The boiler must not produce more vapor than the condenser can handle. In a still, all the heat that's produced must also be removed. Some will be lost if the system isn't insulated well, but let's assume that we have perfect insulation, and all the heat coming from the boiler has to be removed by the condenser. This means that we must know how to match the condenser's capacity to the power of the heater element.
Let's start by assuming that we have a 1 kW heater element (1000 watts).
If you look in Appendix 1, or in an engineering handbook, you will find that 1 Watt of Power will deliver 0.2388 calories per second, 14.3 calories per minute, or 860 calories per hour.
Since 1 calorie raises the temperature of 1 gram of water 1°C, 1 kW of heat will raise one liter of water 14.3°C (25.7°F) per minute. You now we have a good rule of thumb for connecting the power of the heater to the operating parameters of a "perfect" condenser:
1 kW of boiler power will raise the temperature of cooling water flowing at 1 liter per minute by
As long as the cooling water is below the boiling point, the temperature doesn't matter. Cooling water entering at 20°C will come out at 34°C. If it enters at 100°F, it will come out at 126°F.
If you double the power of the heating element, you can double the flow of cooling water to keep the 14°C temperature difference, or accept a temperature rise of 28°C (50°F). The choice is yours, and just a question of how you want to operate.
In practice, you should make a condenser a bit larger than needed for your heating element. If you build a small condenser, you should use a small heater element, but with a large, efficient condenser, you can consider a larger heater element.
This rule of thumb applies to a basic pot still, but more advanced still designs limit the types of condensers and the amount of heat you can use. The next section discusses these different types of and some of the choices you have to make in designing one.
The choice of boiler power also controls how fast you will be able to produce distillate. Let's examine these relationships.
It takes 540 calories of heat to convert 1 gram of water to vapor. This is the Latent Heat of Vaporization (LHV) of water. Each Watt of heat delivers 14.3 calories per minute, so a 1 kW heater will deliver 14,300 calories per minute or 860,000 calories per hour. Dividing these numbers by 540 calories per gram means that a 1 kW heater will produce about 1,600 grams of distilled water every hour, or about 30ml every minute, because pure water's density is 1.0 gram/ml.
The latent heat of vaporization of pure ethanol is 204 calories/gram, so the same 1kW heater will produce 860,000 / 204 = 4,216 grams of ethanol vapor per hour, or about 70 grams per minute. The density of ethanol is 0.791 gm/ml, which means it will produce 5.3 liters of ethanol per hour, or 89 ml every minute.
When distilling a solution of water and ethanol, you can expect to produce between 30 and 89 ml of distillate every minute with a 1 kW heater, depending on the mix you have in the boiler and how well your still concentrates the ethanol.
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