Consider a plane composite wall that is composed of two...

Consider a plane composite wall that is composed of two materials of thermal conductivities $k_{\mathrm{A}}=0.1 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}$ and $k_{\mathrm{B}}=0.04 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}$ and thicknesses $L_{\mathrm{A}}=10 \mathrm{~mm}$ and $L_{\mathrm{B}}=20 \mathrm{~mm}$. The contact resistance at the interface between the two materials is known to be $0.30 \mathrm{~m}^{2} \cdot \mathrm{K} / \mathrm{W}$. Material A adjoins a fluid at $200^{\circ} \mathrm{C}$ for which $h=10$ $\mathrm{W} / \mathrm{m}^{2} \cdot \mathrm{K}$, and material $\mathrm{B}$ adjoins a fluid at $40^{\circ} \mathrm{C}$ for which $h=20 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}$. (a) What is the rate of heat transfer through a wall that is $2 \mathrm{~m}$ high by $2.5 \mathrm{~m}$ wide? (b) Sketch the temperature distribution.

Consider a plane composite wall that is composed of two materials of thermal conductivities $k_{\mathrm{A}}=0.1 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}$ and $k_{\mathrm{B}}=0.04 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}$ and thicknesses $L_{\mathrm{A}}=10 \mathrm{~mm}$ and $L_{\mathrm{B}}=20 \mathrm{~mm}$. The contact resistance at the interface between the two materials is known to be $0.30 \mathrm{~m}^{2} \cdot \mathrm{K} / \mathrm{W}$.
Material A adjoins a fluid at $200^{\circ} \mathrm{C}$ for which $h=10$ $\mathrm{W} / \mathrm{m}^{2} \cdot \mathrm{K}$, and material $\mathrm{B}$ adjoins a fluid at $40^{\circ} \mathrm{C}$ for which $h=20 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}$.
(a) What is the rate of heat transfer through a wall that is $2 \mathrm{~m}$ high by $2.5 \mathrm{~m}$ wide?
(b) Sketch the temperature distribution.

Key Concepts

Thermal Conduction in Solids

Thermal conduction describes the transfer of heat energy through a solid material when there is a temperature gradient. The rate of heat transfer is governed by Fourier’s law, which explains how the thermal conductivity of a material and its thickness determine the temperature drop across it. In composite systems, analyzing conduction involves considering each material's unique properties and dimensions.

Convective Heat Transfer

Convective heat transfer involves the exchange of heat between a solid surface and a fluid in motion. This process is characterized by a convection heat transfer coefficient that quantifies the effectiveness of the convective cooling or heating. In boundary problems, convective resistance at the surface plays a critical role in the overall heat transfer process.

Thermal Contact Resistance

Thermal contact resistance refers to the resistance to heat flow across the interface of two materials in contact, which arises due to microscopic surface roughness and imperfections. This additional resistance can significantly affect the overall rate of heat transfer by introducing a temperature drop at the interface, separate from the conduction resistances of the adjacent layers.

Thermal Resistance Networks

A thermal resistance network is an analytical method used to simplify and solve complex heat transfer problems by analogizing heat transfer to electrical circuits. In this approach, each layer’s conduction resistance, along with convective and contact resistances, is treated as a resistor in series or parallel, allowing for the calculation of the overall resistance and heat transfer rate.

Temperature Distribution Across Composite Walls

The temperature distribution in a composite wall illustrates how temperature changes spatially across different regions, including within each material, at material interfaces, and at fluid boundaries. This concept is important for visualizing the thermal gradient and understanding where significant temperature drops occur due to various resistances such as conduction, convection, and contact resistance.

Transcript

00:02 A rigid container has two rooms a and b, both of which are filled with water, and they have the same volume, but they are different states.

00:10 The partition between the two rooms is then removed, and the water comes to a uniform state.

00:16 And if we know the temperature of this uniform state, we want to find the final pressure and the heat transfer in this process.

00:24 So if we take the tanks a and b to be our control volume, we can write our continuity equation, as the final mass m2 minus the initial mass of a minus the initial mass of b must equal to zero.

00:46 So the final mass is simply the sum of the initial masses of tank a and b.

00:51 And the volume v2 again is equal to the sum of the volumes of a and b.

00:59 B a plus vb and that's two cubic meters.

01:05 The energy equation, and the first law can be written as well.

01:13 And the energy equation tells us that the change in internal energy, u2 minus u1, is equal to the change m2, u2, minus m -a1, minus m -a -1, u -a -1, minus m -b -1, u -b -1, and this must equal to the heat transfer, q, minus the work done.

01:47 But since we have a constant volume, the work done is zero.

01:51 So that's simply equal to the heat transfer.

01:54 So let's first look at our data for state one.

02:03 So state one, actually state one a for tank a, we go to our tables and we can find at that state, the internal energy ua1, since we are given the quality, is 504 .4 .47 plus the quality of 80s...

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