Problem 3
Question
A skin sample has a heat flux meter attached to it in a laboratory. The meter indicates that a heat flux of \(45 \mathrm{~W} / \mathrm{m}^{2}\) passes through a 4 -cm-thick skin section. The temperature on the sides of the skin are \(25^{\circ} \mathrm{C}\) and \(35^{\circ} \mathrm{C}\). Determine the average thermal conductivity of the skin section.
Step-by-Step Solution
Verified Answer
The average thermal conductivity of the skin section is \( 0.18 \, \text{W/m} \, ^{\circ}C \).
1Step 1: Understand the Problem
We are given a heat flux through a skin section with its thickness and the temperatures at each side. We need to calculate the thermal conductivity.
2Step 2: Identify the Formula
The formula to find thermal conductivity \( k \) is based on Fourier's law of heat conduction: \[ q = -k \frac{dT}{dx} \] where \( q \) is the heat flux, \( \frac{dT}{dx} \) is the temperature gradient.
3Step 3: Calculate the Temperature Gradient
The temperature gradient \( \frac{dT}{dx} \) is calculated as the difference in temperature divided by the thickness of the skin section:\[ \frac{dT}{dx} = \frac{T_{2} - T_{1}}{x} = \frac{35^{\circ}C - 25^{\circ}C}{0.04 \, \text{m}} \]This results in:\[ \frac{dT}{dx} = \frac{10 \, ^{\circ}C}{0.04 \, \text{m}} = 250 \, ^{\circ}C/\text{m} \]
4Step 4: Rearrange the Formula
Rearrange Fourier's law to solve for \( k \):\[ k = -\frac{q}{\frac{dT}{dx}} \]
5Step 5: Substitute the Values
Substitute \( q = 45 \mathrm{~W} / \mathrm{m}^{2} \) and \( \frac{dT}{dx} = 250 \, ^{\circ}C/\text{m} \) into the formula:\[ k = -\frac{45}{250} \, \text{W/m}^{ ext{2}} \cdot \, ^{\circ}C/m \]
6Step 6: Calculate Thermal Conductivity
Perform the arithmetic to find \( k \):\[ k = -0.18 \, \text{W/m} \, ^{\circ}C \]The negative sign indicates the direction of heat flow from hot to cold, but the value of thermal conductivity is positive by convention. Thus:\[ k = 0.18 \, \text{W/m} \, ^{\circ}C \]
Key Concepts
Fourier's Law of Heat ConductionHeat FluxTemperature Gradient
Fourier's Law of Heat Conduction
Fourier's Law of Heat Conduction explains how heat energy is transferred through a material. According to this law, the rate of heat transfer through a surface per unit area (heat flux) is proportional to the negative temperature gradient in the direction of heat flow. This means that heat naturally flows from warmer areas to cooler ones. In formulaic terms, Fourier's law is expressed as:\[q = -k \frac{dT}{dx}\]Where:
- \(q\) represents the heat flux, or the rate of heat transfer per unit area.
- k denotes the thermal conductivity of the material, which is a measure of the material's ability to conduct heat.
- \(\frac{dT}{dx}\) is the temperature gradient, representing how temperature changes with respect to a change in position within the material.
Heat Flux
Heat flux is a crucial concept in thermodynamics and refers to the rate at which heat energy passes through a surface area. It is measured in watts per square meter (W/m²). In essence, heat flux provides a snapshot of how much thermal energy is moving through a specific area in a specific time frame.
In practical applications, heat flux meters can be attached to various materials to measure real-time heat transfer, as was done in the exercise with the skin sample. Understanding heat flux is essential:
- To evaluate the effectiveness of insulation materials.
- To assess the efficiency of heating or cooling systems.
- To diagnose potential energy losses in structures and materials.
- To ensure safety by monitoring and managing heat loads.
Temperature Gradient
The concept of a temperature gradient is foundational to understanding how heat transfers through a material. It is defined as the change in temperature over a given distance within a material, typically measured along the direction of heat flow. In mathematical terms, the temperature gradient \( \frac{dT}{dx} \) is the derivative of temperature with respect to position, indicating how quickly the temperature changes along a path. In the context of the exercise, the temperature difference between two sides of the skin section was 10°C, and the distance over which this change happens was 4 cm, or 0.04 meters. Therefore, the temperature gradient was calculated as:\[\frac{dT}{dx} = \frac{35^{\circ}C - 25^{\circ}C}{0.04 \, \text{m}} = 250 \, ^{\circ}C/\text{m}\]This quotient tells us how steep the change in temperature is, with higher gradients indicating a more rapid temperature change. The temperature gradient is an integral part of understanding heat dynamics, as it influences the rate of heat conduction as described by Fourier's law.
Other exercises in this chapter
Problem 1
In a biofluids laboratory, an excised blood vessel has been cannulated to measure the heat transfer properties of this biological tissue. Blood is flowed at a r
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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} \ma
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On a hot day, a fan is blowing warm air over a person. The air is at a temperature of \(45^{\circ} \mathrm{C}\) and the person can be assumed to be represented
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Consider an average person standing in the middle of a room. The various surfaces are found to be at a temperature of \(15^{\circ} \mathrm{C}\) in the winter an
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