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Water Evaporation From Sand (pot in pot refrigerator)

  1. Jan 7, 2014 #1
    Hello,
    i am working about pot in pot refrigerator (zeer) and i have problem with calculation, i have to calculate how many % energy will be lost during T time, i know theory about this effect but i don't know how to calculate it, i have some reference but my english and math skills aren't enough good to handle it, if u can help and explain me how to calculate this.
    thanks and sorry for my bad english (i hope you understood what i said)
     
  2. jcsd
  3. Jan 8, 2014 #2

    maajdl

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    I don't see what you mean by "... i have to calculate how many % energy will be lost during T time ...".
    It is possible to calculate how much heat is needed to evaporate a certain quantity of water.
    This is called the heat of evaporation (which depends on the temperature).

    The rate of evaporation (quantity of water per unit time) depends on the properties of the outer pot material as well as on the ventilation of the pot, the ambient temperature and moisture.
    Therefore, calculation this as a function of some time elapsed in not easy.

    You may get some useful insight by reading about the wet bulb temperature.
     
  4. Jan 8, 2014 #3
    thanks for reply, i typed question incorrectly, it's really hard to calculate this as function, i need to make report about pot in pot refrigerator in 2 days, i have only basic theory how this works and need some calculation about this kind of refrigerator and if u can help me to get some mathematical calculation about it, thanks
     
  5. Jan 8, 2014 #4

    sophiecentaur

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    No idea about a quantitative approach; it's too much of a 'green' thing for Science, perhaps. There are a lot of Google hits on the zeer. None, of course, seem to have any numbers in them. You may find something if you're prepared to do a lot of digging. Also, you could Google 'Swamp coolers', which are air coolers with a bale of wet straw and a fan. These are very effective where the humidity is low and you may find some actual figures there. Of course, your pot in pot cooler will always work better in a draught.
     
  6. Jan 8, 2014 #5
    thanks for reply, i understand what you said but i am looking for more physical explain of this effect but i couldn't found it in google.
     
  7. Jan 8, 2014 #6

    sophiecentaur

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    Yes. It is a problem. Your analysis would need to include a lot of imponderables such as air speed over the surface, area and humidity. The nearest thing I could find (an engineering approach) was some Google hits about Cooling Towers. Wiki does have some figures for their performance but it's a long way from clay pots, I admit.

    I eventually did find this article about evaporation rate, using Langmuir's Evaporation Equation, after googling "evaporation rate". Perhaps that would help for a start.
     
  8. Jan 8, 2014 #7

    maajdl

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    I still advise you to investigate the wet bulb temperature.
    http://en.wikipedia.org/wiki/Wet-bulb_temperature
    http://www.engineeringtoolbox.com/dry-wet-bulb-dew-point-air-d_682.html
    http://apollo.lsc.vsc.edu/classes/met130/notes/chapter4/wet_bulb.html
    http://www.srh.noaa.gov/epz/?n=wxcalc_rh
    ...
    https://www.google.com/?hl=en#hl=en&q=wet+bulb+temperature


    The physics of the wet bulb temperature covers almost everything you need to know.
    In particular, the wet bulb temperature is governed by heat and mass exchange with atmosphere.
    The porosity of the outer clay pot may slow down the process.
    However, qualitatively, it will remain similar.
    You might even consider modifying the wet bulb model to investigate how much the "pot temperature" may differ from the wet bulb temperature.
    The heat flow is included in the wet bulb model.

    Take the time to understand the physics in depth.
    The calculations will take only a fraction of your time.
     
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