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dezoksy [38]
3 years ago
6

A section of highway has a free-flow speed of 55 mph and a capacity of 3300 veh/hr. In a given hour, 2100 vehicles were counted

at a specified point along this highway section. If the linear speed density relationship applies, what would you estimate the space-mean speed of the vehicles to be?
Engineering
1 answer:
alexira [117]3 years ago
5 0

Answer:

Space mean speed = 44 mi/h

Explanation:

Using Greenshield's linear model

q = Uf ( D - D^{2}/Dj )

qcap = capacity flow that gives Dcap

Dcap = Dj/2

qcap = Uf. Dj/4

Where

U = space mean speed

Uf =  free flow speed

D = density

Dj = jam density

now,

Dj = 4 × 3300/55

    = 240v/h

q = Dj ( U - U^{2}/Uf)

2100 = 240 ( U - U^{2}/55)

Solve for U

U = 44m/h

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Water flows steadily through the pipe as shown below, such that the pressure at section (1) and at section (2) are 300 kPa and 1
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Answer:

The velocity at section is approximately 42.2 m/s

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50 \, m + \dfrac{300 \ kPa}{997 \, kg/m^3 \times 9.8 \, m/s^2} + \dfrac{(20 \, m/s)^2}{2 \times 9.8 \, m/s^2} = \dfrac{100 \ kPa}{997 \, kg/m^3 \times 9.8 \, m/s^2} + \dfrac{v_2^2}{2 \times 9.8 \, m/s^2}50 m + 30.704358 m + 20.4081633 m = 10.234786 m + \dfrac{v_2^2}{2 \times 9.8 \, m/s^2}

50 m + 30.704358 m + 20.4081633 m - 10.234786 m = \dfrac{v_2^2}{2 \times 9.8 \, m/s^2}

90.8777353 m = \dfrac{v_2^2}{2 \times 9.8 \, m/s^2}

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v₂ = √(1,781.20361 m²/s²) ≈ 42.204308 m/s

The velocity at section (2), v₂ ≈ 42.2 m/s

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