Water is to be cooled in a packed tower from 330 to 295 K by means of air flowing countercurrently. The liquid flows at the rate of 275 cm³/m² s and the air at 0.7 m³/m² s. The entering air has a temperature of 295 K and a humidity of 20%. Calculate the required height of tower and the condition of

Question

Water is to be cooled in a packed tower from 330 to 295 K by means of air flowing countercurrently. The liquid flows at the rate of 275 cm³/m² s and the air at 0.7 m³/m² s. The entering air has a temperature of 295 K and a humidity of 20%. Calculate the required height of tower and the condition of the air leaving at the top.
The whole of the resistance to heat and mass transfer can be considered as being within the gas phase and the product of the mass transfer coefficient and the transfer surface per unit volume of column ([latex]h_{D}[/latex]a) may be taken as 0.2 [latex]s^{-1}[/latex].

Accepted Answer

Assuming, the latent heat of water at 273 K = 2495 kJ/kg

specific heat capacity of dry air = 1.003 kJ/kg K

specific heat capacity of water vapour = 2.006 kJ/kg K
then the enthalpy of the inlet air stream is:

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

From Fig. 13.4, when θ = 295 K, at 20% humidity, [latex] \mathscr{H} = 0.003[/latex] kg/kg, and:

[latex]H_{G1}[/latex] = (1.003 × 22) + 0.003(2495 + (2.006 × 22)) = 29.68 kJ/kg

In the inlet air, the humidity is 0.003 kg/kg dry air or (0.003/18)/(1/29) = 0.005 kmol/kmol dry air.

Hence the flow of dry air = (1 - 0.005)0.70 = 0.697 m³/m² s.

Density of air at 295 K = (29/22.4)(273/295) = 1.198 kg/m².
and hence the mass flow of dry air = (0.697 × 1.198) = 0.835 kg/m² s
and the mass flow of water = 275 × [latex]10^{-6}[/latex] m³/m² s or (275 × [latex]10^{-6}[/latex] × 1000) = 0.275 kg/m² s.

The slope of the operating line, given by equation 13.37 is:

[latex]G^{\prime} dH_G=L^{\prime}C_Ld heta_L[/latex] (13.37)

[latex]LC_{L}/G[/latex] = (0.275 × 4.18/0.835) = 1.38
The coordinates of the bottom of the operating line are:

[latex] heta _{L1}[/latex] = 295 K and [latex]H_{G1}[/latex] = 29.7 kJ/kg

Hence, on an enthalpy–temperature diagram (Fig. 13a), the operating line of slope 1.38 is drawn through the point (29.7, 295).

The top point of the operating line is given by [latex] heta _{L2}[/latex] = 330 K, and from Fig. 13a, [latex]H_{G2}[/latex] = 78.5 kJ/kg.

From Figs 13.4 and 13.5 the curve representing the enthalpy of saturated air as a function of temperature is obtained and drawn in. This plot may also be obtained by
calculation using equation 13.60.

[latex]\ln \frac{( heta _{G1}- heta _{s})}{( heta _{G2}- heta _{s})} =\frac{h_{G}a}{G^{\prime}s} z[/latex] ( 13.60)

The integral:

[latex]\int dH_{G}/(H_{f}-H_{G})[/latex]

is now evaluated between the limits [latex]H_{G1}[/latex] = 29.68 kJ/kg and [latex]H_{G2}[/latex] = 78.5 kJ/kg, as follows:

[latex]H_{G}[/latex]	θ	[latex]H_{f}[/latex]	([latex]H_{f}-H_{G}[/latex])	1/([latex]H_{f}-H_{G}[/latex])
29.7	295	65	35.5	0.0283
40	302	98	58	0.0173
50	309	137	87	0.0115
60	316	190	130	0.0077
70	323	265	195	0.0051
78.5	330	408	329.5	0.0030

From a plot of 1/([latex]H_{f}-H_{G}[/latex]) and [latex]H_{G}[/latex] the area under the curve is 0.573.

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)

= (0.573 × 0.835)/(0.2 × 1.198)
= 1.997, say 2.0 m

In Fig. 13a, a plot of [latex]H_{G}[/latex] and [latex] heta _{G}[/latex] is obtained using the construction given in Section 13.6.3. and shown in Fig. 13.16. From this plot, the value of [latex] heta _{G2}[/latex] corresponding to [latex]H_{G2}[/latex] = 78.5 kJ/kg is 300 K. From Fig. 13.5 the exit air therefore has a humidity of 0.02 kg/kg which from Fig. 13.4 corresponds to a percentage humidity of 90%.

Question 13.8: Water is to be cooled in a packed tower from 330 to 295 K by...

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