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castortr0y [4]
3 years ago
11

2. The soil borrow material to be used to construct a highway embankment has a mass unit weight of 110.0 pcf, a water content of

6%, and the specific gravity of the soil solids is 2.63. The contract specifications require that the soil be compacted to a dry unit weight of 122 pcf and the water content be held to 5.5%. a. How many cubic yards of borrow are required to construct an embankment having a 245,000-cy net section volume

Engineering
1 answer:
PtichkaEL [24]3 years ago
5 0

Answer:

R1= 288032.53cy

Explanation:

For the Other part of the question, since the water content embankment is less than eater content of borrow material And it doesn't require any extra quantity of water bearing in mind that there is no evaporation, hence no further loss of water.

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Answer:

4

Explanation:

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What is An ampere is
sergij07 [2.7K]

Answer:

the SI base unit of electrical current.

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Consider incompressible flow in a circular channel. Derive general expressions for Reynolds number in terms of (a) volume flow r
Georgia [21]

Answer:

a) Re = \frac{4\cdot \rho \cdot Q}{\pi\cdot \mu\cdot D}, b) Re = \frac{4\cdot \dot m}{\pi\cdot \mu\cdot D}, c) 1600

Explanation:

a) The Reynolds Number is modelled after the following formula:

Re = \frac{\rho \cdot v \cdot D}{\mu}

Where:

\rho - Fluid density.

\mu - Dynamics viscosity.

D - Diameter of the tube.

v - Fluid speed.

The formula can be expanded as follows:

Re = \frac{\rho \cdot \frac{4Q}{\pi\cdot D^{2}}\cdot D }{\mu}

Re = \frac{4\cdot \rho \cdot Q}{\pi\cdot \mu\cdot D}

b) The Reynolds Number has this alternative form:

Re = \frac{4\cdot \dot m}{\pi\cdot \mu\cdot D}

c) Since the diameter is the same than original tube, the Reynolds number is 1600.

8 0
3 years ago
A balanced three phase load is supplied over a three-phase , 60 hz, transmission line with each line have a series impedance of
posledela

Answer:

Explanation:

Given a three-phase system

Frequency f=60Hz

Line impedance Z= 12.84 + j72.76 Ω

Then,

The resistance is R=12.84Ω

And reactance is X=72.76Ω

Z=√(12.84²+72.76²)

Z=73.88

Angle = arctan(X/R)

Angle = arctan(72.76/12.84)

Angle=80°

Then, Z=73.88 < 80° ohms

Load voltage is 132 kV

Load power P=55 MWA

Power factor =0.8lagging

the relation between the sending and receiving end specifications are given using ABCD parameters by the equations below.

Vs = AVr + BIr

Is = CVr + DIr

Where

Vs is sending Voltage

Vr is receiving Voltage

Is is sending current

Ir is receiving current

A is ratio of source voltage to received voltage A=Vs/Vr when Ir=0

B is short circuit resistance

B= Vs/Ir when Vr=0

C is ratio of source current to received voltage C=Is/Vr when Ir=0

D is ratio of source current to received current D=Is/Ir when Vr=0

Now,

The load at 55MVA at 132kV (line to line)

Therefore, load current is

Ir= P/V√3

Ir=55×10^6/(132×10^3×√3)

Ir=240.56 Amps

It has a power factor 0.8 lagging

PF=Cosθ

0.8=Cosθ

θ=arcCos(0.8)

θ=36.87°

Therefore, Ir=240.56 <-36.87°

Vr=V/√3

Vr=132/√3

Vr=76.21 kV. Phase voltage

Vr= 76210 < 0° V

For series impedance,

Using short line approximation

Vs = Vr + IrZ

Vs = 76210 < 0° + (240.56 <-36.87° × 73.88 < 80°)

Using calculator

Vs=76210<0° + 17772.5728<(-36.87°+80°)

Vs=76210<0° + 17772.5728<43.13°

Vs=89970.67<7.7°

Also

Is = Ir = 240.56 <-36.87° Amps

Therefore, the ABCD parameters is

A=Vs/Vr

A= 89970.67 <7.7° / 76210 <0°

A=1.181 <7.7-0

A=1.18 <7.7° no unit

B = Vs/Ir

B = 89970.67 < 7.7° / 240.56 <-36.87°

B = 347.01 < 7.7+36.87

B= 347.01 < 44.57° Ω

C= Is/Vr = 240.56 <-36.87° / 76210 < 0°

C= 0.003157 <-36.87-0

C= 3.157 ×10^-3 < -36.87° /Ω

C= 3.157 ×10^-3 < -36.87° Ω~¹

D= Is/Ir

Since Is=Ir

Then, D = 1 no unit.

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W in liquid = (ρliq)g hliq A where the cross sectional area is constant

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pliq = \frac{hw}{hliq} × 1000 kg /m³ ( density of water) =( \frac{23}{23-1.4}) × 1000 = 1064.8 kg/m³

8 0
3 years ago
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