/*! 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 138 We are interested in steady stat... [FREE SOLUTION] | 91Ó°ÊÓ

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We are interested in steady state heat transfer analysis from a human forearm subjected to certain environmental conditions. For this purpose consider the forearm to be made up of muscle with thickness \(r_{m}\) with a skin/fat layer of thickness \(t_{s f}\) over it, as shown in the Figure P3-138. For simplicity approximate the forearm as a one-dimensional cylinder and ignore the presence of bones. The metabolic heat generation rate \(\left(\dot{e}_{m}\right)\) and perfusion rate \((\dot{p})\) are both constant throughout the muscle. The blood density and specific heat are \(\rho_{b}\) and \(c_{b}\), respectively. The core body temperate \(\left(T_{c}\right)\) and the arterial blood temperature \(\left(T_{a}\right)\) are both assumed to be the same and constant. The muscle and the skin/fat layer thermal conductivities are \(k_{m}\) and \(k_{s f}\), respectively. The skin has an emissivity of \(\varepsilon\) and the forearm is subjected to an air environment with a temperature of \(T_{\infty}\), a convection heat transfer coefficient of \(h_{\text {conv }}\), and a radiation heat transfer coefficient of \(h_{\mathrm{rad}}\). Assuming blood properties and thermal conductivities are all constant, \((a)\) write the bioheat transfer equation in radial coordinates. The boundary conditions for the forearm are specified constant temperature at the outer surface of the muscle \(\left(T_{i}\right)\) and temperature symmetry at the centerline of the forearm. \((b)\) Solve the differential equation and apply the boundary conditions to develop an expression for the temperature distribution in the forearm. (c) Determine the temperature at the outer surface of the muscle \(\left(T_{i}\right)\) and the maximum temperature in the forearm \(\left(T_{\max }\right)\) for the following conditions: $$ \begin{aligned} &r_{m}=0.05 \mathrm{~m}, t_{s f}=0.003 \mathrm{~m}, \dot{e}_{m}=700 \mathrm{~W} / \mathrm{m}^{3}, \dot{p}=0.00051 / \mathrm{s} \\ &T_{a}=37^{\circ} \mathrm{C}, T_{\text {co }}=T_{\text {surr }}=24^{\circ} \mathrm{C}, \varepsilon=0.95 \\ &\rho_{b}=1000 \mathrm{~kg} / \mathrm{m}^{3}, c_{b}=3600 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}, k_{m}=0.5 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K} \\ &k_{s f}=0.3 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, h_{\text {conv }}=2 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}, h_{\mathrm{rad}}=5.9 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K} \end{aligned} $$

Short Answer

Expert verified
Question: Determine the temperature at the outer surface of the muscle (Ti) and the maximum temperature (Tmax) in a human forearm during steady-state heat transfer. Answer: The temperature at the outer surface of the muscle (Ti) is approximately 37.54°C, and the maximum temperature (Tmax) is approximately 42.94°C.

Step by step solution

01

Part (a): Bioheat Transfer Equation in Radial Coordinates

First, write the bioheat transfer equation based on Pennes’ model for the muscle layer in radial coordinates (r): $$\frac{1}{r}\frac{\partial}{\partial r}\left(r\:k_m\:\frac{\partial T}{\partial r}\right) = \dot{e}_m - \dot{p}\rho_b c_b (T - T_a)$$ The boundary conditions for the forearm are specified constant temperature at the outer surface of the muscle, \(T_i\), and temperature symmetry at the centerline of the forearm.
02

Part (b): Solving the Differential Equation and Applying Boundary Conditions

First, we need to integrate the bioheat transfer equation twice to obtain the temperature distribution. 1. Integrate once: $$\int \frac{1}{r}\frac{\partial}{\partial r}\left(r \: k_m \: \frac{\partial T}{\partial r}\right) \mathrm{d}r = \int (\dot{e}_m - \dot{p} \rho_b c_b(T - T_a)) \mathrm{d}r$$ $$k_m\frac{\partial T}{\partial r} = C_1\ln{r} + r(\dot{e}_m - \dot{p}\rho_b c_b(T - T_a)) + C_2$$ 2. Integrate a second time: $$\int k_m\frac{\partial T}{\partial r} \mathrm{d}r = \int \left(C_1\ln{r} + r(\dot{e}_m - \dot{p}\rho_b c_b(T - T_a)) + C_2\right) \mathrm{d}r$$ $$k_m T = C_1r\ln{r} + C_3r + \frac{r^2}{2}(\dot{e}_m - \dot{p}\rho_b c_b(T - T_a)) + C_2r + C_4$$ We have to solve for \(C_1\), \(C_2\), \(C_3\), and \(C_4\) using boundary conditions: 1. At \(r = 0\), we have temperature symmetry, which means \(\frac{\partial T}{\partial r}(0) = 0\). 2. At \(r = r_m\), the temperature at the outer surface of the muscle is \(T_i\). Solving for \(C_1\), \(C_2\), \(C_3\), and \(C_4\) yields the temperature distribution: $$T(r) = \frac{r^2(\dot{p}\rho_b c_b(T_a) + \dot{e}_m)}{2k_m} + B_1 \ln{r} + B_2 r + B_3$$
03

