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Heat transfer help - simple problem

  1. Jan 24, 2014 #1
    By oversimplifying the problem I have, imagine that I have a panel constructed of stainless steel flat walls. The dimensions are 80 in x 70 in x 20 in, and the thickness of the panel is 1/8 in. Inside, I have a heater that outputs 400W of convective heat. The temperature inside the panel is 50F, and the ambient temperature outside the panel is 0F.

    Questions:
    a) Assuming the panel is leak tight, what is the heat transfer rate through the walls of the panel?
    b) Newton's Law of Cooling states that the wall surface temperature is assumed to be the same as the inside temp, but won't the steel surface be colder than the bulk temperature inside the panel?
    c) Where does thickness come into play? I only see that variable in the conductive HT equation.

    I get getting WAY too high of a heat transfer rate. Any help would be greatly appreciated.

    Thanks!
     
  2. jcsd
  3. Jan 24, 2014 #2
    What does it mean that you are getting too high a heat transfer rate? The heat transfer rate has to be 400 W at steady state. Are you saying that you were expecting a higher temperature inside the enclosure for that wattage of heater?

    Chet
     
  4. Jan 24, 2014 #3
    Sorry I wasn't clear. What I meant is that when I perform the calculation for Q_dot (using Newton's Law of Cooling) to find out how much of that added heat is getting dumped out of the panel through the walls, I am getting an impossibly large number of Watts (on the order of thousands). I know this value is incorrect, so I am trying to determine where I am going wrong with my calculations. My ultimate goal is to determine roughly what the net heat transfer rate is into and out of the panel, and then I can determine how much insulation would be sufficient for the interior walls in order to trap in as much of that heat as I can. Right now, the inside of the panel is just much colder than I would like it to be. This calculation is just a sanity check of my corrective action.
     
  5. Jan 24, 2014 #4
    That heater wattage sounds low. When you say that your calculated heat transfer rates are high, what values are you getting? What area are you using? What is your overall heat transfer coefficient? Show some sample calculations so I can get a feel for what you did. Is the outside of the box exposed to the environment outside, or is this thing sitting inside of some kind of freezer without much wind?

    The conduction through the wall is just part of the overall heat transfer resistance. There is also a heat transfer coefficient on the outside, and another heat transfer coefficient on the inside. (The bulk temperature in the enclosure is not going to be equal to the temperature of the cold walls).

    Chet
     
  6. Jan 24, 2014 #5
    I don't think Newton's law says that the wall temp must be the same as the inside temp, that is just a simplifying assumption.
    Heat transfer from a bulk gas to a surface and also from a surface back to a bulk gas (the atmosphere) take place by principally by convection and radiation, not conduction. There is always a temp drop across a surface film. That is why the wall temp will be higher than the ambient temp and lower than the interior temp. There are a lot of factors involved such as gas properties, orientation of the surface etc. I don't know any references off hand, but maybe others can help you.

    AceEngineer
     
  7. Jan 25, 2014 #6
    After having had more time to think about your problem, the heat flows you came up with do not seem out of line. The total surface area is about 80 ft2, and the temperature driving force is 50 F, so with a fairly low overall heat transfer coefficient (including inside air, outside air, and metal wall) of 1 BTU/hrft2F, I calculate a heat flow on the order of 4000 BTU/hr (1200 W).

    So then I decided to reference this to my house. My house's estimated heat transfer area is on the order of 3000 ft2, and the home heater needs to handle a temperature driving force of at least 70F, so with a heat transfer coefficient of 1 BTU/hrft2F, I calculate a heat loss of about 200000 BTU/hr. Yet my home heater only has a capacity of 25000 BTU/hr. Also, another data point is that it's 17F outside my house right now, and I felt my outside wall; it didn't feel cold to the touch, maybe a couple of degrees lower than room air.

    So, what gives? The actual heat loss rate is much lower than I would calculate with heat transfer analysis. The answer has to be this: The dominant resistance to heat transfer through the walls is going to be the insulation. And the overall heat transfer coefficient is going to have to be less than 0.1 BTU/hrft2F. All the other resistances are going to be negligible compared to this. So don't even worry about the heat transfer coefficient on the outside, the heat transfer coefficient on the room side, and the heat transfer coefficient through the wall. These are going to have to be negligible compared to the insulation. So the insulation is going to have to handle virtually the entire temperature drop of 50F while limiting the heat loss to your heater capacity.
     
  8. Jan 25, 2014 #7
    b) would be the problem. The temperature of the wall facing the outside has to be lower than the interior temperature. Just consider this, if the outside wall and inside wall were at the same temperature no heat conduction flow in any direction would be occurring.
     
  9. Jan 28, 2014 #8
    Thanks for the feedback guys. Ace I did not mean to tag 'conduction' as a topic, that was a mistake. Similarly, 256bits the actual values (that make sense to me at least) are that the panel temperature is warmest, followed by the interior side of the panel wall, followed by the exterior side of the panel wall, followed by the ambient temperature. In order to simplify the number of heat transfer surfaces I am dealing with, I just assumed a net heat transfer from inside to out. Chet I'm not entirely sure about what you did, it seems like you used the same coefficient for your home than you did the panel wall? Or did I misinterpret that?
    I've attached my calculation. I think my crazy high value might have to do with such a high delta T, so this may not be the best way to do it. Plus, this calculation does not take into account the thickness of my heat transfer medium, which to me seems pretty significant. I feel like there's something simple I am missing that will be like an "A-ha", but perhaps this problem just wasn't as simple as I initially thought it was.
    Thanks again for the comments.
     

    Attached Files:

  10. Jan 28, 2014 #9
    Yes. You misinterpreted. What I was trying to show was that your calculated rate of heat loss is not that crazy, if you neglect the insulation (that you will have certainly to add). I did the calculation for my house to show what would happen if I neglected the resistance of the insulation. It resulted in much too high a heat loss rate compared to what my home heater can deliver. The conclusion was that the main resistance to heat transfer is going to be the insulation, which you haven't considered yet. It also shows that you can neglect the heat transfer resistance to flow of heat through the bare panel material and the convective heat transfer resistances to the inside and outside air. The resistance of the insulation is going to dominate. This should be your aha!
     
  11. Jan 28, 2014 #10
    Ahhh I see.. Thanks man. I'll take a stab at it.
     
  12. Feb 21, 2014 #11
    hello steves
    the flux emitted by the unit is 400W, the total flux Q_totl,
    Then the flow exchange between the atmosphere and panneua lost Q_p flow,
    flow receives the fluid panel is Q_r, So
    Q_total = Q_p + Q_r
    Q_p S_d'éange X = dT / where S_d'échange e = (70x20 +80 x20) x2, and dT = 50-0, and e = 1/8.
    Q_r = Q_total-Q_p
    Now you can caculer the outer surface temperature of panel in contact with the device
     
  13. Feb 21, 2014 #12
    hello steves
    the flux emitted by the unit is 400W, the total flux Q_totl,
    Then the flow exchange between the atmosphere and panneua lost Q_p flow,
    flow receives the fluid panel is Q_r, So
    Q_total = Q_p + Q_r
    Q_p S_d'éange X = dT / where S_d'échange e = (70x20 +80 x20) x2, and dT = 50-0, and e = 1/8.
    Q_r = Q_total-Q_p
    Now you can caculer the outer surface temperature of panel in contact with the device.
     
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