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Air at \(20^{\circ} \mathrm{C}(1 \mathrm{~atm})\) is flowing over a \(5-\mathrm{cm}\) diameter sphere with a velocity of \(3.5 \mathrm{~m} / \mathrm{s}\). If the surface temperature of the sphere is constant at \(80^{\circ} \mathrm{C}\), determine \((a)\) the average drag coefficient on the sphere and \((b)\) the heat transfer rate from the sphere.

Short Answer

Expert verified
Answer: The average drag coefficient on the sphere is 0.001287, and the heat transfer rate from the sphere is 2.470 W.

Step by step solution

01

Calculate the Reynolds number for the flow

The Reynolds number is a dimensionless quantity that compares the inertial forces to the viscous forces of a flow, and can be calculated using the formula: Re = (蟻 * V * D) / 渭 where 蟻 is the air density, V is the air velocity, D is the sphere's diameter, and 渭 is the air dynamic viscosity. For air at 20掳C (1 atm), we have 蟻 = 1.204 kg/m鲁 and 渭 = 1.815e-5 kg/(m*s). The velocity and diameter are given as V = 3.5 m/s and D = 0.05 m. Calculating the Reynolds number: Re = (1.204 * 3.5 * 0.05) / 1.815e-5 = 18,641
02

Obtain the drag coefficient from the Reynolds number

The drag coefficient (Cd) for a sphere can be obtained from the Reynolds number using empirical correlations. One such correlation is the Schlichting formula: Cd = 24/Re Using the calculated Reynolds number, we can now find the drag coefficient: Cd = 24 / 18,641 = 0.001287 Thus, the average drag coefficient on the sphere is 0.001287.
03

Calculate the Nusselt number for the flow

The Nusselt number is a dimensionless number that represents the ratio of convective heat transfer to conductive heat transfer. For a sphere, an empirical correlation can be used to determine the Nusselt number based on the Reynolds and Prandtl numbers: Nu = 2 + 0.4 * Re^0.5 * Pr^(1/3) For air at 20掳C, the Prandtl number (Pr) can be approximated as 0.7. Using the previously calculated Reynolds number, we can now calculate the Nusselt number: Nu = 2 + 0.4 * 18,641^0.5 * 0.7^(1/3) = 9.982
04

Calculate the heat transfer coefficient from the Nusselt number

The heat transfer coefficient (h) can be obtained from the Nusselt number using the formula: h = Nu * k / D where k is the thermal conductivity of the air and D is the sphere's diameter. For air at 20掳C, k = 0.0262 W/(m*K). Calculating the heat transfer coefficient: h = 9.982 * 0.0262 / 0.05 = 5.228 W/(m虏*K)
05

Calculate the heat transfer rate from the sphere

Finally, the heat transfer rate (Q) can be calculated using the formula: Q = h * A * 螖T where A is the surface area of the sphere and 螖T is the temperature difference between the surface and the air. The surface area of a sphere can be calculated using the formula: A = 4 * 蟺 * (D/2)^2 Substituting the given values: A = 4 * 蟺 * (0.05/2)^2 = 0.007853 m虏 螖T = 80掳C - 20掳C = 60 K Calculating the heat transfer rate: Q = 5.228 * 0.007853 * 60 = 2.470 W In conclusion, the average drag coefficient on the sphere is 0.001287, and the heat transfer rate from the sphere is 2.470 W.

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Key Concepts

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

Reynolds Number
Reynolds Number (\(Re\)) is a crucial dimensionless quantity that helps us determine the flow regime of a fluid over a body, such as a sphere. It is a balance of inertial forces to viscous forces within a fluid flow. The formula for Reynolds number is:
\[Re = \frac{\rho \cdot V \cdot D}{\mu}\]
  • \(\rho\): Density of the fluid (kg/m鲁)
  • \(V\): Velocity of the fluid (m/s)
  • \(D\): Characteristic length, e.g., diameter of a sphere (m)
  • \(\mu\): Dynamic viscosity of the fluid (kg/(m路s))
This dimensionless number tells us whether the flow is laminar or turbulent. In the case of our problem, with a flow over a sphere having a Reynolds number of 18,641, the flow is most likely turbulent. This high Reynolds number indicates that inertial forces dominate over viscous forces, causing eddies and fluctuations in the flow.
Nusselt Number
The Nusselt Number (\(Nu\)) serves to express the enhancement of heat transfer through a fluid due to convection, compared to conduction alone. It is vital in understanding the efficiency of heat transfer from the surface of objects like a sphere. The formula often used for spheres is:
\[Nu = 2 + 0.4 \cdot Re^{0.5} \cdot Pr^{1/3}\]Here, Pr is the Prandtl number, again a dimensionless quantity that represents the ratio of momentum diffusivity to thermal diffusivity.
  • \(Pr\): Prandtl number (dimensionless)
For example, with a Prandtl number of 0.7 and the previously calculated Reynolds number, we find the Nusselt number to be approximately 9.982. This implies that the process enhances the heat transfer significantly beyond conduction alone, a vital insight for engineering applications.
Drag Coefficient
The Drag Coefficient (\(C_d\)) is a dimensionless number that quantifies the resistance of an object in a fluid environment, such as air or water. It measures how "draggy" an object is:
\[C_d = \frac{24}{Re}\]In the case of our sphere, using a Reynolds number of 18,641, we find a remarkably low drag coefficient of 0.001287. Such a low value suggests minimal resistance, ideal for applications where efficiency is critical. The drag coefficient factors in the shape of the body and the nature of the flow, hence its importance in design and analysis tasks for objects moving through fluids.
Heat Transfer Coefficient
The Heat Transfer Coefficient (\(h\)) indicates how effectively heat is transferred from a solid surface to a fluid or vice versa. It is achieved from the Nusselt Number calculation:
\[ h = \frac{Nu \cdot k}{D}\]
  • \(k\): Thermal conductivity of the fluid (W/(m路K))
In this exercise, using an air thermal conductivity of 0.0262 W/(m路K) and a Nusselt number of 9.982 gives us a heat transfer coefficient of 5.228 W/(m虏路K). The heat transfer coefficient is essential for calculating the rate of heat transfer, critical for thermal management in various systems, from automotive engines to pace-makers. Ultimately, the heat transfer coefficient helps us find the heat transfer rate, ensuring we can design systems that balance thermal energy efficiently.

