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Why transmit electricity at HV, and low current

Discussion in 'Electronic Basics' started by Johnny Looser, Mar 1, 2005.

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  1. I am trying to understand why it is more effiecient to transfer
    electrons at HV, and low current, than the reverse. Intuitivly is seems
    very simple that power losses would be much less if there were fewer
    electrons bumping into copper atoms on their way to houses everywhere.
    However when I try to do the math I get lost. The standard explanation I
    get is that P = I^2*R, therefore power losses are proportional to the square
    of the current. So by stepping down the current, savings are realized.
    However, P also = E^2/r. It doen't make sense (to me) that I^2R is always
    < E^2/R.
    Can show me how to prove this mathematically?

    Here is an example I was trying to work out. Say at a house, there slurpy
    machine that operates at 120 VAC, 5A, 600W. Let's say the copper runs a couple
    miles and has a total resistance of 50 Ohms.

    Case A) Just enough V to get by. In order to compensat for line losses, you
    would need 120 VAC + 5A * 50 Ohms = 370 VAC at the genny. This would mean line
    losses would be 250 VAC * 5A = 1250 W.

    This is a lot. However my confusion comes when you start adding transformers
    to the mix.

    Case B) Use a 1:100 step down Xfmr. At secondary attached to the slurpy
    machine you would have 120 Vac, 5A. At the primary attached to the power
    line you would have 120VAC * 100 = 12000 VAC, and 5A/100 = 50 mA. In order to
    compensate for losses, you would have 120000 + .05A * 50 = 120003 VAC at the
    genny. Losses = 0.5A * 3VAC = 1.5W. Or calulated using P = II*R,
    .25A^2 * 50 Ohms = 12.5W. Or yet still using P = EE/R, 9VAC^2 /50 = 1.62W.
    I realize I have a problem because this scenario say the power line is non-ohmic
    in that E!=IR, 3VAC != .05A * 50Ohms. But How do I make everything work out.

    I realize that I do not have a firm understanding of how a transformer works.
    What does it mean to have the secondary producing 100V,at 0.5A into a 33 Ohm load?
    Can you help me. I don't want to be this stupid forever. Please help!

    -Remove @_, when replying via email

  2. I'm sure you can do it yourself once you realize that
    your '= I^2*R' and '= E^2/r' refer to either {different
    resistances} or {different voltages} than you now think.
    Your expression for power versus voltage would be
    correct if the voltage was that dropped along the line
    rather than the 'HV' between the lines. (For purposes
    of this discussion, we can ignore reactance issues.)
    Or it would be correct if 'r' was the load resistance
    rather than the line resistance to which 'R' refers.
  3. mike

    mike Guest

    Isn't reactance the major reason for high voltage transmission lines?
    Yes, copper is expensive, but cost is irrelevant if you can't stuff
    the current into it.

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  4. Uh, no. The major reason for using high voltage in
    transmission lines is to reduce wasted power.
    Reactance certainly has a part in those calculations.
    Don't get me on the wrong side of a "reactance matters"
    debate. But think carefully about the confusion exhibited
    by the OP. Is a long sidetrack into the role of reactance
    going to help him make the distinctions that will resolve
    his confusion? No. Are his expressions for power loss
    correct? Well, not quite, but for many tranmission line
    problems, they are pretty close and certainly sufficient to
    illustrate why it is worthwhile to use expensive HV gear
    to launch/recover power into/from long transmission lines.

