Problem 2 - Biofluid Mechanics: An Introduction to Fluid Mechanics, Macrocirculation, and Microcirculation

Question

A person who has experienced hypothermia is heated steadily from an external source. The thermal conductivity of skin \(0.4 \mathrm{~W} / \mathrm{m}^{\circ} \mathrm{K}\), and heat is being applied to the person over a flat circular surface of diameter \(20 \mathrm{~cm}\) and thickness of \(0.2 \mathrm{~cm}\). Heat is transferred steadily to the person at a rate of \(100 \mathrm{~W}\). If the inner surface of the person is being warmed to \(39^{\circ} \mathrm{C}\), determine the temperature at the outer surface.

Step-by-Step Solution

Answer

The outer surface temperature is approximately $23.08^{\circ} \mathrm{C}$.

Step 1: Understand the setup

We are given a situation where heat is transferred steadily through the skin of a person. We know the thermal conductivity of the skin (\(k = 0.4 \mathrm{~W} / \mathrm{m}^{\circ} \mathrm{K}\)), and that the skin is modeled as a disk with diameter \(20 \mathrm{~cm}\) (radius \(r = 0.1 \mathrm{~m}\)) and thickness \(0.002 \mathrm{~m}\). The rate of heat transfer is \(100 \mathrm{~W}\).

Step 2: Use the heat conduction formula

The formula for steady-state heat conduction through a material is\[Q = \frac{kA(T_{inner}-T_{outer})}{d}\]where \(Q\) is the rate of heat transfer, \(k\) is the thermal conductivity, \(A\) is the area of heat transfer, \(d\) is the thickness, \(T_{inner}\) is the temperature at the inner surface, and \(T_{outer}\) is the temperature at the outer surface.

Step 3: Calculate the area of the heat transfer surface

The area \(A\) of the circular surface can be calculated using the formula for the area of a circle:\[A = \pi r^2\]Substituting the value of \(r = 0.1 \mathrm{~m}\), we get:\[A = \pi (0.1)^2 = 0.0314 \mathrm{~m}^2\]

Step 4: Rearrange the heat conduction formula

The goal is to find \(T_{outer}\), so rearrange the formula:\[T_{outer} = T_{inner} - \frac{Qd}{kA}\]This expression allows us to find the outer surface temperature directly.

Step 5: Substitute and solve for the outer temperature

Using the values \(Q = 100 \mathrm{~W}\), \(d = 0.002 \mathrm{~m}\), \(k = 0.4 \mathrm{~W} / \mathrm{m}^{\circ} \mathrm{K}\), \(T_{inner} = 39^{\circ} \mathrm{C}\), and \(A = 0.0314 \mathrm{~m}^2\), substitute into the formula:\[T_{outer} = 39 - \frac{(100)(0.002)}{(0.4)(0.0314)}\]Performing the calculations, we get:\[T_{outer} = 39 - 15.92 = 23.08^{\circ} \mathrm{C}\]

Step 6: Conclude with the result

The temperature at the outer surface is \(23.08^{\circ} \mathrm{C}\). This is the temperature that the external surface of the skin reaches while the inner temperature is maintained at \(39^{\circ} \mathrm{C}\).

Key Concepts

Thermal Conductivity

Thermal conductivity is a fundamental concept in the study of heat transfer. It defines how well a material can conduct heat. This is particularly important in situations where heat needs to be efficiently transferred through a material or surface. In the context of our exercise, the skin acts as the medium through which heat is being transferred.

Thermal conductivity is denoted by the symbol \(k\) and is measured in watts per meter per degree Celsius \((\mathrm{W}/\mathrm{m}\cdot^{\circ}\mathrm{C})\). A higher thermal conductivity indicates a material that is efficient at conducting heat, whereas a lower value implies insulation properties.

For instance, metals like copper have high thermal conductivity, making them ideal for applications that require effective heat transfer. Conversely, materials like rubber have low thermal conductivity and are used for insulation purposes. In our example, the skin has a thermal conductivity of \(0.4 \mathrm{~W} / \mathrm{m}^{\circ} \mathrm{K}\), suggesting it is a moderate conductor.

Steady-State Conduction

Steady-state conduction is a scenario where the temperature distribution in a material does not change over time. This means that the amount of heat entering a section is equal to the amount of heat leaving it. In our problem, this concept is applied as the heat is being transferred through the skin at a constant rate.

In practical terms, steady-state conduction implies that once equilibrium is reached, temperatures no longer fluctuate. This is crucial for achieving consistent results in thermal management systems. For example, in heating pads or heated clothing, designers ensure that heat is applied in such a manner to maintain steady-state conduction, thereby providing consistent warmth.

The mathematical representation often involves the heat conduction equation, which can be simplified under steady-state conditions, as was done in the solution of this exercise. The formula used is:\[Q = \frac{kA(T_{\text{inner}}-T_{\text{outer}})}{d}\]where each variable has specific roles in modeling the heat transfer accurately.

Temperature Gradient

The temperature gradient is essentially the change in temperature with respect to distance between two points within a material. It is a vector pointing in the direction of greatest temperature change. In simpler terms, it shows how temperature varies spatially within an object.

In our scenario, there is a temperature gradient across the thickness of the skin, from the inner surface to the outer surface. This gradient drives the heat transfer, as heat naturally flows from regions of higher temperature to lower. The equation used in the solution computes these temperature differences, thus illustrating the gradient.

• The gradient can be calculated using the formula:

\[abla T = \frac{T_{\text{inner}} - T_{\text{outer}}}{d}\]By understanding the temperature gradient, engineers and scientists can predict heat flow and design systems to control it efficiently. For instance, in thermal insulation, a large gradient is undesirable because it indicates rapid heat loss.