Question 13.10: In a countercurrent packed column, n-butanol flows down at a......

The first stage is to calculate the enthalpy of the saturated gas by way of the saturated humidity, \mathscr{H}_{0} given by:

The enthalpy is then:

H_{f}=\frac{1}{(1+\mathscr{H}_{0})} x 1.001(\theta _{f} – 273) + \mathscr{H}_{0}[2.5(\theta _{f} – 273) + 590] kJ/kg
where 1.001 kJ/kg K is the specific heat of dry air.
Thus: H_{f}=\frac{1.001\theta_{f}-273.27}{(1+\mathscr{H}_{0})}+\mathscr{H}_{0}(2.5\theta_{f}-92.5) kJ/kg moist air
The results of this calculation are presented in the following table and H_{f} is plotted against \theta _{f} in Figure 13.21.
The modified enthalpy at saturation H^{\prime }_{f} is given by:

where from equation 13.70: λ’ = λ/b = (590/2.34) or 252 kJ/kg

b={\frac{\lambda}{\lambda^{\prime}}}                  (13.70)

∴ H_{f}^{\prime}\,=\,{\frac{(1.001\theta_{f}-273.27)}{(1+\mathscr{H}_{0})}}\,+\mathscr{H}_{0}(2.5\theta_{f}\,-\,430.5) kJ/kg moist air

These results are also given in the following Table and plotted as H^{\prime }_{f} against \theta _{f} in Figure 13.21.

The bottom of the operating line (point a) has coordinates, \theta _{L1} = 295 K and H_{G1}, where:
H_{G1} = 1.05(290 – 273) = 17.9 kJ/kg.
At a mean temperature of, say, 310 K, the density of air is:

and: G’ = (0.70 x 1.140) = 0.798 kg/m² s
Thus, the slope of the operating line becomes:

{\frac{L^{\prime}C_{L}}{G^{\prime}}}\,=\,{\frac{(0.25\times2.5)}{0.798}}\,=\,0.783 kJ/kg K

and this is drawn in as AB in Figure 13.22 and at θ_{L2} = 330 K, H_{G2} = 46 kJ/kg.

From equation 13.77: H^{\prime}_{G} = [H_{G} + (b – 1)s(\theta _{G} – \theta _{0})]/b

H_{G}^{\prime}={\frac{1}{b}}[H_{G}+(b-1)s(\theta_{G}-\theta_{0})]                  (13.77)

∴ H_{G1}^{'}={\frac{17.9+{(2.34-1)}1.05{(290-273)}}{2.34}}=17.87\ \mathrm{kJ/kg}

Point a coincides with the bottom of the column.

A line is drawn through a of slope – \frac{h_{L}}{h_{D}\rho b}=-\left(\frac{3h_{G}}{h_{D}\rho b}\right)\cdot\left(\frac{h_{D}\rho s}{h_{G}}\right)

= -3s   = -3.15 kJ/kg

This line meets curve RS at c (\theta _{f1}, H_{f1}^{\prime }) to give the interface conditions at the bottom of the column. The corresponding air enthalpy is given by point C whose co-ordinates are:
\theta _{f1} = 293 K          H_{f1} = 29.0 kJ/kg
The difference between the ordinates of c and a gives the driving force in terms of the modified enthalpy at the bottom of the column, or:

A similar construction is made at other points along the operating line with the results shown in the following table.

from which:

Substituting in equation 13.78:

\int_{H_{G1}}^{H_{G2}}\frac{\mathrm{d}H_{G}}{(H_{f}^{\prime}-H_{G}^{\prime})}=\frac{b h_{D}a\rho}{G^{\prime}}z                        (13.78)

and: z={\frac{(2.379\times0.798)}{(2.34\times0.1)}} = \underline{\underline{8.1\ m}}

It remains to evaluate the change in gas conditions.
Point e, (\theta _{G1} = 290 K, H_{G1} = 17.9 kJ/kg) represents the condition of the inlet gas. ec is now drawn in, and from equation 13.75, this represents dH_{G}/d\theta _{G}. As for the air-water system, this construction is continued until the gas enthalpy reaches H_{G2}. The final point is given by D at which \underline{\underline{\theta_{G2}=308\ K}}.

\frac{(H_{G}^{\prime}-H_{f}^{\prime})}{(\theta_{G}-\theta_{f})}=\frac{\mathrm{d}H_{G}}{\mathrm{d}\theta_{G}} (cf. equation 13.51)                       (13.75)

\frac{H_{G}-H_{f}}{\theta_{G}-\theta_{f}}=\frac{\mathrm{d}H_{G}}{\mathrm{d}\theta_{G}}                     (13.51)

It is fortuitous that, in this problem, H^{\prime }_{G} = H_{G}. This is not always the case and reference should be made to Section 13.7 for elaboration of this point.