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Chapter 3: Problem 158

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 \(25 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), respectively. (a) What is the heat loss per fin? (b) If 200 such fins are spaced at \(5-\mathrm{mm}\) increments along the tube length, what is the heat loss per meter of tube length?

Short Answer

Expert verified
The heat loss per fin is approximately 8.677 W, and the heat loss per meter of tube length is approximately 1735.4 W/m.

Step by step solution

01

Calculate the Fin Efficiency

To calculate the heat loss per fin, we must first calculate the efficiency of the fin. The fin efficiency can be found using the following formula: \( \eta_{\mathrm{f}} = \dfrac{\tanh(mL)}{mL}\) where \(m\) is the fin parameter and can be calculated as follows: \(m = \sqrt{\dfrac{2h}{k\cdot t}}\) with \(L\), \(t\), \(h\) and \(k\) representing the length, thickness, convection coefficient, and thermal conductivity of the fin, respectively. Given the problem's information, we have: - \(L = 10 \mathrm{~mm} = 0.01 \mathrm{~m}\) - \(t = 1 \mathrm{~mm} = 0.001 \mathrm{~m}\) - \(h = 25 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) - Assume the thermal conductivity of aluminum, \(k = 237 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) Calculating the fin parameter \(m\): \(m = \sqrt{\dfrac{2(25\,\mathrm{W\,m^{-2}\,K^{-1}})}{237\,\mathrm{W\,m^{-1}\,K^{-1}}\cdot 0.001\,\mathrm{m}}} = 137.45\,\mathrm{m^{-1}}\) Now, we can calculate the fin efficiency using the formula: \(\eta_{\mathrm{f}} = \dfrac{\tanh(137.45\,\mathrm{m^{-1}}\cdot 0.01\,\mathrm{m})}{137.45\,\mathrm{m^{-1}}\cdot 0.01\,\mathrm{m}} = 0.987\)
02

Calculate the Heat Loss per Fin

Now that we have the fin efficiency, we can find the heat loss per fin using the formula: \(Q_{\mathrm{f}} = \eta_{\mathrm{f}} \cdot h\cdot A_{\mathrm{f}} \cdot(\Theta_{\mathrm{s}} - \Theta_{\mathrm{\infty}})\) where: - \(Q_{\mathrm{f}}\) is the heat loss per fin, - \(A_{\mathrm{f}}\) is the fin area per fin, - \(\Theta_{\mathrm{s}}\) is the tube surface temperature, and - \(\Theta_{\mathrm{\infty}}\) is the temperature of the ambient fluid. The fin area per fin can be calculated as follows: \(A_{\mathrm{f}} = 2LW = 2\times 0.01\,\mathrm{m}\times (\pi \times (0.0125^2-0.012^2))\,\mathrm{m^2} = 2L(\pi \cdot D_{\mathrm{\text{outside}}}-\pi \cdot D_{\mathrm{\text{inside}}})\) Calculating the fin area per fin: \(A_{\mathrm{f}} = 2(0.01\,\mathrm{m})(\pi(0.0125^2\,\mathrm{m^2} - 0.012^2\,\mathrm{m^2})) = 0.00157\,\mathrm{m^2}\) Now, we can calculate the heat loss per fin: \(Q_{\mathrm{f}}=0.987 \cdot 25\,\mathrm{W\,m^{-2}\,K^{-1}} \cdot 0.00157\,\mathrm{m}^2 (250^{\circ}\,\mathrm{C}-25^{\circ}\,\mathrm{C}) = 8.677\,\mathrm{W}\) So the heat loss per fin is approximately 8.677 W.
03

