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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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Explanation:

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2 years ago
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Assume the availability of an existing class, ICalculator, that models an integer arithmetic calculator and contains: an instanc
shtirl [24]

We connect with computers through coding, often known as computer programming.

<h3>How to code?</h3>
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class ICalculator {

int currentValue;

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return currentValue;

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this.currentValue = currentValue - value;

return currentValue;

}

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this.currentValue = currentValue * value;

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}

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ic.currentValue=5;

System.out.println(ic.add(2));

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3 0
1 year ago
while performing a running compression test how should running compression compare to static compression
algol [13]

Answer:

The idle speed of a running compression should be between 50-75 PSI and that is about half of the static compression.

Explanation:

The Running or Dynamic compression is used to determine how well the cylinder in an engine  is absorbing air, reserving it for the proper length of time, and releasing it to the exhaust. The static or cranking compression test is used to check the sealing of the cylinder. Before performing the running compression test, the static compression test is first performed to rule out other issues like bent valves.

The standard value for the static compression is given by;

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The running compression should always be half of the static compression.

5 0
2 years ago
A cylindrical metal specimen having an original diameter of 11.34 mm and gauge length of 53.3 mm is pulled in tension until frac
WINSTONCH [101]

Answer:

a) 70.29 %

b) 37%

Explanation:

percent reduction can be found from:

PR = 100*(π(do/2)^2-π(df/2)^2)/π(do/2)^2

     = 100*(π(11.34/2)^2-π(6.21/2)^2)/π(11.34/2)^2

     =70.29 %

percent elongation can be found from:

EL =L_f - Lo/Lo*100

    = (73.17 -53.3/53.3)*100

    = 37%

5 0
3 years ago
A gas turbine operates with a regenerator and two stages of reheating and intercooling. Air enters this engine at 14 psia and 60
Rzqust [24]

Answer:

flow(m) = 7.941 lbm/s

Q_in = 90.5184 Btu/lbm

Q_out = 56.01856 Btu/lbm

Explanation:

Given:

- T_1 = 60 F = 520 R

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- Heat ratio for air k = 1.4

- Compression ratio r = 3

- W_net,out = 1000 hp

Find:

mass flow rate of the air

rates of heat addition and rejection

Solution:

- Using ideal gas relation compute T_2, T_4, T_10:

                     T_2 = T_1 * r^(k-1/k)

                     T_2 = T_4 = T_10 = 520*3^(.4/1.4) = 711.744 R

- Using ideal gas relation compute T_7, T_5, T_9:

                     T_7 = T_6 * r^(-k-1/k)

                     T_7 = T_5 = T_9 = 1400*3^(-.4/1.4) = 1022.84 R

- The mass flow rate is obtained by:

                     flow(m) = W_net,out / 2*c_p*(1400-1022.84-711.744+520)

                     flow(m) = 1000*.7068 / 2*0.24*(1400-1022.84-711.744+520)

                     flow(m) = 7.941 lbm/s

- The heat input is as follows:

                     Q_in = c_p*(T_6 - T_5)

                     Q_in = 0.24*(1400 - 1022.84)

                     Q_in = 90.5184 Btu/lbm

- The heat output is as follows:

                     Q_out = c_p*(T_10 - T_1)

                     Q_out = 0.24*(711.744 - 520)

                    Q_out = 56.01856 Btu/lbm

                                           

                     

5 0
3 years ago
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