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9966 [12]
3 years ago
14

A four-lane freeway (two lanes in each direction) is located on rolling terrain and has 12-ft lanes, no lateral obstructions wit

hin 6 ft of the pavement edges, and there are 2 ramps within 3 miles upstream of the segment midpoint and 3 ramps within 3 miles downstream of the segment midpoint. A weekday directional peak-hour volume of 1800 vehicles (familiar users) is observed, with 700 arriving in the most congested 15-min period. If a level of service no worse than C is desired, determine the maximum number of heavy vehicles that can be present in the peak-hour traffic stream.
Consider the freeway and traffic conditions described in Problem 6.3. If 180 of the 1800 vehicles observed in the peak hour were heavy vehicles (assume a 70%/30% SUT/TT split), what would the level of service of this freeway be on a 5-mi, 6% downgrade?

Engineering
1 answer:
guapka [62]3 years ago
7 0

Answer:

Check the explanation

Explanation:

FFS = Free-Flow Speed (km/h) BFFS = the Base Free-Flow Speed (km/h) fLS = the Adjustment for lane and shoulder widths less than 3.65 m and 1.80 m, respectively. fAPD = Adjustment for access points density (km/h) fm = Adjustment for proportion  (km/h) 3.0 Results

Kindly check the below image for the step by step explanation to the above question.

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Oil with a density of 800 kg/m3 is pumped from a pressure of 0.6 bar to a pressure of 1.4 bar, and the outlet is 3 m above the i
Naddik [55]

Answer:

23.3808 kW

20.7088 kW

Explanation:

ρ = Density of oil = 800 kg/m³

P₁ = Initial Pressure = 0.6 bar

P₂ = Final Pressure = 1.4 bar

Q = Volumetric flow rate = 0.2 m³/s

A₁ = Area of inlet = 0.06 m²

A₂ = Area of outlet = 0.03 m²

Velocity through inlet = V₁ = Q/A₁ = 0.2/0.06 = 3.33 m/s

Velocity through outlet = V₂ = Q/A₂ = 0.2/0.03 = 6.67 m/s

Height between inlet and outlet = z₂ - z₁ = 3m

Temperature to remains constant and neglecting any heat transfer we use Bernoulli's equation

\frac {P_1}{\rho g}+\frac{V_1^2}{2g}+z_1+h=\frac {P_2}{\rho g}+\frac{V_2^2}{2g}+z_2\\\Rightarrow h=\frac{P_2-P_1}{\rho g}+\frac{V_2^2-V_1^2}{2g}+z_2-z_1\\\Rightarrow h=\frac{(1.4-0.6)\times 10^5}{800\times 9.81}+\frac{6.67_2^2-3.33^2}{2\times 9.81}+3\\\Rightarrow h=14.896\ m

Work done by pump

W_{p}=\rho gQh\\\Rightarrow W_{p}=800\times 9.81\times 0.2\times 14.896\\\Rightarrow W_{p}=23380.8\ W

∴ Power input to the pump 23.3808 kW

Now neglecting kinetic energy

h=\frac{P_2-P_1}{\rho g}+z_2-z_1\\\Righarrow h=\frac{(1.4-0.6)\times 10^5}{800\times 9.81}+3\\\Righarrow h=13.19\ m\\

Work done by pump

W_{p}=\rho gQh\\\Rightarrow W_{p}=800\times 9.81\times 0.2\times 13.193\\\Rightarrow W_{p}=20708.8\ W

∴ Power input to the pump 20.7088 kW

6 0
3 years ago
Which of the following sensors is used to provide suspension control module with feedback regarding vehicle cornering​ forces?
Readme [11.4K]

Answer:

A

Explanation:

4 0
2 years 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

For rough surfaces, Cfg = 0.0068. Using the same equation:

ΔT = 10.8°C

The tube surface temperature after prolonged service is:

Tprolo = 100 + 10.8 = 110.8°C

8 0
3 years ago
Can someone help me with this maze shown below.
Gnoma [55]
We can’t see the maze
3 0
2 years ago
Consider the following relational database that Best Airlines uses to keep track of its mechanics, their skills, and their airpo
Musya8 [376]

Answer:

Explanation:

A)

SELECT MECHNAME,AGE FROM MECHANIC;

B)

SELECT AIRNAME,SIZE FROM AIRPORT WHERE SIZE>=20 AND STATE='CALIFORNIA' AND YEAROPENED >=1935 ORDER BY SIZE ASC;

C)

SELECT AIRNAME,SIZE FROM AIRPORT WHERE (SIZE>=20 OR YEAROPENED >=1935) AND STATE='CALIFORNIA';

D)

SELECT AVG(SIZE) FROM AIRPORT WHERE STATE='CALIFORNIA' AND YEAROPENED >=1935;

E)

SELECT COUNT(AIRNAME) FROM AIRPORT WHERE STATE='CALIFORNIA' AND YEAROPENED >=1935;

F)

SELECT COUNT(AIRNAME),STATE FROM AIRPORT WHERE YEAROPENED>=1935 GROUP BY STATE;

G)

SELECT COUNT(AIRNAME),STATE FROM AIRPORT WHERE YEAR OPENED>=1935 GROUP BY STATE HAVING COUNT(*)>=5;

H)

SELECT MECHNAME FROM MECHANIC A JOIN AIRPORT B

ON A.AIRNAME=B.AIRNAME AND B.STATE='CALIFORNIA';

I)  

SELECT MECHNAME FROM MECHANIC A

JOIN QUALIFICATION B

ON A.MECHNUM=B.MECHNUM

AND B.PROFRATE=4

JOIN SKILL C

ON B.SKILLNUM=C.SKILLNUM

AND SKILLNAME='FAN BLADE RELACEMENT';

J)  SELECT MECHNAME FROM MECHANIC A

JOIN QUALIFICATION B

ON A.MECHNUM=B.MECHNUM

AND B.PROFRATE=4

JOIN SKILL C

ON B.SKILLNUM=C.SKILLNUM

AND SKILL NAME='FAN BLADE REPLACEMENT'

JOIN AIRPORT D

ON A.AIRNAME=D.AIRNAME

AND STATE='CALIFORNIA';

K)   SELECT SUM(SALARY),CITY FROM MECHANIC A

JOIN AIRPORT B

ON A.AIRNAME=B.AIRNAME

AND STATE='CALIFORNIA'

GROUP BY CITY;

L)   SELECT MAX(SIZE) FROM AIRPORT ;

M)  SELECT MAX(SIZE) FROM AIRPORT WHERE STATE='CALIFORNIA';

6 0
2 years ago
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