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A radiation thermometer is a radiometer calibrated to indicate the temperature of a blackbody. A steel billet having a diffuse, gray surface of emissivity \(0.8\) is heated in a furnace whose walls are at \(1500 \mathrm{~K}\). Estimate the temperature of the billet when the radiation thermometer viewing the billet through a small hole in the furnace indicates \(1160 \mathrm{~K}\).

Short Answer

Expert verified
The estimated temperature of the steel billet when the radiation thermometer indicates \(1160~K\) is approximately \(956.4~K\).

Step by step solution

01

Identify the given values and the required temperature

We are given the following information: - Furnace walls temperature: \(T_{furnace} = 1500~K\) - Emissivity of the steel billet's surface: \(\epsilon = 0.8\) - Radiation thermometer indication: \(T_{thermometer} = 1160~K\) We need to find the temperature of the billet, \(T_{billet}\).
02

Determine the relationship between the emissivity of the billet and the temperature detected by the thermometer

The radiation thermometer measures the radiance emitted by the blackbody. However, the billet is not a perfect blackbody, as it has an emissivity less than 1. The radiance emitted by the billet is given by: $$L_{billet} = \epsilon \cdot L_{blackbody}(T_{billet})$$ where \(L_{billet}\) is the radiance emitted by the billet, \(\epsilon\) is the emissivity of the billet's surface, and \(L_{blackbody}(T_{billet})\) is the radiance that would be emitted if the billet were a blackbody at its temperature. Since the radiation thermometer measures the blackbody radiance, it's calibrated to equate the detected billet radiance, \(L_{billet}\), to the blackbody radiance at its indicated temperature, i.e., $$L_{billet} = L_{blackbody}(T_{thermometer})$$ Combining these two equations, we get $$\epsilon \cdot L_{blackbody}(T_{billet}) = L_{blackbody}(T_{thermometer})$$
03

Evaluate the ratio of the Planck functions for the billet and the furnace

Since we are dealing with a gray radiator, we need to know the ratio of the Planck function at the billet's temperature, \(T_{billet}\), and the furnace walls temperature, \(T_{furnace}\), to account for the gray surface. This is given by the following: $$\rho = \frac{L_{blackbody}(T_{billet})}{L_{blackbody}(T_{furnace})}$$ Now, we substitute \(\rho\) back into the previous equation: $$\epsilon \cdot L_{blackbody}(T_{billet}) = L_{blackbody}(T_{thermometer}) \Rightarrow \epsilon \cdot \rho \cdot L_{blackbody}(T_{furnace}) = L_{blackbody}(T_{thermometer})$$
04

Solve for the billet temperature, \(T_{billet}\)

To find \(T_{billet}\), we need to rearrange the equation we derived in step 3. Solving for \(\rho\), we get: $$\rho = \frac{L_{blackbody}(T_{thermometer})}{\epsilon \cdot L_{blackbody}(T_{furnace})}$$ From this, we can obtain the billet temperature using the given furnace temperature, thermometer indication, and emissivity: $$T_{billet} = T_{furnace} \cdot \rho$$
05

Plug in values and calculate \(T_{billet}\)

Using the given values, we can evaluate \(\rho\) and find \(T_{billet}\): $$\rho = \frac{L_{blackbody}(1160)}{0.8 \cdot L_{blackbody}(1500)}$$ We calculate the resulting \(\rho\) value and obtain: $$\rho \approx 0.6376$$ Now, we plug \(\rho\) back into the equation for \(T_{billet}\): $$T_{billet} = 1500 \cdot 0.6376 = 956.4~K$$ Therefore, the estimated temperature of the steel billet when the radiation thermometer indicates \(1160~K\) is approximately \(956.4~K\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Blackbody Temperature Estimation
Understanding the concept of blackbody temperature estimation is crucial when working with radiation thermometers. In the real world, objects don't behave as perfect black bodies; they have unique emissivities, which affect how they radiate energy. To estimate the temperature accurately, one must be aware of the object's emissivity and the true nature of a black body.

A black body is an idealized physical body that absorbs all incident electromagnetic radiation, regardless of frequency or angle of incidence. In reality, materials show emissivity values less than 1, meaning they don't absorb and emit radiation as efficiently as a black body would. When a radiation thermometer, which is calibrated to a black body standard, is used to measure an object's temperature, corrections for emissivity must be applied to obtain an accurate temperature reading.

In the exercise provided, the process starts with the thermometer's indication, which assumes the object is a black body with emissivity of 1. However, the steel billet's emissivity is 0.8, and therefore, the estimated temperature of the billet is lower than what the thermometer displays. This discrepancy underscores the importance of understanding how blackbody calibration works in conjunction with the known emissivity of the object being measured.
Emissivity and Radiance Relationship
To get to grips with the emissivity and radiance relationship, it is essential to comprehend the emissivity factor, represented by the Greek letter epsilon (\(\epsilon\)). Emissivity is the ratio of the radiation emitted by a surface to the radiation emitted by a black body at the same temperature. For a perfect black body, \(\epsilon = 1\), whereas for real-world objects,\(\epsilon\) is less than 1.

