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Photograph of transformers installed in transmission lines.
Power is distributed over large distances at high voltage to reduce power loss in the transmission lines. The voltages generated at the power plant are stepped up by passive devices called transformers (see Transformers ) to 330,000 volts (or more in some places worldwide). At the point of use, the transformers reduce the voltage transmitted for safe residential and commercial use. (Credit: GeorgHH, Wikimedia Commons)

Power losses are less for high-voltage transmission

(a) What current is needed to transmit 100 MW of power at 200 kV? (b) What is the power dissipated by the transmission lines if they have a resistance of 1 . 00 Ω size 12{1 "." "00" %OMEGA } {} ? (c) What percentage of the power is lost in the transmission lines?

Strategy

We are given P ave = 100 MW size 12{P rSub { size 8{"ave"} } ="100"`"MW"} {} , V rms = 200 kV size 12{V rSub { size 8{"rms"} } ="200"`"kV"} {} , and the resistance of the lines is R = 1 . 00 Ω size 12{R=1 "." "00"` %OMEGA } {} . Using these givens, we can find the current flowing (from P = IV size 12{P = ital "IV"} {} ) and then the power dissipated in the lines ( P = I 2 R size 12{P = I rSup { size 8{2} } R} {} ), and we take the ratio to the total power transmitted.

Solution

To find the current, we rearrange the relationship P ave = I rms V rms size 12{P rSub { size 8{"ave"} } = I rSub { size 8{"rms"} } V rSub { size 8{"rms"} } } {} and substitute known values. This gives

I rms = P ave V rms = 100 × 10 6 W 200 × 10 3 V = 500 A . size 12{I rSub { size 8{"rms"} } = { {P rSub { size 8{"ave"} } } over {V rSub { size 8{"rms"} } } } = { {"100 " times " 10" rSup { size 8{6} } " W"} over {"200 " times " 10" rSup { size 8{3} } " V"} } =" 500 A"} {}

Solution

Knowing the current and given the resistance of the lines, the power dissipated in them is found from P ave = I rms 2 R size 12{P rSub { size 8{"ave"} } = I rSub { size 8{"rms"} } rSup { size 8{2} } R} {} . Substituting the known values gives

P ave = I rms 2 R = ( 500 A ) 2 ( 1 . 00 Ω ) = 250 kW . size 12{P rSub { size 8{"ave"} } = I rSub { size 8{"rms"} } rSup { size 8{2} } R = \( "500 A" \) rSup { size 8{2} } \( 1 "." "00 " %OMEGA \) =" 250 kW"} {}

Solution

The percent loss is the ratio of this lost power to the total or input power, multiplied by 100:

% loss= 250 kW 100 MW × 100 = 0 . 250 % . size 12{%" loss=" { {"250"" kW"} over {"100"" MW"} } ´"100"=0 "." "250 %"} {}

Discussion

One-fourth of a percent is an acceptable loss. Note that if 100 MW of power had been transmitted at 25 kV, then a current of 4000 A would have been needed. This would result in a power loss in the lines of 16.0 MW, or 16.0% rather than 0.250%. The lower the voltage, the more current is needed, and the greater the power loss in the fixed-resistance transmission lines. Of course, lower-resistance lines can be built, but this requires larger and more expensive wires. If superconducting lines could be economically produced, there would be no loss in the transmission lines at all. But, as we shall see in a later chapter, there is a limit to current in superconductors, too. In short, high voltages are more economical for transmitting power, and AC voltage is much easier to raise and lower, so that AC is used in most large-scale power distribution systems.

It is widely recognized that high voltages pose greater hazards than low voltages. But, in fact, some high voltages, such as those associated with common static electricity, can be harmless. So it is not voltage alone that determines a hazard. It is not so widely recognized that AC shocks are often more harmful than similar DC shocks. Thomas Edison thought that AC shocks were more harmful and set up a DC power-distribution system in New York City in the late 1800s. There were bitter fights, in particular between Edison and George Westinghouse and Nikola Tesla, who were advocating the use of AC in early power-distribution systems. AC has prevailed largely due to transformers and lower power losses with high-voltage transmission.

Questions & Answers

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s. Reply
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s.
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Yeah, it is a pain to say the least. You basically have to heat the substarte up to around 1000 degrees celcius then pass phosphene gas over top of it, which is explosive and toxic by the way, under very low pressure.
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Graphene has a hexagonal structure
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I start with an easy one. carbon nanotubes woven into a long filament like a string
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AMJAD
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Yes, Nanotechnology has a very fast field of applications and their is always something new to do with it...
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Stotaw
In this morden time nanotechnology used in many field . 1-Electronics-manufacturad IC ,RAM,MRAM,solar panel etc 2-Helth and Medical-Nanomedicine,Drug Dilivery for cancer treatment etc 3- Atomobile -MEMS, Coating on car etc. and may other field for details you can check at Google
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Source:  OpenStax, College physics ii. OpenStax CNX. Nov 29, 2012 Download for free at http://legacy.cnx.org/content/col11458/1.2
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