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Natali [406]
2 years ago
13

when electric power is transmitted over long distance, line losses can be reduced by generating AC rather than DC voltage and by

transmitting?
Engineering
1 answer:
Tanya [424]2 years ago
3 0

When electric power is transferred over a long distance then, line loss can be reduced by high voltage and low current. Thus, option A is correct.

<h3>What is electric power?</h3>

The complete question is attached to the image below.

The electric circuit transfers the electric current then it is called electric power. The loss in the line over a larger distance is reduced by transferring AC instead of DC voltage.

The AC transmitted is of high voltage, and the power is low. The equation for current is given,

P = VI

The increase in the voltage causes the decrease in the current and in turn the resistance also reduces resulting in the loss of the power of the circuit to decrease.

Therefore, the loss is reduced by transferring high voltage and, low current AC.

Learn more about high volatge AC here:

brainly.com/question/18069892

#SPJ1

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In an experiment, the local heat transfer over a flat plate were correlated in the form of local Nusselt number as expressed by
Stells [14]

Answer:

\dfrac{\bar{h}}{h}=\dfrac{5}{4}

Explanation:

Given that

Nu_x=0.035Re_x^{0.8} Pr^{1/3}

We know that

Rex=ρvx/μ

So

Nu_x=0.035Re_x^{0.8} Pr^{1/3}

Nu_x=0.035\times\left(\dfrac{\rho vx}{\mu}\right)^{0.8}Pr^{1/3}

All other quantities are constant only x is a variable in the above equation .so lets take all other quantities as a constant C

Nu_x=C.x^{0.8}=C.x^{4/5}

We also know that

Nux=hx/K

C.x^{4/5}=\dfrac{hx}{k}

m is the constant

h=mx^{-1/5}

This is local heat transfer coefficient

The average value of h given as

\bar{h}=\dfrac{\int_{0}^{L}hdx}{L}

\bar{h}=\dfrac{5m}{4}\times\dfrac{L^{4/5}}{L}

\bar{h}=\dfrac{5m}{4}L^{-1/5}             ---------1

The value of local heat transfer coefficient at x=L

h=mx^{-1/5}

h=mL^{-1/5}            -----------2

From 1 and 2 we can say that

\dfrac{\bar{h}}{h}=\dfrac{5}{4}

8 0
3 years ago
What do you understand by the term phase angle?<br>​
aleksandrvk [35]

Answer:

The angle between the earth and the sun as seen from a planet is called phase angle.

5 0
2 years ago
Tadpoles raised in water with atrazine levels of 0.1 ppb should produce a higher percentage of male frogs with gonadal abnormali
Kobotan [32]

Answer:

correct option is c. a testable prediction leading to design of an experiment

Explanation:

The results of raising tadpoles were estimated to be water with an atrazine level of 0.1 ppb compared to those grown in pure water. So, this is not the question. If this assumption can now be tested, an experiment can be made in which a certain number of tadpoles can be raised in pure water and the same number of tadpoles can be raised in water with a 0.1ppb atrazine level can. The difference between the two populations can be estimated or compared.

8 0
3 years ago
For some transformation having kinetics that obey the Avrami equation, the parameter n is known to have a value of 2. If, after
kotegsom [21]

This question is incomplete, the complete question is;

For some transformation having kinetics that obey the Avrami equation, the parameter n is known to have a value of 2. If, after 100 s, the reaction is 40% complete, how long (total time in seconds) will it take the transformation to go to 95% completion

y = 1 - exp( -ktⁿ )

Answer: the time required for 95% transformation is 242.17 s

Explanation:

First, we calculate the value of k which is the dependent variable in Avrami equation

y = 1 - exp( -ktⁿ )

exp( -ktⁿ ) = 1 - y

-ktⁿ = In( 1 - y )

k = - In( 1 - y ) / tⁿ

now given that; n = 2, y = 40% = 0.40, and t = 100 s

we substitute

k = - In( 1 - 0.40 ) / 100²

k = - In(0.60) / 10000

k = 0.5108 / 10000

k = 0.00005108 ≈ 5.108 × 10⁻⁵

Now calculate the time required for 95% transformation

tⁿ = - In( 1 - y ) / k

t = [- In( 1 - y ) / k ]^1/n

n = 2, y = 95% = 0.95 and k = 5.108 × 10⁻⁵

we substitute our values

t = [- In( 1 - 0.95 ) / 5.108 × 10⁻⁵ ]^1/2

t = [2.9957 / 5.108 × 10⁻⁵]^1/2

t = [ 58647.22 ]^1/2

t = 242.17 s

Therefore the time required for 95% transformation is 242.17 s

8 0
3 years ago
Two production methods are being compared. One manual and the other automated. The manual method produces 10 pc per hour and req
Genrish500 [490]

Answer:

Check the explanation

Explanation:

Kindly check the attached image below to see the step by step explanation to the question above.

6 0
3 years ago
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