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Chapter 7: Problem 80

Copper spheres of \(20-\mathrm{mm}\) diameter are quenched by being dropped into a tank of water that is maintained at \(280 \mathrm{~K}\). The spheres may be assumed to reach the terminal velocity on impact and to drop freely through the water. Estimate the terminal velocity by equating the drag and gravitational forces acting on the sphere. What is the approximate height of the water tank needed to cool the spheres from an initial temperature of \(360 \mathrm{~K}\) to a center temperature of \(320 \mathrm{~K}\) ?

Short Answer

Expert verified
The terminal velocity of the copper spheres can be found by equating the drag and gravitational forces as \(v = \sqrt{\frac{2mg}{蟻_w C_D A}}\). The heat transfer required for cooling is calculated as \(Q = mc_p (T_i - T_f)\). The time required to reach the desired center temperature can be found using Newton's law of cooling, as \(t = \frac{Q}{hA_s (T_s - T_w)}\). Finally, the height of the water tank can be determined using the formula \(h = vt\).

Step by step solution

01

Calculate gravitational force

Let's first determine the gravitational force acting on the copper sphere. The weight of the copper sphere can be calculated using the following formula: \[F_g = mg\] Where \(F_g\) is gravitational force, \(m\) is mass, and \(g = 9.81\,m/s^2\) is the acceleration due to gravity. To find the mass, we need the density of copper (\(蟻_c = 8960\,kg/m^3\)) and the sphere's volume: \[V = (4/3)蟺r^3\] Now, calculate the mass using density and volume: \[m = 蟻_c V\] Finally, calculate the gravitational force.
02

Calculate drag force

To calculate the drag force on the sphere, we will use the following formula: \[F_d = (1/2) 蟻_w C_D A v^2\] Where \(F_d\) is the drag force, \(蟻_w = 1000\,kg/m^3\) is the density of water, \(C_D\) is the drag coefficient (assume \(C_D = 0.5\) for a sphere), \(A\) is the sphere's cross-sectional area (which can be calculated as \(蟺r^2\)), and v is the terminal velocity we need to find.
03

Equate gravitational and drag forces

Since the sphere reaches terminal velocity, the drag force equals the gravitational force. Therefore, we can equate drag and gravitational forces and solve for terminal velocity: \[(1/2) 蟻_w C_D A v^2 = mg\] \[v = \sqrt{\frac{2mg}{蟻_w C_D A}}\] Now, calculate the terminal velocity. 2. Determine heat transfer.
04

Heat transfer required

We need to calculate the heat transfer required to cool the spheres from the initial temperature of \(360\,K\) to a center temperature of \(320\,K\). This can be determined using the specific heat of copper (\(c_p = 385\,J/(kg 鈰� K)\)): \[Q = mc_p (T_i - T_f)\] Calculate the heat transfer required. 3. Calculate the time to reach the desired center temperature.
05

Newton's law of cooling

To find the time needed, we can use Newton's law of cooling: \[Q = hA_s(t) (T_s - T_w)\] Where \(Q\) is the heat transfer calculated in step 2, \(h\) is the heat transfer coefficient (For spheres in water at low Re, an approximation \(h 鈮� 6.12(Re)^{0.5205}\) can be used; more specific values require more information), \(A_s\) is the surface area of the sphere, \(t\) is the time needed, \(T_s\) is the center temperature (\(320\,K\)), and \(T_w\) is the water temperature (\(280\,K\)). Rearrange the formula to solve for time: \[t = \frac{Q}{hA_s (T_s - T_w)}\] Calculate the time needed to reach the desired center temperature. 4. Estimate the water tank height.
06

Water tank height

Finally, we can find the height of the water tank needed by using the terminal velocity \(v\) found in step 1 and the time \(t\) found in step 3, with the formula: \[h = vt\] Calculate the height of the water tank. For more accurate results, students may iterate the process, considering heat transfer coefficient changes, as well as buoyancy and other minor factors that are not accounted for in this simplified approach.

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Key Concepts

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

Terminal Velocity Calculation
When an object falls through a fluid, such as water, it experiences a balance of forces that ultimately leads to terminal velocity. This is the constant speed the object maintains as it moves through the fluid. The gravitational force pulling it downward is counteracted by the drag force from the fluid, and when these forces equate, the object stops accelerating and continues to fall at a steady speed.

