Transcript Example 2.15 in 3rd edition Example 2.14 Consider the situation shown in Figure 2.5 where on side A of the membrane we.

Example 2.15 in 3rd edition
Example 2.14 Consider the situation shown in Figure 2.5 where on side A of the
membrane we have a solution that has a water mole fraction of 0.99. On side B of the
membrane we have pure water. If the temperature is 25 oC and side B is at 1 atmosphere
pressure, estimate the pressure needed on side A to stop the osmosis of water from region
A. What would happen if the pressure on side A was increased to a value above this
pressure?
SOLUTION Assuming the solution in region A is an ideal solution we can use Equation
2.147 to calculate the osmotic pressure of region A, ie.
1atm
Pa m 3
8.314
x
298K
gmol K 101,325Pa
  PA  PB  
x 0.01  13.6 atm
3
-6 m
18 x 10
gmol
Therefore the pressure in region A would have to be 14.6 atm to prevent the osmosis of
water from region B into region A. If the pressure in region A is greater than 14.6 atm,
then water will move from region A into region B and this process is called reverse
osmosis.
Since small pressures are easy to measure, the osmotic pressure is also useful for
determining the molecular weight of macromolecules. For example, letting msolute
represent the mass concentration (g/liter) of the solute in the solution, then from Equation
2.147 we can write the molecular weight of the solute (MWsolute) in terms of the osmotic
pressure of the solution as follows
MWsolute 
RT m solute
(2.149)

Example 2.16 in 3rd edition
Example 2.15 The osmotic pressure of a solution containing a macromolecule is
equivalent to the pressure exerted by 8 cm of water. The mass concentration of
the protein in the solution is 15 g/liter. Estimate the molecular weight of this
macromolecule.
SOLUTION We can use Equation 2.149 to estimate the molecular weight as
follows.
100cm 1atm
m 3 Pa
15g
1liter
8.314
x 298K x
x
x
3
molK
liter 1000cm
101,325Pa
m3

1in
1ft
1atm
8 cm H 2 O x
x
x
2.54cm 12in 33.91ft H 2 O
3
MWsolute
MWsolute = 47400 g/mol
Example 3.1 Calculate the filtration flowrate (cm3/sec) of a pure fluid across a 100
cm2 membrane. Assume the viscosity () of the fluid is 1.8 cP. The porosity of the
membrane is 40% and the thickness of the membrane is 500 microns. The pores run
straight through the membrane and these pores have a radius of 0.225 microns. The
pressure drop applied across the membrane is 75 psi. ( Recall that 14.7 psi = 1 atm =
101325 Pa (or N m-2) , that 1 N = 1 kg m sec-2, and from Chapter 4 we have for the
viscosity that 1 cP = 0.001 N sec m-2 = 0.001 Pa sec).
SOLUTION To find the filtration flowrate we will use Equations 3.4 and 3.7. Since
there are no retained solutes there are no osmotic effects and the effective pressure
drop across the membrane is equal to the applied pressure drop of 75 psi. From the
information about the membrane we first calculate the value of the hydraulic
conductance as shown below.
Ap
S
 0.4 , R  0.225 m icrons  2.25  105 cm , t m  500 m icrons  0.05 cm ,
  1.8 cP  0.0018 Pa sec , and P  75 psi 
1 atm
101325Pa

 5.17  105 Pa
14.7 psi
1 atm
now substituting these values in Equation 3.7 :


2
0.40  2.25  105 cm
cm
LP 
 2.813  107
8  0.0018 Pa sec  0.05cm
Pa sec
and from Equation3.4
Q  2.813  107
Q  14.54
cm3
sec
cm
 100cm 2  5.17  105 Pa
Pa sec
Example 3.2 in 3rd edition