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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

A steam pipe of \(0.12-\mathrm{m}\) outside diameter is insulated with a layer of calcium silicate. (a) If the insulation is \(20 \mathrm{~mm}\) thick and its inner and outer surfaces are maintained at \(T_{s, 1}=800 \mathrm{~K}\) and \(T_{s, 2}=490 \mathrm{~K}\), respectively, what is the heat loss per unit length \(\left(q^{\prime}\right)\) of the pipe? (b) We wish to explore the effect of insulation thickness on the heat loss \(q^{\prime}\) and outer surface temperature \(T_{s, 2}\), with the inner surface temperature fixed at \(T_{s, 1}=\) \(800 \mathrm{~K}\). The outer surface is exposed to an airflow \(\left(T_{\infty}=25^{\circ} \mathrm{C}\right)\) that maintains a convection coefficient of \(h=25 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) and to large surroundings for which \(T_{\text {sur }}=T_{\infty}=25^{\circ} \mathrm{C}\). The surface emissivity of calcium silicate is approximately \(0.8\). Compute and plot the temperature distribution in the insulation as a function of the dimensionless radial coordinate, \(\left(r-r_{1}\right) /\left(r_{2}-r_{1}\right)\), where \(r_{1}=0.06 \mathrm{~m}\) and \(r_{2}\) is a variable \(\left(0.06

Consider a tube wall of inner and outer radii \(r_{i}\) and \(r_{o}\), whose temperatures are maintained at \(T_{i}\) and \(T_{o}\), respectively. The thermal conductivity of the cylinder is temperature dependent and may be represented by an expression of the form \(k=k_{o}(1+a T)\), where \(k_{o}\) and \(a\) are constants. Obtain an expression for the heat transfer per unit length of the tube. What is the thermal resistance of the tube wall?

Rows of the thermoelectric modules of Example \(3.13\) are attached to the flat absorber plate of Problem 3.108. The rows of modules are separated by \(L_{\text {sep }}=0.5 \mathrm{~m}\) and the backs of the modules are cooled by water at a temperature of \(T_{w}=40^{\circ} \mathrm{C}\), with \(h=45 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Determine the electric power produced by one row of thermoelectric modules connected in series electrically with a load resistance of \(60 \Omega\). Calculate the heat transfer rate to the flowing water. Assume rows of 20 immediately adjacent modules, with the lengths of both the module rows and water tubing to be \(L_{\text {row }}=20 W\) where \(W=54 \mathrm{~mm}\) is the module dimension taken from Example 3.13. Neglect thermal contact resistances and the temperature drop across the tube wall, and assume that the high thermal conductivity tube wall creates a uniform temperature around the tube perimeter. Because of the thermal resistance provided by the thermoelectric modules, it is no longer appropriate to assume that the temperature of the absorber plate directly above a tube is equal to that of the water.

As seen in Problem \(3.109\), silicon carbide nanowires of diameter \(D=15 \mathrm{~nm}\) can be grown onto a solid silicon carbide surface by carefully depositing droplets of catalyst liquid onto a flat silicon carbide substrate. Silicon carbide nanowires grow upward from the deposited drops, and if the drops are deposited in a pattern, an array of nanowire fins can be grown, forming a silicon carbide nano-heat sink. Consider finned and unfinned electronics packages in which an extremely small, \(10 \mu \mathrm{m} \times 10 \mu \mathrm{m}\) electronics device is sandwiched between two \(d=100\)-nm-thick silicon carbide sheets. In both cases, the coolant is a dielectric liquid at \(20^{\circ} \mathrm{C}\). A heat transfer coefficient of \(h=1 \times 10^{5} \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) exists on the top and bottom of the unfinned package and on all surfaces of the exposed silicon carbide fins, which are each \(L=300 \mathrm{~nm}\) long. Each nano-heat sink includes a \(200 \times 200\) array of nanofins. Determine the maximum allowable heat rate that can be generated by the electronic device so that its temperature is maintained at \(T_{t}<85^{\circ} \mathrm{C}\) for the unfinned and finned packages.

A \(0.20\)-m-diameter, thin-walled steel pipe is used to transport saturated steam at a pressure of 20 bars in a room for which the air temperature is \(25^{\circ} \mathrm{C}\) and the convection heat transfer coefficient at the outer surface of the pipe is \(20 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). (a) What is the heat loss per unit length from the bare pipe (no insulation)? Estimate the heat loss per unit length if a 50 -mm-thick layer of insulation (magnesia, \(85 \%\) is added. The steel and magnesia may each be assumed to have an emissivity of \(0.8\), and the steam-side convection resistance may be neglected. (b) The costs associated with generating the steam and installing the insulation are known to be \(\$ 4 / 10^{9} \mathrm{~J}\) and \(\$ 100 / \mathrm{m}\) of pipe length, respectively. If the steam line is to operate \(7500 \mathrm{~h} / \mathrm{yr}\), how many years are needed to pay back the initial investment in insulation?
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