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Thread: Formula to cool water with air

  1. #1
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    Confused Formula to cool water with air

    I need to determine how much Air is needed to cool a copper tube. The air will be blowing on the outside surface of the tube as there will be water running through the tube.

    The starting temperature of the water will be 70C and the goal end temperature is 35C...all done in less than 2 minutes. I already know that it will take ~250 W to do this; however, how would I determine the amount of air needed? Am I looking for the velocity of the air to cool it? If so, what equation should I use to solve for this?

    I won't be able to have room temperature water available to cool the solution, neither will I be able to use ice. This needs to be repeated on command, so I'll need a fan that will kick on to cool the piping as the 70C starts flowing through.

    As you can see I'm quite rusty on my heat transfer and thermodynamics. It doesn't help that I can't find my textbook from when I was in school either!!

  2. #2
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    What does this tube look like. In order to do this , you need some minimum length and surface area of the tube, more like a heat exchanger;
    similar to the radiator in your car.

  3. #3
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    Sorry about that...

    The tube is placed in the block (similar to a heat exchanger). It is shaped in a labyrinth manner right now, but obviously can be changed and can be run outside the block and on its own.. Right now the tube is ~3 feet long with 3/8" diameter.

    Quote Originally Posted by zeke View Post
    What does this tube look like. In order to do this , you need some minimum length and surface area of the tube, more like a heat exchanger;
    similar to the radiator in your car.

  4. #4
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    Quote Originally Posted by Vig16 View Post
    the tube is ~3 feet long with 3/8" diameter.
    You have pretty much answered your own question, right there.

    Surface area.
    Thermal conductivity per second/minute/hour.
    V1 T1 = V2 T2 and V will be constant.

    You should not need any "text books."

    Dave

  5. #5
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    Thanks, Dave!!

    Quote Originally Posted by PinkertonD View Post
    You have pretty much answered your own question, right there.

    Surface area.
    Thermal conductivity per second/minute/hour.
    V1 T1 = V2 T2 and V will be constant.

    You should not need any "text books."

    Dave

  6. #6
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    I would buy a heat exchanger from a mfr who could advise you better. This is not simple.since the conductance at the surface of the block/tube depends on air flow and the internal film coefficient depends on water flow rate, etc.

  7. #7
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    Zeke,

    Not sure why, but I guessed this was non-critical application since he was trying to just cool the pipe. Another alternative could be a transmission cooler from an Auto wrecker and a couple of muffin fans. That's what I use on one of my Lasers here. Total cost about 20-bucks.

    Dave
    Generally, I will not give you the answer to your question, but I **will** guide you into discovering how to solve this yourself.

  8. #8
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    Dave,

    Since he mentions cooling water from 75 to 35 deg C , it looks like a heat exchanger problem, not merely cooling pipe
    He's pretty inexperienced and I wanted to show that it is not a trivial problem.

  9. #9
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    hi the amount of heat is to be extracted is 250w, and temperature limits are 70 to 35
    u can use the formula " Q=ha(t1-t2)"
    where q= amount of heat transferd in ur problem it is 250
    h=convective heat transfer co efficient of air
    a = surface area
    t1= intial temperature of the air
    t2 = final temperature of the air

    with that u will know the value of t2

    then substute it in Q=mc(t1-t2)
    where m= mass of air required
    c= sp.heat of air
    from this u will get mass of air

  10. #10
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    Thanks for your reply! I still have a question though. If the temperature of the environment is ~23C, I still don't know what my t1 and t2 are. Wouldn't they be the same?

    Quote Originally Posted by harsha.mech360 View Post
    hi the amount of heat is to be extracted is 250w, and temperature limits are 70 to 35
    u can use the formula " Q=ha(t1-t2)"
    where q= amount of heat transferd in ur problem it is 250
    h=convective heat transfer co efficient of air
    a = surface area
    t1= intial temperature of the air
    t2 = final temperature of the air

    with that u will know the value of t2

    then substute it in Q=mc(t1-t2)
    where m= mass of air required
    c= sp.heat of air
    from this u will get mass of air

  11. #11
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    Quote Originally Posted by Vig16 View Post
    I still don't know what my t1 and t2 are. Wouldn't they be the same?
    Only if there was no heat transfer. Think about it. You want to transfer heat from one thing to another.

    Dave
    Generally, I will not give you the answer to your question, but I **will** guide you into discovering how to solve this yourself.

  12. #12
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    Quote Originally Posted by Vig16 View Post
    Thanks for your reply! I still have a question though. If the temperature of the environment is ~23C, I still don't know what my t1 and t2 are. Wouldn't they be the same?
    it is t2... with this u can get t1

  13. #13
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    Oh, OK...thanks everyone!

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