Solve Thermal Rad. Homework: Box Painted Black, 500 W Light

In summary: I could plug that in and solve for ε.In summary, at 500W, the box gives off 1990 J of energy and the air inside the box reaches equilibrium at 1.05 K.
  • #1
Vector1962
61
8

Homework Statement


A wooden box is painted completely with black paint. The interior dimensions are 18”x24”x3.5”. The bottom and sides are made with 9 pieces of 2x4 douglas-fir-larch dimension wood lumber. The top of the box is open and is covered with 0.118” thick clear acrylic. The acrylic is sealed to the box and only allows light to enter the box. The air surrounding the box has negligible velocity.

The box and the ambient air temperature are 47 degrees F. At time, t = 0, t a 500 W light bulb that is suspended 6 feet above the box is turned on. Thermal equilibrium of the air and box occurs 10 minutes later. What is the temperature of the air inside the box at this time?

Black Paint ε = 0.80
500 W light ε = 0.95
T light = 3000 K

CWood 1700 J/kg*K
CAir 1009 J/kg*K

ρ air = 1.225 kg/m3
ρ wood = 530 kg/m3

σ = 5.670373E -8 J/S*m2*K4

Homework Equations



E = A ε σ T4
Q = m c ΔT

The Attempt at a Solution


To heat the Air:
Vair = 18x24*3.5 in3 = 0.02477 m3
ma = ρ airVair = 0.0304 kg
Qair = ma cair ΔT

To heat the Wood:
Vwood = 3.5x1.5x 24 in3 = 0.0021 m3
mw = ρ woodVwood = 1.094 kg
Qwood = mw cwood ΔT

To heat the wood/air:
Qtotal = ma cair ΔT + mw cwood ΔT

Energy from 500 W light:
Distribute the 500 W over a sphere with a 6 ft (1.8288 m) radius, A = 4π r2 = 42.028 m2 500 W/ 42.028 m2 = 11.9 w/m2 (at the box from the light)

The box is 18x24 in 2 (0.2787 m2) so it receives 3.317 W on the surface at that distance. 3.317 W applied for 10 minutes (600 seconds) = 1990 J applied to the air and box.

1990 J = ma cair ΔT + mw cwood ΔT

solve for ΔT = 1.05 K ?
 
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  • #2
What processes are going on once thermal equilibrium is reached?
 
  • #3
Just intuitively, the box is giving off as much energy as it is taking in because at that point the temp quits changing so the air and the box are in equilibrium with each other, however, there is a ΔT between the box and surrounding air so convection (minimal) and radiation (thermal) are going on. I'm more interested in the process I applied, more specifically, is it correct to distribute the 500W to the "inside surface area" of the sphere as I have done?
 
  • #4
Vector1962 said:
the box is giving off as much energy as it is taking in because at that point the temp quits changing so the air and the box are in equilibrium with each other, however, there is a ΔT between the box and surrounding air so convection (minimal) and radiation (thermal) are going on.
Right. So I would think this is what you need to analyse in order to arrive at the equilibrium temperature.
Vector1962 said:
is it correct to distribute the 500W to the "inside surface area" of the sphere
Yes, and I agree with the 3.3W. (It is only an approximation, though. To get it exact you would need to compute the solid angle the rectangle subtends at the centre of the sphere, which is quite tricky.)

However, you cannot simply multiply that by 600 seconds to find the added energy. As you note, it will be losing energy increasingly as the temperature rises.
If you assume the only losses are radiative, and that the principal radiating surfaces are all at the internal temperature (bit of a stretch) can you figure out the effective emissivity?
 
  • #5
haruspex said:
To get it exact you would need to compute the solid angle the rectangle subtends at the centre of the sphere, which is quite tricky.)
For now, the simpler approach is better. If necessary, I can refine once I get a better understanding of the procedure.
haruspex said:
If you assume the only losses are radiative, and that the principal radiating surfaces are all at the internal temperature (bit of a stretch) can you figure out the effective emissivity?
I'm not sure I can answer your question. While at steady state, air in the box and the box are at the same temperature and this "system" is receiving 3.3W hence, it is giving off 3.3 W otherwise the temp would go up. If I knew the final temp I can get ε. I know A so I could get ε using E = A ε σ T4 . BUT, I don't know Tf.
I could use the ΔT from my "first pass", redo the calculation, and see if Tf converges? I'm afraid I'm guessing because I don't see a direct route.
3.3 =A ε σ T4 has 2 unknowns.
 
  • #6
Vector1962 said:
3.3 =A ε σ T4 has 2 unknowns.
Sorry, I forgot - you are given the emissivities. So you just need to get the equilibrium equation right. Don't forget that without the light it was in equilibrium at Ti, and the light is an additional input of power.
 
