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Chapter 3: Problem 139

Consider two long, slender rods of the same diameter but different materials. One end of each rod is attached to a base surface maintained at \(100^{\circ} \mathrm{C}\), while the surfaces of the rods are exposed to ambient air at \(20^{\circ} \mathrm{C}\). By traversing the length of each rod with a thermocouple, it was observed that the temperatures of the rods were equal at the positions \(x_{\mathrm{A}}=0.15 \mathrm{~m}\) and \(x_{\mathrm{B}}=0.075 \mathrm{~m}\), where \(x\) is measured from the base surface. If the thermal conductivity of rod \(\mathrm{A}\) is known to be \(k_{\mathrm{A}}=70 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\), determine the value of \(k_{\mathrm{B}}\) for rod B.

Short Answer

Expert verified
The thermal conductivity of rod B is \(140 \mathrm{~W/m\cdot K}\).

Step by step solution

01

Write down the Fourier's law of heat conduction

Fourier's law of heat conduction states that the rate of heat transfer (Q) through a material is proportional to the temperature gradient and the material's thermal conductivity: \(Q = -kA\frac{dT}{dx}\) Where: - Q is the rate of heat transfer (W) - k is the thermal conductivity of the material (W/m*K) - A is the cross-sectional area (m虏) - \( \frac{dT}{dx} \) is the temperature gradient (K/m) Since the rods A and B have equal temperatures at the positions \(x_A\) and \(x_B\), we can assume the rate of heat transfer is also equal for both rods. This assumption allows us to relate the thermal conductivities of the rods.
02

Equate the rate of heat transfer for both rods

As mentioned earlier, due to equal temperatures at the specified positions, we can assume an equal rate of heat transfer for both rods A and B. Therefore, we can write the equation: \(Q_A = Q_B\) Applying Fourier's law to both rods, we get: \(-k_AA\frac{dT_A}{dx_{A}} = -k_BA\frac{dT_B}{dx_{B}}\) The rods have the same diameter, meaning they have the same cross-sectional area (A). We can simplify the equation: \(\frac{k_A}{k_B} = \frac{\frac{dT_A}{dx_{A}}}{\frac{dT_B}{dx_{B}}}\)
03

Calculate temperature gradients

To find the temperature gradients, we can use the information provided in the problem, which states that the base surface is maintained at \(100^{\circ}\textrm{C}\), while the ambient air temperature is \(20^{\circ}\textrm{C}\). Thus, the temperature difference for both rods is the same: \(\Delta T = 100 - 20 = 80\textrm{K}\). For rod A, the temperature gradient can be calculated as: \(\frac{dT_A}{dx_{A}} = \frac{\Delta T}{x_{A}} = \frac{80}{0.15} = 533.33\mathrm{~K/m}\) Similarly, for rod B, the temperature gradient can be calculated as: \(\frac{dT_B}{dx_{B}} = \frac{\Delta T}{x_{B}} = \frac{80}{0.075} = 1066.67\mathrm{~K/m}\) Now we can substitute these values back into our equation from Step 2.
04

Determine the thermal conductivity of rod B

Plugging the temperature gradients and given \(k_A\) value into our equation from Step 2, we can solve for \(k_B\): \(\frac{70}{k_B} = \frac{533.33}{1066.67}\) Cross-multiplying and solving for \(k_B\): \(k_B = \frac{70\times1066.67}{533.33} = 140\mathrm{~W/m\cdot K}\) Thus, the thermal conductivity of rod B is \(140 \mathrm{~W/m\cdot K}\).

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Key Concepts

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

Fourier's Law of Heat Conduction
Fourier's Law of Heat Conduction is a fundamental principle that helps us understand how heat travels through a material. This law states that heat flows from regions of high temperature to regions of low temperature and that the rate at which this transfer happens is proportional to the temperature gradient and the thermal conductivity of the material. The mathematical expression for Fourier's law is:
  • \( Q = -kA\frac{dT}{dx} \)
Here:
  • \( Q \) represents the rate of heat transfer in watts.
  • \( k \) is the thermal conductivity, measured in watts per meter-kelvin \( \mathrm{(W/m \cdot K)} \).
  • \( A \) is the cross-sectional area through which heat is conducted \( \mathrm{(m^2)} \).
  • \( \frac{dT}{dx} \) is the temperature gradient \( \mathrm{(K/m)} \), which tells us how the temperature changes as we move through the material.
Fourier's law helps us to calculate how much heat energy is transferred within a certain time frame. Understanding this law is crucial for solving problems related to heat conduction, like calculating thermal conductivity or temperature changes in materials.
Temperature Gradient
The temperature gradient is a measure of how temperature changes as you move through a material. It is essentially the rate of change of temperature with respect to distance. In mathematical terms, it is represented as the derivative of temperature with respect to the position, or \( \frac{dT}{dx} \).In the context of the original exercise, the temperature gradient is critical to calculating how quickly heat moves through the rods. For rod A and rod B, the temperature gradients were calculated as:
  • Rod A: \( \frac{dT_A}{dx_A} = 533.33\, \mathrm{K/m} \)
  • Rod B: \( \frac{dT_B}{dx_B} = 1066.67\, \mathrm{K/m} \)
These values indicate how rapidly the temperature drops as you move along each rod. A higher temperature gradient means that there is a steeper change in temperature, which can affect the rate at which heat is transferred. The difference in temperature gradients for rods A and B plays a fundamental role in determining the thermal conductivity of rod B.
Heat Transfer Rate
The heat transfer rate, denoted as \( Q \), refers to the amount of heat energy transferred per unit time through a material. It is a vital concept that helps evaluate the efficiency of materials in conducting heat.In practice, this rate depends on several factors, such as the thermal conductivity of the material, the cross-sectional area, and the temperature gradient. According to Fourier's Law, the heat transfer rate is calculated using the formula:
  • \( Q = -kA\frac{dT}{dx} \)
In the original problem, by equating the heat transfer rates of rods A and B at certain positions, we determined that:
  • The heat transfer rates were equal at specific lengths, allowing us to set the expressions equal to each other: \( Q_A = Q_B \).
  • This equivalence provided a method to solve for the unknown thermal conductivity of rod B.
By thoroughly understanding this concept, you can optimize processes that involve heat exchange, whether in engineering, everyday appliances, or cutting-edge technology.