    Do you see a purpose of the discussion (prior to your
    diversion) that would merit bringing in reactance other
    than to recognize that it is being ignored?
  5. This is the first error. In order to keep your calculations correct you have
    to add 2.5V
    Second error. 50mA= 0.05A. So 0.05 * 2.5 = 0.125W
    Third error. As I=0.05A, I^2*R = 0.05^2 * 50 = 0.0025 * 50 = 0.125W
    Using E^2/R: 2.5^2 / 50 = 6.25/50 = 0.125W
    It all fits you see. Maybe you have to train your math.

    petrus bitbyter
  6. Inductive reactance, like resistance, in a transmission line (a short or
    medium-short one, as in most well below .25 wavelength or most near or
    below .1 wavelength) becomes less of a problem when you use higher voltage
    in order to send less current.
    As long as you don't use so much voltage and so little current that
    capacitance across the line becomes more significant than inductive
    reactance (as in load impedance exceeds the "characteristic impedance" of
    the line, which is both inductive reactance and capacitive reactance at
    the frequency at which they are equal according to extrapolation from low
    frequency, and that frequency has the line length 1 "electrical radian" or
    1/(2*pi) of a wavelength at the "velocity factor" of the transmission line.

    ("Velocity factor" is a factor to multiply by the speed of light. This
    is the square root of the product of dielectric constant and permeability
    of materials exposed to electric and magnetic fields. Where more than
    one material is exposed to such fields, you have to give weighting by the
    square of field intensity in order to achieve a "relevant average" -
    worse when both electric and magnetic fields combined with both
    permeability and dielectric constant are factors. At least many coax
    cables have velocity factor ratings and that is from the square root of
    the dielectric constant of the insulation between the shield and the
    conductor. For wire transmission lines and coax caples with straight
    nonmagnetic conductors the velocity factor is usually .6 to 1, and for
    transmission lines with mostly air between the wires the velocity factor
    is usually near or above .9 but does not exceed 1.)

    (Most coaxial cables and many other shieled cables as well as most
    "twinlead" cable has a rated "characteristic impedance". Transmission to
    a resistive load of such impedance has largely minimized losses as well as
    largely minimized effects of reactances regardless of frequency or
    wavelength of the cable for the frequency in question. Other wire
    transmission lines have such a characteristic impedance - that which
    [according to extrapolation at low frequencies] is inductive and
    capacitive reactance at the frequency where these are equal to each other.

    - Don Klipstein ()
  7. Beeper

    Beeper Guest

    And I thought it was because it would not be practical to have lines big
    enough to carry high currents?
  8. John Fields

    John Fields Guest

    A transformer works by using a varying magnetic field to transfer
    power from one winding to another, but the situation you cite is
    impossible in that, according to Ohm's law,

    E = IR

    and if you plugged your numbers into the equation it wouldn't come out

    E = IR = 0.5A * 33 = 16.5V,

    _not_ 100V. If we were to look at a realistic example, say a 120V
    lamp drawing half an amp from the secondary of a transformer, then
    the resistance of the lamp would be

    E 120V
    R = --- = ------ = 240 ohms
    I 0.5A

    and the power the lamp would dissipate would be

    P = IE = 0.5A * 120V = 60 watts

    OK, but what about the transformer?

    Let's say it's what's called an 'isolation transformer' with a 120V
    input and a 120V output and that the winding driving the load is
    called the 'secondary', while the driven winding is called the

    Since it's transferring _power_ from the primary to the secondary and
    then to the load, to get 60 watts out of the secondary you'd have to
    put 60 watts into the primary. Actually a little bit more than that
    to make up for the losses in the transformer, but we'll ignore that
    for the moment and assume the transformer is perfect.

    Now, Since the transformer is transferring power and it was designed
    with a 120V primary and a 120V secondary, if the lamp takes half an
    amp out of the secondary, the power company has to put half an amp
    into the primary to make the lamp light. That'll make the power in
    equal to the power out and everything will work out just fine.

    A transformer like the one we're talking about is made by winding the
    primary and the secondary wires around a number of thin silicon steel
    sheets which are stacked together to form the core of the transformer
    which is used to couple the magnetic field created in the primary into
    the secondary.

