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Basic Thermo confusion

  1. Aug 20, 2009 #1
    Hello... first post here so take it easy on me.

    From the Book: "Aquatic Systems Engineering: Devices and How They Function, Selection Installation Operation" by P.R. Escobal

    In the section about heat loss (section 12.9) the author gives the basic conductivity equation in the form:

    Q = (A * 2.54) * k * T * time/d

    Where:
    Q = calories
    A = Area in square inches (using 2.54 to convert to centimeters)
    k = cal cm/sec in Celcius
    T = Temperature differential between Hot and Cold side of panel in degrees Celcius
    d = thickness in inches

    So using an example of
    A = 3505 square inches
    k = .002 (thermal conductivity of glass)
    T = 4.44 degress celcious (8 degrees F)
    t = 1 second
    d = .375 inches

    Q = 3505 * 2.54 * .002 * 4.44 * (1/.375)
    Q = 251.85

    So far so good. The result is consistent with any other form of the conductivity equation I find. I tried a few different variation and got the same result.

    So here is where I am confused. The equation outputs 251.85 calories. I presume that because I chose 1 second as the TIME, then thermal loss to the aquarium is 251.84 calories/sec?

    1 Watt = 4.184 calories/sec
    So 251.84 calories/sec = ~1054 Watts

    HOWEVER:
    The author indicates that to convert the result to Watts, you must multiply Q by 0.23889, giving a result of 60.16 Watts!

    If I use the conductivity equation in the form found here:
    http://hyperphysics.phy-astr.gsu.edu/Hbase/thermo/heatcond.html
    I get the result of 1055.5 Watts or 3601.62 btu/h There is no "time" input. I would assume that it is an HOUR based on the BTU/h form fields.

    That exactly matches the results from the authors equation before his conversion to Watts.

    What am I missing here? Where is the authors multiplier to get Watts coming from?

    Oddly, if I divide the BTU/h result by 60 I get the authors Wattage number. But I don't understand where the factor of 60 is coming from. The authors equation is in calories/sec and the reference equation from the website is in btu/h both give the same answer that is a factor of 60 from the authors Wattage.

    Somebody please enlighten me!
     
    Last edited: Aug 20, 2009
  2. jcsd
  3. Aug 20, 2009 #2

    sylas

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    one calorie is 4.184 Joules.

    Therefore 1 Watt is 1/4.184 = 0.239 cal/sec

    Added in edit: but I note you've gone back to a correct form in the second line. I'll look at it all again...

    Added again: I note you are using 2.54 to go from square inches to square centimeters. That should be 2.542

    Cheers -- sylas
     
  4. Aug 21, 2009 #3

    sylas

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    Starting over. Let's be real scientists and put everything in SI units. :approve:

    One thing is a bit odd. The units for thermal conductivity are energy per time per distance per temperature. Glass is typically around 1 W/(m.K), which is 0.00239 cal/(s.cm.K); not far off your value. But note that the the units are an inverse distance, where you've written cal cm/sec per K.

    Anyhow, moving on:

    A = 3505*0.02542 m2 = 2.261 m2
    k = 0.002 * 4.184 * 100 = 0.837 Js-1m-1K-1
    T = 4.44 K
    t = 1 s
    d = 0.375 * 0.0254 = 0.009525 m

    Hence Q = A k T t / d = 2.261 * 0.837 * 4.44 * 1 / 0.009525 = 882 J

    which is 210.8 cal

    You've already got the t = 1 in there, so remove that to get the same number with units cal/sec

    Cheers -- sylas

    PS. There also seems to be an error in your multiplication.
     
    Last edited: Aug 21, 2009
  5. Aug 21, 2009 #4
    Forgive the stupidity, but I am still lost.

    The author lists Q as calories. The time entered into the equation is 1 second. The result is Q = 251.85 calories in one second. That equates to 1053.698 Watts (unless I am not understanding something).

    0.239 cal/sec = 1 Watt
    251.85 cal/sec = 1053.698 Watts
    The result is exactly confirmed by the calculator found here:
    http://www.unitconversion.org/power/watts-to-calories-th--per-second-conversion.html

    Again, this also EXACTLY matches the output from another form of the conductivity equation found here: Note that the outout is in both Watts and BTU/h
    http://hyperphysics.phy-astr.gsu.edu/Hbase/thermo/heatcond.html

    What am I missing?
     
  6. Aug 21, 2009 #5

    sylas

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    What result is given in the book?

