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Chapter 3: Problem 137

Circular copper rods of diameter \(D=1 \mathrm{~mm}\) and length \(L=25 \mathrm{~mm}\) are used to enhance heat transfer from a surface that is maintained at \(T_{s, 1}=100^{\circ} \mathrm{C}\). One end of the rod is attached to this surface (at \(x=0\) ), while the other end \((x=25 \mathrm{~mm})\) is joined to a second surface, which is maintained at \(T_{s, 2}=0^{\circ} \mathrm{C}\). Air flowing between the surfaces (and over the rods) is also at a temperature of \(T_{\infty}=0^{\circ} \mathrm{C}\), and a convection coefficient of \(h=100 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) is maintained. (a) What is the rate of heat transfer by convection from a single copper rod to the air? (b) What is the total rate of heat transfer from a \(1 \mathrm{~m} \times 1 \mathrm{~m}\) section of the surface at \(100^{\circ} \mathrm{C}\), if a bundle of the rods is installed on 4 -mm centers?

Short Answer

Expert verified
The rate of heat transfer by convection from a single copper rod to the air is approximately 1.5708 W. The total rate of heat transfer from a 1m x 1m section of the surface with a bundle of rods installed on 4mm centers is approximately 98,175 W.

Step by step solution

01

Calculate the Surface Area of a Rod

First, we need to find the surface area of a single rod. Since it's a cylindrical shape, we need to calculate the lateral area. The formula for the lateral area of a cylinder is: \(A = 2 \pi rL\) where: - \(A\) is the lateral area of the cylinder (in m虏). - \(r\) is the radius of the cylinder (in m). - \(L\) is the length of the cylinder (in m). We are given the diameter of the rod, which is \(D = 1mm\), so the radius, in meters, is: \(r = \frac{D}{2} = \frac{0.001m}{2} = 0.0005m\) Now, we can calculate the surface area of a single rod: \(A = 2\pi(0.0005m)(0.025m) = 0.00015708\, m^2\)
02

Calculate the Rate of Heat Transfer by Convection

Now that we have the surface area, we can use the given values to calculate the rate of heat transfer by convection for a single rod. \(q = hA(T_{s, 1} - T_\infty)\) Using the given values: - \(h = 100 \, \mathrm{W/m^2K}\) - \(T_{s, 1} = 100^{\circ} \mathrm{C} = 373.15 K\) - \(T_\infty = 0^{\circ} \mathrm{C} = 273.15 K\) Calculate the rate of heat transfer by convection: \(q = (100 \, \mathrm{W/m^2K})(0.00015708 \, m^2)(373.15 \, K - 273.15 \, K) = 1.5708 \, W\) The rate of heat transfer by convection from a single copper rod to the air is approximately 1.5708 W. #b) Total Rate of Heat Transfer from a 1m x 1m Section#
03

Calculate the Number of Installed Rods

We are given that a bundle of rods is installed on 4mm centers. Since the dimensions of the surface are \(1m\times1m\), we can calculate the number of rods in each row and column of the grid: \(\textrm{Rods per row} = 1m / 0.004m = 250 \, \textrm{rods}\) \(\textrm{Rods per column} = 1m / 0.004m = 250 \, \textrm{rods}\) Now, we find the total number of rods installed on the surface: \(\textrm{Total rods} = 250 \times 250 = 62,500\)
04

Calculate the Total Rate of Heat Transfer

Now that we have the number of rods and the rate of heat transfer by convection per rod, we can calculate the total rate of heat transfer from the 1m x 1m section: \(\textrm{Total rate of heat transfer} = \textrm{Total rods} \times \textrm{Rate of heat transfer per rod} = 62,500 \times 1.5708 \, W = 98,175 \, W\) The total rate of heat transfer from a 1m x 1m section of the surface is approximately 98,175 W.

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Key Concepts

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

Thermal Conduction
Thermal conduction is the transfer of heat through a material due to the direct contact of its particles. In materials like copper rods, heat energy is passed along from the hot end to the cooler end at the molecular level. This transfer occurs because particles at the higher temperature end have more kinetic energy and vibrate more vigorously, which in turn transfers energy to adjacent, less energetic particles. In the context of our exercise, although the primary mechanism we're discussing is convection, conduction within the rods is vital for ensuring the heat from the hot surface reaches the surface area where convection can occur.
Rate of Heat Transfer
The rate of heat transfer refers to the amount of heat energy moving from one place to another per unit time. Our exercise shows how to calculate this rate using the formula
\( q = hA(T_{s, 1} - T_\infty) \),
where \( q \) represents the heat transfer rate in watts (W), \( h \) the convection coefficient (W/m虏K), \( A \) the area through which the heat is being transferred (m虏), and \( (T_{s, 1} - T_\infty) \) the temperature difference between the surface and the fluid (K). Convection allows heat to be transferred from the rod to the air, and the 'rate' gives us a quantitative measurement of this process.
Convection Coefficient
The convection coefficient, symbolized by the letter \( h \), is a measure of how effectively a fluid (in our case, air) can remove heat from a surface. The higher the convection coefficient, the more efficient the heat transfer process. In the exercise, a convection coefficient of \( h = 100 W/m^2K \) is given. This value, combined with the surface area and the temperature difference, allows us to calculate the convection-based heat transfer for a single copper rod. The knowledge of \( h \) is critical because it directly affects the heat transfer rate.
Cylindrical Surface Area
Cylindrical surface area is crucial when determining how much heat is transferred in objects like the copper rods mentioned in our exercise. Since the rod is a cylinder, its lateral surface area represents the contact area with the air through which the heat is being transferred. The larger this area, the greater the potential for heat transfer. The cylinder's lateral surface area is given by the formula
\( A = 2\pi rL \),
where \( r \) is the radius and \( L \) is the length of the cylinder. Properly calculating the cylindrical surface area ensures that we can accurately determine the rate of heat transfer via convection.
Temperature Gradient
The temperature gradient is the change in temperature with distance. It is essentially what drives the heat transfer; in the case of our exercise, it's the change in temperature from the surface of the rod at \( T_{s, 1} \) to the air at \( T_\infty \). A greater temperature gradient generally means a greater potential for heat transfer. It's crucial to accurately measure or calculate the temperatures at both ends for a precise evaluation of the heat transfer rate. In practice, the gradient dictates how quickly the heat will flow from the copper rod to the surrounding air, assuming all other factors are held constant.

