/*! 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 31 A hole of diameter \(D=0.25 \mat... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 4: Problem 31

A hole of diameter \(D=0.25 \mathrm{~m}\) is drilled through the center of a solid block of square cross section with \(w=1 \mathrm{~m}\) on a side. The hole is drilled along the length, \(l=2 \mathrm{~m}\), of the block, which has a thermal conductivity of \(k=150 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). The four outer surfaces are exposed to ambient air, with \(T_{\infty, 2}=25^{\circ} \mathrm{C}\) and \(h_{2}=4 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), while hot oil flowing through the hole is characterized by \(T_{\infty, 1}=300^{\circ} \mathrm{C}\) and \(h_{1}=50 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Determine the corresponding heat rate and surface temperatures.

Short Answer

Expert verified
\(A_c = 1^2 - \frac{\pi (0.25)^2}{4} = 0.8036\) m虏 \(R_{cond} = \frac{2}{150(0.8036)} = 0.0166\) K/W #tag_title# Step 2: Calculate the thermal resistance due to convection #tag_content# The thermal resistance due to the convection at the hole's surface can be calculated as: \(R_{conv1} = \frac{1}{h_1 A_1}\) and the thermal resistance due to convection at the outer surfaces of the block can be calculated as: \(R_{conv2} = \frac{1}{h_2 A_2}\) where: \(h_1\) = Convection heat transfer coefficient of the hot oil (50 W/m虏路K) \(A_1\) = Area of the hole (circumference of the hole times the length) \(h_2\) = Convection heat transfer coefficient of the ambient air (4 W/m虏路K) \(A_2\) = Area of the outer surfaces of the block Calculating the areas and the thermal resistances due to convection: \(A_1 = \pi D L = \pi(0.25)(2) = 1.571\) m虏 \(R_{conv1} = \frac{1}{50(1.571)} = 0.0127\) K/W \(A_2 = 4wL = 4(1)(2) = 8\) m虏 \(R_{conv2} = \frac{1}{4(8)} = 0.0313\) K/W #tag_title# Step 3: Calculate the temperature difference and the heat rate #tag_content# The net thermal resistance can be calculated as the sum of the individual resistances: \(R_{net} = R_{cond} + R_{conv1} + R_{conv2}\) Calculating the net thermal resistance: \(R_{net} = 0.0166 + 0.0127 + 0.0313 = 0.0606\) K/W The temperature difference between the hot oil and ambient air is: \(\Delta T = T_{\infty, 1} - T_{\infty, 2} = 300 - 25 = 275\) K We can now determine the heat rate using Ohm's law for thermal circuits: \(Q = \frac{\Delta T}{R_{net}}\) Calculating the heat rate: \(Q = \frac{275}{0.0606} = 4534.98\) W #tag_title# Step 4: Calculate the surface temperatures #tag_content# Finally, we can calculate the surface temperatures. Let's denote the temperature at the hole's surface as \(T_s1\) and the temperature at the outer surface of the block as \(T_s2\). We can use the heat rate and the thermal resistances to find these temperatures: \(T_s1 = T_{\infty, 1} - Q R_{conv1}\) \(T_s2 = T_{\infty, 2} + Q R_{conv2}\) Calculating the surface temperatures: \(T_s1 = 300 - 4534.98(0.0127) = 242.22^\circ \textrm{C}\) \(T_s2 = 25 + 4534.98(0.0313) = 167.21^\circ \textrm{C}\) #Answer# The heat rate is 4534.98 W. The surface temperature at the hole's surface is \(242.22^\circ \textrm{C}\), and the surface temperature at the outer surface of the block is \(167.21^\circ \textrm{C}\).

Step by step solution

01

Calculate the thermal resistance due to conduction in the block

The block is a square prism with a square cross section, so we can treat it as a one-dimensional problem: \(R_{cond} = \frac{L}{kA_c}\) where: \(R_{cond}\) = Thermal resistance due to conduction \(L\) = Length of the block (2 m) \(k\) = Thermal conductivity of the block (150 W/ m路K) \(A_c\) = Area of the conduction (Cross-sectional area - Hole area) The cross-sectional area is: \(A_c =w^2 - \frac{\pi D^2}{4}\), where \(w = 1\) m and \(D = 0.25\) m. Calculating the area and the thermal resistance due to conduction:

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.

Thermal Conductivity
Thermal conductivity is a material鈥檚 ability to conduct heat. It signifies how easily heat can pass through a material. Take, for example, a solid block of metal with a certain thermal conductivity. This value, often represented by the symbol \( k \) and measured in \( \text{W/m}\cdot\text{K} \) (watts per meter-kelvin), tells us how effectively the block can transfer heat from its hot end to its cold end.

Using metals as our primary examples, let鈥檚 consider the exercise where the block has a thermal conductivity of \( k=150 \text{W/m}\cdot\text{K} \). This is quite high, indicating that metal is an excellent conductor, capable of allowing a large amount of heat to pass through it. In practical applications, like heat sinks or cooking utensils, high thermal conductivity is desired because it helps in the efficient transfer of heat.

Understanding this concept is crucial when attempting to control or utilize heat transfer in engineering systems, such as the flow of hot oil through a drilled hole in our block.
Thermal Resistance
Thermal resistance is a measure of a material鈥檚 opposition to the flow of heat. The greater the resistance, the slower the heat transfer. It's analogous to electrical resistance, but instead of impeding electrical current, it inhibits the flow of thermal energy. This value is represented by \( R \) and is given by the formula: \( R_{\text{cond}} = \frac{L}{kA_c} \), where \( L \) is the object鈥檚 thickness, \( k \) is the thermal conductivity, and \( A_c \) is the cross-sectional area available for heat flow.

