beekeeper

6 hours ago, studiot said:

Well spotted, the full paper is a 3.6MB PDF  - not to bad, and better it's free. +1

I have sent a copy to my friend, who is chair of the Exmoor Beekeepers Association, for comment.

 

I'd love to here your friends comment .

It does seem to work though , I added 12 grams of Trehalose as a solution ,  to a liter of honey that had been crystallized earlier and gently heated to bring it to a liquid again , I did one control batch as well , the one with the added Trehalose is showing no signs of crystallization , the control batch is almost solid again , this was in a period of two weeks. 

I used Canola honey which is known to crystallize very rapidly , the farms around me grow Canola every four years or so , and it makes it hard to sell when if it's not liquid.

At 12 grams per liter it is not financially feasible though , that's why I've been trying to determine the proper amounts.

I also don't want to alter the honey too much either , I want to keep it as natural as possible , 1/2 gram or so per liter shouldn't be too bad as it is found in varying amounts in all honey anyway.

Thanks again for all the help.

12 hours ago, studiot said:

Well, 2% solution means 2g / 100 ml as the measures are given in mls.

So 12 to 15 ml contain

2 x 12/100    to    2 x 15/100 g

or

0.24 to 0.30 g added per kg.

Also since you have confirmed they were working in millimoles (added to 100g of honey)

We have from my previous calculation

0.024  to  0.03 g added to 100g

or

0.24  to  0.3 g per 1000g or 1kg.

 

this is all consistent.

 

So I would suggest this piece suffers in translation

It is not clear if an additional 1.2  to 1.5 ml of solution was added to a 500g jar, in addition to the trehalose treatment already made.

This is what the text literally says.

 

 

So I would suggest you need to try 0.24 to 0.3 g /kg.

This would explain why the trehalose was added as a solution, not a quarter gramme 'pinch', which might not dissolve and distribute evenly.

Thank you , .3 g/kg it is then.

 Your input is greatly appreciated.

The highlighted text in my post was copied and pasted from the article , they said mmol per 100 g of honey , so I'm assuming that's what they meant , millimoles.

But there's something amiss here , later in the article they say : Control and treated samples weighing 500 g were stored in glass jars in the dark at a temperature of 14-16 °C. In each treated sample, 1.2–1.5 mL of 2% trehalose solution was added.

But when I found their patent application online , it states 12 - 15 ml of 2% Trehalose solution was added to 1 kg of Honey. so ten times the amount originally stated.

The study was done in Romania and translated to English , so it may be screwed up in the translation . I would expect the Patent application to be right as it is a legal document.

Thank you for your responses , much appreciated.

Anybody ?

A study done on Honey crystallization says this :  The paper aims to promote a new method to prevent honey crystallization by using trehalose, a disaccharide found in all types of honey in different proportions. Two batches of samples, a control one and a trehalose-added one, were analyzed. In each sample with the addition of trehalose were added between 0.0701 and 0.087 mmol of trehalose per 100 g honey.

Then later in the article it says this : Control and treated samples weighing 500 g were stored in glass jars in the dark at a temperature of 14-16 °C. In each treated sample, 1.2–1.5 mL of 2% trehalose solution was added.

What I need to know , is how many grams of Trehalose would be added to a Kg of honey , to duplicate this experiment at home.

Trehalose is approximately 4 grams per teaspoon , I need some realistic measures to work with , I come up with 4.8 grams per Kg , does that sound right ?

It doesn't to me.

Thanks for any input.