Question 13.9: Water is to be cooled in a small packed column from 330 to 2...

As in Problem 13.8, assuming the relevant latent and specific heat capacities:

From Fig. 13.4, at θ = 295 and 60% humidity, \mathscr{H} = 0.010 kg/kg and hence:

H_{G1} = (1.003 × 22) + 0.010(2495 + 44.13) = 47.46 kJ/kg

In the inlet air, water vapour = 0.010 kg/kg dry air or (0.010/18)/(1/29) = 0.016 kmol/kmol dry air.

Thus the flow of dry air = (1 – 0.016)3.0 = 2.952 m³/m²s.

Density of air at 295 K = (29/22.4)(273/293) = 1.198 kg/m³.

and mass flow of dry air = (1.198 × 2.952) = 3.537 kg/m²s.

Liquid flow = 1.4 × 10^{-3} m³/m²s

and mass flow of liquid = (1.4 × 10^{-3} × 1000) = 1.4 kg/m²s.

The slope of the operating line is thus: LC_{L}/G = (1.40 × 4.18)/3.537 = 1.66 and the coordinates of the bottom of the line are:

\theta _{L1} = 285  K, H_{G1} = 47.46 kJ/kg

From these data, the operating line may be drawn in as shown in Fig. 13b and the top
point of the operating line is:

\theta _{L2} = 330  K, H_{G2} = 122 kJ/kg

Again as in Problem 13.8, the relation between enthalpy and temperature at the interface H_{f} vs. \theta _{f} is drawn in Fig. 13b. It is seen that the operating line cuts the saturation curve, which is clearly an impossible situation and, indeed, it is not possible to cool the water to 285 K under these conditions. As discussed in Section 13.6.1, with mechanical draught towers, it is possible, at the best, to cool the water to within, say, 1 deg K of the wet bulb temperature. From Fig. 13.4, at 295 K and 60% humidity, the wet-bulb temperature of the inlet air is 290 K and at the best water might be cooled to 291 K. In the present case, therefore, 291 K will be chosen as the water outlet temperature.

Thus an operating line of slope: LC_{L}/G = 1.66 and bottom coordinates: \theta _{L1} = 291 K and H_{G1} = 47.5 kJ/kg is drawn as shown in Fig. 13c. At the top of the operating line:

\theta _{L2} = 330 K and H_{G2} = 112.5 kJ/kg
As an alternative to the method used in Problem 13.8, the approximate method of Carey and Williamson (equation 13.54) is adopted.

At the bottom of the column:

H_{G1} = 47.5 kJ/kg, H_{f1} = 52.0 kJ/kg ∴ ΔH_{1} = 4.5 kJ/kg

At the top of the column:

H_{G2} = 112.5 kJ/kg, H_{f2} = 382 kJ/kg ∴ ΔH_{2} = 269.5 kJ/kg

At the mean water temperature of 0.5(330 + 291) = 310.5 K:

H_{Gm} = 82.0 kJ/kg, H_{fm} = 152.5 kJ/kg ∴  ΔH_{m} = 70.5 kJ/kg

∴ ΔH_{m}/ΔH_{1} = 15.70 and ΔH_{m}/ΔH_{2} = 0.262

and from Fig. 13.17: f = 0.35 (extending the scales)

Thus:

height of packing, z = \int_{H_{G1}}^{H_{G2}}[dH_{G}/H_{f}-H_{G1})]G/h_{D}a \rho           (equation 13.53)

= (0.35 × 3.537)/(2.0 × 1.198) = 0.52 m

Due to the close proximity of the operating line to the line of saturation, the gas will be saturated on leaving the column and will therefore be at 100% humidity. From Fig. 13c the exit gas will be at 306 K.