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Chapter 3: Problem 69

One modality for destroying malignant tissue involves imbedding a small spherical heat source of radius \(r_{e}\) within the tissue and maintaining local temperatures above a critical value \(T_{z}\) for an extended period. Tissue that is well removed from the source may be assumed to remain at nomal 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 \(r_{e}=0.5 \mathrm{~mm}\), what heat rate must be supplied to maintain a tissue temperature of \(T \geq T_{e}=42^{\circ} \mathrm{C}\) in the domain \(0.5 \leq r \leq\) \(5 \mathrm{~mm}\) ? The tissue thermal cceductivity is approximately \(0.5 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). Assume negligible perfusion.

Short Answer

Expert verified
The heat rate needed is approximately 0.00157 W to maintain the temperature.

Step by step solution

01

Understanding the Problem

We need to find a general expression for the radial temperature distribution around a spherical heat source in tissue under steady-state conditions. The heat source has a radius of \(r_e=0.5\ mm\) and dissipates heat at a rate \(q\). We aim to maintain the temperature of the tissue at \(T_e=42^{\circ} C\) within the domain \(0.5 \leq r \leq 5\ mm\). The tissue's thermal conductivity is given as \(k=0.5\ W/m \cdot K\).
02

Using Heat Conduction Equation

The steady-state heat conduction in spherical coordinates without internal generation and negligible perfusion can be described by the equation:\[abla^2 T = 0 \]Where \( abla^2 T \) is the Laplacian of the temperature \( T(r) \). For spherical coordinates, this simplifies to:\[\frac{d}{dr} \left( r^2 \frac{dT}{dr} \right) = 0\]
03

Solving the Differential Equation

The differential equation from Step 2, \( \frac{d}{dr} \left( r^2 \frac{dT}{dr} \right) = 0 \), implies that:\[ r^2 \frac{dT}{dr} = C_1 \]Integrating gives:\[ \frac{dT}{dr} = \frac{C_1}{r^2} \]Integrating again provides:\[ T(r) = -\frac{C_1}{r} + C_2\]
04

Applying Boundary Conditions

The boundary conditions are: 1. \(T(r_e) = T_e = 42^{\circ}C\)2. As \(r \to \infty\), \(T \to T_b = 37^{\circ}C\) (or a large \(r\) to consider tissue sufficiently far away to be at body temperature). Using \(T(r_e) = T_e\):\[42 = -\frac{C_1}{0.0005} + C_2\]For a large \(r\), assuming \(T = T_b\):\[37 = 0 + C_2\]Solving gives \(C_2 = 37\).
05

Solving for C_1 and Expression for T(r)

Substituting \(C_2 = 37\) into \(42 = -\frac{C_1}{0.0005} + 37\):\[42 - 37 = -\frac{C_1}{0.0005}\]Solving gives:\[\frac{C_1}{0.0005} = -5\]\[C_1 = -5 \times 0.0005 = -0.0025\]The expression for \(T(r)\) is then:\[T(r) = \frac{0.0025}{r} + 37\]
06

Determining Heat Rate q

The rate of heat dissipation, \( q \, [W]\), from the sphere is related by the surface area and the temperature gradient:\[ q = -k \cdot 4 \pi r_e^2 \frac{dT}{dr}\bigg|_{r=r_e} \]Where \( \frac{dT}{dr} = \frac{0.0025}{r^2} \). Thus,\[ q = -0.5 \cdot 4 \pi (0.0005)^2 \frac{0.0025}{(0.0005)^2} = 0.5 \cdot 4 \pi \times 0.0025 = 0.5 \pi \cdot 0.01 = 0.0005 \pi \] This gives \( q = 0.00157 \) W.

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Key Concepts

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

Steady-State Temperature Distribution
In biological tissues, maintaining a consistent temperature distribution is crucial to ensure effective processes like therapeutic heat treatments. When we talk about steady-state temperature distribution, we're referring to a scenario where the temperature at any given point in the tissue does not change over time.
For example, when a spherical heat source maintains its heat at a consistent level, the temperature away from the source will eventually reach a balance or steady state. Under these conditions, the temperature distribution becomes stable, allowing reliable predictions of temperature at various radial distances from the source.
In the exercise, this concept is used to find how well heat dissipates from the spherical source into the surrounding tissue and how the tissue maintains a consistent temperature to achieve desired therapeutic effects.
Spherical Heat Conduction
Heat transfer in tissues often adheres to the principles of spherical conduction when utilizing a spherical heat source. This concept involves how heat moves from the center of the sphere outward through the tissue.
Unlike in flat surfaces where heat moves in straight lines, in spherical conduction, heat dissipates radially outward, resembling the layers of an onion. The mathematical representation involves solving a differential equation in spherical coordinates to understand the heat distribution.
To put this into context, the equation used in our exercise simplifies the heat conduction behavior into a form that shows how heat disperses symmetrically from the point source to the surrounding tissue, which helps to understand how effectively the treatment will engage the tissue areas involved.
Thermal Conductivity of Tissue
Thermal conductivity is a measure of a material's ability to conduct heat, which is a critical property when calculating heat transfer in biological tissues. In the provided example, the tissue has a known thermal conductivity of 0.5 W/m路K.
This value helps determine how much heat can move through the tissue, influencing how quickly or slowly the temperature reaches steady state. A higher thermal conductivity means heat spreads faster through the tissue, useful for treatments meant to cover larger areas.
Therefore, understanding thermal conductivity is essential for calculating the necessary heat input to achieve desired temperature distributions, ensuring effective heat-based treatments like thermal ablation.
Boundary Conditions in Heat Conduction
Boundary conditions are vital for solving heat conduction problems because they set the stage for how temperature changes across distances. In this exercise, specific boundary conditions are applied: the temperature at the heat source boundary and the temperature at a point far from the source.
These conditions anchor the solution of the differential equation, making it possible to determine how temperature varies from the center outward. Specifically, the boundary condition of the tissue at body temperature far away allows for modeling real-life scenarios where tissues at a distance return to normal body conditions.
By incorporating these conditions into the solution, we achieve practical and accurate calculations that reflect real-life heat transfer scenarios in biological systems. This understanding ensures that heat treatments are precisely controlled and targeted to affect only desired tissue areas.

