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Murrr4er [49]
2 years ago
10

A signalized intersection approach has three lanes with no exclusive left or right turning lanes. The approach has a 40-second g

reen out of a 75-second cycle. The yellow plus all-red intervals for phase total 4.0 seconds. If the start-up lost time is 2.3s/phase, the clearance lost time is 1.1s/phase, and thesaturation headway is 2.48 s/veh under prevailing conditions, what is the capacity of the intersection approach?
Engineering
1 answer:
ipn [44]2 years ago
6 0

Answer:

786.01

Explanation:

Calculate the saturation flow rate by using the following equation

S = 3600/h, where h is the saturation headway

S = 3600/2.48 = 1452 veh/hg

<u>Calculate the effective green time</u>

G(i) = G(i) + Y(i) – t(Li), where G(i) is the actual green time for movement, Y(i) is the sum of yellow and all red intervals for movement, and t(Li), is the total lost time for movement

G(i) = 40, Y(i) = 4, t(Li) = 2.3 + 1.1 = 3.4

G(i) = 40 + 4 – 3.4 = 40.6

<u>Calculate the capacity of an intersection approach by using the following equation</u>

C(i) = s(i) [g(i)/C], where s(i) the saturation flow rate, g(i) is the effective green time, and C is the cycle length

s(i) = 1452, g(i) = 40.6, C = 75

C(i) = 1452 x (40.6/75) = 786.01 veh/h/ln

The capacity of intersection approach is 786.01

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A glycerin pump is powered by a 5-kW electric motor. The pressure differential between the outlet and the inlet of the pump at f
Mila [183]

Answer:

Explanation:

Given that:

The pressure differential between the outlet and the inlet of the pump at full load is measured to be ΔP is 211 kPa.

the flow rate through the pump is 18 L/s

The Formula of flow rate.

{Q_f} = \frac{{{P_o}}}{{\Delta P}}

Rearrange

{P_o} = \Delta P\left( {{Q_f}} \right)

Substitute 211{\rm{ kPa}}\, for\, \Delta P \,and\, 18{\rm{ L/s}}\, for\, {Q_f}

\begin{array}{c}\\{P_o} = \left( {211{\rm{ kPa}}} \right)\left( {\left( {18{\rm{ L/s}}} \right)\left( {\frac{{{{10}^{ - 3}}{\rm{ }}{{\rm{m}}^{\rm{3}}}{\rm{/s}}}}{{1\;{\rm{L/s}}}}} \right)} \right)\\\\ = 3.8{\rm{ kW}}\\\end{array}

​The formula for the overall efficiency of the pump.

\eta = \frac{{{P_o}}}{{{{\rm{P}}_{in}}}} \times 100

input the values 3.8kW for {P_o} and 5kW for {P_i}

\begin{array}{c}\\\eta = \left( {\frac{{3.8{\rm{ kW}}}}{{5{\rm{ kW}}}}} \right) \times 100\\\\ = 76\% \\\end{array}

The overall efficiency of the glycerin pump is 76% .

6 0
2 years ago
Two kilograms of oxygen fills the cylinder of a piston-cylinder assembly. The initial volume and pressure are 2 m3 and 1 bar, re
valentinak56 [21]

Answer: Heat transfer (Q) is 521 kJ.

Explanation: In the piston-cilinder assembly, we can suppose that the oxygen act as an ideal gas, so, it can be used the General Gas Equation:

PV=\frac{m}{M}RT, where:

P is pressure;

V is volume;

m is mass;

M is molar mass;

R is a constant: R = 8.314.10^{-5} m³bar.K⁻¹.mol⁻¹;

T is temperature;

Using this equation, find the intial temperature:

PV = \frac{m}{M}RT

1.2 = \frac{2}{16}.8.314.10^{-5}.T

T = 1.924.10^{5} K

To determine the final temperature, use Combined Gas Law:

\frac{P_{i} . V_{i} }{T_{i} } = \frac{P.V }{T}, in which, the left side of the equality is related to the initial values and the right side, to the final values.

As pressure is constant:

\frac{V_{i} }{T_{i} } =\frac{V}{T}

T = \frac{V.T_{i} }{V_{i} }

T = \frac{4.1.924.10^{-5} }{2}

T = 3.85.10^{5} K

With the temperatures, calculate the heat transfer of the process:

Q = m.k.ΔT, where:

k is heat constant

ΔT = T - T_{i}

Q = m.k.ΔT

Q = 2.1.35.(3.85 - 1.92).10^{5}

Q = 521 kJ

The heat transfer in the process is 521 kJ.

6 0
2 years ago
Read 2 more answers
If the current on your power supply exceeds 500 mA it can damage the supply. Suppose the supply is set for 37 V. What is the sma
lilavasa [31]

Answer:

74 Ω

Explanation:

Data provided in the question:

Maximum value of the current that can be provided = 500 mA

= 500 × 10⁻³ A  

Applied voltage set for the system = 37 V

Now,  

The smallest resistance that can be measured    

= [ Applied voltage ] ÷ [ Maximum current that can be applied ]

= 37 ÷ (  500 × 10⁻³ )

= 74 Ω

4 0
2 years ago
In a parallel one-dimensional flow in the positive x direction, the velocity varies linearly from zero at y = 0 to 32 m/s at y =
monitta

Answer:

Ψ = 10(y^2) + c

<em><u>y = 1.067m</u></em>

Explanation:

since the flow is one dimensional in positive X direction, the only velocity component is in X, which is denoted by u

while u is a function of y

we find the u in terms of y; u varies linearly wih y

we use similiraty to find the relation

32/1.6 =<em>u/y</em>

<em><u>u = 20y</u></em>

<em><u>Ψ = ∫20ydy</u></em>

<em><u>Ψ = 10(y^2) + c</u></em>

<em><u>(b)</u></em>

<em><u>the flow is half below y = 1.6*(2/3)=1.067 m</u></em>

<em><u>this is because at two third of the height of a triangle lies the centroid of triangle. since the velocity profile forms a right angled triangle , its height is 1.6 m . the flow is halved at y = 1.067m</u></em>

3 0
2 years ago
Identify the four engineering economy symbols and their values from the following problem statement. Use a question mark with th
Vilka [71]

Answer:

The company will have $311,424 in its investment set-aside account.

Explanation:

To determine the amount of money that the company will have after 7 years with an interest rate of 11% per year, we must calculate the price to start with an increase of 11%, and then repeat the operation until reaching seven years:

Year 0: 150,000

Year 1: 150,000 x 1.11 = 166,500

Year 2: 166,000 x 1.11 = 184,815

Year 3: 184,815 x 1.11 = 205,144.65

Year 4: 205,144.65 x 1.11 = 227,710.56

Year 5: 227,710.56 x 1.11 = 252,758.72

Year 6: 252,758.72 x 1.11 = 280,562.18

Year 7: 280,562.18 x 1.11 = 311,424

3 0
2 years ago
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