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In a solar pond, the absorption of solar energy can be modeled as heat generation and can be approximated by \(\dot{e}_{\text {gen }}=\dot{e}_{0} e^{-b x}\), where \(\dot{e}_{0}\) is the rate of heat absorption at the top surface per unit volume and \(b\) is a constant. Obtain a relation for the total rate of heat generation in a water layer of surface area \(A\) and thickness \(L\) at the top of the pond.

Short Answer

Expert verified
The final expression for the total heat generation rate (Q) in a solar pond for a water layer of surface area A and thickness L at the top of the pond is given by: \(Q = A \frac{-\dot{e}_0}{b}(e^{-bL} -1)\)

Step by step solution

01

Determine the variables

In this problem, the given variables are: - \(\dot{e}_{\text {gen }}\): Heat generation rate per unit volume - \(\dot{e}_{0}\): Rate of heat absorption at the top surface per unit volume - \(b\): Constant - \(A\): Surface area of the water layer - \(L\): Thickness of the water layer at the top of the pond
02

Integrate the heat generation model over the thickness of the water layer

The heat generation model \(\dot{e}_{gen}=\dot{e}_0 e^{-bx}\) needs to be integrated over the thickness of the water layer (\(0\) to \(L\)) to find the total heat generation rate per unit volume in the layer. Let's call this integral result as \(I\): \(I = \int_0^L \dot{e}_0 e^{-bx} dx\)
03

Solve the integral

Solving the integral for \(I\), we get: \(I = \frac{-\dot{e}_0}{b}(e^{-bL} -1)\)
04

Compute the total heat generation rate for the water layer of area A

To find the total heat generation rate for the entire water layer of area \(A\), we'll multiply the result from the integral (\(I\)) by the surface area (\(A\)). Let's call the total heat generation rate \(Q\): \(Q = A \cdot I\)
05

Find the final expression for the total heat generation rate

Plug in the value of \(I\) into the expression for \(Q\): \(Q = A \frac{-\dot{e}_0}{b}(e^{-bL} -1)\) This is the final expression for the total heat generation rate in a solar pond for a water layer of surface area \(A\) and thickness \(L\) at the top of the pond.

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Key Concepts

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

Heat Generation Model
In the context of a solar pond, the heat generation model is a way to describe how solar energy is absorbed and converted into heat within the pond's layers. The model provided in the exercise is expressed as \( \dot{e}_{\text{gen}} = \dot{e}_0 e^{-bx} \), where \( \dot{e}_{\text{gen}} \) represents the rate of heat generation per unit volume. Here are some key points to understand this model:
  • \( \dot{e}_0 \): This is the initial rate of heat absorption per unit volume at the surface. It indicates how much heat is absorbed at the very top of the water layer.

  • \( b \): This constant determines how quickly the heat absorption decreases as the depth \( x \) in the layer increases. A higher value of \( b \) means the absorption diminishes faster.

  • \( e^{-bx} \): This exponential decay function shows that as you go deeper into the pond, the heat generated decreases. This is because solar radiation is absorbed and scattered as it penetrates the water.

Understanding this model helps in predicting how much heat is being generated at different depths of the pond, which is crucial for applications like energy storage or heating.
Integration in Heat Transfer
Integration is a fundamental tool in understanding heat transfer, particularly when dealing with varying conditions over a distance or surface. In this exercise, the integration of \( \dot{e}_0 e^{-bx} \) over the thickness \( L \) of the water layer is crucial. The integration process can be seen in these steps:
  • The function \( \dot{e}_0 e^{-bx} \) is integrated over the limits from \( 0 \) to \( L \). This range represents the total thickness of the water layer where heat is absorbed, taking into account its variation with depth.

  • Solving this integral yields \( I = \frac{-\dot{e}_0}{b} (e^{-bL} - 1) \). This result gives the cumulative heat generation per unit volume throughout the depth \( L \).

  • The method of integration captures the diminishing effect of heat generation as depth increases, providing a total rather than just point-specific understanding.

