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4.6. A distillation column is a process unit in which a feed mixture is separated by multiple partial vapor izations and condensations to form two or more product streams. The overhead product stream is rich in the most volatile components of the feed mixture (the ones that vaporize most readily), and the bottom product stream is rich in the least volatile components The following flowchart shows a distillation column with two feed streams and three product streams: i (kg/h) 0.03 kg Bkg 0.97 kg Ck 1200 kg/h 0.70 kg /kg yakg B/kg 4kg Ckg) 5300 kgh kg Ake yzkg B/kg) ig(kg/h) 0.60 kg Bkg 0.40 kg Ckg (a) How many independent material balances may be written for this system? (b) How many of the unknown flow rates and/or mole fractions must be specified before the othens may be calculated? (See Example 4.3-4. Also, remember what you know about the component mole fractions of a mixture-for example, the relationship between x2 and 2.) Briefly explain your answer (c) Suppose values are given for mand x2.Give a series of equations, each involving only a single unknown, for the remaining variables. Circle the variable for which you would solve. (Once a variable has been calculated in one of the equations, it may appear in subsequent equations without being counted as an unknown)

Explanation / Answer

Ans 4.6

Part a

Number of independent material balance equations

Overall material balance

m1 + 5300 = m3 + 1200 + m5 ............... Eq1

Component A balance

m1 * 0 + 5300 * x2 = m3 + 1200*0.70 + m5*0

5300*x2 = m3 + 840 ................ Eq2

Component B balance

m1 * 0.03 + 5300 * y2 = 1200*y4 + m5*0.60 ............ Eq3

Component C balance

m1 * 0.97 + 5300 * 0 = 1200*z4 + m5*0.40

0.97*m1 = 1200*z4 + m5*0.40 .......... Eq4

Relation between mass fractions

Sum of mass fraction of all components = 1

x2 + y2 = 1 ........ Eq5

Part a

Total Number of independent material balance equations = 5

Number of unknowns = 7

Part b

Degrees of freedom = Number of unknowns - Total Number of independent material balance equations

= 7 - 5 = 2

We need to fix 2 variables to fix the overall system

One mass flow rate from m1, m3, m5

one mass fraction from x2, y2, y4, z4

Part C

Let m1 and x2 is known, we can calculate the remaining unknowns m3, m5, y2, y4, z4

from eq1

m3 + m5 = m1 + 4100........... Eq4

From eq2

m3 = 5300*x2 - 840 .......... Eq5

Put the value of m3 from eq5 into eq4 and calculate m5

m5 = m1 + 4100 - (5300*x2 - 840)............ Eq6

From eq3, calculate y4

m1 * 0.03 + 5300 * y2 - m5*0.60 = 1200*y4

y4 = (1/1200)[0.03m1 + 5300(1-x2) - 0.60{m1 + 4100 - (5300*x2 - 840)} ]......... Eq7

z4 = 1 - 0.70 - y4 ....... Eq8

We calculated the unknowns m3 (from eq5) , m5 (from eq6) , y2 (from eq5) , y4 (from eq7), z4 (from Eq 8)

(with known m1 & x2)

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