/*! 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 111 Copper tubing is joined to a sol... [FREE SOLUTION] | 91影视

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Chapter 3: Problem 111

Copper tubing is joined to a solar collector plate of thickness \(t\), and the working fluid maintains the temperature of the plate above the tubes at \(T_{o}\). There is a uniform net radiation heat flux \(q_{\text {rad }}^{\prime \prime}\) to the top surface of the plate, while the bottom surface is well insulated. The top surface is also exposed to a fluid at \(T_{\infty}\) that provides for a uniform convection coefficient \(h\). (a) Derive the differential equation that governs the temperature distribution \(T(x)\) in the plate. (b) Obtain a solution to the differential equation for appropriate boundary conditions.

Short Answer

Expert verified
The temperature distribution, \(T(x)\), in the solar collector plate is given by the following equation: \[ T(x) = T_\infty + (T_o - T_\infty)e^{-\eta x} \] where \(\eta = \frac{h}{k}\), \(h\) is the convection coefficient, \(k\) is the thermal conductivity of the plate, \(T_o\) is the temperature of the plate above the tubes, and \(T_\infty\) is the temperature of the fluid surrounding the plate.

Step by step solution

01

Energy Balance in the Solar Collector Plate

The first thing we need to do is to write an energy balance equation for the solar collector plate. As the plate is receiving radiation heat flux from the top surface and has convection heat loss, the energy balance can be written as: \[ q_{\text{rad}}^{\prime \prime} = h[T(x) - T_\infty] \]
02

Derive the Governing Differential Equation

For a steady-state temperature distribution in one dimension, Fourier's Law of heat conduction (considering the plate's thickness to be in the x-direction) can be written as: \[ q_x = -k \frac{dT(x)}{dx} \] Considering the heat transfer from one face to the other; we can write: \[ q_{\text{rad}}^{\prime \prime} - q_x = 0 \] Substitute the Fourier's Law equation into the energy balance equation: \[-k\frac{dT(x)}{dx}= h[T(x) - T_\infty] \] To simplify the equation, we can divide both sides by \(k\) and introduce a new variable: \[ \frac{dT(x)}{dx} = -\frac{h}{k}[T(x) - T_\infty] \] Let \(\eta = \frac{h}{k}\): \[ \frac{dT(x)}{dx} = -\eta[T(x) - T_\infty] \] This is the first-order linear differential equation that governs the temperature distribution, \(T(x)\), in the plate.
03

Solve the Differential Equation with Boundary Conditions

To solve the differential equation, we will first find the integrating factor, which is \(e^{\int \eta dx} = e^{\eta x}\). Now, multiply the given differential equation by the integrating factor: \[ e^{\eta x} \frac{dT(x)}{dx} = -\eta e^{\eta x}(T(x) - T_\infty) \] Integrate both sides with respect to x: \[ \int e^{\eta x} \frac{dT(x)}{dx} dx = -\eta \int e^{\eta x}(T(x) - T_\infty) dx \] This can be further simplified by substitution: Let \(u = T(x) - T_\infty\), \(du = \frac{dT(x)}{dx} dx\), and \(v = e^{\eta x}\), \(dv = \eta e^{\eta x} dx\). Then we can write: \[ \int udv = -\eta \int vdu \] Using integration by parts: \[ uv - \int vdu = -\eta \int u dv \] Substitute back u, du, and v: \[ (T(x) - T_\infty)e^{\eta x} - \int (T(x) - T_\infty)\eta e^{\eta x} dx = -\eta \int (T(x) - T_\infty) dx \] Apply boundary conditions: at \(x = 0\), \(T(0) = T_o\), and at \(x = t\), \(T(t) = T_o\). Integrating and simplifying, we get the temperature distribution as: \[ T(x) = T_\infty + (T_o - T_\infty)e^{-\eta x} \] The temperature distribution, \(T(x)\), in the solar collector plate has been obtained using the governing differential equation and boundary conditions.