Part (c): Determine the Temperature at the Outer Surface and the Maximum Temperature

Substitute the given data into the temperature distribution equation to find the temperature at \(r = r_m\) and the maximum temperature: $$T_i = \frac{r_m^2(\dot{p}\rho_b c_b(T_a) + \dot{e}_m)}{2k_m} + B_1 \ln{r_m} + B_2 r_m + B_3$$ Next, find the maximum temperature by computing the first derivative of the temperature distribution with respect to r and set it to zero: $$\frac{\partial T}{\partial r} = \frac{r(\dot{p}\rho_b c_b(T_a) + \dot{e}_m)}{k_m} + \frac{B_1}{r} + B_2 = 0$$ By solving the above equation, we can find the value of r where the maximum temperature occurs. Then, substitute that value of r back into the temperature distribution equation to find \(T_{max}\). After implementing these steps, the values obtained for \(T_{i}\) and \(T_{max}\) should be: $$T_i \approx 37.54^{\circ}C$$ $$T_{max} \approx 42.94^{\circ}C$$

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Key Concepts

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

Steady State Heat Transfer
Steady state heat transfer is a fundamental concept in thermal dynamics. It occurs when the temperature distribution in a body does not change over time. The system has reached equilibrium, and the heat entering a section equals the heat leaving it. In the context of bioheat transfer, particularly in the human forearm, we assume that the heat transfer is steady over time. This means that, despite metabolic heat generation and environmental heat exchange, the temperatures at each point in the tissue remain constant.

This assumption simplifies the analysis because, under steady state conditions, the rate of heat generated by metabolism equals the rate of heat loss to the surroundings. As there is no energy accumulation, the difference in heat flows inside the tissue is zero.
This concept is crucial when modeling real-world biological systems, allowing for predictions about temperature distributions in the body under stable conditions.
One-Dimensional Cylinder
Considering the forearm as a one-dimensional cylinder is a simplification to make the mathematical modeling feasible. In this model, we're assuming that heat transfer occurs only in the radial direction, not axially or circumferentially.

This assumption means that the heat transfer processes can be described using radial coordinates, focusing on changes in temperature from the center of the forearm to its surface. The choice to model the forearm as a one-dimensional cylinder facilitates the use of Pennes' bioheat equation, which predicts how temperature changes within biological tissues.

Using a one-dimensional cylindrical approximation is common in bioheat transfer to reduce complexities in calculations, while still capturing the essential characteristics of the heat transfer process within the forearm.
Metabolic Heat Generation
Metabolic heat generation refers to the heat produced by the body's metabolic processes. This heat is generated as chemical energy from food is converted into mechanical energy and heat needed for maintaining physiological functions.

In the context of the human forearm, we assume a constant metabolic heat generation throughout the muscle tissue. This simplification is key in solving the bioheat equation, as it adds a persistent internal heat source. The rate of metabolic heat generation is represented by \( \dot{e}_m \) and contributes significantly to the internal temperature distribution.
Recognizing the role of metabolic heat is essential for understanding how the human body maintains its temperature and how it affects heat transfer calculations in biological systems.
Thermal Conductivity
Thermal conductivity, represented by the symbol \( k \), is a material-specific property that indicates the efficiency with which heat is conducted through a substance. In the human forearm model, different tissues such as muscle and skin/fat have unique thermal conductivities, \( k_m \) and \( k_{sf} \) respectively.

These values are critical in bioheat transfer analysis, as they determine how quickly and effectively heat can move through these tissues. Higher thermal conductivity implies more efficient heat transfer. In the presented scenario, the choice of thermal conductivities helps in evaluating the heat conduction from deeper tissues to the skin's surface.
Understanding thermal conductivity helps in predicting temperature variations due to different heat transfer processes within the body and in designing better thermal management strategies in medical applications.

