Solution

Equation (7.39) is used to find the diameter of the tower. Iw should be 0.08 or above, but for stripping column it is very difficult to obtain such a high Iw value and therefore it will be attempted to select Iw = 0.04. It implies that a cross sectional area of 4 m2 should be used according to equation (7.39), giving L = 2500 kg/h/m2, provided that a is about 60 1/ m. It is the case for 4 inch raschig rings (see Table 7.4), which are chosen. The minimum ratio air to water is about 3000, which is selected. It will correspond to 30 000 m3 / h air or 36 000 kg / h. It gives a flow rate 9 000 kg/h/m2 or 7 500 m / h, corresponding to about 2 m / s, which is fully acceptable see Table 7.4.

The flooding point is found from equation (7.40), using Figure 7.14. Q is found to be:

which will give a Z value of about 5. As pL is 0,001 kg / m*s and dh3/2 is 0,01 (see Table 7.4), Iw is therefore 0.05 or slightly more than found above, which is acceptable.

HtG is found from equation (7.38), as the constants are found in Table 7.3:

Sc for air at 15° C can be found from the the viscosity of air (0.0648 kg / m • h ), the diffusion coefficient ( 0.0392 m2 / h) and the specific gravity (1.2 kg / m3) to be 1.37. HtG is now found from equation (7.38):

R is found from equation (7.42).

Henry's constant i found from (7.11) to be 0.69 bar. Therefore R = 1.66, which by use of Fig. 7.16 is translated to 3 transfer units, as the fraction 0.9 is removed.

The height of the tower is calculated to be 3.9* 3 = 11. 7 m.

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