Calculates the vapor above and the reflux below a partial condenser and the rectification resulting from the partial condensation.
Based on 100g/min vapor below the reflux condenser:
optional input:
hPa ambient pressure (75-10000 hPa)thermometer error
Result:
The reflux condenser in systems such as CM (cooling management) is designed to condense part of the vapor in a controlled manner and direct it downwards.
More or less depending on the water supply setting.
Such a condenser is often called a dephlegmator.
In other systems, the vapor is completely condensed and returned as reflux. The division into reflux and product takes place either afterwards (e.g. with LM) or before (e.g. with VM).
If only a part is condensed, the question arises as to whether the condensate has a different alcohol strength than the remaining vapor, just as vapor from an alcohol-containing liquid has a different alcohol strength than the liquid. The boiling point diagram provides the basic answer to this question:
On the left is the 0% reflux limit case, on the right the 100% reflux limit case and in the middle an example of the area in between.
The x-axis is the alcohol strength, the y-axis is the boiling or vapor temperature. The lower curve refers to liquid, the upper curve to vapor. The red point is the temperature and alcohol strength of the vapor below the reflux condenser, i.e. before condensation. The two blue dots are the temperature and alcohol strength of the reflux and the vapor above the reflux condenser.
In the diagram, the vapor above the reflux condenser is at the same height as the reflux itself, so both have the same temperature. However, the alcohol strength of the vapor above the reflux condenser is one boiling point higher than that of the reflux. Just as the vapor of a liquid has the same temperature as the liquid, but the alcohol strength is one step higher.
A reflux condenser therefore behaves similarly to a physical plate and can also add up to one theoretical plate of rectification to the still. However, as with a physical plate, this is only theoretical, as 100% reflux is required for this, meaning that no distillate is produced. With just under 100% reflux, however, just under one additional theoretical plate should be possible.
A scientific text on this can be found in the "VDI-Wärmeatlas", in the chapter "J2 Kondensation von Mehrstoffgemischen" by Reiner Numrich. It's in German language. The page of interest for this topic can be downloaded here: VDI-Wärmeatlas, p.1120
In practice, however, we did not succeed in generating one or just under one theoretical plate with a partial condenser. We only achieved around 0.7 theoretical plates. We can only speculate as to why this is the case. It is plausible that the very fluctuating distillate flow in a CM with high reflux is to blame. If you try to set over 90% reflux, there is alternately no distillate at all and a gush. This means that you do not get distillate with a high reflux. You only get distillate in the moments with less reflux.
And since there is no physical barrier between the vapor below and above the reflux condenser, the stronger upper vapor can easily be diluted by the weaker lower vapor. Stills do not run perfectly evenly; a certain amount of mixing cannot be avoided.
But this also corresponds to the results of physical plates. This is because even physical plates are not normally 100% efficient, i.e. they cannot produce one theoretical plate at 100% reflux. This is called plate efficiency. In practice, plate efficiencies of less than 70% are often achieved.
This calculator is therefore intended more to explain the theory than for practical use. However, there is a slider to simulate this efficiency. 50% efficiency means that only half of the concentration is achieved in terms of wt%. So assuming the vapor below the reflux condenser has 60%abw (alcohol by weight) and the vapor above it has 70%abw at 100% efficiency, then the vapor above it has only 65%abw at 50% efficiency.
Exactly which % reflux in the range between 0 and 100% reflux means which locations in the boiling diagram can only be calculated iteratively. The following rules must be followed:
If one follows these rules, there is only one mathematical solution, which is calculated iteratively.
The calculator presupposes 100g/min of vapor under the reflux condenser. Depending on the alcohol strength, this means different heating rates. And depending on the alcohol strength, different amounts of cooling water are required to achieve the specified % reflux. This is because much less energy is required to vaporize the same mass of alcohol as water. And in the same way, the cooling water has to absorb much less energy to condense this mass.
The consideration of the air pressure has a significant influence on the temperatures. If nothing is entered, the calculator assumes the local atmospheric pressure 1013.25 hPa.
Since almost no one has an absolutely accurate thermometer, an additional "thermometer error" can be specified. This can be determined with the help of the calculator Thermometer Error. Temperatures calculated here are then those displayed on this thermometer, not the real ones.
Information about our boiling point data and the influence of atmospheric pressure
In other systems, the vapor is completely condensed and returned as reflux. The division into reflux and product takes place either afterwards (e.g. with LM) or before (e.g. with VM).
