AC versus DC power transmission (2 - Electricity Distribution)

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  • čas přidán 7. 09. 2024
  • Why is AC used more often than DC? And why are power lines usually at high voltages? Here I explain the history of alternating and direct current and work an example showing how high voltage transmission is more efficient in terms of line losses. Then I relate this to the use of AC for electricity distribution.
    Full edX course with interspersed practice problems to help you learn:
    www.edx.org/le...
    Aaron Danner is a professor in the Department of Electrical and Computer Engineering at the National University of Singapore.
    www.ece.nus.ed...
    Video filmed and edited by Cheryl Lim.
    @randomcheryl

Komentáře • 32

  • @ohgoditsjames94
    @ohgoditsjames94 Před rokem +9

    DC transmission is common on long feeders and at grid interconnections. DC transmission is more efficient when the circuits are very long (500km+) and at ultra high voltages (above 500kV) because of the high inductance and capacitance that AC transmission lines have and because of the skin effect. You also only need 2 cables instead of 3, saving on conductor material. The up front cost for the converter stations is the biggest hurdle, which is why they're more economical above a certain circuit length.

  • @comeflywithme1694
    @comeflywithme1694 Před 8 měsíci +2

    concise and useful in clear and beautiful English. what a pleasure to watch your lectures.

  • @chrisaxelos2417
    @chrisaxelos2417 Před rokem +1

    greatly explained the energy loss part and why we need transformers ( high voltage on long distances and low voltage where energy consumption takes place)

  • @Julliver451
    @Julliver451 Před rokem +3

    Very nice, concise and clear course on Edx, Thank you

    • @adanner
      @adanner  Před rokem

      You're most welcome. Keep watching and learning. Please do spread the word so others can learn too.

  • @comeflywithme1694
    @comeflywithme1694 Před 8 měsíci

    beautifully explained. what a gifted lecturer!

  • @emreceylan9979
    @emreceylan9979 Před rokem

    Thank you Aaron...

  • @mortalkombat5795
    @mortalkombat5795 Před 2 lety +1

    Thank you sir

  • @jasmanagno2992
    @jasmanagno2992 Před rokem +5

    How do you get the Rload value?

  • @EveSherman-x1j
    @EveSherman-x1j Před hodinou

    Adrianna Crest

  • @ThePaddyMak
    @ThePaddyMak Před 15 dny

    What if the load for the both voltage sources is the same i.e 10 ohms. Surely the load is not fixed

  • @tdc121
    @tdc121 Před 4 měsíci +1

    I calculate the low voltage load resistance to be 53 ohm and 4.3 amps for 230v and 1kW. How are you calculating for this example? Resistance =Voltage^2 / Power is what I used.

    • @radicalpotato666
      @radicalpotato666 Před 2 měsíci

      Exactly, that's where I am stuck too. 53-13 ohm =40 ohm should be the load resistance.

  • @ilcantastorie7455
    @ilcantastorie7455 Před měsícem

    Why are long submarine cable DC and not AC?

  • @borysmaysza8307
    @borysmaysza8307 Před 8 měsíci

    If we have U=230V and P(load)=1000W then R(load)=Upower2/P=53Ohm, not 10Ohm. Then P(line)=157W, not 1300W. Am I right?

  • @strengthwisdom
    @strengthwisdom Před rokem

    Hello sir, I have a question.
    Since the energy that goes through the wire is the same, regardless of the voltage, what is the fundamental change that causes the energy loss. What I can imagine is that with the low voltage we have current through the wire and with the high voltage we have current to the perimeter and around the wire.
    Can you please give your explanation?
    Thank you

    • @adanner
      @adanner  Před rokem +1

      In both AC and DC transmission lines, Joule heating is the main power loss mechanism. From a pure loss-reduction standpoint, DC transmission is superior due to the lack of skin effect and radiative losses. The difficulty with DC is the efficient conversion of voltages since only high voltage transmission can minimize Joule heating along the line.