    The relationship between the primary and secondary voltages is
    determined by the ratio of the number of turns on the primary and the
    secondary and is given by:

    Vp Np
    ---- = ----
    Vs Ns

    where Vp is the primary voltage
    Np is the number of turns on the primary
    Vs is the secondary voltage, and
    Ns is the number of turns on the secondary


    is called the "turns ratio", and in our case, since we have

    Vp 120V
    ---- = ------ = 1
    Vs 120V

    our turns ratio must also be 1. That is, if

    ---- = 1

    then Np and Ns must be equal to each other, whatever they are, and
    the reason the input and output voltages are identical in our
    transformer is because the number of primary turns (the number of
    times the primary wire has been wrapped around the core) is equal to
    the number of secondary turns.

    Now let's say that we have a 240V source and a 120V 60W lamp that we'd
    like to be able to light using that source. We could do it if we had a
    transformer, and we know that we'd need a 240V primary and a 120V
    primary, and we know that since

    Vp 240V
    ---- = ------ = 2
    Vs 120V

    we'll need a transformer with twice the number of turns on the primary
    than on the secondary. Interestingly, since we're moving _power_
    through the transformer, we'll still have

    P 60W
    I = --- = ------ = 0.5A
    E 120V

    for the secondary, but we'll have:

    P 60W
    I = --- = ------ = 0.25A
    E 240V

    or half the current running through the primary than for the 1:1
    transformer with the same load on the secondary!

    Now envision this:

    Let's say that you live 10 miles from the power company and that they
    can only make 120V electricity, and that at the end of the 20 mile run
    of wire (10 miles out and ten miles back) you've got your 120V 60W
    lamp burning. If we say that they've got #10AWG wire strung between
    the power plant and your house, that comes out to about 1 ohm per 1000
    feet, so that's about 106 ohms, and the circuit looks like this:

    [60W LAMP]

    After substituting 50 ohm resistors, since that's what I've got on
    hand, and firing the circuit up, I got:

    <-- 0.3A -->
    120HOT>------[50R]---------+ <--------+
    | |
    [60W LAMP] 57.4V
    | |
    120NEUT>-----[50R]---------+ <--------+

    Which means that your 60W lamp would only be using about

    P = IE =57.4V * 0.3A ~ 17 watts

    and the 20 miles of transmission line would be wasting

    P = 0.3A * (120V - 57.4V) 19 watts!

    One way to fix that problem would be to replace the transmission lines
    with larger diameter (lower resistance) wire, but in lieu of that, if
    we could send less current down the old lines, there'd be less of a
    voltage drop across the lines and more power would get to your lamp
    instead of just heating up the transmission lines on the way to your
    house. Solution? Send out high voltage, low current stuff, and then
    use a transformer close to where your lamp is to transform it down to
    a lower voltage and a higher current. Remember the example where a
    transformer with a 240V primary only needed a quarter of an amp in to
    get half an amp out? If you wound the primary for 480V you'd only need
    125mA in, and so on with the current getting lower and lower as the
    voltage got higher and higher.

    Using the 10 mile example, if we raise the transmission voltage to
    1200V and put a transformer at your house, it'll look like this:

    1200VAC>-----[50R]-----|P S|-----------+
    | | |
    | | [60W LAMP]
    | | |
    1200VAC>-----[50R]-----|P S|-----------+

    Now the lamp will be drawing close to 0.5A, but the current in the
    primary of the transformer will be 0.05A, and the power being wasted
    in the transmission lines will be

    P = 2*(I²R) = 2 * (0.05A² *50) = 0.25W

    instead of 17 watts, and the voltage dropped across the lines will be:

    E = 2Ir = 2 * 0.05 * 50 = 5V

    Which will be a 0.5V decrease in voltage out of the primary.

    So that's why it's more efficient to transmit power over the grid at
    high voltages and low currents. :)
  9. Johny Looser

    Johny Looser Guest

    Ah Thanks!. I am just retartded. I should have checked my work a lot
  10. Johny Looser

    Johny Looser Guest

    Thanks For all the help everyone gave me! I realize now, I
    misunderstood what I was looking at, and made a whole bunch careless
    math mistakes!

    Many Thanks! Especially to Mr Fields

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