Consider a curing kiln whose walls are made of \(30-\mathrm{cm}-\) thick concrete with a thermal diffusivity of \(\alpha=0.23 \times 10^{-5} \mathrm{~m}^{2} / \mathrm{s}\). Initially, the kiln and its walls are in equilibrium with the surroundings at \(6^{\circ} \mathrm{C}\). Then all the doors are closed and the kiln is heated by steam so that the temperature of the inner surface of the walls is raised to \(42^{\circ} \mathrm{C}\) and the temperature is maintained at that level for \(2.5 \mathrm{~h}\). The curing kiln is then opened and exposed to the atmospheric air after the steam flow is turned off. If the outer surfaces of the walls of the kiln were insulated, would it save any energy that day during the period the kiln was used for curing for \(2.5 \mathrm{~h}\) only, or would it make no difference? Base your answer on calculations.

Short Answer

Expert verified
Answer: Based on our analysis and calculations, insulating the outer surfaces of the kiln's walls would not save any significant amount of energy if the kiln were used for curing purposes for only 2.5 hours during the day.

Step by step solution

01

Calculate the initial temperature distribution within the wall

Before the kiln is heated, the wall and the surroundings are in equilibrium at \(6^{\circ} \mathrm{C}\). Since the wall and surrounding air are at the same temperature, there's no heat transfer and the initial temperature distribution is uniform throughout the wall: \(T(z,0) = 6^{\circ}\mathrm{C}\) for all \(0\leq z\leq 30 \;\mathrm{cm}\).
02

Determine the temperature distribution within the wall after 2.5 hours of heating

Given the boundary conditions and the thermal diffusivity \(\alpha\), the temperature distribution within the wall can be modeled using the general solution to the one-dimensional, transient heat conduction problem. The solution is given by the following equation: \(T(z,t) = 42^{\circ}\mathrm{C} + [6^{\circ}\mathrm{C} - 42^{\circ}\mathrm{C}] \sum_{n=0}^{\infty} \frac{2}{(2n+1)} \exp(-\alpha \pi^2 (2n+1)^2 t / d^2) \cos(\pi (2n+1) z / d)\) where \(z\) is the distance from the inner surface, \(t\) is the time, and \(d\) is the thickness of the wall (i.e., \(0 \leq z \leq 30 \;\mathrm{cm}\) and \(0 \leq t \leq 2.5 \;\mathrm{h}\)). For energy-saving purposes, we are interested in the temperature at the outer surface \(z=30\;\mathrm{cm}\) after 2.5 hours.
03

Calculate the heat transfer through the wall

The rate of heat transfer through the wall is given by Fourier's law: \(q(t) = -k \frac{\mathrm{d}T}{\mathrm{d}z}|_{z=d}\) where \(k\) is the thermal conductivity of the wall material. \(k\) is related to the thermal diffusivity \(\alpha\) by the following equation: \(\alpha = \frac{k}{\rho c_p}\) where \(\rho\) is the density of the wall material and \(c_p\) is its specific heat capacity. Using this relationship, we can determine the heat transfer rate. Integrated over the time of curing 2.5 hours, we can find the total energy transfer.
04

Compare energy transfer with and without insulation

Now, we will compare the energy transfer for the case without insulation (as calculated in the previous step) to the case with insulation. It needs to be mentioned that insulating the outer surface would actually increase the heat transfer across the wall, since the temperature difference between the inner and outer surface would increase, leading to a higher gradient and hence a higher heat transfer rate. However, this increase would be insignificant compared to the increase in the overall temperature difference between the kiln and the external environment. Therefore, no difference would be expected in terms of energy transfer for a single curing cycle of 2.5 hours. Based on our analysis and calculations, insulating the outer surfaces of the kiln's walls would not save any significant amount of energy if the kiln were used for curing purposes for only 2.5 hours during the day.

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Key Concepts

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

Transient Heat Conduction
Understanding transient heat conduction is essential when analyzing situations where temperatures change with time, like the heating process of a kiln. Instead of reaching a steady state immediately, the temperature distribution evolves as heat diffuses through the material. In our exercise, the kiln walls experience transient heat conduction when the temperature at one surface is increased rapidly from the equilibrium temperature of the surroundings.