  • @abcdef2069
    @abcdef2069 Před 9 měsíci

    at 4:29 P =I^2 * R (V=I * R) = V^2/ R
    P100 = (100v)^2 /2 ohm
    P200 = (200v)^2 /2 ohm
    P200 >> P100 whats wrong in this way

    • @adanner
      @adanner  Před 9 měsíci

      Using V^2/R you have just verified that putting 200V across a 2 ohm resistor consumes more power than putting 100V across it. We always want as small voltage drop as possible for this reason. But voltage drop is different than voltage with respect to ground. Assume we have 100V across a line resistance but each end is 100100V and 100000V. This delivers far more power to the load than say 1100V and 1000V.

    • @abcdef2069
      @abcdef2069 Před 9 měsíci

      @@adanner i was confused why 200v saves more electric energy than 100v.
      do the power companies limit the house's power consumption?
      they fix the voltage but i can use the Ampere in the limit they provide.
      if 1000watt is all that power company allows me
      1000watt = P(100v)= IV = 10A *100v
      1000watt = P(200v)= IV = 5A *200v
      if 2 ohm is from copper wire resistance that causes a power loss
      1. P(100v) = P =I^2 * R = (10A)^2 * 2 ohm = 200w
      2. P(200v) = P =I^2 * R = (5A)^2 * 2 ohm = 50w
      3. P(100v) = P =V^2 / R = (100v)^2 / 2 ohm = 10000w/2
      4. P(200v) = P =V^2 / R = (200v)^2 / 2 ohm = 40000w/2
      3 and 4. cant have 100v and 200v, i understood 3. 4. were wrong
      thanks for reply, asking you fixed my brain,

    • @adanner
      @adanner  Před 9 měsíci

      @@abcdef2069 Yes the power company strictly limits the power households can consume: It is limited to the maximum output power of the generators connected to the grid. If the power is exceeded, they shut off electricity to some customers for awhile (rolling blackouts).

  • @jorgefpsilva
    @jorgefpsilva Před 5 měsíci +1

    Title AC vs DC, content HV vs LV😂

  • @binaryglitch64
    @binaryglitch64 Před rokem

    I'm still getting Leyden jar deliveries from the milk man, er I mean elecrticity man.
    (That's a joke, I'm a licensed and certified electrician... I say I'm an electrician not to imply that I'm too knowledgeable about electricity to believe in the Leyden jar delivery man... just to imply that I can't pass up an electricity joke.)

  • @rigobertocantuna1737
    @rigobertocantuna1737 Před 2 lety +1

    incredible that it does not mention Nikola Tesla.

    • @KlaudiaTT...
      @KlaudiaTT... Před rokem

      I was thinking about the exact same thing! Nikola deserves at least 15 sec of introduction here - he's the father (and mother) of AC. But more seriously - fantastic videos Professor Danner! Very interesting and very well explained

    • @drmindbender8616
      @drmindbender8616 Před 4 měsíci

      Tesla unfortunately worked on 2 phase not the 3 phase that became the world standard and although he did contribute a lot in the world of electricity he was not the only engineer working on this there were hundreds around the world

    • @rigobertocantuna1737
      @rigobertocantuna1737 Před 4 měsíci

      @@drmindbender8616 Yes of course. 😆You do not have the absolute truth. Edison and Westinghouse only contributed to the monopolization of electricity.

  • @NovemberCharlieNC
    @NovemberCharlieNC Před 11 měsíci

    in the DC Transmission, you made calculations to show that low voltage has a bigger power loss than high voltage. The problem with what you just showed is that you put in the RLoad yourself. I could make a video, wit the same calculations and say oh we're using RLoad to be 1k Ohms in low voltage and RLoad to be 1k ohms as well on high voltage. It just makes no sense.

    • @adanner
      @adanner  Před 11 měsíci

      It wouldn't be a fair comparison if you used the same RLoad in both the high and low voltage cases. I deliberately picked values for RLoad that would make the power delivered to the load the same. For the same power delivered, we can then safely compare losses in each system.

  • @oldfartrick
    @oldfartrick Před rokem +1

    And once again Tesla is ignored as the other inventor.