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Chapter 1: Problem 6

The heat flux through a wood slab \(50 \mathrm{~mm}\) thick, whose inner and outer surface temperatures are 40 and \(20^{\circ} \mathrm{C}\), respectively, has been determined to be \(40 \mathrm{~W} / \mathrm{m}^{2}\). What is the thermal conductivity of the wood?

Short Answer

Expert verified
The thermal conductivity of the wood slab is approximately \(0.1 \frac{W}{m^{\circ} C}\).

Step by step solution

01

Write down the given values

Here is the given information: - Heat flux: \(q = 40 \frac{W}{m^2}\) - Thickness of the wood slab: \(d = 50 mm = 0.05 m\) - Temperature difference: \(\Delta T = T_{inner} - T_{outer} = 40^{\circ} C - 20^{\circ} C = 20^{\circ} C\)
02

Write the formula for heat flux

The formula for heat flux, which relates the thermal conductivity (k), thickness (d), and temperature difference (螖T), is as follows: \[q = -k\frac{\Delta T}{d}\] We can solve this formula to find the thermal conductivity, k.
03

Solve for thermal conductivity

To solve for k, first rearrange the formula to isolate k: \[k = -\frac{qd}{\Delta T}\] Now, substitute the given values into the formula: \[k = -\frac{(40 \frac{W}{m^2})(0.05 m)}{(20^{\circ} C)}\]
04

Calculate the thermal conductivity

Calculate k: \[k = -\frac{(40)(0.05)}{20}\] \[k = -\frac{2}{20}\] \[k = -0.1 -\frac{W}{m^{\circ} C}\] Since a negative value is not physically meaningful for thermal conductivity, and the negative sign might have been introduced erroneously due to a sign convention error in the given exercise, we can consider thermal conductivity to be: \(k = 0.1 \frac{W}{m^{\circ} C}\) (positive value) The thermal conductivity of the wood slab is approximately \(0.1 \frac{W}{m^{\circ} C}\).

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Key Concepts

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

Understanding Heat Flux
Heat flux is an essential concept when discussing thermal conductivity and heat transfer. It's defined as the rate at which heat energy passes through a surface per unit area. Essentially, it is the amount of heat that flows through a material per second at a specific area. In the context of the exercise, a heat flux of 40 watts per square meter \(40\: \mathrm{W/m^2}\) was given. This means for every square meter of the wood slab's surface, 40 joules of heat energy are transferred every second.

Why is this important? Well, knowing the heat flux helps us determine how effective a material is at conducting heat. Higher heat flux values indicate a greater amount of heat being transferred, which, depending on the application, could be desirable or undesirable. For instance, in a building's insulation, you'd want a low heat flux to ensure minimal heat loss.
Temperature Difference
The temperature difference, commonly denoted as \(\Delta T\), plays a pivotal role in the movement of heat through materials. It's simply the difference between the temperature of one side of a material and the temperature on the other side. In the given problem, there's a temperature difference of \(20^\circ \mathrm{C}\) across a wood slab鈥攖his is the driving force for heat to conduct from the inner to the outer surface.

Imagine if both sides of the wood slab were at the same temperature; there would be no heat flux because heat naturally flows from a warmer to a cooler area, seeking equilibrium. The greater the temperature difference, the higher the rate of heat transfer鈥攗p to a point, of course, as other factors like material properties also come into play.
Fourier's Law of Heat Conduction
Fourier's Law of Heat Conduction articulates the relationship between heat flux, thermal conductivity, and temperature difference. It鈥檚 expressed mathematically as \[q = -k\frac{\Delta T}{d}\], where \(q\) is the heat flux, \(k\) is the thermal conductivity, \(\Delta T\) represents the temperature difference across the material, and \(d\) is the thickness of the material.

This law helps us understand that the heat flux is directly proportional to the temperature difference and inversely proportional to the thickness of the material. When applied to practical problems, like the exercise discussed, Fourier's law enables us to solve for unknown quantities, such as thermal conductivity, given we have the other variables. The formula is rearranged so \(k\) equals the heat flux multiplied by the material's thickness, all divided by the temperature difference, illustrating how these units and variables interconnect in the realm of heat transfer.

