drying time x diameter

[cashmkr2001]

Guys,
Any graphs/sheets out there showing the approximated drying time for a certain salami diameter?

[wheels]

Certainly, there are FDA specs in the USA for fermentation times etc. Is that what you're after?

Edit: Are you in the US, if not my apologies

[grisell]

Guys,
Any graphs/sheets out there showing the approximated drying time for a certain salami diameter?

Let's make a mathematical model:



We assume a cylindrical geometry and that the salami is oblong enough so that evaporation takes place through the sides only (for the sake of simplicity):

Side surface area = 2πrh ~ r where r ≡ radius and h ≡ length
Volume = πr^2h ~ r^2

If h is kept constant, the volume (and the amount of water to be evaporated) is proportional to the radius squared. The area through which evaporation takes place is proportional to the radius.

Therefore, the volume is proportional to the area (and radius) squared and the drying time is proportional to the radius squared. T ~ r^2



(Note: Similarly, for an arbitrary three-dimensional geometry, volume is proportional to area^(3/2) and T ~ r^(3/2))


Example: A 36 mm salami will take (36/22)^2 = 2.68 times longer to dry than a 22 mm salami; all other parameters being equal.

This is in theory only. In practice, the parameters are never equal.

As for absolute numbers of drying times, it's impossible to give any. So many factors come into play; humidity, temperature, air flow, mould, permeability of casing, case hardening effects, coarseness of meat, original water content, salt content, final weight loss, fat content etc etc etc...

However, I can give you a hint. Looking at my notes I have the following approximate drying times for coarsely ground salami at 12 C/54 F, 80-85% relative humidity, negligible air flow, 35% weight loss:

22 mm lamb casing: 4 days
36 mm hog casing: 10 days
50 mm beef casing: 25 days
120 mm beef bung: 95 days

The correspondence with the mathematical model is quite good, confirming it's basically correct. I'll leave the calculations to the interested person.

[Yannis]

Volume to Area ratio equals to 0.5r so if drying time depends only on this ratio when you double r you double drying time.

Obviously this is false so I think this mathematical model is incomplete. There must be some more coefficients, maybe for moisture moving from center of the meat to its perimeter.

edit

Ofcourse I don't argue your experimental results showing that drying time is proportional to r^2.

[grisell]

Correct Yannis! Thanks for telling me. Something is wrong with the math above. I will take a good look tomorrow and edit it.

[grisell]

Edited. The result was correct, but the calculations were wrong. I think I got it right this time(?).

[Yannis]

Well the model has not changed. If drying time is proportional to r^2 and inversely proportional to r then the combined result is proportonal to r (if you divide πr^2h by 2πrh you get 0.5r). That's why I think that there must be another coefficient (factor) which includes r so when multiplied with volume to area ratio you get drying time proportional to r^2. Anyway this is well beyond my knowledge on the subject so I will keep the result and won't care about the model

[grisell]

What the...

Yeah, you're right...! I'm glad there is a descendant to Pythagoras on this forum.

Okay, let's skip the strict algebra. If one reasons less strictly and more intuitively, the result should be correct though.

Increasing the diameter by two in a cylinder (neglecting the top and bottom sides) will increase the area by two and the volume by four. Increasing it by three will increase the volume by nine etc. Thus, drying time is proportional to the diameter squared. T ~ d^2.

Increasing the diameter by two in a sphere will increase the area by four and the volume by eight. Increasing it by three will increase the area by nine and the volume by 27 etc. Increasing the area by two will increase the volume by sqrt(2)^3. Thus drying time is proportional to the square root of the area raised to three, or the diameter raised to 3/2. T ~ d^(3/2).

Does this sound right?

[grisell]

It seems that the original poster left the discussion long ago...

[Yannis]

It seems that the original poster left the discussion long ago...

Well I think that from a sausage maker point of view your experimental results is more than enough

Anyway back to the problem you still say the same.

Increasing the diameter by two in a cylinder (neglecting the top and bottom sides) will increase the area by two and the volume by four. Increasing it by three will increase the volume by nine etc. Thus, drying time is proportional to the diameter squared. T ~ d^2.

When you increase diameter by 2 you still get a volume to area ratio 4/2=2 which is proportional to diameter and not diameter^2
The same with sphere 8/4=2.

What I think that may be the solution is to multiply the volume to area ratio by kr where k has to do with something like meat to fat ratio or how coarse the meat is ground and r with the distance that moisture have to travel from the center to edge of sausage and the final formula could be something like 0.5r * kr =0.5*k*r^2.

Ofcourse if my assumption is correct then finding k is very long story

[grisell]

[---]
When you increase diameter by 2 you still get a volume to area ratio 4/2=2 which is proportional to diameter and not diameter^2
The same with sphere 8/4=2.
[---]

Nope. Area scale is length scale^2, volume scale is length scale^3 and hence volume scale is area scale^(3/2).

For a cylinder: Increase diameter by e.g. 10 and mantle area will increase by 10, volume by 100. V ~ r^2 ~ A^2.

For a sphere: Increase diameter by 10 and area will increase by 100, volume by 1000. V ~ r^3 ~ A^(3/2)

Edit: It's still not satisfactory. I will look over it again.

[BriCan]

I am glad to see two find minds working for a solution that has been perplexing most if not all on the drying time per inch/centimetre of a sausage/salami

Could you fine gentlemen (?) please remember to put your findings in layman’s terms as some of us country folks av no fine education

Me; I’m as common as muck, taught by the old’uns and use the squeeze test then weigh

[grisell]

[---]
Could you fine gentlemen (?) please remember to put your findings in layman’s terms as some of us country folks av no fine education
[---]

Yes, I/we will, once we get it right. Something is still fishy about it...

[grisell]

[---]
What I think that may be the solution is to multiply the volume to area ratio by kr where k has to do with something like meat to fat ratio or how coarse the meat is ground and r with the distance that moisture have to travel from the center to edge of sausage and the final formula could be something like 0.5r * kr =0.5*k*r^2.

Ofcourse if my assumption is correct then finding k is very long story

Yes, why not? I will look into it.

[cashmkr2001]

hm.. interesting, get back to the forum, see a post and think "interesting question!" the I look deeper into it, it turns out I ASKED the question. And completely forgot about right after that. I should see a neurologist.
Thanks for the answers quys, and pardon my impoliteness