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A flat-plate solar collector is used to heat water by having water flow through tubes attached at the back of the thin solar absorber plate. The absorber plate has an emissivity and an absorptivity of \(0.9\). The top surface \((x=0)\) temperature of the absorber is \(T_{0}=35^{\circ} \mathrm{C}\), and solar radiation is incident on the absorber at \(500 \mathrm{~W} / \mathrm{m}^{2}\) with a surrounding temperature of \(0^{\circ} \mathrm{C}\). Convection heat transfer coefficient at the absorber surface is \(5 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), while the ambient temperature is \(25^{\circ} \mathrm{C}\). Show that the variation of temperature in the absorber plate can be expressed as \(T(x)=-\left(\dot{q}_{0} / k\right) x+T_{0}\), and determine net heat flux \(\dot{q}_{0}\) absorbed by the solar collector.

Short Answer

Expert verified
Question: Show that the temperature variation in the absorber plate can be expressed as \(T(x)=-\left(\dot{q}_{0} / k\right) x+T_{0}\) and determine the net heat flux \(\dot{q}_{0}\) absorbed by the solar collector. Answer: The temperature variation in the absorber plate can be expressed as \(T(x)=-\left(\dot{q}_{0} / k\right) x+T_{0}\). The net heat flux absorbed \(\dot{q}_{0}\) can be found by calculating the incoming solar radiation absorbed by the absorber plate and the heat loss due to convection to the ambient air using the formula \(\dot{q}_0 = q_{in} - h(T_0 - T_a)\).

Step by step solution

01

Calculate absorber's solar radiation absorptance

First, calculate the amount of solar radiation absorbed by the absorber plate. This can be found by multiplying the incoming solar radiation by the absorber plate's absorptivity. $$ q_{in} = \alpha \times I $$ where, \(\alpha\) is the absorptivity of the absorber plate, \(I\) is the incident solar radiation.
02

Calculate the total heat absorbed

Next, we need to calculate the total heat absorbed by the plate by summing the incoming absorbed solar radiation and the heat loss due to convection to the ambient air: $$ \dot{q}_0 = q_{in} - h(T_0 - T_a) $$ where, \(h\) is the convective heat transfer coefficient, \(T_0\) is the top surface temperature of the absorber plate, and \(T_a\) is the ambient temperature.
03

Calculate the temperature profile

Now, plug in the provided values and solve for the net heat flux: $$ \dot{q}_0 = 0.9 \times 500 \mathrm{~W} / \mathrm{m}^{2} - 5 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K} \times (35^{\circ} \mathrm{C} - 0^{\circ} \mathrm{C}) $$ This will give the value for \(\dot{q}_{0}\).
04

Express the temperature variation in the absorber plate

By plugging the net heat flux \(\dot{q}_{0}\) into the following equation, we can express the temperature variation in the absorber plate as a function of \(x\). $$ T(x) = -\left(\frac{\dot{q}_{0}}{k}\right) x+T_{0}, $$ where \(k\) is the thermal conductivity of the absorber plate material. Note that \(k\) is not given in the problem statement, so we cannot find an exact expression for \(T(x)\) in this case.

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Key Concepts

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

Solar Collector
Solar collectors are devices designed to capture solar radiation and convert it into heat. These collectors are commonly used in heating systems, such as for domestic water heating or in solar thermal power plants. By absorbing sunlight, the collector heats a fluid, which can then be used directly or stored for later use.

In the context of a flat-plate solar collector, it typically consists of an absorber plate, tubes, insulation, and a cover. The absorber plate, often with high emissivity and absorptivity, plays a crucial role in absorbing as much solar energy as possible. The absorbed radiation then heats a liquid, like water, passing through the attached tubes.

The efficiency of these collectors largely depends on their ability to maximize absorption and minimize heat losses to the surroundings. Thus, materials and designs that retain heat and facilitate efficient heat transfer to the fluid are crucial in these systems.
Convection Heat Transfer
Convection is one of the primary modes of heat transfer and is crucial in the operation of solar collectors. It occurs when heat is transferred between the absorber plate and the surrounding air. Heat transfer by convection involves the movement of fluid, which in this case could be air or water, around a surface.

The efficiency of heat transfer in convection depends on the convective heat transfer coefficient, denoted as \( h \). This coefficient defines how easily heat is transferred from the surface of the absorber plate to the moving fluid.

