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aleksandr82 [10.1K]
3 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]3 years ago
4 0

Answer:

40.04pc/mile/hour

Explanation:

Check the attachment

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How much metal can be removed from a cracked drum to restore surface
Alisiya [41]

Answer:sc

Explanation:

sc

7 0
3 years ago
Read 2 more answers
A four-cylinder, four-stroke internal combustion engine operates at 2800 RPM. The processes within each cylinder are modeled as
Ulleksa [173]

Answer:

1) 287760.4 Hp

2) 18410899.5 kPa

Explanation:

The parameters given are;

p₁ = 14.7 lbf/in² = 101325.9 Pa

v₁ = 0.0196 ft³ = 0.00055501 m³

T₁ = 80°F = 299.8167 K

k = 1.4

Assumptions;

1) Air standard conditions are appropriate

2) There are negligible potential and kinetic energy changes

3) The air behaves as an ideal gas and has constant specific heat capacities of temperature and pressure

1) Process 1 to 2

Isentropic compression

T₂/T₁ = (v₁/v₂)^(1.4 - 1) = 10^0.4

p₂/p₁ = (v₁/v₂)^(1.4)

p₂ = p₁×10^0.4 =  101325.9*10^0.4 = 254519.153 Pa

T₂ = 299.8167*10^0.4 = 753.106 K

p₃ = 1080 lbf/in² = 7,446,338 Pa

Stage 2 to 3 is a constant volume process

p₃/T₃ = p₂/T₂

7,446,338/T₃ =   254519.153/753.106

T₃ = 7,446,338/(254519.153/753.106) = 22033.24 K

T₃/T₄ = (v₁/v₂)^(1.4 - 1) = 10^0.4

T₄ = 22033.24/(10^0.4) = 8771.59 K

The heat supplied, Q₁ = cv(T₃ - T₂) = 0.718*(22033.24 -753.106) = 15279.14 kJ

The heat rejected = cv(T₄ - T₁) = 0.718*(8771.59 - 299.8167) = 6082.73 kJ

W(net) = The heat supplied - The heat rejected = (15279.14 - 6082.73) = 9196.41 kJ

The power = W(net) × RPM/2*1/60 = 9196.41 * 2800/2*1/60 = 214582.9 kW

The power by the engine = 214582.9 kW = 287760.4 Hp

2) The mean effective pressure, MEP  = W(net)/(v₁ - v₂)

v₁ = 0.00055501 m³

v₁/v₂ = 10

v₂ = v₁/10 = 0.00055501/10 = 0.000055501

MEP  = 9196.41/(0.00055501 -  0.000055501) = 18410899.5 kPa

4 0
3 years ago
Coal fire burning at 1100 k delivers heat energy to a reservoir at 500 k. Find maximum efficiency.
Marizza181 [45]

Answer:

<em>55%</em>

Explanation:

hot reservoir = 1100 K

cold reservoir = 500 K

<em>This is a Carnot system</em>

For a Carnot system, maximum efficicency of the system is given as

Eff = 1 - \frac{Tc}{Th}

where Tc = temperature of cold reservoir = 500K

Th = temperature of hot reservoir = 1100 K

Eff = 1 - \frac{500}{1100}

Eff = 1 - 0.45 = 0.55 or<em> 55%</em>

7 0
3 years ago
Developed an automated program in any language which take 12 dependent variable and corresponding independent variables and show
otez555 [7]

Answer:

Explanation:

var generator = new Random(1);

// Now the nextGaussian() function returns a normal distribution of random numbers with the following parameters: a mean of zero and a standard deviation of one

var draw = function() {

var num = generator.nextGaussian();

var standardDeviation = 60;

var mean = 2003;

// Multiply by the standard deviation and add the mean.

var x = standardDeviation * num + mean;

noStroke();

fill(214, 159, 214, 10);

ellipse(x, 200, 16, 16); };

Hope this will be helpful

3 0
3 years ago
A river has an average rate of water flow of 59.6 M3/s. This river has three tributaries, tributary A, B and C, which account fo
Fiesta28 [93]

Answer:

50421.6 m³

Explanation:

The river has an average rate of water flow of 59.6 m³/s.

Tributary B accounts for 47% of the rate of water flow. Therefore the rate of water flow through tributary B is:

Flow rate of water through tributary B = 47% of 59.6 m³/s = 0.47 * 59.6 m³/s = 28.012 m³/s

The volume of water that has been discharged through tributary B = Flow rate of water through tributary B * time taken

time = 30 minutes = 30 minutes * 60 seconds / minute = 1800 seconds

The volume of water that has been discharged through tributary B in 30 seconds = 28.012 m³/s * 1800 seconds = 50421.6 m³

3 0
3 years ago
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