Question 1
A company wants to make O–W emulsions by means of membrane emulsification.
They decide to use a ‘‘membrane’’ with pores of 1mm diameter (d). Can you think of
some necessary conditions for making this endeavor a success, mainly by using
knowledge gained from Chapter 10? Tip: Make a simple drawing of the situation. You
may assume the pores to be cylindrical and at a right angle to the membrane surface.
Answer
First, the material of which the membrane is made should preferentially be wetted by
the aqueous phase, since otherwise no drops would be formed at the end of the pores,
but an oil layer spreading over the membrane. Ideal would be a zero contact angle as
measured in the aqueous phase. Second, the oil should be pressed through the pores,
and the pressure should be larger than the Laplace pressure. The latter will be at
maximum when the oil is making a half sphere at the end of a pore, and it then equals
4 g/d, which may amount to (see Table 10.1) 4 6 0.01/10^6 ¼40 kPa or 0.4 bar.
(Actually, the pressure will have to be substantially larger, to realize a reasonable
flow rate of the oil through the membrane.) Third, the pores should be at sufficient
distance from each other to avoid contact between the drops emerging from the
pores, since otherwise the coalescence of the drops may readily occur. Experience
shows that droplet diameter obtained is at least 3 times the pore diameter, which
implies that the mutual distance must be at least 3mm.
Question 2
You are whipping 200 ml of a protein solution with a small kitchen beater. After 2
minutes you have obtained 400 ml of foam. Microscopic observation shows that the
bubbles are about 0.2 mm in diameter. Can you make a rough estimate of the
fraction of the net energy applied that has been used to create the A–W surface?
Answer
A small kitchen beater has an electric motor consuming, e.g., 150 W. Assuming that
the net power uptake then would be 50 W, this would imply a net energy uptake of
5062660 ¼6000 W?s or J. The increase in surface area would beA¼ 6 j=d¼
660 : 5 = 2? 10 ^4 ¼ 15 ;000 m^2 =m^3 or 6 m^2 in 400 ml. Assumingg¼ 0 :05 N?m^1 , the
total amount of surface free energy added due to whipping would be 6 6 0.05¼0.3 J.
This would equal 0.005%of the net energy input.
Note This example illustrates the very poor efficiency of such a beating
method with respect to energy use. In industrial machines for foam or
emulsion making a higher efficiency can be obtained, though rarely over
0.3%.