Calculating time to reduce alcohol in wine using heating method

[ArtZ]

TL;DR Summary

Using rate kinetics to calculate the time to reach final CF of alcohol knowing starting C0 and heating T

Hi,

This is my first post on PF. I am engaged in a project seeking to find a simple way to reduce ethanol content in wine for those wishing to minimize alcohol intake for health or personal reasons. The alcohol-reduced wine will be used for cooking. Seems heating would do the trick however, the temperature must be kept as low as reasonable to minimize degradation to the wine flavor.

Keep in mind that my field is EE not CE so I am feeling this out as I go.

So as I understand it, the rate of alcohol removal in wine is dependent on several factors, including temperature, time, and alcohol concentration. One common model used to describe the kinetics of alcohol removal is the first-order reaction model, which assumes that the rate of reaction is proportional to the concentration of alcohol remaining in the wine.

The general equation for the first-order reaction model is:

dC/dt = -kC

where:
- dC/dt is the rate of change of alcohol concentration with time
- k is the reaction rate constant
- C is the concentration of alcohol in the wine at any given time

Assuming that the reaction follows first-order kinetics, seems that we can use the following equation to calculate the concentration of alcohol remaining in the wine after a given time:

C/C0 = exp(-kt)

where:
- C0 is the initial concentration of alcohol in the wine
- t is the time (in minutes)
- exp is the exponential function

I've defined the reaction rate constant, k, using the following:

k = A * exp(-Ea/RT)

where:

A is the pre-exponential factor or frequency factor, A = 10^13 s^-1
Ea is the activation energy, Ea = 65 kJ/mol = 65000 J/mol
R is the gas constant, R = 8.314 J/(mol*K)
T is the absolute temperature used in heating, T = 356.45 K

Substituting the values of A, Ea, R, and T into the equation, we get:
k = (10^13 s^-1) * exp(-65000 J/mol / (8.314 J/(mol*K) * 356.45 K))

k = 2.982 x 10^3 s^-1

When I solved for t using the above:

t = -ln(C/C0)/k

I found that the time, t, value returned, was too short given the Cf desired and the C0 of the unheated wine. The first-order reaction model does not account for the energy needed to overcome the heat of evaporation making this an incomplete solution to dC/dt.

As I understand it, the heat of vaporization or heat of evaporation, is the amount of energy that must be added to a liquid substance, to transform a quantity of that substance into a gas.

The heat of vaporization of ethanol (hfg) is (0.826 kJ/g). To incorporate we'll convert the heat of vaporization from kJ/g to J/mol. The molar mass of ethanol is 46.07 g/mol, so: hfg = 0.826 kJ/g * 1000 J/kJ * 1 mol/46.07 g = 17930 J/mol

In both equations below, C0 represents the starting alcohol concentration, Cf represents the desired final alcohol concentration, k is the rate constant for ethanol at the given temperature (in units of s^-1), hfg is the heat of vaporization for ethanol (in J/mol), and R is the gas constant (8.314 J/mol*K).

With some algebraic manipulation I arrived at these canonical forms of the t, and T equations, where t represents time (in seconds) and T represents temperature (in Kelvin):

To solve for time (given T):t = -ln(Cf/C0) / (k*(1 + (hfg/R*T)*k))

To solve for temperature (given t):T = hfg/(Rk) * (1/t - 1/(kln(Cf/C0)))

In both equations below, C0 represents the starting alcohol concentration, Cf represents the desired final alcohol concentration, k is the rate constant for ethanol at the given temperature (in units of s^-1), hfg is the heat of vaporization for ethanol (in J/mol), and R is the gas constant (8.314 J/mol*K).

Hopefully this makes sense so far. Like I said, am an EE not a CE. The first example that I'd like to solve for is using the expression for t above given T:

Given:

• Heating temperature of wine = 83.3C • Initial alcohol concentration, C0= = 0.15 (15%) • Final concentration desired, Cf = 0.04 (4%) • Reaction rate constant = 2.982 x 10^3 s^-1 • Heat of evaporization for ethanol = 17930 J/mol • Boiling point of ethanol =78.5C

An answer that I arrived at that seemed reasonable = 51 minutes to reduce the alcohol concentration from 15% to 4%. Everything I've presented seems reasonable, however, I don't get agreement between Excel and Mathcad which I use for units consistency checking.

