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hjlf
2 years ago
7

The correct statement about the relation between the shear stress (force) at the surface for laminar and turbulent flow over a f

lat plate is. a.The laminar shear stress is greater because the fluid viscosity is not important in turbulent flow. b.The turbulent shear stress is greater because the mixing dissipates more energy, producing a greater stress at the wall. c.The laminar and turbulent shear stresses are equal at the same position (x) along the plate, but since turbulence occurs farther down the plate it produces a higher shear stress. d.The laminar shear stress is less because the boundary is thinner in the laminar regime.
Engineering
1 answer:
-Dominant- [34]2 years ago
7 0

Answer:

b.The turbulent shear stress is greater because the mixing dissipates more energy, producing a greater stress at the wall.

Explanation:

For a fluid, shear stress is related to the velocity gradient (derivative), not the absolute velocity.

Following the Poiseuille model, flow profile will appear in a channel. In that case, the flow velocity is maximum in the center, but the gradient will be zero in the center. At the wall, the flow will be zero, but the gradient will have a maximum value, therefore maximum shear stress.

Since stress depends highly on the velocity gradient near to the wall of the plate, and turbulence occurs in very high velocity flows, with high Reynold's number, then the turbulent shear stress is greater because the mixing dissipates more energy, producing a greater stress at the wall.

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Initialize the tuple team_names with the strings 'Rockets', 'Raptors', 'Warriors', and 'Celtics' (The top-4 2018 NBA teams at th
Drupady [299]

Answer:

#Initialise a tuple

team_names = ('Rockets','Raptors','Warriors','Celtics')

print(team_names[0])

print(team_names[1])

print(team_names[2])

print(team_names[3])

Explanation:

The Python code illustrates or printed out the tuple team names at the end of a season.

The code displayed is a function that will display these teams as an output from the program.

4 0
2 years ago
A horizontal, cylindrical, tank, with hemispherical ends, is used to store liquid chlorine at 10 bar. The vessel is 4 m internal
sp2606 [1]

Answer: 0.021818m =2.18x10^-²m

Explanation: Using Laplace principle

Design pressure is =12 bar= 1200000N/m²

T = Wall thickness

Design stress=110Mn/m²= 110,000,000N/m²

Radius= diameter/2 =4/2=2m

Design stress=Hoop stress =Pr/t where p=internal pressure or internal pressure, r=radius and t= wall thickness.

As Laplace equation stated.

110,000,000= 1,2000,00 x 2/t

t= 2,400,000/110,000,000.

t= 0.021818m

t=2.18x10^-2m.

3 0
2 years ago
The extruder head in a fused- deposition modeling setup has a diameter of 1.25 mm (0.05 in) and produces layers that are 0.25mm
Angelina_Jolie [31]

Answer:

The time taken will be "1 hour 51 min". The further explanation is given below.

Explanation:

The given values are:

Number of required layers:

= \frac{38}{0.25}

= 152 \ layers

Diameter (d):

= 1.25 mm

Velocity (v):

= 40 mm/s

Now,

The area of one layer will be:

= 38\times 38 \ mm^2

= 1444 \ mm^2

The area covered every \second will be:

= d\times v

= 1.25\times 40

= 50 \ mm^2

The time required to deposit one layer will be:

= \frac{1444}{50}

= 28.88 \ sec

The time required for one layer will be:

= 15 \ sec

∴ Total times required for one layer will be:

= 15+28.88

= 43.88 \ sec

So,

Number of layers = 152

Therefore,

Total time will be:

= 152\times 43.88

= 6669.76 \ sec

= 1 \ hour \ 51 \ min

6 0
2 years ago
To unload a bound stack of plywood from a truck, the driver first tilts the bed of the truck and then accelerates from rest. Kno
azamat

Answer:

a) The truck must have an acceleration of at least 3.92 * cos(θ)^2 for the stack to start sliding.

b) a = 2.94 * cos(θ)^2 + 0.32 * L * cos(θ)

Explanation:

The stack of plywood has a certain mass. The weight will depend on that mass.

w = m * g

There will be a normal reaction between the stack and the bed of the truck, this will be:

nr = -m * g * cos(θ)

Being θ the tilt angle of the bed.

The static friction force will be:

ffs = - m * g * cos(θ) * fJK

The dynamic friction force will be:

ffd = - m * g * cos(θ) * fik

These forces would produce accelerations

affs = -g * cos(θ) * fJK

affd = -g * cos(θ) * fik

affs = -9.81 * cos(θ) * 0.4 = -3.92 * cos(θ)

affd = -9.81 * cos(θ) * 0.3 = -2.94 * cos(θ)

These accelerations oppose to movement and must be overcome by another acceleration to move the stack.

The acceleration of the truck is horizontal, the horizontal component of these friction forces is:

affs = -3.92 * cos(θ)^2

affd = -2.94 * cos(θ)^2

The truck must have an acceleration of at least 3.92 * cos(θ)^2 for the stack to start sliding.

Assuming the bed has a lenght L.

The horizontal movement will be over a distance cos(θ) * L because L is tilted.

Movement under constant acceleration:

X(t) = X0 + V0 * t + 1/2 * a * t^2

In this case X0 = 0, V0 = 0, and a will be the sum of the friction force minus the acceleration of the truck.

If we set the frame of reference with the origin on the initial position of the stack and the positive X axis pointing backwards, the acceleration of the truck will now be negative and the dynamic friction acceleration positive.

L * cos(θ) = 1/2 * (2.94 * cos(θ)^2 - a) * 0.4^2

2.94 * cos(θ)^2 - a = 2 * 0.16 * L * cos(θ)

a = 2.94 * cos(θ)^2 + 0.32 * L * cos(θ)

6 0
2 years ago
Determine the deflection at the center of the beam. Express your answer in terms of some or all of the variables LLL, EEE, III,
Rom4ik [11]

Answer:

See explanations for step by step procedures to get answer.

Explanation:

Given that;

Determine the deflection at the center of the beam. Express your answer in terms of some or all of the variables LLL, EEE, III, and M0M0M_0. Enter positive value if the deflection is upward and negative value if the deflection is downward.

4 0
2 years ago
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