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Most popular questions from this chapter

A thermopane window consists of two pieces of glass \(7 \mathrm{~mm}\) thick that enclose an air space \(7 \mathrm{~mm}\) thick. The window separates room air at \(20^{\circ} \mathrm{C}\) from outside ambient air at \(-10^{\circ} \mathrm{C}\). The convection coefficient associated with the inner (room-side) surface is \(10 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). (a) If the convection coefficient associated with the outer (ambient) air is \(h_{o}=80 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), what is the heat loss through a window that is \(0.8 \mathrm{~m}\) long by \(0.5 \mathrm{~m}\) wide? Neglect radiation, and assume the air enclosed between the panes to be stagnant. (b) Compute and plot the effect of \(h_{o}\) on the heat loss for \(10 \leq h_{o} \leq 100 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Repeat this calculation for a triple-pane construction in which a third pane and a second air space of equivalent thickness are added.

The energy transferred from the anterior chamber of the eye through the cornea varies considerably depending on whether a contact lens is worn. Treat the eye as a spherical system and assume the system to be at steady state. The convection coefficient \(h_{o}\) is unchanged with and without the contact lens in place. The cornea and the lens cover one-third of the spherical surface area. Values of the parameters representing this situation are as follows: \(\begin{array}{ll}r_{1}=10.2 \mathrm{~mm} & r_{2}=12.7 \mathrm{~mm} \\\ r_{3}=16.5 \mathrm{~mm} & T_{\infty, o}=21^{\circ} \mathrm{C} \\ T_{\infty \infty, i}=37^{\circ} \mathrm{C} & k_{2}=0.80 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K} \\ k_{1}=0.35 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K} & h_{o}=6 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K} \\ h_{i}=12 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K} & \end{array}\) (a) Construct the thermal circuits, labeling all potentials and flows for the systems excluding the contact lens and including the contact lens. Write resistance elements in terms of appropriate parameters. (b) Determine the heat loss from the anterior chamber with and without the contact lens in place. (c) Discuss the implication of your results.

Gold is commonly used in semiconductor packaging to form interconnections (also known as interconnects) that carry electrical signals between different devices in the package. In addition to being a good electrical conductor, gold interconnects are also effective at protecting the heat-generating devices to which they are attached by conducting thermal energy away from the devices to surrounding, cooler regions. Consider a thin film of gold that has a cross section of \(60 \mathrm{~nm} \times 250 \mathrm{~nm}\). (a) For an applied temperature difference of \(20^{\circ} \mathrm{C}\), determine the energy conducted along a \(1-\mu \mathrm{m}\) long, thin-film interconnect. Evaluate properties at \(300 \mathrm{~K}\). (b) Plot the lengthwise (in the \(1-\mu \mathrm{m}\) direction) and spanwise (in the thinnest direction) thermal conductivities of the gold film as a function of the film thickness \(L\) for \(30 \leq L \leq 140 \mathrm{~nm}\).

A spherical Pyrex glass shell has inside and outside diameters of \(D_{1}=0.1 \mathrm{~m}\) and \(D_{2}=0.2 \mathrm{~m}\), respectively. The inner surface is at \(T_{s, 1}=100^{\circ} \mathrm{C}\) while the outer surface is at \(T_{s, 2}=45^{\circ} \mathrm{C}\). (a) Determine the temperature at the midpoint of the shell thickness, \(T\left(r_{m}=0.075 \mathrm{~m}\right)\). (b) For the same surface temperatures and dimensions as in part (a), show how the midpoint temperature would change if the shell material were aluminum.

A brass rod \(100 \mathrm{~mm}\) long and \(5 \mathrm{~mm}\) in diameter extends horizontally from a casting at \(200^{\circ} \mathrm{C}\). The rod is in an air environment with \(T_{\infty}=20^{\circ} \mathrm{C}\) and \(h=30\) \(\mathrm{W} / \mathrm{m}^{2} \cdot \mathrm{K}\). What is the temperature of the rod 25,50 , and \(100 \mathrm{~mm}\) from the casting?
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