Maybe my approach is way off base or just needs some fine tuning. Is there someone in the Forum with a better background in this area who could point me in the right direction?

Any help is greatly appreciated.

Thanks,

Art

 

[Hyperfine]

The alcohol-reduced wine will be used for cooking. Seems heating would do the trick however, the temperature must be kept as low as reasonable to minimize degradation to the wine flavor.

The cooking process itself should achieve your objectives starting with normal wine having normal alcohol concentrations.

 

[ArtZ]

Thanks for your response. One might think that but it's not correct. Tasters are still able to detect the alcohol signature when starting alcohol concentrations are relatively high (13-16%) especially if the food is cooked for a short time. (stir fry) I am using Chinese Shaozing wine which comes in at about 15% ABV. What most people don't understand is that the reduction of alcohol in the cooking process is not as complete as is widely thought. The attached shows a USDA study for residual nutrients and alcohol after heating.

 

[Hyperfine]

Fair enough. But I doubt that this situation should be treated as a problem in chemical kinetics.

Nor do I think that I would appreciate my wine being heated to 83^oC for long enough reduce the ethanol concentration to the desired level.

Obvious questions: What volume of wine do you need to deal with? How often do you need do this?

 

[Tom.G]

This sounds like a question for @Chestermiller, our resident thermodynamics expert; and this mention should alert him to take a look.

p.s. Interesting question

 

[Borek]

A is the pre-exponential factor or frequency factor, A = 10^13 s^-1
Ea is the activation energy, Ea = 65 kJ/mol = 65000 J/mol

You never defined what the reaction is, you never stated where these values come from. It is about as random as possible.

But in general, such processes are very difficult to describe as they typically heavily depend on poorly defined and hard to control factors like mixing and/or diffusion. That in turn means it is easier to test the result experimentally than to calculate what will happen.

Alcohol from beer is removed by reverse osmosis, which doesn't require elevated temperatures, that would be probably a good starting point.

 

[Chestermiller]

Your theoretical development is way off. What you are dealing with here is a solution of alcohol and water, and when you heat the mixture, both alcohol and water go into the gas phase, and in different proportions than in the liquid phase. So what you are dealing with is the vapor-liquid equilibrium behavior of a binary solution of alcohol and water. The process being considered is distillation. See Treybel, Mass Transfer Operations. Google VLE Alcohol Water. Consider using vacuum distillation to lower the temperature of the operation.

 

[Hyperfine]

Consider using vacuum distillation to lower the temperature of the operation.

A simple rotary evaporator might work well.

 

[ArtZ]

Fair enough. But I doubt that this situation should be treated as a problem in chemical kinetics.

Nor do I think that I would appreciate my wine being heated to 83^oC for long enough reduce the ethanol concentration to the desired level.

Obvious questions: What volume of wine do you need to deal with? How often do you need do this?

I'll start with a single 750mL bottle. Typical recipes only call out maybe a tablespoon or two at most so a single 750mL bottle should last a while.

 

[ArtZ]

You never defined what the reaction is, you never stated where these values come from. It is about as random as possible.

But in general, such processes are very difficult to describe as they typically heavily depend on poorly defined and hard to control factors like mixing and/or diffusion. That in turn means it is easier to test the result experimentally than to calculate what will happen.

Alcohol from beer is removed by reverse osmosis, which doesn't require elevated temperatures, that would be probably a good starting point.

One common model used to describe the kinetics of alcohol removal is the first-order reaction model, which assumes that the rate of reaction is proportional to the concentration of alcohol remaining in the wine.

 

[ArtZ]

You never defined what the reaction is, you never stated where these values come from. It is about as random as possible.

But in general, such processes are very difficult to describe as they typically heavily depend on poorly defined and hard to control factors like mixing and/or diffusion. That in turn means it is easier to test the result experimentally than to calculate what will happen.

Alcohol from beer is removed by reverse osmosis, which doesn't require elevated temperatures, that would be probably a good starting point.

I looked into the RO process and found a company that does this for wine makers. They require 1 case of wine and the processing fee is $850 USD. Too rich for a small time tinkerer.

 

[ArtZ]

A simple rotary evaporator might work well.

Too complicated. A large pot should work fine.

