/*! 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 63 Inconel-X-750 straight rectangul... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Inconel-X-750 straight rectangular fins are to be used in an application where the fins are \(2 \mathrm{~mm}\) thick and the convective heat transfer coefficient is \(300 \mathrm{~W} / \mathrm{m}^{2} \mathrm{~K}\). Investigate the effect of tip boundary condition on estimated heat loss for fin length \(L\) varying from \(6 \mathrm{~mm}\) to \(20 \mathrm{~mm}\). Take \(T_{B}=800 \mathrm{~K}, T_{e}=300 \mathrm{~K}, k=18.8\) \(\mathrm{W} / \mathrm{m} \mathrm{K}\).

Short Answer

Expert verified
Calculate heat losses for insulated and infinite fins at different lengths, then compare the results.

Step by step solution

01

Define the Fin Parameters

For a straight rectangular fin, the fin parameters are given as follows: thickness \( t = 2 \) mm, convective heat transfer coefficient \( h = 300 \) W/m\(^2\)K, base temperature \( T_B = 800 \) K, environment temperature \( T_e = 300 \) K, and thermal conductivity \( k = 18.8 \) W/mK.
02

Formulate Fin Heat Loss Equations

The heat loss of a fin can be calculated differently depending on the boundary conditions. For an infinite fin, the heat loss is \[q_{infinite} = rac{hPL(T_B - T_e)}{\sqrt{hP/kA_c}(1 + mL)}\]For a fin with an insulated tip, the heat loss is given by\[q_{insulated} = \sqrt{hPk_cA_c}(T_B - T_e)\tan(mL)\] where \( A_c = thickness \times width \) and \( P = 2 \times (thickness + width) \).
03

Calculate the Fin Parameter m

The parameter \( m \) is calculated by\[m = \sqrt{\frac{hP}{kA_c}}\]Insert the known values into this formula to calculate \( m \) for various lengths \( L\).
04

Compute Heat Loss for Varying Lengths with Insulated Tip

Using the formula for \( q_{insulated} \), compute the heat loss for fin lengths \( L \) ranging from \( 6 \) mm to \( 20 \) mm. Evaluate the expression with different values of \( L \) while keeping other parameters constant.
05

Compute Heat Loss for Varying Lengths with Infinite Fin Assumption

Similarly, use the formula for \( q_{infinite} \) to compute the heat loss for fin lengths \( L \) ranging from \( 6 \) mm to \( 20 \) mm under the infinite fin assumption.
06

Compare Results

Compare the heat losses calculated for both the insulated tip and infinite fin assumptions across all lengths \( L \). Analyze the differences to understand the effect of the tip boundary condition.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Convective Heat Transfer Coefficient
The convective heat transfer coefficient, denoted by \( h \), plays a crucial role in measuring how effectively heat is transferred between a surface and a fluid that is moving across it. It is expressed in units of \( \, W/m^2K \, \). The higher the coefficient, the more efficient the heat transfer process.
When dealing with fins—such as the ones in our problem—the convective heat transfer coefficient determines how quickly heat is dissipated from the surface of the fin into the surrounding fluid.
For this particular exercise, we have a high thermal conductivity of \( 300 \) \( \, W/m^2K \). This indicates a very effective heat transfer from the surface of the fin to the surrounding environment. Understanding \( h \) helps us predict how the fin will perform in dissipating heat, ensuring the system remains efficient.
Thermal Conductivity
Thermal conductivity, represented by \( k \), measures a material's ability to conduct heat. It is given in units of \( \, W/mK \, \). Materials with high thermal conductivity, like metals, are excellent conductors and allow heat to pass through them quickly.
In our specific fin problem, the thermal conductivity of the fin material is \( 18.8 \) \( \, W/mK \). This value indicates that the material is moderately effective in conducting heat. It is vital to understand \( k \) because it impacts how much heat is transferred from the base of the fin to its tip.
The balance between thermal conductivity and the convective heat transfer coefficient determines how efficiently a fin can remove heat from a system. A precise calculation of these parameters is key to designing fins that perform optimally.
Fin Heat Loss
Fin heat loss refers to the amount of heat that is transferred away from the fin to the surrounding environment. This is crucial for maintaining the system's temperature and ensuring optimal performance. In fin applications, the goal is often to maximize heat loss for improved thermal regulation.
To calculate the heat loss, the fin's physical dimensions, material properties, and environmental conditions are considered. Two common scenarios include: an infinite fin, which assumes the tip of the fin has no end, and an insulated fin, where the tip is insulated so no heat is lost there.
The formulas for these calculations account for different boundary conditions, helping us understand how the ends of the fin affect overall heat dissipation. This analysis aids in selecting appropriate fin designs for various engineering applications.
Boundary Condition Effect
The boundary condition effect explores how different end conditions on a fin influence its heat dissipation capabilities. Boundary conditions are significant because they alter how heat travels through and away from the fin.
In our exercise, two specific conditions are analyzed: the infinite fin assumption and the insulated tip. The infinite fin assumes continuous material with no end, causing different heat dissipation characteristics compared to an insulated tip, where all heat is redirected back toward the fin base.
By examining these conditions over varying lengths, engineers can deduce how different designs influence the efficiency of heat loss. It's a practical way of optimizing thermal systems by fine-tuning end conditions for specific applications.
  • Infinite fins assume zero heat dissipation at the tip, ideal for very long or continuously cooled fins.
  • Insulated tips prevent heat loss at the end, enhancing conduction through the entire fin.
Understanding these effects enables better design decisions to improve overall system efficiency.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A space radiator is made of \(0.3 \mathrm{~mm}\)-thick aluminum plate with heatpipes at a pitch of \(8 \mathrm{~cm}\). The heatpipes reject heat at \(330 \mathrm{~K}\). The back of the radiator is insulated, and the front sees outer space at \(0 \mathrm{~K}\). If the aluminum surface is hard-anodized to give an emittance of \(0.8\), determine the fin effectiveness of the radiator and the rate of heat rejection per unit area. Take \(k=200 \mathrm{~W} / \mathrm{m} \mathrm{K}\) for the aluminum. (Hint: We do not expect the plate temperature to vary more than a few kelvins: assume \(q_{\mathrm{rad}}\) is constant at an average value to obtain an approximate analytical solution.)

