/*! 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 78 One modality for destroying mali... [FREE SOLUTION] | 91Ó°ÊÓ

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One modality for destroying malignant tissue involves imbedding a small spherical heat source of radius \(r_{o}\) within the tissue and maintaining local temperatures above a critical value \(T_{c}\) for an extended period. Tissue that is well removed from the source may be assumed to remain at normal body temperature \(\left(T_{b}=37^{\circ} \mathrm{C}\right)\). Obtain a general expression for the radial temperature distribution in the tissue under steady- state conditions for which heat is dissipated at a rate \(q\). If \(r_{o}=0.5 \mathrm{~mm}\), what heat rate must be supplied to maintain a tissue temperature of \(T \geq T_{c}=42^{\circ} \mathrm{C}\) in the domain \(0.5 \leq r \leq\) \(5 \mathrm{~mm}\) ? The tissue thermal conductivity is approximately \(0.5 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). Assume negligible perfusion.

Short Answer

Expert verified
The radial temperature distribution in the malignant tissue under steady-state conditions is given by \(T(r) = -\frac{-q r_o^2}{k r} + T_b\), where \(q\) is the heat rate, \(r_o\) is the radius of the heat source, \(k\) is the tissue thermal conductivity, and \(T_b\) is the body temperature. To maintain a tissue temperature of \(T \geq T_c = 42^{\circ} \mathrm{C}\) in the domain \(0.5 \leq r \leq 5\,\mathrm{mm}\), the required heat rate is approximately \(420\,\mathrm{W/m}\).

Step by step solution

01

Apply the heat conduction equation in spherical coordinates

We apply the heat conduction equation in spherical coordinates for the radial temperature distribution. \[ \frac{1}{r^2} \frac{d}{dr} \left( r^2 \frac{dT}{dr} \right) = 0 \]
02

Simplify and solve for the radial temperature distribution

Solve the differential equation to get the temperature distribution function. \[ \frac{d}{dr} \left( r^2 \frac{dT}{dr} \right) = 0 \] Integrating once with respect to r, \[ r^2 \frac{dT}{dr} = C_1 \] Where \(C_1\) is a constant of integration. Divide both sides by \(r^2\), \[ \frac{dT}{dr} = \frac{C_1}{r^2} \] Now, we integrate once again with respect to r to get \(T(r)\), \[ T(r) = -\frac{C_1}{r} + C_2 \] Here, \(C_2\) is another constant of integration.
03

Apply boundary conditions

We have two boundary conditions to determine the constants of integration \(C_1\) and \(C_2\). 1. At the surface of the heat source \(r = r_o\), the heat flux is \(q\). 2. At the location far away from the heat source, the temperature is \(T_b = 37^{\circ} \mathrm{C}\). First, apply the heat flux boundary condition. Heat flux is given by: \[ q = -k \frac{dT}{dr} \] At \(r = r_o\), \[ q = -k \frac{C_1}{r_o^2} \] Solving for \(C_1\), \[ C_1 = -q r_o^2 / k \] Now, apply the second boundary condition at the location far away from the heat source, where \(r \to \infty\) and \(T = T_b\), \[ T_b = -\frac{(-q r_o^2 / k)}{\infty} + C_2 \] As \(\frac{1}{\infty} \to 0\), we get \[ C_2 = T_b \] So, the radial temperature distribution is \[ T(r) = -\frac{-q r_o^2}{k r} + T_b \]
04

Calculate heat rate q

Given values: \[ T_c = 42^{\circ}C, \quad r_o = 0.5\,\mathrm{mm}, \quad k = 0.5\,\mathrm{W/m\cdot K} \] We need to maintain temperature \(T\geq T_c=42^{\circ}C\) in the domain \(0.5\leq r\leq 5\,\mathrm{mm}\). So, at the boundary \(r=5\,\mathrm{mm}\), we can set \[ T(5\,\mathrm{mm}) = T_c \] Plug the values into the radial temperature distribution equation, \[ 42 = -\frac{(-q (0.5\cdot 10^{-3})^2}{0.5\cdot(5\cdot 10^{-3})} + 37 \] Solve for \(q\), \[ q \approx 420\,\mathrm{W/m} \] Therefore, the heat rate required to maintain a tissue temperature of \(T \geq T_c = 42^{\circ} \mathrm{C}\) in the domain \(0.5 \leq r \leq 5\,\mathrm{mm}\) is approximately \(420\,\mathrm{W/m}\).

