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Anna [14]
3 years ago
10

In fully developed laminar flow in a circular pipe the velocity at R/2 (mid-way between the wall surface and the centerline) is

measured to be 13 m/s. Determinethe velocity at the center of the pipe.

Engineering
2 answers:
Butoxors [25]3 years ago
6 0

Answer:

u_{max} = 17.334\,\frac{m}{s}

Explanation:

Let consider that velocity profile inside the circular pipe is:

u(r) = 2\cdot U_{avg} \cdot \left(1 - \frac{r^{2}}{R^{2}}  \right)

The average speed at r = \frac{1}{2} \cdot R is:

U_{avg} = \frac{13\,\frac{m}{s} }{2\cdot \left(1-\frac{1}{4}  \right)}

U_{avg} = 8.667\,\frac{m}{s}

The velocity at the center of the pipe is:

u_{max} = 2\cdot U_{avg}

u_{max} = 17.334\,\frac{m}{s}

kumpel [21]3 years ago
6 0

Answer:

17.3m/s

Explanation:

Detailed explanation and calculation is shown in the image

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Summary

Students learn about the variety of materials used by engineers in the design and construction of modern bridges. They also find out about the material properties important to bridge construction and consider the advantages and disadvantages of steel and concrete as common bridge-building materials to handle compressive and tensile forces.

This engineering curriculum aligns to Next Generation Science Standards (NGSS).

Engineering Connection

When designing structures such as bridges, engineers carefully choose the materials by anticipating the forces the materials (the structural components) are expected to experience during their lifetimes. Usually, ductile materials such as steel, aluminum and other metals are used for components that experience tensile loads. Brittle materials such as concrete, ceramics and glass are used for components that experience compressive loads.

Learning Objectives

After this lesson, students should be able to:

List several common materials used the design and construction of structures.

Describe several factors that engineers consider when selecting materials for the design of a bridge.

Explain the advantages and disadvantages of common materials used in engineering structures (steel and concrete).

Educational Standards

NGSS: Next Generation Science Standards - Science

Common Core State Standards - Math

International Technology and Engineering Educators Association - Technology

State Standards

Suggest an alignment not listed above

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Worksheets and Attachments

Strength of Materials Worksheet (doc)

Strength of Materials Worksheet (pdf)

Strength of Materials Worksheet Answers (doc)

Strength of Materials Worksheet Answers (pdf)

Strength of Materials Math Worksheet (doc)

Strength of Materials Math Worksheet (pdf)

Strength of Materials Math Worksheet Answers (doc)

Strength of Materials Math Worksheet Answers (pdf)

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MIDDLE SCHOOL Activity

Breaking the Mold

Explanation:

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8 0
2 years ago
If the maximum allowable shear stress is 70 MPa, find the shaft diameter needed to transmit 40 kW when the shaft speed is 250 rp
victus00 [196]

Answer:

The diameter is 50mm

Explanation:

The answer is in two stages. At first the torque (or twisting moment) acting on the shaft and needed to transmit the power needs to be calculated. Then the diameter of the shaft can be obtained using another equation that involves the torque obtained above.

T=(P×60)/(2×pi×N)

T is the Torque

P is the the power to be transmitted by the shaft; 40kW or 40×10³W

pi=3.142

N is the speed of the shaft; 250rpm

T=(40×10³×60)/(2×3.142×250)

T=1527.689Nm

Diameter of a shaft can be obtained from the formula

T=(pi × SS ×d³)/16

Where

SS is the allowable shear stress; 70MPa or 70×10⁶Pa

d is the diameter of the shaft

Making d the subject of the formula

d= cubroot[(T×16)/(pi×SS)]

d=cubroot[(1527.689×16)/(3.142×70×10⁶)]

d=0.04808m or 48.1mm approx 50mm

7 0
4 years ago
6.
zhenek [66]
The answer is b i did the same thing and i got it right
5 0
3 years ago
Read 2 more answers
What classification is an E7018?<br><br> will give brainliest
Hunter-Best [27]

Answer: E7018 is a low-hydrogen iron powder type electrode that produces high quality x-ray welds. It can be used in all positions on AC or DC reverse polarity welding current.

Explanation: Hope this helps

3 0
3 years ago
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