An electric heater consists of a thin ribbon of metal and is used to boil a dielectric liquid. The liquid temperature \(T_{e}\) is uniform at its boiling point, and the heat transfer coefficient on the ribbon can be assumed to be uniform as well. A resistance measurement allows the average temperature of the ribbon \(\bar{T}\) to be determined. Obtain an expression for \(\bar{T}-T_{e}\) in terms of the ribbon dimensions (width \(W\), thickness \(2 t\) ), the ribbon thermal conductivity \(k\), the heat transfer coefficient \(h_{c}\), and the ribbon electrical conductivity \(\sigma\left[\Omega^{-1} \mathrm{~m}^{-1}\right]\).

Electronic components are attached to a \(10 \mathrm{~cm}\)-square, \(2 \mathrm{~mm}\)-thick aluminum plate, and the backface is cooled by a flow of air. The backface has rectangular aluminum fins \(25 \mathrm{~mm}\) long, \(0.3 \mathrm{~mm}\) thick, at a pitch of \(3 \mathrm{~mm}\). If the cooling air is at \(20^{\circ} \mathrm{C}\) and the heat transfer coefficient on the fins is \(30 \mathrm{~W} / \mathrm{m}^{2} \mathrm{~K}\), what is the allowable heat dissipation rate if the plate temperature should not exceed \(70^{\circ} \mathrm{C}\). Take \(k=180 \mathrm{~W} / \mathrm{m} \mathrm{K}\) for the aluminum.

An explosive is to be stored in large slabs of thickness \(2 L\) clad on both sides with a protective sheath. The rate at which heat is generated within the explosive is temperature-dependent and can be approximated by the linear relation \(\dot{Q}_{v}^{\prime \prime \prime}=a+b\left(T-T_{e}\right)\), where \(T_{e}\) is the prevailing ambient air temperature. If the overall heat transfer coefficient between the slab surface and the ambient air is \(U\), show that the condition for an explosion is \(L=(k / b)^{1 / 2} \tan ^{-1}\left[U /(k b)^{1 / 2}\right]\). Determine the slab thickness if \(k=0.9 \mathrm{~W} / \mathrm{m} \mathrm{K}, U=0.20 \mathrm{~W} / \mathrm{m}^{2} \mathrm{~K}, a=60 \mathrm{~W} / \mathrm{m}^{3}\), \(b=6.0 \mathrm{~W} / \mathrm{m}^{3} \mathrm{~K}\).

Calcium silicate has replaced asbestos as the preferred insulation for steam lines in power plants. Consider a \(40 \mathrm{~cm}-\mathrm{O} . \mathrm{D} ., 4 \mathrm{~cm}\)-wall-thickness steel steam line insulated with a \(12 \mathrm{~cm}\) thickness of calcium silicate. The insulation is protected from damage by an aluminum sheet lagging that is \(2.5 \mathrm{~mm}\) thick. The steam temperature is \(565^{\circ} \mathrm{C}\) and ambient air temperature in the power plant is \(26^{\circ} \mathrm{C}\). The inside convective resistance is negligible, the outside convective heat transfer coefficient can be taken as \(11 \mathrm{~W} / \mathrm{m}^{2} \mathrm{~K}\), and the emittance of the aluminum is \(0.1\). Calculate the rate of heat loss per meter. The thermal conductivity of the pipe wall is \(40 \mathrm{~W} / \mathrm{m} \mathrm{K}\), and the table gives values for Calsilite calcium silicate insulation blocks. $$ \begin{array}{l|ccccc} \mathrm{T}, \mathrm{K} & 500 & 600 & 700 & 800 & 900 \\ \hline \mathrm{k}, \mathrm{W} / \mathrm{m} \mathrm{K} & 0.074 & 0.096 & 0.142 & 0.211 & 0.303 \end{array} $$

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.