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Chapter 2: Problem 18

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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Most popular questions from this chapter

Consider uniform heat generation in a cylinder and a sphere of equal radius made of the same material in the same environment. Which geometry will have a higher temperature at its center? Why?

A spherical container, with an inner radius of \(1 \mathrm{~m}\) and a wall thickness of \(5 \mathrm{~mm}\), has its inner surface subjected to a uniform heat flux of \(7 \mathrm{~kW} / \mathrm{m}^{2}\). The outer surface of the container is maintained at \(20^{\circ} \mathrm{C}\). The container wall is made of a material with a thermal conductivity given as \(k(T)=k_{0}(1+\beta T)\), where \(k_{0}=1.33 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, \beta=0.0023 \mathrm{~K}^{-1}\), and \(T\) is in \(\mathrm{K}\). Determine the temperature drop across the container wall thickness.

Is the thermal conductivity of a medium, in general, constant or does it vary with temperature?

What is the difference between an ordinary differential equation and a partial differential equation?

Consider a large plane wall of thickness \(L=0.3 \mathrm{~m}\), thermal conductivity \(k=2.5 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\), and surface area \(A=\) \(12 \mathrm{~m}^{2}\). The left side of the wall at \(x=0\) is subjected to a net heat flux of \(\dot{q}_{0}=700 \mathrm{~W} / \mathrm{m}^{2}\) while the temperature at that surface is measured to be \(T_{1}=80^{\circ} \mathrm{C}\). 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 temperature of the right surface of the wall at \(x=L\).
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