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Gennadij [26K]
2 years ago
13

A traffic flow has density 61 veh/km when the speed is 59 veh/hr. If a flow has a jam density of 122 veh/km, what is the maximum

flow in veh/hr (in whole number)? Use Greenshield Model.
Engineering
1 answer:
antoniya [11.8K]2 years ago
6 0

Since this traffic flow has a jam density of 122 veh/km, the maximum flow is equal to 3,599 veh/hr.

<u>Given the following data:</u>

  • Density = 61 veh/km.
  • Speed = 59 km/hr.
  • Jam density = 122 veh/km.

<h3>How to calculate the maximum flow.</h3>

According to Greenshield Model, maximum flow is given by this formula:

q_{max}=\frac{V_f \times K_i}{4}

<u>Where:</u>

  • V_f is the free flow speed.
  • K_i is the Jam density.

In order to calculate the free flow speed, we would use this formula:

V_f =2 V\\\\V_f =2\times 59\\\\V_f=118\;km/hr

Substituting the parameters into the model, we have:

q_{max}=\frac{118 \times 122}{4}\\\\q_{max}=\frac{14396}{4}

Max flow = 3,599 veh/hr.

Read more on traffic flow here: brainly.com/question/15236911

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Answer:

Writing an excellent problem statement will not help guide you through the rest of the process and steer you towards the BEST solution.

False

Explanation:

An excellent problem statement sets the overall tone for the rest of the engineering process, whether it be at the analysis, design, or implementation stages.  This is why a problem statement must be focused, clear, and specific.  An excellent problem statement contains the problem definition, method for solving the problem (the claim proposed), purpose, statement of objectives, and scope.  For an excellent problem statement to be effective, it must also show the gap that is to be closed to achieve the intended objective.

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3 years ago
Steam at 20 bars is in the saturated vapor state (call this state 1) and contained in a pistoncylinderdevice with a volume of 0.
saul85 [17]

Answer:

Explanation:

Given that:

<u>At state 1:</u>

Pressure P₁ = 20 bar

Volume V₁ = 0.03 \mathbf{m^{3}}

From the tables at saturated vapour;

Temperature T₁ = 212.4⁰ C  ; v_1 = vg_1 = 0.0996 \mathbf{m^{3}} / kg

The mass inside the cylinder is m = 0.3 kg, which is constant.

The specific internal energy u₁ = ug₁ = 2599.2 kJ/kg

<u>At state 2:</u>

Temperature T₂ = 200⁰ C

Since the 1 - 2 occurs in an isochoric process v₂ = v₁ = 0.099 \mathbf{m^{3}} / kg

From temperature T₂ = 200⁰ C

v_f_2 = 0.0016 \ m^3/kg  

vg_2 = 0.127 \ m^3/kg  

Since  vf_2 < v_2 , the saturated pressure at state 2 i.e. P₂ = 15.5 bar

Mixture quality x_2 = \dfrac{v_2-vf_2}{vg_2 -vf_2}

x_2 = \dfrac{(0.099-0.0016)m^3/kg}{(0.127 -0.0016) m^3/kg}

x_2 = \dfrac{(0.0974)m^3/kg}{(0.1254) m^3/kg}

\mathsf{x_2 =0.78}

At temperature T₂, the specific internal energy u_f_2 = 850.6 \ kJ/kg , also ug_2 = 2594.3 \ kJ/kg

Thus,

u_2 = uf_2 + x_2 (ug_2 -uf_2)

u_2 =850.6  +0.78 (2594.3 -850.6)

u_2 =850.6  +1360.086

u_2 =2210.686 \ kJ/kg

<u>At state 3:</u>

Temperature T_3=T_2 = 200 ^0 C ,

V_3 = 2V_1 = 0.06 \ m^3

Specific volume v_3 = 0.2  \ m^3/kg

Thus; vg_3 =vg_2 = 0.127 \ m^3/kg ,

SInce v_3 > vg_3, therefore, the phase is in a superheated vapour state.

From the tables of superheated vapour tables; at v_3 = 0.2  \ m^3/kg and T₃ = 200⁰ C

The pressure = 10 bar and v =0.206 \ m^3/kg

The specific internal energy u_3 at the pressure of 10 bar = 2622.3 kJ/kg

The changes in the specific internal energy is:

u_2-u_1

= (2210.686 - 2599.2) kJ/kg

= -388.514 kJ/kg

≅ - 389 kJ/kg

u_3-u_2

= (2622.3 - 2210.686)  kJ/kg

= 411.614 kJ/kg

≅ 410 kJ/kg  

We can see the correct sketches of the T-v plot showing the diagrammatic expression in the image attached below.

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photoshop1234 [79]

Answer:

\eta = 70.711\,\%

Explanation:

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\dot W  = \dot U_{g}

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\dot W = \left(26\,\frac{persons}{min}\right)\cdot (124\,lbm)\cdot \left(32.174\,\frac{ft}{s^{2}}\right)\cdot \left(\frac{1\,lbf}{32.174\,\frac{lbm\cdot ft}{s^{2}} } \right)\cdot (27.5\,ft)

\dot W = 88660\,\frac{lbf\cdot ft}{min}\,\left(2.687\,hp\right)

The mechanical efficiency of the escalator is:

\eta = \frac{2.687\,hp}{3.8\,hp}\times 100\,\%

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Answer:

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F=ma for the weight of an object (which is a force) we have that the acceleration experienced by that object is equal to the gravitational acceleration, obtaining that  W = mg

For simplicity we work with g =9.807 \frac{m}{s^{2}} despiting the effect of the height above sea level. In this problem, we've been asked by the height above sea level that makes the weight of an object 0.30% more lighter.

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m(a-bz) = 0.997ma now we group like terms on the same sides ma(1-0.997) = mbz we cancel equal tems on both sides and obtain that z = \frac{a}{b} (0.003) = \frac{9.807 \frac{m}{s^{2} } }{3.32*10^{-6} s^{-2} } (0.003) = 8861.75 m

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