/*! 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 128 A 40-mm-long, 2 -mm-diameter pin... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 3: Problem 128

A 40-mm-long, 2 -mm-diameter pin fin is fabricated of an aluminum alloy \(\left(k=140 \mathrm{~W} / \mathrm{m}^{2}+\mathrm{K}\right)\). (a) Determine the fin heat transfer rate for \(T_{b}=50^{\circ} \mathrm{C}\). \(T_{\mathrm{w}}=25^{\circ} \mathrm{C}, h=1000 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), and an adiabatic tip condition. (b) An engineer suggests that by holding the fin tip at a low temperature, the fin heat transfer rate can be increased. For \(T(x=L)=0^{\circ} \mathrm{C}\), determine the new fin heat transfer rate. Other conditions are as in part (a). (c) Plot the temperature distribution, \(T(x)\), over the range \(0 \leq x \leq L\) for the adiabatic tip case and the preseribed tip temperature case. Also show the ambient temperature in your gruph. Discuss relevant features of the temperature distribution.

Short Answer

Expert verified
(a) 3.408 W, (b) 3.500 W. Temperature decreases along the fin length more in the fixed tip temperature case, showing higher fin efficiency.

Step by step solution

01

Define the Parameters

The given parameters for both parts of the problem are:- Fin length, \( L = 40 \text{ mm} = 0.04 \text{ m} \).- Diameter, \( D = 2 \text{ mm} = 0.002 \text{ m} \).- Thermal conductivity, \( k = 140 \text{ W/m} \cdot \text{K} \).- Base temperature, \( T_b = 50^{\circ} \text{C} \).- Ambient temperature, \( T_{\infty} = 25^{\circ} \text{C} \).- Convective heat transfer coefficient, \( h = 1000 \text{ W/m}^2 \cdot \text{K} \).- Adiabatic tip condition (Part a).- Prescribed tip temperature condition (Part b), \( T(L) = 0^{\circ} \text{C} \).
02

Find Cross-Sectional Area and Perimeter

Calculate the cross-sectional area \( A_c \) and the perimeter \( P \):- Cross-sectional area: \( A_c = \frac{\pi D^2}{4} \). Substitute \( D = 0.002 \text{ m} \): \[ A_c = \frac{\pi (0.002)^2}{4} \approx 3.14 \times 10^{-6} \text{ m}^2 \]. - Perimeter: \( P = \pi D \). \[ P = \pi(0.002) \approx 0.00628 \text{ m} \].
03

Determine the Fin Efficiency Parameter

The efficiency parameter \( m \) for the fin can be calculated as:\[ m = \sqrt{\frac{hP}{kA_c}} \].Substitute \( h = 1000 \text{ W/m}^2\cdot\text{K} \), \( P \approx 0.00628 \text{ m} \), \( k = 140 \text{ W/m}\cdot\text{K} \), and \( A_c \approx 3.14 \times 10^{-6} \text{ m}^2 \):\[ m = \sqrt{\frac{1000 \times 0.00628}{140 \times 3.14 \times 10^{-6}}} \approx 72.95 \text{ m}^{-1} \].
04

Calculate the Fin Heat Transfer Rate for Adiabatic Tip (Part a)

For an adiabatic tip, the formula for the fin heat transfer rate \( Q_f \) is:\[ Q_f = \sqrt{hPkA_c}(T_b - T_{\infty}) \tanh(mL) \].Substitute the known values:- \( mL = 72.95 \times 0.04 = 2.918 \).Using \( \tanh(2.918) \approx 0.993 \), the fin heat transfer rate is:\[ Q_f = \sqrt{1000 \times 0.00628 \times 140 \times 3.14 \times 10^{-6}} \times (50 - 25) \times 0.993 \approx 3.408 \text{ W} \].
05

Calculate the Fin Heat Transfer Rate with Prescribed Tip Temperature (Part b)

For a prescribed tip temperature \( T(L) = 0^{\circ} \text{C} \), the fin heat transfer rate is given by:\[ Q_f = \sqrt{hPkA_c}(T_b - T_\infty) \frac{\sinh(mL) + (T(L) - T_{\infty})/\Delta T \cosh(mL)}{\cosh(mL) + (T(L) - T_{\infty})/\Delta T \sinh(mL)} \].Since \( T(L) = 0 \text{ C} \) and \( T_{\infty} = 25 \text{ C} \):- \( \Delta T = T_b - T_{\infty} = 50 - 25 = 25 \).- Plug in known values into the expression:\[ Q_f = 3.408 \times \frac{\sinh(2.918)}{\cosh(2.918)} \approx 3.500 \text{ W} \].
06

Plot Temperature Distribution (Part c)

To plot the temperature distribution over the fin, we use the fin temperature equations:- For the adiabatic tip:\[ T(x) = T_{\infty} + (T_b - T_{\infty}) \frac{\cosh(m(L-x))}{\cosh(mL)} \].- For the prescribed tip temperature:\[ T(x) = T_{\infty} + (T_b - T_{\infty}) \frac{\cosh(m(L-x))}{\cosh(mL) + (T(L) - T_{\infty})/\Delta T \sinh(mL)} \].By plotting these equations from \( x=0 \) to \( x=L \), we observe that the temperature for adiabatic tip decreases less drastically compared to the prescribed tip 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 91影视!

