Page 306 - Separation process engineering

where B is the bottoms flow rate of the organic phase. Since F = 10, x = 0.15, and z C10 = 0.5, we
bot
can solve for n C10,vapor and B. Eq. (8-18) gives n once T boiler is known. Since the feed, bottoms, and
w
vapor are all at T boiler , the energy balance simplifies to

λ
n C10 C10 = (moles of water condensed in still)λ w
D. Do it. a. Perry and Green (1997) give the following n-decane vapor pressure data (vapor pressure
in mm Hg and T in °C):

A very complete table of water vapor pressures is given in that source (see Problem 8.D10). As a
first guess, try 95.5°C, where (VP) = 645.7 mm.Hg. Then Eq. (8-15),
w
(VP) C10 C10 in org + (VP) x = p tot
x
w w in w
becomes

(60)(0.15) + (645.7)(1.0) = 654.7 < 745 = p tot

where we have assumed completely immiscible phases so that x = 1.0. This temperature is too low.
w
Approximate solution of Eq. (8-15) eventually gives T boiler = 99°C and (VP) C10 = 66 mm Hg.

b. Solving the mass balances, Eqs. (8-19), the kmol/h of vapor are

(8-20)

which is 586.2 kg/h. Eq. (8-18) becomes

or

which is 5347.1 kg/h.
c. The moles of water to vaporize the decane is

where λ = H vap —h and the saturated vapor and liquid enthalpies at 99 °C are interpolated from
liq
Tables 2-249 and 2-352 in Perry and Green (1997) for decane and water, respectively.

E. Check. A check for complete immiscibility is advisable since all the calculations are based on this
assumption.