/*! 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 155 Determine the percentage increas... [FREE SOLUTION] | 91影视

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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

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 \(T_{o}=25^{\circ} \mathrm{C} .\)

A firefighter's protective clothing, referred to as a turnout coat, is typically constructed as an ensemble of three layers separated by air gaps, as shown schematically. The air gaps between the layers are \(1 \mathrm{~mm}\) thick, and heat is transferred by conduction and radiation exchange through the stagnant air. The linearized radiation coefficient for a gap may be approximated as, \(h_{\text {rad }}=\sigma\left(T_{1}+T_{2}\right)\left(T_{1}^{2}+T_{2}^{2}\right) \approx 4 \sigma T_{\text {avg }}^{3}\), where \(T_{\text {avg }}\) represents the average temperature of the surfaces comprising the gap, and the radiation flux across the gap may be expressed as \(q_{\text {rad }}^{\prime \prime}=h_{\text {rad }}\left(T_{1}-T_{2}\right)\). (a) Represent the turnout coat by a thermal circuit, labeling all the thermal resistances. Calculate and tabulate the thermal resistances per unit area \(\left(\mathrm{m}^{2}\right.\). \(\mathrm{K} / \mathrm{W}\) ) for each of the layers, as well as for the conduction and radiation processes in the gaps. Assume that a value of \(T_{\mathrm{avg}}=470 \mathrm{~K}\) may be used to approximate the radiation resistance of both gaps. Comment on the relative magnitudes of the resistances. (b) For a pre-ash-over fire environment in which firefighters often work, the typical radiant heat flux on the fire-side of the turnout coat is \(0.25 \mathrm{~W} / \mathrm{cm}^{2}\). What is the outer surface temperature of the turnout coat if the inner surface temperature is \(66^{\circ} \mathrm{C}\), a condition that would result in burn injury?

The outer surface of a hollow sphere of radius \(r_{2}\) is subjected to a uniform heat flux \(q_{2}^{\prime \prime}\). The inner surface at \(r_{1}\) is held at a constant temperature \(T_{s, 1}\). (a) Develop an expression for the temperature distribution \(T(r)\) in the sphere wall in terms of \(q_{2}^{\prime \prime}, T_{s, 1}, r_{1}, r_{2}\), and the thermal conductivity of the wall material \(k\). (b) If the inner and outer tube radii are \(r_{1}=50 \mathrm{~mm}\) and \(r_{2}=100 \mathrm{~mm}\), what heat flux \(q_{2}^{\prime \prime}\) is required to maintain the outer surface at \(T_{s, 2}=50^{\circ} \mathrm{C}\), while the inner surface is at \(T_{s, 1}=20^{\circ} \mathrm{C}\) ? The thermal conductivity of the wall material is \(k=10 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\).

The temperature of a flowing gas is to be measured with a thermocouple junction and wire stretched between two legs of a sting, a wind tunnel test fixture. The junction is formed by butt-welding two wires of different material, as shown in the schematic. For wires of diameter \(D=125 \mu \mathrm{m}\) and a convection coefficient of \(h=700 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), determine the minimum separation distance between the two legs of the sting, \(L=L_{1}+L_{2}\), to ensure that the sting temperature does not influence the junction temperature and, in turn, invalidate the gas temperature measurement. Consider two different types of thermocouple junctions consisting of (i) copper and constantan wires and (ii) chromel and alumel wires. Evaluate the thermal conductivity of copper and constantan at \(T=300 \mathrm{~K}\). Use \(k_{\mathrm{Ch}}=19 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) and \(k_{\mathrm{Al}}=29 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) for the thermal conductivities of the chromel and alumel wires, respectively.

Consider the conditions of Problem \(3.149\) but now allow for a tube wall thickness of \(5 \mathrm{~mm}\) (inner and outer diameters of 50 and \(60 \mathrm{~mm}\) ), a fin-to-tube thermal contact resistance of \(10^{-4} \mathrm{~m}^{2} \cdot \mathrm{K} / \mathrm{W}\), and the fact that the water temperature, \(T_{w}=350 \mathrm{~K}\), is known, not the tube surface temperature. The water-side convection coefficient is \(h_{w}=2000 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Determine the rate of heat transfer per unit tube length \((\mathrm{W} / \mathrm{m})\) to the water. What would be the separate effect of each of the following design changes on the heat rate: (i) elimination of the contact resistance; (ii) increasing the number of fins from four to eight; and (iii) changing the tube wall and fin material from copper to AISI 304 stainless steel \((k=20\) \(\mathrm{W} / \mathrm{m} \cdot \mathrm{K})\) ?
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