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larisa [96]
3 years ago
15

A high-voltage direct-current (dc) transmission line between Celilo, Oregon and Sylmar, California is 845 mi in length. The line

contains four conductors, each weighing 2460 lbper 1000 ft.How many kilograms of conductor are in the line?
Engineering
1 answer:
Gennadij [26K]3 years ago
4 0

Answer:

<u>Total Mass of Conductors = 19913661.3 kg</u>

Explanation:

The total length is given as:

Total Length = 845 mi

Converting it to ft:

Total length = (845 mi)(5280 ft/1 mi)

Total length = 4,461,600 ft

The weight of one conductor per unit length is:

Mass of one conductor per unit length = 2460 lb/1000 ft

Mass of one conductor per unit length = 2.46 lb/ft

Since there are 4 conductors, therefore, total weight of all conductors per unit length will be:

Mass of all four conductors per unit length = 4 x 2.46 lb/ft

Mass of all four conductors per unit length = (9.84 lb/ft)(0.453592 kg/1 lb)

Mass of all four conductors per unit length = 4.463 kg/ft

Therefore, total mass of conductors in the line will be:

Total Mass of Conductors = (Mass of all four conductors per unit length)(Total Length)

Total Mass of Conductors = (4.463 kg/ft)(4,461,600 ft)

<u>Total Mass of Conductors = 19913661.3 kg</u>

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Answer: 78.89%

Explanation:

Given : Sample size : n=  1200

Sample mean : \overline{x}=2.45

Standard deviation : \sigma=0.07

We assume that it follows Gaussian distribution (Normal distribution).

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Using formula, z=\dfrac{x-\mu}{\sigma}, the z-value corresponds to 2.39 will be :-

z=\dfrac{2.39-2.45}{0.07}\approx-0.86

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z=\dfrac{2.60-2.45}{0.07}\approx2.14

Using the standard normal table for z, we have

P-value = P(-0.86

=P(z

Hence, the percentage of the diameter of the total shipment of shafts will fall between 2.39 inch and 2.60 inch = 78.89%

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The autorotation spin characteristics of a straight-wing aircraft are induced by Group of answer choices
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Sometimes autorotation takes place in rotating parachutes, kite tails. Etc.

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Water at a pressure of 3 bars enters a short horizontal convergent channel at 3.5 m/s. The upstream and downstream diameters of
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Answer:

The pressure reduces to 2.588 bars.

Explanation:

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\frac{P}{\gamma _{w}}+\frac{V^{2}}{2g}+z=constant

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\frac{P_{u}}{\gamma _{w}}+\frac{V-{u}^{2}}{2g}+z_{u}=\frac{P_{L}}{\gamma _{w}}+\frac{V{L}^{2}}{2g}+z_{L}

Now by continuity equation we have

A_{u}v_{u}=A_{L}v_{L}\\\\\therefore v_{L}=\frac{A_{u}}{A_{L}}\times v_{u}\\\\v_{L}=\frac{d^{2}_{u}}{d^{2}_{L}}\times v_{u}\\\\\therefore v_{L}=\frac{2500}{900}\times 3.5\\\\\therefore v_{L}=9.72m/s

Applying the values in the Bernoulli's equation we get

\frac{P_{L}}{\gamma _{w}}=\frac{300000}{\gamma _{w}}+\frac{3.5^{2}}{2g}-\frac{9.72^{2}}{2g}(\because z_{L}=z_{u})\\\\\frac{P_{L}}{\gamma _{w}}=26.38m\\\\\therefore P_{L}=258885.8Pa\\\\\therefore P_{L}=2.588bars

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3 years ago
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