Key Concepts

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

Pin Fin
Pin fins are slender rods attached to surfaces to enhance heat transfer from the surface to the surrounding fluid. These fins are commonly used in applications like electronics cooling, where removing excess heat is crucial.
Pin fins maximize surface area without adding too much weight or volume. This small size makes them suitable for environments where space is limited.
  • Structure: Pin fins are usually cylindrical in shape, which allows for uniform distribution of heat along the fin.
  • Function: The main purpose of pin fins is to increase the rate of heat transfer through conduction and convection processes.
  • Material: Metal is often used because of its high thermal conductivity, which helps efficient heat distribution.
A pin fin's performance is often characterized by its efficiency and effectiveness. It depends on factors such as the fin's material, size, and the environment it's operating in.
Thermal Conductivity
Thermal conductivity characterizes a material's ability to conduct heat. It's a crucial property that affects how quickly or slowly heat can be transferred through a material.
The higher the thermal conductivity, the faster heat moves through the material. This property is defined as the rate at which heat is conducted through a material at a specified temperature gradient. For instance, in the given exercise, we use an aluminum alloy with a thermal conductivity of 140 W/m路K.
  • Importance: Materials with high thermal conductivity, like metals, are excellent for heat transfer applications because they rapidly transfer heat.
  • Factors Affecting: The internal structure of the material and the temperature difference across it.
Measuring thermal conductivity helps engineers select appropriate materials for applications involving heat transfer.
Convective Heat Transfer
Convective heat transfer refers to the movement of heat from a surface into a fluid, like air or liquid, moving over it. It combines the effects of conduction within the fluid and the motion of the fluid itself.
Convective heat transfer is greatly influenced by the velocity and properties of the fluid and the surface area. In our exercise, a convective heat transfer coefficient of 1000 W/m虏路K is used.
  • Types: Natural convection involves fluid motion caused by buoyancy differences, while forced convection involves external forces like fans or pumps.
  • Coefficient: The convective heat transfer coefficient quantifies the resistance to heat transfer between a solid surface and a moving fluid.
Understanding convective heat transfer is key for designing efficient thermal systems and is particularly important in the field of heat exchangers and radiators.

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

In a manufacturing process, as transparent film is being bonded to a substrate as shown in the sketch. To cure the bond at a temperature \(T_{\text {ty }}\) a radiant source is used to provide a heat flux \(q_{i}^{\pi}\left(\mathrm{W} / \mathrm{m}^{2}\right)\), all of which is absorbed at the bonded surface. The back of the substrate is maintained at \(T_{1}\) while the free surface of the film is exposed to air at \(T_{m}\) and a convection heat transfer coefficient \(h\). (a) Show the thermal circuit representing the steady-state heat transfer situation, Be sure to label all elements, nodes, and heat rates. Leave in symbolic form. (b) Assume the following conditions: \(T_{w}=20^{\circ} \mathrm{C}, h=\) \(50 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}_{\text {, }}\) and \(T_{1}=30^{\circ} \mathrm{C}\). Calculate the heat flux \(q_{0}^{\prime}\) that is required to maintain the bonded surface at \(T_{0}=60^{\circ} \mathrm{C}\). (c) Compute and plot the required heat flux as a function of the film thickness for \(0 \leq L_{y} \leq 1 \mathrm{~mm}\). (d) If the film is not transparent and all of the radiant heat flux is absorbed at its upper surface, determine the heat flux required to achieve bonding. Plot your results as a function of \(L_{f}\) for \(0 \leq L_{f} \leq 1 \mathrm{~mm}\).

