/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 44 A device used to measure the sur... [FREE SOLUTION] | 91Ó°ÊÓ

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A device used to measure the surface temperature of an object to within a spatial resolution of approximately \(50 \mathrm{~nm}\) is shown in the schematic. It consists of an extremely sharp-tipped stylus and an extremely small cantilever that is scanned across the surface. The probe tip is of circular cross section and is fabricated of polycrystalline silicon dioxide. The ambient temperature is measured at the pivoted end of the cantilever as \(T_{\infty}=\) \(25^{\circ} \mathrm{C}\), and the device is equipped with a sensor to measure the temperature at the upper end of the sharp tip, \(T_{\text {sen. }}\). The thermal resistance between the sensing probe and the pivoted end is \(R_{t}=5 \times 10^{6} \mathrm{~K} / \mathrm{W}\). (a) Determine the thermal resistance between the surface temperature and the sensing temperature. (b) If the sensing temperature is \(T_{\text {sen }}=28.5^{\circ} \mathrm{C}\), determine the surface temperature. Hint: Although nanoscale heat transfer effects may be important, assume that the conduction occurring in the air adjacent to the probe tip can be described by Fourier's law and the thermal conductivity found in Table A. \(4 .\)

Short Answer

Expert verified
The thermal resistance between the surface temperature and the sensing temperature, \(R_s\), can be expressed as: \(R_s = (T_{\infty} - T_{\text{sen}}) / q - R_t\) Due to the lack of information about the probe's dimensions, we cannot find the exact value of \(R_s\) and subsequently the surface temperature, \(T_s\). In practical situations, the heat transfer, \(q\), would be calculated using the probe's dimensions, and the surface temperature could be determined using the equation: \(T_s = T_{\text{sen}} + q \cdot R_s\)

Step by step solution

01

Determine thermal resistance between surface and sensing temperature

We are looking for the thermal resistance between the surface temperature and the sensing temperature, \(R_{s}\). From the given information, we know the temperature at the pivoted end of the cantilever is \(T_{\infty}\), and the temperature at the upper end of the sharp tip is \(T_{\text {sen }}\). The thermal resistance between the sensing probe and the pivoted end is given by \(R_{t}\). The heat transfer between these points can be represented by the following equation: \(q = (T_{\infty} - T_{\text {sen }}) / (R_{t} + R_{s})\) We need to find the value of \(R_{s}\) which is the thermal resistance between the surface temperature and the sensing temperature. We will do this by rearranging the above equation.
02

Rearrange the equation for \(R_s\)

We can rewrite the equation in step 1 to isolate \(R_s\) as follows: \(R_{s} = (T_{\infty} - T_{\text {sen }}) / q - R_{t}\) However, we still don't have the heat transfer, \(q\), but we are given thermal resistance between the sensing probe and the pivoted end \(R_t\) as 5×10^6 K/W.
03

Calculate the heat transfer, \(q\)

From Fourier's law and the thermal conductivity of air (0.026 W/mK), we can express the heat transfer in terms of the temperature difference and the thermal resistance as follows: \(q = k \cdot A \cdot (T_{\infty} - T_{\text {sen }}) / L\) Where \(k\) is the thermal conductivity of air, \(A\) is the cross-sectional area of the probe, and \(L\) is the length of the probe. We are not given the exact values for \(A\) and \(L\), so we cannot calculate the heat transfer \(q\) in this exercise. However, for larger variations in temperature, this equation would be used to calculate the heat transfer amount.
04

Calculate the surface temperature

We are asked to calculate the surface temperature when the sensing temperature is \(T_{\text {sen }} = 28.5 ^{\circ}C\). The heat flow in part (b) of the problem can be given by: \(q = (T_s - T_{\text {sen }}) / R_s\) Where \(T_s\) is the surface temperature. We are given \(T_{\text {sen }}\) and have calculated the thermal resistance, so we can easily find the surface temperature by rearranging the equation: \(T_s = T_{\text {sen }} + q \cdot R_s\) However, as we were unable to calculate the heat transfer \(q\) due to missing information about the probe's dimensions, we cannot find the exact surface temperature in this exercise. In practical situations, the heat transfer can be calculated using the probe's dimensions, and this equation would be used to find the surface temperature.

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Key Concepts

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

Heat Transfer
Heat transfer is a discipline of thermal engineering that concerns the generation, use, conversion, and exchange of thermal energy (heat) between physical systems. Heat transfer is classified into various mechanisms, such as thermal conduction, thermal convection, thermal radiation, and phase change. One possible scenario is the thermal analysis of a device measuring surface temperatures, as outlined in the textbook exercise. When dealing with such devices, understanding heat transfer is crucial to predict and measure how thermal energy moves through and between materials.

For instance, in the device mentioned in the exercise, the transfer of heat occurs from the surface to the sharp-tipped stylus, and then to the cantilever where the ambient temperature is measured. Thermal resistance plays a key role here; it quantifies the temperature difference across a material when a certain amount of heat is passing through it. Lower thermal resistance implies better heat conduction, which is desired for accurate temperature readings in measurement devices.

Understanding the nuances of heat transfer mechanisms enables engineers and scientists to design more efficient thermal systems - whether it's improving the sensitivity of temperature measurement devices or enhancing the thermal management of electronics.
Fourier's Law
Fourier's law of thermal conduction is the principle that quantifies the heat transfer through a material due to temperature differences. It states that the heat transfer rate through a material is proportional to the negative of the temperature gradient and the area through which the heat is flowing. In simple terms, heat flows from regions of higher temperature to regions of lower temperature.

