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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.

Step by step solution

01

Determine the areas and interface resistance

Firstly, calculate the area of the entire wall (\(A_W\)) and the area of the contact (\(A_F\)) using the provided dimensions of the wall. Then, determine the interface resistance (R_contact) using the given contact resistance value. \[A_W = 2 \text{ m} × 2.5 \text{ m} = 5 \text{ m}^2\] \[A_F = 2 \text{ m} × 2 \text{ m} = 4 \text{ m}^2\] \[R_{contact} = \frac{0.30 \text{ m}^2 \cdot \text{K} / \text{W}}{A_F} = 0.075 \text{ K} / \text{W}\]

02

Calculate the convection and conduction resistances

Calculate the convection resistances on both fluid boundaries (R_conv_A and R_conv_B) using the given convective heat transfer coefficients: \[R_{conv_A} = \frac{1}{h_A A_W} = \frac{1}{(10 \text{ W/m}^2 \text{K})(5 \text{ m}^2)} = 0.02 \text{ K/W}\] \[R_{conv_B} = \frac{1}{h_B A_W} = \frac{1}{(20 \text{ W/m}^2 \text{K})(5 \text{ m}^2)} = 0.01 \text{ K/W}\] Next, calculate the conduction resistances for both materials (R_cond_A and R_cond_B): \[R_{cond_A}= \frac{L_A}{k_A A_W} = \frac{0.01 \text{ m}}{(0.1 \text{ W/m K})(5 \text{ m}^2)} = 0.02 \text{ K/W}\] \[R_{cond_B}= \frac{L_B}{k_B A_W} = \frac{0.02 \text{ m}}{(0.04 \text{ W/m K})(5 \text{ m}^2)} = 0.1 \text{ K/W}\]

03

Calculate the total thermal resistance and heat transfer

Add up all the resistances to find the total thermal resistance (R_total): \[R_{total} = R_{conv_A} + R_{cond_A} + R_{contact} + R_{cond_B} + R_{conv_B} = 0.02 + 0.02 + 0.075 + 0.1 + 0.01 = 0.225 \, \text{K/W}\] Now, calculate the overall heat transfer rate (Q) using the temperature difference between the hot and cold fluids and the total thermal resistance: \[Q = \frac{T_{hot} - T_{cold}}{R_{total}} = \frac{200 - 40}{0.225} = \frac{160}{0.225} \approx 711.11 \, \text{W}\]

04

Sketch the temperature distribution

To sketch the temperature distribution, plot the temperatures at the interfaces between convection and conduction, and conduction and contact resistance. Remember that heat transfer is continuous, meaning that the temperatures have to show a continuous decrease from the hot fluid to the cold fluid along the wall. | | fluid at 200°C | ----> 0 Convection resistance | Material A| | ----> 0 Conduction resistance | | Interface (contact resistance) | Material B| | ----> 0 Conduction resistance | | fluid at 40°C V The heat transfer rate through the wall is found to be 711.11 W, and this graph provides a visual representation of the temperature distribution across the composite wall.

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Key Concepts

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

Thermal Conductivity

Thermal conductivity, denoted as k, is a measure of a material's ability to conduct heat. It quantifies how easily heat can flow through a material due to a temperature gradient. In the exercise, two different materials, A and B, are given distinct thermal conductivity values, k_A = 0.1 W/m·K and k_B = 0.04 W/m·K. Higher k values indicate better heat conduction. When thermal conductivities are mixed in a composite wall, understanding each material’s k helps us assess how they affect overall heat transfer.

Convection Resistance

Convection resistance, represented by R_conv, relates to the opposition a fluid provides to heat flow due to convection. The inverse of the convective heat transfer coefficient, h, and the surface area A, the formula is R_conv = 1 / (h·A). In the exercise, the wall experiences convection on two sides, with coefficients h_A = 10 W/m²·K and h_B = 20 W/m²·K. The resistance to heat transfer by convection is lower where h is higher, as seen with R_{conv_B} being half the value of R_{conv_A}.

Conduction Resistance

Conduction resistance is a measure of how strongly a material opposes the flow of heat through its thickness. Represented by R_cond, it is determined by the thickness of the material L, its thermal conductivity k, and the surface area A, following the equation R_cond = L / (k·A). Materials A and B exhibit different conduction resistances in the exercise, reflecting their distinct thicknesses and thermal conductivities. Conduction resistance is critical for understanding how heat travels across solid materials and the impact of varying material properties on overall heat flow.

Contact Resistance

Contact resistance arises when two materials converge and there is a thermal barrier at their interface. This resistance is due to surface roughness and imperfections that restrict the flow of heat. In the given exercise, the contact resistance at the interface between materials A and B is 0.30 m²·K/W, assuming perfect contact over the interface area. The actual resistance to heat flow, R_contact, is obtained by dividing the contact resistance per unit area by the actual surface area of contact. Contact resistance often acts as a bottleneck in heat transfer across composite structures.

Heat Transfer Rate

The heat transfer rate, denoted by Q, describes the amount of heat flowing through a material per unit time. It is calculated using the overall temperature difference and the total thermal resistance of the heat path. In the exercise, the formula Q = (T_hot - T_cold) / R_total is used, where T_hot and T_cold are the temperatures of the hot and cold fluids, respectively. The heat transfer rate is essential in engineering applications as it defines the efficiency of thermal systems and plays a crucial role in thermal management and control.

Temperature Distribution

Temperature distribution refers to how temperature varies across a system from one location to another. In the context of a composite wall, it is vital to understand how temperature reduces from the hotter side to the cooler side and how various resistances affect this gradient. Conduction, convection, and contact resistance each cause a 'drop' in temperature akin to voltage drop in electrical circuits. The solution's sketch visualizes the continuous decrease in temperature across different media in the system. Understanding temperature distribution allows for predictions and enhancements of material performance and energy efficiency in thermal systems.