# Heat transfer between enclosed parallel horizontal plates

1. Jan 23, 2014

### manderson99

1. The problem statement, all variables and given/known data

I am curious about the situation of heat transfer between parallel horizontal plates in an enclosed space (particularly with the lower plate being higher temperature and subjected to a steady heat flux) with the enclosure itself being composed of a fictional "perfect insulator". The self-teaching heat transfer text I have mentions how to calculate Grb and the Nusselt number based on Grb, but it only covers situations when the dimensions of the two plates are identical.

How should I handle situations where the lengths and widths of the plates are not identical? For example, what if the top plate is 1 meter x 1 meter, but the bottom plate is only .5 meters x .5 meters? Consider the bottom plate to be centered inside a perfect insulator whose surface area, when added to that of the bottom plate itself, is 1 m2.

Also, should I use the Nusselt number (and, by logical extension, h) calculated from the standard horizontal plate/bottom plate hotter equation if one or both of the plates are equipped with, say, fins to improve surface area? Or should I calculate h locally for the fins according to equations that pertain to them individually?

2. Relevant equations

Grb = (gβ-1(T1 - T2)b3)/v2

For Grb > 4x105, Nub = .068(Grb)(1/3)
(note that the situation I'm currently working with appears to have a Grb > 4x105, so I have omitted the other two equations for Nub)

3. The attempt at a solution

Lacking any other guidance, I decided to solve the problem of the bottom hot plate being smaller by adjusting the value of b. I used an average of the distance between the midpoint of the bottom plate and the midpoint of a "virtual" plate of identical dimensions embedded within the top plate and moved to the furthest extremes the virtual plate could be moved without encroaching on the perfectly insulated sides of the enclosure. This resulted in a much higher value of b. I am still a bit unsettled as to how to calculate q in this circumstance - by using the surface area of the top plate or the bottom plate. With the adjusted value of b, I would think I'd use the surface area value of the top plate, but . . .

I also attempted a different tack involving a double integral, but I do not think my approach was correct. It resulted in an integral I was not skilled enough to solve on my own, which is another matter altogether. I could dig up that work if it is deemed at all relevant to this discussion.

When adding fins to the situation, I simply used Nub to calculate h, though I am not sure if doing so is appropriate. I am also still unsettled as to which array of plates I should use to calculate q (the ones on the top plate, or the bottom) when using an adjusted value for b; obviously, this situation is further confused by the question of whether or not an adjusted value of b is even appropriate.

2. Jan 23, 2014

### Staff: Mentor

Can you give a sketch of the setup? What is between the plates?

In general, those problems get too hard for an analytic solution and you need some approximations. Heat flow is not restricted to the vertical direction, this makes the calculations tricky.

3. Jan 23, 2014

### Staff: Mentor

Let's take a step backward first. Suppose there was no natural convection (either because the top plate is hot and the bottom plate is cold, or because the system is in outer space where there is no gravity). Also, instead of using square plates, lets use unequal diameter circular plates, so that we have axial symmetry. What in your judgement would the temperature profiles and the heat flow lines look like without the natural convection present (I presume you are assuming that the spacing of the places is small compared to their lateral dimensions)? What would be your best estimate of the heat transfer rate, in terms (algebraically) of the physical parameters? What assumptions did you use to make this estimate? Looking at this problem will give us entry into your more complicated natural convection problem.

Chet

4. Jan 28, 2014

### manderson99

My son wishes to apologize for being unable to log in and respond in a timely manner. I am posting the responses he wanted to make in his stead. I hope this is not a problem. The remainder is his response.

Okay! They're crude, but I've included both examples from my initial post as examples #1 and #2. #1 is just the square plates of unequal size, while #2 include fins on both plates. The uh, spacing isn't exactly perfect, especially with the spacing of the fins in #2. My MS Paint skills aren't that impressive. The diagonal slashes you see in the sketch are supposed to represent "perfect" insulation.

Oh, and I neglected to mention the air between the plates. Sorry! I'm assuming air between the plates, with a T of (T1-T2)/2.

Let's just say that the top plate is hotter than the bottom so that we can assume Nub = 1.
I put the small plate at the top to make sure the small plate is still the hotter of the two.

Okay, that might make things simpler.

I included two representations of what I think heat flow lanes would look like. I did my best to make it look like flux plotting, though as I mentioned above to mfb, my Paint skills just aren't there.

Yeah, b << .5 meters.

Stepping back and looking at it this way, I could tackle this problem in at least three ways:

1). Calculate h based on the actual vertical distance between the plates (h = k/b, given Nub = 1), then calculate q (q = hA(T1 - T2)), using A = As).
2). Produce an average distance between the edge of the smaller plate and the remaining surface area of the larger plate, use that for my b value in calculating h, and then produce another q value with A = Al - As.
3). Add both q values to produce the total heat transfer.

Or, I could:

1). Calculate the average difference in distance between any two proportionately similar points on each plate using a similarity-of-triangles situation (minimum distance would be between the centers of each plate, maximum distance would be from the edge of one plate to the edge of the other, add them and divide by two). Use this for b, and calculate h.
2). Use Al in q = hA(T1 - T2) to find the rate of heat transfer.

