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Chapter 3: Problem 155

Determine the percentage increase in heat transfer associated with attaching aluminum fins of rectangular profile to a plane wall. The fins are \(50 \mathrm{~mm}\) long, \(0.5 \mathrm{~mm}\) thick, and are equally spaced at a distance of \(4 \mathrm{~mm}\) ( 250 fins \(/ \mathrm{m})\). The convection coefficient associated with the bare wall is \(40 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), while that resulting from attachment of the fins is \(30 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\).

Short Answer

Expert verified
The percentage increase in heat transfer associated with attaching aluminum fins of rectangular profile to a plane wall is \(\frac{30 \times (A_{fins} + A_{base})}{40 \times 0.05} - 1\), where \(A_{fins}\) and \(A_{base}\) are calculated in Steps 1 and 2. After calculating the surface area of one fin (\(A_{fin}\)), the total heat transfer area in the finned case (\(A_{total}\)), and the heat transfer rates for both the bare and finned cases, this expression will provide the required percentage increase in heat transfer.

Step by step solution

01

Calculate the surface area of the fins

First, calculate the surface area of one fin. Since the fin is of rectangular profile, the surface area can be calculated as the product of the length and perimeter of the fin. The perimeter of a rectangular profile fin is given by the formula 2l + 2w, where l is the length and w is the width (or thickness) of the fin. Surface area of one fin = length 脳 perimeter \(A_{fin} = L \times (2l + 2w)\) Taking the given dimensions of the fin: Length \(L = 50\) mm = \(0.05\) m Width (thickness) \(w = 0.5\) mm = \(0.0005\) m Fins are equally spaced at a distance of 4 mm (250 fins/m), and so the length of a cycle is 4 mm. \(A_{fin} = 0.05 \times (2 \times 0.0005 + 2 \times 0.004)\)
02

Calculate the total surface area for heat transfer in the finned case

Next, calculate the total surface area for heat transfer in the finned case, which includes both the surface area of fins and the area of the base per one-meter length of the wall. Area of fin per meter of wall (assuming 250 fins/m): \(A_{fins} = 250 \times A_{fin}\) Area of the base per one-meter length of the wall (considering fin dimensions): \(A_{base} = (1 - 250 \times 0.0005) \times 0.05\) Total surface area for heat transfer in the finned case: \(A_{total} = A_{fins} + A_{base}\)
03

Calculate the heat transfer rates for bare and finned cases

To calculate the heat transfer rate, we use the following formula: heat_transfer_rate = convection_coefficient 脳 surface_area 脳 螖T (temperature difference). We are given convection coefficients for the bare and finned cases, and we can assume a constant temperature gradient between the wall and the environment for both cases. The surface area for the bare wall case (1 m length) would be 0.05 m since it doesn't have any fins. Heat transfer rate for the bare wall case: \(q_{bare} = h_{bare} \times A_{bare} \times \Delta T\) Heat transfer rate for the finned wall case: \(q_{finned} = h_{finned} \times A_{total} \times \Delta T\) To find the percentage increase in heat transfer, calculate the ratio of finned to bare heat transfer rates and subtract 1 (or 100%).
04

Calculate the percentage increase in heat transfer

The percentage increase in heat transfer rate can be found by dividing the finned case heat transfer rate by the bare case heat transfer rate and then subtracting 1. Percentage increase in heat transfer: \(\frac{q_{finned}}{q_{bare}} - 1 \) Combine the equations from Steps 2 and 3, and plug in the given values: \(\frac{30 \times (A_{fins} + A_{base})}{40 \times 0.05} - 1\) Now, calculate the percentage increase in heat transfer by substituting the values obtained in Steps 1 and 2. This percentage value will be the solution to the problem.

