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The Biot number can be thought of as the ratio of (a) The conduction thermal resistance to the convective thermal resistance. (b) The convective thermal resistance to the conduction thermal resistance. (c) The thermal energy storage capacity to the conduction thermal resistance. (d) The thermal energy storage capacity to the convection thermal resistance. (e) None of the above.

Short Answer

Expert verified
Answer: The Biot number can be thought of as the ratio of the conduction thermal resistance to the convective thermal resistance.

Step by step solution

01

Definition of Biot number

The Biot number (Bi) is a dimensionless quantity used in heat transfer calculations. It is defined as the ratio of the conductive thermal resistance inside the object to the convective thermal resistance at the surface of the object.
02

Option (a) Analysis

The Biot number can be thought of as the ratio of the conduction thermal resistance to the convective thermal resistance. This option is consistent with the definition of Biot number.
03

Option (b) Analysis

The Biot number can be thought of as the ratio of the convective thermal resistance to the conduction thermal resistance. This option is the reciprocal of the definition of Biot number, so this option is incorrect.
04

Option (c) Analysis

The Biot number can be thought of as the ratio of the thermal energy storage capacity to the conduction thermal resistance. This option involves thermal energy storage capacity, which is not part of the Biot number definition. Thus, this option is incorrect.
05

Option (d) Analysis

The Biot number can be thought of as the ratio of the thermal energy storage capacity to the convection thermal resistance. This option also involves thermal energy storage capacity, which is not part of the Biot number definition. Thus, this option is incorrect.
06

Option (e) Analysis

None of the above. Since option (a) is correct, this option is incorrect.
07

Conclusion

Based on the analysis of each option, we can conclude that the Biot number can be thought of as the ratio of the conduction thermal resistance to the convective thermal resistance, which corresponds to option (a).

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Key Concepts

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

Conductive Thermal Resistance
When we talk about conductive thermal resistance, we are referring to the opposition to heat flow through a solid material. Imagine you're holding a metal rod with one end in a flame. The heat doesn't instantly reach your hand because the metal resists the flow of heat to some degree. That resistance is measured by how easily heat is conducted, which depends on the material's thermal conductivity and thickness.

For example, in the case of a wall, the conductive resistance is calculated as the wall thickness divided by the product of its area and thermal conductivity. In mathematical terms, this is expressed as: \[ R_{conductive} = \frac{L}{kA} \] where \( L \) is the thickness, \( k \) is the thermal conductivity, and \( A \) is the area through which heat is being transferred. Higher thermal resistance means the material is a better insulator, hindering the flow of heat.
Convective Thermal Resistance
On the other side, we have convective thermal resistance, which relates to the resistance of heat transfer due to convection between a solid surface and a fluid (like air or water) in motion. Unlike conduction that occurs within the material itself, convection involves the movement of the fluid, which carries heat away.

The resistance to heat flow by convection can be characterized by the heat transfer coefficient. A low coefficient means high resistance and vice versa. The convective thermal resistance can be described by the equation: \[ R_{convective} = \frac{1}{hA} \] where \( h \) is the convective heat transfer coefficient, and \( A \) is the area of the object's surface. A large surface area or a high heat transfer coefficient indicates lower thermal resistance, leading to more efficient heat removal from the surface.
Heat Transfer Calculations
Moving on to heat transfer calculations, these are crucial for engineers and scientists when designing systems to manage thermal energy effectively. Calculations typically involve determining the rate at which heat is transferred, which in turn requires knowledge about both conductive and convective thermal resistances, among other factors.

In order to calculate the total thermal resistance in a system, one could sum up individual resistances, akin to electrical resistances in a series circuit. Engineers then use the concept of thermal resistance to determine how much heat can be transferred and at what rate, applying formulas like Fourier's law of heat conduction or Newton's law of cooling for convection. Clear understanding of these calculations is key for the optimization of heating, cooling, and insulation systems.
Dimensionless Quantity
Lastly, let's talk about the dimensionless quantity. 'Dimensionless' means that the quantity has no physical units associated with it; it's simply a ratio of two like units that cancel each other out. The Biot number is an excellent example of this.

Dimensionless numbers are used to help compare different systems and to scale experiments from laboratory size to full-scale applications. They also allow us to generalize results from specific cases, simplifying complex phenomena into more manageable forms. In the context of heat transfer, they guide the choice of appropriate models and are essential in the correlations that predict heat transfer rates. Understanding these numbers is significant for a unified and scale-independent analysis of heat transfer systems.

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Most popular questions from this chapter

How can we use the transient temperature charts when the surface temperature of the geometry is specified instead of the temperature of the surrounding medium and the convection heat transfer coefficient?

Consider a spherical shell satellite with outer diameter of \(4 \mathrm{~m}\) and shell thickness of \(10 \mathrm{~mm}\) is reentering the atmosphere. The shell satellite is made of stainless steel with properties of \(\rho=8238 \mathrm{~kg} / \mathrm{m}^{3}, c_{p}=468 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\), and \(k=13.4 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). During the reentry, the effective atmosphere temperature surrounding the satellite is \(1250^{\circ} \mathrm{C}\) with convection heat transfer coefficient of \(130 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). If the initial temperature of the shell is \(10^{\circ} \mathrm{C}\), determine the shell temperature after 5 minutes of reentry. Assume heat transfer occurs only on the satellite shell.

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.

What is the effect of cooking on the microorganisms in foods? Why is it important that the internal temperature of a roast in an oven be raised above \(70^{\circ} \mathrm{C}\) ?

4-115 A semi-infinite aluminum cylinder \((k=237 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\), \(\left.\alpha=9.71 \times 10^{-5} \mathrm{~m}^{2} / \mathrm{s}\right)\) of diameter \(D=15 \mathrm{~cm}\) is initially at a uniform temperature of \(T_{i}=115^{\circ} \mathrm{C}\). The cylinder is now placed in water at \(10^{\circ} \mathrm{C}\), where heat transfer takes place by convection with a heat transfer coefficient of \(h=140 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Determine the temperature at the center of the cylinder \(5 \mathrm{~cm}\) from the end surface 8 min after the start of cooling. 4-116 A 20-cm-long cylindrical aluminum block \((\rho=\) \(2702 \mathrm{~kg} / \mathrm{m}^{3}, c_{p}=0.896 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}, k=236 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\), and \(\alpha=\) \(\left.9.75 \times 10^{-5} \mathrm{~m}^{2} / \mathrm{s}\right), 15 \mathrm{~cm}\) in diameter, is initially at a uniform temperature of \(20^{\circ} \mathrm{C}\). The block is to be heated in a furnace at \(1200^{\circ} \mathrm{C}\) until its center temperature rises to \(300^{\circ} \mathrm{C}\). If the heat transfer coefficient on all surfaces of the block is \(80 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), determine how long the block should be kept in the furnace. Also, determine the amount of heat transfer from the aluminum block if it is allowed to cool in the room until its temperature drops to \(20^{\circ} \mathrm{C}\) throughout.

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