Water is to be cooled in a small packed column from 330 to 285 K by means of air flowing countercurrently. The rate of flow of liquid is 1400 cm³/m²s and the flowrate of the air, which enters at 295 K with a humidity of 60% is 3.0 m³/m²s. Calculate the required height of tower if the whole of the

Question

Water is to be cooled in a small packed column from 330 to 285 K by means of air flowing countercurrently. The rate of flow of liquid is 1400 cm³/m²s and the flowrate of the air, which enters at 295 K with a humidity of 60% is 3.0 m³/m²s. Calculate the required height of tower if the whole of the resistance to heat and mass transfer can be considered as being in the gas phase and the product of the mass transfer coefficient and the transfer surface per unit volume of column is 2 [latex]s^{-1}[/latex]. What is the condition of the air which leaves at the top?

Accepted Answer

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

[latex]H_{G1} = 1.003(295 - 273) + \mathscr{H} (2495 + 2.006(295 - 273))[/latex]

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

[latex]H_{G1}[/latex] = (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 × [latex]10^{-3}[/latex] m³/m²s

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

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

[latex] heta _{L1} = 285 K, H_{G1}[/latex] = 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:

[latex] heta _{L2} = 330 K, H_{G2}[/latex] = 122 kJ/kg

Again as in Problem 13.8, the relation between enthalpy and temperature at the interface [latex]H_{f}[/latex] vs. [latex] heta _{f}[/latex] 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: [latex]LC_{L}/G[/latex] = 1.66 and bottom coordinates: [latex] heta _{L1}[/latex] = 291 K and [latex]H_{G1}[/latex] = 47.5 kJ/kg is drawn as shown in Fig. 13c. At the top of the operating line:

[latex] heta _{L2}[/latex] = 330 K and [latex]H_{G2}[/latex] = 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:

[latex]H_{G1}[/latex] = 47.5 kJ/kg, [latex]H_{f1}[/latex] = 52.0 kJ/kg ∴ Δ[latex]H_{1}[/latex] = 4.5 kJ/kg

At the top of the column:

[latex]H_{G2}[/latex] = 112.5 kJ/kg, [latex]H_{f2}[/latex] = 382 kJ/kg ∴ Δ[latex]H_{2}[/latex] = 269.5 kJ/kg

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

[latex]H_{Gm}[/latex] = 82.0 kJ/kg, [latex]H_{fm}[/latex] = 152.5 kJ/kg ∴ Δ[latex]H_{m}[/latex] = 70.5 kJ/kg

∴ Δ[latex]H_{m}[/latex]/Δ[latex]H_{1}[/latex] = 15.70 and Δ[latex]H_{m}[/latex]/Δ[latex]H_{2}[/latex] = 0.262

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

Thus:

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

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

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