## Info

(6) Since the light nonkeys appear to ail go to the distillate, and the heavy nonkeys all to the bottom, only the keys needs to be considered. In addition

Calculations are shown in Table 3.4. The propane in the bottom diminishes to 0.55/40.1 = 1.4 percent, while the butane in the distillate diminishes to (17 -16.57)/59.9 = 0.7 percent. The calculated value of B is 40.1, which equals the specified 40.1; a second trial is not necessary.

3.2.8 The analytical x-y diagram: Smoker's equation

The Smoker equation (59) is convenient to use in binary separations with a large number of stages. The equation assumes constant relative volatility and constant molar overflow. The main application is in superfractionators such as ethylene-ethane and isobutane-n-butane separations. The Smoker equation is essentially an analytical solution of the x-y diagram. The Smoker equation is

log-

xM = [(R + 1)2 + (q - 1)*d]/CR + q) (3.34) and k is the root of the quadratic equation (0 < k < 1)

The Smoker equation must be applied individually for the rectifying and stripping sections. The values used for the rectifying section are m = Rl(R + 1) (3.36c)

x„ = (3.39a) Similarly, for the stripping section

Hohmann (60) presents an extension of the Smoker equation to complex columns.

Example 3.7 Solve Example 2.1 using Smoker's method. Assume an average relative volatility of 2.49.

solution (a) Rectifying section:

*int = [(3 + D0.4 + (0.75 - 1)0.951/(3 + 0.75) = 0.363 (3.34)

0.75(2.49 - l)k2 + [0.75 + 0.2375(2.49 - 1) - 2.49)A + 0.2375 = 0 (3.35) Solving 1.1175£2 - 1.3861& + 0.2375 = 0,

, 1.3861 ± V(1.3861)2 - 4 x 1.1175 x 0.2375 A --2TTU75-= 1.035 or 0.2053

Since 0 < k < 1, use k = 0.2053, x'0 = 0.95 - 0.2053 = 0.7447 (3.31)

Numerator of the top log term in Eq. (3.30) is

0.7447(1 - [0.75 x 1.3060(2.49 - 1)0.1577/(2.49 - 0.75 x 1.30602)]} = 0.6031 Denominator of the top log term in Eq. (3.30) is

0.1577{1 - [0.75 x 1.3060(2.49 - 1)0.7447/(2.49 - 0.75 x 1.30602)]} = 0.0161 Bottom term in Eq. (3.30) is 2.49/(0.75 x 1.3062) = 1.9465

Nr = [log (0.6031/0.0161)]/log 1.9465 = 5.44 lb) Similarly for the stripping section

b = (3 + 1)0.4 + (0.75 - 1)0.95 - (3 + 0.75)0.10 = "°-°557 (3"376)

= 0.10 (3.396) 1.5514(2.49 - 1)fc2 + [1.5514 - 0.0557(2,49 - 1) - 2.49]fe - 0.0557 = 0 (3.35) 2.3116A2 - 1.0216& - 0.0557 = 0

1,0216 ± V(1.0216)2 + 4 x 2.3116 x 0,0557 *--2x 2.3116-= °"491 °r "°-0491

Numerator of the top by term in Eq. (3.30) is

-0.1277{1 - [1.5514 x 1.7316(2.49 - l)(-0.3910)/(2.49 - 1.5514 1.73162)]} =