answer.
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]
2 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]2 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>

Guest1 year ago
0 0

thank you a lit brother

You might be interested in
Explain why failure of this garden hose occurred near its end and why the tear occurred along its length. Use numerical values t
alukav5142 [94]

Answer:

  • hoop stress
  • longitudinal stress
  • material used

all this could led to the failure of the garden hose and the tear along the length

Explanation:

For the flow of water to occur in any equipment, water has to flow from a high pressure to a low pressure. considering the pipe, water is flowing at a constant pressure of 30 psi inside the pipe which is assumed to be higher than the allowable operating pressure of the pipe. but the greatest change in pressure will occur at the end of the hose because at that point the water is trying to leave the hose into the atmosphere, therefore the great change in pressure along the length of the hose closest to the end of the hose will cause a tear there. also the other factors that might lead to the failure of the garden hose includes :

hoop stress ( which acts along the circumference of the pipe):

αh = \frac{PD}{2T}     EQUATION 1

and Longitudinal stress ( acting along the length of the pipe )

αl = \frac{PD}{4T}       EQUATION 2

where p = water pressure inside the hose

          d = diameter of hose, T = thickness of hose

we can as well attribute the failure of the hose to the material used in making the hose .

assume for a thin cylindrical pipe material used to be

\frac{D}{T} ≥  20

insert this value into equation 1

αh = \frac{20 *30}{2}  = 60/2 = 30 psi

the allowable hoop stress was developed by the material which could have also led to the failure of the garden hose

8 0
2 years ago
Consider a normal shock wave in air. The upstream conditions are given by M1=3, p1 = 1 atm, and r1 = 1.23 kg/m3. Calculate the d
mart [117]

Answer and Explanation:

The answer is attached below

7 0
2 years ago
If you have power steering and you are able to __________, you should have your vehicle checked out by a qualified technician.
svp [43]

Answer: drive

Explanation:

The best word that would fit this sentence is drive. A vehicle owner should know how to drive, and they can get their vehicle checked by a qualified technician. The best word that would fit this sentence is drive. If you have power steering and you are able to <u>drive</u>, you should have your vehicle checked by a qualified technician.

8 0
2 years ago
Water is flowing in a metal pipe. The pipe OD (outside diameter) is 61 cm. The pipe length is 120 m. The pipe wall thickness is
Yuki888 [10]

Answer:

1113kN

Explanation:

The ouside diameter OD of the pipe is 61cm and the thickness T is 0.9cm, so the inside diameter ID will be:

Inside Diameter = Outside Diameter - Thickness

Inside Diameter = 61cm - 0.9cm = 60.1cm

Converting this diameter to meters, we have:

60.1cm*\frac{1m}{100cm}=0.601m

This inside diameter is useful to calculate the volume V of water inside the pipe, that is the volume of a cylinder:

V_{water}=\pi  r^{2}h

V_{water}=\pi (\frac{0.601m}{2})^{2}*120m

V_{water}=113.28m^{3}

The problem gives you the water density d as 1.0kg/L, but we need to convert it to proper units, so:

d_{water}=1.0\frac{Kg}{L}*\frac{1L}{1000cm^{3}}*(\frac{100cm}{1m})^{3}

d_{water}=1000\frac{Kg}{m^{3}}

Now, water density is given by the equation d=\frac{m}{V}, where m is the water mass and V is the water volume. Solving the equation for water mass and replacing the values we have:

m_{water}=d_{water}.V_{water}

m_{water}=1000\frac{Kg}{mx^{3}}*113.28m^{3}

m_{water}=113280Kg

With the water mass we can find the weight of water:

w_{water}=m_{water} *g

w_{water}=113280kg*9.8\frac{m}{s^{2}}

w_{water}=1110144N

Finally we find the total weight add up the weight of the water and the weight of the pipe,so:

w_{total}=w_{water}+w_{pipe}

w_{total}=1110144N+2500N

w_{total}=1112644N

Converting this total weight to kN, we have:

1112644N*\frac{0.001kN}{1N}=1113kN

7 0
2 years ago
A 150-lbm astronaut took his bathroom scale (a spring scale) and a beam scale (compares masses) to the moon where the local grav
kozerog [31]

Answer:

a)Wt =25.68 lbf

b)Wt = 150 lbf

F= 899.59 N

Explanation:

Given that

g = 5.48 ft/s^2.

m= 150 lbm

a)

Weight on the spring scale(Wt) = m g

We know that

1\ lbf=32.17 \ lmb.ft/s^2

Wt = 150 x 5.48/32 lbf

Wt =25.68 lbf

b)

On the beam scale

This is scale which does not affects by gravitational acceleration.So the wight on the beam scale will be 150 lbf.

Wt = 150 lbf

If the plane is moving upward with acceleration 6 g's then the for F

F = m a

We know that

1\ ft/s^2= 0.304\ m/s^2

5.48\ ft/s^2= 1.66\ m/s^2

a=6 g's

a=9.99\ m/s^2

So

F = 90 x 9.99 N

F= 899.59 N

3 0
2 years ago
Other questions:
  • As shown, a load of mass 10 kg is situated on a piston of diameter D1 = 140 mm. The piston rides on a reservoir of oil of depth
    9·1 answer
  • Steam flows at steady state through a converging, insulated nozzle, 25 cm long and with an inlet diameter of 5 cm. At the nozzle
    11·1 answer
  • Prompt the user to enter five numbers, being five people's weights. Store the numbers in an array of doubles. Output the array's
    11·2 answers
  • A three-story school has interior column bays that are spaced 25 ft apart in both directions. If the loading on the flat roof is
    7·1 answer
  • A 5-cm-diameter shaft rotates at 4500 rpm in a 15-cmlong, 8-cm-outer-diameter cast iron bearing (k = 70 W/m·K) with a uniform cl
    10·1 answer
  • A piece of corroded metal alloy plate was found in a submerged ocean vessel. It was estimated that the original area of the plat
    6·1 answer
  • A rectification column is fed 100 kg mol/h of a mixture of 50 mol % benzene and 50 mol % toluene at 101.32 kPa abs pressure. The
    5·1 answer
  • Discuss your interpretation of the confidence-precision trade-off, and provide a few examples of how you might make a choice in
    14·1 answer
  • The BARO sensor informs the PCM about changes in weather and ____________________.
    14·1 answer
  • The rate of flow through an ideal clarifier is 8000m3 /d, the detention time is 1h and the depth is 3m. If a full-length movable
    5·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!