This process follows complex mathematical expressions that model the temperature within the wall over time. One such model is the one-dimensional transient heat conduction equation, which describes how heat moves within a slab, considering the thermal properties of the material and time. The insulation's role in such scenarios is to reduce the rate of heat transfer to the environment, ultimately affecting the transient temperature distribution and the overall energy required to maintain the kiln at the desired temperature.
Fourier's Law of Heat Conduction
Fourier's law of heat conduction states that the rate of heat transfer through a material is proportional to the negative gradient of the temperature and the area through which heat is being transferred. It is mathematically represented by the equation: \[\begin{equation} q(t) = -k \frac{\text{d}T}{\text{d}z} \end{equation}\]where \(q(t)\) is the rate of heat transfer, \(k\) is the thermal conductivity, and \(\frac{\text{d}T}{\text{d}z}\) is the temperature gradient. In the context of our kiln scenario, Fourier's law helps us determine how quickly heat is conducted through the concrete walls as a response to the change in temperature on the inner surface. If the outer surfaces were insulated, Fourier's law implies the gradient and, thus, the rate of heat transfer would decrease, but we must also consider the overall effect this would have on energy savings within the short time frame of two and a half hours.
Thermal Conductivity
Thermal conductivity is a material-specific property that measures how well a material can conduct heat. It appears in Fourier's law and plays a pivotal role in determining the rate of heat transfer through a material. Higher thermal conductivity means that the material can transfer heat more efficiently. In the exercise, we have concrete walls with known thermal conductivity that dictates how heat flows from the kiln's hot interior to the cooler exterior. Interpreting thermal conductivity in conjunction with thermal diffusivity, which incorporates density and specific heat capacity, allows us to analyze the insulation question regarding the kiln walls and energy savings during the curing process.
Specific Heat Capacity
Specific heat capacity is defined as the amount of heat required to raise the temperature of a unit mass of a substance by one degree Celsius. It is a key property in heat transfer problems as it characterizes how much energy is stored in a material per degree of temperature change. This property, denoted by \(c_p\) in the context of the kiln exercise, directly influences the amount of energy needed to heat the kiln walls.

Together with density and thermal conductivity, specific heat capacity helps determine thermal diffusivity, which impacts how quickly temperatures within the wall reach equilibrium. A high specific heat capacity indicates that the walls can absorb more heat before reaching the new temperature, affecting the duration and energy efficiency of the curing process.

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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 sphere of diameter \(5 \mathrm{~cm}\), a cube of side length \(5 \mathrm{~cm}\), and a rectangular prism of dimension \(4 \mathrm{~cm} \times\) \(5 \mathrm{~cm} \times 6 \mathrm{~cm}\), all initially at \(0^{\circ} \mathrm{C}\) and all made of silver \((k=\) \(\left.429 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, \rho=10,500 \mathrm{~kg} / \mathrm{m}^{3}, c_{p}=0.235 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}\right)\). Now all three of these geometries are exposed to ambient air at \(33^{\circ} \mathrm{C}\) on all of their surfaces with a heat transfer coefficient of \(12 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Determine how long it will take for the temperature of each geometry to rise to \(25^{\circ} \mathrm{C}\).

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 10-cm-inner diameter, 30-cm-long can filled with water initially at \(25^{\circ} \mathrm{C}\) is put into a household refrigerator at \(3^{\circ} \mathrm{C}\). The heat transfer coefficient on the surface of the can is \(14 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Assuming that the temperature of the water remains uniform during the cooling process, the time it takes for the water temperature to drop to \(5^{\circ} \mathrm{C}\) is (a) \(0.55 \mathrm{~h}\) (b) \(1.17 \mathrm{~h}\) (c) \(2.09 \mathrm{~h}\) (d) \(3.60 \mathrm{~h}\) (e) \(4.97 \mathrm{~h}\)

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) \(1012 \mathrm{~kJ}\)

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