/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 108 The use of rock pile thermal ene... [FREE SOLUTION] | 91影视

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Chapter 7: Problem 108

The use of rock pile thermal energy storage systems has been considered for solar energy and industrial process heat applications. A particular system involves a cylindrical container, \(2 \mathrm{~m}\) long by \(1 \mathrm{~m}\) in diameter, in which nearly spherical rocks of \(0.03-\mathrm{m}\) diameter are packed. The bed has a void space of \(0.42\), and the density and specific heat of the rock are \(\rho=2300 \mathrm{~kg} / \mathrm{m}^{3}\) and \(c_{p}=879 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\), respectively. Consider conditions for which atmospheric air is supplied to the rock pile at a steady flow rate of \(1 \mathrm{~kg} / \mathrm{s}\) and a temperature of \(90^{\circ} \mathrm{C}\). The air flows in the axial direction through the container. If the rock is at a temperature of \(25^{\circ} \mathrm{C}\), what is the total rate of heat transfer from the air to the rock pile?

Short Answer

Expert verified
The total rate of heat transfer from the air to the rock pile in the given rock pile thermal energy storage system is \(65.35 \mathrm{~kW}\).

Step by step solution

01

Calculate Mass Flow Rate of Air

We are given the mass flow rate of atmospheric air, which is equal to \(1 \mathrm{~kg} / \mathrm{s}\).
02

Calculate the Heat Capacity Rate of Air

The heat capacity rate of the flowing air is given by the product of its mass flow rate and specific heat capacity at constant pressure. For atmospheric air, we can assume the specific heat capacity at constant pressure (\(c_{p, air}\)) to be approximately \(1005 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\). The heat capacity rate of the air (\(C_{air}\)) can be calculated using the formula: \(C_{air} = \dot{m}_{air} \cdot c_{p, air}\) Where: \(\dot{m}_{air} = 1 \mathrm{~kg/s}\) \(c_{p, air} = 1005 \mathrm{~J/kg \cdot K}\) \(C_{air} = 1 \cdot 1005 = 1005 \mathrm{~W/K}\)
03

Calculate the Heat Capacity Rate of the Rock Pile

First, we need to calculate the mass of the rocks packed in the container. We are given the container's dimensions, rock diameter, and rock density. Volume of the cylindrical container (\(V_{container}\)): \(V_{container} = \pi r^2 h\) Where \(r = 0.5 \mathrm{~m}\) and \(h = 2 \mathrm{~m}\) \(V_{container} = \pi (0.5)^2 \cdot 2 = 1.571 \mathrm{~m^3}\) Volume of the rocks in the container (\(V_{rocks}\)): \(V_{rocks} = V_{container} \cdot (1 - \text{void space})\) Where the void space is 0.42 \(V_{rocks} = 1.571 \cdot (1 - 0.42) = 0.910 \mathrm{~m^3}\) Now, let's calculate the mass of the rock pile: \(m_{rocks} = V_{rocks} \cdot \rho\) Where \(\rho = 2300 \mathrm{~kg/m^3}\) \(m_{rocks} = 0.910 \cdot 2300 = 2093 \mathrm{~kg}\) Next, calculate the heat capacity rate of the rock pile (\(C_{rocks}\)): \(C_{rocks} = m_{rocks} \cdot c_{p, rocks}\) Where \(c_{p, rocks} = 879 \mathrm{~J/kg \cdot K}\) \(C_{rocks} = 2093 \cdot 879 = 1839517 \mathrm{~W/K}\)
04

Calculate Total Rate of Heat Transfer

Using the conservation of energy principle, we can determine the total rate of heat transfer (\(\dot{Q}\)) as: \(\dot{Q} = C_{air} \cdot (T_{air, in} - T_{rocks})\) Where \(T_{air, in} = 90^{\circ} \mathrm{C}\) and \(T_{rocks} = 25^{\circ} \mathrm{C}\) \(\dot{Q} = 1005 \cdot (90 - 25) = 65350 \mathrm{~W}\) The total rate of heat transfer from the air to the rock pile is \(65.35 \mathrm{~kW}\).

