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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.

Short Answer

Expert verified
Answer: The key steps to analyze the effect of heating time on the temperature at different points within a refractory brick wall are: 1. Prepare for calculation in a compatible software. 2. Input given values and create variables for unknowns. 3. Calculate the temperature at different distances. 4. Vary the heating time and calculate temperatures. 5. Plot the temperatures as a function of heating time.

Step by step solution

01

Prepare for calculation in the software

Open your chosen software and ensure you are familiar with entering data and equations, as well as plotting data.
02

Input given values and create variables for unknowns

Enter the given values of thermal conductivity (k), thermal diffusivity (α), initial temperature (T_initial), and heat flux (q) into the software. Also, create variables for the temperature at different distances (T0, T1, T5) and heating times (t).
03

Calculate the temperature at different distances

Using the software's equation solver, apply the appropriate heat conduction equation to calculate the temperature at each distance (x = 0, 1, and 5 cm). You will need to modify the equation depending on the distance from the surface.
04

Vary the heating time and calculate temperatures

Now, create a loop (or similar function) in the software to vary the heating time from 10 to 3600 seconds, with a reasonable step size (such as 10 seconds). In each iteration of the loop, calculate the temperature at the given distances for the current heating time and store the results in a table or array.
05

Plot the temperatures as a function of heating time

Once the loop finishes and you have the calculated temperatures at each distance for every heating time, plot the results. Create a plot with heating time on the x-axis and temperature on the y-axis. Plot T0, T1, and T5 as separate curves to compare their temperature evolution over time. With these steps, you can study the effect of heating time on the temperature at the wall surface and at different distances from the surface. The plot will show how the temperature changes with heating time at each location.

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Key Concepts

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

Thermal Conductivity
Thermal conductivity refers to a material's ability to conduct heat. It is denoted by the symbol \( k \) and is measured in watts per meter-kelvin (W/m·K).
High thermal conductivity means heat passes through the material quickly, while low thermal conductivity means the material acts as an insulator.
  • In the given exercise, the thermal conductivity of the refractory brick wall is \( 1.0 \, \mathrm{W/m \cdot K} \).
  • This indicates a moderate ability to transfer heat, playing a crucial role in how the temperature changes within the wall over time.
Knowing \( k \) helps us understand how efficiently the wall will spread the heat applied to its surface, especially under constant heat flux conditions.
Heat Flux
Heat flux is the rate of heat energy transfer through a given surface per unit area. It's expressed in units of watts per square meter (W/m²).
In our exercise, the wall surface is subjected to a uniform heat flux of \( 20 \, \mathrm{kW/m}^2 \).
  • This represents a powerful and consistent amount of energy entering the wall.
  • It's important because the heat flux tells us how aggressive the heating process is.
Understanding heat flux is key to determining how quickly different parts of the wall will increase their temperature.
With constant heat flux, the temperature rise is mainly dependent on how the material distributes the incoming energy.
Temperature Distribution
Temperature distribution refers to how temperature varies across different points within a material over time.
In the context of this exercise, we are interested in how temperatures at the wall surface and at depths of 1 cm and 5 cm change as a function of heating time.
  • The temperature at the surface (\( x=0 \)) changes fastest as it directly receives the heat flux.
  • At \( x=1 \) and \( x=5 \) cm, the temperature changes more slowly due to the travel distance of heat energy through the material.
By plotting these temperature changes against time, we can visualize how heat penetrates the wall, which is crucial for applications involving heat treatment and insulation design.
Thermal Diffusivity
Thermal diffusivity is a measure of how quickly heat spreads through a material. Represented by \( \alpha \), it combines the effects of thermal conductivity, density, and specific heat capacity. The unit is square meters per second (m²/s).
For the refractory brick wall, \( \alpha = 5.08 \times 10^{-7} \, \mathrm{m}^2/\mathrm{s} \).
  • This low value indicates that heat spreads slowly, aligning with the idea of the wall being an insulator.
  • Thermal diffusivity is useful when predicting how temperature changes at various depths of a material respond to surface heating.
It helps in understanding the delay between the surface heating and the temperature change experienced deeper in the wall.
In processes where quick temperature responses are needed, materials with higher \( \alpha \) are preferred.

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Most popular questions from this chapter

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.

It is claimed that beef can be stored for up to two years at \(-23^{\circ} \mathrm{C}\) but no more than one year at \(-12^{\circ} \mathrm{C}\). Is this claim reasonable? Explain.

A potato may be approximated as a 5.7-cm-diameter solid sphere with the properties \(\rho=910 \mathrm{~kg} / \mathrm{m}^{3}, c_{p}=4.25 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}\), \(k=0.68 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\), and \(\alpha=1.76 \times 10^{-1} \mathrm{~m}^{2} / \mathrm{s}\). Twelve such potatoes initially at \(25^{\circ} \mathrm{C}\) are to be cooked by placing them in an oven maintained at \(250^{\circ} \mathrm{C}\) with a heat transfer coefficient of \(95 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). The amount of heat transfer to the potatoes during a 30-min period is (a) \(77 \mathrm{~kJ}\) (b) \(483 \mathrm{~kJ}\) (c) \(927 \mathrm{~kJ}\) (d) \(970 \mathrm{~kJ}\) (e) \(1012 \mathrm{~kJ}\)

Long aluminum wires of diameter \(3 \mathrm{~mm}(\rho=\) \(2702 \mathrm{~kg} / \mathrm{m}^{3}, c_{p}=0.896 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}, k=236 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\), and \(\alpha=\) \(9.75 \times 10^{-5} \mathrm{~m}^{2} / \mathrm{s}\) ) are extruded at a temperature of \(350^{\circ} \mathrm{C}\) and exposed to atmospheric air at \(30^{\circ} \mathrm{C}\) with a heat transfer coefficient of \(35 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). (a) Determine how long it will take for the wire temperature to drop to \(50^{\circ} \mathrm{C}\). (b) If the wire is extruded at a velocity of \(10 \mathrm{~m} / \mathrm{min}\), determine how far the wire travels after extrusion by the time its temperature drops to \(50^{\circ} \mathrm{C}\). What change in the cooling process would you propose to shorten this distance? (c) Assuming the aluminum wire leaves the extrusion room at \(50^{\circ} \mathrm{C}\), determine the rate of heat transfer from the wire to the extrusion room.

Stainless steel ball bearings \(\left(\rho=8085 \mathrm{~kg} / \mathrm{m}^{3}, k=\right.\) \(15.1 \mathrm{~W} / \mathrm{m} \cdot{ }^{\circ} \mathrm{C}, c_{p}=0.480 \mathrm{~kJ} / \mathrm{kg} \cdot{ }^{\circ} \mathrm{C}\), and \(\left.\alpha=3.91 \times 10^{-6} \mathrm{~m}^{2} / \mathrm{s}\right)\) having a diameter of \(1.2 \mathrm{~cm}\) are to be quenched in water. The balls leave the oven at a uniform temperature of \(900^{\circ} \mathrm{C}\) and are exposed to air at \(30^{\circ} \mathrm{C}\) for a while before they are dropped into the water. If the temperature of the balls is not to fall below \(850^{\circ} \mathrm{C}\) prior to quenching and the heat transfer coefficient in the air is \(125 \mathrm{~W} / \mathrm{m}^{2} \cdot{ }^{\circ} \mathrm{C}\), determine how long they can stand in the air before being dropped into the water.

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