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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

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.

Consider a flat plate subject to parallel flow (top and bottom) characterized by \(u_{\infty}=5 \mathrm{~m} / \mathrm{s}, T_{\infty}=20^{\circ} \mathrm{C}\). (a) Determine the average convection heat transfer coefficient, convective heat transfer rate, and drag force associated with an \(L=2\)-m-long, \(w=2-\mathrm{m}\) wide flat plate for airflow and surface temperatures of \(T_{s}=50^{\circ} \mathrm{C}\) and \(80^{\circ} \mathrm{C}\). (b) Determine the average convection heat transfer coefficient, convective heat transfer rate, and drag force associated with an \(L=0.1\)-m-long, \(w=0.1\)-m-wide flat plate for water flow and surface temperatures of \(T_{s}=50^{\circ} \mathrm{C}\) and \(80^{\circ} \mathrm{C}\).

Consider the velocity boundary layer profile for flow over a flat plate to be of the form \(u=C_{1}+C_{2} y\). Applying appropriate boundary conditions, obtain an expression for the velocity profile in terms of the boundary layer thickness \(\delta\) and the free stream velocity \(u_{\infty}\). Using the integral form of the boundary layer momentum equation (Appendix G), obtain expressions for the boundary layer thickness and the local friction coefficient, expressing your result in terms of the local Reynolds number. Compare your results with those obtained from the exact solution (Section 7.2.1) and the integral solution with a cubic profile (Appendix \(G\) ).

Dry air at atmospheric pressure and \(350 \mathrm{~K}\), with a free stream velocity of \(25 \mathrm{~m} / \mathrm{s}\), flows over a smooth, porous plate \(1 \mathrm{~m}\) long. (a) Assuming the plate to be saturated with liquid water at \(350 \mathrm{~K}\), estimate the mass rate of evaporation per unit width of the plate, \(n_{\mathrm{A}}^{\prime}(\mathrm{kg} / \mathrm{s} \cdot \mathrm{m})\). (b) For air and liquid water temperatures of 300,325 , and \(350 \mathrm{~K}\), generate plots of \(n_{\mathrm{A}}^{\prime}\) as a function of velocity for the range from 1 to \(25 \mathrm{~m} / \mathrm{s}\).

A long, cylindrical, electrical heating element of diameter \(D=10 \mathrm{~mm}\), thermal conductivity \(k=240 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\), density \(\rho=2700 \mathrm{~kg} / \mathrm{m}^{3}\), and specific heat \(c_{p}=900 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\) is installed in a duct for which air moves in cross flow over the heater at a temperature and velocity of \(27^{\circ} \mathrm{C}\) and \(10 \mathrm{~m} / \mathrm{s}\), respectively. (a) Neglecting radiation, estimate the steady-state surface temperature when, per unit length of the heater, electrical energy is being dissipated at a rate of \(1000 \mathrm{~W} / \mathrm{m}\). (b) If the heater is activated from an initial temperature of \(27^{\circ} \mathrm{C}\), estimate the time required for the surface temperature to come within \(10^{\circ} \mathrm{C}\) of its steady-state value.

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