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Chapter 12: Problem 58

A radiometer can be used to determine the position of an approaching hot object by measuring the amount of irradiation it detects. Consider a radiometer placed at a distance \(H=0.5 \mathrm{~m}\) from the \(x\)-axis is used to measure the position of an approaching small blackbody object. If the radiometer is measuring \(80 \%\) of the irradiation corresponding to the position of the object directly under the radiometer \((x=0)\), determine the position of the object \(L\).

Short Answer

Expert verified
Answer: The position of the hot object along the \(x\)-axis is approximately \(0.25\) meters.

Step by step solution

01

Relate irradiation to the position of the object

We note that the fraction of irradiation detected by the radiometer when the object is at position \(x=L\) is directly proportional to the inverse square of the distance between the object and the radiometer. Let the fraction of irradiation be denoted by \(f\), which is given by: \[f = \frac{1}{(H^2 + L^2)}\]
02

Use given information to find a relationship with \(L\)

The problem statement says that the radiometer is measuring \(80 \%\) of the irradiation corresponding to the position of the object directly under the radiometer \((x=0)\). The fraction when the object is directly under the radiometer is \(f_{0} = \frac{1}{H^2}\). So we have the equation: \[0.8 f_{0} = \frac{1}{(H^2 + L^2)}\]
03

Solve the equation for \(L\)

We need to solve the equation from Step 2 for \(L\). Rewriting the equation, we get: \[\frac{1}{(H^2 + L^2)} = \frac{0.8}{H^2}\] Now, we will cross-multiply and then rearrange the equation to isolate \(L^2\): \[(H^2 + L^2) = \frac{H^2}{0.8}\] Subtract \(H^2\) from both sides: \[L^2 = \frac{H^2}{0.8} - H^2\] Finally, take the square root of both sides to get the position \(L\): \[L = \sqrt{\frac{H^2}{0.8}-H^2}\]
04

Calculate \(L\) using the given value of \(H\)

Now that we have the formula for \(L\), we can plug in the given value for the distance \(H=0.5\) m: \[L = \sqrt{\frac{(0.5)^2}{0.8}-(0.5)^2}\] Calculate the value of \(L\): \[L \approx 0.25 \mathrm{~m}\] Hence, the position of the object \(L\) is approximately \(0.25\) meters from the radiometer.

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Most popular questions from this chapter

The electromagnetic spectrum that lies between \(0.40\) and \(0.76 \mu \mathrm{m}\) is what we call visible light. Within this spectrum, the color violet has the shortest wavelength while the color red has the longest wavelength. Determine which of these colors, violet \((\lambda=0.40 \mu \mathrm{m})\) or red \((\lambda=0.76 \mu \mathrm{m})\), propagates more photon energy.

A long metal bar \(\left(c_{p}=450 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}, \rho=\right.\) \(7900 \mathrm{~kg} / \mathrm{m}^{3}\) ) is being conveyed through a water bath to be quenched. The metal bar has a cross section of \(30 \mathrm{~mm} \times 15 \mathrm{~mm}\), and it enters the water bath at \(700^{\circ} \mathrm{C}\). During the quenching process, \(500 \mathrm{~kW}\) of heat is released from the bar in the water bath. In order to prevent thermal burn on people handling the metal bar, it must exit the water bath at a temperature below \(45^{\circ} \mathrm{C}\). A radiometer is placed normal to and at a distance of \(1 \mathrm{~m}\) from the bar to monitor the exit temperature. The radiometer receives radiation from a target area of \(1 \mathrm{~cm}^{2}\) of the bar surface. Irradiation signal detected by the radiometer is used to control the speed of the bar being conveyed through the water bath so that the exit temperature is safe for handing. If the radiometer detects an irradiation of \(0.015 \mathrm{~W} / \mathrm{m}^{2}\), determine the speed of the bar bar can be approximated as a blackbody.

Define the properties emissivity and absorptivity. When are these two properties equal to each other?

The variation of the spectral transmissivity of a \(0.6\)-cm-thick glass window is as given in Fig. P12-80. Determine the average transmissivity of this window for solar radiation \((T \approx 5800 \mathrm{~K})\) and radiation coming from surfaces at room temperature \((T \approx 300 \mathrm{~K})\). Also, determine the amount of solar radiation transmitted through the window for incident solar radiation of \(650 \mathrm{~W} / \mathrm{m}^{2}\).

A surface is exposed to solar radiation. The direct and diffuse components of solar radiation are 350 and \(250 \mathrm{~W} / \mathrm{m}^{2}\), and the direct radiation makes a \(35^{\circ}\) angle with the normal of the surface. The solar absorptivity and the emissivity of the surface are \(0.24\) and \(0.41\), respectively. If the surface is observed to be at \(315 \mathrm{~K}\) and the effective sky temperature is \(256 \mathrm{~K}\), the net rate of radiation heat transfer to the surface is (a) \(-129 \mathrm{~W} / \mathrm{m}^{2}\) (b) \(-44 \mathrm{~W} / \mathrm{m}^{2}\) (c) \(0 \mathrm{~W} / \mathrm{m}^{2}\) (d) \(129 \mathrm{~W} / \mathrm{m}^{2}\) (e) \(537 \mathrm{~W} / \mathrm{m}^{2}\)
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