katmar

At 95%RH, 120 deg C and 5 bar the water constitutes about 38% of the gas. This is surely enough for the specific heat difference to become significant. I don't think this is comparable with CaptainPanic's conditions.

Hawkeyes, sorry to reply so late - I hope you are still following this thread. Designing a siphon system is not easy, and I would recommend that you engage a professional engineering consultant before you spend any money on piping and equipment. But I can at least give you the basic principles here. The strategy that you have to follow is to get the driving force causing the liquid to flow to match up with the factors that counter that driving force. The driving force is simply the static pressure difference between where the water is coming from (i.e. the surface of the water in the filter) and where it is going to (i.e the discharge point of the siphon pipe if it is not submerged). This pressure difference is calculated by Pressure = density x gravity x height. Obviously the water must flow from a high point to a low point. There are two factors that "consume" this driving force (these were mentioned by insane_alien in 2 separate posts). The first is the friction of the water flowing through the pipe and fittings, and the second is the acceleration of the water. The acceleration term is usually called the velocity head or the exit loss. A free program that will enable you to calculate the friction loss is available at http://emka.xs4all.nl/dP/ The practical implementation of the Bernoulli equation which incorporates the friction effects is called the Darcy-Weisbach equation. You need to determine what flowrate you require to achieve the backwashing of your filter. Then you need to work out what size of siphon pipe will give you this flow with the available driving force. That is the simple version. To actually design the system you need to look at it in more detail to ensure that the siphon pipe remains full and how you will start and stop the siphon. You could make some expensive mistakes here and it would be best to get professional help.

Captain, it may well be that psychology is more important than physics here! I cannot speak from experience because here in Durban, South Africa, we simply do not have the cold weather you speak of. But if I had to look for a physical reason I would look more to the human body than the air for the cause. In summer it is the inability of the body's sweat to evaporate and cool the person that makes muggy hot weather feel worse than dry hot weather. Perhaps in winter if the air is dry it will cause a thin layer of dry skin to form on the body, making a thin insulating layer? In wet weather the skin will stay moist and be a better conductor of the body's heat out to the air, making it feel colder. This would make it the effect of the air's humidity on the body that impacts on how cold we feel, rather than the heat transfer properties of the air itself. But this is just conjecture.

Captain, You are on the correct track with your calculations based on the Perry data. I will come back to this. The problem is that mixing ethanol with water is the microscopic equivalent of a builder mixing a bucket of sand with a bucket of stone and not finishing up with 2 buckets of mixture. In a similar way as the sand fits between the stones, the ethanol and water molecules are able to pack together more tightly. Let's look at the definition of volume %. It is not the volume after separating the components as you suggested. A 40% ethanol by volume solution is made up by (say) taking 40 ml of ethanol and then adding however much water is required to give 100 ml of mixture. If you did this at 20 deg C you would find that you needed to add 63.34 ml of water to bring the total volume to 100 ml. So we have the weird situation where a solution can have 40 vol% of one component and 63.34 vol% of the other. It's simply a function of the way vol % is defined. Getting back to your calculation - As you correctly calculated when you added 1 liter of ethanol to 1 liter of water, you get 1.9289 liters of mix with a mass of 1.787 kg and the mass % is 44.15%. Now that we have the definition of vol% we can work this out to be (100 * 1/1.9289 ) = 51.84 vol%. This is one point on your conversion table and you could work all the way through the Perry data and build up the complete table you need. In fact, it is even more complicated than this. A 50 vol% mixture at 20 deg C would have a strength of 50.91 vol% at 80 deg C because ethanol has a higher coefficient of thermal expansion than water. If you do find a conversion table just be aware of the temperature to which it applies. How do I know all this useless info? I have worked a lot in ethanol distillery design where it is much easier to do all the calculations based on mass%, but for historical reasons the clients always want the answers in vol%. I do not know the site rules here about whether I am allowed to say it here, but I have written and I sell a computer program that does these conversions. The program will work in evaluation mode for 30 days for free, so if you want to download it and draw up the table for yourself it is available from http://katmarsoftware.com Follow the links to AlcoDens. Alternatively you could type the Perry data into a spreadsheet and do it that way.