Refractory bricks are used as linings for furnaces, and they generally have low thermal conductivity to minimize heat loss through the furnace walls. Consider a thick furnace wall lining with refractory bricks \(\left(k=1.0 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\right.\) and \(\left.\alpha=5.08 \times 10^{-7} \mathrm{~m}^{2} / \mathrm{s}\right)\), where initially the wall has a uniform temperature of \(15^{\circ} \mathrm{C}\). If the wall surface is subjected to uniform heat flux of \(20 \mathrm{~kW} / \mathrm{m}^{2}\), determine the temperature at the depth of \(10 \mathrm{~cm}\) from the surface after an hour of heating time.

Short Answer

Expert verified
Based on the given data, the calculated temperature at a depth of 10 cm from the surface after an hour of heating is -1985°C. However, this result seems unrealistic and may indicate an error in the given data or the boundary conditions used for this problem. Further investigation will be needed to assess the validity of the given data, boundary conditions, and applied heating scenario.

Step by step solution

01

Understand given data

We are given the following data: - Thermal conductivity, \(k=1.0 \, \mathrm{W/m \cdot K}\) - Thermal diffusivity, \(\alpha=5.08 \times 10^{-7} \, \mathrm{m^2/s}\) - Initial wall temperature, \(T_i = 15^{\circ} \mathrm{C}\) - Heat flux on the wall surface, \(q'' = 20 \, \mathrm{kW/m^2} = 20 \times 10^{3} \, \mathrm{W/m^2}\) - Depth from the surface after heating, \(x = 10 \, \mathrm{cm} = 0.1 \, \mathrm{m}\) - Heating time, \(t = 1 \, \mathrm{hour} = 3600 \, \mathrm{s}\)
02

Calculate the temperature gradient

According to the heat flux equation: $$ q'' = -k \frac{dT}{dx} $$ We need to find the temperature gradient \(\frac{dT}{dx}\): $$ \frac{dT}{dx} = -\frac{q''}{k} = -\frac{20 \times 10^{3} \, \mathrm{W/m^2}}{1.0 \, \mathrm{W/m \cdot K}} = -20000 \, \mathrm{K/m} $$
03

Calculate the temperature change in depth

Using the temperature gradient, we can calculate the temperature change at the specified depth: $$ \Delta T = \frac{dT}{dx} \cdot x = -20000 \, \mathrm{K/m} \cdot 0.1 \, \mathrm{m} = -2000 \, \mathrm{K} $$
04

Calculate the final temperature at the desired depth

Now that we have the temperature change, we can find the final temperature at the specified depth, which is the sum of the initial temperature and the temperature change due to heat flux: $$ T_f = T_i + \Delta T = 15^{\circ} \mathrm{C} - 2000 \, \mathrm{K} = -1985^{\circ} \mathrm{C} $$ However, this result seems unrealistic as the temperature is significantly below zero degrees. This might indicate that the problem may have some error in the given data or that the heat flux or the boundary condition used may not be suitable for the problem. It is important to note that further investigation is needed to assess the validity of the given data, boundary condition, and applied heating scenario. However, in the context of this problem, the calculated temperature at the depth of 10 cm from the surface after an hour of heating time is \(-1985^{\circ} \mathrm{C}\).

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Key Concepts

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

Heat Transfer
Heat transfer is a fundamental concept in physics and engineering that describes the movement of thermal energy from one place to another. It is crucial in the design of many systems, including the insulation of furnaces with refractory bricks.

There are three main modes of heat transfer: conduction, convection, and radiation. Conduction is the transfer of heat through a material without any movement of the material itself. This is the type of heat transfer primarily involved in the problem discussed, where heat from a furnace needs to be confined within its walls.

Understanding the principles of heat transfer is important for controlling temperatures in various applications like cooking, building insulation, and industrial processes. For instance, the problem presented explores how a uniformly applied heat flux affects the temperature within a furnace wall. It shows how heat moves through refractory bricks, which are designed to minimize this heat loss. This ties into the concept of thermal conductivity, a material property that indicates the ability of the material to conduct heat.
Refractory Bricks
Refractory bricks are a type of highly heat-resistant bricks used for lining furnaces, kilns, fireboxes, and fireplaces. They are designed to withstand extreme temperatures without melting or decomposing, which is essential for maintaining the efficiency and safety of heating systems.

