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Chapter 4: Problem 1

What is the physical significance of the Biot number? Is the Biot number more likely to be larger for highly conducting solids or poorly conducting ones?

Short Answer

Expert verified
Answer: A large Biot number indicates that the internal resistance to heat conduction within the object is higher than the convective resistance at the surface, making convective heat transfer more dominant in the heat transfer process. This is typically observed in poorly conducting solids.

Step by step solution

01

Definition of Biot Number

The Biot number (Bi) is a dimensionless quantity that helps in determining the importance of conduction within an object compared to convection on its surface. It is defined as the ratio of the internal thermal resistance of the object to the convective thermal resistance at the surface. Mathematically, it can be expressed as: Bi = \(\frac{hL_c}{k}\) where: - h is the heat transfer coefficient due to convection at the surface (W/m²·K) - Lc is the characteristic length of the object (m) - k is the thermal conductivity of the solid material (W/m·K)
02

Physical Significance

The physical significance of the Biot number lies in its ability to give information about the dominant mode of heat transfer within a system. A small Biot number (typically less than 0.1) indicates that the internal resistance to heat conduction within the object is much smaller than the convective resistance at the surface, which means that conduction is more dominant. In the case of large Biot numbers (typically greater than 10), the thermal resistance within the object is higher than the convective resistance at the surface, making convection more dominant in the heat transfer process.
03

Biot Number for Highly and Poorly Conducting Solids

Now, let's analyze the relationship between the Biot number and the thermal conductivity of solids. From the formula, we can see that the Biot number is inversely proportional to the thermal conductivity (k). As the thermal conductivity increases, the Biot number decreases and vice versa. For highly conducting solids, the thermal conductivity (k) is large, resulting in a smaller Biot number. This implies that heat conduction is more dominant in highly conducting solids as the internal resistance is lower compared to convective heat transfer at the surface. On the other hand, for poorly conducting solids, the thermal conductivity (k) is small, leading to a larger Biot number. This implies that convective heat transfer at the surface is more dominant in these materials, and the internal resistance to heat conduction is higher. To conclude, the Biot number is more likely to be larger for poorly conducting solids than for highly conducting ones.

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Key Concepts

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

Thermal Resistance
This resistance becomes particularly important when considering the insulation properties of materials. Higher thermal resistance implies better insulation, meaning less heat is lost through the material. In the context of the Biot number, it's the internal thermal resistance that we are concerned with—which represents the resistance to heat conduction within a solid object. The comparison between this internal resistance and the resistance posed by convection at the surface of the object helps to predict which mode of heat transfer—conduction or convection—will be more significant.
Convection Heat Transfer
The rate of convective heat transfer is characterized by the heat transfer coefficient, denoted by 'h'. A larger 'h' value means the surface is more effective at transferring heat through convection. This coefficient is affected by several factors, including fluid velocity, viscosity, and the temperature difference between the surface and the fluid.
Thermal Conductivity
In quantitative terms, thermal conductivity is the rate at which heat passes through a material with a given area and temperature gradient. In the context of our problem, understanding thermal conductivity is vital. It directly influences the Biot number and thus impacts whether convection or conduction is the predominant mode of heat transfer in a system.
Dimensionless Numbers in Heat Transfer
There are other important dimensionless numbers in heat transfer as well, such as the Reynolds number which indicates the flow regime of a fluid, and the Prandtl number, which relates the thickness of the thermal boundary layer to the velocity boundary layer. These numbers are used extensively in designing and analyzing systems where heat transfer is crucial, such as radiators, heat exchangers, and cooling systems in electronics.

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Most popular questions from this chapter

A 6-mm-thick stainless steel strip \((k=21 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\), \(\rho=8000 \mathrm{~kg} / \mathrm{m}^{3}\), and \(\left.c_{p}=570 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right)\) exiting an oven at a temperature of \(500^{\circ} \mathrm{C}\) is allowed to cool within a buffer zone distance of \(5 \mathrm{~m}\). To prevent thermal burn to workers who are handling the strip at the end of the buffer zone, the surface temperature of the strip should be cooled to \(45^{\circ} \mathrm{C}\). If the air temperature in the buffer zone is \(15^{\circ} \mathrm{C}\) and the convection heat transfer coefficient is \(120 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), determine the maximum speed of the stainless steel strip.

The walls of a furnace are made of \(1.2\)-ft-thick concrete \(\left(k=0.64 \mathrm{Btu} / \mathrm{h} \cdot \mathrm{ft} \cdot{ }^{\circ} \mathrm{F}\right.\) and \(\left.\alpha=0.023 \mathrm{ft}^{2} / \mathrm{h}\right)\). Initially, the furnace and the surrounding air are in thermal equilibrium at \(70^{\circ} \mathrm{F}\). The furnace is then fired, and the inner surfaces of the furnace are subjected to hot gases at \(1800^{\circ} \mathrm{F}\) with a very large heat transfer coefficient. Determine how long it will take for the temperature of the outer surface of the furnace walls to rise to \(70.1^{\circ} \mathrm{F}\). Answer: \(116 \mathrm{~min}\)

A person puts a few apples into the freezer at \(-15^{\circ} \mathrm{C}\) to cool them quickly for guests who are about to arrive. Initially, the apples are at a uniform temperature of \(20^{\circ} \mathrm{C}\), and the heat transfer coefficient on the surfaces is \(8 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Treating the apples as 9 -cm-diameter spheres and taking their properties to be \(\rho=840 \mathrm{~kg} / \mathrm{m}^{3}, c_{p}=3.81 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}, k=0.418 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\), and \(\alpha=1.3 \times 10^{-7} \mathrm{~m}^{2} / \mathrm{s}\), determine the center and surface temperatures of the apples in \(1 \mathrm{~h}\). Also, determine the amount of heat transfer from each apple. Solve this problem using analytical one-term approximation method (not the Heisler charts).

During a fire, the trunks of some dry oak trees (es) \(\left(k=0.17 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\right.\) and \(\left.\alpha=1.28 \times 10^{-7} \mathrm{~m}^{2} / \mathrm{s}\right)\) that are initially at a uniform temperature of \(30^{\circ} \mathrm{C}\) are exposed to hot gases at \(520^{\circ} \mathrm{C}\) for a period of \(5 \mathrm{~h}\), with a heat transfer coefficient of \(65 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) on the surface. The ignition temperature of the trees is \(410^{\circ} \mathrm{C}\). Treating the trunks of the trees as long cylindrical rods of diameter \(20 \mathrm{~cm}\), determine if these dry trees will ignite as the fire sweeps through them. Solve this problem using analytical one-term approximation method (not the Heisler charts).

A long 18-cm-diameter bar made of hardwood \(\left(k=0.159 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, \alpha=1.75 \times 10^{-7} \mathrm{~m}^{2} / \mathrm{s}\right)\) is exposed to air at \(30^{\circ} \mathrm{C}\) with a heat transfer coefficient of \(8.83 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). If the center temperature of the bar is measured to be \(15^{\circ} \mathrm{C}\) after a period of 3-hours, the initial temperature of the bar is (a) \(11.9^{\circ} \mathrm{C}\) (b) \(4.9^{\circ} \mathrm{C}\) (c) \(1.7^{\circ} \mathrm{C}\) (d) \(0^{\circ} \mathrm{C}\) (e) \(-9.2^{\circ} \mathrm{C}\)
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