solution (a) First estimate tn and tm using Eq. (3.29) tn = (70 + 211)/2 = 140.5°F fm(308 4 211)/2 = 259.5°F Using the material balance in Example 2.4

D = 59.9, B = 40.1 lb-mole/h Assuming constant molar overflow,

From the problem definition, N = 20, M = 12. Due to the partial condenser, N = 20 + 1 = 21. Since the feed is mostly vapor, is calculated using Eq. (3.286).

Calculations for the first trial are shown in Table 3.4. The calculated value of B is 40.2, which approximates the specified value of 40.1 in the material balance in Example 2.4. The propane in the bottoms is 0.73/40.2 = 1.8 percent, while the /¡-butane in the distillate is (17 - 16.46)/59.8 = 0.9 percent. These values are higher than those used in Examples 2.4 and 3.4.

Depending on the initial guesses, the calculated value of B may not equal the specified B. A calculated B higher than specified indicates that tm and /„ are too low, and that the bottom product contains excessive lights. An additional trial is then needed, at higher t„ and/or tm. Usually, the estimated feed temperature is raised; this raises both tn and tm. Since the top and bottom temperatures are fixed by the products dew and bubble points, they are best maintained constant until Bxb adds up to the specified amount. Conversely, if the calculated Bxb is lower than specified, the estimated feed temperature should be lowered. The above is analogous to an operator actuating the control temperature to obtain a desired split.

table 3.4 Calculation for Example 3.6





Component K,

0 0

Post a comment