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Spheres \(\mathrm{A}\) and \(\mathrm{B}\) are initially at \(800 \mathrm{~K}\), and they are simultancously quenched in large constant temperature baths. each having a temperature of \(320 \mathrm{~K}\). The follow. ing parameters are associated with each of the spheres and their cooling processes. $$ \begin{array}{lcc} & \text { Sphere A } & \text { Sphere B } \\ \hline \text { Diameter }(\mathrm{mm}) & 300 & 30 \\ \text { Density }\left(\mathrm{kg} / \mathrm{m}^{2}\right) & 1600 & 400 \\ \text { Specific heat }(\mathrm{N} / \mathrm{kg} \cdot \mathrm{K}) & 0.400 & 1.60 \\ \text { Thermal conductivity }(\mathrm{W} / \mathrm{m} \cdot \mathrm{K}) & 170 & 1.70 \\ \text { Convection coefficient }\left(\mathrm{W} / \mathrm{m}^{2}-\mathrm{K}\right) & 5 & 50 \\ \hline \end{array} $$ (a) Show in a qualitative manner, on \(T\) versus \(t\) coordinates, the temperatures at the center and at the surface for each sphere as a function of time. Briefly explain the reasening by which you determine the relative positions of the curves, (b) Calculate the time required for the surface of each sphere to reach \(415 \mathrm{~K}\). (c) Determine the energy that has been gained by each of the baths during the process of the spheres cooling to \(415 \mathrm{~K}\).

Step by step solution

01

Understanding the Heat Transfer Mechanism

Both spheres are being quenched in a 320 K bath, indicating they undergo a cooling process governed by the heat transfer mechanisms. The heat transfer involves both conduction within the sphere and convection at the surface. For these spheres, the Biot number \( \text{Bi} = \frac{hL}{k} \) will help determine whether the assumptions of lumped capacitance methods apply or if we need to consider more complex conduction equations.

02

Qualitative Temperature-Time Graph

For Sphere A, with a larger diameter and lower convection coefficient, its cooling curve will be slower than Sphere B. The center will take longer to cool compared to the surface due to the large size. For Sphere B, being smaller and with a higher convection coefficient, the temperature both at the surface and center will decrease more rapidly, and the center approaches the surface temperature quicker compared to Sphere A.

03

Calculate Biot Number for Each Sphere

Calculate Biot number: \( \text{Bi} = \frac{hL}{k} \), where \( h \) is the convection coefficient, \( L \) is a characteristic length (radius), and \( k \) is the thermal conductivity. For Sphere A: \( L = \frac{300}{2} = 150 \ mm = 0.15 \ m \), hence \( \text{Bi}_A = \frac{5 \times 0.15}{170} \). For Sphere B: \( L = \frac{30}{2} = 15 \ mm = 0.015 \ m \), hence \( \text{Bi}_B = \frac{50 \times 0.015}{1.7} \).

04

Lumped Capacitance Assumption Validity

Check if \( \text{Bi} < 0.1 \) to use lumped capacitance. If so, the temperature change is uniform throughout the sphere. If \( \text{Bi} > 0.1 \), this approach isn't valid, and more detailed transient conduction solutions are needed. Based on calculations, only Sphere A qualifies for the lumped capacitance model.

05

Temperature-Time Equation for Sphere Surface Cooling

Using Newton's law of cooling, the formula \[ T(t) = T_{ ext{bath}} + (T_0 - T_{ ext{bath}})e^{-ht/( ho c_p V)} \] predicts temperature change. Compute using \( \rho \), \( c_p \) (specific heat), and \( V \) (volume based on sphere's diameter). Calculate time \( t \) to reach \( 415 \ \mathrm{K} \) for each sphere's surface.

06

Energy Transfer to Baths Calculation

Energy transferred \( Q \) can be calculated with \( Q = m c_p (T_{ ext{initial}} - T_{ ext{final}}) \). Calculate mass \( m \) (density \( \rho \) times volume \( V \)), then determine energy using specific heat \( c_p \), initial temperature \( 800 \ \mathrm{K} \), and final temperature \( 415 \ \mathrm{K} \). Perform these calculations for both spheres.

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Key Concepts

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

Cooling Process

The cooling process is essentially the decrease in temperature of an object as it releases heat to its surroundings. When spheres A and B are transferred into the cooler baths, their initial high temperatures begin to drop. The driving force behind this is the temperature difference between the sphere and the environment. This cooling continues until thermal equilibrium is achieved, meaning the temperature of the spheres matches the bath temperature of 320 K. In our exercise, the spheres initially have a temperature of 800 K, so this difference is significant.

During the cooling process, heat transfer mechanisms such as conduction and convection are active. These play crucial roles in how quickly each sphere reaches 415 K. Different materials and sphere sizes affect how heat is transferred, with larger volumes or low conductivity materials often cooling slower. Specific parameters of the sphere like diameter and thermal properties are crucial in determining how fast or slow this cooling process happens.

Heat Conduction

Heat conduction is the transfer of thermal energy within the sphere itself. This means heat moves from the hotter inner regions to cooler outer parts, trying to balance internal temperatures. In our scenario, we see this in Sphere A and B as they cool in their respective baths.

The rate of conduction depends on the material's thermal conductivity, denoted as 'k'. This property tells us how well heat travels through a material. For instance, Sphere A, with a thermal conductivity of 170 W/m·K, will conduct heat differently compared to Sphere B, which has a much lower conductivity of 1.70 W/m·K.

• Higher conductivity results in faster heat transfer within the object.
• Lower conductivity implies slower internal heat transfer, requiring more time for the entire object to cool uniformly.

Understanding conduction helps predict temperature differences between the center and surface of the spheres.

Convection

Convection is the process of heat transfer from the sphere's surface to the surrounding bath. It's an external mechanism, involving the movement of fluid particles around the sphere, which transfer the heat away.

This process is influenced by the convection coefficient 'h'. In our exercise, Sphere B has a higher convection coefficient of 50 W/m²·K compared to Sphere A's 5 W/m²·K. This implies:

• Sphere B will experience heat loss more swiftly due to a more efficient heat transfer to its surroundings.
• Sphere A, with a lower convection coefficient, cools slower as less heat per unit area is being transferred away at the same rate.

Convection's efficiency directly affects how quickly the outer temperature of a sphere adjusts to its new, cooler environment.

Biot Number

The Biot number is a dimensionless parameter critical for understanding heat transfer in the spheres. It helps determine whether assumptions such as the lumped capacitance model are valid for predicting heat transfer behavior.

The Biot number is calculated using the formula:\[\text{Bi} = \frac{hL}{k}\]where:

• h
is the convection coefficient,
• L
is a characteristic length, generally the radius for spherical objects, and
• k
is the material's thermal conductivity.

For instance, if the Biot number is less than 0.1, the lumped capacitance model can be used, indicating the temperature throughout the sphere changes uniformly. In our scenario:

• Sphere A, with Bi < 0.1, meets this requirement, enabling simpler calculations using uniform temperature assumptions.
• Sphere B has a Bi > 0.1, indicating that non-uniform temperature distribution needs to be considered for accurate predictions.

Therefore, calculating the Biot number aids in selecting the right heat transfer modeling approach.