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Tree Bark

& Soil

Insulative Property Model

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Lesson Overview

A physical model will be used to simulate the insulative properties of tree bark and soil.
Bark and soil are two materials that can provide living things with protection from the heat of fires.

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Background Information

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Background Information

“Top-kill”
“Resprouting”
Cambium death

60°C (140°F)

Fire Scar

Burned areas often appear desolate because the aboveground parts of plants have been injured or consumed by flames. But it must be remembered that most native vegetation is adapted to fire and will even recover from intense blazes. Click A plant is described as “top-killed” when aboveground plant tissues have been destroyed but underground parts from which the plant can produce new stems and leaves have survived. Click Resprouting from belowground vegetative structures is an important adaptation for most native vegetation which allows an area to recover from a fire even when aboveground plant tissues have been “top-killed.” Click Click A temperature of 60°C (140°F) is considered adequate to break down cambium tissue which results in cambium death (described in the next slide). Click If bark can reduce heat transfer to the cambium layer so that its temperature does not reach 60°C, or Click Click only damage to a small area of cambium occurs, a tree will survive. Click

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Simplified Tree Trunk Cross Section

Bark

Phloem

Xylem

Cambium

This diagram shows three, interior tissue layers that are enclosed by a layer of protective bark. The cambium layer Click Click is essential for growth in plants; cambium produces phloem and xylem. Click Water and dissolved minerals and nutrients flow through the xylem. Click Click Phloem is a conducting tissue in vascular plants that carries sugars and other manufactured “food” from their source of production (usually leaves) to stems and roots that do not photosynthesize but require sugars for metabolic processes. When cambium death occurs, phloem and xylem can no longer be produced in a tree where the cambium tissue has been broken down. If bark can reduce heat transfer to the cambium layer so that its temperature does not reach 60°C, or only damage to a small area of cambium occurs, a tree will survive. Click

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Heat Transfer from a Passing Fire

Y

Fire will generally pass through an area rather quickly. Thus, plants are usually exposed to extreme heat for only a short period of time. The heat transfer mechanism of convection Click causes most of the heat from a fire to rise into the air (the heat/smoke plume). Click Less heat is transferred into the soil primarily by conduction. In most cases fires will not damage underground parts of vegetation, such as buried buds. Click Resprouting is an important adaptation for most native vegetation which allows an area to recover from a fire event. Click

“Resprouting image” from:

Stevens, Russell. “Prescribed Burning During the Growing Season.” Ignis (Newsletter of

the Oak Woodlands and Forests Fire Consortium), vol. 6, issue 4, October, 2017. (https://oakfirescience.com/download/2017-newsletters/OFS-NL-10_2017.pdf)

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Thermal Conductivity

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Thermal Conductivity

Here is an equation that represents the schematic from the previous slide. Click “Q divided by delta t” represents heat transferred over time, the rate of heat transfer. What would happen to the rate at which heat is transferred (Q/∆t) if the cross-sectional area (A) is increased? Give students a chance to respond. The rate of heat transfer would increase. Mathematically, since A is in the numerator, if it’s value increases the value of Q/∆t would also increase. At the same pressure, more water can pass through a pipe with a larger diameter. Likewise, more heat can pass through a larger cross-sectional surface. What would happen if the thickness (L) increased? Give students an opportunity to respond. Mathematically, since L is in the denominator, if its value increases the value of Q/∆T would decrease which indicates that heat would transfer more slowly. For example, ice inside of a cooler will take longer to melt if the cooler is constructed of thicker styrofoam. Finally, what would happen to the rate of heat transfer if the “hot” and “cold” temperatures were exactly the same? Give students a chance to answer. ∆T = Temp_hot-Temp_cold so ∆T would equal “0” resulting in Q/∆t = 0 which indicates that heat has no tendency to “move” since both sides of the material have the same temperature. Thus, if the temperatures are the same, there is no net transfer of heat. What happens to the rate of heat transfer as the difference in temperatures increases? Since ∆T is in the numerator, as it increases so will the rate at which heat is transferred. In this activity, the material of the simulated tree bark or soil will be the same, so the k value will not change between situations. Therefore, k is not important when making comparisons between experimental trials. Likewise, the cross-sectional area will be approximately the same for all tree trunk simulations and for all soil simulations, so “area” will not be important for comparisons. Click A simplified proportionality can be used to understand how temperatures and thickness affect the rate of heat transfer. The “∝” signifies that the equation is a proportionality. To replace the “∝” with a “=” the values of “k” and “A” would also need to be known. This proportionality can be found on the worksheet and will be helpful in answering questions. Click

