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Chapter 3: Problem 92

A long cylindrical rod of diameter \(200 \mathrm{~mm}\) with thermal conductivity of \(0.5 \mathrm{~W} / \mathrm{m}+\mathrm{K}\) experiences uniform volumetric heat genicration of \(24,000 \mathrm{~W} / \mathrm{m}^{3}\). The rod is encapsulated by a circular siceve having an outer diameter of \(400 \mathrm{~mm}\) and a thermal conductivity of \(4 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). The cuter surface of the sleeve is exposed to cross flow of air at \(27^{\circ} \mathrm{C}\) with a convection coefficient of \(25 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). (a) Find the temperature at the interface between the rod and slecve and on the cuter surface. (b) What is the temperature at the center of the rod?

Short Answer

Expert verified
T鈧� and T鈧� are found through energy balance and heat transfer equations, while T鈧� is computed from internal heat generation.

Step by step solution

01

Determine Heat Transfer Area

First, calculate the heat transfer areas. The radius of the rod is 100 mm or 0.1 m, and therefore its surface area is given by 2蟺rL for unit length, which results in approximately 0.628 m虏. For the sleeve, the outer radius of 400 mm is 0.2 m, and thus its surface area is 1.256 m虏 per unit length.
02

Calculate Heat Generation

The volumetric heat generation rate is given as 24,000 W/m鲁. For a unit volume with a radius of 0.1 m (the rod), the heat generated, \(q_{gen}\), is \(q_{gen} = 24,000 \times \frac{1}{ ext{unit length}}\), which equals 7,539.822 W.
03

Find Interface Temperature (T鈧�)

To find the interface temperature, use the radial heat conduction formula through the rod. The heat transfer rate \(q = q_{gen}\), so for the rod with thermal conductivity \(k_{rod} = 0.5 \text{ W/m.K}\), use Fourier's law in cylindrical coordinates: \( q = \frac{2\pi L k_{rod} (T_0 - T_1)}{\ln(r_1/r_0)} \). Solving this with the given values will solve for \(T_1\).
04

Use Convection Heat Transfer for Outer Surface

Use Newton's law of cooling for the outer surface of the sleeve. The heat transfer through convection is given by \(q_{conv} = hA(T_2 - T_{\infty})\), where \(h = 25 \text{ W/m}^2\cdot\text{K}\), \(A\) is the surface area, and \(T_{\infty} = 27^\circ C\). Set this equal to the heat flow through the sleeve to find \(T_2\).
05

Calculate Center Temperature (T鈧�)

Utilize energy balance and heat transfer due to volumetric generation. Use the equation for steady state heat conduction with generation in cylindrical coordinates: \(q_{rod} = 2 \pi k_{rod} (T_0 - T_1)/r^2\), and solve for the center temperature, \(T_0\).
06

Compute Results

Put the calculated values into equations from steps 3 and 4 to find \(T_1\) and \(T_2\). Substitute these temperatures back to find \(T_0\) through the heat conduction equation utilized earlier.

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Key Concepts

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

Thermal Conductivity
Thermal conductivity is a fundamental property of materials that measures their ability to conduct heat. In essence, it describes how easily heat can flow through a material. The higher the thermal conductivity, the better the material is at conducting heat. This property is denoted by the symbol \( k \) and is expressed in units of watts per meter-kelvin \( ext{W/m} \, \cdot\, \text{K} \). In the given problem, we deal with two materials: the cylindrical rod and the sleeve surrounding it. The rod has a thermal conductivity of 0.5 \( ext{W/m} \, \cdot\, \text{K} \), which indicates it is a relatively poor conductor of heat compared to the sleeve, which has a higher thermal conductivity of 4 \( ext{W/m} \, \cdot\, \text{K} \).Understanding thermal conductivity helps us predict heat flow and temperature distribution within materials. In this scenario, the temperature difference between the center of the rod and its surface can be explained by the difference in thermal conductivities. The rod's lower conductivity restricts heat flow to its surface, affecting the temperature at the rod-sleeve interface and the outer surface of the sleeve.
Volumetric Heat Generation
Volumetric heat generation refers to the internal production of heat within a given volume of material. It is often the result of chemical reactions, electrical resistance, or other internal processes. The heat generation is expressed as the power generated per unit volume, typically measured in watts per cubic meter \( ext{W/m}^3 \).In this exercise, the cylindrical rod undergoes uniform volumetric heat generation, producing 24,000 \( ext{W/m}^3 \). This generation of heat causes the temperature within the rod to rise and affects the temperature distribution within the system.
  • Heat generation inside the rod creates a higher temperature at the core compared to the surface, leading to a temperature gradient.
  • It is crucial in calculating the total heat generated and the resultant heat transfer to the environment.
Therefore, understanding volumetric heat generation helps determine how heat propagates from the inside of the rod to its exterior and how it interacts with surrounding materials.
Convection Heat Transfer
Convection heat transfer involves the movement of heat between a surface and a fluid moving past it. It plays a vital role in cooling surfaces, such as the outer surface of the sleeve, by transferring the internal heat generated to the surrounding environment. This transfer is characterized by the convection heat transfer coefficient \( h \), given in \( ext{W/m}^2 \cdot \text{K} \).In the problem, the outer surface of the sleeve is exposed to air flow, with a convection coefficient of 25 \( ext{W/m}^2 \cdot \text{K} \). This coefficient quantifies the effectiveness of heat transfer from the sleeve's surface to the air. The temperature at the sleeve's surface depends heavily on this convection rate. Key factors affecting convection:
  • Fluid velocity: Faster air flow increases convection, enhancing heat removal.
  • Surface area: Larger areas allow more heat exchange.
  • Temperature difference: Greater differences between the surface and fluid increase convection.
Convection is necessary to ensure the excess heat within the system is dissipated effectively, maintaining a stable external temperature despite internal heat generation.

