1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
cupoosta [38]
3 years ago
15

Consider the freeway in Problem 1. At one point along this freeway there is a 4% upgrade with a directional hourly traffic volum

e of 5,435 vehicles. If all other conditions are as described in Problem 1, how long can this grade be without the freeway LOS dropping to F?
Engineering
2 answers:
ryzh [129]3 years ago
6 0

Answer:

The Question is incomplete, the complete question is as follows:

<em>Consider the freeway in Problem 1. At one point along this freeway there is a 4% upgrade with a directional hourly traffic volume of 5,435 vehicles. If all other conditions are as described in Problem 1, how long can this grade be without the freeway LOS dropping to F? </em>

A six-lane rural freeway (three lanes in each direction) has regular weekday users and currently operates at maximum LOS C conditions. The base free-flow speed is 65 mi/h, lanes are 11 ft wide, the right-side shoulder is 4 ft wide, and the interchange density is 0.25 per mile. The highway is one rolling terrain with 10% large trucks and buses (no recreational vehicles), and the peak-hour factor is 0.90. Determine the hourly volume for these conditions

Explanation:

<em>Make the assumption Base continuous flow velocity (BFFS)= 65 mph. </em>

Pitch width= 11 ft.

Decrease in lane width pace,fLW= 1.9 mph.

Complete Lateral clearance= 4 ft. Lateral clearance speed reduction, fLC= 0.8 mph.

Complete Width of the Ramp= 0.25 mile.

Velocity reduction proportional to the ramp height, f ID= 0 mph.

Assume lane number to be = 3.

Reduction in speed corresponding to no. of lanes, fN = 3 mph

Free Flow Speed (FFS) = BFFS – fLW – fLC – fN – fID = 65 – 1.9 – 0.8 – 3 – 0 = 59.3 mph

Peak Flow, V veh/hr

Peak-hour factor = 0.90

Trucks = 10%

Rolling Terrain

fHV = 1/ (1 + 0.10 (2.5-1)) = 1/1.15 = 0.8696

fP = 1.0

Peak Flow Rate, Vp = V / (PHV*n*fHV*fP) = V/ (0.90*3*0.8696*1.0) = 0.426V veh/hr/ln

Average speed of vehicles, S = FFS = 59.3 mph

Level of service C

Density of LOS C lies between 18 - 25 veh/mi/ln

Maximum density = 25 veh/mi/ln

Density = V​​​​​​p /S = 25

0.426V = 25 * 59.3

V = 3480 veh/hr

b) V = 5435 veh/hr

LOS dropping to F

Max density = 45 veh/mi/ln

Density = Vp/S = 45

V​​​​​​p = 45 * 59.3 = 2668.5 veh/hr/ln

V/(PHF * n * f​​​​​​HV * f​​​​​​P​​​) = 2668.5

f​​​​​​HV = 5435/(0.9*3*2668.5*1.0) = 0.754

1/(1+0.10 (E​​​​​​T -1)) = 0.754

E​​​​​​T = 4.26 ~ 3.5

<em>For 4% upgrade and 10% trucks with E​​​​​​T = 3.5, length of the grade is Greater than 1.0 miles</em>

yazid2 years ago
0 0

thank you a lit brother

You might be interested in
Water vapor at 10 MPa, 600°C enters a turbine operating at steady state with a volumetric flow rate of 0.36 m3/s and exits at 0.
elena-s [515]

Answer:

Hook's law holds good up to. A elastic limit. B. plastic limit. C.yield point. D.Breaking point

8 0
3 years ago
Using OOP, write a C++ program that will read in a file of names. The file is called Names.txt and should be located in the curr
Tanzania [10]
Not really sure sorry for not being able to help
5 0
3 years ago
Read 2 more answers
Which investigative process is most helpful for learning about past societies?
tatuchka [14]

Answer: think it A

Explanation: makes

6 0
3 years ago
Q.17) A 50-acre catchment containing cropland is converted ot a Qatar mail
Ghella [55]

Answer:

Option D

Explanation:

A post development hydrograph will have lower concentration time and lower infiltration losses and hence sooner peak and higher peak and more runoff or higher area under graph. Therefore, all the answers are correct hence option D

3 0
2 years ago
The link acts as part of the elevator control for a small airplane. If the attached aluminum tube has an inner diameter of 25 mm
aksik [14]

Answer:

Tmax=14.5MPa

Tmin=10.3MPa

Explanation:

T = 600 * 0.15 = 90N.m

T_max =\frac{T_c}{j}  = \frac{x}{y}  = \frac{90 \times 0.0175}{\frac{\pi}{2} \times (0.0175^4-0.0125^4)}

=14.5MPa

T_{min} =\frac{T_c}{j}  = \frac{x}{y}  = \frac{90 \times 0.0125}{\frac{\pi}{2} \times (0.0175^4-0.0125^4)}

=10.3MPa

7 0
3 years ago
Other questions:
  • 1. Looking at the case study provided under the Companion Material section, what is the main problem that is addressed in this c
    13·1 answer
  • 10. Which of these requires a wheel alignment after replacement?
    11·2 answers
  • Prompt the user to enter five numbers, being five people's weights. Store the numbers in an array of doubles. Output the array's
    11·1 answer
  • (d) Suppose two students are memorizing a list according to the same model dL dt = 0.5(1 − L) where L represents the fraction of
    6·1 answer
  • The cars of a roller-coaster ride have a speed of 30 km / h as they pass over the top of the circular track. Neglect any frictio
    9·1 answer
  • Consider the following signal:
    8·1 answer
  • The value 100 MW is equivalent to (a) 100×10^6 w (b) 100 x 10^-6 w (c) 100 x 10^-3 w (d) 100 x 10^3 w
    14·1 answer
  • Steam in a heating system flows through tubes whose outer diameter is 5 cm and whose walls are maintained at a temperature of 19
    13·1 answer
  • Suppose to build RSA crypto system you picked primes "p" and "q" as 3 and 7 and "e" as 5 what are the public and private keys? W
    11·1 answer
  • What is a problem that technology can help solve that problem?
    12·2 answers
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!