Consider the multi-unit process shown below. Dashed lines are shown around diffe

Question

Consider the multi-unit process shown below. Dashed lines are shown around different Map subsystems around which you can perform balances. mol C/min n Map Distillation nh mol/min i Separator ru mo/min Column x4 molAlr 141 mmoVmin 0.55 mol Amol 0.19 mol BWmol (1-x) moll Bymol (1-x4-y) mold Carrol 0.26 mol C/mol 167 mol/min 153 mol Bmin 0.39 mol Arrol 0.61 mol C/mol What is the maximum number of independent balances that may be written around each subsystem? Number Number Distillation column: O Separator: 0 (Scroll down for more questions) Number Number Mixing Point: Overall Process: e Previous Give Up & View solution Check Answer Next Exit

Explanation / Answer

Input= out put is the key to all these balances

Acrosss the separator, there are two components, two independent equations can be written.

One is overall balance, n1= n2+153 ( Overall balance) (1)

Writing A balance, n1*x1= n2*x2( No A in 153 moles/min of B with drawn). (2)

Writing B balance can be obtained by Eq.1-Eq.2 Which is not independent

Across the mixing point, there are 3 components and three independent equations can be written.

One overall balance n2+n3=n4, (3)

n2*x2= n4x4 ( A balance) (4)

, n2*(1-x2)= n4y4 ( B balance) (5)

Eq.3- (Eq.4+Eq.5) gives the C balance equation.

Across the distillation column, there are three components and hence three material balance equations

n4= 141+167 ( Overall balance) (6),

n4x4= 167*0.39+141*0.55 ( A balance)- (7) and n4*y4= 167*0+141*0.19 (8)

Eq,6- (Eq.7+Eq.8) gives the balance equation for component C. Hence not independent equation.

Overall process, there are 3 components, A, B and C. Hence three indpenent equations can be written

Overall balance n1+n3=153+167+141 ( Overall balance) (9)

Component A balance , n1*x1= 167*0.39+141*0.55 ( A balance)- 10

Component B balance, n1*(1-x1)= 153+141*0.19 ( B balance)- 11

Overall process, there are 3 components, A, B and C. Hence three indpenent equations can be written

Overall balance n1+n3=153+167+141 ( Overall balance) (9)

Component A balance , n1*x1= 167*0.39+141*0.55 ( A balance)- 10

Component B balance, n1*(1-x1)= 153+141*0.19 ( B balance)- 11

Flow rates of outputs across the distillation column are given, 1st

, n4= 141+167=308 can be determined.

C is entering vide n3, hence C balance gives n3, n3= 167*0.61+141*0.26= 138.53 mol/min

Overall balance gives n1, n1+138.53= 153+167+141, n1= 322.47 moles/min

Then component A balance gives 322.47*x1 = 167*0.39+141*0.55, this will give x1.

Now n2 can be separated as n1=n2+153, n2=n1-153, since n2 is known, component balance across the separator gives x1, n1*x1=n2*x2, x2= (n1/n2)x2.

Now the balance of A across the separator n2x2= n4x4, x4= n2x2/n4.

B balance across the separator n2*(1-x2)= n4*y4, y4= n2*(1-x2)/n4.

Order-4 is the correct one.

Related Questions

Navigate

• Browse (All)
• Browse C
• Subjects

---