5.10: Untitled Page 108
- Page ID
- 18241
Chapter 5
Special Condition:
( x )
0
(5‐75)
A o
so that Eq. 5‐74 takes the form
( x ) M
( y ) M
( y ) M
(5‐76)
A 2
A 1
A 3
At this point our objective is to determine ( y ) in terms of ( y ) , and we begin A 1
A 3
Figure 5‐12. Two‐unit extraction process
by arranging this result in the form
( y ) ( x )
M
M
( y )
(5‐77)
A 1
A 2
A 3
Here we note that the process equilibrium relation is given by Process equilibrium relation:
( y )
K
( x )
(5‐78)
A 2
eq,A
A 2
and use of this result allows us to simplify the mole balance for species A to ( y ) ( y )
M
M
K
( y )
(5‐79)
A 1
A 2
eq,A
A 3
We now use Eq. 5‐73 to determine ( y ) so that Eq. 5‐79 provides the following A 2
result
( y ) ( y ) A 1 A ( y ) (5‐80)
A 1
A 1
A 3
which can be solved for ( y ) to obtain
A 1
( y )
Two Equilibrium Stages:
A 3
( y )
(5‐81)
A 1
2
1 A A
Two‐Phase Systems & Equilibrium Stages
197
We are now ready to determine ( y ) when three equilibrium stages are A 1
employed as illustrated in Figure 5‐13.
Figure 5‐13. Three‐unit extraction process
In this case the molar balance for species A can be expressed as ( x ) M
( y ) M
( x ) M
( y ) M
(5‐82)
A 3
A 1
A o
A
4
molar flow of species A
molar flow of species A
out of the control volume
into the control volume
and we continue to impose the condition
Special Condition:
( x )
0
(5‐83)
A o
so that Eq. 5‐82 takes the form
( x ) M
( y ) M
( y ) M
(5‐84)
A 3
A 1
A 4
At this point our objective is to determine ( y ) as a function of ( y ) , and we A 1
A 4
begin by arranging this result in the form
( y ) ( x )
M
M
( y )
(5‐85)
A 1
A 3
A 4
Here we note that the process equilibrium relation is given by Process equilibrium relation:
( y )
K
( x )
(5‐86)
A 3
eq,A
A 3
and use of this result allows us to simplify the mole balance for species A to ( y ) ( y )
M
M
K
( y )
(5‐87)
A 1
A 3
eq,A
A 4