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Most popular questions from this chapter

Jakob (1949) suggests the following correlation be used for square tubes in a liquid cross-flow situation: $$ \mathrm{Nu}=0.102 \mathrm{Re}^{0.675} \mathrm{Pr}^{1 / 3} $$ Water \((k=0.61 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, \operatorname{Pr}=6)\) at \(50^{\circ} \mathrm{C}\) flows across a \(1-\mathrm{cm}\) square tube with a Reynolds number of 10,000 and surface temperature of \(75^{\circ} \mathrm{C}\). If the tube is \(2 \mathrm{~m}\) long, the rate of heat transfer between the tube and water is (a) \(6.0 \mathrm{~kW}\) (b) \(8.2 \mathrm{~kW}\) (c) \(11.3 \mathrm{~kW}\) (d) \(15.7 \mathrm{~kW}\) (e) \(18.1 \mathrm{~kW}\)

Air \((\operatorname{Pr}=0.7, k=0.026 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) at \(200^{\circ} \mathrm{C}\) flows across 2-cm-diameter tubes whose surface temperature is \(50^{\circ} \mathrm{C}\) with a Reynolds number of 8000 . The Churchill and Bernstein convective heat transfer correlation for the average Nusselt number in this situation is $$ \mathrm{Nu}=0.3+\frac{0.62 \mathrm{Re}^{0.5} \mathrm{Pr}^{0.33}}{\left[1+(0.4 / \mathrm{Pr})^{0.67}\right]^{0.25}} $$ (a) \(8.5 \mathrm{~kW} / \mathrm{m}^{2}\) (b) \(9.7 \mathrm{~kW} / \mathrm{m}^{2}\) (c) \(10.5 \mathrm{~kW} / \mathrm{m}^{2}\) (d) \(12.2 \mathrm{~kW} / \mathrm{m}^{2}\) (e) \(13.9 \mathrm{~kW} / \mathrm{m}^{2}\)

A 5-m-long strip of sheet metal is being transported on a conveyor at a velocity of \(5 \mathrm{~m} / \mathrm{s}\), while the coating on the upper surface is being cured by infrared lamps. The coating on the upper surface of the metal strip has an absorptivity of \(0.6\) and an emissivity of \(0.7\), while the surrounding ambient air temperature is \(25^{\circ} \mathrm{C}\). Radiation heat transfer occurs only on the upper surface, while convection heat transfer occurs on both upper and lower surfaces of the sheet metal. If the infrared lamps supply a constant heat flux of \(5000 \mathrm{~W} / \mathrm{m}^{2}\), determine the surface temperature of the sheet metal. Evaluate the properties of air at \(80^{\circ} \mathrm{C}\).

Air at \(15^{\circ} \mathrm{C}\) and \(1 \mathrm{~atm}\) flows over a \(0.3\)-m-wide plate at \(65^{\circ} \mathrm{C}\) at a velocity of \(3.0 \mathrm{~m} / \mathrm{s}\). Compute the following quantities at \(x=0.3 \mathrm{~m}\) : (a) Hydrodynamic boundary layer thickness, \(\mathrm{m}\) (b) Local friction coefficient (c) Average friction coefficient (d) Total drag force due to friction, \(\mathrm{N}\) (e) Local convection heat transfer coefficient, W/m虏 \(\mathbf{K}\) (f) Average convection heat transfer coefficient, W/m虏 \(\mathrm{K}\) (g) Rate of convective heat transfer, W

Air at \(25^{\circ} \mathrm{C}\) flows over a 5 -cm-diameter, 1.7-m-long smooth pipe with a velocity of \(4 \mathrm{~m} / \mathrm{s}\). A refrigerant at \(-15^{\circ} \mathrm{C}\) flows inside the pipe and the surface temperature of the pipe is essentially the same as the refrigerant temperature inside. The drag force exerted on the pipe by the air is (a) \(0.4 \mathrm{~N}\) (b) \(1.1 \mathrm{~N}\) (c) \(8.5 \mathrm{~N}\) (d) \(13 \mathrm{~N}\) (e) \(18 \mathrm{~N}\) (For air, use \(\nu=1.382 \times 10^{-5} \mathrm{~m}^{2} / \mathrm{s}, \rho=1.269 \mathrm{~kg} / \mathrm{m}^{3}\) )

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