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aleksandr82 [10.1K]
4 years ago
11

A six-lane multilane highway (three lanes in each direction) has a peak-hour factor of 0.90, 11-ft lanes with a 4-ft right-side

shoulder, and a two-way left-turn lane in the median. The directional peak-hour traffic flow is 4000 vehicles with 14% SUTs and 6% TTs. What will the level of service of this highway be on a 4% upgrade that is 1.5 miles long if the speed limit is 55 mi/h and there are 15 access points per mile

Engineering
1 answer:
Dennis_Churaev [7]4 years ago
4 0

Answer:

40.04pc/mile/hour

Explanation:

Check the attachment

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In a diesel engine, the fuel is ignited by (a) spark (c) heat resulting from compressing air that is supplied for combustion (d)
-Dominant- [34]

Answer:

(c) heat resulting from compressing air that is supplied for combustion

Explanation:

In a diesel engine the combustion of the fuel takes place due to the adiabatic compression which leads to elevated temperatures in the cylinder. First the air is compressed by the piston in the cylinder which raises the temperature, then the atomized fuel is put in the cylinder causing the ignition of fuel. Diesel engines have the highest thermal efficiency.

Spark causes ignition in petroleum engine.

7 0
3 years ago
True or False; If I was trying to find the Voltage of my computer, and I was given the Watts and Amps it uses, I would use Watt'
defon

Answer:

true

Explanation:

8 0
3 years ago
Very thin films are usually deposited under vacuum conditions to prevent contamination and ensure that atoms can fly directly fr
katrin [286]

Answer:

a. 9947 m

b. 99476 times

c. 2*10^11 molecules

Explanation:

a) To find the mean free path of the air molecules you use the following formula:

\lambda=\frac{RT}{\sqrt{2}\pi d^2N_AP}

R: ideal gas constant = 8.3144 Pam^3/mol K

P: pressure = 1.5*10^{-6} Pa

T: temperature = 300K

N_A: Avogadros' constant = 2.022*10^{23}molecules/mol

d: diameter of the particle = 0.25nm=0.25*10^-9m

By replacing all these values you obtain:

\lambda=\frac{(8.3144 Pa m^3/mol K)(300K)}{\sqrt{2}\pi (0.25*10^{-9}m)^2(6.02*10^{23})(1.5*10^{-6}Pa)}=9947.62m

b) If we assume that the molecule, at the average, is at the center of the chamber, the times the molecule will collide is:

n_{collision}=\frac{9947.62m}{0.05m}\approx198952\  times

c) By using the equation of the ideal gases you obtain:

PV=NRT\\\\N=\frac{PV}{RT}=\frac{(1.5*10^{-6}Pa)(\frac{4}{3}\pi(0.05m)^3)}{(8.3144Pa\ m^3/mol\ K)(300K)}=3.14*10^{-13}mol\\\\n=(3.14*10^{-13})(6.02*10^{23})\ molecules\approx2*10^{11}\ molecules

5 0
3 years ago
What is the role of the architects in modern development​
pshichka [43]

Answer:

to plan, design, and oversee the construction of a building

7 0
3 years ago
Steam enters a turbine at 120 bar, 508oC. At the exit, the pressure and quality are 50 kPa and 0.912, respectively. Determine th
Archy [21]

Answer:

The turbine produces <u>955.53 KW</u> power.

Explanation:

Taking the turbine as a system. Applying Law of Conservation of Energy:

m(h₁ - h₂) - Heat Loss = P

where,

m = mass flow rate of steam = 1.31 kg/s

h₁ = enthalpy at state 1 (120 bar and 508°C)

h₂ = enthalpy at state 2 (50 KPa and x = 0.912)

Heat Loss = 225 KW

P = Power generated by turbine = ?

First, we find h₁ from super steam tables:

At,

T = 508°C

P = 120 bar = 12000 KPa = 12 MPa

we find that steam is in super-heated state with enthalpy:

Due to unavailibility of values in chart we approximate the state to 500° C and 12.5 MPa:

h₁ = 3343.6 KJ/kg

Now, for state 2, we have:

P = 50 KPa and x = 0.912

From saturated steam table:

h₂ = hf₂ + x(hfg₂) = 340.54 KJ/kg + (0.912)(2304.7 KJ/kg)

h₂ = 2442.4 KJ/kg

Now, using values in the conservation equation:

(1.31 kg/s)(3343.6 KJ/kg - 2442.4 KJ/kg) - 225 KW = P

<u>P = 955.53 KW</u>

5 0
3 years ago
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