What is the Correct Way to Calculate Emissivity and Absorptivity?

In summary, the problem involves an impenetrable flat horizontal plate that is radiated with 3000 W/m^2, where 500 W/m^2 is reflected. The surface temperature of the plate is 200 °C and it emits 500 W/m^2. The air flowing over the plate has a temperature of 25 °C and the heat transfer coefficient is 20 W/m^2*K. The goal is to calculate the emissivity and absorptivity of the plate. However, the approach taken by the student, which involves applying Kirchoff's law and assuming a black body, is incorrect. This is because the power per area radiated by an object is dependent on its temperature and emissivity,
  • #1
Kqwert
160
3

Homework Statement


An impenetrable flat horizontal plate is radiated with 3000 W/m^2, where 500 W/m^2 is reflected. The surface temperature of the plate is 200 °C and it emits 500 W/m^2. Air with temperature 25 °C flows over the plate, and the heat transfer coefficient due to this is 20 W/m^2*K.

Calculate the (i) emissivity, (ii) absorptivity

Homework Equations

The Attempt at a Solution


(i) A black body absorbs all incoming radiation, and all absorbed radiation is emitted. Therefore, I would say that a black plate would emit 3000 W/m^2, and the emissivity of my plate would be 500/3000 = 0.166. This is wrong, could anyone please explain me what is wrong?(ii) Using Kirchoff's law I would assume that emissivity = absorptivity, but this is also wrong according to the solutions manual. Could anyone correct my thinking?
 
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  • #2
Kqwert said:

Homework Statement


An impenetrable flat horizontal plate is radiated with 3000 W/m^2, where 500 W/m^2 is reflected. The surface temperature of the plate is 200 °C and it emits 500 W/m^2. Air with temperature 25 °C flows over the plate, and the heat transfer coefficient due to this is 20 W/m^2*K.

Calculate the (i) emissivity, (ii) absorptivity

Homework Equations

The Attempt at a Solution


(i) A black body absorbs all incoming radiation, and all absorbed radiation is emitted. Therefore, I would say that a black plate would emit 3000 W/m^2, and the emissivity of my plate would be 500/3000 = 0.166. This is wrong, could anyone please explain me what is wrong?(ii) Using Kirchoff's law I would assume that emissivity = absorptivity, but this is also wrong according to the solutions manual. Could anyone correct my thinking?
I can see several problems in your argument.
The first is that power per area radiated by an object depends on its temperature and emissivity. Since the plate is not a blackbody you cannot apply Kirchoff's law to it.
The second is that you have not considered the heat that the plate is losing to the air around it.
The third is that you have not considered the reflected radiation.
The problem does not say that the system is in thermal equilibrium, but I think it would be difficult to determine the emissivity without knowing the rate of change of temperature. If it is in equilibrium, then you should be able to write an energy balance equation with the heat gains on one side and the heat losses on the other, and then solve for the emissivity and absorptivity. When I try to do that, I end up with more heat losses than heat gains, even without considering radiative emission, so perhaps I have made a mistake somewhere. I hope you have better luck when you try it.
 
  • #3
tnich said:
I can see several problems in your argument.
The first is that power per area radiated by an object depends on its temperature and emissivity. Since the plate is not a blackbody you cannot apply Kirchoff's law to it.
The second is that you have not considered the heat that the plate is losing to the air around it.
The third is that you have not considered the reflected radiation.
The problem does not say that the system is in thermal equilibrium, but I think it would be difficult to determine the emissivity without knowing the rate of change of temperature. If it is in equilibrium, then you should be able to write an energy balance equation with the heat gains on one side and the heat losses on the other, and then solve for the emissivity and absorptivity. When I try to do that, I end up with more heat losses than heat gains, even without considering radiative emission, so perhaps I have made a mistake somewhere. I hope you have better luck when you try it.
But doesn't Kirchoff´s law apply to non-black bodies as well? And will the heat lost through convection influence the emissivity/absorbtivity?
 