If only a part is condensed, the question arises as to whether the condensate has a different alcohol strength than the remaining vapor, just as vapor from an alcohol-containing liquid has a different alcohol strength than the liquid. The boiling point diagram provides the basic answer to this question:
The x-axis is the alcohol strength, the y-axis is the boiling or vapor temperature. The lower curve refers to liquid, the upper curve to vapor. The red point is the temperature and alcohol strength of the vapor below the reflux condenser, i.e. before condensation. The two blue dots are the temperature and alcohol strength of the reflux and the vapor above the reflux condenser.
In the diagram, the vapor above the reflux condenser is at the same height as the reflux itself, so both have the same temperature. However, the alcohol strength of the vapor above the reflux condenser is one boiling point higher than that of the reflux. Just as the vapor of a liquid has the same temperature as the liquid, but the alcohol strength is one step higher.
A reflux condenser therefore behaves similarly to a physical plate and can also add up to one theoretical plate of rectification to the still. However, as with a physical plate, this is only theoretical, as 100% reflux is required for this, meaning that no distillate is produced. With just under 100% reflux, however, just under one additional theoretical plate should be possible.
A scientific text on this can be found in the "VDI-Wärmeatlas", in the chapter "J2 Kondensation von Mehrstoffgemischen" by Reiner Numrich. It's in German language. The page of interest for this topic can be downloaded here: VDI-Wärmeatlas, p.1120
In practice, however, we did not succeed in generating one or just under one theoretical plate with a partial condenser. We only achieved around 0.7 theoretical plates. We can only speculate as to why this is the case. It is plausible that the very fluctuating distillate flow in a CM with high reflux is to blame. If you try to set over 90% reflux, there is alternately no distillate at all and a gush. This means that you do not get distillate with a high reflux. You only get distillate in the moments with less reflux.
And since there is no physical barrier between the vapor below and above the reflux condenser, the stronger upper vapor can easily be diluted by the weaker lower vapor. Stills do not run perfectly evenly; a certain amount of mixing cannot be avoided.
But this also corresponds to the results of physical plates. This is because even physical plates are not normally 100% efficient, i.e. they cannot produce one theoretical plate at 100% reflux. This is called plate efficiency. In practice, plate efficiencies of less than 70% are often achieved.
This calculator is therefore intended more to explain the theory than for practical use. However, there is a slider to simulate this efficiency. 50% efficiency means that only half of the concentration is achieved in terms of wt%. So assuming the vapor below the reflux condenser has 60%abw (alcohol by weight) and the vapor above it has 70%abw at 100% efficiency, then the vapor above it has only 65%abw at 50% efficiency.
Exactly which % reflux in the range between 0 and 100% reflux means which locations in the boiling diagram can only be calculated iteratively. The following rules must be followed:
1. that the reflux plus the vapor above the reflux condenser together give the vapor below the reflux condenser.
2. that the reflux and the vapor above the reflux condenser are exactly one boiling point diagramm step apart at 100% efficiency.
3. and of course that the % reflux after the calculation is as set on the slider.
For this, you must first decide how % reflux is defined.
Because if reflux and vapor above the reflux condenser have a different alcohol strength, half the volume, for example, no longer means half the weight or half the molar quantity.
So it no longer doesn't matter what this refers to.
With our calculators, % reflux always refers to the weight.
So based on 100g/min of vapor under the reflux condenser, you have 70g/min of reflux at 70% reflux and 30g/min of vapor above the reflux condenser.2. that the reflux and the vapor above the reflux condenser are exactly one boiling point diagramm step apart at 100% efficiency.
3. and of course that the % reflux after the calculation is as set on the slider.
If one follows these rules, there is only one mathematical solution, which is calculated iteratively.
The calculator presupposes 100g/min of vapor under the reflux condenser. Depending on the alcohol strength, this means different heating rates. And depending on the alcohol strength, different amounts of cooling water are required to achieve the specified % reflux. This is because much less energy is required to vaporize the same mass of alcohol as water. And in the same way, the cooling water has to absorb much less energy to condense this mass.
The consideration of the air pressure has a significant influence on the temperatures. If nothing is entered, the calculator assumes the local atmospheric pressure 1013.25 hPa.
Since almost no one has an absolutely accurate thermometer, an additional "thermometer error" can be specified. This can be determined with the help of the calculator Thermometer Error. Temperatures calculated here are then those displayed on this thermometer, not the real ones.
Information about our boiling point data and the influence of atmospheric pressure