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Thermodynamics challenge

  1. May 8, 2008 #1
    hey guys,
    I am trying to figure out how many btu
    will be put out by a coil of copper pipe
    half inch by 20 feet painted black to heat
    my pool. Is there any way to figure this.
    I heard that a meter of sun produces 1000
    watts but don't know how to figure
     
  2. jcsd
  3. May 9, 2008 #2
    additional info

    I also have a gas heater that is 335,000 btu's
    but I want to use it as little as possible, as you could imagine.

    also about 25,000 gallons
     
  4. May 9, 2008 #3

    russ_watters

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    Staff: Mentor

    Is this coil of pipe part of a solar collector or is it just a black piece of pipe? How big is the collector? If it's just a black piece of pipe, it won't collect much (find the surface area).

    How about a solar collector cover for the pool?
     
  5. May 9, 2008 #4
    well...

    I was thinking of using half inch copper coiled pipe
    like you would find at home depot.
    I was going to paint it black.
    copper,because metal transfers heat better (I think)
    and black for obvious reasons.

    Half inch because I figured a smaller diameter
    pipe would put more water in contact with
    pipe walls.

    I was going to lay it on a black sheet of tin
    pump water in one end and back to pool

    I thought if anyone could help it would be here
    thanks for your help
     
  6. May 9, 2008 #5
    and possibly enclosing in a glass topped box.

    i guess i need to know how much surface area collects how many btu's or watts and how much will be transfered into the water

    anyone????????????/
     
  7. May 10, 2008 #6
    is no one up to the challenge??
     
  8. May 10, 2008 #7
    well, let me pose this:

    when they say that a square meter of sunlight produces 1000 watts, do they mean per hour??

    if so... then i guess i could deduct that if i had 1 sq meter of surface area exposed i would get 3413 btu's/hr and to raise my temp 1 degree ....

    1 btu raises 1 lb of h2o 1 deg.
    24000 gallons = ~200,000 lbs
    200,000 / 3413= 58 hours of direct sunlight
    of course I would have heat loss constantly especially night

    but what benefit then does painting it black or using copper rather than plastic have??

    very challenging
     
  9. May 10, 2008 #8

    Mapes

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    Science Advisor
    Homework Helper
    Gold Member

    1000 W is 1000 J (~1 BTU) per second. The watt is a measure of power, or energy per unit time.
     
  10. May 10, 2008 #9
    so how does that apply in my problem
    Are you saying that it would raise the water temp
    1 degree per second???
    I'm confused
     
  11. May 10, 2008 #10

    russ_watters

    User Avatar

    Staff: Mentor

    Ok, sorry, I didn't have much time to spend before to help...

    If you have 1000 w/m^2 or 317 btu/ft^2. Now, what is confusing about a btu is that a btu in common language isn't really a btu, it's a btu-h. 1btu for one hour. So 1 square foot of peak sunlight can raise 317 lb (about 37 gallons) of water 1 degree in one hour.

    Now, for your particular problem, we need a guestimate of your efficiency. A very good solar collector can have an efficiency of 60-80%: http://www.thermomax.com/effi.htm If you get 25%, consider yourself talented. So now we're at about 10 gallons of water, 1 degree per hour per sq ft.

    Now, unless your rig tracks the sun, you'll only collect perhaps 1/3 of the energy you otherwise could, and for about 10 hours a day, in the summer. So, 33 gallons, 1 degree per day per sq foot.

    The 1000 w/m^2 is based on not having an atmosphere. Depending on where you live, you'll get an average of around 40-60%. We'll split that and call it 50%. So now we're around 16 gallons, 1 degree per day per square foot.

    A pool that's 40'x20'x5' contains about 30,000 gallons of water, so to have any noticeable impact on the temperature (say, enough for a 1 degree per day rise), you'll need around 1,900 square feet of collector.

    Now, this ignores the heat loss. I can tell you that the biggest form of heat loss is via evaporation and as a result, the equilibrium temperature of a pool is somewhere around the average dew point at your location. A solar cover helps heat the pool, but it also eliminates evaporation, potentially saving much more energy than the cover actually collects. Ie, if you have a heater, a cover will reduce the usage of the heater, but mostly because it reduces the evaporation rate, not because it collects more heat. This calculation is kinda complicated and has been discussed before: https://www.physicsforums.com/showthread.php?t=158353&highlight=pool+heat+loss+evaporation

    That's enough for now - I'll let you chew on that for a while, and then come back later and work on making sure that previous thread got the calculations correct....
     
  12. May 11, 2008 #11
    wow.
    thats impressive and very appreciated.
    I thank you for saving me the time. I dont think I could make a 1900 sq ft collector

    is there any way that a hommade solar heater could work?



    thanks again
     
  13. Mar 2, 2009 #12
    Re: well...

    Your intuition is good! The smaller the diameter, the larger the ratio of surface area to volume, and the more efficient use of your space.

    But, there's another thing you need to consider. The smaller the diameter, the less light from the sun hits the pipe, so the less heat is transferred.

    You could have the best of both worlds by bending this pipe back and forth like the coil in this picture:
    [​IMG]

    That way, you get the most surface area in the least space and you get the largest exposed area to the sun.

    In fact that's why they use similar configurations for coils in refrigerators and heating units. Not the "exposure to the sun" part, but the "maximum heat exchange in the least space" part.
     
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