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Bark & Soil Model

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Bark & Soil Model

1 hair dryer	1 temperature probe system with 2 probes
5 felt fabric sheets (9” x 12”)	1 stopwatch (smartphone)
1 piece of PVC pipe (3” diameter; 9” length)	1 ruler or “distance regulator”
3 rubber bands

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Bark & Soil Model

Model	Layers of Felt	PVC Pipe Used	Rubber Bands Used
Bark - Blackjack Oak	5	Yes	Yes
Bark - Eastern Redcedar	1	Yes	Yes
Soil - “Deep” Subsurface Vegetative Structures	5	No	No
Soil - “Shallow” Subsurface Vegetative Structures	1	No	No

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Bark & Soil Model

Model	Layers of Felt
Bark - Blackjack Oak	5
Bark - Eastern Redcedar	1

Blackjack

Oak

Eastern

Redcedar

5 Layers

1 Layer

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Bark & Soil Model

Model	Layers of Felt
Soil - “Deep” Vegetative Subsurface Structures	5
Soil - “Shallow” Vegetative Subsurface Structures	1

5 Layers

1 Layer

“Shallow”

“Deep”

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Activity

Group Members’ Tasks/Responsibilities:

Timer	Calls out the elapsed time at 10-second intervals and also calls out “Heat off!” after 30 seconds.
Heater	Controls the hair dryer. Aims the hair dryer towards the tips of the temperature probes while holding the hair dryer 15 cm from the probes/surface using a ruler or the “distance regulator.” When “Heat off!” is called, aim the hair dryer away from the probes and turn it off.
Recorder #1	Records the temperature of probe #1 at 10-second intervals.
Recorder #2	Records the temperature of probe #2 at 10-second intervals.

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Activity

To complete the activity:

The “Data Collection” table must be filled.
The “Temperature vs. Time” graph must be completed according to the guidelines on the worksheet.
All “Questions/Analysis” prompts must have responses.

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Activity

This graph could be used for class discussions.

From this activity’s lesson plan:

The steeper the slope of the temperature versus time curve/line the greater the rate of heat transfer. Whether a curve/line has a positive or negative slope is important. A positive slope indicates heat is being transferred into a material; it is heating up. A negative slope indicates that the material is cooling off as heat is transferred away. In Figure 4, it is evident that, during the first 35 seconds, the temperature probe on the surface of a tree trunk (red) heats up more quickly than the probe beneath the bark (blue) as a fire passes, and, thus, the surface probe reaches a higher maximum temperature. The surface probe has no insulating “thickness,” L, to slow the transfer of heat to it, whereas the “beneath” probe does. Since the rate of heat transfer is inversely proportional to L (Equation 3), the greater the L, the slower the transfer of heat will be. A “cambium death” temperature of 60°C has been included in Figure 4. The “Figure 4: Tree Bark Example” shows that temperature of the “Bark Surface” probe surpasses the “cambium death” temperature. The “Beneath Bark” probe, where the cambium layer is located in a tree, does not reach 60°C, and, thus, the cambium layer would not have been damaged.

Another interesting trend to note is the rate at which the probes lose heat to their surroundings or cool off. In Figure 4, cooling off (negative slope) for both temperature probes begins after 35 seconds and continues until data collection ends. Since the probe beneath the bark is surrounded by insulation, it cools down slowly; its negative slope is less than that of the exposed probe on the tree trunk’s surface. The rate of “cooling off” is also influenced by temperature difference, ∆T (temp_hot - temp_cold), as shown in Equation 3. After the fire passes, the surface probe’s temperature is much higher than air temperature resulting in a high ∆T so the rate of heat transfer is fast resulting in a steep slope. The temperature of the probe beneath the bark has a temperature very close to that of the bark, so the rate of heat transfer during cooling is slow because ∆T is small.