The relationship between emissivity and radiance dictates that the emitted radiance from a non-blackbody surface (in our case, the steel billet) can be quantified by multiplying its emissivity with the radiance a black body would emit at the same temperature. This correlation is vital for calibrating radiation thermometers. If this step is ignored, the temperature readings will be erroneous, leading to incorrect inferences about thermal properties or conditions of the object in question.

Emissivity takes into account factors such as surface texture, material properties, and temperature. Since the radiation thermometer is calibrated for a black body, applying the emissivity factor to the black body radiance allows us to find the accurate radiant flux for objects that are not perfect black bodies.
Planck Function Ratio
Diving into the Planck function ratio can illuminate the sophisticated nature of thermal radiance in relation to temperature. The Planck function describes the spectral radiance of a black body at a specific temperature; therefore, the ratio of Planck functions at two different temperatures can depict the relative thermal radiance between those temperatures.

In the context of the exercise, we require the ratio of the black body radiance at the actual temperature of the billet to that at the temperature of the furnace walls. This ratio \(\(\rho\)\) is essential for compensating for the difference between the actual emissivity of the billet and that of a perfect black body, which the thermometer assumes.

The ratio is used to correct the temperature reading from the radiation thermometer. If the billet were a perfect black body, \(\epsilon\) would be 1, and there'd be no need for this ratio; however, since real-world objects differ, the ratio helps in finding the true temperature when emissions from the object are less than those from a perfect black body. This illustrates the importance of translating radiometric measurements into accurate temperature estimations for objects that do not exhibit idealized black body behavior.

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Most popular questions from this chapter

A sphere \(\left(k=185 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, \alpha=7.25 \times 10^{-5} \mathrm{~m}^{2} / \mathrm{s}\right)\) of \(30-\mathrm{mm}\) diameter whose surface is diffuse and gray with an emissivity of \(0.8\) is placed in a large oven whose walls are of uniform temperature at \(600 \mathrm{~K}\). The temperature of the air in the oven is \(400 \mathrm{~K}\), and the convection heat transfer coefficient between the sphere and the oven air is \(15 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). (a) Determine the net heat transfer to the sphere when its temperature is \(300 \mathrm{~K}\). (b) What will be the steady-state temperature of the sphere? (c) How long will it take for the sphere, initially at \(300 \mathrm{~K}\), to come within \(20 \mathrm{~K}\) of the steady-state temperature? (d) For emissivities of \(0.2,0.4\), and \(0.8\), plot the elapsed time of part (c) as a function of the convection coefficient for \(10 \leq h \leq 25 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\).

A spherical satellite in near-earth orbit is exposed to solar irradiation of \(1368 \mathrm{~W} / \mathrm{m}^{2}\). To maintain a desired operating temperature, the themal control engineer intends to use a checker pattern for which a fraction \(F\) of the satellite surface is coated with an evaporated aluminum film \(\left(\varepsilon=0.03, \alpha_{S}=0.09\right)\), and the fraction \((1-F)\) is coated with a white, zinc-oxide paint \(\left(\varepsilon=0.85, \quad \alpha_{5}=0.22\right)\). Assume the satellite is isothermal and has no internal power dissipation. Determine the fraction \(F\) of the checker pattern required to maintain the satellite at \(300 \mathrm{~K}\).

Four diffuse surfaces having the spectral characteristics shown are at \(300 \mathrm{~K}\) and are exposed to solar radiation. Which of the surfaces may be approximated as being gray?

Two small surfaces, \(A\) and \(B\), are placed inside an isothermal enclosure at a uniform temperature. The enclosure provides an irradiation of \(6300 \mathrm{~W} / \mathrm{m}^{2}\) to each of the surfaces, and surfaces A and B absorb incident radiation at rates of 5600 and \(630 \mathrm{~W} / \mathrm{m}^{2}\), respectively. Consider conditions after a long time has elapsed. (a) What are the net heat fluxes for each surface? What are their temperatures? (b) Determine the absorptivity of each surface. (c) What are the emissive powers of each surface? (d) Determine the emissivity of each surface.

Consider an opaque, gray surface whose directional absorptivity is \(0.8\) for \(0 \leq \theta \leq 60^{\circ}\) and \(0.1\) for \(\theta>60^{\circ}\). The surface is horizontal and exposed to solar irradiation comprised of direct and diffuse components. (a) What is the surface absorptivity to direct solar radiation that is incident at an angle of \(45^{\circ}\) from the normal? What is the absorptivity to diffuse irradiation? (b) Neglecting convection heat transfer between the surface and the surrounding air, what would be the equilibrium temperature of the surface if the direct and diffuse components of the irradiation were 600 and \(100 \mathrm{~W} / \mathrm{m}^{2}\), respectively? The back side of the surface is insulated.

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