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Key Concepts

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

Thermal Energy Storage Systems
When delving into the world of thermal energy storage systems, we look at methods designed to store energy in a thermal reservoir for later use. These systems are critical for balancing the demand and supply of energy, especially in contexts where energy is generated unevenly, like with solar power during daylight hours. In our exercise scenario, a rock pile functions as the thermal reservoir, where the specific properties of the material, such as the density and specific heat capacity of the rocks, are taken into account to calculate energy storage capabilities.

The concept relies on the principle that certain materials can store heat when they are heated up and release it when the temperature drops. This stored thermal energy can be utilized later, providing an efficient way to manage energy resources, reduce waste, and save costs in industrial applications or renewable energy systems.
Specific Heat Capacity
The term specific heat capacity describes how much energy is required to raise the temperature of a unit mass of a substance by one degree Kelvin or Celsius. It's an intrinsic property of materials that plays a key role in the exercise we explored. Knowing the specific heat capacity of the rocks (\(879 \text{ J/kg} \text{K}\)) allowed us to calculate the amount of heat the rock pile can store or release.

It's worth understanding that materials with a high specific heat capacity can absorb a lot of heat energy without a significant rise in temperature, making them ideal for thermal storage. This property is harnessed in designing systems like the rock pile discussed in the exercise, allowing for the storage of excess thermal energy and its retrieval when needed.
Conservation of Energy
The conservation of energy is a fundamental concept in physics stating that energy cannot be created or destroyed, only transferred or changed from one form to another. In the context of our exercise, this principle is utilised to determine the total rate of heat transfer from the air to the rocks.

When the air, heated to a high temperature, moves through the rock pile, it transfers some of its thermal energy to the rocks, which are at a lower temperature. This heat transfer will continue until thermal equilibrium is reached鈥攚hen the rocks and the air have the same temperature. At equilibrium, according to the conservation of energy, the energy lost by the air will equal the energy gained by the rock pile, showing us that energy is conserved in this process.
Mass Flow Rate
The concept of mass flow rate is a measure of the mass of substance that passes through a given surface per unit time. In our exercise, the mass flow rate of air was given as \(1 \text{ kg/s}\). It's a pivotal variable that, combined with the specific heat capacity, allowed us to calculate the heat capacity rate of the air stream.

Understanding and calculating the mass flow rate is essential in various applications beyond thermal storage systems, such as in engineering systems involving fluids or gases, where control of the mass flow rate impacts system performance and efficiency. It's especially critical in the design and analysis of heating and cooling systems, as we've seen in the exercise, as it directly relates to the amount of thermal energy transferred over time.

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Most popular questions from this chapter

A flat plate of width \(1 \mathrm{~m}\) is maintained at a uniform surface temperature of \(T_{s}=150^{\circ} \mathrm{C}\) by using independently controlled, heat-generating rectangular modules of thickness \(a=10 \mathrm{~mm}\) and length \(b=50 \mathrm{~mm}\). Each module is insulated from its neighbors, as well as on its back side. Atmospheric air at \(25^{\circ} \mathrm{C}\) flows over the plate at a velocity of \(30 \mathrm{~m} / \mathrm{s}\). The thermophysical properties of the module are \(k=5.2 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, c_{p}=320 \mathrm{~J} / \mathrm{kg}+\mathrm{K}\), and \(\rho=2300 \mathrm{~kg} / \mathrm{m}^{3}\). (a) Find the required power generation, \(\dot{q}\left(\mathrm{~W} / \mathrm{m}^{3}\right)\), in a module positioned at a distance \(700 \mathrm{~mm}\) from the leading edge. (b) Find the maximum temperature \(T_{\max }\) in the heatgenerating module.

