Why optical pyrometer gives a low reading?

[vcsharp2003]

Homework Statement

For the question as in screenshot, explain why the optical pyrometer gives a too low reading.

Relevant Equations

$\frac {\Delta Q} {\Delta t} = \epsilon \sigma A T^4$ where $\epsilon$ is emissivity of the body at temperature $T$ having a surface area of A and $\sigma$ is a constant for all bodies called Stefan Boltzmann constant..

From my knowledge, the optical pyrometer determines temperature of a body based on the visible part of the heat radiation spectrum emitted by a body. Now the iron piece was red hot inside the furnace as well as outside the furnace, so it is emitting heat radiation in the visible part both while in and out of the furnace. My only guess is that when red hot iron piece is outside, the heat radiation in the visible part of light becomes very small as compared to when it was inside the furnace, and so the optical pyrometer senses a too low temperature of the iron piece outside the furnace.

If above explanation is correct then I still don't understand why the visible part of the heat radiation spectrum becomes very less when iron body is placed outside the furnace, even though it's still red hot outside.

 

[jbriggs444]

Why is it important that the pyrometer be calibrated for an ideal black body? Can you explain what a "black body" is?

No, the emitted glow from the iron is not dramatically reduced due to removal from the furnace. The answer that I have in mind has nothing to do with rapid surface cooling or any other magical reduction in the glow as the iron is removed from the furnace.

 

[vcsharp2003]

Why is it important that the pyrometer be calibrated for an ideal black body?

I don't know the reason for this, but it seems if it's calibrated for black body then using emissivity of a regular body one can determine the amount of heat radiation and therefore it's temperature.

Can you explain what a "black body" is?

A black body absorbs all incident radiation and also emits the same amount of absorbed radiation. A normal body will not absorb all incident radiation and therefore it will also emit less radiation compared to a black body at the same temperature.

 

[jbriggs444]

A black body absorbs all incident radiation and also emits the same amount of absorbed radiation. A normal body will not absorb all incident radiation and therefore it will also emit less radiation compared to a black body at the same temperature.

What happens to the incident radiation that is not absorbed?

 

[vcsharp2003]

What happens to the incident radiation that is not absorbed?

It is reflected.

 

[jbriggs444]

It is reflected.

For instance, what happens to the incident radiation onto a piece of iron in a furnace?

 

[vcsharp2003]

For instance, what happens to the incident radiation onto a piece of iron in a furnace?

My guess is that most of it is absorbed since inside the furnace the iron object is closer to acting like a black body.

 

[jbriggs444]

My guess is that most of it is absorbed since inside the furnace the iron object is closer to acting like a black body.

Most. But not all. So what about the portion that is not absorbed? What happens to that portion?

 

[vcsharp2003]

Most. But not all. So what about the portion that is not absorbed? What happens to that portion?

That's a good question. That small part which is not absorbed is reflected back into the furnace space. Probably, the red part of incident light is reflected, which makes the iron appear red hot.

 

[jbriggs444]

That's a good question. That small part which is not absorbed is reflected back into the furnace space. Probably, the red part of incident light is reflected.

Good. Now take it the rest of the way. When the optical pyrometer looks at the iron in the furnace, what will it see?

 

[vcsharp2003]

Good. Now take it the rest of the way. When the optical pyrometer looks at the iron in the furnace, what will it see?

Red part of the light. But I think the emitted radiation is also detected by the pyrometer, which means it sense both the red part plus the emitted visible part of radiation.

 

[jbriggs444]

Red part of the light.

That is not what I was aiming at.

Yes, it is looking (we assume) for the intensity of the red portion of the spectrum. And it will see red light coming from the surface of the iron sample. Some of that red light will be emitted. Some of that red light will be reflected. The total measured by the pyrometer will be the sum of the two, right?

 

[vcsharp2003]

That is not what I was aiming at.

Yes, it is looking (we assume) for the intensity of the red portion of the spectrum. And it will see red light coming from the surface of the iron sample. Some of that red light will be emitted. Some of that red light will be reflected. The total measured by the pyrometer will be the sum of the two, right?

Yes. I added that to my last message.

If this is true inside the furnace then outside the furnace the same object will emit less red part. Right?

 

[jbriggs444]

Yes. I added that to my last message.

I am not sure that I understand that last part:

But I think the emitted radiation is also detected by the pyrometer, which means it sense both the red part plus the emitted visible part of radiation.

 

[vcsharp2003]

I am not sure that I understand that last part:

I meant there is going to be some visible (i.e. VIBGYOR) wavelengths in the emitted radiation including red light, over and above the reflected red light.

 

[jbriggs444]

I meant there is going to be some visible (i.e. VIBGYOR) wavelengths in the emitted radiation including red light, over and above the reflected red light.

OK. I could not decide whether you were trying to distinguish between red light and visible light or between emitted light and reflected light.

In any case, we agree that what comes off of the hunk of iron in the furnace is a mix between reflected and emitted light.

What happens when the hunk of iron is removed from the furnace? What does the pyrometer see now? You did turn off the light switch, right?

 

[vcsharp2003]

What happens when the hunk of iron is removed from the furnace? What does the pyrometer see now?

There should be less red light incident when it's outside, since the fuel in the furnace was emitting red radiation. This results in less total red light being detected by the pyrometer when it's outside.

 

[Frabjous]

@jbriggs444
Wouldn’t the oven measurement overestimate T, because the walls are probably hotter than the bar? Are we just looking to measure where we are minimizing ΔT between item and environment?
Do you know the magnitude of this effect at 1000C or some other convenient high temperature?
Would focusing optics minimize the effect? Is there a better way?
Is there something subtle about the detector being blackbody (not greybody) calibrated?

 

[jbriggs444]

@jbriggs444
Wouldn’t the oven measurement overestimate T, because the walls are probably hotter than the bar?

Possible. I am an ivory tower academic. My assumption is that an oven will be set for the desired equilibrium temperature. But perhaps they are set higher than that.

Are we just looking to measure where we are minimizing ΔT between item and environment?

Ideally, I would think the goal is to measure the temperature of something that is not at equilibrium with its environment. So one needs to beware of the confounding fact of the target's illumination and reflectivity in the relevant frequency band(s). Not all objects are ideally black.

Do you know the magnitude of this effect at 1000C or some other convenient high temperature?

Nope. Just an ivory tower academic here.

Would focusing optics minimize the effect? Is there a better way?

I do not see how focusing could help.

Is there something subtle about the detector being blackbody (not greybody) calibrated?

Not subtle, no. But if your body is strongly reflective in a particular band at a certain temperature, it follows by the second law of thermodynamics that it must be poorly emissive in the same band. If the detector is sampling in such a band (strongly reflective, poorly emissive) and is calibrated for an ideal black body then it will indicate the wrong temperature -- even after you turn out the lights.

The meter will expect a strongly emissive black body. It will see a weakly emissive colored body and misjudge the temperature as a result.

Take this explanation with a certain amount of caution. I am not a subject matter expert. I do not know how optical pyrometers are tuned to mitigate this problem.

 

[et_al]

Red-hot iron is not actually an ideal blackbody. For this non-ideal blackbody, the monochromatic emissivity $\varepsilon_\lambda$ is defined as the ratio of the monochromatic emissive power of the real surface to the monochromatic emissive power of a black body. It is less than one. If the temperature of the object in the furnace is observed from the small hole in the furnace door, then the model is very close to the black body, so the measured temperature is accurate