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Annette [7]
3 years ago
9

In the hydrodynamic entrance region of a pipe with a steady flow of an incompressible liquid

Engineering
2 answers:
Yakvenalex [24]3 years ago
6 0

Answer:

D. The maximum velocity decreases with distance from the entrance.

Explanation:

This is because over time, the pressure with with the incompressible liquid enters decreases with distance from the entrance

Contact [7]3 years ago
6 0

Answer:

C. The maximum velocity increases with distance from the entrance

Explanation:

As the fluid particles moves into the pipe, the layer of fluid in contact with the surface of the pipe come to a complete stop. This layer also causes the fluid

particles in the adjacent layers to gradually slow down as a result of friction between fluid molecules, leaving the fluid at the center of the pipe with the maximum velocity.

Since the fluid is incompressible, to make up for this velocity reduction, the velocity of the fluid at the mid-

section of the pipe has to increase to keep the mass flow rate through the

pipe constant. As a result, a velocity gradient develops along the pipe and the maximum velocity which is at the center of the pipe increases with distance from entrance.

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(a) 3.455

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3 years ago
7.4 A pretimed four-timing-stage signal has critical lane group flow rates for the first three timing stages of 200, 187, and 21
Irina18 [472]

Answer:

16 seconds

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Let's first find the fourth stage critical lane group ratio \frac{v}{s} , using the formula:

C = \frac{1.5L +5}{1 - ( \frac{200}{1800} + \frac{187}{1800} + \frac{210}{1800}) + ( \frac{v}{s})}

60 = \frac{1.5*16 + 5}{1 - ( \frac{200}{1800} + \frac{187}{1800} + \frac{210}{1800}) + ( \frac{v}{s})}

60 = \frac{24+5}{1 - (0.332 + ( \frac{v}{s}))}

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Let's also calculate the volume capacity ratio X,

X = (\frac{200}{1800} + \frac{187}{1800} + \frac{210}{1800} + 0.185)(\frac{60}{60-16}

X = 0.704

For the the estimated effective green time of the fourth timing stage, we have:

g_4 = (\frac{v}{s}) (\frac{C}{X})

Substituting figures in the equation, we now have:

g_4 = (0.185) (\frac{60}{0.704})

g_4 = 15.78 seconds

15.78 ≈ 16 seconds

The estimated effective green time of the fourth timing stage is 16 seconds

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Answer:

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3 0
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