/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 19 Two cylindrical rods are identic... [FREE SOLUTION] | 91Ó°ÊÓ

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Two cylindrical rods are identical, except that one has a thermal conductivity \(k_{1}\) and the other has a thermal conductivity \(k_{2} .\) As the drawing shows, they are placed between two walls that are maintained at different temperatures \(T_{\mathrm{W}}\) (warmer) and \(T_{\mathrm{C}}\) (cooler). When the rods are arranged as in part \(a\) of the drawing, a total heat \(Q^{\prime}\) flows from the warmer to the cooler wall, but when the rods are arranged as in part \(b\) , the total heat flow is \(Q\) . Assuming that the conductivity \(k_{2}\) is twice as great as \(k_{1}\) and that heat flows only along the lengths of the rods, determine the ratio \(Q^{\prime} / Q .\)

Short Answer

Expert verified
The ratio \(\frac{Q'}{Q}\) is \(\frac{9}{2}\).

Step by step solution

01

Understand Heat Flow in Parallel Arrangement

When the rods are in a parallel arrangement, heat flows through both rods simultaneously. The total heat flow is the sum of the heat flows through each rod. Using the formula for heat conduction \(Q = \frac{kA(T_1 - T_2)t}{L}\), where \(A\) is the cross-sectional area and \(L\) is the length of the rod, we find the individual heat flows through the rods as \(Q_1' = \frac{k_1A(T_W - T_C)t}{L}\) and \(Q_2' = \frac{k_2A(T_W - T_C)t}{L}\). Thus, the total heat flow is \(Q' = Q_1' + Q_2' = \left(\frac{k_1 + k_2}{L}\right)A(T_W - T_C)t\).
02

Understand Heat Flow in Series Arrangement

For rods in series, the total heat flow rate is the same through each rod due to continuity. The effective thermal resistance is the sum of individual resistances: \(R_1 = \frac{L}{k_1A}\) and \(R_2 = \frac{L}{k_2A}\). Total resistance \(R = R_1 + R_2 = \frac{L}{k_1A} + \frac{L}{k_2A}\). The heat flow through rods in series is \(Q = \frac{(T_W - T_C)}{R}\), simplifying to \(Q = \frac{A(T_W - T_C)t}{\frac{L}{k_1} + \frac{L}{k_2}}\).
03

Calculate the Ratio \(\frac{Q'}{Q}\)

Now we find the ratio of the heat flows \(\frac{Q'}{Q}\). From Step 1, \(Q' = \left(\frac{k_1 + k_2}{L}\right)A(T_W - T_C)t\) and from Step 2, \(Q = \frac{A(T_W - T_C)t}{\frac{L}{k_1} + \frac{L}{k_2}}\). This ratio becomes \(\frac{Q'}{Q} = \frac{\frac{k_1 + k_2}{L}}{\frac{k_1k_2}{k_1 + k_2}}\). Simplifying gives \(\frac{Q'}{Q} = \frac{(k_1 + k_2)^2}{k_1k_2}\).
04

Substitute Known Values and Calculate Final Ratio

Given \(k_2 = 2k_1\), substitute into the expression from Step 3, \(\frac{Q'}{Q} = \frac{(k_1 + 2k_1)^2}{k_1 \times 2k_1} = \frac{9k_1^2}{2k_1^2} = \frac{9}{2}\).

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Key Concepts

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

Heat Flow
Heat flow represents the transfer of thermal energy from an area of higher temperature to an area of lower temperature. In the context of thermal conductivity, it is crucial to understand that this flow is driven by the temperature difference between two points. This process occurs naturally, aiming to reach thermal equilibrium, where the temperatures are balanced.
Using the formula for heat conduction, we can calculate the heat flow through a rod: \[Q = \frac{kA(T_1 - T_2)t}{L}\]Where:
  • \(k\) is the thermal conductivity of the material.
  • \(A\) is the cross-sectional area of the rod.
  • \(T_1\) and \(T_2\) are the temperatures at either end of the rod (\(T_1 > T_2\)).
  • \(t\) is the time for which the heat flow occurs.
  • \(L\) is the length of the rod.

