z$	≈ 4.4 m

Check

Because the equilibrium line is curved, the overall coefficient $1/K^{\prime}_y=1/k^{\prime}_y+m/k^{\prime}_x$ uses the secant slope of the tie line, not the tangent $dy^*/dx$; substituting the tangent slope overstates the height (here by ~50%). The rigorous interface (tie-line) integration used above avoids that error.

Concept & Theory

Topic. Packed-tower height with individual film coefficients and a curved equilibrium line.

Key relations. $z=\dfrac{G}{k^{\prime}_ya}\displaystyle\int_{y_2}^{y_1}\dfrac{dy}{y-y_i}$; interface via $k^{\prime}_ya(y-y_i)=k^{\prime}_xa(x_i-x)$, $y_i=5.3x_i^{1.07}$.

Why this works. The two film fluxes must match at the interface, which forces the interface state onto the equilibrium curve along a line of slope $-k^{\prime}_xa/k^{\prime}_ya$. Integrating the local gas-film driving force $(y-y_i)$ over the column gives the true number of transfer units even when a single overall coefficient cannot be defined.

Common pitfall. Factoring $z=H_{OG}N_{OG}$ with a constant $m$ when the equilibrium line is curved, or reading the equilibrium at the interface liquid composition instead of solving the tie line simultaneously.

Source. Geankoplis §10.6 (film coefficients, curved equilibrium, Example 10.6-2); Treybal §5.4.

Question 5 of 7: Height of a Cyanogen Absorption Column

Question 5: Height of a Cyanogen Absorption Column (equal value)