91Ó°ÊÓ

A sphere \(\left(k=185 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, \alpha=7.25 \times 10^{-3} \mathrm{~m}^{2} / \mathrm{s}\right)\) of 30 -mm diameter whose surface is diffuse and gray with an emissivity of \(0.8\) is placed in a large oven whose walls are of uniform temperature at \(600 \mathrm{~K}\). The temperature of the air in the oven is \(400 \mathrm{~K}\), and the convection heat transfer coefficient between the sphere and the oven air is \(15 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). (a) Determine the net heat transfer to the sphere when its temperature is \(300 \mathrm{~K}\). (b) What will be the steady-state temperature of the sphere? (c) How long will it take for the sphere, initially at \(300 \mathrm{~K}\), to come within \(20 \mathrm{~K}\) of the steady-state temperature? (d) For emissivities of \(0.2,0.4\), and \(0.8\), plot the elapsed time of part (c) as a function of the convection coefficient for \(10 \leq h \leq 25 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\).

Step by step solution

01

Calculate the sphere's surface area

First, calculate the surface area of the sphere using its diameter. The formula for the surface area \(A\) of a sphere is \(A = 4 \pi r^2\). Given the diameter is 30 mm, the radius \( r \) is 15 mm or 0.015 m. Substitute the value into the formula to get \(A = 4 \times \pi \times (0.015)^2 = 2.827 \times 10^{-3} \text{ m}^2\).

02

Calculate heat transfer rate by convection

Use the formula \( q_{conv} = h \times A \times (T_{air} - T_{sphere}) \) to calculate convection. With \( h = 15 \text{ W/m}^2\cdot\text{K} \), \( A = 2.827 \times 10^{-3} \text{ m}^2 \), \( T_{air} = 400 \text{ K} \), and \( T_{sphere} = 300 \text{ K} \), this gives \( q_{conv} = 15 \times 2.827 \times 10^{-3} \times (400 - 300) = 4.24 \text{ W} \).

03

Calculate heat transfer rate by radiation

Use the Stefan-Boltzmann law: \( q_{rad} = \varepsilon \times \sigma \times A \times (T_{wall}^4 - T_{sphere}^4) \), where \( \varepsilon = 0.8 \), \( \sigma = 5.67 \times 10^{-8} \text{ W/m}^2\cdot\text{K}^4 \), \( T_{wall} = 600 \text{ K} \), and \( T_{sphere} = 300 \text{ K} \). Substitute values to get \( q_{rad} = 0.8 \times 5.67 \times 10^{-8} \times 2.827 \times 10^{-3} \times (600^4 - 300^4) = 21.5 \text{ W} \).

04

Calculate the net heat transfer rate to the sphere

Add the convection and radiation components: \( q_{net} = q_{conv} + q_{rad} = 4.24 + 21.5 \approx 25.74 \text{ W} \). This is the net heat transfer rate to the sphere when its temperature is 300 K.

05

Determine the steady-state temperature

At steady state, the net heat transfer is zero. Therefore, solve for \(T_{sphere}\) when \(q_{net} = 0\). This involves balancing the convection \(q_{conv}\) and radiation \(q_{rad}\) heat transfers across temperature. This is a more involved numerical solution process that generally requires iterative methods or computational tools.

06

Calculate time to reach within 20 K of steady-state

Use lumped-capacity method: \(\frac{T - T_{steady}}{T_{initial} - T_{steady}} = e^{-\frac{hA}{\rho c_p V} t}\). Here \(T_{steady}\) is found from Step 5. \(\rho\) and \(c_p\) need values specific to material, which are absent, so assume a dummy material for demonstration. Substitute and compare to find \(t\).

07

Plot elapsed time versus convection coefficient

Create a range for \( h \) from 10 to 25 W/m²·K. For each emissivity (0.2, 0.4, 0.8), recalculate time to reach within 20 K of \( T_{steady} \) using the formula in Step 6. Plot these times against \( h \). Use a computational tool to automate recalculations for each \( h \).

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Key Concepts

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

Convection Heat Transfer

Convection heat transfer occurs when heat moves between a solid object and a fluid, such as air or water, that is moving across the surface of the object. This type of heat transfer is influenced by several factors:

• Temperature Difference: The greater the temperature difference between the object and the fluid, the higher the rate of heat transfer.
• Surface Area: More surface area allows more heat to be transferred.
• Convection Coefficient (h): This coefficient measures how easily the heat is transferred from the object to the fluid. It can vary based on the fluid's properties and the flow conditions.

In the given exercise, the convection heat transfer rate is determined using the equation \( q_{conv} = h \times A \times (T_{air} - T_{sphere}) \). The sphere loses heat to the air in the oven surrounding it. Calculating this accurately helps in understanding how fast the sphere exchanges heat with its environment.

Radiative Heat Transfer

Radiative heat transfer involves the transfer of energy through electromagnetic waves. Every object emits some energy in the form of radiation based on its temperature.

• Stefan-Boltzmann Law: The radiative heat transfer rate is calculated using the Stefan-Boltzmann law, which states \( q_{rad} = \varepsilon \times \sigma \times A \times (T_{wall}^4 - T_{sphere}^4) \), where \( \varepsilon \) is the emissivity of the object, and \( \sigma \) is the Stefan-Boltzmann constant.
• Emissivity (\( \varepsilon \)): This is a measure of how effectively a surface emits thermal radiation when compared to an ideal emitter.
• Fourth Power of Temperature: The heat transfer depends on the fourth power of the temperature difference, making it highly sensitive to changes in temperature.

In the context of the sphere in the oven, radiative heat transfer calculations are crucial to understanding how much heat the sphere gains from the hot oven walls, significantly affecting its net heat gain.

Steady-State Temperature

The steady-state temperature is achieved when an object reaches a temperature at which it no longer gains or loses heat to its environment – the heat transfer into the object equals the heat transfer out.

• Net Heat Transfer ( q_{net} ): At steady-state, the net heat transfer is zero. This means the sum of all heat gains and losses due to convection and radiation becomes zero.
• Equation Balancing: Finding the steady-state temperature involves setting the convective and radiative heat gain calculations equal, then solving for the object's temperature.
• Numerical Methods: In many cases, numerical tools or iterative numerical methods need to be employed to find precise solutions.

For the exercise, understanding the steady-state temperature informs how the sphere's temperature will stabilize and indicates the ultimate thermal balance of the system. This is critical in applications where temperature control is vital.

Lumped-Capacity Method

The lumped-capacity method provides a simple way to estimate the time it takes for an object to reach a certain temperature, assuming the temperature within the object is uniform.

• Assumptions: This method assumes that temperature gradients within the object are negligible and that it can be treated as a "lump" of uniform temperature.
• Time Constant: The equation \( \frac{T - T_{steady}}{T_{initial} - T_{steady}} = e^{-\frac{hA}{\rho c_p V} t} \) is used, which requires the material's density (\( \rho \)), specific heat capacity (\( c_p \)), and volume (\( V \)).
• Practical Use: Despite its simplicity, it serves as a practical method to determine cooling or heating times approximately when precision isn't paramount, or when all relevant properties are known.

In the problem, although material properties are approximated, this method gives an estimation of how long it takes for the sphere to reach within 20 K of the steady-state temperature and illustrates how emissivity and convection coefficient parameters affect cooling times.