 

[ArtZ]

Your theoretical development is way off. What you are dealing with here is a solution of alcohol and water, and when you heat the mixture, both alcohol and water go into the gas phase, and in different proportions than in the liquid phase. So what you are dealing with is the vapor-liquid equilibrium behavior of a binary solution of alcohol and water. The process being considered is distillation. See Treybel, Mass Transfer Operations. Google VLE Alcohol Water. Consider using vacuum distillation to lower the temperature of the operation.

I am not distilling anything. If anything, this process would be called reverse distillation.

 

[Borek]

One common model used to describe the kinetics of alcohol removal is the first-order reaction model, which assumes that the rate of reaction is proportional to the concentration of alcohol remaining in the wine.

Removal by what means and in what conditions?

Sorry, but what you write is completely meaningless without context. The question is "how long does it take to get from point A to point B" and you answer with "my car is capable of 100 mph and we drive on the right". There is no way to connect the dots.

I am not distilling anything. If anything, this process would be called reverse distillation.

Reverse distillation is - if anything - just a distillation in which you keep not what evaporated, but what was left. It is exactly the same process and it is described by exactly the same models and math.

 

[Hyperfine]

I'll start with a single 750mL bottle. Typical recipes only call out maybe a tablespoon or two at most so a single 750mL bottle should last a while.

Simple. Add to a small pan over low heat and reduce to taste.

Cooking, not chemistry.

 

[ArtZ]

Simple. Add to a small pan over low heat and reduce to taste.

Cooking, not chemistry.

I sort of plan to do that. My wife has hypersensitivity to residual alcohol in cooked food. The plan is, this week if I get to it, is to reduce a known quantity of the Shaoxzing wine (15%) at a known temperatue and then test with a wine makers hydrometer that I bought online. I figure that a final concentration of 4% won't be noticed in a 4 serving cooked dish.

 

[DaveE]

Why don't you just buy some non-alcoholic wine? It's readily available, and done by people with better equipment than you are likely to have.

 

[hutchphd]

Absolutely @ArtZ you are overthinking this. Put the wine in a container and heat it to a low boil. The alcohol will preferentially vaporize. Reduce the volume by 20% and there will be far less than 4% residual ethanol. Probably a 10% reduction would get you to 4% residual. The Shaozing wine flavor is pretty robust and it is being cooked anyway so not a problem. Enjoy.

 

[ArtZ]

Absolutely @ArtZ you are overthinking this. Put the wine in a container and heat it to a low boil. The alcohol will preferentially vaporize. Reduce the volume by 20% and there will be far less than 4% residual ethanol. Probably a 10% reduction would get you to 4% residual. The Shaozing wine flavor is pretty robust and it is being cooked anyway so not a problem. Enjoy.

I may overthinking this, but I wanted to have some technical fun. Since I retired, I don't do much of what I did before. Since the alcohol BP is 78.5C (173.3 F) I figure that if I keep the wine at a simmer but below the boiling point of the water constituent, I should be OK. Also will need to replenish the remaining volume with water to restore the original volume.

 

[DaveE]

If you really want to remove all of the ethanol, you will essentially have to boil the wine. Heat will preferentially remove ethanol, but only to a certain concentration, as shown in the phase diagram below.

 

[Chestermiller]

I am not distilling anything. If anything, this process would be called reverse distillation.

You don’t know much about distillation,, do you? As a ChE, my advice to you as an EE is to stick to Ohm's law.

 

[ArtZ]

Only what I've learned from watching Moonshiners.

 

[ArtZ]

If you really want to remove all of the ethanol, you will essentially have to boil the wine. Heat will preferentially remove ethanol, but only to a certain concentration, as shown in the phase diagram below.

View attachment 323574

I don't expect to remove 100% of the alcohol, just enough so it can't be tasted in a prepared food dish. All online cooking references agree with your point. It's of course more evident when high alcohol spirits are used. Can you explain what the chart you posted is telling us?

 

[berkeman]

I figure that if I keep the wine at a simmer but below the boiling point of the water constituent, I should be OK.

Have you tried this? What does it do to the taste, bouquet, aftertaste, etc?

https://www.nytimes.com/1982/01/20/garden/the-tasting-ritual-4-basic-elements.html

 

[Chestermiller]

If you really want to remove all of the ethanol, you will essentially have to boil the wine. Heat will preferentially remove ethanol, but only to a certain concentration, as shown in the phase diagram below.