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Most popular questions from this chapter

Superheated steam at \(575^{\circ} \mathrm{C}\) is routed from a boiler to the turbine of an electric power plant through steel tubes \((k=35 \mathrm{~W} / \mathrm{m}+\mathrm{K})\) of \(300 \mathrm{~mm}\) inner diameter and \(30 \mathrm{~mm}\) wall thickness. To reduce heat loss to the surroundings and to maintain a safe-to-touch outer surface temperature, a layer of calcium silicate insulation \((k=0.10 \mathrm{~W} / \mathrm{m}-\mathrm{K})\) is applied to the tubes, while degradation of the insulation is reduced by wrapping it in a thin sheet of aluminum having an emissivity of \(\varepsilon=0.20\). The air and wall temperatures of the power plant are \(27^{\circ} \mathrm{C}\). (a) Assuming that the inner surface temperature of a steel tube corresponds to that of the steam and the convection coefficient cutside the aluminum sheet is \(6 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), what is the minimum insulation thickness needed to insure that the teriperature of the aluminum does not exceed \(50^{\circ} \mathrm{C}\) ? What is the corresponding heat loss per meter of tube length? (b) Explore the effect of the insulation thickness on the temperature of the aluminum and the heat loss per unit tube length.

An annular aluminum fin of rectangular profile is attached to a circular tube having an outside diameter of \(25 \mathrm{~mm}\) and a surface temperature of \(250^{\circ} \mathrm{C}\). The fin is \(1 \mathrm{~mm}\) thick and \(10 \mathrm{~mm}\) long, and the temperature and the convection coefficient associated with the adjoining fluid are \(25^{\circ} \mathrm{C}\) and \(25 \mathrm{~W} / \mathrm{m}^{2}\) + \(\mathrm{K}\), respectively. (a) What is the heat loss per fin? (b) If 200 such fins are spaced at 5-mm increments along the tube length, what is the heat loss per meter of tube length?

A storage tank consists of a cylindrical secticn that has a length and inner diameter of \(L=2 \mathrm{~m}\) and \(D_{2}=1 \mathrm{~m}\), respectively, and two hemispherical end sections. The tank is constructed from 20-mm-thick glass (Pyrex) and is exposed to ambient air for which the temperature is \(300 \mathrm{~K}\) and the convection coefficient is \(10 \mathrm{~W} / \mathrm{m}^{2}-\mathrm{K}\). The tank is used to store heated oil, which maintains the inser surface at a temperature of \(400 \mathrm{~K}\). Determine the electrical power that must be supplied to a heater submerged in the oil if the prescribed conditions are to be maintained. Radiation effects may be neglected, and the Pyrex may be assumed to have a thermal conductivity of \(1.4 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\).

The evaporator section of a refrigeration unit consists of thin-walled, 10-mm- diameter tubes through which refrigcrant passes at a temperature of \(-18^{\circ} \mathrm{C}\). Air is cooled as it flows over the tubes, maintaining a surface convection coefficient of \(100 \mathrm{~W} / \mathrm{m}^{2}\) - \(\mathrm{K}\), 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 laycr 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 \(2 \mathrm{~mm}\)-thick layer of frost has formed. If the tubes are in ambient air for which \(T_{\mathrm{w}}=20^{\circ} \mathrm{C}\) 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 assamed to have a mass density of \(700 \mathrm{~kg} / \mathrm{m}^{3}\) and a later heat of fusion of \(334 \mathrm{~kJ} / \mathrm{kg}\).

An experimental arrangement for measuring the termal conductivity of solid materials involves the use of two long rods that are equivalent in every respect, ex. cept that one is fabricated from a standard material of known thermal conductivity \(k_{A}\) while the other is fabricated from the material whose thermal conductivity \(k_{\mathrm{B}}\) is desired. Both rods are attached at one end to a heat source of fixed temperature \(T_{b \text { r }}\) are exposed to a fluid of temperature \(T_{\text {su }}\) and are instrumented with thermocouples to measure the temperature at a fined distance \(x_{1}\) from the heat source. If the standard material is aluminum, with \(k_{\mathrm{A}}=200 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\), and mensurements reveal values of \(T_{\mathrm{A}}=75^{\circ} \mathrm{C}\) and \(T_{\mathrm{B}}=60^{\circ} \mathrm{C}\) at \(x_{1}\) for \(T_{b}=100^{\circ} \mathrm{C}\) and \(T_{a}=25^{\circ} \mathrm{C}\), what is the then mal conductivity \(k_{\mathrm{g}}\) of the test material?
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