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Key Concepts

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

Radial Temperature Distribution
Understanding radial temperature distribution is key in solving heat conduction problems in spherical coordinates, especially in applications like medical treatments where a heat source affects surrounding tissues. In systems with spherical symmetry, the temperature distribution emanating from a central heat source is dependent on the radial distance from the center. This means we only need to consider how temperature changes with respect to the radius. When dealing with a steady-state condition, there are no changes in temperature over time, allowing us to only focus on spatial temperature variation. Mathematically, the radial temperature distribution can be determined using the heat equation for spherical coordinates:\[ \frac{1}{r^2} \frac{d}{dr} \left( r^2 \frac{dT}{dr} \right) = 0 \] Solving this differential equation provides us with the temperature as a function of the radial position. This solution tells you how the temperature decreases as you move away from the heat source, driven by the nature of heat spreading out into the surrounding medium.
Thermal Conductivity
Thermal conductivity is a material's ability to conduct heat and plays a crucial role in determining how quickly or effectively heat spreads. In the context of the exercise involving a small spherical heat source in tissue, the thermal conductivity of the tissue dictates the rate at which heat is absorbed and spread around. A higher thermal conductivity means that heat spreads faster through the material, while a lower thermal conductivity results in slower heat propagation.The given thermal conductivity value (\[ k = 0.5 \mathrm{W/m \cdot K} \]) tells us how well the tissue transfers heat. In our calculations, it directly influences the heat rate \( q \), required to maintain the critical temperature around the heat source. This is expressed in the heat conduction equation:\[ q = -k \frac{dT}{dr} \]By determining \( q \), we can understand what heat input is needed under steady-state conditions to maintain the critical temperature for medical applications like tumor destruction.
Spherical Coordinates
Spherical coordinates are a natural choice for problems involving radially symmetric situations, such as heat conduction in spheres or around spherical objects. This coordinate system allows us to simplify the analysis and solve problems involving complex geometries by focusing on radial variations without worrying about angular changes in the horizontal and vertical directions. The primary coordinates are the radius \( r \), polar angle \( \theta \), and azimuthal angle \( \phi \).In heat conduction, since we're mainly concerned with radial variations, the problem reduces to dealing solely with the radial coordinate \( r \). This dramatically simplies the governing differential equations, making it easier to derive solutions, like the radial temperature distribution equation used in the exercise: \[ \frac{1}{r^2} \frac{d}{dr} \left( r^2 \frac{dT}{dr} \right) = 0 \]By focusing on spherically symmetric situations, we can accurately and efficiently model the heat distribution in tissues around a spherical heat source, considering both material properties and desired temperature thresholds.

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Most popular questions from this chapter

A spherical tank of \(3-\mathrm{m}\) diameter contains a liquifiedpetroleum gas at \(-60^{\circ} \mathrm{C}\). Insulation with a thermal conductivity of \(0.06 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) and thickness \(250 \mathrm{~mm}\) is applied to the tank to reduce the heat gain. (a) Determine the radial position in the insulation layer at which the temperature is \(0^{\circ} \mathrm{C}\) when the ambient air temperature is \(20^{\circ} \mathrm{C}\) and the convection coefficient on the outer surface is \(6 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). (b) If the insulation is pervious to moisture from the atmospheric air, what conclusions can you reach about the formation of ice in the insulation? What effect will ice formation have on heat gain to the LP gas? How could this situation be avoided?