For a solid sphere like the copper sphere in our exercise, the terminal velocity calculation involves equating the gravitational force (\( F_g = mg \)) and the drag force (\( F_d = \frac{1}{2} \rho_w C_D A v^2 \)). The drag force depends on factors such as the density of the fluid (\( \rho_w \)), the drag coefficient (\( C_D \)), cross-sectional area (\( A = \text{\textpi} r^2 \)), and the velocity (\( v \) for terminal velocity). By solving the equation \( \frac{1}{2} \rho_w C_D A v^2 = mg \) for velocity, students can find the terminal velocity for any spherical object.

Understanding the forces involved in reaching terminal velocity is crucial when determining how objects move through various mediums. Whether in water, air, or other fluids, this concept is particularly important for engineers and physicists working with motion dynamics.
Newton's Law of Cooling
Newton's law of cooling describes the rate at which an object changes temperature through radiation, stating that the rate of heat loss of a body is proportional to the difference in temperatures between the body and its surroundings. This can be expressed by the formula \( Q = hA_s(t) (T_s - T_w) \), where \( Q \) is the heat transfer, \( h \) is the heat transfer coefficient, \( A_s \) is the surface area of the object, \( t \) is time, \( T_s \) is the object's temperature, and \( T_w \) is the ambient temperature.

In the context of our exercise, Newton's law of cooling allows us to calculate how long it will take for the copper spheres to cool from their initial temperature to the center temperature in the water. This law is widely used in various engineering applications, such as designing cooling systems, and in everyday scenarios, such as estimating how quickly a cup of coffee will cool down in a room. Knowledge of this law is instrumental for students aiming to understand heat transfer in real-world situations.
Quenching Process
Quenching is a rapid cooling process used to alter the microstructure of materials, such as metals, to enhance their mechanical properties like hardness and strength. In industrial settings, quenching often involves immersing a hot metal object into a liquid, most commonly water or oil. The quenching process is characterized by different cooling rates throughout the material, causing transformations in its crystal structure which result in the desired alterations.

Specifically, the exercise mentions quenching copper spheres by dropping them into a tank of water. Quenching to a certain temperature, like from 360K to a center temperature of 320K, requires understanding the material properties, such as heat capacity and the heat transfer coefficient. The heat removed (\( Q \)) is calculated based on the copper's specific heat and the temperature change, according to the formula \( Q = mc_p (T_i - T_f) \). The quenching process is a critical step in manufacturing and metallurgy and a fundamental concept for students studying material science and thermodynamics.

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Most popular questions from this chapter

A stream of atmospheric air is used to dry a series of biological samples on plates that are each of length \(L_{i}=0.25 \mathrm{~m}\) in the direction of the airflow. The air is dry and at a temperature equal to that of the plates \(\left(T_{\infty}=T_{s}=50^{\circ} \mathrm{C}\right)\). The air speed is \(u_{\infty}=9.1 \mathrm{~m} / \mathrm{s}\). (a) Sketch the variation of the local convection mass transfer coefficient \(h_{m x}\) with distance \(x\) from the leading edge. Indicate the specific nature of the \(x\) dependence. (b) Which of the plates will dry the fastest? Calculate the drying rate per meter of width for this plate \((\mathrm{kg} / \mathrm{s}+\mathrm{m})\). (c) At what rate would heat have to be supplied to the fastest drying plate to maintain it at \(T_{s}=50^{\circ} \mathrm{C}\) during the drying process?

Steel (AISI 1010) plates of thickness \(\delta=6 \mathrm{~mm}\) and length \(L=1 \mathrm{~m}\) on a side are conveyed from a heat treatment process and are concurrently cooled by atmospheric air of velocity \(u_{\infty}=10 \mathrm{~m} / \mathrm{s}\) and \(T_{x}=20^{\circ} \mathrm{C}\) in parallel flow over the plates. For an initial plate temperature of \(T_{i}=300^{\circ} \mathrm{C}\), what is the rate of heat transfer from the plate? What is the corresponding rate of change of the plate temperature? The velocity of the air is much larger than that of the plate.

