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: $$ \begin{aligned} &r_{m}=0.05 \mathrm{~m}, t_{s f}=0.003 \mathrm{~m}, \dot{e}_{m}=700 \mathrm{~W} / \mathrm{m}^{3}, \dot{p}=0.00051 / \mathrm{s} \\ &T_{a}=37^{\circ} \mathrm{C}, T_{\text {co }}=T_{\text {surr }}=24^{\circ} \mathrm{C}, \varepsilon=0.95 \\ &\rho_{b}=1000 \mathrm{~kg} / \mathrm{m}^{3}, c_{b}=3600 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}, k_{m}=0.5 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K} \\ &k_{s f}=0.3 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, h_{\text {conv }}=2 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}, h_{\mathrm{rad}}=5.9 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K} \end{aligned} $$

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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