Hello flaminang,
This is the first time that I have heard of separating a methanol/water mixture by reverse osmosis. When you mention a 'max of 20% separation' does this mean a beginning mixture of 50:50 methanol:water will have a output of 100% methanol, but the other output tube (from the RO unit) will have a mixture of 20:80 methanol:water, or does it become a 10:90 methanol:water mixture (50% x 20/100 = 0.10 or 10% methanol)?
If the 10:90 mixture is attainable, then could you re-feed the 10:90 mixture back into the feed pipe of the RO unit to separate more methanol from the stream?
Just a thought.
Cheers,
Eugene