/*! 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 137 Circular copper rods of diameter... [FREE SOLUTION] | 91影视

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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

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 \(h_{i}=h_{o}=5 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\).

A new building to be located in a cold climate is being designed with a basement that has an \(L=200\)-mm-thick wall. Inner and outer basement wall temperatures are \(T_{i}=20^{\circ} \mathrm{C}\) and \(T_{o}=0^{\circ} \mathrm{C}\), respectively. The architect can specify the wall material to be either aerated concrete block with \(k_{\mathrm{ac}}=0.15 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\), or stone mix concrete. To reduce the conduction heat flux through the stone mix wall to a level equivalent to that of the aerated concrete wall, what thickness of extruded polystyrene sheet must be applied onto the inner surface of the stone mix con-crete wall? Floor dimensions of the basement are \(20 \mathrm{~m} \times 30 \mathrm{~m}\), and the expected rental rate is \(\$ 50 / \mathrm{m}^{2} /\) month. What is the yearly cost, in terms of lost rental income, if the stone mix concrete wall with polystyrene insulation is specified?

A device used to measure the surface temperature of an object to within a spatial resolution of approximately \(50 \mathrm{~nm}\) is shown in the schematic. It consists of an extremely sharp-tipped stylus and an extremely small cantilever that is scanned across the surface. The probe tip is of circular cross section and is fabricated of polycrystalline silicon dioxide. The ambient temperature is measured at the pivoted end of the cantilever as \(T_{\infty}=\) \(25^{\circ} \mathrm{C}\), and the device is equipped with a sensor to measure the temperature at the upper end of the sharp tip, \(T_{\text {sen. }}\). The thermal resistance between the sensing probe and the pivoted end is \(R_{t}=5 \times 10^{6} \mathrm{~K} / \mathrm{W}\). (a) Determine the thermal resistance between the surface temperature and the sensing temperature. (b) If the sensing temperature is \(T_{\text {sen }}=28.5^{\circ} \mathrm{C}\), determine the surface temperature. Hint: Although nanoscale heat transfer effects may be important, assume that the conduction occurring in the air adjacent to the probe tip can be described by Fourier's law and the thermal conductivity found in Table A. \(4 .\)

A thin electrical heater is wrapped around the outer surface of a long cylindrical tube whose inner surface is maintained at a temperature of \(5^{\circ} \mathrm{C}\). The tube wall has inner and outer radii of 25 and \(75 \mathrm{~mm}\), respectively, and a thermal conductivity of \(10 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). The thermal contact resistance between the heater and the outer surface of the tube (per unit length of the tube) is \(R_{t, c}^{\prime}=\) \(0.01 \mathrm{~m} \cdot \mathrm{K} / \mathrm{W}\). The outer surface of the heater is exposed to a fluid with \(T_{\infty}=-10^{\circ} \mathrm{C}\) and a convection coefficient of \(h=100 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Determine the heater power per unit length of tube required to maintain the heater at \(T_{o}=25^{\circ} \mathrm{C} .\)

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 \(334 \mathrm{~kJ} / \mathrm{kg}\).
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