2) Acrylic acid will be extracted from its aqueous mixture, containing 5 mole percent acrylic acid,
by using pure di-isopropyl ether (DIISO-01) through stage wise contact in a counter-current
extraction system at 25 °C.
a) Draw a block diagram of the system containing N stages, and please mark and label
all the streams in it.
b) If the acrylic acid content in the raffinate product is 10 mole percent of that in the
raffinate feed, what will be the minimum solvent flow rate required for the raffinate
feed flow rate of 20 kgmol/h?
c) When the solvent flowrate used is 1.5 times the minimum solvent flowrate, what will be
the number of stages needed?
d) Show the flow rates and compositions of the raffinate and extract phases around the
second stage?
e) Calculate the recovery in the system.
Solubility and distribution diagrams of acrylic acid, water and di-isopropyl ether ternary system
at 25 °C is given in Figure Q 2.
pMmin = 0.0475 0.0497
Smin = 1.0455
Mmin = 21.0455
Stages R xA xB E yA yB
1 18 0.0335 0.0044 4 0.2275 0.3799
2 18 0.0235 0.0037 3 0.1838 0.5398
3 18 0.0168 0.0032 2 0.1441 0.6649
4 18 0.0118 0.0029 2 0.1045 0.7701
5 18 0.0086 0.0028 2 0.0663 0.8533
6 18 0.0048 0.0025 2 0.037 0.905
Solubility Curves of
AcrylicAcid in Water and in DIISO-01
at 25 °C
$x_{DIISO-01}$ or $y_{DIISO-01}$
$x_{AcrylicAcid}$ or $y_{AcrylicAcid}$
Distribution Curve of AcrylicAcid
between Water and DIISO-01
at 25 °C
$y_{AcrylicAcid}$ mol fraction of AcrylicAcid in DIISO-01
$x_{AcrylicAcid}$ mol fraction of AcrylicAcid in Water
Figure Q 2 Solubility and distribution diagrams of acrylic acid, water, and di-isopropyl ether
ternary system at 25 °C.