SME
Separation Processes
Membranes
Use this for “Brackish water holds 5 g/L NaCl at 20 °C. At 15 bar and Aw = 2 L/(m²·h·bar), what water flux crosses the membrane?”
Diagram
This is an example. Type your own values to start a new problem.
π (bar)
=
i
×
c (g/L)
M (g/mol)
×
0.08314
×
(
T (°C)
+
273.15)
Jw (L/(m²·h))
=
Aw (L/(m²·h·bar))
×
(
ΔP (bar)
−
π (bar)
)
Formulas
π
=
i
×
(
c
M
)
×
0.08314
×
(
T
+
273.15)
29.6915
=
2
×
(
35
58.44
)
×
0.08314
×
(25
+
273.15)
Jw
=
Aw
×
(ΔP
−
π
)
30.3085 = 1 × (60 − 29.6915)
Assumptions
•
The permeate is nearly salt-free, and the dilute van ’t Hoff form gives π.
•
No concentration polarization: the membrane face sees the feed’s own salt level.
•
R = 0.08314 L·bar/(mol·K); T is typed in °C and turned to kelvins.
Step-by-step
WE KNOW
c = 35 g/L, M = 58.44 g/mol, i = 2, T = 25 °C, ΔP = 60 bar, Aw = 1 L/(m²·h·bar)
FIND
osmotic pressure (π), water flux (Jw)
Step 1 · Find osmotic pressure (π)
Use π = i × (c ÷ M) × 0.08314 × (T + 273.15)
Van
’t
Hoff:
the
molar
concentration
of
ions,
i
×
c
M
,
times
RT
with
T
in
kelvins.
π
=
2
×
(
35
58.44
)
×
0.08314
×
(25
+
273.15)
π
=
2
×
0.5989
×
0.08314
×
298.15
π
=
29.6915
bar
Step 2 · Find water flux (Jw)
Use Jw = Aw × (ΔP − π)
Water
crosses
with
the
pressure
left
once
π
is
overcome,
times
the
permeability.
Jw = 1 × (60 − 29.6915)
Jw
=
30.3085
L/(
m
2
·h)
Check
Put the numbers back in. Both sides should match.
29.6915
=
2
×
(
35
58.44
)
×
0.08314
×
(25
+
273.15)
✓
30.3085 = 1 × (60 − 29.6915) ✓
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Ideal selectivity and permeate purity
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