    Where did you get 251.85? In the original post you wrote:
    Q = 3505 * 2.54 * .002 * 4.44 * (1/.375)​

    When I do that multiplication, I get 210.816.

    That has units of cal/sec, given units for the five numbers, which is 882 Watts.

    Cheers -- sylas
     
  7. Aug 21, 2009 #6
    Sorry, I rounded in my posts... I used more decimal places in the spreadsheet for k (0.00238933) . Small changes in k equate to fairly large swings in heat flow :)

    The book does not give a result, only the equation and the additional instructions that:



    His outcome for a 3/8" thick 12" x 24 " 60" (WHL) tank with an insulated bottom and 1/8" glass top and a 15F delta T is 169.25 Watts. Using the equation he posted, the result is fairly close to that (he takes into account the insulative properties of the air gap between the top of the water and the lid for the example) So the 169.25 Watts makes sense with regard to his text and that unexplaned Watts conversion.

    My Calcs
    Using K = .002 for the sides and top and k= .0005 for the bottom (wood stand) with the straight thermal conductivity equation, I get a little over 3000 Watts (per hour?) of thermal loss to the room. Not even in the ballpark of his answer. Again, the results from his base equation match that of the .edu link I posted above.

    This author is a well respected engineer who has authored several physics texts.

    SORRY for all of the edits... it is 2:30 AM and the brain is a bit slow!
     
    Last edited: Aug 21, 2009
  8. Aug 21, 2009 #7
    I have made serious edits to the post above. I at first used the 4.44 DeltaT instead of the new 8.3 DeltaT

    So with the listed glass tank and a 15 Degree F (8.3C) temperature differential between the tank and room, he predicts a 169 Watt thermal loss to the room and the standard conductivity equation predicts 3000 Watt thermal loss to the room. Somebody has to be WAAAAY off.
     
  9. Aug 21, 2009 #8
    please move this back to the physic forum! I am not a student, nor is this coursework or homework. I am an adult seeking help understanding a physics concept, not a student looking for homework help
     
  10. Aug 21, 2009 #9

    sylas

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    I found an article online which uses numbers similar to your example. See Thermodynamics for the Reef Aquarist.

    This example uses:
    A = 3505 in2 (same as you have used)
    d = 0.375 in (same as you have used)
    T = 8 °F = 4.44 °C (same as you gave originally)
    k = 0.578 BTU/hr /ft /°F

    Converting units. 1 BTU/hr = 0.293 W
    1 ft = 0.3048 m
    1 °F = 0.555... °C (for a temperature difference)

    Hence k = 0.578 * 0.293 / 0.048 * 1.8 = 1 W m-1 K-1

    which is 1/4.184/100 = 0.00239 cal s-1 cm-1 K-1

    So they are using a value of k that is the same as yours -- though in your first post you used 0.002 which was 20% too small. I am thinking this must be a standard example.

    Mathematically, you should be aware that in fact there ISN'T a big dependence on k. The problem is that you rounded it by 20%, which makes a 20% error. When multiplying, the important thing is to have the same number of significant figures; not the same number of decimal places. So rounding 0.0024 to 0.002 is just as bad as rounding 240 to 200.

    The article calculates a heat loss as 3595.79 BTU/hr, which is 1053 W. Same as you and I obtained.

    So your calculation is correct. I do not know what the book is doing. Perhaps that are looking at some other terms for heat loss, such as an evaporation loss; or allowing for the heat input from a pump. I am just guessing.

    Cheers -- sylas

    PS. It doesn't make much difference which sub-forum this is in. A question like this is well suited to the homework forums, and you get useful input here. It's all still physicsforums.

    PPS. I suddenly get why you had 0.002. You transcribed numbers from your spreadsheet, which was set to display 3 decimal places. You did explain this, but I was slow to see what you mean. So your calculation is correct.
     
    Last edited: Aug 21, 2009
  11. Aug 21, 2009 #10
    Lets take a step back here because I think the very basic question and reasoning is being lost. I don't wish to sound ungrateful, but I am struggling to understand this information and this exchange has only confused the issue.

    My question is not being answered and you are isntead distracted by rather insignificant details. With all due respect, if you could answer my technical question, I could learn. If you are unable to, Then I will start a new thread, as this one is somewhat sidetracked and tossed in the homework forum.

    In hopes of getting back to the question at hand, can we please forget about significant digits, rounding and improper order of written units.

    Please carefully consider the points here.