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Key Concepts

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

Fourier's Law of Heat Conduction
Understanding the way heat moves through materials is crucial when analyzing the efficiency of solar collector plates. Fourier's Law of Heat Conduction is a principle that helps us predict how heat energy transfers within a conductor like a metal plate. It states that the rate of heat transfer through a material is proportional to the negative gradient of the temperature and the area through which the heat is flowing. Mathematically, it can be expressed as:
\[ q_x = -k \frac{dT(x)}{dx} \]
where \( q_x \) is the heat transfer rate per unit area (also known as heat flux), \( k \) is the thermal conductivity of the material, and \( \frac{dT(x)}{dx} \) is the temperature gradient in the direction of the heat flow. In the context of a solar collector, the implication is that heat flows from the hotter region near the tubes maintaining temperature \( T_o \) down the temperature gradient to the cooler exterior surface of the collector plate.
Radiation Heat Flux
Solar collector plates absorb energy primarily through radiation, which is then converted into heat. Radiation heat flux, denoted as \( q_{\text{rad}}^{\prime \prime} \), refers to the rate at which radiative energy is being transferred per unit area. This energy comes from the sun and is a critical component in heating the plate. Unlike conduction and convection, radiation does not require any medium to transfer energy, it can occur in a vacuum. The net radiation heat flux for the top surface of the collector is the energy absorbed from the sun minus any radiative losses the plate might have back to the environment. In the energy balance for the collector plate, the radiation heat flux is a source term, adding heat to the system.
Convection Heat Transfer
While radiation accounts for heat gain, convection is the process by which heat is lost from the collector plate to the surrounding air. It involves the movement of molecules within fluids (which includes gases such as air), resulting in heat transfer. Convection heat transfer is quantified by the convection heat transfer coefficient \( h \), which represents the amount of heat transferred per unit area per unit temperature difference between the surface and the fluid. The equation for the heat loss due to convection from the solar collector plate's surface to the air is:
\[ h[T(x) - T_{\infty}] \]
In this formula, \( T(x) \) is the temperature at a given point x on the plate, and \( T_{\infty} \) is the temperature of the surrounding fluid (air). Convection plays an integral part in determining the temperature profile of the solar collector as it directly influences the rate of heat loss from the plate to the environment.
Steady-State Temperature Distribution
The steady-state temperature distribution of a solar collector plate is the temperature profile that develops over time and does not change as long as the conditions remain constant. This implies that all the energy inputs and outputs are balanced and the temperature at any given point in the material remains constant over time. In other words, in the steady-state, the amount of heat entering any section of the collector plate equals the amount of heat leaving it. It's important to clarify that 'steady-state' does not mean that the temperature is uniform across the plate; rather, it means that the temperature distribution is fixed and unchanging while the system is under consistent, non-fluctuating conditions.
Governing Differential Equation
The governing differential equation for the temperature distribution in the solar collector plate represents a mathematical model that describes how the temperature changes within the plate based on position and time. This differential equation is derived from combining Fourier's Law of Heat Conduction with the energy balance which includes convection and radiation terms. For our solar collector plate, where the conditions are steady (do not change over time), we formulate the equation as a function of position, \( x \), across the thickness of the plate. The specific equation for our scenario is a first-order linear differential equation:
\[ \frac{dT(x)}{dx} = -\eta[T(x) - T_{\infty}] \]
Here, \( \eta \) is a constant that equals \( \frac{h}{k} \). The negative sign indicates that the temperature decreases along the direction of increasing \( x \). This equation embodies all the mechanisms of heat transfer at play and encapsulates the physics governing the temperature behaviour in the collector plate. Solving this differential equation with the appropriate boundary conditions allows us to find the temperature distribution \( T(x) \), which is crucial for optimizing the performance and efficiency of solar collector systems.

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Most popular questions from this chapter

A nanolaminated material is fabricated with an atomic layer deposition process, resulting in a series of stacked, alternating layers of tungsten and aluminum oxide, each layer being \(\delta=0.5 \mathrm{~nm}\) thick. Each tungsten-aluminum oxide interface is associated with a thermal resistance of \(R_{t, i}^{\prime \prime}=3.85 \times 10^{-9} \mathrm{~m}^{2} \cdot \mathrm{K} / \mathrm{W}\). The theoretical values of the thermal conductivities of the thin aluminum oxide and tungsten layers are \(k_{\mathrm{A}}=1.65 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) and \(k_{\mathrm{T}}=6.10 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\), respectively. The properties are evaluated at \(T=300 \mathrm{~K}\). (a) Determine the effective thermal conductivity of the nanolaminated material. Compare the value of the effective thermal conductivity to the bulk thermal conductivities of aluminum oxide and tungsten, given in Tables A.1 and A.2. (b) Determine the effective thermal conductivity of the nanolaminated material assuming that the thermal conductivities of the tungsten and aluminum oxide layers are equal to their bulk values.

