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Long distance transmission of power

  1. Sep 17, 2009 #1
    The reason given for the transmission of power by way of ac rather than by dc is that the ac voltage can be stepped up so that the current can be proportionately reduced. In that case the power loss is calculated by using the formula I^2 R. But my long standing doubt is how the resistance of the transmission cable wire is calculated? We know that the resistance of a wire increases with length. In that case, if the distance between the power plant and the receiving point, say a town, is 100 km, will not the resitance of a 100 km wire be extraordinasrily high? So, even if the current is very low, when the square of the current is multiplied by the resistance, will not the power loss increse with the increase in the distance? But, in all model calculations, resistance of the wire is kept constant. Certainly the quantity in the power formula is not the specific resistance but only the resistance. For a person well versed in this concept this question may be trivial in nature. But, I couldn't get a convincing explantion for the above doubt. Will any body thow light on this subject?
     
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  3. Sep 17, 2009 #2

    negitron

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    Assuming the wire stays consistent, the resistance increases linearly with length. If the resistance of a transmission wire is 1 ohm per 1000/ft, then the wire to a town 100 km away will be 100 ohms. On the other hand, by inspection of I2R, you can see the watts losses increase with the square of the current. Therefore, its easier (and more cost-effective) to increase efficiency by lowering the transmitted current than trying to decrease the wire resistance.
     
  4. Sep 17, 2009 #3

    dlgoff

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    Just to add to negitrons reason for resistive loss;
    http://en.wikipedia.org/wiki/Electric_power_transmission#Losses"
     
    Last edited by a moderator: Apr 24, 2017
  5. Sep 17, 2009 #4
    Hi ananthu-
    I believe some of the lowest-loss long-distance power lines are now HVDC, and not HVAC. Although I am no expert in this area, two of the reasons for this are:

    1) The high voltage limit for a power line is the peak voltage. The higher the rms voltage, the lower the current and the I^2 R losses. For HVDC power transmission, the peak voltage is the rms voltage, while for HVAC power transmission, the rms voltage is only 0.707 times the peak voltage. So the rms current is lower for a HVDC transmission line. But as everyone knows, transformers (and ac power) are required to step up to HV power transmission line voltages, and down from HV power transmission lines voltages. Nilola Tesla knew this more than a century ago. So efficient and economical ac/dc and dc/ac conversion in HV power transmission systems is very important.

    2) AC power transmission systems radiate power. Although 60 Hz is a very low frequency, and a very long wavelength, interstate power lines are antennas; half-wavelength is 250 Km at 60 Hz. Satellites measure significant 60 Hz power electromagnetic radiation over North America, and 50 Hz over western Europe. HVDC transmission would eliminate most of this.

    Finally, I think the resististive (I^2 R) power losses for HVAC and HVDC lines is calculated and compared per Km of length, and not for theentire length, e.g., foe 100 Km in your example.
    Bob S
     
  6. Sep 18, 2009 #5

    mgb_phys

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    I think the main reason for using DC for long distance links is that you don't need the two distant grids to be in phase, this is especially important if the grids at each end are further linked to other grids.

    I don't think radiation losses are significant compared to resistive an capacitive losses.
     
  7. Sep 18, 2009 #6

    Averagesupernova

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    Aren't you forgetting a zero?
     
  8. Sep 30, 2009 #7
    Thank you all for your kind replies. But my doubts remain mostly unanswered. If the reisistance increases with the increase in length of the wire, then the amount of current will all reduce with the length. For example a person at a distance of 100 km from the power station will draw more current than one at a distance of 200 km. But all draw the same constant rms current of 5 ampere only irrespective of their location. Hence, I think that the notion of the resistance increasing with the length does not hold water. Also, when we draw current for our domestic usage, we take one lead from the phase line and the other from the neutral line. Will it mean that every point on the entire length of the wire is at the same potential say, 220 V? In that case no current should flow throgh the wire at all as current can not flow between two points which are at the same potential! The current flows only when the phase wire and neutral wire are connected. These things terribly confuse me and the entire concepts seem vague. Do any body have a simple explanation for the mechanisms involved in the transmission of current from the generating point and drawing of the current in the receiving points?
     
  9. Sep 30, 2009 #8
    yes, my error. Should be 2500 Km. Bob S.

    To Anathu
    The I2R losses are roughly 10 % of the power consumed by the end user. If the AC transmission line is say 680 kV rather than 340 kV, then twice as much power can be transmitted on the same line with the same I2R losses. As for the potential between the lines, it is essentially the integral form of Faraday's law, where there is a magnetized rotor on an alternator at a coal-fired power plant creating a voltage between the hot and neutral wires (actually 3 hot wires and neutral), and upconverting it to 340 kV using a transformer.
    Bob S
     
  10. Sep 30, 2009 #9

    mgb_phys

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    No - current is the same along the wire, the current you put in must come out at the end.
    The voltage does reduce along the wire.
    If the wire has a resistance of eg 0.1ohm/Km then with a current of 100A there will be a voltage drop of 10V/km. So the voltage (above ground) at 100km will be 1000V less than at the power station and at 200km it will be 2000V less.


    In a domestic single phase the neutral line is at 0v - it's connected tp earth at the fuse board. The 220V is between the live phase and ground.

    Does this help http://en.wikipedia.org/wiki/Three-phase_electric_power
     
  11. Sep 30, 2009 #10

    stewartcs

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    Multiply the resistance per meter of wire times the length of the transmission line.

    Yes, the resistance will be high.

    No. The power loss is simply the square of the current (flowing in the transmission line) times the total resistance from point A to point B of the transmission line.

    The resistance of the wire is constant since the length from point A to point B (i.e. the transmission line length) does not change. Again, we are talking about the total length and not an incremental length.

    Hope this helps.

    CS
     
  12. Sep 30, 2009 #11
    Hi Anathu
    Here is something for you to mull over.
    A large fraction of end-use power is used for electric motors, especially induction motors. If the voltage on an induction motor drops by say 10%, the current INCREASES by roughly the same amount. So an I2R loss in a transmission line often leads to an INCREASE of current demand.
    Bob S
     
    Last edited: Sep 30, 2009
  13. Sep 30, 2009 #12

    stewartcs

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    The load is what determines the amount of current drawn. So if the same load is connected at 100 km as is at 200 km then the current in the line will be the same (i.e. the current flowing in the transmission line is constant for each scenario). However, the voltage drop (and I^2R losses) will be less at 100 km than at 200 km.

    No. The voltage is dropped along the wire. The current is constant.

    CS
     
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