Or, I could rely on radial symmetry to treat this as a two dimensional heat transfer situation when setting up a numerical analysis grid (assuming steady-state conduction through the air with a known - and constant- value for k). Instead of q/L = (kAΔT)/(LΔS), I would use q = (kAΔT)/ΔS, where ΔS = Δy, and A = Δx
* pi(a2 - b2), where a equals the distance from the center axis to the edge of the heat transfer lane furthest from the axis, and b equals the distance from the center axis to the edge of the heat flow lane closest to the axis (both a and b would be assessed at the middle y position of the segment of the heat lane in question). This would only be an approximation of the total "surface" area covered by the heat flow lane, since each lane would actually be a large cone, and each segment of the lane would be a small conic section. My approximation would be more accurate for a cylinder.

The main assumption was that there would be heat transfer from every part of the smaller plate into every part of the larger plate. There would probably be heat flow lanes set up roughly following the geometric similarity of the two plates. Any heat flow lanes into the smaller, cooler plate not directly superimposed under the larger, hotter plate would be travelling through more of the quiescent air, so the effective h value for that transfer would be lower. Or, to put it differently, there would almost have to be more heat flow in this situation than between two small (.5 meter diameter) plates of identical dimensions, but not necessarily that much more.

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5. Jan 28, 2014

### Staff: Mentor

For conduction without convection, image 3 has the right idea. For b<<.5 meters the outer part is negligible for the total heat flow. Or, to be more precise, only a part with a width proportional to b is relevant in the overlap region.

6. Jan 28, 2014

### Staff: Mentor

Manderson,

You did a very nice job of doping things out. In my judgement, you have a great future as an engineering modeler (that's a compliment).

I would just like to iterate a little on what you said. Most of my iteration applies to diagram #3 you provided. The center region, with the vertical heat flow lines, is right on target. The region outboard to the small disc needs modification. The flow line from the very rim of the small disk to the rim of the large disk takes a much different path than you showed. We are dealing with potential heat flow here (i.e., the temperature satisfies Laplace's equation), and, for this situation, the outer flow line goes radially outward from the small disk, reaches the edge, and makes a right angle turn downward to the rim of the large disk. The entire region outboard of the small disk is filled with heat flow lines, although they are further apart than those in the central region. Knowing what the outer flow line is doing should give you an idea of what you need to do to interpolate the paths of the heat flow lines outboard of the small disk.

Another feature of the figure that you need to reconsider is that, as you said, the figure is not to scale. A key aspect of the particular system you are analyzing is the very small gap between the plates, compared to the radii of the plates (i.e., the small aspect ratio of the system). If the figure were drawn closer to scale (both in terms of the small gap and the considerably larger radius of the large disk compared to the small disk), it would have been much easier to recognize the nature of the flow lines and isotherms outboard of the small disk. I would like to elaborate a little on what mfb alluded to. Because of the features of the geometry that I just discussed, if you actually solved Laplace's equation for the temperature distribution and the heat flow distribution in the system, you would find

(1) beyond a distance about 1 channel height b inboard of the edge of the small disk, the flow lines would all be vertical, and the isotherms would all be horizontal

(2) beyond a distance of about 1 channel height b outboard of the edge of the small disk, the temperature would be virtually uniform, and equal to the temperature of the larger disk. This region would contribute negligibly to the flow of heat between the plates.

(3) So essentially, all the heat flow would occur in the central region inboard of the rim of the smaller disk, and the heat transfer rate, to a great approximation, would be kAsΔT/b.

Now, with this knowledge, lets return to your natural convection situation. The same kind of features with regard to the aspect ratio would apply. So, inboard to the small disk rim, you would have Benard cells present and enhancing the heat transfer. Outboard of the rim of the small disk, you would have very little heat transfer. So to do the problem with natural convection, you would get the Benard heat transfer coefficient for the case of two infinite parallel plates and apply that to the region inboard of the small disk rim. Outboard of that region, the temperature would be equal to the temperature of the larger plate, and the contribution to heat flow between the plates would be negligible.

Chet

7. Feb 7, 2014

### manderson99

That makes sense, especially in light of what Chestermiller has to say below.

Thanks! There is still more work and learning to be done. That, and I'm going to need to graduate beyond Paintshop at some point. Heh.

Okay, that does make sense.

Hmm, okay. So increasing the diameter of the smaller disk and/or increasing the value of b would allow more of the larger disk to become a factor in heat transfer (with the former being of greater effect). Interesting. And yes, drawing to scale would have made things a bit more evident. Using a more exact value of b would also have been useful here, methinks.

That makes sense as well. With reference to my question of adding fins to both plates involved, I am leaning towards the conclusion that the best h value to use here would likely be determined by the overall situation minus the fins. The spacing and temperature differential between the plates plays such a prominent role in determining what, if any motion of air there is inside this enclosure that conditions local to fins would be comparatively irrelevant, so long as the fins are not crammed too close together. Or so I would think.

8. Feb 7, 2014

### Staff: Mentor

Sure. If the whole space between the plates was filled with fins that are touching one another, they would constitute a solid metal plate which would provide great heat conduction.