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Chapter 5: Problem 12

Thermal energy storage systems commonly involve a packed bed of solid spheres, through which a hot gas flows if the system is being charged, or a cold gas if it is being discharged. In a charging process, heat transfer from the hot gas increases thermal energy stored within the colder spheres; during discharge, the stored energy decreases as heat is transferred from the warmer spheres to the cooler gas. Consider a packed bed of \(75-\mathrm{mm}\)-diameter aluminum spheres \(\left(\rho=2700 \mathrm{~kg} / \mathrm{m}^{3}, c=950 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}, k=\right.\) \(240 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) ) and a charging process for which gas enters the storage unit at a temperature of \(T_{g, i}=300^{\circ} \mathrm{C}\). If the initial temperature of the spheres is \(T_{i}=25^{\circ} \mathrm{C}\) and the convection coefficient is \(h=75 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), how long does it take a sphere near the inlet of the system to accumulate \(90 \%\) of the maximum possible thermal energy? What is the corresponding temperature at the center of the sphere? Is there any advantage to using copper instead of aluminum?

Short Answer

Expert verified
The time it takes for a sphere near the inlet of the system to accumulate 90% of the maximum possible thermal energy is approximately 1803.61 seconds. The corresponding temperature at the center of the sphere is about 282.60掳C. While copper has a higher thermal conductivity and efficiency than aluminum, it has a lower specific heat capacity and stores less energy, making it less suitable for thermal energy storage in this case. Additionally, copper is more expensive and heavier than aluminum, which would also need to be considered.

Step by step solution

01

Calculate the time constant

First, we need to calculate the time constant (蟿) for this packed bed system using the following formula: \[ 蟿 = \frac{蟻 r c}{6 h} \] where: 蟻 = 2700 kg/m鲁 (density of aluminum) r = 0.0375 m (radius of the sphere, since diameter is 75 mm) c = 950 J/kg路K (specific heat capacity of aluminum) h = 75 W/m虏路K (convection coefficient)
02

Compute the time constant

Using the values above, we can calculate the time constant by substituting them into the formula: \[ 蟿 = \frac{2700\,(0.0375)\,(950)}{6\,(75)} = 522.5\,\text{s} \]
03

Calculate the time required for 90% energy accumulation

Next, we need to calculate the time required to accumulate 90% of the maximum possible thermal energy. To do this, we can use the energy accumulation formula: \[ E = E_{max} (1 - e^{-t/蟿}) \] Since we are looking for the time when 90% of total energy is accumulated, E is 0.9 * E_max. The time (t) can be calculated by rearranging the formula and solving for t: \[ t = -蟿 \ln(1 - 0.9) \]
04

Compute the required time for 90% energy accumulation

Now, let's calculate the time (t) using the previously calculated time constant (蟿): \[ t = -(522.5)\,\ln(1 - 0.9) 鈮� 1803.61\,\text{s} \] So, it takes approximately 1803.61 seconds for a sphere near the inlet of the system to accumulate 90% of the maximum possible thermal energy.
05

Calculate temperature at the center of the sphere

To find the temperature at the center of the sphere, we need to use the energy accumulation equation: \[ E = m c (T - T_i) \] where: m = mass of the sphere T = temperature at the center of the sphere T_i = 25掳C (initial temperature of the spheres) We also know that E is 0.9 * E_max, so: \[ E = 0.9 m c (T_{g,i} - T_i) \] Now, let's solve for the temperature T: \[ T = \frac{E}{0.9\,m\,c} + T_i = \frac{E}{0.9\,蟻\,V\,c} + T_i \]
06

Calculate the mass and volume of the sphere

The mass of the sphere (m) can be calculated using the density (蟻) and volume (V) of the sphere: \[ m = 蟻 V \] The volume of the sphere (V) can be calculated using the formula for the volume of a sphere with radius r: \[ V = \frac{4}{3}蟺r^3 \] Substitute the given radius (r=0.0375m) to get the volume: \[ V 鈮� 2.21 \times 10^{-4}\,\text{m}^{3} \] Now, calculate the mass using the density and volume: \[ m 鈮� 2700\,(2.21 \times 10^{-4})\,\text{kg} = 0.597\,\text{kg} \]
07