A high-temperature, gas-cooled nuclear reactor consists of a composite cylindrical wall for which a thorium fuel element \((k \approx 57 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) is encased in graphite \((k \approx 3\) \(\mathrm{W} / \mathrm{m} \cdot \mathrm{K})\) and gaseous helium flows through an annular coolant channel. Consider conditions for which the helium temperature is \(T_{\infty}=600 \mathrm{~K}\) and the convection coefficient at the outer surface of the graphite is \(h=2000 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). (a) If thermal energy is uniformly generated in the fuel element at a rate \(\dot{q}=10^{8} \mathrm{~W} / \mathrm{m}^{3}\), what are the temperatures \(T_{1}\) and \(T_{2}\) at the inner and outer surfaces, respectively, of the fuel element? (b) Compute and plot the temperature distribution in the composite wall for selected values of \(\dot{q}\). What is the maximum allowable value of \(\dot{q}\) ?

Consider a power transistor encapsulated in an aluminum case that is attached at its base to a square aluminum plate of thermal conductivity \(k=240 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\), thickness \(L=6 \mathrm{~mm}\), and width \(W=20 \mathrm{~mm}\). The case is joined to the plate by screws that maintain a contact pressure of 1 bar, and the back surface of the plate transfers heat by natural convection and radiation to ambient air and large surroundings at \(T_{\infty}=T_{\text {sur }}=\) \(25^{\circ} \mathrm{C}\). The surface has an emissivity of \(\varepsilon=0.9\), and the convection coefficient is \(h=4 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). The case is completely enclosed such that heat transfer may be assumed to occur exclusively through the base plate. (a) If the air-filled aluminum-to-aluminum interface is characterized by an area of \(A_{c}=2 \times 10^{-4} \mathrm{~m}^{2}\) and a roughness of \(10 \mu \mathrm{m}\), what is the maximum allowable power dissipation if the surface temperature of the case, \(T_{s, c}\), is not to exceed \(85^{\circ} \mathrm{C}\) ? (b) The convection coefficient may be increased by subjecting the plate surface to a forced flow of air. Explore the effect of increasing the coefficient over the range \(4 \leq h \leq 200 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\).

An electrical current of 700 A flows through a stainless steel cable having a diameter of \(5 \mathrm{~mm}\) and an electrical resistance of \(6 \times 10^{-4} \mathrm{\Omega} / \mathrm{m}\) (i.e., per meter of cable length). The cable is in an environment having a temperature of \(30^{\circ} \mathrm{C}\), and the total coefficient associated with convection and radiation between the cable and the environment is approximately \(25 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). (a) If the cable is bare, what is its surface temperature? (b) If a very thin coating of electrical insulation is applied to the cable, with a contact resistance of \(0.02 \mathrm{~m}^{2} \cdot \mathrm{K} / \mathrm{W}\), what are the insulation and cable surface temperatures? (c) There is some concern about the ability of the insulation to withstand elevated temperatures. What thickness of this insulation \((k=0.5 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) will yield the lowest value of the maximum insulation temperature? What is the value of the maximum temperature when this thickness is used?

The energy transferred from the anterior chamber of the eye through the cornea varies considerably depending on whether a contact lens is worn. Treat the eye as a spherical system and assume the system to be at steady state. The convection coefficient \(h_{o}\) is unchanged with and without the contact lens in place. The cornea and the lens cover one-third of the spherical surface area. Values of the parameters representing this situation are as follows: \(\begin{array}{ll}r_{1}=10.2 \mathrm{~mm} & r_{2}=12.7 \mathrm{~mm} \\\ r_{3}=16.5 \mathrm{~mm} & T_{\infty, o}=21^{\circ} \mathrm{C} \\ T_{\infty \infty, i}=37^{\circ} \mathrm{C} & k_{2}=0.80 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K} \\ k_{1}=0.35 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K} & h_{o}=6 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K} \\ h_{i}=12 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K} & \end{array}\) (a) Construct the thermal circuits, labeling all potentials and flows for the systems excluding the contact lens and including the contact lens. Write resistance elements in terms of appropriate parameters. (b) Determine the heat loss from the anterior chamber with and without the contact lens in place. (c) Discuss the implication of your results.

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