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A heat transfer mystery

October 22, 2013

My Intro Physics class started the year with energy, and we incorporated the possibility of heat transfer between the system and surroundings from the very beginning. Our first experiment was to measure the energy content of a battery and a peanut. I’m going to be writing much more about this very soon.

Since then, we’ve been studying heat transfer more closely and I’ve reached a bit of a mystery on an experiment I’d like some help with. We’ve already done a few experiments to compare the thermal conductivities of different materials, and now we are measuring the conductances of various building materials by constructing these light boxes—a 100 W lightbulb surrounded by 5 different types of building materials (wood, plexiglass, various types of foam insulation, and sheet metal-zinc coated steel).

IMG 1180

The mystery comes when we try to measure the conductance of the sheet metal. We are measuring the temperature of the air inside the box and the inner and outer walls of the sheet metal using a Vernier Surface Temperature Probe, taped to the plate using white masking tape. The measurement we get for the inner wall temperature is 49.7^{\circ}\textrm{C}) while the outer wall has a temperature of 42.6^{\circ}\textrm{C}), for a difference of 7.1 degrees.

IMG 1338

To calculate the heat flowing through the plate, we need the conductivity of steel  (43\frac{\textrm{W}}{\textrm{m}\cdot \textrm{K}}) the area of the plate (0.09\textrm{m}^2), the thickness ( 0.0005 \textrm{m}) and the temperature difference ( 7.1^{\circ}).

Putting all of this together I get:

\textrm{Heat flow} = 43\frac{\textrm{W}}{\textrm{m}\cdot \textrm{K}}\cdot 0.09\textrm{m}^2\cdot \frac{1}{ 0.0005 \textrm{m}}\cdot 7.1\textrm{K}=5.4\times10^4 \;\textrm{W}.

This is obviously wrong, since we’ve only got a 100 W lightbulb inside the box. How can 54,000 W of energy be flowing through 1 of the 5 faces?

One explanation my colleague Mark Hammond came up with is that the surface temperature sensors aren’t reading just the temperature of the surface of the metal. The inside sensor is reading higher than it should because it is also partially reading the temperature of the the air. Likewise, the outside sensor is partially measuring the temperature of the outside air which is lower than the outside surface of the sheet metal. Both of these effects would combine to produce a larger temperature difference than the actual value and skew us toward a larger heat transfer value. Still, it doesn’t seem like this could account for how wildly off we are.

Shouldn’t it be that when we are in steady state all of the heat transfers through the 5 faces should be close to 90W, which is the rate at which thermal energy is being produced by the 100W lightbulb?

I’d love any ideas you might have for how to resolve this mystery.

6 Comments leave one →
  1. Andy "SuperFly" Rundquist permalink
    October 22, 2013 7:17 pm

    how much variation is there in temperature if you put the sensors toward the corner?

    • October 22, 2013 10:43 pm

      I’m thinking this must be the primary problem. I’m going to check this out tomorrow to be sure.

  2. October 22, 2013 9:39 pm

    Could the temperature sensor be absorbing more of the radiation than the steel plate, or is it shielded? What’s the gauge of the sheet metal ( )?

    • October 22, 2013 10:42 pm

      It’s 28 gauge, but I think the zinc coating adds significantly to the thickness.

  3. October 22, 2013 11:33 pm

    There should be almost no temperature difference between the sides of the metal sheet. I think you have measurement errors.

    My first thought was the same as Brian’s—that the inside sensor was being heated directly by radiation from the bulb, not by contact with the metal. What happens if you try using other temperature sensors (like a contactless IR thermometer), or if you shade the interior sensor?

  4. sorin permalink
    December 5, 2013 4:30 am

    You didn’t specified the measurement method. Are both thermocouples simetrically installed on both sides of the wall? I recommand you to reload the measurement, by covering of the inner thermocouple with an Al foil. In this case the heat transfer through radiation isn’t accounted. In the same manner, the outer thermocouple to be covered the same, to avoid unwanted convection due to the outer environment.

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