This integration approach is straightforward but highlights the exponential distribution of heat, a common phenomenon in environments where heat absorption is affected by depth or material properties.
Surface Area and Thickness in Energy Calculations
The surface area (\( A \)) and thickness (\( L \)) of a water layer are critical when calculating total energy or heat generation. Here, they dictate the effectiveness and extent to which solar heat is absorbed and generated in the solar pond:
  • Surface Area \( A \): This factor scales the quantity of heat generated, implying how much solar energy is initially captured. A larger surface area means more energy can be absorbed.

  • Thickness \( L \): It affects how far down the heat penetrates and influences total heat integration. A thicker layer allows analysis over a greater volume, capturing more variability in heat generation.

  • The combination: When calculating the total heat generation, \( Q = A \times I \), both area and thickness work together. The expression \( Q = A \frac{-\dot{e}_0}{b} (e^{-bL} - 1) \) reflects how surface geometries play a role in capturing integrated heat values.

Understanding these spatial factors helps in designing and optimizing solar ponds for maximum efficiency, ensuring that as much energy is absorbed and utilized as possible.

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Most popular questions from this chapter

In a food processing facility, a spherical container of inner radius \(r_{1}=40 \mathrm{~cm}\), outer radius \(r_{2}=41 \mathrm{~cm}\), and thermal conductivity \(k=1.5 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) is used to store hot water and to keep it at \(100^{\circ} \mathrm{C}\) at all times. To accomplish this, the outer surface of the container is wrapped with a 800 -W electric strip heater and then insulated. The temperature of the inner surface of the container is observed to be nearly \(120^{\circ} \mathrm{C}\) at all times. Assuming 10 percent of the heat generated in the heater is lost through the insulation, \((a)\) express the differential equation and the boundary conditions for steady one-dimensional heat conduction through the container, \((b)\) obtain a relation for the variation of temperature in the container material by solving the differential equation, and \((c)\) evaluate the outer surface temperature of the container. Also determine how much water at \(100^{\circ} \mathrm{C}\) this tank can supply steadily if the cold water enters at \(20^{\circ} \mathrm{C}\).

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.

The variation of temperature in a plane wall is determined to be \(T(x)=52 x+25\) where \(x\) is in \(\mathrm{m}\) and \(T\) is in \({ }^{\circ} \mathrm{C}\). If the temperature at one surface is \(38^{\circ} \mathrm{C}\), the thickness of the wall is (a) \(0.10 \mathrm{~m}\) (b) \(0.20 \mathrm{~m}\) (c) \(0.25 \mathrm{~m}\) (d) \(0.40 \mathrm{~m}\) (e) \(0.50 \mathrm{~m}\)

A spherical container of inner radius \(r_{1}=2 \mathrm{~m}\), outer radius \(r_{2}=2.1 \mathrm{~m}\), and thermal conductivity \(k=30 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) is filled with iced water at \(0^{\circ} \mathrm{C}\). The container is gaining heat by convection from the surrounding air at \(T_{\infty}=25^{\circ} \mathrm{C}\) with a heat transfer coefficient of \(h=18 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Assuming the inner surface temperature of the container to be \(0^{\circ} \mathrm{C},(a)\) express the differential equation and the boundary conditions for steady one- dimensional heat conduction through the container, \((b)\) obtain a relation for the variation of temperature in the container by solving the differential equation, and \((c)\) evaluate the rate of heat gain to the iced water.

A solar heat flux \(\dot{q}_{s}\) is incident on a sidewalk whose thermal conductivity is \(k\), solar absorptivity is \(\alpha_{s}\), and convective heat transfer coefficient is \(h\). Taking the positive \(x\) direction to be towards the sky and disregarding radiation exchange with the surroundings surfaces, the correct boundary condition for this sidewalk surface is (a) \(-k \frac{d T}{d x}=\alpha_{s} \dot{q}_{s}\) (b) \(-k \frac{d T}{d x}=h\left(T-T_{\infty}\right)\) (c) \(-k \frac{d T}{d x}=h\left(T-T_{\infty}\right)-\alpha_{s} \dot{q}_{s}\) (d) \(h\left(T-T_{\infty}\right)=\alpha_{s} \dot{q}_{s}\) (e) None of them

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