A firefighter's protective clothing, referred to as a humout cout, is typically constructed as an ensemble of three layexs separated by air gaps, as shown schematically. Representative dimensions and thermal conductivities for the layers are as follows. \begin{tabular}{lcc} \hline Layer & Thickness (mu) & \(k(\mathrm{~W} / \mathrm{m}-\mathrm{K})\) \\ \hline Shell (s) & \(0.8\) & \(0.047\) \\ Moisture barricr \((\mathrm{mb})\) & \(0.55\) & \(0.012\) \\ Thermal liner (t) & \(3.5\) & \(0.038\) \\ \hline \end{tabular} The air gaps between the layers are \(1 \mathrm{~mm}\) thick, and heat is transferred by conduction and radiation ex. change through the stagnant air. The linearized radiation coefficient for a gap may be approximated as, \(h_{\text {rat }}=\sigma\left(T_{1}+T_{2}\right)\left(T_{1}^{2}+T_{2}^{2}\right)=4 o T_{m z}^{3}\), where \(T_{\text {ax }}\) represents the average temperature of the surfaces comprising the gap, and the radiation flux across the gap may be expressed as \(q_{n d}^{*}=h_{\text {ad }}\left(T_{1}-T_{2}\right)\). (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 \(T_{\text {es }}=470 \mathrm{~K}\) may be used to approximate the radiation resistance of both gaps. Comment on the relative magnitudes of the resistances. (b) For a pre-flash-over fire environment in which firefighters often work, the typical radiant heat flux on the fire-side of the tumout coat is \(0.25 \mathrm{~W} / \mathrm{cm}^{2}\). What is the outer surface temperature of the tumout coat if the inner surface temperature is \(66^{\circ} \mathrm{C}\), a condition that would result in bum injury?

The walls of a refrigerator are typically constructed by sandwiching a layer of insulation between shect metal pancls. Consider a wall made from fiberglass insulation of thermal conductivity \(k_{1}=0.046 \mathrm{~W} / \mathrm{m}-\mathrm{K}\) and thickness \(I_{-}=50 \mathrm{~mm}\) and steel parels, each of thermal conductivity \(k_{p}=60\) W/m \(: \mathrm{K}\) and thickness \(L_{7}=3 \mathrm{~mm}\). If the wall separates refrigerated air at \(T_{x, j}=4^{\circ} \mathrm{C}\) from ambient air at \(T_{2,0}=25^{\circ} \mathrm{C}\), what is the heat gain per unit surface area? Cocfficicnts associated with natural convection at the inner and outer surfaces may be approximated as \(h_{y}=h_{g}=5 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\).

A spherical vessel used as a reactor for producing pharmaceuticals has a 10 -mam-thick stainless steel wall \((k=\) \(17 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) and an inner diameter of \(1 \mathrm{~m}\). The exteria surface of the vessel is exposed to ambient air \(\left(T_{z}=\right.\) \(25^{\circ} \mathrm{C}\) ) for which a convection coefficient of \(6 \mathrm{~W} / \mathrm{m}^{2}+\mathrm{K}\) may be assumed. (a) During steady-state operation, an inner surfact temperature of \(50^{\circ} \mathrm{C}\) is maintained by enengy geneated within the reactor. What is the heat loss from the vessel? (b) If a 20-mm-thick layer of fiberglass insulation \((k=\) \(0.040 \mathrm{~W} / \mathrm{m}\) - K) is applied to the exterior of the ve sel and the rate of thermal energy generation is inchanged, what is the inner surface temperature of the vessel?

A plane wall of thickness \(2 L\) and thermal conductivity \(k\) experiences a uniform volumetric generation rate 4. As shown in the sketch for Case I. the surface at \(x=-L\) is perfectly insulated, while the other surface is maintained at a uniform, constant temperature \(T_{a}\). For Case 2, a very thin dielectric strip is inserted at the midpoint of the wall \((x=0)\) in order to electrically isolate the two sections, \(A\) and B. The thermal resistance of the strip is \(R_{7}^{*}=0.0005 \mathrm{~m}^{2} \cdot \mathrm{K} / \mathrm{W}\). The parameters associated with the wall are \(k=50 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, L=20 \mathrm{~mm}\), \(q=5 \times 10^{6} \mathrm{~W} / \mathrm{m}^{3}\), and \(T_{e}=50^{\circ} \mathrm{C}\). (a) Sketch the temperature distribution for case I on \(T-x\) coordinates. Deseribe the key features of this distribution. Identify the location of the maxirmum temperature in the wall and calculate this temperature. (b) Sketch the temperature distribution for Cuse 2 on the same \(T-x\) coordinates. Describe the key features of this distribution. (c) What is the temperature difference between the two walls at \(x=0\) for Case 2? (d) What is the location of the maximum temperature in the composite wall of Case 2? Calculate this temperature.
See all solutions

Recommended explanations on Physics Textbooks

Energy Physics

Read Explanation

Solid State Physics

Read Explanation

Dynamics

Read Explanation

Quantum Physics

Read Explanation

Radiation

Read Explanation

Translational Dynamics

Read Explanation
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.