  • #4
Kqwert said:
But doesn't Kirchoff´s law apply to non-black bodies as well? And will the heat lost through convection influence the emissivity/absorbtivity?
The simple answer is no. As I understand it, Kirchoff's law says that absorptivity is equal to emissivity for a blackbody in equilibrium. For other objects, they are not necessarily equal.
Convective heat loss should not influence emissivity or absorptivity, but in this problem it should help you determine how much heat is lost by radiative emission, from which you can determine the emissivity.
Are you sure that the values of the constants are correct in the problem statement?
 
  • #5
tnich said:
m which you can determine the emissivity.
Are you sure that the values of the constants are correct in the problem statement?
Yeah, everything is correct. I am wondering if the emitted radiation from the plate is "independent" of the incoming radiation, could that be an explanation?
 
  • #6
Kqwert said:
Yeah, everything is correct. I am wondering if the emitted radiation from the plate is "independent" of the incoming radiation, could that be an explanation?
It depends only on temperature and emissivity, so yes, in that sense it is independent of incoming radiation.
 
  • #7
tnich said:
As I understand it, Kirchoff's law says that absorptivity is equal to emissivity for a blackbody in equilibrium.
My reading is that if you know the absorption coefficient of the body at a given wavelength and the body's temperature then you can determine the emission power at that wavelength. See https://en.m.wikipedia.org/wiki/Kirchhoff's_law_of_thermal_radiation.
 
  • #8
Kqwert said:
Yeah, everything is correct. I am wondering if the emitted radiation from the plate is "independent" of the incoming radiation, could that be an explanation?
In principle, the solution should consist of:
Assuming thermal equilibrium (because there is not enough info otherwise)
Calculating the losses from conduction and reflectance
Deducing from that the heat lost by radiation
Comparing that with the radiation power of a black body at the same temperature.

However, something seems to be wrong with the data. What would you calculate for the heat lost by conduction to the air?
 
  • #9
haruspex said:
My reading is that if you know the absorption coefficient of the body at a given wavelength and the body's temperature then you can determine the emission power at that wavelength. See https://en.m.wikipedia.org/wiki/Kirchhoff's_law_of_thermal_radiation.
Yes, I think you are right. Unfortunately, that will still not help in this problem since we don't know the wavelength distribution of incident radiation or how the absorptivity/emissivity varies with wavelength.
 
Last edited:

1. What is "Radiation onto a flat plate"?

"Radiation onto a flat plate" refers to the process of electromagnetic radiation being directed onto a flat surface, such as a metal plate. This can occur through natural processes, like the sun's rays hitting the Earth's surface, or through artificial means, like using a laser to direct radiation onto a plate.

2. How does radiation onto a flat plate work?

Radiation onto a flat plate works through the transfer of energy from the radiation source to the flat plate. This transfer of energy occurs through the emission of electromagnetic waves, which carry energy and can be absorbed by the plate. The amount of energy transferred depends on factors such as the intensity and wavelength of the radiation, as well as the properties of the plate itself.

3. What are the applications of radiation onto a flat plate?

The applications of radiation onto a flat plate are vast and diverse. Some common uses include heating and drying processes in industrial settings, sterilization of medical equipment, and radiation therapy for cancer treatment. It can also be used in scientific research, such as in spectroscopy to analyze the composition of materials.

4. What are the potential risks associated with radiation onto a flat plate?

The potential risks depend on the type and intensity of radiation being used. Some forms of radiation, such as ultraviolet and ionizing radiation, can be harmful to living organisms if proper precautions are not taken. It is important to follow safety protocols and use protective equipment when working with radiation onto a flat plate to minimize any potential risks.

5. How is radiation onto a flat plate measured?

Radiation onto a flat plate can be measured using devices such as radiometers, which detect and measure the intensity of electromagnetic radiation. Other methods include using thermometers to measure the temperature change on the surface of the plate or using specialized equipment to measure the amount of energy absorbed by the plate. These measurements can help scientists understand the effects of radiation onto a flat plate and how it can be utilized in various applications.

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