View attachment 323574

I think you are interpreting this wrong. It seems to me you are looking at the wrong side of the phase diagram. If you start off with 15 % by volume of ethanol, the first vapor to come off at 195 F will be 60 % ethanol. If you raise the temperature from 195 F to 202 F and allow the vapor and liquid to equilibrate, you will have lowered the concentration of ethanol in the liquid to 7.5 volume %, and decreased the amount of liquid by only a small amount. This is all at a pressure of 1 atm.

Working with a phase diagram like this is very basic stuff to us Chemical Engineers.

 

[berkeman]

just enough so it can't be tasted in a prepared food dish.

What prepared dishes? Doesn't the cooking process already remove most of the EtOH anyway?

 

[Hyperfine]

Change the menu!

 

[berkeman]

just enough so it can't be tasted in a prepared food dish.

BTW, EtOH is tasteless.

Spoiler

Just like folks who've had way too much EtOH...

 

[Chestermiller]

Only what I've learned from watching Moonshiners.

So moonshiners don’t have low concentrations of alcohol in their reboilers?

 

[DaveE]

I think you are interpreting this wrong. It seems to me you are looking at the wrong side of the phase diagram. If you start off with 15 % by volume of ethanol, the first vapor to come off at 195 F will be 60 % ethanol. If you raise the temperature from 195 F to 202 F and allow the vapor and liquid to equilibrate, you will have lowered the concentration of ethanol in the liquid to 7.5 volume %, and decreased the amount of liquid by only a small amount. This is all at a pressure of 1 atm.

Working with a phase diagram like this is very basic stuff to us Chemical Engineers.

Hum... Now I'm confused. Isn't it true that at 1atm, 202F, EtOH concentration in the liquid can't fall below 7.5% in equilibrium?

Now, if you blew fresh air in you wouldn't be at equilibrium. Does that mean you can reduce the concentration lower, or are you just removing H2O and EtOH in proportion to maintain the same concentration?

 

[DaveE]

You don’t know much about distillation,, do you? As a ChE, my advice to you as an EE is to stick to Ohm's law.

Working with a phase diagram like this is very basic stuff to us Chemical Engineers.

One of the things I really like about PF is the ability to learn about subjects in which I lack expertise. I very much appreciate all of you that take the time to educate people outside of your area of expertise.

OTOH, you don't have to help others if you don't want to. But if you do want to teach, it's not a great approach to imply they shouldn't ask questions or shouldn't risk being wrong. It's not a crime to talk about "reverse distillation", even though that's not the description that ChEs would use.

In any case, we are aware we aren't Chemical Engineers, but thanks for the reminder.

 

[Chestermiller]

Hum... Now I'm confused. Isn't it true that at 1atm, 202F, EtOH concentration in the liquid can't fall below 7.5% in equilibrium?

Yes, that’s what I said. At 202, it is 7.5% in the liquid.

Now, if you blew fresh air in you wouldn't be at equilibrium. Does that mean you can reduce the concentration lower, or are you just removing H2O and EtOH in proportion to maintain the same concentration?

This diagram is for the binary system water and alcohol; no air.

 

[DaveE]

This diagram is for the binary system water and alcohol; no air.

Oh. Yes you're right, I didn't know that part.
At a basic level, what difference would a bunch of Nitrogen molecules make?
Can you treat vapor pressures separately (non-reactive, of course), like superposition in linear systems?

 

[hutchphd]

This diagram is for the binary system water and alcohol; no air.

Oh. Yes you're right, I didn't know that part.

But if the purpose is to remove ethanol from the liquid why would you maintain the the alcohol vapor eqilibrium??? Take the lid off the boiling pot and let the ethanol diffuse away. Let the volume in the kettle decrease until 11% of the liquid is gone and/or the temperature has reached the value stipulated on the chart Equilibrium neither required nor requested.
I don't like to mix efficient engineering with physics pondering although I love them equally.

 

[DaveE]

But if the purpose is to remove ethanol from the liquid why would you maintain the the alcohol vapor eqilibrium??? Take the lid off the boiling pot and let the ethanol diffuse away. Let the volume in the kettle decrease until 11% of the liquid is gone and/or the temperature has reached the value stipulated on the chart Equilibrium neither required nor requested.
I don't like to mix efficient engineering with physics pondering although I love them equally.

Yes, which leads to my next/previous question.
At the liquid-vapor interface, is it always in equilibrium at the molecular level, no matter how much you try to blow fresh air onto it?