As a means of preventing ice formation on the wings of a small, private aircraft, it is proposed that electric resistance heating elements be installed within the wings. To determine representative power requirements, consider nominal flight conditions for which the plane moves at 100 m/s in air that is at a temperature of -23 degree C. If the characteristic length of the airfoil is L = 2 m and wind tunnel measurements indicate an average friction coefficient of of C_f = 0.0025 for the nominal conditions, what is the average heat flux needed to maintain a surface temperature of T_s = 5 degree C?

Answers

Answer 1

Answer:

Answer:

Average heat flux=3729.82 W/ $m^(2)$

Explanation:

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The dam cross section is an equilateral triangle, with a side length, L, of 50 m. Its width into the paper, b, is 100 m. The dam material has a specific gravity, SG, of 3.1. You may assume that the dam is loosely attached to the ground at its base, though there is significant friction to keep it from sliding.Is the weight of the dam sufficient to prevent it from tipping around its lower right corner?
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It is proposed to deposit a 5 μm thick nickel coating uniformly on all surfaces of a ceramic strip measuring 15 cm x 5 cm x 2 cm by employing a vapor-phase deposition (evaporation-condensation) technique. The vapor pressure-temperature relationship for liquid Ni is of the following form: ln p (atm) = -(51,590/T) – 2.01 ln T + 32.40.

The normal melting point and boiling point of nickel are 1453°C and 2730°C, respectively, and the density and atomic weights of Ni are 8.91 g.cm^-3 and 58.71 atomic mass units respectively. Calculate the energy in joules needed to evaporate the required quantity of nickel.

Answers

Answer:

Check the explanation

Explanation:

Let’s take for instance, when an object with a mass of 10 kg (m = 10 kg) is moving at a 5 meters per second (v = 5 m/s) velocity rate, the kinetic energy is equal to 125 Joules

Kindly check the attached images below to get the step by step explanation to the question above.

You are analyzing an open-return wind tunnel that intakes air at 20 m/s and 320K. When the flow exits the wind tunnel it is moving at a speed of 250 m/s. What is the temperature of the flow exiting that wind tunnel

Answers

The solution is in the attachment

Answer:

please find attached.

Explanation:

Define a homogeneous material. O Material has temperature dependent refractive index.O Material exhibits both elastic and plastic behavior. O Material exhibits little or no yielding before failure. O Material has uniform properties throughout.

Answers

Answer:

Option D

Material has uniform properties throughout.

Explanation:

A homogeneous material is a material that exhibits uniform properties throughout and these properties cannot be separated. These materials can be metals, ceramics or alloys that exhibit similar properties throughout. The similar properties may include evaporation point, density and other properties.

For steels, we can assume the endurance limit measured by rotating beam tests is the 50% of the ultimate tensile strength (UTS) as long as the UTS is equal to or less than 200 ksi (1400 MPa).True/False

Answers

Answer:

True

Explanation:

For steels, we can assume the endurance limit measured by rotating beam tests is the 50% of the ultimate tensile strength (UTS) as long as the UTS is equal to or less than 200 kpsi (1400 MPa).

Also, It is a simplistic rule of thumb that, for steels having a UTS less than 160 kpsi, the endurance limit for the material will be approximately 45 to 50% of the UTS.

An equal-tangent sag vertical curve is designed for 45 mi/h. The low point is 237 ft from the PVC at station 112 37 and the final offset at the PVT is 19.355 ft. If the PVC is at station 110 00, what is the elevation difference between the PVT and a point on the curve at station 111 00

Answers

Answer:

18.722 ft

Explanation:

The elevation difference between the PVT and a point on the curve at station

111 + 00

attached below is a detailed solution to the problem

Δelevation ( elevation difference )

= Yt - Y

= 19.355 - 0.632 = 18.722 ft

A solid cylindrical workpiece made of 304 stainless steel is 150 mm in diameter and 100 mm is high. It is reduced in height by 50%, at room temperature, by open-die forging with flat dies. Assume that the coefficient of friction is 0.2. Calculate the forging force at the end of the stroke.

Answers

The answer is "45.3 NM".

There at end of the movement, the forging force is given by

$\to F = Y * \pi * r^2 * [1 + ((2 \mu r)/(3h))]$

h is the final height.

$\to h = (100)/(2)= 50 \ mm$

The ultimate radius is determined by following a volume constancy law, which states that volumes before deformation measured amount following distortion.

$\to \pi * 75^2 * 2 * 100 = \pi * r^2 * 2 * 50\n\n\to 75^2 * 2 = r^2\n\n\to r^2 = 11250\n\n\to r = √(11250)\n\n\to r = 106 \ mm\n\n\to E = \In((100)/(50))\n\n\to E = 0.69\n\n$

You may deduce from the graph flow that $Y = 1000\ MPa$ , thus we use the formula.

$= 1000 * 3.14 * 0.106^2 * [1 + (( 2 * 0.2 * 0.106)/(3 * 0.05))]\n\n= 1000 * 3.14 * 0.011236 * [1 + (( 0.0424)/(0.15))]\n\n= 35.3 * 1.2826\n\n = 45.3 \ MN\n\n\n$