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Re: [Distillers] Re: what to build ?

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  • RLB
    All I can say is that the so called experts state increasing temperature, surface area, and decreasing pressure does increase rate of evaporation.  With that
    Message 1 of 11 , Aug 16, 2013
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      All I can say is that the so called experts state increasing temperature, surface area, and decreasing pressure does increase
      rate of evaporation.  With that said, I will not comment on this subject further.



      From: tgfoitwoods <zymurgybob@...>
      To: Distillers@yahoogroups.com
      Sent: Friday, August 16, 2013 12:24 PM
      Subject: [Distillers] Re: what to build ?

       
      Sorry to be a bubble-buster, but for a constant temperature, the the evaporation rate is solely a function of system energy input and latent heats of evaporation of the volatile liquids involved, and has nothing to do with surface area or decreased pressure.

      The heat of vaporization of water is 2260 joules per gram, and for ethanol that number is 841 joules per gram. The effective heat of vaporization of a mixture of those 2 liquids, depending on their concentrations, will be somewhere between those 2 numbers.

      "But", you might say, "if I lower the pressure, I can lower the boiling point" and that's true, but when it boils at the lower boiler point, every gram of water still requires 2260 joules, or watt-seconds to vaporize, and the ethanol still requires 841 joules.

      For the other part of the discussion, we all know we can force a bit of evaporation by passing air over the liquid, without adding any heat. This is exactly what we do to hot soup when we blow on it to cool the soup, so if you force evaporation without adding the necessary heat of evaporation, the temperature of the liquid goes down, and the next time you blow on that soup it doesn't evaporate as much because it's now cooler. Applied to a still, this mode of operation would quickly shut you down.

      So, if you want to control vapor output rate in a still (assuming good boiler insulation) your only choice is to control the heat input. 2.26kW will evaporate 1 gram of water per second and .841 kW will evaporate 1 gram of ethanol per second, period.

      Zymurgy Bob, a simple potstiller Making Fine Spirits

    • Paul
      Thanks Bob for the clarification. The important thing, I guess, to remember is that distillation is a function of boiling and not evaporation. The latent heat
      Message 2 of 11 , Aug 18, 2013
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        Thanks Bob for the clarification. The important thing, I guess, to remember is that distillation is a function of boiling and not evaporation. The latent heat constant is interesting, and not immediately obvious. Reducing system pressure (ie vacuum) would decrease specific heat.

        Do many people insulate keg-type boilers to reduce heat losses and costs?
        Paul


        From: tgfoitwoods
        To: Distillers@yahoogroups.com
        Sent: Saturday, 17 August 2013 4:24 AM
        Subject: [Distillers] Re: what to build ?

         
        Sorry to be a bubble-buster, but for a constant temperature, the the evaporation rate is solely a function of system energy input and latent heats of evaporation of the volatile liquids involved, and has nothing to do with surface area or decreased pressure.

        The heat of vaporization of water is 2260 joules per gram, and for ethanol that number is 841 joules per gram. The effective heat of vaporization of a mixture of those 2 liquids, depending on their concentrations, will be somewhere between those 2 numbers.

        "But", you might say, "if I lower the pressure, I can lower the boiling point" and that's true, but when it boils at the lower boiler point, every gram of water still requires 2260 joules, or watt-seconds to vaporize, and the ethanol still requires 841 joules.

        For the other part of the discussion, we all know we can force a bit of evaporation by passing air over the liquid, without adding any heat. This is exactly what we do to hot soup when we blow on it to cool the soup, so if you force evaporation without adding the necessary heat of evaporation, the temperature of the liquid goes down, and the next time you blow on that soup it doesn't evaporate as much because it's now cooler. Applied to a still, this mode of operation would quickly shut you down.

        So, if you want to control vapor output rate in a still (assuming good boiler insulation) your only choice is to control the heat input. 2.26kW will evaporate 1 gram of water per second and .841 kW will evaporate 1 gram of ethanol per second, period.

      • tgfoitwoods
        You re very welcome, Paul. Actually, what RLB says is is true in any given instant, but it doesn t allow for the fact that if you take more energy out of a
        Message 3 of 11 , Aug 18, 2013
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          You're very welcome, Paul.

          Actually, what RLB says is is true in any  given instant, but it doesn't allow for the fact that if you take more energy out of a system (as evaporation does in the numbers I mentioned) than you put into it, the system's energy, and therefore the system's temperature (as long as there's no phase change) goes down.

          So if 2 lakes have the same volume but greatly different surface areas, the one with the larger area will evaporate at a greater rate than the smaller area lake in the first instant, as long as water temperature and ambient pressure are equal. But unless equal energy, often solar, is applied to both lakes, the larger area lake will cool more quickly than the smaller area lake, slowing its evaporation rate, and this is ok for lakes.

