Protect your lawn

Classical Mechanics Level 5

Bob wanted to have a campfire in the middle of his lawn, but he needed to protect the grass. So he obtained a thick steel plate, 6 feet by 6 feet in area, and placed 5 bricks on the ground to support it at its four corners and in the center. Then he made the fire on the steel plate. The next morning, he found most of the grass under the fire dried out due to the heat.

Upon learning the outcome, Alice said to Bob, "You could have reduced the heat flow a lot, if only you had placed a thin sheet of stainless steel on the first five bricks, put another five bricks on it, and then placed the steel plate on top of that. This way, the stainless steel sheet would have been half-way between the grass and the plate that held the fire."

What is the ratio of the heat flow to the ground in Alice's fire relative to Bob's fire? Give your answer in percentage.

Assumptions:

The dominant mode of heat transfer is radiation.
Neglect the heat loss at the edges.
The temperature of the steel plate is $600\, ^\circ\text{C},$ and the temperature of the ground is $15\, ^\circ\text{C}.$
The steel plate and the ground can be treated as black bodies.
The emissivity of the stainless steel sheet is $\epsilon=0.1,$ i.e. it reflects 90% of the radiation.

The answer is 6.125.

1 solution

Laszlo Mihaly
Sep 4, 2018

Let us denote the temperatures of the thick plate, the thin sheet and the ground by $T_1$ , $T_2$ and $T_3$ , respectively. According to the Stefan–Boltzmann law, the heat flow radiated by the thick plate is $Q_1=A\sigma T_1^4$ , where $A$ is the area. If there is nothing between the ground and the plate, the heat flow towards the ground is $Q_1$ .

The heat flow radiated by the ground is $A\sigma T_3^4$ , upwards. The thin sheet reflects and radiates in both directions. The reflected radiation in the upwards direction is $(1-\epsilon)Q_1=(1-\epsilon)A\sigma T_1^4$ . In the downwards direction it is $A(1-\epsilon)\sigma T_3^4$ . The emitted radiation in the upwards direction is $A\epsilon\sigma T_2^4$ ; the same amount is emitted downwards. The total energy flow downwards is $Q_2=A(1-\epsilon)\sigma T_3^4+A\epsilon\sigma T_2^4$ . We are interested in the $Q_2/Q_1$ ratio.

The condition for thermal equilibrium for the thin sheet is that the total incident energy flow is equal to the outgoing energy flow:

$A\sigma T_1^4+A\sigma T_3^4=A (1-\epsilon)\sigma T_1^4+A(1-\epsilon)\sigma T_3^4+2A\epsilon\sigma T_2^4$

This yields

$T_2^4=\frac{T_1^4+T_3^4}{2}$

Notice that this number is independent of the emissivity. With the numbers in the problem $T_2=736K=463^{\circ}$ C. The ratio of the heat flows is

$\frac{Q_2}{Q_1}=\frac{A(1-\epsilon)\sigma T_3^4+A\epsilon\sigma T_2^4}{A\sigma T_1^4}= \frac{\epsilon}{2}+ \left(1-\frac{\epsilon}{2}\right)\left(\frac{T_3}{T_1}\right)^4= 0.06125$

The answer is 6.125%.

Comments:

AT 600 $^{\circ}$ C the steel plate is already loosing strength, but if it is thick enough it will support the weight of the fire. The temperature of the thin sheet, $463^{\circ}$ C is so high that aluminum would not work.
When the thin sheet is inserted, the temperature of the thick plate will probably go up and that will increase the energy flow from it. On the other hand, the air circulation around the thin sheet will cause a reduction of its temperature.
I actually tried this with a real campfire, and the grass was just fine (I watered the grass before, and the evaporation also helped to keep it cool).

Protect your lawn

The answer is 6.125.

1 solution

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