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The Biot number during a heat transfer process between a sphere and its surroundings is determined to be \(0.02\). Would you use lumped system analysis or the transient temperature charts when determining the midpoint temperature of the sphere? Why?

Short Answer

Expert verified
Answer: Given the Biot number of 0.02, which is less than 0.1, we should use lumped system analysis for the heat transfer process. This is because the rate of heat conduction within the sphere is much greater than the rate of heat transfer on the surface, making the temperature gradients within the sphere negligible.

Step by step solution

01

Understanding the Biot number

The Biot number (Bi) is a dimensionless parameter used in heat transfer problems to characterize the relative importance of heat conduction within an object as compared to heat transfer across the object's surface. It is defined as the ratio of the thermal resistance of heat conduction within an object to the thermal resistance of heat convection on the object's surface. Mathematically, the Biot number is given by: Bi = \(hL_c/k\), where \(h\) is the convective heat transfer coefficient, \(L_c\) is the characteristic length (one common choice for spheres is the radius), and \(k\) is the thermal conductivity of the material.
02

Criteria for lumped system analysis

Lumped system analysis assumes that the temperature inside the object is uniform at any given time, and that the heat transfer takes place only at the object's surface. This simplifies the analysis by reducing the problem to a single time-dependent variable (the temperature of the object) rather than a time- and space-dependent variable. In general, lumped system analysis can be used when the Biot number is small (Bi \(<\) 0.1), which implies that the rate of heat conduction within the object is much greater than the rate of heat transfer on the surface. In other words, the temperature gradients inside the object are insignificant compared to the gradients at the surface.
03

Criteria for transient temperature charts

Transient temperature charts, also known as Heisler charts, are used to characterize heat transfer in objects where the temperature does vary meaningfully within the object at a given time. These charts are used to estimate the temperature profiles inside an object during the heat transfer process. Transient temperature charts are typically used when the Biot number is greater than 0.1, which indicates that the temperature gradients within the object are not negligible and must be considered for an accurate analysis.
04

Solution based on the Biot number

In this exercise, the given Biot number during the heat transfer process between the sphere and its surroundings is 0.02. Since the Biot number is less than 0.1, we would use lumped system analysis to determine the midpoint temperature of the sphere. This is because the rate of heat conduction within the sphere is much greater than the rate of heat transfer on the surface, and therefore the temperature gradients within the sphere are negligible.

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Key Concepts

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

Biot Number
The Biot Number is a crucial aspect in understanding heat transfer in solids. It essentially helps us determine how well heat from the surrounding environment can penetrate an object and change its internal temperature. The Biot Number is a dimensionless parameter defined by the formula: \[ \text{Bi} = \frac{hL_c}{k} \] where:
  • \( h \) is the convective heat transfer coefficient
  • \( L_c \) is the characteristic length of the object
  • \( k \) is the thermal conductivity of the object
For spheres, the characteristic length is often the radius. A smaller Biot number (less than 0.1) indicates that heat conduction inside the object is much faster than heat transfer at the surface. This implies a relatively uniform temperature within the object, making detailed spatial analyses unnecessary for understanding its thermal behavior.
Heat Transfer
Heat transfer is the movement of thermal energy from one physical system to another. It can occur in three primary modes: conduction, convection, and radiation. For solids, especially those involved in environments where they are heated or cooled, conduction and convection are generally the most relevant forms. Conduction occurs within solids or stationary fluids where heat is transferred through molecular agitation and free electron diffusion. In your exercise, the primary focus is on conduction within the sphere and convection at its surface.
Convection is the process where heat is carried away from the surface of an object due to the motion of a fluid (like air or water) across its surface. The efficiency of this process is partially described by the convective heat transfer coefficient \( h \), which is used in calculating the Biot Number.
Transient Temperature Charts
Transient Temperature Charts, or Heisler Charts, are tools used to estimate the temperature distribution within solids over time during heat transfer situations. These charts are especially useful when analyzing objects where temperature gradients inside the material cannot be ignored.
Heisler Charts provide a graphical solution to the transient heat conduction equations for simple geometries such as infinite plates, cylinders, and spheres. When the Biot number is above 0.1, implying significant temperature variation inside the object, Heisler Charts come into play to simplify solving for the temperature at various points and times.
Sphere Heat Transfer
Sphere heat transfer analysis is different from that of other geometric shapes due to its symmetry. In a spherical object, heat conduction processes are often simplified because the path for heat flow is identical in all radial directions.
The symmetry ensures a more uniform distribution of temperature change when exposed to an external thermodynamic process, such as heating. When utilizing the lumped system analysis for spheres (when Bi < 0.1), it assumes that the temperature throughout the sphere is effectively uniform. This significantly simplifies calculations by treating the sphere as a whole with a single temperature variable, rather than having to consider complex internal thermal profiles.
This symmetry, combined with a low Biot number, means temperature can be treated as if it changes uniformly across the entire volume of the sphere, easing the analysis considerably.

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Most popular questions from this chapter

Consider a 7.6-cm-long and 3-cm-diameter cylindrical lamb meat chunk \(\left(\rho=1030 \mathrm{~kg} / \mathrm{m}^{3}, c_{p}=3.49 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}\right.\), \(\left.k=0.456 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, \alpha=1.3 \times 10^{-7} \mathrm{~m}^{2} / \mathrm{s}\right)\). Fifteen such meat chunks initially at \(2^{\circ} \mathrm{C}\) are dropped into boiling water at \(95^{\circ} \mathrm{C}\) with a heat transfer coefficient of \(1200 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). The amount of heat transfer during the first 8 minutes of cooking is (a) \(71 \mathrm{~kJ}\) (b) \(227 \mathrm{~kJ}\) (c) \(238 \mathrm{~kJ}\) \(\begin{array}{ll}\text { (d) } 269 \mathrm{~kJ} & \text { (e) } 307 \mathrm{~kJ}\end{array}\)

Consider a cubic block whose sides are \(5 \mathrm{~cm}\) long and a cylindrical block whose height and diameter are also \(5 \mathrm{~cm}\). Both blocks are initially at \(20^{\circ} \mathrm{C}\) and are made of granite \((k=\) \(2.5 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) and \(\left.\alpha=1.15 \times 10^{-6} \mathrm{~m}^{2} / \mathrm{s}\right)\). Now both blocks are exposed to hot gases at \(500^{\circ} \mathrm{C}\) in a furnace on all of their surfaces with a heat transfer coefficient of \(40 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Determine the center temperature of each geometry after 10,20 , and \(60 \mathrm{~min}\).

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.

A thick wood slab \((k=0.17 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) and \(\alpha=1.28 \times\) \(10^{-7} \mathrm{~m}^{2} / \mathrm{s}\) ) that is initially at a uniform temperature of \(25^{\circ} \mathrm{C}\) is exposed to hot gases at \(550^{\circ} \mathrm{C}\) for a period of \(5 \mathrm{~min}\). The heat transfer coefficient between the gases and the wood slab is \(35 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). If the ignition temperature of the wood is \(450^{\circ} \mathrm{C}\), determine if the wood will ignite.

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 \(70^{\circ} \mathrm{C}\) ?

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