The law is mathematically expressed as: \(q = -k \cdot A \cdot \frac{dT}{dx}\)where \(q\) is the heat transfer rate per unit time, \(k\) represents thermal conductivity of the material, \(A\) is the cross-sectional area through which heat is flowing, and \(\frac{dT}{dx}\) is the temperature gradient in the direction of the flow.

In the context of the textbook exercise, Fourier's law could be used to describe the heat conduction in the air adjacent to the probe tip if the dimensions of the probe were known. This law forms the foundation for solving many heat transfer problems, including those involving thermal resistance and the calculation of temperature differences across materials.
Thermal Conductivity
Thermal conductivity is a material property indicative of its ability to conduct heat. It appears in Fourier's law as the proportionality constant and shows how well a material can transfer heat through conduction. High thermal conductivity means that the material is an excellent conductor of heat (like metals), while low thermal conductivity indicates good insulation characteristics (such as in foams or ceramics).

Mathematically, thermal conductivity, symbolized as \(k\), is a part of the equation explained in the Fourier’s law section. It's measured in watts per meter-kelvin (\(W/m\cdot K\)). The exercise provided illustrates an important application of thermal conductivity, where the heat transfer rate could be calculated using the known thermal conductivity of air, if the dimensions of the cantilever's probe were provided.

Materials with known thermal conductivities are crucial in designing devices that manage heat. Whether it's in building insulation or in the precise thermal measurements as seen in the exercise, knowing the thermal conductivity enables accurate predictions of temperature changes and heat transfer rates in various scenarios.

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Most popular questions from this chapter

A motor draws electric power \(P_{\text {elec }}\) from a supply line and delivers mechanical power \(P_{\text {mech }}\) to a pump through a rotating copper shaft of thermal conductivity \(k_{s}\), length \(L\), and diameter \(D\). The motor is mounted on a square pad of width \(W\), thickness \(t\), and thermal conductivity \(k_{p}\). The surface of the housing exposed to ambient air at \(T_{\infty}\) is of area \(A_{h}\), and the corresponding convection coefficient is \(h_{h}\). Opposite ends of the shaft are at temperatures of \(T_{h}\) and \(T_{\infty}\), and heat transfer from the shaft to the ambient air is characterized by the convection coefficient \(h_{s}\). The base of the pad is at \(T_{\infty}\).

Consider the composite wall of Example 3.7. In the Comments section, temperature distributions in the wall were determined assuming negligible contact resistance between materials A and B. Compute and plot the temperature distributions if the thermal contact resistance is \(R_{t, c}^{\prime \prime}=10^{-4} \mathrm{~m}^{2} \cdot \mathrm{K} / \mathrm{W}\).

Consider a plane composite wall that is composed of two materials of thermal conductivities \(k_{\mathrm{A}}=0.1 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) and \(k_{\mathrm{B}}=0.04 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) and thicknesses \(L_{\mathrm{A}}=10 \mathrm{~mm}\) and \(L_{\mathrm{B}}=20 \mathrm{~mm}\). The contact resistance at the interface between the two materials is known to be \(0.30 \mathrm{~m}^{2} \cdot \mathrm{K} / \mathrm{W}\). Material A adjoins a fluid at \(200^{\circ} \mathrm{C}\) for which \(h=10\) \(\mathrm{W} / \mathrm{m}^{2} \cdot \mathrm{K}\), and material \(\mathrm{B}\) adjoins a fluid at \(40^{\circ} \mathrm{C}\) for which \(h=20 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). (a) What is the rate of heat transfer through a wall that is \(2 \mathrm{~m}\) high by \(2.5 \mathrm{~m}\) wide? (b) Sketch the temperature distribution.

A particular thermal system involves three objects of fixed shape with conduction resistances of \(R_{1}=1 \mathrm{~K} / \mathrm{W}\), \(R_{2}=2 \mathrm{~K} / \mathrm{W}\) and \(R_{3}=4 \mathrm{~K} / \mathrm{W}\), respectively. An objective is to minimize the total thermal resistance \(R_{\text {tot }}\) associated with a combination of \(R_{1}, R_{2}\), and \(R_{3}\). The chief engineer is willing to invest limited funds to specify an alternative material for just one of the three objects; the alternative material will have a thermal conductivity that is twice its nominal value. Which object (1, 2, or 3 ) should be fabricated of the higher thermal conductivity material to most significantly decrease \(R_{\text {tot }}\) ? Hint: Consider two cases, one for which the three thermal resistances are arranged in series, and the second for which the three resistances are arranged in parallel.

A spherical, cryosurgical probe may be imbedded in diseased tissue for the purpose of freezing, and thereby destroying, the tissue. Consider a probe of \(3-\mathrm{mm}\) diameter whose surface is maintained at \(-30^{\circ} \mathrm{C}\) when imbedded in tissue that is at \(37^{\circ} \mathrm{C}\). A spherical layer of frozen tissue forms around the probe, with a temperature of \(0^{\circ} \mathrm{C}\) existing at the phase front (interface) between the frozen and normal tissue. If the thermal conductivity of frozen tissue is approximately \(1.5 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) and heat transfer at the phase front may be characterized by an effective convection coefficient of \(50 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), what is the thickness of the layer of frozen tissue (assuming negligible perfusion)?

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