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3 years ago

6

A 3-phase, 50 Hz, 110 KV overhead line has conductors placed in a horizontal plane 3 m apart. Conductor diameter is 2.5 cm. If t

he line length is 220 km, determine the charging current per phase assuming complete transposition. (6 Marks)

1 answer:

Bas_tet [7]3 years ago

6 0

Answer:

A 3-phase, 50 Hz, 110 KV overhead line has conductors

Explanation:

hope it will helps you

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Air flows through a device such that the stagnation pressure is 0.4 MPa, the stagnation temperature is 400°C, and the velocity i

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To solve this problem it is necessary to apply the concepts related to temperature stagnation and adiabatic pressure in a system.

The stagnation temperature can be defined as

$T_0 = T+\frac{V^2}{2c_p}$

Where

T = Static temperature

V = Velocity of Fluid

$c_p =$ Specific Heat

Re-arrange to find the static temperature we have that

$T = T_0 - \frac{V^2}{2c_p}$

$T = 673.15-(\frac{528}{2*1.005})(\frac{1}{1000})$

$T = 672.88K$

Now the pressure of helium by using the Adiabatic pressure temperature is

$P = P_0 (\frac{T}{T_0})^{k/(k-1)}$

Where,

$P_0$ = Stagnation pressure of the fluid

k = Specific heat ratio

Replacing we have that

$P = 0.4 (\frac{672.88}{673.15})^{1.4/(1.4-1)}$

$P = 0.399Mpa$

Therefore the static temperature of air at given conditions is 72.88K and the static pressure is 0.399Mpa

<em>Note: I took the exactly temperature of 400 ° C the equivalent of 673.15K. The approach given in the 600K statement could be inaccurate.</em>

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3 years ago

A civil engineer is asked to design a curved section of roadway that meets the following conditions: With ice on the road, when

lianna [129]

Answer:

1. $3.4^{o}$

2. 163.3 m

Explanation:

Static friction between road and rubber, μs =0.06

The maximum speed of the car, v = 50 km/h

= (50)(1000/3600) m/s

= 13.89 m/s

The acceleration due to gravity, $g = 9.81 m/s^{2}$

The frictional force, f = μsN ...... (1)

The component mg cosθ which balance the normal reaction N

The component mg sinθ acts in an opposite direction to the frictional force f.

ΣF = mg sinθ-f = 0 ...... (2)

Substitute the equation (1) in equation (2), we get

ΣF = mgsinθ-μsN = 0

mgsinθ-μsmgcosθ =0

μs = sinθ/cosθ

tanθ = μs

θ = tan-1( μs) = tan-1(0.06) = $3.4^{o}$

(b)The vertical component of the force is

N cosθ = fsinθ+mg

N cosθ = μsNsinθ+mg

N[cosθ- μs sinθ] = mg ...... (3)

The horizontal component of the force along the motion of the car is

Nsinθ+fcosθ = ma (Centripetal acceleration, $a = \frac {v^{2}}{r}$

Nsinθ+fcosθ = $m(\frac {v^{2}}{r})$

Nsinθ+μsNcosθ = $m(\frac {v^{2}}{r})$

N[sinθ+μs cosθ] = $m(\frac {v^{2}}{r})$ ...... (4)

Dividing the equation (4) with equation (3),

[sinθ+μscosθ]/[cosθ- μs sinθ] = $\frac {v^{2}}{rg}$

cosθ[sinθ/cosθ+μs]/cosθ[1- μs sinθ/cosθ] = $\frac {v^{2}}{rg}$

[tanθ+μs]/[1-μs tanθ] = $\frac {v^{2}}{rg}$

From part (1), tanθ = μs

Then the above equation becomes

$\frac {(\mu_s+\mu_s]}{[1-\mu_s^{2}]} =\frac {v^{2}}{rg}$

$\frac {(2\mu_s]}{[1-\mu_s^{2}]} =\frac {v^{2}}{rg}$

Therefore, the minimum radius of the curvature of the curve is

$r = \frac {v^{2}}{{2 \mu_s/[1-\mu_s^{2}]}g}$

= $\frac {v^{2}[1-\mu_s^{2}]}{2\mu_s g}$

= $\frac {(13.89 m/s)^{2}[1-(0.06)^{2}]}{(2)(0.06)(9.81)}$

= 163.3 m

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Answer:

See explaination

Explanation:

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