Calculate the Heat Loss per Meter of Tube Length

We're given that there are 200 fins spaced at 5-mm increments. Therefore, the total finned length of the tube is: \(L_{\mathrm{\text{total}}} = 200 \times 0.005\,\mathrm{m} = 1\,\mathrm{m}\) Knowing the heat loss per fin, we can find the heat loss per meter of tube length by multiplying the heat loss per fin by the total number of fins per meter of tube length: \(Q_{\mathrm{\text{per meter}}} = Q_{\mathrm{f}} \cdot N\) where: - \(Q_{\mathrm{\text{per meter}}}\) is the heat loss per meter of tube length, and - \(N\) is the number of fins per meter of tube length (200 fins in this case). Calculating the heat loss per meter of tube length: \(Q_{\mathrm{\text{per meter}}} = 8.677\,\mathrm{W} \cdot 200 = 1735.4\,\mathrm{W/m}\) The heat loss per meter of tube length is approximately 1735.4 W/m.

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Key Concepts

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

Fin Efficiency
Fin efficiency is a measure of how effectively a fin can transfer heat from a surface into its surroundings. When a fin attached to a heated surface is exposed to a cooler fluid, it is supposed to dissipate heat effectively. However, not all fins are equal. Fin efficiency, denoted by \( \eta_{\text{f}} \), is calculated using the formula:
  • \( \eta_{\mathrm{f}} = \frac{\tanh(mL)}{mL} \)
Here, \( m \) is a parameter defined as:
  • \( m = \sqrt{\frac{2h}{k \cdot t}} \)
In these equations:
  • \( h \) is the convection heat transfer coefficient.
  • \( k \) is the thermal conductivity of the fin material.
  • \( t \) is the thickness of the fin.
  • \( L \) is the length of the fin.
A value near 1 indicates a highly efficient fin, which means it loses very little usable heat due to conduction losses within the fin itself.
Thermal Conductivity
Thermal conductivity \( k \) is a material's ability to conduct heat. It plays a significant role in determining how well a material will perform as a heat spreader between different temperatures. Simply put, it's about how fast the heat can travel through a material. Metals, like aluminum used in heat sinks and fins, usually have high thermal conductivities, meaning they are excellent at moving heat around.
  • For aluminum, the typical thermal conductivity is about 237 W/m路K.
Thermal conductivity is critical in calculating the fin parameter \( m \), affecting the overall fin efficiency. In engineering, selecting the right material with the suitable thermal conductivity ensures effective and efficient heat transfer, which is crucial in applications like cooling devices and systems.
Heat Convection
Heat convection is the mechanism of heat transfer where heat moves away from a surface through a fluid like air or water. It happens when the molecules near a hot surface heat up and move away, being replaced by cooler ones, thus carrying heat away. The efficiency of this process is governed by the convection coefficient \( h \), which depends on the properties of the fluid, the nature of the surface and the flow conditions.
  • For example, in natural convection, the coolant often has a lower \( h \) than in forced convection settings such as fans or pumps enhancing the flow.
  • In this problem, the coefficient \( h \) is given as 25 W/m虏路K.
An effective heat convection contributes to efficient heat dissipation by the fins, preventing overheating in thermal management systems.
Annular Fin
An annular fin is a fin designed to dissipate heat from a circular tube or cylindrical object. These fins are especially effective in increasing the surface area available for heat dissipation in annular or ring-shaped devices. The presence of the annular fins allows for enhanced heat loss from surfaces that might have otherwise been inefficient due to their shape.
  • The fin is described by its geometry: length, thickness, and inside and outside diameters.
  • In the current problem, the annular fin is an aluminum piece, simply 1 mm thick and stretches 10 mm away from the tube, creating an effective heat dissipation structure.
Annular fins are vital in many industrial applications, including heating, ventilation, and air conditioning (HVAC) systems, power electronics, and engine components to manage heat effectively by increasing the heat transfer surface area.