A spherical thermocouple junction \(1.0 \mathrm{~mm}\) in diameter is inserted in a combustion chamber to measure the temperature \(T_{\infty}\) of the products of combustion. The hot gases have a velocity of \(V=5 \mathrm{~m} / \mathrm{s}\). (a) If the thermocouple is at room temperature, \(T_{i}\), when it is inserted in the chamber, estimate the time required for the temperature difference, \(T_{\infty}-T\), to reach \(2 \%\) of the initial temperature difference, \(T_{\infty}-T_{i}\). Neglect radiation and conduction through the leads. Properties of the thermocouple junction are approximated as \(k=100 \mathrm{~W} / \mathrm{m}+\mathrm{K}, c=385 \mathrm{~J} / \mathrm{kg}+\mathrm{K}\), and \(\rho=8920 \mathrm{~kg} / \mathrm{m}^{3}\), while those of the combustion gases may be approximated as \(k=0.05 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\), \(\nu=50 \times 10^{-6} \mathrm{~m}^{2} / \mathrm{s}\), and \(\operatorname{Pr}=0.69\). (b) If the thermocouple junction has an emissivity of \(0.5\) and the cooled walls of the combustor are at \(T_{c}=400 \mathrm{~K}\), what is the steady- state temperature of the thermocouple junction if the combustion gases are at \(1000 \mathrm{~K}\) ? Conduction through the lead wires may be neglected. (c) To determine the influence of the gas velocity on the thermocouple measurement error, compute the steady-state temperature of the thermocouple junction for velocities in the range \(1 \leq V \leq 25 \mathrm{~m} / \mathrm{s}\). The emissivity of the junction can be controlled through application of a thin coating. To reduce the measurement error, should the emissivity be increased or decreased? For \(V=5 \mathrm{~m} / \mathrm{s}\), compute the steadystate junction temperature for emissivities in the range \(0.1 \leq \varepsilon \leq 1.0\).

Latent heat capsules consist of a thin-walled spherical shell within which a solid-liquid, phase-change material \((\mathrm{PCM})\) of melting point \(T_{\mathrm{mp}}\) and latent heat of fusion \(h_{s f}\) is enclosed. As shown schematically, the capsules may be packed in a cylindrical vessel through which there is fluid flow. If the \(\mathrm{PCM}\) is in its solid state and \(T_{\mathrm{mp}}T_{i}\), energy is released from the \(\mathrm{PCM}\) as it freezes and heat is transferred to the fluid. In either situation, all of the capsules within the packed bed would remain at \(T_{\mathrm{mp}}\) through much of the phase change process, in which case the fluid outlet temperature would remain at a fixed value \(T_{o^{*}}\). Consider an application for which air at atmospheric pressure is chilled by passing it through a packed bed \((\varepsilon=0.5)\) of capsules \(\left(D_{c}=50 \mathrm{~mm}\right)\) containing an organic compound with a melting point of \(T_{\text {mp }}=4^{\circ} \mathrm{C}\). The air enters a cylindrical vessel \(\left(L_{v}=D_{v}=0.40 \mathrm{~m}\right)\) at \(T_{i}=25^{\circ} \mathrm{C}\) and \(V=1.0 \mathrm{~m} / \mathrm{s}\). (a) If the PCM in each capsule is in the solid state at \(T_{\text {map }}\) as melting occurs within the capsule, what is the outlet temperature of the air? If the density and latent heat of fusion of the \(\mathrm{PCM}\) are \(\rho=\) \(1200 \mathrm{~kg} / \mathrm{m}^{3}\) and \(h_{s f}=165 \mathrm{~kJ} / \mathrm{kg}\), what is the mass rate \((\mathrm{kg} / \mathrm{s})\) at which the \(\mathrm{PCM}\) is converted from solid to liquid in the vessel? (b) Explore the effect of the inlet air velocity and capsule diameter on the outlet temperature. (c) At what location in the vessel will complete melting of the PCM in a capsule first occur? Once complete melting begins to occur, how will the outlet temperature vary with time and what is its asymptotic value?

Consider laminar, parallel flow past an isothermal flat plate of length \(L\), providing an average heat transfer coefficient of \(\bar{h}_{L^{-}}\)If the plate is divided into \(N\) smaller plates, each of length \(L_{N}=L / N\), determine an expression for the ratio of the heat transfer coefficient averaged over the \(N\) plates to the heat transfer coefficient averaged over the single plate, \(\bar{h}_{L, N} / \bar{h}_{L, 1}\).
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