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Key Concepts

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

Fins
Fins are extended surfaces that are added to a base surface to enhance heat transfer. When fins are attached to a surface, they increase the overall area available for inducing heat transfer, which can improve heat dissipation. This is particularly useful in applications where the base surface alone cannot provide sufficient cooling or heating. Fins work by conducting heat from the base surface and then transferring it to the surrounding fluid environment through convection. The efficiency of a fin depends on factors such as its material, shape, and size, as well as the thermal properties of the surrounding fluid. Common applications of fins include:
  • Cooling in electronic devices to prevent overheating.
  • Heat exchangers in HVAC systems to maintain desired temperatures.
  • Automotive radiators and other vehicle components to manage engine heat.
In this exercise, aluminum fins with a rectangular profile are attached to a wall. This setup aims to increase the heat transfer capability by increasing the surface area that is exposed to a cooler environment. This, in turn, reduces the overall thermal resistance between the wall and its surroundings.
Convection Coefficient
The convection coefficient, denoted typically by 'h', is a critical parameter in the study of heat transfer, specifically in convection processes. It quantifies the rate at which heat is transferred from a solid surface to a fluid or from fluid to fluid via convection. The value of the convection coefficient depends on various factors, such as:
  • The nature of the fluid (e.g., air, water) and its properties such as viscosity and thermal conductivity.
  • The velocity of the fluid over the surface, which affects turbulence and mixing.
  • The temperature difference between the surface and the fluid.
  • The surface roughness and geometry, which can perturb flow patterns.
In the original problem, the convection coefficient for a bare wall was given as 40 W/m虏路K, while the presence of fins altered it to 30 W/m虏路K. This change occurs because the addition of fins influences the boundary layer characteristics, slightly decreasing the effective convection coefficient but still resulting in a net increase in heat transfer due to the larger surface area.
Surface Area Calculation
Surface area calculation is fundamental in determining the potential for heat transfer across a surface. The larger the surface area interacting with the environment, the more heat can be transferred, assuming all other factors remain constant. For fins, calculating the surface area involves understanding their geometry. In the given problem, the fin is rectangular, and its perimeter can be calculated as:\[ P = 2L + 2w \]where \(L\) is the length and \(w\) is the thickness of the fin.The surface area of one fin is therefore:\[ A_{fin} = L \times P \]When numerous fins (250 fins/m) are added to a base wall, the total finned area for a one-meter section of the wall becomes significant. This calculation encompasses both the fin surface area and any remaining wall areas that contribute to heat flow:
  • Fin area: Multiply the area of one fin by the number of fins per meter.
  • Base area: Account for the unoccupied wall area between fins.
By considering the total increased area, one can determine the enhanced heat transfer capacity, which translates to efficiency improvements in thermal systems.

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Most popular questions from this chapter

An experimental arrangement for measuring the thermal conductivity of solid materials involves the use of two long rods that are equivalent in every respect, except that one is fabricated from a standard material of known thermal conductivity \(k_{\mathrm{A}}\) 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 \(T_{b}\), are exposed to a fluid of temperature \(T_{\infty}\), and are instrumented with thermocouples to measure the temperature at a fixed 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 measurements 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 \(T_{\infty}=25^{\circ} \mathrm{C}\), what is the thermal conductivity \(k_{\mathrm{B}}\) of the test material?

A radioactive material of thermal conductivity \(k\) is cast as a solid sphere of radius \(r_{o}\) and placed in a liquid bath for which the temperature \(T_{\infty}\) and convection coefficient \(h\) are known. Heat is uniformly generated within the solid at a volumetric rate of \(\dot{q}\). Obtain the steadystate radial temperature distribution in the solid, expressing your result in terms of \(r_{o}, \dot{q}, k, h\), and \(T_{\infty}\).

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.

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?

A composite cylindrical wall is composed of two materials of thermal conductivity \(k_{\mathrm{A}}\) and \(k_{\mathrm{B}}\), which are separated by a very thin, electric resistance heater for which interfacial contact resistances are negligible. Liquid pumped through the tube is at a temperature \(T_{\infty, i}\) and provides a convection coefficient \(h_{i}\) at the inner surface of the composite. The outer surface is exposed to ambient air, which is at \(T_{\infty, o}\) and provides a convection coefficient of \(h_{o^{*}}\) Under steady-state conditions, a uniform heat flux of \(q_{h}^{n}\) is dissipated by the heater. (a) Sketch the equivalent thermal circuit of the system and express all resistances in terms of relevant variables. (b) Obtain an expression that may be used to determine the heater temperature, \(T_{h+}\). (c) Obtain an expression for the ratio of heat flows to the outer and inner fluids, \(q_{o}^{\prime} / q_{i}^{\prime}\). How might the variables of the problem be adjusted to minimize this ratio?
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