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Most popular questions from this chapter

The thermal characteristics of a small, dormitory refrigerator are determined by performing two separate experiments, each with the door closed and the refrigerator placed in ambient air at \(T_{\infty}=25^{\circ} \mathrm{C}\). In one case, an electric heater is suspended in the refrigerator cavity, while the refrigerator is unplugged. With the heater dissipating \(20 \mathrm{~W}\), a steady-state temperature of \(90^{\circ} \mathrm{C}\) is recorded within the cavity. With the heater removed and the refrigerator now in operation, the second experiment involves maintaining a steady-state cavity temperature of \(5^{\circ} \mathrm{C}\) for a fixed time interval and recording the electrical energy required to operate the refrigerator. In such an experiment for which steady operation is maintained over a 12-hour period, the input electrical energy is \(125,000 \mathrm{~J}\). Determine the refrigerator's coefficient of performance (COP).

An electrical current of 700 A flows through a stainless steel cable having a diameter of \(5 \mathrm{~mm}\) and an electrical resistance of \(6 \times 10^{-4} \mathrm{\Omega} / \mathrm{m}\) (i.e., per meter of cable length). The cable is in an environment having a temperature of \(30^{\circ} \mathrm{C}\), and the total coefficient associated with convection and radiation between the cable and the environment is approximately \(25 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). (a) If the cable is bare, what is its surface temperature? (b) If a very thin coating of electrical insulation is applied to the cable, with a contact resistance of \(0.02 \mathrm{~m}^{2} \cdot \mathrm{K} / \mathrm{W}\), what are the insulation and cable surface temperatures? (c) There is some concern about the ability of the insulation to withstand elevated temperatures. What thickness of this insulation \((k=0.5 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) will yield the lowest value of the maximum insulation temperature? What is the value of the maximum temperature when this thickness is used?

Consider cylindrical and spherical shells with inner and outer surfaces at \(r_{1}\) and \(r_{2}\) maintained at uniform temperatures \(T_{s, 1}\) and \(T_{s, 2}\), respectively. If there is uniform heat generation within the shells, obtain expressions for the steady-state, one-dimensional radial distributions of the temperature, heat flux, and heat rate. Contrast your results with those summarized in Appendix 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 \(334 \mathrm{~kJ} / \mathrm{kg}\).

As a means of enhancing heat transfer from highperformance logic chips, it is common to attach a heat \(\sin k\) to the chip surface in order to increase the surface area available for convection heat transfer. Because of the ease with which it may be manufactured (by taking orthogonal sawcuts in a block of material), an attractive option is to use a heat sink consisting of an array of square fins of width \(w\) on a side. The spacing between adjoining fins would be determined by the width of the sawblade, with the sum of this spacing and the fin width designated as the fin pitch \(S\). The method by which the heat sink is joined to the chip would determine the interfacial contact resistance, \(R_{t, c^{*}}^{n}\) Consider a square chip of width \(W_{c}=16 \mathrm{~mm}\) and conditions for which cooling is provided by a dielectric liquid with \(T_{\infty}=25^{\circ} \mathrm{C}\) and \(h=1500 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). The heat \(\operatorname{sink}\) is fabricated from copper \((k=400 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\), and its characteristic dimensions are \(w=0.25 \mathrm{~mm}\), \(S=0.50 \mathrm{~mm}, L_{f}=6 \mathrm{~mm}\), and \(L_{b}=3 \mathrm{~mm}\). The prescribed values of \(w\) and \(S\) represent minima imposed by manufacturing constraints and the need to maintain adequate flow in the passages between fins. (a) If a metallurgical joint provides a contact resistance of \(R_{t, c}^{\prime \prime}=5 \times 10^{-6} \mathrm{~m}^{2} \cdot \mathrm{K} / \mathrm{W}\) and the maximum allowable chip temperature is \(85^{\circ} \mathrm{C}\), what is the maximum allowable chip power dissipation \(q_{c} ?\) Assume all of the heat to be transferred through the heat sink. (b) It may be possible to increase the heat dissipation by increasing \(w\), subject to the constraint that \((S-w) \geq 0.25 \mathrm{~mm}\), and/or increasing \(L_{f}\) (subject to manufacturing constraints that \(L_{f} \leq 10 \mathrm{~mm}\) ). Assess the effect of such changes.
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