/*! 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

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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:

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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.

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Most popular questions from this chapter

A straight fin of uniform cross section is fabricated from a material of thermal conductivity \(50 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\), thickness \(w=6 \mathrm{~mm}\), and length \(L=48 \mathrm{~mm}\), and it is very long in the direction normal to the page. The convection heat transfer coefficient is \(500 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) with an ambient air temperature of \(T_{\infty}=30^{\circ} \mathrm{C}\). The base of the \(\mathrm{f}\) is maintained at \(T_{b}=100^{\circ} \mathrm{C}\), while the fi tip is well insulated. (a) Using a finite-difference method with a space increment of \(4 \mathrm{~mm}\), estimate the temperature distribution within the fin. Is the assumption of onedimensional heat transfer reasonable for this fin? (b) Estimate the fin heat transfer rate per unit length normal to the page. Compare your result with the one-dimensional, analytical solution, Equation \(3.81\). (c) Using the finite-difference mesh of part (a), compute and plot the fin temperature distribution for values of \(h=10,100,500\), and \(1000 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Determine and plot the fin heat transfer rate as a function of \(h\).

An electrical heater \(100 \mathrm{~mm}\) long and \(5 \mathrm{~mm}\) in diameter is inserted into a hole drilled normal to the surface of a large block of material having a thermal conductivity of \(5 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). Estimate the temperature reached by the heater when dissipating \(50 \mathrm{~W}\) with the surface of the block at a temperature of \(25^{\circ} \mathrm{C}\).

A hot fluid passes through circular channels of a cast iron platen (A) of thickness \(L_{\mathrm{A}}=30 \mathrm{~mm}\) which is in poor contact with the cover plates (B) of thickness \(L_{\mathrm{B}}=7.5 \mathrm{~mm}\). The channels are of diameter \(D=15 \mathrm{~mm}\) with a centerline spacing of \(L_{o}=60 \mathrm{~mm}\). The thermal conductivities of the materials are \(k_{\mathrm{A}}=20 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) and \(k_{\mathrm{B}}=75 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\), while the contact resistance between the two materials is \(R_{t, c}^{r}=2.0 \times 10^{-4} \mathrm{~m}^{2} \cdot \mathrm{K} / \mathrm{W}\). The hot fluid is at \(T_{i}=150^{\circ} \mathrm{C}\), and the convection coeftient is \(1000 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). The cover plate is exposed to ambient air at \(T_{\infty}=25^{\circ} \mathrm{C}\) with a convection coeftient of \(200 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). The shape factor between one channel and the platen top and bottom surfaces is 4.25.

A rod of \(10-\mathrm{mm}\) diameter and \(250-\mathrm{mm}\) length has one end maintained at \(100^{\circ} \mathrm{C}\). The surface of the rod experiences free convection with the ambient air at \(25^{\circ} \mathrm{C}\) and a convection coefficient that depends on the difference between the temperature of the surface and the ambient air. Specifically, the coefficient is prescribed by a correlation of the form, \(h_{\mathrm{fc}}=2.89[0.6+0.624\) \(\left.\left(T-T_{\infty}\right)^{1 / 6}\right]^{2}\), where the units are \(h_{\mathrm{fc}}\left(\mathrm{W} / \mathrm{m}^{2} \cdot \mathrm{K}\right)\) and \(T(\mathrm{~K})\). The surface of the rod has an emissivity \(\varepsilon=0.2\) and experiences radiation exchange with the surroundings at \(T_{\text {sur }}=25^{\circ} \mathrm{C}\). The fi tip also experiences free convection and radiation exchange. Assuming one-dimensional conduction and using a finite-difference method representing the fin by five nodes, estimate the temperature distribution for the fin. Determine also the fin heat rate and the relative contributions of free convection and radiation exchange. Hint: For each node requiring an energy balance, use the linearized form of the radiation rate equation, Equation 1.8, with the radiation coefficient \(h_{r}\), Equation \(1.9\), evaluated for each node. Similarly, for the convection rate equation associated with each node, the free convection coefficient \(h_{\mathrm{fc}}\) must be evaluated for each node.

Small-diameter electrical heating elements dissipating \(50 \mathrm{~W} / \mathrm{m}\) (length normal to the sketch) are used to heat a ceramic plate of thermal conductivity \(2 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). The upper surface of the plate is exposed to ambient air at \(30^{\circ} \mathrm{C}\) with a convection coeftient of \(100 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), while the lower surface is well insulated. (a) Using the Gauss-Seidel method with a grid spacing of \(\Delta x=6 \mathrm{~mm}\) and \(\Delta y=2 \mathrm{~mm}\), obtain the temperature distribution within the plate. (b) Using the calculated nodal temperatures, sketch four isotherms to illustrate the temperature distribution in the plate. (c) Calculate the heat loss by convection from the plate to the fluid. Compare this value with the element dissipation rate. (d) What advantage, if any, is there in not making \(\Delta x=\Delta y\) for this situation? (e) With \(\Delta x=\Delta y=2 \mathrm{~mm}\), calculate the temperature field within the plate and the rate of heat transfer from the plate. Under no circumstances may the temperature at any location in the plate exceed \(400^{\circ} \mathrm{C}\). Would this limit be exceeded if the airfw were terminated and heat transfer to the air were by natural convection with \(h=10 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) ?
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