Heat flow is faster in materials with higher thermal conductivity and greater temperature differences. Understanding heat flow is essential for addressing real-world engineering problems, such as insulation, refrigeration, and heating systems. Hence, thermal conduction is a fundamental mechanism that governs many natural and technological processes.
Parallel Arrangement
When rods are arranged in parallel between two temperatures, they both conduct heat simultaneously. This setup ensures the total heat conducted is the sum of each rod's heat contribution. This parallel arrangement allows for efficient thermal management since it distributes the heat over multiple paths.
In a parallel arrangement, we use the heat conduction formula for each rod separately. For example, the heat conducted through the first rod, \(Q_1'\), is given by:\[Q_1' = \frac{k_1 A (T_W - T_C)t}{L}\]while for the second rod, \(Q_2'\), is:\[Q_2' = \frac{k_2 A (T_W - T_C)t}{L}\]Thus, the total heat flow in a parallel arrangement, \(Q'\), is:\[Q' = Q_1' + Q_2' = \left(\frac{k_1 + k_2}{L}\right)A(T_W - T_C)t\]
This setup is useful in scenarios where a quick heat dissipation is required, such as in cooling systems and heat exchangers. The ability to add the conductivity of each path allows for flexibility and optimization in designs where thermal properties need adjustments.
Series Arrangement
In a series arrangement, heat transfers through each rod one after the other. This setting implies that the heat flow rate through both rods remains the same due to continuity. Hence, the heat flow is dictated by each rod's combined thermal resistance.
The key factor in series is thermal resistance, calculated as:\[R_1 = \frac{L}{k_1 A}\]\[R_2 = \frac{L}{k_2 A}\]Where:
  • \(R_1\) and \(R_2\) are the thermal resistances for rods 1 and 2 respectively.
The total thermal resistance, \(R\), in series is:\[R = R_1 + R_2 = \frac{L}{k_1 A} + \frac{L}{k_2 A}\]The heat flow through the series arrangement is:\[Q = \frac{A(T_W - T_C)t}{R}\]Thus, the flow decreases with increased resistance. This setup is often applied where controlled heat flow is needed, as in thermal insulation and protective barriers against heat escape.
Series arrangement is an efficient method to regulate the total heat flow by merely altering the properties of the materials used.
Thermal Resistance
Thermal resistance is a property that quantifies a material's resistance to heat flow. It is instrumental in both series and parallel arrangements. High thermal resistance implies low heat flow for a given temperature difference.
Thermal resistance is computed with the formula:\[R = \frac{L}{k A}\]Where:
  • \(L\) is the length of the material.
  • \(k\) is the material's thermal conductivity.
  • \(A\) is the cross-sectional area.
Greater resistance is achieved by increasing the length or reducing the thermal conductivity in a series configuration. In parallel arrangements, thermal resistance provides insight into how efficiently multiple paths can conduct heat.
Understanding thermal resistance allows for designing materials that either promote or retard heat flow, depending on the application. It's crucial in the development of thermal insulators in buildings and the optimization of electronic devices to prevent overheating. Hence, mastering the concept of thermal resistance is central to a wide range of practical applications.

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Most popular questions from this chapter

A cubical piece of heat-shield tile from the space shuttle measures 0.10 \(\mathrm{m}\) on a side and has a thermal conductivity of 0.065 \(\mathrm{J} /\left(\mathrm{s} \cdot \mathrm{m} \cdot \mathrm{C}^{\circ}\right) .\) The outer surface of the tile is heated to a temperature of \(1150^{\circ} \mathrm{C},\) while the inner surface is maintained at a temperature of \(20.0^{\circ} \mathrm{C}\) . (a) How much heat flows from the outer to the inner surface of the tile in five minutes? \((\mathrm{b})\) If this amount of heat were transferred to two liters \((2.0 \mathrm{kg})\) of liquid water, by how many Celsius degrees would the temperature of the water rise?

The Kelvin temperature of an object is \(T_{1},\) and the object radiates a certain amount of energy per second. The Kelvin temperature of the object is then increased to \(T_{2},\) and the object radiates twice the energy per second that it radiated at the lower temperature. What is the ratio \(T_{2} / T_{1} ?\)

A solid sphere has a temperature of 773 \(\mathrm{K}\) . The sphere is melted down and recast into a cube thas the same emissivity and emits the same radiant power as the sphere. What is the cube's temperature?

Sirius \(B\) is a white star that has a surface temperature (in kelvins) that is four times that of our sun. Sirius \(\mathrm{B}\) radiates only 0.040 times the power radiated by the sun. Our sun has a radius of \(6.96 \times 10^{8} \mathrm{m} .\) Assuming that Sirius \(\mathrm{B}\) has the same emissivity as the sun, find the radius of Sirius \(\mathrm{B}\) .

Multiple-Concept Example 8 reviews the approach that is used in problems such as this. A person eats a dessert that contains 260 Calories. This "Calorie" unit, with a capital \(C,\) is the one used by nutritionists; 1 Calorie \(=4186 \mathrm{J}\) . See Section \(12.7 .\) The skin temperature of this individual is \(36^{\circ} \mathrm{C}\) and that of her environment is \(21^{\circ} \mathrm{C}\) . The emissivity of her skin is 0.75 and its surface area is 1.3 \(\mathrm{m}^{2}\) How much time would it take for her to emit a net radiant energy from her body that is equal to the energy contained in this dessert?

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