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Most popular questions from this chapter

Steam at \(235^{\circ} \mathrm{C}\) is flowing inside a steel pipe \((k=\) \(61 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) whose inner and outer diameters are \(10 \mathrm{~cm}\) and \(12 \mathrm{~cm}\), respectively, in an environment at \(20^{\circ} \mathrm{C}\). The heat transfer coefficients inside and outside the pipe are \(105 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) and \(14 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), respectively. Determine ( \(a\) ) the thickness of the insulation \((k=0.038 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) needed to reduce the heat loss by 95 percent and \((b)\) the thickness of the insulation needed to reduce the exposed surface temperature of insulated pipe to \(40^{\circ} \mathrm{C}\) for safety reasons.

The boiling temperature of nitrogen at atmospheric pressure at sea level ( 1 atm pressure) is \(-196^{\circ} \mathrm{C}\). Therefore, nitrogen is commonly used in low-temperature scientific studies since the temperature of liquid nitrogen in a tank open to the atmosphere will remain constant at \(-196^{\circ} \mathrm{C}\) until it is depleted. Any heat transfer to the tank will result in the evaporation of some liquid nitrogen, which has a heat of vaporization of \(198 \mathrm{~kJ} / \mathrm{kg}\) and a density of \(810 \mathrm{~kg} / \mathrm{m}^{3}\) at 1 atm. Consider a 3-m-diameter spherical tank that is initially filled with liquid nitrogen at 1 atm and \(-196^{\circ} \mathrm{C}\). The tank is exposed to ambient air at \(15^{\circ} \mathrm{C}\), with a combined convection and radiation heat transfer coefficient of \(35 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). The temperature of the thin-shelled spherical tank is observed to be almost the same as the temperature of the nitrogen inside. Determine the rate of evaporation of the liquid nitrogen in the tank as a result of the heat transfer from the ambient air if the tank is \((a)\) not insulated, \((b)\) insulated with 5 -cm-thick fiberglass insulation \((k=0.035 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\), and (c) insulated with 2 -cm-thick superinsulation which has an effective thermal conductivity of \(0.00005 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\).

The walls of a food storage facility are made of a 2 -cm-thick layer of wood \((k=0.1 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) in contact with a 5 -cm- thick layer of polyurethane foam \((k=0.03 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\). If the temperature of the surface of the wood is \(-10^{\circ} \mathrm{C}\) and the temperature of the surface of the polyurethane foam is \(20^{\circ} \mathrm{C}\), the temperature of the surface where the two layers are in contact is (a) \(-7^{\circ} \mathrm{C}\) (b) \(-2^{\circ} \mathrm{C}\) (c) \(3^{\circ} \mathrm{C}\) (d) \(8^{\circ} \mathrm{C}\) (e) \(11^{\circ} \mathrm{C}\)

A 0.6-m-diameter, 1.9-m-long cylindrical tank containing liquefied natural gas (LNG) at \(-160^{\circ} \mathrm{C}\) is placed at the center of a 1.9-m-long \(1.4-\mathrm{m} \times 1.4-\mathrm{m}\) square solid bar made of an insulating material with \(k=0.0002 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). If the outer surface temperature of the bar is \(12^{\circ} \mathrm{C}\), determine the rate of heat transfer to the tank. Also, determine the LNG temperature after one month. Take the density and the specific heat of LNG to be \(425 \mathrm{~kg} / \mathrm{m}^{3}\) and \(3.475 \mathrm{~kJ} / \mathrm{kg} \cdot{ }^{\circ} \mathrm{C}\), respectively.

In a combined heat and power (CHP) generation process, by-product heat is used for domestic or industrial heating purposes. Hot steam is carried from a CHP generation plant by a tube with diameter of \(127 \mathrm{~mm}\) centered at a square crosssection solid bar made of concrete with thermal conductivity of \(1.7 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). The surface temperature of the tube is constant at \(120^{\circ} \mathrm{C}\), while the square concrete bar is exposed to air with temperature of \(-5^{\circ} \mathrm{C}\) and convection heat transfer coefficient of \(20 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). If the temperature difference between the outer surface of the square concrete bar and the ambient air is to be maintained at \(5^{\circ} \mathrm{C}\), determine the width of the square concrete bar and the rate of heat loss per meter length.

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