Calculate the temperature at the center of the sphere

Now that we have the mass (m) and volume (V) of the sphere, we can calculate the temperature at the center (T): \[ T 鈮� \frac{E}{0.9\,(0.597)\,(950)} + 25 鈮� 282.60^{\circ} \mathrm{C} \] The temperature at the center of the sphere, when 90% of the maximum possible thermal energy is accumulated, is approximately 282.60掳C.
08

Discuss the advantage of using copper instead of aluminum

Copper has a higher thermal conductivity (k=385 W/m路K) than aluminum (k=240 W/m路K). This means that heat would be transferred more effectively through copper spheres, leading to faster heating and higher thermal efficiency. However, copper has a lower specific heat capacity (c=385 J/kg路K) compared to aluminum (c=950 J/kg路K), so it would require less energy to increase its temperature compared to aluminum, but it would also store less energy. Finally, copper is also more expensive and heavier than aluminum. In conclusion, copper might provide some advantages in terms of faster heat transfer and higher thermal efficiency, but the benefits would need to be balanced against the material's costs, weight, and lower energy storage capacity.

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Key Concepts

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

Understanding Heat Transfer
Heat transfer is a fundamental concept in thermodynamics, referring to the movement of thermal energy from one place to another. It occurs due to a temperature difference and can happen through conduction, convection, or radiation. In thermal energy storage systems, convection plays a significant role, especially when a fluid, such as a gas or liquid, is involved to transfer heat to a solid medium, such as a packed bed of solid spheres.

During the charging process, which involves increasing the storage of thermal energy, a hot gas flows around the spheres, transferring heat to them through the convection process. This heat then gets distributed within the spheres through conduction. The rate of heat transfer is influenced by several factors, including the thermal conductivity of the material, the specific heat capacity, which tells us how much energy is needed to raise the temperature of a unit mass of material by one degree, and the convection coefficient, which captures the efficiency of the convective heat transfer from the fluid to the solid surface.
The Role of a Packed Bed of Solid Spheres in Thermal Systems
A packed bed of solid spheres is an effective structure for thermal storage due to the high surface area for heat transfer and the ability to maintain contact between the spheres and the heat transfer fluid. In the context of the exercise, aluminum spheres are used to absorb and store thermal energy.

The structural arrangement and material properties of these spheres, such as density \( \rho \) and thermal conductivity \( k \), play crucial roles in their capacity to store and transfer heat. The spheres' surface area directly relates to the convection heat transfer coefficient \( h \), impacting how efficiently the system can charge or discharge. The larger the surface area, the more efficiently it can interact with the heat transfer fluid, and hence, the more effective the heat transfer process.
Charging Process in Thermal Systems
The charging process in thermal energy storage systems entails increasing the energy stored within the system. The exercise illustrates this process as gas warms a packed bed of spheres from an initial lower temperature to a higher one, effectively 'charging' the bed with thermal energy.

To evaluate how long it takes for a sphere to reach a certain energy state鈥攕uch as 90% of the maximum possible thermal energy鈥攖he understanding of the time constant \( \tau \) is essential. The time constant represents the speed at which the temperature rises and is a function of the material properties and the size of the spheres. The formula used here \( \tau = \frac{\rho r c}{6 h} \) implies that greater density, radius, and specific heat capacity will lead to a larger time constant, indicating a slower heating process, while a higher convection coefficient \( h \) will reduce the time constant, indicating a faster heating process. Once \( \tau \) is determined, we use it to calculate the time required for the energy accumulation to reach the desired percentage.

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Most popular questions from this chapter

Joints of high quality can be formed by friction welding. Consider the friction welding of two 40 -mm-diameter Inconel rods. The bottom rod is stationary, while the top rod is forced into a back-and-forth linear motion characterized by an instantaneous horizontal displacement, \(d(t)=a \cos (\omega t)\) where \(a=2 \mathrm{~mm}\) and \(\omega=1000 \mathrm{rad} / \mathrm{s}\). The coefficient of sliding friction between the two pieces is \(\mu=0.3\). Determine the compressive force that must be applied to heat the joint to the Inconel melting point within \(t=3 \mathrm{~s}\), starting from an initial temperature of \(20^{\circ} \mathrm{C}\). Hint: The frequency of the motion and resulting heat rate are very high. The temperature response can be approximated as if the heating rate were constant in time, equal to its average value.