  • #7
At the beginning of the problem, the box is in equilibrium with the surroundings all at a temperature of 47 degrees F (281.483 K). Which means, again, it is giving up as much as it is taking in. ΔE = Ef- Ei = A ε σ (Tf4 -Ti4) = 3.317 W → Tf = 284.21 K (51.9° F). ?
 
  • #8
Vector1962 said:
At the beginning of the problem, the box is in equilibrium with the surroundings all at a temperature of 47 degrees F (281.483 K). Which means, again, it is giving up as much as it is taking in. ΔE = Ef- Ei = A ε σ (Tf4 -Ti4) = 3.317 W → Tf = 284.21 K (51.9° F). ?
That's the right formula. Haven't checked your numbers... what did you get for A?
I note you are given ε=0.95 for the light, i.e. you do not get the full 500W.

In reality, the outsides of the box would not be at the same temperature as the inside. There would be a temperature gradient through the wood. Were you given a conductivity for the wood?
 
  • #9
haruspex said:
That's the right formula. Haven't checked your numbers... what did you get for A?
I used the area of the box, A = 0.2787 m2 (18"x24") in the equation to arrive at Tf = 284.21 K.

Now I have another question... shoot... If the box is radiating and absorbing at the same rate AND some it is lost to heat in conduction, say H = h A ΔT then I need to subtract H from the 3.317W being delivered?:

3.317W - h A (Tf - TI) = A ε σ (Tf4 -Ti4) (in this case the area conducting = area radiating).
 
  • #10
Vector1962 said:
AND some it is lost to heat in conduction,
I wasn't sure what you were calculating as A. I see that you only used the area of the acrylic, so ignored losses through the wood.
That's probably ok.
To allow for conduction is a bit tricky because you have to set the outside temperature of the box to some unknown, Tb, and match the conduction through the box over that temperature gradient to the radiation from the outside of the box at that temperature. That involves solving a quartic.
 
  • #11
Understood.

I think for now I'm good with just considering the radiation portion. I'm sure this discussion, which I enjoyed and appreciated your leadership in, can be further refined and tweaked to a more precise answer, but for now, I think I have something close that I can work with comfortably.

Here's the rub. I built this dumb box and tested it. Don't ask why, I just did... I was getting between 4 and 5 degrees F change and was really having trouble finding how to calculate it. I appreciate the help getting to the equation in post #7 (this thread) and I did have a good laugh with the result of that calculation, 47 to 51.9, so the measured temps seem to be in reasonable line with that calculation. Who knows, if time allows, I'll make another one of different size and repeat the that calculation to confirm.
 
  • #12
Vector1962 said:
Understood.

I think for now I'm good with just considering the radiation portion. I'm sure this discussion, which I enjoyed and appreciated your leadership in, can be further refined and tweaked to a more precise answer, but for now, I think I have something close that I can work with comfortably.

Here's the rub. I built this dumb box and tested it. Don't ask why, I just did... I was getting between 4 and 5 degrees F change and was really having trouble finding how to calculate it. I appreciate the help getting to the equation in post #7 (this thread) and I did have a good laugh with the result of that calculation, 47 to 51.9, so the measured temps seem to be in reasonable line with that calculation. Who knows, if time allows, I'll make another one of different size and repeat the that calculation to confirm.
Congratulations on getting a result in such good agreement between theory and practice.
 

FAQ: Solve Thermal Rad. Homework: Box Painted Black, 500 W Light

What is thermal radiation?

Thermal radiation is a type of electromagnetic radiation that is emitted by objects due to their temperature. It is also known as heat radiation and is one of the ways that heat is transferred between objects.

Why is the box painted black?

The box is painted black because black is the best color for absorbing thermal radiation. This means that the box will absorb more of the 500 W light that is shining on it, increasing its temperature and causing it to emit its own thermal radiation.

How does the 500 W light affect the box?

The 500 W light affects the box by transferring energy to it in the form of thermal radiation. The box absorbs some of this energy and heats up, causing it to emit its own thermal radiation. This process continues until the box reaches thermal equilibrium with its surroundings.

What factors affect the amount of thermal radiation emitted by the box?

The amount of thermal radiation emitted by the box is affected by several factors, including its temperature, surface area, and emissivity (a measure of how well it emits thermal radiation). In this scenario, the box's temperature and surface area are fixed, so the only variable is its emissivity.

How can we calculate the amount of thermal radiation emitted by the box?

The amount of thermal radiation emitted by the box can be calculated using the Stefan-Boltzmann law, which states that the power emitted by a blackbody (such as the box in this scenario) is directly proportional to the fourth power of its temperature and its surface area. In this scenario, the power emitted by the box can be calculated using the formula P = σA(T^4), where σ is the Stefan-Boltzmann constant, A is the surface area of the box, and T is its temperature in Kelvin.

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