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Most popular questions from this chapter

Consider a composite wall that includes an 8-mm-thick hardwood siding, 40 -mm by 130 -mm hardwood studs on \(0.65-\mathrm{m}\) centers with glass fiber insulation (paper faced, \(28 \mathrm{~kg} / \mathrm{m}^{3}\) ), and a 12 -mm layer of gypsum (vermiculite) wall board. What is the thermal resistance associated with a wall that is \(2.5 \mathrm{~m}\) high by \(6.5 \mathrm{~m}\) wide (having 10 studs, each \(2.5 \mathrm{~m}\) high)? Assume surfaces normal to the \(x\)-direction are isothermal.

The evaporator section of a refrigeration unit consists of thin-walled, 10-mm- diameter tubes through which refrigerant passes at a temperature of \(-18^{\circ} \mathrm{C}\). Air is cooled as it flows over the tubes, maintaining a surface convection coefficient of \(100 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), and is subsequently routed to the refrigerator compartment. (a) For the foregoing conditions and an air temperature of \(-3^{\circ} \mathrm{C}\), what is the rate at which heat is extracted from the air per unit tube length? (b) If the refrigerator's defrost unit malfunctions, frost will slowly accumulate on the outer tube surface. Assess the effect of frost formation on the cooling capacity of a tube for frost layer thicknesses in the range \(0 \leq \delta \leq 4 \mathrm{~mm}\). Frost may be assumed to have a thermal conductivity of \(0.4 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). (c) The refrigerator is disconnected after the defrost unit malfunctions and a 2-mm-thick layer of frost has formed. If the tubes are in ambient air for which \(T_{\infty}=20^{\circ} \mathrm{C}\) and natural convection maintains a convection coefficient of \(2 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), how long will it take for the frost to melt? The frost may be assumed to have a mass density of \(700 \mathrm{~kg} / \mathrm{m}^{3}\) and a latent heat of fusion of \(334 \mathrm{~kJ} / \mathrm{kg}\).

A very long rod of \(5-\mathrm{mm}\) diameter and uniform thermal conductivity \(k=25 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) is subjected to a heat treatment process. The center, 30 -mm-long portion of the rod within the induction heating coil experiences uniform volumetric heat generation of \(7.5 \times 10^{6} \mathrm{~W} / \mathrm{m}^{3}\). The unheated portions of the rod, which protrude from the heating coil on either side, experience convection with the ambient air at \(T_{\infty}=20^{\circ} \mathrm{C}\) and \(h=10 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Assume that there is no convection from the surface of the rod within the coil. (a) Calculate the steady-state temperature \(T_{o}\) of the rod at the midpoint of the heated portion in the coil. (b) Calculate the temperature of the rod \(T_{b}\) at the edge of the heated portion.

The walls of a refrigerator are typically constructed by sandwiching a layer of insulation between sheet metal panels. Consider a wall made from fiberglass insulation of thermal conductivity \(k_{i}=0.046 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) and thickness \(L_{i}=50 \mathrm{~mm}\) and steel panels, each of thermal conductivity \(k_{p}=60 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) and thickness \(L_{p}=3 \mathrm{~mm}\). If the wall separates refrigerated air at \(T_{\infty, i}=4^{\circ} \mathrm{C}\) from ambient air at \(T_{\infty, o}=25^{\circ} \mathrm{C}\), what is the heat gain per unit surface area? Coefficients associated with natural convection at the inner and outer surfaces may be approximated as \(h_{i}=h_{o}=5 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\).

Aluminum fins of triangular profile are attached to a plane wall whose surface temperature is \(250^{\circ} \mathrm{C}\). The fin base thickness is \(2 \mathrm{~mm}\), and its length is \(6 \mathrm{~mm}\). The system is in ambient air at a temperature of \(20^{\circ} \mathrm{C}\), and the surface convection coefficient is \(40 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). (a) What are the fin efficiency and effectiveness? (b) What is the heat dissipated per unit width by a single fin?
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