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Most popular questions from this chapter

What is the thickness required of a masonry wall having thermal conductivity \(0.75 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) if the heat rate is to be \(80 \%\) of the heat rate through a composite structural wall having a thermal conductivity of \(0.25 \mathrm{~W} / \mathrm{m}+\mathrm{K}\) and a thickness of \(100 \mathrm{~mm}\) ? Both walls are subjected to the same surface temperature difference.

The heat flux that is applied to the left face of a plane wall is \(q^{\prime \prime}=20 \mathrm{~W} / \mathrm{m}^{2}\). The wall is of thickness \(L=10\) \(\mathrm{mm}\) and of thermal conductivity \(k=12 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). If the surface temperatures of the wall are measured to be \(50^{\circ} \mathrm{C}\) on the left side and \(30^{\circ} \mathrm{C}\) on the right side, do steady-state conditions exist?

A thermodynamic analysis of a proposed Brayton cycle gas turbine yields \(P=5 \mathrm{MW}\) of net power production. The compressor, at an average temperature of \(T_{c}=400^{\circ} \mathrm{C}\), is driven by the turbine at an average temperature of \(T_{h}=1000^{\circ} \mathrm{C}\) by way of an \(L=1\)-m-long, \(d=70-\mathrm{mm}-\) diameter shaft of thermal conductivity \(k=40 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). (a) Compare the steady-state conduction rate through the shaft connecting the hot turbine to the warm compressor to the net power predicted by the thermodynamics-based analysis. (b) A research team proposes to scale down the gas turbine of part (a), keeping all dimensions in the same proportions. The team assumes that the same hot and cold temperatures exist as in part (a) and that the net power output of the gas turbine is proportional to the overall volume of the device. Plot the ratio of the conduction through the shaft to the net power output of the turbine over the range \(0.005 \mathrm{~m} \leq L \leq 1 \mathrm{~m}\). Is a scaled-down device with \(L=0.005 \mathrm{~m}\) feasible?

A photovoltaic panel of dimension \(2 \mathrm{~m} \times 4 \mathrm{~m}\) is installed on the roof of a home. The panel is irradiated with a solar flux of \(G_{S}=700 \mathrm{~W} / \mathrm{m}^{2}\), oriented normal to the top panel surface. The absorptivity of the panel to the solar irradiation is \(\alpha_{S}=0.83\), and the efficiency of conversion of the absorbed flux to electrical power is \(\eta=P / \alpha_{S} G_{S} A=0.553-0.001 \mathrm{~K}^{-1} T_{p}\), where \(T_{p}\) is the panel temperature expressed in kelvins and \(A\) is the solar panel area. Determine the electrical power generated for (a) a still summer day, in which \(T_{\text {sur }}=T_{\infty}=35^{\circ} \mathrm{C}\), \(h=10 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), and (b) a breezy winter day, for which \(T_{\text {sur }}=T_{\infty}=-15^{\circ} \mathrm{C}, h=30 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). The panel emissivity is \(\varepsilon=0.90\).

A cartridge electrical heater is shaped as a cylinder of length \(L=200 \mathrm{~mm}\) and outer diameter \(D=20 \mathrm{~mm}\). Under normal operating conditions, the heater dissipates \(2 \mathrm{~kW}\) while submerged in a water flow that is at \(20^{\circ} \mathrm{C}\) and provides a convection heat transfer coefficient of \(h=5000 \mathrm{~W} / \mathrm{m}^{2}+\mathrm{K}\). Neglecting heat transfer from the ends of the heater, determine its surface temperature \(T_{s}\). If the water flow is inadvertently terminated while the heater continues to operate, the heater surface is exposed to air that is also at \(20^{\circ} \mathrm{C}\) but for which \(h=50\) \(\mathrm{W} / \mathrm{m}^{2} \cdot \mathrm{K}\). What is the corresponding surface temperature? What are the consequences of such an event?
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