From Problem 1.71, consider the wire leads connecting the transistor to the circuit board. The leads are of thermal conductivity \(k\), thickness \(t\), width \(w\), and length \(L\). One end of a lead is maintained at a temperature \(T_{c}\) corresponding to the transistor case, while the other end assumes the temperature \(T_{b}\) of the circuit board. During steady-state operation, current flow through the leads provides for uniform volumetric heating in the amount \(\dot{q}\), while there is convection cooling to air that is at \(T_{\infty}\) and maintains a convection coefficient \(h\). (a) Derive an equation from which the temperature distribution in a wire lead may be determined. List all pertinent assumptions. (b) Determine the temperature distribution in a wire lead, expressing your results in terms of the prescribed variables.

The evaporator section of a refrigeration unit consists of thin-walled, 10-mm- diameter tubes through which refrigerant passes at a temperature of \(-18^{\circ} \mathrm{C}\). Air is cooled as it flows over the tubes, maintaining a surface convection coefficient of \(100 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), and is subsequently routed to the refrigerator compartment. (a) For the foregoing conditions and an air temperature of \(-3^{\circ} \mathrm{C}\), what is the rate at which heat is extracted from the air per unit tube length? (b) If the refrigerator's defrost unit malfunctions, frost will slowly accumulate on the outer tube surface. Assess the effect of frost formation on the cooling capacity of a tube for frost layer thicknesses in the range \(0 \leq \delta \leq 4 \mathrm{~mm}\). Frost may be assumed to have a thermal conductivity of \(0.4 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). (c) The refrigerator is disconnected after the defrost unit malfunctions and a 2-mm-thick layer of frost has formed. If the tubes are in ambient air for which \(T_{\infty}=20^{\circ} \mathrm{C}\) and natural convection maintains a convection coefficient of \(2 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), how long will it take for the frost to melt? The frost may be assumed to have a mass density of \(700 \mathrm{~kg} / \mathrm{m}^{3}\) and a latent heat of fusion of \(334 \mathrm{~kJ} / \mathrm{kg}\).

A radioactive material of thermal conductivity \(k\) is cast as a solid sphere of radius \(r_{o}\) and placed in a liquid bath for which the temperature \(T_{\infty}\) and convection coefficient \(h\) are known. Heat is uniformly generated within the solid at a volumetric rate of \(\dot{q}\). Obtain the steadystate radial temperature distribution in the solid, expressing your result in terms of \(r_{o}, \dot{q}, k, h\), and \(T_{\infty}\).

Approximately \(10^{6}\) discrete electrical components can be placed on a single integrated circuit (chip), with electrical heat dissipation as high as \(30,000 \mathrm{~W} / \mathrm{m}^{2}\). The chip, which is very thin, is exposed to a dielectric liquid at its outer surface, with \(h_{o}=1000 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) and \(T_{\infty, 0}=20^{\circ} \mathrm{C}\), and is joined to a circuit board at its inner surface. The thermal contact resistance between the chip and the board is \(10^{-4} \mathrm{~m}^{2} \cdot \mathrm{K} / \mathrm{W}\), and the board thickness and thermal conductivity are \(L_{b}=5 \mathrm{~mm}\) and \(k_{b}=1 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\), respectively. The other surface of the board is exposed to ambient air for which \(h_{i}=40\) \(\mathrm{W} / \mathrm{m}^{2} \cdot \mathrm{K}\) and \(T_{\infty, i}=20^{\circ} \mathrm{C}\). (a) Sketch the equivalent thermal circuit corresponding to steady-state conditions. In variable form, label appropriate resistances, temperatures, and heat fluxes. (b) Under steady-state conditions for which the chip heat dissipation is \(q_{c}^{\prime \prime}=30,000 \mathrm{~W} / \mathrm{m}^{2}\), what is the chip temperature? (c) The maximum allowable heat flux, \(q_{c, m}^{\prime \prime}\), is determined by the constraint that the chip temperature must not exceed \(85^{\circ} \mathrm{C}\). Determine \(q_{c, m}^{\prime \prime}\) for the foregoing conditions. If air is used in lieu of the dielectric liquid, the convection coefficient is reduced by approximately an order of magnitude. What is the value of \(q_{c, m}^{\prime \prime}\) for \(h_{o}=100 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) ? With air cooling, can significant improvements be realized by using an aluminum oxide circuit board and/or by using a conductive paste at the chip/board interface for which \(R_{t, c}^{n}=10^{-5} \mathrm{~m}^{2} \cdot \mathrm{K} / \mathrm{W}\) ?
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