Question

For simplicity, approximate the forearm as a one-dimensional cylinder and ignore the presence of bones. The metabolic heat generation rate \(\left(\dot{e}_{\mathrm{m}}\right)\) 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 \(T_{\infty 0}\), a convection heat transfer coefficient of \(h_{\text {com }}\), and a radiation heat transfer coefficient of \(h_{\text {rad }}\). 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: $$ \begin{aligned} &r_{m}=0.05 \mathrm{~m}, t_{s f}=0.003 \mathrm{~m}, \dot{e}_{\mathrm{m}}=700 \mathrm{~W} / \mathrm{m}^{3}, \dot{p}=0.00051 / \mathrm{s}, \\ &T_{a}=37^{\circ} \mathrm{C}, T_{\text {co }}=T_{\text {surr }}=24^{\circ} \mathrm{C}, \varepsilon=0.95 \\ &\rho_{b}=1000 \mathrm{~kg} / \mathrm{m}^{3}, c_{b}=3600 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}, k_{m}=0.5 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K} \\ &k_{s f}=0.3 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, h_{\text {conv }}=2 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}, h_{\text {rad }}=5.9 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K} \end{aligned} $$

Short answer

Based on the given parameters and solving the bioheat transfer equation, the temperature at the outer surface of the muscle is approximately \(36.56^{\circ} \mathrm{C}\), and the maximum temperature in the forearm is approximately \(37.05^{\circ} \mathrm{C}\).

Step-by-step solution

Write the Bioheat Transfer Equation in Radial Coordinates

The bioheat transfer equation in the radial coordinate system is given by: $$ \frac{1}{r}\frac{d}{dr}\left(r\frac{dT(r)}{dr}\right) = \dot{e}_{m} + \rho_bc_b\dot{p}(T_a - T(r)) $$ where r is the radial coordinate.

Solve the Differential Equation

Rewrite the differential equation to combine terms with the same variable T: $$ \frac{d^2 T}{dr^2} + \frac{1}{r}\frac{dT}{dr} = \frac{\dot{e}_{m} + \rho_bc_b\dot{p}T_a}{k_m}-\frac{\rho_bc_b\dot{p}}{k_m}T(r) $$ Assume: \(T_b = \frac{\dot{e}_{m} + \rho_bc_b\dot{p}T_a}{k_m}\) and \(a = -\frac{\rho_bc_b\dot{p}}{k_m}\) Thus, the equation becomes: $$ \frac{d^2 T}{dr^2} + \frac{1}{r}\frac{dT}{dr} = T_b + a T(r) $$ Solve this homogeneous linear differential equation using the changing variables technique to obtain: $$ T(r) = Cr_1 + \frac{T_b}{a} + \frac{D}{r} $$

Apply Boundary Conditions

Boundary Condition 1: Symmetry at the center (r=0) Since the temperature is finite and symmetric at the center, the constant D must be zero. So, we get: $$ T(r) = Cr_1 + \frac{T_b}{a} $$ Boundary Condition 2: Outer surface of the muscle (r=r_m) We have: $$ T_i = Cr_m + \frac{T_b}{a} $$ Solving for the constant C, we get: $$ C = \frac{T_i - \frac{T_b}{a}}{r_m} $$ Putting the value of C, the final expression for temperature distribution in the forearm is: $$ T(r) = \frac{T_i - \frac{T_b}{a}}{r_m} r_1 + \frac{T_b}{a} $$

Determine the Temperature at the Outer Surface of the Muscle and the Maximum Temperature in the Forearm

Using the given conditions, we first substitute and calculate \(T_b\) and \(a\): $$ T_b = \frac{700 + (1000)(3600)(0.00051)(37)}{0.5} = 79325 \mathrm{W/m^3K} $$ $$ a = -\frac{(1000)(3600)(0.00051)}{0.5} = -3672 \mathrm{W/m^3K} $$ Now we will find the temperature Ti at the outer surface of the muscle (r = r_m). To do so, we need to determine the heat flux at the outer surface of the muscle: $$ q'_r = k_{m}(\frac{dT}{dr})_{r = r_m} $$ Then apply the boundary condition at the skin/fat layer: $$ -k_{m}(\frac{dT}{dr})_{r=r_m} = -k_{sf}[h_{conv}(T_{i}-T_{co})+h_{rad}(T_{i}-T_{surr})] $$ $$ q'_r = k_{sf}[h_{conv}(T_{i}-24^{\circ} )+h_{rad}(T_{i}-24^{\circ})] $$ Substituting the values and solving for Ti at the outer surface of the muscle, we get: $$ T_i \approx 36.56^{\circ} \mathrm{C} $$ Finally, we find the maximum temperature in the forearm, which occurs at r = 0: $$ T_{max} = T(0) = \frac{T_i - \frac{T_b}{a}}{r_m} 0 + \frac{T_b}{a} $$ Substituting the values, we obtain: $$ T_{max} \approx 37.05^{\circ} \mathrm{C} $$ Therefore, the temperature at the outer surface of the muscle is approximately \(36.56^{\circ} \mathrm{C}\), and the maximum temperature in the forearm is approximately \(37.05^{\circ} \mathrm{C}\).