For example, the solar collector in the given exercise has a convection heat transfer coefficient of \(5 \, ext{W/m}^2 \, \cdot \text{K}\), which influences the net heat flux. By calculating the difference between the heat absorbed by the absorber plate and the heat lost through convection, you can determine the effectiveness of the collector in transferring heat to the fluid inside the tubes.
Temperature Profile
The temperature profile in a solar collector describes how temperature changes across the absorber plate. In mathematical terms, this variation can be modeled as a linear equation, showing how temperature decreases from the top surface deeper into the plate.

In the problem, the temperature profile expression is given as \( T(x) = -\left(\frac{\dot{q}_{0}}{k}\right)x+T_{0} \). Here, \( T_{0} \) represents the temperature at the top surface of the absorber plate, and \( x \) denotes the position along the thickness of the plate. The slope of this equation is dependent on the net heat flux \( \dot{q}_{0} \) and the thermal conductivity \( k \).

Understanding this profile is vital as it helps in designing more efficient solar collectors. If the temperature drop across the plate is too steep, it indicates significant heat loss to the environment, suggesting potential improvements in insulation or materials.
Thermal Conductivity
Thermal conductivity, denoted as \( k \), is a material property that describes how well heat is conducted through a material. It plays a critical role in heat transfer processes, including those in solar collectors.

A material with high thermal conductivity can transfer heat more efficiently from the absorber plate to the fluid in the tubes, which is desirable in solar collectors to quickly heat water. However, too high a thermal conductivity might also mean more heat is lost to the surrounding environment if not properly insulated.

In the exercise, the precise value of \( k \) is not provided. However, it is crucial because it is used in the equation that describes the temperature profile. The thermal conductivity will affect how heat is distributed through the absorber plate and ultimately impact the overall performance of the solar collector. Materials like copper and aluminum often feature high thermal conductivity and are typically used in constructing such plates.

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Most popular questions from this chapter

Consider steady one-dimensional heat conduction in a plane wall in which the thermal conductivity varies linearly. The error involved in heat transfer calculations by assuming constant thermal conductivity at the average temperature is \((a)\) none, \((b)\) small, or \((c)\) significant.

Heat is generated in a \(3-\mathrm{cm}\)-diameter spherical radioactive material uniformly at a rate of \(15 \mathrm{~W} / \mathrm{cm}^{3}\). Heat is dissipated to the surrounding medium at \(25^{\circ} \mathrm{C}\) with a heat transfer coefficient of \(120 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). The surface temperature of the material in steady operation is (a) \(56^{\circ} \mathrm{C}\) (b) \(84^{\circ} \mathrm{C}\) (c) \(494^{\circ} \mathrm{C}\) (d) \(650^{\circ} \mathrm{C}\) (e) \(108^{\circ} \mathrm{C}\)

How do you recognize a linear homogeneous differential equation? Give an example and explain why it is linear and homogeneous.

Consider a large plane wall of thickness \(L=0.4 \mathrm{~m}\), thermal conductivity \(k=1.8 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\), and surface area \(A=\) \(30 \mathrm{~m}^{2}\). The left side of the wall is maintained at a constant temperature of \(T_{1}=90^{\circ} \mathrm{C}\) while the right side loses heat by convection to the surrounding air at \(T_{\infty}=25^{\circ} \mathrm{C}\) with a heat transfer coefficient of \(h=24 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Assuming constant thermal conductivity and no heat generation in the wall, \((a)\) express the differential equation and the boundary conditions for steady one-dimensional heat conduction through the wall, \((b)\) obtain a relation for the variation of temperature in the wall by solving the differential equation, and \((c)\) evaluate the rate of heat transfer through the wall. Answer: (c) \(7389 \mathrm{~W}\)

Consider a long rectangular bar of length \(a\) in the \(x-\) direction and width \(b\) in the \(y\)-direction that is initially at a uniform temperature of \(T_{i}\). The surfaces of the bar at \(x=0\) and \(y=0\) are insulated, while heat is lost from the other two surfaces by convection to the surrounding medium at temperature \(T_{\infty}\) with a heat transfer coefficient of \(h\). Assuming constant thermal conductivity and transient two-dimensional heat transfer with no heat generation, express the mathematical formulation (the differential equation and the boundary and initial conditions) of this heat conduction problem. Do not solve.

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