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Ilia_Sergeevich [38]
3 years ago
6

Airplanes typically have a Pitot-static probe located on the underside that measures relative wind speed as the airplane flies.

Explain your answers to the below questions. a. Does the Pitot-static probe provide information for an Eulerian or Lagrangian approach? b. If the Pitot-static probe was mounted on the roof of a building would it provide information for an Eulerian or Lagrangian approach?
Engineering
1 answer:
densk [106]3 years ago
7 0

Answer:

a:- Pitot or Pitot static tube is an open ended tube where moving fluid flows in order to measure the stagnation pressure . It is usually mounted under the wings of an aircraft   It is widely used to determine the airspeed of an aircraft and gas flow velocities in certain industrial applications. The pitot static probe provide information for both  Eulerian and Lagrangian approach

b:-Eulerian method is observing fluid properties as a function of time and space..Here the pitot tube mounted on roof of building use an approach of Eulerian method..In Eulerian approach measurement is done based on air flow through pitot static probe which can be referred as control volume.And the measurement is on the basis of control volume flows ad of tracking individual wind particle.

Explanation:

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A three-phase wye-connected synchronous generator supplies a network through a transmission line. The network can absorb or deli
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Answer:

the graph and the answer can be found in the explanation section

Explanation:

Given:

Network rated voltage = 24 kV

Impedance of network = 0.07 + j0.5 Ω/mi, 8 mi

Rn = 0.07 * 8 = 0.56 Ω

Xn = 0.5 * 8 = 4 Ω

If the alternator terminal voltage is equal to network rated voltage will have

Vt = 24 kV/√3 = 13.85 kV/phase

The alternative current is

I_{a} =\frac{40x10^{6} }{\sqrt{3}*24x10^{3}  } =926.2A

X_{s} =0.85\frac{13.85}{926.2} =12.7ohm

The impedance Zn is

\sqrt{0.56^{2}+4^{2}  } =4.03ohm

The voltage drop is

I_{a} *Z_{n} =926.2*4.03=3732.58V

r_{dc} =\frac{voltage}{2*current} =\frac{13.85}{2*926.2} =7.476ohm

rac = 1.2rdc = 1.2 * 7.476 = 8.97 Ω

The effective armature resistance is

Z_{s} =\sqrt{R_{a}^{2}+X_{s}^{2}    } =\sqrt{8.97^{2}+12.7^{2}  } =15.55ohm

The induced voltage for leading power factor is

E_{F} ^{2} =OB^{2} +(BC-CD)^{2}

if cosθ = 0.5

E_{F} =\sqrt{(13850*0.5)^{2}+(\frac{3741}{2}-926.2*12.7)^{2}   } =11937.51V

if cosθ= 0.6

EF = 12790.8 V

if cosθ = 0.7

EF = 13731.05 V

if cosθ = 0.8

EF = 14741.6 V

if cosθ = 0.9

EF = 15809.02 V

if cosθ = 1

EF = 13975.6 V

The voltage regulation is

\frac{E_{F}-V_{t}  }{V_{t} } *100

For each value:

if cosθ = 0.5

voltage regulation = -13.8%

if cosθ = 0.6

voltage regulation = -7.6%

if cosθ = 0.7

voltage regulation = -0.85%

if cosθ = 0.8

voltage regulation = 6.4%

if cosθ = 0.9

voltage regulation = 14%

if cosθ = 1

voltage regulation = 0.9%

the graph is shown in the attached image

for 10% of regulation the power factor is 0.81

8 0
3 years ago
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