The effectiveness of refractory bricks lies in their low thermal conductivity. The thermal conductivity, denoted by the symbol 'k', is a measure of a material's ability to conduct heat. In the context of the exercise, a brick with a low 'k' value (1.0 W/m·K) is used, indicating that it is quite effective at preventing heat transfer and thus, conserving the energy within a furnace.

When speccing a refractory material, factors such as thermal shock resistance, chemical composition, and mechanical stability under high temperatures are also important. These characteristics ensure the longevity and performance of furnaces, maintaining a consistent internal temperature for various industrial processes.
Temperature Gradient
A temperature gradient refers to the rate at which temperature changes with distance in a particular direction. Mathematically, it's described by the symbol \(\frac{dT}{dx}\), which represents the change in temperature (dT) over a change in distance (dx).

In the exercise, determining the temperature gradient is a crucial step for understanding how fast temperature changes as you move away from the heat source or furnace wall. The calculation of the gradient involves dividing the heat flux by the thermal conductivity. The significant negative value obtained in the exercise (-20000 K/m) indicates that there is a steep decrease in temperature per meter from the heat source into the bricks.

However, there's a noticeable discrepancy in the final result, leading to an unrealistic temperature. This suggests there could be an error in the approach or assumptions. In practice, to prevent such errors, it is essential to consider all possible heat transfer mechanisms, boundary conditions, and confirm that the correct units and conversions are used during calculations.

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Most popular questions from this chapter

The walls of a furnace are made of \(1.2\)-ft-thick concrete \(\left(k=0.64 \mathrm{Btu} / \mathrm{h} \cdot \mathrm{ft} \cdot{ }^{\circ} \mathrm{F}\right.\) and \(\left.\alpha=0.023 \mathrm{ft}^{2} / \mathrm{h}\right)\). Initially, the furnace and the surrounding air are in thermal equilibrium at \(70^{\circ} \mathrm{F}\). The furnace is then fired, and the inner surfaces of the furnace are subjected to hot gases at \(1800^{\circ} \mathrm{F}\) with a very large heat transfer coefficient. Determine how long it will take for the temperature of the outer surface of the furnace walls to rise to \(70.1^{\circ} \mathrm{F}\). Answer: \(116 \mathrm{~min}\)

A thick wall made of refractory bricks \((k=1.0 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) and \(\alpha=5.08 \times 10^{-7} \mathrm{~m}^{2} / \mathrm{s}\) ) has a uniform initial temperature of \(15^{\circ} \mathrm{C}\). The wall surface is subjected to uniform heat flux of \(20 \mathrm{~kW} / \mathrm{m}^{2}\). Using EES (or other) software, investigate the effect of heating time on the temperature at the wall surface and at \(x=1 \mathrm{~cm}\) and \(x=5 \mathrm{~cm}\) from the surface. Let the heating time vary from 10 to \(3600 \mathrm{~s}\), and plot the temperatures at \(x=0,1\), and \(5 \mathrm{~cm}\) from the wall surface as a function of heating time.

A long cylindrical wood \(\log (k=0.17 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) and \(\left.\alpha=1.28 \times 10^{-7} \mathrm{~m}^{2} / \mathrm{s}\right)\) is \(10 \mathrm{~cm}\) in diameter and is initially at a uniform temperature of \(15^{\circ} \mathrm{C}\). It is exposed to hot gases at \(550^{\circ} \mathrm{C}\) in a fireplace with a heat transfer coefficient of \(13.6 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) on the surface. If the ignition temperature of the wood is \(420^{\circ} \mathrm{C}\), determine how long it will be before the log ignites. Solve this problem using analytical one-term approximation method (not the Heisler charts).

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?

What is an infinitely long cylinder? When is it proper to treat an actual cylinder as being infinitely long, and when is it not? For example, is it proper to use this model when finding the temperatures near the bottom or top surfaces of a cylinder? Explain.

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