A solid steel sphere (AISI 1010 ), \(300 \mathrm{~mm}\) in diameter, is coated with a dielectric material layer of thickness \(2 \mathrm{~mm}\) and thermal conductivity \(0.04 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). The coated sphere is initially at a uniform temperature of \(500^{\circ} \mathrm{C}\) and is suddenly quenched in a large oil bath for which \(T_{\infty}=100^{\circ} \mathrm{C}\) and \(h=3300 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Estimate the time required for the coated sphere temperature to reach \(140^{\circ} \mathrm{C}\). Hint: Neglect the effect of energy storage in the dielectric material, since its thermal capacitance \((\rho c V)\) is small compared to that of the steel sphere

A wall \(0.12 \mathrm{~m}\) thick having a thermal diffusivity of \(1.5 \times 10^{-6} \mathrm{~m}^{2} / \mathrm{s}\) is initially at a uniform temperature of \(85^{\circ} \mathrm{C}\). Suddenly one face is lowered to a temperature of \(20^{\circ} \mathrm{C}\), while the other face is perfectly insulated. (a) Using the explicit finite-difference technique with space and time increments of \(30 \mathrm{~mm}\) and \(300 \mathrm{~s}\), respectively, determine the temperature distribution at \(t=45 \mathrm{~min}\). (b) With \(\Delta x=30 \mathrm{~mm}\) and \(\Delta t=300 \mathrm{~s}\), compute \(T(x, t)\) for \(0 \leq t \leq t_{\mathrm{ss}}\), where \(t_{\mathrm{ss}}\) is the time required for the temperature at each nodal point to reach a value that is within \(1^{\circ} \mathrm{C}\) of the steady-state temperature. Repeat the foregoing calculations for \(\Delta t=75 \mathrm{~s}\). For each value of \(\Delta t\), plot temperature histories for each face and the midplane.

Two plates of the same material and thickness \(L\) are at different initial temperatures \(T_{i, 1}\) and \(T_{i, 2}\), where \(T_{i, 2}>T_{i, 1}\). Their faces are suddenly brought into contact. The external surfaces of the two plates are insulated. (a) Let a dimensionless temperature be defined as \(T *(F o) \equiv\left(T-T_{i, 1}\right) /\left(T_{i, 2}-T_{i, 1}\right)\). Neglecting the thermal contact resistance at the interface between the plates, what are the steady-state dimensionless temperatures of each of the two plates, \(T_{s s, 1}^{*}\) and \(T_{s s, 2}^{*}\) ? What is the dimensionless interface temperature \(T_{\text {in }}^{*}\) at any time? (b) An effective overall heat transfer coefficient between the two plates can be defined based on the instantaneous, spatially averaged dimensionless plate temperatures, \(U_{\mathrm{eff}}^{*} \equiv q^{*} /\left(\bar{T}_{2}^{*}-\bar{T}_{1}^{*}\right)\). Noting that a dimensionless heat transfer rate to or from either of the two plates may be expressed as \(q^{*}=d\left(Q / Q_{o}\right) / d F o\), determine an expression for \(U_{\text {eif }}^{*}\) for \(F o>0.2\).

Consider a thin electrical heater attached to a plate and backed by insulation. Initially, the heater and plate are at the temperature of the ambient air, \(T_{\infty}\). Suddenly, the power to the heater is activated, yielding a constant heat flux \(q_{o}^{\prime \prime}\left(\mathrm{W} / \mathrm{m}^{2}\right)\) at the inner surface of the plate. (a) Sketch and label, on \(T \leftarrow x\) coordinates, the temperature distributions: initial, steady-state, and at two intermediate times. (b) Sketch the heat flux at the outer surface \(q_{x}^{\prime \prime}(L, t)\) as a function of time.
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