A long, cylindrical, electrical heating element of diameter \(D=10 \mathrm{~mm}\), thermal conductivity \(k=240 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\), density \(\rho=2700 \mathrm{~kg} / \mathrm{m}^{3}\), and specific heat \(c_{p}=900 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\) is installed in a duct for which air moves in cross flow over the heater at a temperature and velocity of \(27^{\circ} \mathrm{C}\) and \(10 \mathrm{~m} / \mathrm{s}\), respectively. (a) Neglecting radiation, estimate the steady-state surface temperature when, per unit length of the heater, electrical energy is being dissipated at a rate of \(1000 \mathrm{~W} / \mathrm{m}\). (b) If the heater is activated from an initial temperature of \(27^{\circ} \mathrm{C}\), estimate the time required for the surface temperature to come within \(10^{\circ} \mathrm{C}\) of its steady-state value.

Consider the following fluids, each with a velocity of \(V=5 \mathrm{~m} / \mathrm{s}\) and a temperature of \(T_{\infty}=20^{\circ} \mathrm{C}\), in cross flow over a 10-mm-diameter cylinder maintained at \(50^{\circ} \mathrm{C}\) : atmospheric air, saturated water, and engine oil. (a) Calculate the rate of heat transfer per unit length, \(q^{\prime}\), using the Churchill-Bernstein correlation. (b) Generate a plot of \(q^{\prime}\) as a function of fluid velocity for \(0.5 \leq V \leq 10 \mathrm{~m} / \mathrm{s}\).

The roof of a refrigerated truck compartment is of composite construction, consisting of a layer of foamed urethane insulation \(\left(t_{2}=50 \mathrm{~mm}, k_{i}=0.026 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\right)\) sandwiched between aluminum alloy panels \(\left(t_{1}=5 \mathrm{~mm}\right.\), \(\left.k_{p}=180 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\right)\). The length and width of the roof are \(L=10 \mathrm{~m}\) and W \(=3.5 \mathrm{~m}\), respectively, and the temperature of the inner surface is \(T_{s, i}=-10^{\circ} \mathrm{C}\). Consider conditions for which the truck is moving at a speed of \(V=105 \mathrm{~km} / \mathrm{h}\), the air temperature is \(T_{\infty}=32^{\circ} \mathrm{C}\), and the solar irradiation is \(G_{S}=750 \mathrm{~W} / \mathrm{m}^{2}\). Turbulent flow may be assumed over the entire length of the roof. (a) For equivalent values of the solar absorptivity and the emissivity of the outer surface \(\left(\alpha_{S}=\varepsilon=0.5\right)\), estimate the average temperature \(T_{s, o}\) of the outer surface. What is the corresponding heat load imposed on the refrigeration system? (b) A special finish \(\left(\alpha_{S}=0.15, \varepsilon=0.8\right)\) may be applied to the outer surface. What effect would such an application have on the surface temperature and the heat load? (c) If, with \(\alpha_{S}=\varepsilon=0.5\), the roof is not insulated \(\left(t_{2}=0\right)\), what are the corresponding values of the surface temperature and the heat load?

Dry air at atmospheric pressure and \(350 \mathrm{~K}\), with a free stream velocity of \(25 \mathrm{~m} / \mathrm{s}\), flows over a smooth, porous plate \(1 \mathrm{~m}\) long. (a) Assuming the plate to be saturated with liquid water at \(350 \mathrm{~K}\), estimate the mass rate of evaporation per unit width of the plate, \(n_{\mathrm{A}}^{\prime}(\mathrm{kg} / \mathrm{s} \cdot \mathrm{m})\). (b) For air and liquid water temperatures of 300,325 , and \(350 \mathrm{~K}\), generate plots of \(n_{\mathrm{A}}^{\prime}\) as a function of velocity for the range from 1 to \(25 \mathrm{~m} / \mathrm{s}\).
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