In the context of our exercise, calculating the thermal resistance due to conduction in a solid block involves considering both the properties of the material (the \( k \) value) and the geometry (the area \( A_c \) through which heat is conducted). The presence of the hole in the middle of the block alters the effective cross-sectional area, which, in turn, modifies the thermal resistance when compared to a completely solid block.
Conduction in Solids
Conduction in solids is the process of transferring heat through a solid material without any actual movement of the material itself. It occurs on a microscopic scale as vibrating atoms and free electrons transfer kinetic energy from one to the next. The rate at which this energy is transferred is governed by the thermal conductivity of the solid.

For a solid like the square block in our exercise, conduction occurs from the hot oil in the center, passing through the block, to the cooler ambient air on the outside. It鈥檚 important when considering conduction in solids to look at the entire geometry of the heat transfer path including any variations like holes or indentations.

The hole drilled through the center of the block introduces complexity into this heat transfer scenario by reducing the cross-sectional area through which heat can be conducted. This highlights the importance of understanding the entire structure when predicting heat flow in engineering and design contexts.

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

Conduction within relatively complex geometries can sometimes be evaluated using the finite-difference methods of this text that are applied to subdomains and patched together. Consider the two-dimensional domain formed by rectangular and cylindrical subdomains patched at the common, dashed control surface. Note that, along the dashed control surface, temperatures in the two subdomains are identical and local conduction heat fluxes to the cylindrical subdomain are identical to local conduction heat fluxes from the rectangular subdomain. Calculate the heat transfer per unit depth into the page, \(q^{\prime}\), using \(\Delta x=\Delta y=\Delta r=10 \mathrm{~mm}\) and \(\Delta \phi=\pi / 8\). The base of the rectangular subdomain is held at \(T_{h}=20^{\circ} \mathrm{C}\), while the vertical surface of the cylindrical subdomain and the surface at outer radius \(r_{o}\) are at \(T_{c}=0^{\circ} \mathrm{C}\). The remaining surfaces are adiabatic, and the thermal conductivity is \(k=10 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\).

An electronic device, in the form of a disk \(20 \mathrm{~mm}\) in diameter, dissipates \(100 \mathrm{~W}\) when mounted flush on a large aluminum alloy (2024) block whose temperature is maintained at \(27^{\circ} \mathrm{C}\). The mounting arrangement is such that a contact resistance of \(R_{t, c}^{\prime \prime}=5 \times 10^{-5}\) \(\mathrm{m}^{2} \cdot \mathrm{K} / \mathrm{W}\) exists at the interface between the device and the block. (a) Calculate the temperature the device will reach, assuming that all the power generated by the device must be transferred by conduction to the block. (b) To operate the device at a higher power level, a circuit designer proposes to attach a finned heat sink to the top of the device. The pin fins and base material are fabricated from copper \((k=400 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) and are exposed to an airstream at \(27^{\circ} \mathrm{C}\) for which the convection coefficient is \(1000 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). For the device temperature computed in part (a), what is the permissible operating power?

A long constantan wire of \(1-\mathrm{mm}\) diameter is butt welded to the surface of a large copper block, forming a thermocouple junction. The wire behaves as a fin, permitting heat to flow from the surface, thereby depressing the sensing junction temperature \(T_{j}\) below that of the block \(T_{\sigma}\). Copper block, \(T_{o}\) (a) If the wire is in air at \(25^{\circ} \mathrm{C}\) with a convection coefficient of \(10 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), estimate the measurement error \(\left(T_{j}-T_{o}\right)\) for the thermocouple when the block is at \(125^{\circ} \mathrm{C}\). (b) For convection coefficients of 5,10 , and \(25 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), plot the measurement error as a function of the thermal conductivity of the block material over the range 15 to \(400 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). Under what circumstances is it advantageous to use smaller diameter wire?

Consider heat transfer in a one-dimensional (radial) cylindrical coordinate system under steady-state conditions with volumetric heat generation. (a) Derive the finite-difference equation for any interior node \(m\). (b) Derive the finite-difference equation for the node \(n\) located at the external boundary subjected to the convection process \(\left(T_{\infty 0}, h\right)\).

The top surface of a plate, including its grooves, is maintained at a uniform temperature of \(T_{1}=200^{\circ} \mathrm{C}\). The lower surface is at \(T_{2}=20^{\circ} \mathrm{C}\), the thermal conductivity is \(15 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\), and the groove spacing is \(0.16 \mathrm{~m}\). (a) Using a finite-difference method with a mesh size of \(\Delta x=\Delta y=40 \mathrm{~mm}\), calculate the unknown nodal temperatures and the heat transfer rate per width of groove spacing \((w)\) and per unit length normal to the page. (b) With a mesh size of \(\Delta x=\Delta y=10 \mathrm{~mm}\), repeat the foregoing calculations, determining the temperature field and the heat rate. Also, consider conditions for which the bottom surface is not at a uniform temperature \(T_{2}\) but is exposed to a fluid at \(T_{\infty}=20^{\circ} \mathrm{C}\). With \(\Delta x=\Delta y=10 \mathrm{~mm}\), determine the temperature field and heat rate for values of \(h=5,200\), and 1000 \(\mathrm{W} / \mathrm{m}^{2} \cdot \mathrm{K}\), as well as for \(h \rightarrow \infty\).
See all solutions

Recommended explanations on Physics Textbooks

Magnetism and Electromagnetic Induction

Read Explanation

Conservation of Energy and Momentum

Read Explanation

Nuclear Physics

Read Explanation

Space Physics

Read Explanation

Thermodynamics

Read Explanation

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.