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madreJ [45]
3 years ago
15

An approach to a signalized intersection has a saturation flow rate of 1800 veh/h. At the beginning of an effective red, there a

re 6 vehicles in queue and vehicles arrive at 900 veh/h. The signal has a 60-second cycle with 25 seconds of effective red. What isthe total vehicle delay after one cycle?
Engineering
1 answer:
Tcecarenko [31]3 years ago
4 0

Answer: The total vehicle delay is

39sec/veh

Explanation: we shall define only the values that are important to this question, so that the solution will be very clear for your understanding.

Effective red time (r) = 25sec

Arrival rate (A) = 900veh/h = 0.25veh/sec

Departure rate (D) = 1800veh/h = 0.5veh/sec

STEP1: FIND THE TRAFFIC INTENSITY (p)

p = A ÷ D

p = 0.25 ÷ 0.5 = 0.5

STEP 2: FIND THE TOTAL VEHICLE DELAY AFTER ONE CYCLE

The total vehicle delay is how long it will take a vehicle to wait on the queue, before passing.

Dt = (A × r^2) ÷ 2(1 - p)

Dt = (0.25 × 25^2) ÷ 2(1 - 0.5)

Dt = 156.25 ÷ 4 = 39.0625

Therefore the total vehicle delay after one cycle is;

Dt = 39

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Answer and Explanation:

The criteria defined for the instruments that changes rapidly with time, ae called dynamic characteristics. These characteristics are namely

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It is the speed with which a measurement system responds to changes in the measured quantity.

Fidelity

It is the degree to which a measurement system indicates changes in the measured quantity without dynamic error.

Dynamic error

It is the difference between the true value of the quantity changing with time and the value indicated by the measurement system if no static error is assumed. It is also known as measurement error.

Measuring lag

It is the delay in the response of a measurement system to changes in the measured quantity. It is divided into two as follows.

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3 years ago
Match the words with their definitions
Pavlova-9 [17]

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Explanation:

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3 years ago
What is -4 (negative 4) in a 2's complement of 8 bits?
Pachacha [2.7K]

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5 0
3 years ago
Derive an expression for the specific heat difference of a substance whose equation of state is 1 2 ( ) RT a P b b T ν ν ν = − −
sergij07 [2.7K]

Answer:

Given data:

Equation of the state p=\frac{RT}{v-b}-\frac{a}{v(v+b) T^{1/2} }

Where p = pressure of fluid, pα

T = Temperature of fluid, k

V = Specific volume of fluid m^{3} / k g

R = gas constant , j/k g k

a, b = Constants

Solution:

Specific heat difference, \begin{array}{c}c_{p}-c_{v}=-T\left(\frac{\partial v}{\partial T}\right)^{2} p \\\left(\frac{\partial P}{\partial v}\right)_{r}\end{array}

According to cyclic reaction

\left(\frac{\ dv}{\ dT}\right)_{p}=-\frac{\left(\frac{\ d P}{\ d T}\right)_{v}}{\left(\frac{\ d P}{\ d v}\right)_{v}}

Hence specific heat difference is

c_{p}-c_{v}=\frac{-T\left(\frac{\ d v}{\ d T}\right)_{p}^{2}}{\left(\frac{\ d p}{\ dv}\right)_{v}}

Equation of state, p=\frac{R T}{v-b}-\frac{a}{v(v+b)^{\ 1/2}}

Differentiating the equation of state with respect to temperature at constant volume,

\(\left(\frac{\ d P}{\ d T}\right)_{v}=\frac{R}{v-b}-\frac{1}{2}- \frac{a}{v(v+b)^} T^{\frac{-1}{2}}\)

\begin{aligned}&\left(\frac{\ dP}{\ dT}\right)_{V}=\frac{R}{v-b}+\frac{a}{2 v(v+b) T^{3 / 2}}\end{aligned}

Differentiating the equation of the state with respect to volume at constant temperature.

\(\left(\frac{\ dP}{\ dv}\right)_{\gamma}=+(-1) \times R T(v-b)^{-1-1}+\frac{a}{b T^{1 / 2}}\left(\frac{1}{v^{2}}-\frac{1}{(v+b)^{2}}\right)\)\\\(\left(\frac{\ dP}{\ dv}\right)_{r}=-\frac{R T}{(v-b)^{2}}+\frac{a}{T^{1 / 2}}\left(\frac{2 v+b}{v^{2}(v+b)^{2}}\right)\)

Substituting both eq (3) and eq (4) in eq (2)

We get,

       {cp{} - } c_{v}=\frac{T\left(\frac{R}{v-b}+\frac{a}{2 v(v+b) T^{3 / 2}}\right)^{2}}{\left(\frac{R T}{(v-b)^{2}}-\frac{a(2 v+b)}{T^{1 / 2} v^{2}(v+b)^{2}}\right)}

Specific heat difference equation,

\(c_{p} -c_{v}}=\frac{T\left(\frac{R}{v-b}+\frac{a}{2 v(v+b)^{T}^{3 / 2}}\right)^{2}}{\left(\frac{R T}{(v-b)^{2}}-\frac{a(2 v+b)}{T^{1 / 2} v^{2}(v+b)^{2}}\right)}\)

 

     

7 0
3 years ago
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WITCHER [35]

Answer:

Explanation:

A multipurpose transformer can act as step up as well as step down transformer according to the desired setting by a user.

When the voltage at the output is greater than the voltage at the input of the transformer then it acts as step-up transformer and vice-versa acting is a step down transformer.

Given that:

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\frac{N_o}{230}=\frac{5.60}{220}

N_o=5.85\approx 6 turns compensating the losses

  • When the output voltage is 12.0 V:

V_o=12.0~V

\frac{N_i}{N_o} =\frac{V_i}{V_o}

\frac{N_o}{230}=\frac{12.0}{220}

N_o=12.45\approx 13 turns compensating the losses

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V_o=480~V

\frac{N_i}{N_o} =\frac{V_i}{V_o}

\frac{N_o}{230}=\frac{480}{220}

N_o=501.8\approx 502 turns compensating the losses

5 0
3 years ago
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