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Most popular questions from this chapter

The energy transferred from the anterior chamber of the eye through the cornea varies considerably depending on whether a contact lens is worm. Treat the eye as a splerical system and assume the system to be at steady state. The convection coefficient \(h_{e}\) 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: \(\begin{array}{ll}r_{1}=10.2 \mathrm{~mm} & r_{2}=12.7 \mathrm{~mm} \\\ r_{3}=16.5 \mathrm{~mm} & T_{-a}=21^{\circ} \mathrm{C} \\ T_{s 1}=37^{\circ} \mathrm{C} & k_{2}=0.80 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K} \\\ k_{1}=0.35 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K} & h_{n}=6 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K} \\ h_{4}=12 \mathrm{~W} / \mathrm{m}^{2}-\mathrm{K} & \end{array}\) (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.

A thin clectrical heater is inserted between a long circular rod and a concentric tube with inner and outer radii of 20 and \(40 \mathrm{~mm}\). The rod (A) has a thermal conductivity of \(k_{A}=0.15 \mathrm{~W} / \mathrm{m}+\mathrm{K}\). while the tube (B) has a thermal conductivity of \(k_{\mathrm{B}}=1.5 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) and its cuter surface is subjected to convection with a fluid of temperature \(T_{w}=-15^{\circ} \mathrm{C}\) and heat transfer coefficient \(50 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). The thermal contact resistance betueen the cylinder surfaces and the heater is negligible. (a) Determine the electrical power per unit length of the cylinders (W/m) that is required to maintain the outer surface of eylinder \(\mathrm{B}\) at \(5^{\circ} \mathrm{C}\). (b) What is the temperature at the center of cylinder A?

Radinactive wastes \(\left(k_{1 w}=20 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\right)\) are stored in a spherical, stainless steel \(\left(k_{\mathrm{m}}=15 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\right)\) container of inner and outer radii equal to \(r_{i}=0.5 \mathrm{~m}\) and \(r_{e}=0.6 \mathrm{~m}\). Heat is genersed volumetrically within the wastes at a unifom rate of \(\hat{q}=10^{8} \mathrm{~W} / \mathrm{m}^{3}\), and the outer surface of the container is exposed to a water flow for which \(h=\) \(1000 \mathrm{~W} / \mathrm{m}^{2}-\mathrm{K}\) and \(T_{\mathrm{s}}=25^{\circ} \mathrm{C}\).

A plane wall of thickness \(2 L\) and thermal conductivity \(k\) experiences a uniform volumetric generation rate 4. As shown in the sketch for Case I. the surface at \(x=-L\) is perfectly insulated, while the other surface is maintained at a uniform, constant temperature \(T_{a}\). For Case 2, a very thin dielectric strip is inserted at the midpoint of the wall \((x=0)\) in order to electrically isolate the two sections, \(A\) and B. The thermal resistance of the strip is \(R_{7}^{*}=0.0005 \mathrm{~m}^{2} \cdot \mathrm{K} / \mathrm{W}\). The parameters associated with the wall are \(k=50 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, L=20 \mathrm{~mm}\), \(q=5 \times 10^{6} \mathrm{~W} / \mathrm{m}^{3}\), and \(T_{e}=50^{\circ} \mathrm{C}\). (a) Sketch the temperature distribution for case I on \(T-x\) coordinates. Deseribe the key features of this distribution. Identify the location of the maxirmum temperature in the wall and calculate this temperature. (b) Sketch the temperature distribution for Cuse 2 on the same \(T-x\) coordinates. Describe the key features of this distribution. (c) What is the temperature difference between the two walls at \(x=0\) for Case 2? (d) What is the location of the maximum temperature in the composite wall of Case 2? Calculate this temperature.

An air heater may be fabricated by coiling Nichfome wire and passing air in cross flow over the wire. Consider a heater fabricated from wire of diameter \(D=\) \(1 \mathrm{~mm}\), electrical resistivity \(p_{e}=10^{-6} \mathrm{n} \cdot \mathrm{m}\), thernal conductivity \(k=25 \mathrm{~W} / \mathrm{m}=\mathrm{K}\), and emissivity \(\varepsilon=0.2\). The heater is designed to deliver air at a temperatiue of \(T_{m}=50^{\circ} \mathrm{C}\) under flow conditions that provide a convection coefficient of \(h=250 \mathrm{~W} / \mathrm{m}^{2}\). \(\mathrm{K}\) for the wire. The temperature of the housing that encloses the wire and through which the air flows is \(T_{\text {er }}=50^{\circ} \mathrm{C}\). If the maximum allowable temperature of the wire is \(T_{\max }=1200^{\circ} \mathrm{C}\), what is the maximum allowable clextric current \(n\) ? If the maximum available voltage is \(\Delta E=110 \mathrm{~V}\), what is the corresponding length \(L\) of wire that may be used in the heater and the power raing of the heater? Himt: In your solution, assume negligible temperature variations within the wire, but after obtaining the desired results, assess the validity of this arsumption.
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