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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

The free convection heat transfer coefficient on a thin hot vertical plate suspended in still air can be determined from observations of the change in plate temperature with time as it cools. Assuming the plate is isothermal and radiation exchange with its surroundings is negligible, evaluate the convection coefficient at the instant of time when the plate temperature is \(225^{\circ} \mathrm{C}\) and the change in plate temperature with time \((d T / d t)\) is \(-0.022 \mathrm{~K} / \mathrm{s}\). The ambient air temperature is \(25^{\circ} \mathrm{C}\) and the plate measures \(0.3 \times 0.3 \mathrm{~m}\) with a mass of \(3.75 \mathrm{~kg}\) and a specific heat of \(2770 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\).

A thin electrical heating element provides a uniform heat flux \(q_{o}^{\prime \prime}\) to the outer surface of a duct through which airflows. The duct wall has a thickness of \(10 \mathrm{~mm}\) and a thermal conductivity of \(20 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). (a) At a particular location, the air temperature is \(30^{\circ} \mathrm{C}\) and the convection heat transfer coefficient between the air and inner surface of the duct is \(100 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). What heat flux \(q_{o}^{\prime \prime}\) is required to maintain the inner surface of the duct at \(T_{i}=85^{\circ} \mathrm{C}\) ? (b) For the conditions of part (a), what is the temperature \(\left(T_{o}\right)\) of the duct surface next to the heater? (c) With \(T_{i}=85^{\circ} \mathrm{C}\), compute and plot \(q_{o}^{\prime \prime}\) and \(T_{o}\) as a function of the air-side convection coefficient \(h\) for the range \(10 \leq h \leq 200 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Briefly discuss your results.

A glass window of width \(W=1 \mathrm{~m}\) and height \(H=2 \mathrm{~m}\) is \(5 \mathrm{~mm}\) thick and has a thermal conductivity of \(k_{g}=\) \(1.4 \mathrm{~W} / \mathrm{m}=\mathrm{K}\). If the inner and outer surface temperatures of the glass are \(15^{\circ} \mathrm{C}\) and \(-20^{\circ} \mathrm{C}\), respectively, on a cold winter day, what is the rate of heat loss through the glass? To reduce heat loss through windows, it is customary to use a double pane construction in which adjoining panes are separated by an air space. If the spacing is \(10 \mathrm{~mm}\) and the glass surfaces in contact with the air have temperatures of \(10^{\circ} \mathrm{C}\) and \(-15^{\circ} \mathrm{C}\), what is the rate of heat loss from a \(1 \mathrm{~m} \times 2 \mathrm{~m}\) window? The themal conductivity of air is \(k_{a}=0.024 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\).

An overhead 25-m-long, uninsulated industrial steam pipe of \(100-\mathrm{mm}\) diameter is routed through a building whose walls and air are at \(25^{\circ} \mathrm{C}\). Pressurized steam maintains a pipe surface temperature of \(150^{\circ} \mathrm{C}\), and the coefficient associated with natural convection is \(h=10\) \(\mathrm{W} / \mathrm{m}^{2} \cdot \mathrm{K}\). The surface emissivity is \(\varepsilon=0.8\). (a) What is the rate of heat loss from the steam line? (b) If the steam is generated in a gas-fired boiler operating at an efficiency of \(\eta_{f}=0.90\) and natural gas is priced at \(C_{g}=\$ 0.02\) per \(\mathrm{MJ}\), what is the annual cost of heat loss from the line?

An electric resistance heater is embedded in a long cylinder of diameter \(30 \mathrm{~mm}\). When water with a temperature of \(25^{\circ} \mathrm{C}\) and velocity of \(1 \mathrm{~m} / \mathrm{s}\) flows crosswise over the cylinder, the power per unit length required to maintain the surface at a uniform temperature of \(90^{\circ} \mathrm{C}\) is \(28 \mathrm{~kW} / \mathrm{m}\). When air, also at \(25^{\circ} \mathrm{C}\), but with a velocity of \(10 \mathrm{~m} / \mathrm{s}\) is flowing, the power per unit length required to maintain the same surface temperature is \(400 \mathrm{~W} / \mathrm{m}\). Calculate and compare the convection coefficients for the flows of water and air.
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