Why transmit electricity at HV, and low current

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!


Thanks!
-Remove @_, when replying via email

 

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.

 

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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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?

 

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

 

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 ()

 

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

 

---
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
right.


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
'primary'.

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


Np
----
Ns

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


Np
---- = 1
Ns

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:


120HOT>------[53R]---------+
|
[60W LAMP]
|
120NEUT>-----[53R]---------+


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.

 

Ah Thanks!. I am just retartded. I should have checked my work a lot
better.

 

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

-Regards