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I Air velocity inside a pipe (natural convection)

  1. Jan 30, 2017 #1
    Hi guys,

    I'm doing some research in papers and literature to solve this problem but I've not found anything useful. I want to know the air velocity inside a pipe by the only effect of natural convection. I've attached a picture of the system right here:

    attachment.php?attachmentid=12218&d=1485776110.jpg
    I hope you can help me to find the equation that let me know the velocity in this system.

    Thanks in advance,

    Salus
     
  2. jcsd
  3. Jan 30, 2017 #2

    Nidum

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  4. Jan 31, 2017 #3
    This equation will help you to solve this:

    V= Qa/60Pi(d/2)^2

    Where, d : Pipe Inner Diameter (m)
    Qa : Air Flow Rate (Actual) (m³/min)
    v : Air Velocity (m/s)
     
  5. Jan 31, 2017 #4
    What do you think you have omitted from this description?
     
  6. Jan 31, 2017 #5
    Thank you Chadi B Ghaith but there is a problem with your equation, I need to know the vertical speed of the air generated by the natural convection. Applying your equation in my problem Qa is also an unknown variable.
     
  7. Jan 31, 2017 #6
    Well, the internal diameter is not shown, but I only need the expression that allows me to know the flow rate or the vertical speed due to convection. ¿Could you help me with this?
     
  8. Jan 31, 2017 #7
    Is the tube cooled, or is it adiabatic? What causes the temperature to be colder at the top than at the bottom? Is the pressure at the top the same as at the bottom? Don't you think that knowing the answers to these questions would help to give us an idea what we are dealing with here?
     
  9. Jan 31, 2017 #8

    russ_watters

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    The diagram isn't even correct for convection unless there is another tube or downward airflow path we aren't being shown. It violates continuity. So we need more details about the system.

    Also, the source of the heat is more important than you may think. The temperature difference in the air drives the convection, but without heat sources/sinks the convection stops before it fully develops.

    So please describe the system in a lot more detail.
     
  10. Jan 31, 2017 #9
    Well, first of all thanks for your interest and your help, and secondly I'm sorry for the missing information but I'm not physicist. Let's take a look to the things that you've mentioned:
    • Diameter: 81 mm
    • Pipe type: PVC pipe 3 mm thick
    • Pipe length: 30 cm
    • Heater: 36 W resistance set at the bottom
    • Pressure: I think there should be a gradient of pressure due to the difference in temperature, that is the reason why the air flows up, isn't it? But I don't know the exact pressure. The top of the pipe is covered then the whole system is confined inside the pipe.
    • Temperature difference: 21ºC Between T1 and T2, where T1 > T2
    I thing that the system could be simplified as an adiabatic system because this process last 45s aprox. Thus, the conduction loss through the PVC probably won't need to be taken into account.

    Is anything else missing?

    Thanks for all!

    Salus
     
  11. Jan 31, 2017 #10
    Thanks for your suggestions, I have given more details in the Chestermiller answer. I hope you it could help.
     
  12. Jan 31, 2017 #11

    russ_watters

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    Is the pipe open or closed? If open, what is the temperature outside?
    [Edit: misread] if the pipe is closed, what is cooling the top? The diagram can't be correct; air has to flow down somewhere. Convection is circular.
     
  13. Jan 31, 2017 #12
    Thank you! Here I post another block diagram that is probably closer to the reality.

    dda24f3fc2.jpg
    I hope it would be clarifying!

    PD: I agree with you regarding the speed profile. I think it would be like I have drawn in the center of the cylinder and oposite in both sides. In 3D i could imaging the vectorial field like an elongated toroid. The center air moves up and the external part down.
     
  14. Feb 1, 2017 #13

    Nidum

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    From your description so far I'm not convinced that anything much is going to happen at all other than the contained air will get a bit warmer and it's pressure may rise a little .

    What is your reason for wanting information on this - are you working on a specific project that you can tell us about ?
     
  15. Feb 1, 2017 #14
    I'm working in a project but I cannot give you more information about that. I'm sure that something happened because I've registered this convection. The thing is that I want to justify what is happening using maths, but at the moment I've had no success.

    Thank you for the comment.
     
  16. Feb 1, 2017 #15

    Nidum

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    In the problem as described all that I would expect to observe would be some transient air movements as the system warmed up . It may be that during the very short duration of your experiment these transients had not died away .

    Really though unless you can tell us more about the design and use of this system we are just guessing what might be going on .
     
  17. Feb 1, 2017 #16
    Well, that is totally true. The data is collected during the transient period. I've been checking the data and I've seen that in 45s, beeing the lower temperature 48ºC, the higher has only increased 0.3ºC (24.2ºC -> 24.5ºC) . Is this information helpful?
     
  18. Feb 1, 2017 #17

    Nidum

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    What is the warm up time of the heater ?
     
  19. Feb 1, 2017 #18
    Nidium, I've been using the post you shared like this but I'm not sure about how to proceed:

    That is my system:
    589fb11c65.jpg

    I've supposed that the air is using the half pipe diameter to go up and the other half to go down, then I have these equations:$$\rho_{bottom} = \frac{353}{273 + T} \rightarrow \rho_{bottom} = \frac{353}{273 + 48} = 1.1 \frac{kg}{m^3}$$
    $$\rho_{top} = \frac{353}{273 + T} \rightarrow \rho_{top} = \frac{353}{273 + 21} = 1.201 \frac{kg}{m^3}$$
    $$v=\sqrt{\frac{(2g(\rho_{top}-\rho_{bottom})h)}{\frac{\lambda h\rho_{bottom}}{dh+\sum \varepsilon \rho_{bottom}}}}\rightarrow v=\sqrt{\frac{(2*9.81(1.201-1.1)0.3)}{\frac{\lambda *0.3*1.1}{0.04+1.1}}}=?$$

    There is still missing how to calculate##\lambda## for the Air friction coefficient.

    EDIT: I thought I've used your link but not... I've used the equations of this one: http://www.engineeringtoolbox.com/natural-draught-ventilation-d_122.html
     
    Last edited: Feb 1, 2017
  20. Feb 1, 2017 #19
    I couldn't imagine how this information could help you to know how to calculate the final speed. Anyway, it lasts like 20' to be in a steady state, but the thing is that this hot air is suddenly applied to the system, there is a valve that puts in contact this hot air with the cold one. I suppose, this could be more useful. ;-)

    Thanks Nidum!
     
  21. Feb 1, 2017 #20

    Nidum

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    Your description of this problem keeps changing .
     
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