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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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Answer:

45.3 MN

Explanation:

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75² * 2 = r²

r² = 11250

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E = 0.69

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F = Y.π.r².[1 + (2μr/3h)]

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F = 35.3 * [1 + 0.2826]

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3 years ago
What are three types of land reform ​
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Abolition of intermediaries (rent collectors under the pre-Independence land revenue system); Tenancy regulation (to improve the contractual terms including the security of tenure); A ceiling on landholdings (to redistributing surplus land to the landless);

5 0
2 years ago
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Steam at 75 kPa and 8 percent quality is contained in a spring-loaded piston–cylinder device, as shown in Figure, with an initia
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The heat transferred to and the work produced by the steam during this process  is 13781.618 kJ/kg

<h3>​How to calcultae the heat?</h3>

The Net Change in Enthalpy will be:

= m ( h2 - h1 ) = 11.216 ( 1755.405 - 566.78 ) = 13331.618 kJ/kg

Work Done (Area Under PV curve) = 1/2 x (P1 + P2) x ( V1 - V2)

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From the First Law of Thermodynamics, Q = U + W

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6 0
1 year ago
Water at atmospheric pressure boils on the surface of a large horizontal copper tube. The heat flux is 90% of the critical value
masya89 [10]

Answer:

The tube surface temperature immediately after installation is 120.4°C and after prolonged service is 110.8°C

Explanation:

The properties of water at 100°C and 1 atm are:

pL = 957.9 kg/m³

pV = 0.596 kg/m³

ΔHL = 2257 kJ/kg

CpL = 4.217 kJ/kg K

uL = 279x10⁻⁶Ns/m²

KL = 0.68 W/m K

σ = 58.9x10³N/m

When the water boils on the surface its heat flux is:

q=0.149h_{fg} \rho _{v} (\frac{\sigma (\rho _{L}-\rho _{v})}{\rho _{v}^{2} }  )^{1/4} =0.149*2257*0.596*(\frac{58.9x10^{-3}*(957.9-0.596) }{0.596^{2} } )^{1/4} =18703.42W/m^{2}

For copper-water, the properties are:

Cfg = 0.0128

The heat flux is:

qn = 0.9 * 18703.42 = 16833.078 W/m²

q_{n} =uK(\frac{g(\rho_{L}-\rho _{v})     }{\sigma })^{1/2} (\frac{c_{pL}*deltaT }{c_{fg}h_{fg}Pr  } \\16833.078=279x10^{-6} *2257x10^{3} (\frac{9.8*(957.9-0.596)}{0.596} )^{1/2} *(\frac{4.127x10^{3}*delta-T }{0.0128*2257x10^{3}*1.76 } )^{3} \\delta-T=20.4

The tube surface temperature immediately after installation is:

Tinst = 100 + 20.4 = 120.4°C

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ΔT = 10.8°C

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