    1) The author of the book is a well respected engineer. One of us is clearly wrong. I would imagine it is me, but many sources on the web would indicated otherwise.

    2) The author has not taken any heat gain into consideration and has also not considered other forms of heat loss such as evaporation, radiation. So please, lets ignore that they exist. We are talking about CONDUCTION from a GLASS tank full of water into an air filled room and the Wattage needed to keep it at a constant temperatre in that room. No more, no less.

    2) The equation he uses is the Fourier equation in the form (Q = (A * 2.54) * k * T * time/d). He uses it in a form that outputs cal/sec. The links posted above use a form of the equation that outputs btu/hr and Watts. They all provide the EXACT same result.


    3) The author claims that the RESULT of the equation Q = (A * 2.54) * k * T * time/d needs to be multiplied by 0.23889 to give an output in WATTS. That is to size a heater to keep that tank at the target temperature, it would need to be W watts where W = 0.23889 (Q = A k T t / d )


    So let me ask very clearly

    WHY IS THE AUTHOR MULTIPLYING BY 0.23889?

    Please note: I have not used color to denote anything more attention to my actual question. I am not attempting to be rude or flip, just clear.
     
  12. Aug 21, 2009 #11

    sylas

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    He's just converting units, from cal/sec to Watts. 0.23889 is 1/4.186

    Have a look at my very first reply, number 2. This was the point I singled out for you there. I have seen both 4.184 and 4.186 used as the conversion factor; I think 4.186 is correct. (see postscript)

    Cheers -- sylas

    PS. There are different notions of "calorie" in use. 4.184 is a "thermochemical calorie" 4.1868 is an "Internation Table" calorie. Definitions described at NIST http://physics.nist.gov/Pubs/SP811/footnotes.html#f09".
     
    Last edited by a moderator: Apr 24, 2017
  13. Aug 21, 2009 #12
    So again, what I am missing here?

    You would multiply cal/sec by 4.186 to get watts, not by 0.23889

    1 Watt = 0.23889 cal/sec = 3.41 btu/h
    1050 Watt = 251 cal/sec = 3583 btu/h

    That matches every calculator and example I can find online. Yet, the author multiplies by 0.23990 instead of 4.186 and gets an estremely small number that does not match any of the online calculators. Is the author wrong or are the online calculators wrong, or am i simply missing something very obvious?

    P.S. I was aware that there are "thermochemical (TH)" and "International (IT)" constants differ sligtly... and also that when we talk food we talk kcal even though we call it "Calories" :)
     
  14. Aug 22, 2009 #13

    sylas

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    Quite right... to get from cal/s to Watts you should divide by 0.23899, or else multiply by 4.184.

    That's actually what I did in my units conversion, and what is done in the article I quoted above. The number they get is correct. If you calculate a value in cal/s, you multiply by 4.184 to get Watts.

    I don't think you are missing anything. If your book takes a value that is in cal/s, and multiplies by 0.23899, then it is wrong.

    Cheers -- sylas
     
  15. Aug 22, 2009 #14
    That leads me to question the entire validity of the information in the book. How could the author make such a blatant error and use that error to derive an entire formula for sizing heaters and other equipment? I am simply stunned.

     
  16. Aug 22, 2009 #15

    sylas

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    It's not that unusual for a genius to slip up in a problem. Most textbooks will usually have a couple of errors in them that slip past review; I don't think there is such a thing as a perfect text book free of simple errors.

    Doing a division rather than a multiplication might be a basic error; but it is not because of a lack of understanding, so much as a slip of focus. We all make basic mistakes from time to time, and my guess is that this is what happened here.

    Cheers -- sylas
     
  17. Aug 22, 2009 #16
    Fully understood... but he used the basic premise to develop a rather complex integration to calculate heat loss and heater sizing, devoting an entire chapeter to the subject and using the results as the basis for fundamentals in other chapeters.

    There is a second edition to the book and I am curious to know if any of the errors have been corrected. Both the 1st and 2nd edition are long out of print and cost upwards of $100 at used book outlets on the internet. Nedless to say, I am not spending $100 to find out :)

    Thank you for taking the time to respond to this thread. I thought my understanding of these matters was fairly informed and asked the question here to ensure that I was not fully ignorant of the basic physics. It would appear that I have stumbled upon an error by a well regarded name in the hobby. His work is referenced any time "physics" with regard to aquatic systems is the subect. I shudder to think that so many people have been misled due to a basic error in converting units.
     
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