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3 years ago

8

an existing highway-railway at-grade crossing is being redesigned as grade separated to improve traffic operations. The railway

must remain at the same elevation. The highway is being reconstructed to travel under the railway. The underpass will be a sag curve that has an initial grade of -2% and a final grade of 2%. The PVI of the sag curve will be centered under the railway (a symmetrical alignment). The sag curve design speed is 45 mi/h. How many feet below the railway should the curve PVI be located

1 answer:

Gekata [30.6K]3 years ago

7 0

Answer:

Please find attached solution

Explanation:

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In the well-insulated trans-Alaska pipeline, the high viscosity of the oil and long distances cause significant pressure drops,

s2008m [1.1K]

Our values are defined by,

$L=150*10^3m$

$\Phi = 1.2m$

$\dot{m}=700kg/s$

$\rho_o=900km/m^3$

$Sp=2000J/KgK$

$\mu_k = 0.765Ns/m^2$

With this data we can easily calculate first the mean velocity of the flow,

The velocity is given by,

$\nu=\frac{\dot{m}}{\rho_o A_c}$

$\nu = \frac{\dot{m}}{\rho_o \pi/4\Phi^2}$

$\nu = \frac{4*700}{900\pi(1.2)^4}$

$\nu=0.688m/s$

With the mean velocity is now possible determine if the flow is Laminar or turbulent, for that we need Number Reynolds.

Remember that a Laminar Flow is given when Reynolds number is minor to 2300, then

$Re_x = \frac{4\dot{m}}{\pi D\mu_k}$

$Re_x = \frac{4*700}{\pi*1.2*0.765}$

$Re_x = 970.8798$

$Re_x < 2300$ , then the flow is laminar.

For this section we will consider the Pressure drop to calculate the flow Work and the Total Energy conservation of this flow.

To calculate the Pressure drop we need the Friction factor, which is given by,

$f= \frac{64}{Re}=\frac{64}{970.8789}$

$f= 0.06591$

Then the Pressure drop is given by,

$\Delta P = \frac{f\rho_o \nu^2 L}{2D}$

$\Delta P = \frac{0.06591*900*(0.688)^2*183*10^3}{2*1.2}$

$\Delta P = 2.141*10^6Pa$

We can now calculate the Flow Work given by,

$W_f = \frac{\Delta P\dot{m}}{\rho_o}$

$W_f = \frac{2.141*10^6*700}{900}$

$W_f = 1.6651*10^6J$

From energy conservation we have

$W_f=q$

$W_f = \dot{m}c_p\Delta T$

Re-arrange for $\Delta T$

$\Delta T = \frac{W_f}{\dot{m}c_p}$

$\Delta T = \frac{1.56*10^6}{700*2000}$

$\Delta T = 1.1894\°c$

5 0

3 years ago

An extruder barrel has a diameter of 4.22 inches and a length of 75 inches. The screw rotates at 65 revolutions per minute. The

Vlada [557]

Answer:

volume flow rate Q = 53.23 in³/s

Explanation:

given data

diameter = 4.22 inches

length = 75 inches

screw rotates = 65 revolutions per minute

depth = 0.23 in

flight angle = 21.4 degrees

head pressure = 705 lb/in²

viscosity = 145 x $10^{-4}$ lb·sec/in²

solution

we get here volume flow rate of platstic in barrel that is express as

volume flow rate Q = volume flow rate of die - volume flow of extruder barrel ................1

here

volume flow rate extruder barrel is

flow rate = $\frac{\pi \times 705 \times 4.22 \times 0.23\times sin21.4 }{12\times 145 \times 10^{-4}\times 75 }$

flow rate = 60.10

and

volume flow rate of die is express as

volume flow rate = 0.5 × π² × D² × Ndc × sinA × cosA .............2

put here value and w eget

volume flow rate = 0.5 × π² × 4.22² × 0.23 × 1 × sin21.4 × cos21.4

volume flow rate = 6.866

so put value in equation 1 we get

volume flow rate Q = 60.10 - 6.866

volume flow rate Q = 53.23

7 0

3 years ago

A gas has an initial volume o.25m^3, and absolute pressure 100kPa. Its initial temperature is 290k. The gas is compressed into a

dezoksy [38]

Answer:

<h2>698.3Kpa</h2>

Explanation:

Step one:

given data

V1=0.25m^3

T1=290k

P1=100kPa

V2=0.5m^2

T2=405k

P2=? final pressure

Step two:

The combined gas equation is given as

P1V1/T1=P2V2/T2

Substituting we have

(100*0.25)/290=P2*0.05/405

25/290=0.5P2/405

0.086=0.05P2/405

cross multiply

0.086*405=0.05P2

34.9=0.05P2

divide both sides by 0.05

P2=34.9/0.05

P2=698.3Kpa

<u>Therefore the new pressure is 698.3Kpa when the gas is compressed</u>

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3 years ago

Which word from the passage best explains what the web in the passage symbolizes

frez [133]

Answer:

I cant see the passage

Explanation:

I cant see the passage

6 0

3 years ago

Elements of parallel computing

Leya [2.2K]

$\huge{\orange}\fcolorbox{purple}{cyan}{\bf{\underline{\green{\color{pink}Answer}}}}$

<h3><u>E</u><u>lements of parallel </u><u>computing:</u></h3>

Computer systems organization.
Computing methodologies.
General and reference.
Networks.
Software and its engineering.
Theory of computation.

6 0

2 years ago