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Vadim26 [7]
2 years ago
6

Find the inductive reactance per mile of a single-phase overhead transmission line operating at 60 Hz, given the conductors to b

e Patridge and the spacing between centers to be 20 ft
Engineering
1 answer:
spin [16.1K]2 years ago
6 0

Answer:

The inductive reactance is 0.8281 Ω/mile

Explanation:

Given;

frequency, f = 60 Hz

space between the center, GMD = 20 ft

The inductive reactance per mile is calculated as;

X_L =X_a + X_d\\\\X_L = 2.022*10^{-3} *f *ln\frac{1}{GMR} + 2.022*10^{-3} *f *ln\ GMD

Where;

GMR is geometric mean radius (obtained from manufacturer's table)

GMD is geometric mean distance

For Partridge conductor, GMR = 0.0217 ft;

Xa = 2.022 x 10⁻³ x 60 x ln (1 / 0.0217)

Xa = 0.4647 Ω/mile

X_d = 2.022 x 10⁻³ x f x ln GMD

X_d = 2.022 x 10⁻³ x 60 x ln(20)

X_d = 0.3634 Ω/mile

Inductive reactance;

X_L = X_a + X_d\\\\X_L = 0.4647 \ ohms/mile \ +  0.3634 \ ohms/mile\\\\X_L = 0.8281 \ ohms/mile

Therefore, the inductive reactance is 0.8281 Ω/mile

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Fofino [41]

Answer:

the minimum shaft diameter is 35.026 mm

the maximum shaft diameter is 35.042mm

Explanation:

Given data;

D-maximum = 35.020mm and d-minimum = 35.000mm

we have to go through Tables "Descriptions of preferred Fits using the Basic Hole System" so from the table, locational interference fits H7/p6

so From table, Selection of International Trade Grades metric series

the grade tolerance are;

ΔD = IT7(0.025 mm)

Δd = IT6(0.016 mm)

Also from Table "Fundamental Deviations for Shafts" metric series

Sf = 0.026

so  

D-maximum

Dmax = d + Sf + Δd

we substitute

Dmax = 35 + 0.026 + 0.016

Dmax = 35.042 mm

therefore the maximum diameter of shaft is 35.042mm

d-minimum

Dmin = d + Sf

Dmin = 35 + 0.026

Dmin = 35.026 mm

therefore the minimum diameter of shaft is 35.026 mm

8 0
2 years ago
What forces are not present in space
ss7ja [257]

Answer:

C.) Weight and distance I believe

Explanation:

7 0
3 years ago
A 50 mm diameter shaft is subjected to a static axial load of 160 kN. If the yield stress of the material is 350 MPa, the ultima
zvonat [6]

In order to develop this problem it is necessary to take into account the concepts related to fatigue and compression effort and Goodman equation, i.e, an equation that can be used to quantify the interaction of mean and alternating stresses on the fatigue life of a materia.

With the given data we can proceed to calculate the compression stress:

\sigma_c = \frac{P}{A}

\sigma_c = \frac{160*10^3}{\pi/4*0.05^2}

\sigma_c = 81.5MPa

Through Goodman's equations the combined effort by fatigue and compression is expressed as:

\frac{\sigma_a}{S_e}+\frac{\sigma_c}{\sigma_u}=\frac{1}{Fs}

Where,

\sigma_a=Fatigue limit for comined alternating and mean stress

S_e =Fatigue Limit

\sigma_c=Mean stress (due to static load)

\sigma_u = Ultimate tensile stress

Fs =Security Factor

We can replace the values and assume a security factor of 1, then

\frac{\sigma_a}{320}+\frac{81.5}{400}=\frac{1}{1}

Re-arrenge for \sigma_a

\sigma_a = 254.8Mpa

We know that the stress is representing as,

\sigma_a = \frac{M_c}{I}

Then,

Where M_c=Max Moment

I= Intertia

The inertia for this object is

I=\frac{\pi d^4}{64}

Then replacing and re-arrenge for M_c

M_c = \frac{\sigma_a*\pi*d^3}{32}

M_c = \frac{260.9*10^6*\pi*0.05^3}{32}

M_c = 3201.7N.m

Thereforethe moment that can be applied to this shaft so that fatigue does not occur is 3.2kNm

5 0
3 years ago
how is friction losses in pipes reduced? a. decrease the pipe diameter b. increase the length of the pipes. c. decrease the leng
Citrus2011 [14]

Friction losses in pipes can be reduced by decreasing the length of the pipes, reducing the surface roughness of the pipes, and increasing the pipe diameter. Thus, options (c),(e), and (f) hold correct answers.

Friction loss is a measure of the amount of energy a piping system loses because flowing fluids meet resistance. As fluids flow through the pipes, they carry energy with them. Unfortunately, whenever there is resistance to the flow rate, it diverts fluids, and energy escapes. These opposing forces result in friction loss in pipes.

Friction loss in pipes can decrease the efficiency of the functions of pipes. These are a few ways by which friction loss in pipes can be reduced and the efficiency of the piping system can be boosted:

  • <u><em>Decrease the length of the pipes</em></u>: By decreasing pipe lengths and avoiding the use of sharp turns, fittings, and tees, whenever possible result in a more natural path for fluids to flow.
  • <u><em>Reduce the surface roughness of the pipes</em></u>:  By reducing the interior surface roughness of pipes, a smooth and clearer path is provided for liquids to flow.
  • <u><em>Increase the pipe diameter: </em></u>By widening the diameters of pipes, it is ensured that fluids squeeze through pipes easily.

You can learn more about friction losses at

brainly.com/question/13348561

#SPJ4

3 0
1 year ago
Three point bending is better than tensile for evaluating the strength of ceramics. a)-True b)- False
amid [387]

Answer:

a)-True

Explanation:

Three point bending is better than tensile for evaluating the strength of ceramics. It is got a positive benefit to tensile for evaluating the strength of ceramics.

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