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BartSMP [9]
3 years ago
5

A six-lane freeway (three lanes in each direction) currently operates at maximum LOS C conditions. The lanes are 11 ft wide, the

right-side shoulder is 4 ft wide, and there are two ramps within three miles upstream of the segment midpoint and one ramp within three miles downstream of the segment midpoint. The highway is on rolling terrain with 10% heavy vehicles, and the peak hour factor is 0.90. Determine the hourly volume for these conditions.
Engineering
1 answer:
Gelneren [198K]3 years ago
7 0

Answer:

4.071 veh/h.

Explanation:

Step one: calculate the estimated free flow speed by using the formula below;

= 75.4 - F(L) - F(C) - 3.22T^ 0.84.

The value of F(L) = 1.9 m/h and F(C) = 0.8 m/h.

Hence,

free flow speed = 75.4 - 1.9 - 0.8 - 3.22(3/6)^0.84.

free flow speed= 75.4 - 1.9 - 0.8 - 1.799

free flow speed= 70.901 mi/h = 70 mi/h.

Step two: determine the adjustment factor by using the formula below;

The adjustment factor = 1/ [1 + Pm (Em - 1) + Pn (En - 1)].

The adjustment factor = 1/[ 1+ 10/100 ( 2.5 - 1) + 0(2.0 - 1] = 0.869.

Step three;

Calculate the hourly volume;

The value corresponding to the LOS C conditions and free flow speed at 70 mi/h is 1735.

Therefore,

hourly volume, V = 1735 × 0.9 × 0.869 × 3.

Hourly volume, V = 4.071 veh/h.

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Given:

Assuming the transition to turbulence for flow over a flat plate happens at a Reynolds number of 5x105, determine the following for air at 300 K and engine oil at 380 K. Assume the free stream velocity is 3 m/s.

To Find:

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Calculation:

The transition from the lamina to turbulent begins when the critical Reynolds

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(a).  \;\text{Rex}_{cr}=5 \times 10^5\\\\\frac{\rho\;vx}{\mu}=5 \times 10^5\\\text{density of of air at}\;300K=1.16  \frac{kg}{m\cdot s}\\\text{viscosity of of air at}\;300K=1.846 \times 10^{-5} \frac{kg}{m\cdot s} \\v=3m/s\\\Rightarrow x=\frac{5\times 10^5 \times 1.846 \times 10^{-5} }{1.16 \times 3} =2.652 \;m \;\text{for air}\\(\text{similarly for engine oil at 380 K for given}\; \rho \;\text{and} \;\mu)\\

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2 years ago
Air enters the compressor of an air-standard Brayton cycle with a volumetric flow rate of 60 m3/s at 0.8 bar, 280 K. The compres
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Considering this:

h_{i} =T_{i}C_{pair}=T_{i}1.005\frac{KJ}{Kg K}

\mu_{comp}=\frac{h_{2S}-h_{1}}{h_{2}-h_{1}}

\mu_{comp}=\frac{h_{3}-h_{4}}{h_{3}-h_{4S}}

G_{m} =\frac{PMG_{v}}{TR} =59.73\frac{Kg}{s}

Now we can calculate the enthalpy of each work point:

h₁=281.4KJ/Kg

h₂=695.41KJ/Kg

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The rate of heat:

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2 years ago
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According to the question of the pulsating brake pedal, both A and B are correct.

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Answer:

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8 0
3 years ago
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