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Chapter 3: Problem 78

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

Approximately \(10^{6}\) discrete electrical components can be placed on a single integrated circuit (chip), with electrical heat dissipation as high as \(30,000 \mathrm{~W} / \mathrm{m}^{2}\). The chip, which is very thin, is exposed to a dielectric liquid at its outer surface, with \(h_{o}=1000 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) and \(T_{\infty, 0}=20^{\circ} \mathrm{C}\), and is joined to a circuit board at its inner surface. The thermal contact resistance between the chip and the board is \(10^{-4} \mathrm{~m}^{2} \cdot \mathrm{K} / \mathrm{W}\), and the board thickness and thermal conductivity are \(L_{b}=5 \mathrm{~mm}\) and \(k_{b}=1 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\), respectively. The other surface of the board is exposed to ambient air for which \(h_{i}=40\) \(\mathrm{W} / \mathrm{m}^{2} \cdot \mathrm{K}\) and \(T_{\infty, i}=20^{\circ} \mathrm{C}\). (a) Sketch the equivalent thermal circuit corresponding to steady-state conditions. In variable form, label appropriate resistances, temperatures, and heat fluxes. (b) Under steady-state conditions for which the chip heat dissipation is \(q_{c}^{\prime \prime}=30,000 \mathrm{~W} / \mathrm{m}^{2}\), what is the chip temperature? (c) The maximum allowable heat flux, \(q_{c, m}^{\prime \prime}\), is determined by the constraint that the chip temperature must not exceed \(85^{\circ} \mathrm{C}\). Determine \(q_{c, m}^{\prime \prime}\) for the foregoing conditions. If air is used in lieu of the dielectric liquid, the convection coefficient is reduced by approximately an order of magnitude. What is the value of \(q_{c, m}^{\prime \prime}\) for \(h_{o}=100 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) ? With air cooling, can significant improvements be realized by using an aluminum oxide circuit board and/or by using a conductive paste at the chip/board interface for which \(R_{t, c}^{n}=10^{-5} \mathrm{~m}^{2} \cdot \mathrm{K} / \mathrm{W}\) ?

A new building to be located in a cold climate is being designed with a basement that has an \(L=200\)-mm-thick wall. Inner and outer basement wall temperatures are \(T_{i}=20^{\circ} \mathrm{C}\) and \(T_{o}=0^{\circ} \mathrm{C}\), respectively. The architect can specify the wall material to be either aerated concrete block with \(k_{\mathrm{ac}}=0.15 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\), or stone mix concrete. To reduce the conduction heat flux through the stone mix wall to a level equivalent to that of the aerated concrete wall, what thickness of extruded polystyrene sheet must be applied onto the inner surface of the stone mix con-crete wall? Floor dimensions of the basement are \(20 \mathrm{~m} \times 30 \mathrm{~m}\), and the expected rental rate is \(\$ 50 / \mathrm{m}^{2} /\) month. What is the yearly cost, in terms of lost rental income, if the stone mix concrete wall with polystyrene insulation is specified?

Aluminum fins of triangular profile are attached to a plane wall whose surface temperature is \(250^{\circ} \mathrm{C}\). The fin base thickness is \(2 \mathrm{~mm}\), and its length is \(6 \mathrm{~mm}\). The system is in ambient air at a temperature of \(20^{\circ} \mathrm{C}\), and the surface convection coefficient is \(40 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). (a) What are the fin efficiency and effectiveness? (b) What is the heat dissipated per unit width by a single fin?

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 rod of diameter \(D=25 \mathrm{~mm}\) and thermal conductivity \(k=60 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) protrudes normally from a furnace wall that is at \(T_{w}=200^{\circ} \mathrm{C}\) and is covered by insulation of thickness \(L_{\text {ins }}=200 \mathrm{~mm}\). The rod is welded to the furnace wall and is used as a hanger for supporting instrumentation cables. To avoid damaging the cables, the temperature of the rod at its exposed surface, \(T_{o}\), must be maintained below a specified operating limit of \(T_{\max }=100^{\circ} \mathrm{C}\). The ambient air temperature is \(T_{\infty}=\) \(25^{\circ} \mathrm{C}\), and the convection coefficient is \(h=15 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). (a) Derive an expression for the exposed surface temperature \(T_{o}\) as a function of the prescribed thermal and geometrical parameters. The rod has an exposed length \(L_{o}\), and its tip is well insulated. (b) Will a rod with \(L_{o}=200 \mathrm{~mm}\) meet the specified operating limit? If not, what design parameters would you change? Consider another material, increasing the thickness of the insulation, and increasing the rod length. Also, consider how you might attach the base of the rod to the furnace wall as a means to reduce \(T_{o}\).
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