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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/m虏路K) - Lc is the characteristic length of the object (m) - k is the thermal conductivity of the solid material (W/m路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鈥攚hich 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鈥攃onduction or convection鈥攚ill 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

Spherical glass beads coming out of a kiln are allowed to \(c o o l\) in a room temperature of \(30^{\circ} \mathrm{C}\). A glass bead with a diameter of \(10 \mathrm{~mm}\) and an initial temperature of \(400^{\circ} \mathrm{C}\) is allowed to cool for 3 minutes. If the convection heat transfer coefficient is \(28 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), determine the temperature at the center of the glass bead using \((a)\) Table 4-2 and \((b)\) the Heisler chart (Figure 4-19). The glass bead has properties of \(\rho=\) \(2800 \mathrm{~kg} / \mathrm{m}^{3}, c_{p}=750 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\), and \(k=0.7 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\).

The chilling room of a meat plant is \(15 \mathrm{~m} \times 18 \mathrm{~m} \times\) \(5.5 \mathrm{~m}\) in size and has a capacity of 350 beef carcasses. The power consumed by the fans and the lights in the chilling room are 22 and \(2 \mathrm{~kW}\), respectively, and the room gains heat through its envelope at a rate of \(14 \mathrm{~kW}\). The average mass of beef carcasses is \(220 \mathrm{~kg}\). The carcasses enter the chilling room at \(35^{\circ} \mathrm{C}\), after they are washed to facilitate evaporative cooling, and are cooled to \(16^{\circ} \mathrm{C}\) in \(12 \mathrm{~h}\). The air enters the chilling room at \(-2.2^{\circ} \mathrm{C}\) and leaves at \(0.5^{\circ} \mathrm{C}\). Determine \((a)\) the refrigeration load of the chilling room and \((b)\) the volume flow rate of air. The average specific heats of beef carcasses and air are \(3.14\) and \(1.0 \mathrm{~kJ} / \mathrm{kg} \cdot{ }^{\circ} \mathrm{C}\), respectively, and the density of air can be taken to be \(1.28 \mathrm{~kg} / \mathrm{m}^{3}\).

A stainless steel slab \((k=14.9 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) and \(\alpha=\) \(\left.3.95 \times 10^{-6} \mathrm{~m}^{2} / \mathrm{s}\right)\) and a copper slab \((k=401 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) and \(\alpha=117 \times 10^{-6} \mathrm{~m}^{2} / \mathrm{s}\) ) are placed under an array of laser diodes, which supply an energy pulse of \(5 \times 10^{7} \mathrm{~J} / \mathrm{m}^{2}\) instantaneously at \(t=0\) to both materials. The two slabs have a uniform initial temperature of \(20^{\circ} \mathrm{C}\). Using \(\mathrm{EES}\) (or other) software, investigate the effect of time on the temperatures of both materials at the depth of \(5 \mathrm{~cm}\) from the surface. By varying the time from 1 to \(80 \mathrm{~s}\) after the slabs have received the energy pulse, plot the temperatures at \(5 \mathrm{~cm}\) from the surface as a function of time.

A steel casting cools to 90 percent of the original temperature difference in \(30 \mathrm{~min}\) in still air. The time it takes to cool this same casting to 90 percent of the original temperature difference in a moving air stream whose convective heat transfer coefficient is 5 times that of still air is (a) \(3 \mathrm{~min}\) (b) \(6 \mathrm{~min}\) (c) \(9 \mathrm{~min}\) (d) \(12 \mathrm{~min}\) (e) \(15 \mathrm{~min}\)

Carbon steel balls ( \(\rho=7830 \mathrm{~kg} / \mathrm{m}^{3}, k=64 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\), \(\left.c_{p}=434 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right)\) initially at \(150^{\circ} \mathrm{C}\) are quenched in an oil bath at \(20^{\circ} \mathrm{C}\) for a period of 3 minutes. If the balls have a diameter of \(5 \mathrm{~cm}\) and the convection heat transfer coefficient is \(450 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). The center temperature of the balls after quenching will be (Hint: Check the Biot number). (a) \(27.4^{\circ} \mathrm{C}\) (b) \(143^{\circ} \mathrm{C}\) (c) \(12.7^{\circ} \mathrm{C}\) (d) \(48.2^{\circ} \mathrm{C}\) (e) \(76.9^{\circ} \mathrm{C}\)
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