          In stills, however, we don't want our wash to cool while we are running this fairly closed system, because once we stop boiling, we pretty much stop distilling, so we input the amount of energy we need to get the amount of vapor we want.

          As you say, if you lower the pressure in the still, you'll lower the boiling point of the wash, and you will therefore lower the amount of energy it takes to get the wash to a boil, and that has economic significance, but once you're boiling, vapor output rate is a direct function of energy input rate. That's why we speak of a condenser "being able to knock down X number of watts", identical to the number of watts input.

          It was not my intent to be hard on you RLB, but these are some very basic physical laws, and the subject comes up often as to whether these laws apply to stills, and they do.

          Zymurgy Bob, a simple potstiller Making Fine Spirits

          --- In Distillers@yahoogroups.com, Paul wrote:
          >
          > Thanks Bob for the clarification. The important thing, I guess, to remember is that distillation is a function of boiling and not evaporation. The latent heat constant is interesting, and not immediately obvious. Reducing system pressure (ie vacuum) would decrease specific heat.
          >
          > Do many people insulate keg-type boilers to reduce heat losses and costs?
          > Paul
          >
          >
          >
          > ________________________________
          > From: tgfoitwoods
          > To: Distillers@yahoogroups.com
          > Sent: Saturday, 17 August 2013 4:24 AM
          > Subject: [Distillers] Re: what to build ?
          >
          >
          >
          >  
          > Sorry to be a bubble-buster, but for a constant temperature, the the evaporation rate issolely a function of system energy input and latent heats of evaporation of the volatile liquids involved, and has nothing to do with surface area or decreased pressure.
          ----snip----
        • Eddie Hoskin
          Ah yes, but you are confusing two distinct mechanisms--evaporation and boiling. Evaporation connotates a state below the boiling temperature.  In such a case,
          Message 4 of 11 , Aug 18, 2013
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            Ah yes, but you are confusing two distinct mechanisms--evaporation and boiling.

            Evaporation connotates a state below the boiling temperature.  In such a case, yes, increasing the temperature/ surface area and decreasing pressure will indeed increase the amount of liquid turning into vapor.

            However, at the temperature of boiling, you cannot change the temperature.  Instead, the energy you input into the liquid will convert the liquid into a gas.  Only when all the liquid has been evaporated will you rise above the boiling point.

            Of course, in ethanol/water distillation, it's a touch more complicated.  As the stilling run progresses, the alcohol % decreases, thereby increasing the water/ethanol mixture's boiling point.  But at any given point in time, you cannot exceed the mixture's boiling point.

            Surface area has no impact on this--more heat, more boiling.

            Pressure does have an impact on this, by altering the boiling temperature of the mixture.  Again, the rate of boiling liquid into vapor will be directly proportional to the heat input, but this will occur at a lower temp.

            Hope that helps,
            Radical Ed



            From: RLB <last2blast@...>
            To: "Distillers@yahoogroups.com" <Distillers@yahoogroups.com>
            Sent: Friday, August 16, 2013 2:11 PM
            Subject: Re: [Distillers] Re: what to build ?

             
            All I can say is that the so called experts state increasing temperature, surface area, and decreasing pressure does increase
            rate of evaporation.  With that said, I will not comment on this subject further.



            From: tgfoitwoods <zymurgybob@...>
            To: Distillers@yahoogroups.com
            Sent: Friday, August 16, 2013 12:24 PM
            Subject: [Distillers] Re: what to build ?

             
            Sorry to be a bubble-buster, but for a constant temperature, the the evaporation rate is solely a function of system energy input and latent heats of evaporation of the volatile liquids involved, and has nothing to do with surface area or decreased pressure.

            The heat of vaporization of water is 2260 joules per gram, and for ethanol that number is 841 joules per gram. The effective heat of vaporization of a mixture of those 2 liquids, depending on their concentrations, will be somewhere between those 2 numbers.

            "But", you might say, "if I lower the pressure, I can lower the boiling point" and that's true, but when it boils at the lower boiler point, every gram of water still requires 2260 joules, or watt-seconds to vaporize, and the ethanol still requires 841 joules.

            For the other part of the discussion, we all know we can force a bit of evaporation by passing air over the liquid, without adding any heat. This is exactly what we do to hot soup when we blow on it to cool the soup, so if you force evaporation without adding the necessary heat of evaporation, the temperature of the liquid goes down, and the next time you blow on that soup it doesn't evaporate as much because it's now cooler. Applied to a still, this mode of operation would quickly shut you down.

            So, if you want to control vapor output rate in a still (assuming good boiler insulation) your only choice is to control the heat input. 2.26kW will evaporate 1 gram of water per second and .841 kW will evaporate 1 gram of ethanol per second, period.

            Zymurgy Bob, a simple potstiller Making Fine Spirits



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