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 = (ρ * V * D) / μ where ρ is the air density, V is the air velocity, D is the sphere's diameter, and μ is the air dynamic viscosity. For air at 20°C (1 atm), we have ρ = 1.204 kg/m³ and μ = 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°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°C, k = 0.0262 W/(m*K). Calculating the heat transfer coefficient: h = 9.982 * 0.0262 / 0.05 = 5.228 W/(m²*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 * ΔT where A is the surface area of the sphere and ΔT is the temperature difference between the surface and the air. The surface area of a sphere can be calculated using the formula: A = 4 * π * (D/2)^2 Substituting the given values: A = 4 * π * (0.05/2)^2 = 0.007853 m² ΔT = 80°C - 20°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.

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 Vaia!

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

Oil at \(60^{\circ} \mathrm{C}\) flows at a velocity of \(20 \mathrm{~cm} / \mathrm{s}\) over a \(5.0\)-m-long and \(1.0-\mathrm{m}\)-wide flat plate maintained at a constant temperature of \(20^{\circ} \mathrm{C}\). Determine the rate of heat transfer from the oil to the plate if the average oil properties are: \(\rho=880 \mathrm{~kg} / \mathrm{m}^{3}, \mu=0.005 \mathrm{~kg} / \mathrm{m} \cdot \mathrm{s}, k=0.15 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\), and \(c_{p}=2.0 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}\).

Engine oil at \(105^{\circ} \mathrm{C}\) flows over the surface of a flat plate whose temperature is \(15^{\circ} \mathrm{C}\) with a velocity of \(1.5 \mathrm{~m} / \mathrm{s}\). The local drag force per unit surface area \(0.8 \mathrm{~m}\) from the leading edge of the plate is (a) \(21.8 \mathrm{~N} / \mathrm{m}^{2}\) (b) \(14.3 \mathrm{~N} / \mathrm{m}^{2}\) (c) \(10.9 \mathrm{~N} / \mathrm{m}^{2}\) (d) \(8.5 \mathrm{~N} / \mathrm{m}^{2}\) (e) \(5.5 \mathrm{~N} / \mathrm{m}^{2}\) (For oil, use \(\nu=8.565 \times 10^{-5} \mathrm{~m}^{2} / \mathrm{s}, \rho=864 \mathrm{~kg} / \mathrm{m}^{3}\) )

Solar radiation is incident on the glass cover of a solar collector at a rate of \(700 \mathrm{~W} / \mathrm{m}^{2}\). The glass transmits 88 percent of the incident radiation and has an emissivity of \(0.90\). The entire hot water needs of a family in summer can be met by two collectors \(1.2 \mathrm{~m}\) high and \(1 \mathrm{~m}\) wide. The two collectors are attached to each other on one side so that they appear like a single collector \(1.2 \mathrm{~m} \times 2 \mathrm{~m}\) in size. The temperature of the glass cover is measured to be \(35^{\circ} \mathrm{C}\) on a day when the surrounding air temperature is \(25^{\circ} \mathrm{C}\) and the wind is blowing at \(30 \mathrm{~km} / \mathrm{h}\). The effective sky temperature for radiation exchange between the glass cover and the open sky is \(-40^{\circ} \mathrm{C}\). Water enters the tubes attached to the absorber plate at a rate of \(1 \mathrm{~kg} / \mathrm{min}\). Assuming the back surface of the absorber plate to be heavily insulated and the only heat loss to occur through the glass cover, determine \((a)\) the total rate of heat loss from the collector, \((b)\) the collector efficiency, which is the ratio of the amount of heat transferred to the water to the solar energy incident on the collector, and \((c)\) the temperature rise of water as it flows through the collector.

A \(20 \mathrm{~mm} \times 20 \mathrm{~mm}\) silicon chip is mounted such that the edges are flush in a substrate. The substrate provides an unheated starting length of \(20 \mathrm{~mm}\) that acts as turbulator. Airflow at \(25^{\circ} \mathrm{C}(1 \mathrm{~atm})\) with a velocity of \(25 \mathrm{~m} / \mathrm{s}\) is used to cool the upper surface of the chip. If the maximum surface temperature of the chip cannot exceed \(75^{\circ} \mathrm{C}\), determine the maximum allowable power dissipation on the chip surface.

Consider a person who is trying to keep cool on a hot summer day by turning a fan on and exposing his body to air flow. The air temperature is \(32^{\circ} \mathrm{C}\), and the fan is blowing air at a velocity of \(5 \mathrm{~m} / \mathrm{s}\). The surrounding surfaces are at \(40^{\circ} \mathrm{C}\), and the emissivity of the person can be taken to be \(0.9\). If the person is doing light work and generating sensible heat at a rate of \(90 \mathrm{~W}\), determine the average temperature of the outer surface (skin or clothing) of the person. The average human body can be treated as a 30 -cm-diameter cylinder with an exposed surface area of \(1.7 \mathrm{~m}^{2}\). Evaluate the air properties at film temperature of \(35^{\circ} \mathrm{C}\) and \(1 \mathrm{~atm}\).

See all solutions

Recommended explanations on Physics Textbooks

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.

Sign-up for free