Why high voltage for power transmission?
Stephen-I-am
How do you get less loss in the same transmission line with higher
voltage? Wouldn't the loss scale as the square of the voltage?
Please set me straight. ;)
Stephen
Luhan
As you increase the voltage, more people get fried trying to steal
power. This decreases your loses.
Luhan
ian field
"Stephen-I-am" <ste...@theboulets.net> wrote in message
news:1152301746.5...@m73g2000cwd.googlegroups.com...
If the entire power distribution system operated at 120/240V the conductors
would have to be several feet thick!!!
Abstract Dissonance
"Stephen-I-am" <ste...@theboulets.net> wrote in message
news:1152301746.5...@m73g2000cwd.googlegroups.com...
P = I*V
and I = V/R <==> V = IR
So
P = I^2*R
or
P = V^2/R
So the second power dissipation equation is reduced by a factor of R while
the first is multiplied by it. This helps you with the idea if you want to
send X volts or X amps along the line with some fixed R. Always choose X
volts because you will be wasted less power. i.e.
P1 = X^2*R
P2 = X^2/R
Then P1 = R^2*P2
i.e., you will be wasting R^2 times the amount of power by choosing to send
X amps instead of X volts. for R > 1 this is huge.
e.g.,
Say you want to transfer power of 100Watts. You can do this by choosing any
I and V such that I*V = 100. Or I = 100/V.
choosing a real large number for V, say, 1000, gives I = 100/1000 = 1/10 A
and the power we are trying to send is 100W(obviously, because this is what
we started with). Or we could choose I = 1000A while V = 1/10V.
Now, we have a dissipation in the line so what is our actual power at the
other end, say, of a 10Ohm line?
100W - 1/10A*10 = 99W
VS
100W - 1000A*10 = -99900W
i.e., in the first case of using 1000V we only lost 1W and in the second
case we lost all of our power.
Why does this happen though? You have to look into why ohms law exists. It
has to do with how electrons flow in wire. As electrons flow they are
bumping into atoms of the wire.
An analogy is like a freeway. If you just have a few cars going really fast
then its usually not a problem... but what happens if you have a lot of cars
going slow? Current is like the number of cars and voltage is there speed.
(and just note that they are inversely proportional) If you want a lot of
cars then they all must go slow and going slow makes people mad(power
dissipation). If you just allow a few fast cars then no one is mad. The
total power behind the cars is like momentum. If you have 10 cars going 1
MPH or 1 car going 10 MPG then you still have the same momentum. But we
loose a little momentum from people getting off the freeway from being
pissed because its to slow... i.e., electrons hitting atoms and loosing
energy to that atom creating heat.
The idea is simple though. Sending one electron through a wire at a high
speed is much less likely to collide with an atom. Sending a large amount of
electrons through a wire, even at a very slow speed will have many more
collisions. The reason has to do with the interaction between the electrons
themselfs too. Its kinda like a whole crowd of people trying to walk over a
man hole but avoid falling in. Its much harder to do the more people you
have. If you walking towards it and you think you can jump over it but get
bumped in the process you might fall in.
Joel Kolstad
news:1152301746.5...@m73g2000cwd.googlegroups.com...
> voltage? Wouldn't the loss scale as the square of the voltage?
It's scales with the square of the voltage *dropped across the transmission
line*, which is (hopefully) quite small compared to the voltage dropped across
the load. To avoid this confusion, people normally talk about "I-squared R"
losses in transmission lines, since the same current flows through the
transmission line as through the load.
As such, if you double the input voltage, the current required by the load for
the same delivered power is halved. Therefore, the power lost in the line --
I^2*Rline -- has been quartered.
This is why high voltages are used for power distribution.
---Joel Kolstad
John Fields
<ste...@theboulets.net> wrote:
---
Power goes as the square of the current.
View in Courier:
100V
/
1KV / 10A-->
/ +--------+
+-----+ / 1A--> R1 +-----+ | |
| ~|--------[0.5R]---------| |-----+ |
| | | | [10R]
| ~|--------[0.5R]---------| |-----+ |
+-----+ R2 +-----+ | |
GEN XFMR +--------+
LOAD
R1 and R2 represent the transmission line impedance, so with a 1kV
source pushing 1 ampere through the line the loss is:
P = I²(R1+R2) = 1A² * 1R = 1 watt
With a 10kV source pushing 100mA through the line, however:
100V
/
10kV / 10A-->
/ +--------+
+-----+ /0.1A--> R1 +-----+ | |
| ~|--------[0.5R]---------| |-----+ |
| | | | [10R]
| ~|--------[0.5R]---------| |-----+ |
+-----+ R2 +-----+ | |
GEN XFMR +--------+
LOAD
P = I²(R1+R2) = 0.1A² * 1R = 0.01 watt,
Which is a hundred-fold decrease in loss for a ten-fold increase in
line voltage.
--
John Fields
Professional Circuit Designer
John B
"Luhan" <luh...@yahoo.com> wrote in message
news:1152302187.6...@75g2000cwc.googlegroups.com...
John B
Your losses are primarily through resistive causes, and they are
proportional to the CURRENT (not voltage) flowing through the wire. So you
want to minimize CURRENT in the transmission line.
Power is the product of voltage times current. Use a transformer to step
voltage UP at the generating station. This allows the amount of CURRENT
pumped down the wire to be LOWER, by the same ratio that you multiplied
voltage.
Now move your electric power several hundred miles.
At the receiving end, reverse the voltage change, and voila! You've got
power ready for toasters, computers, etc.
Of course, there are losses in these transformers. But you expect these to
be small, compared to the power you saved along the transmission path.
Saves on COPPER (or aluminum) too, and that is a big deal.
I believe this AC transformation process was the undoing of Thomas Edison's
foray into the electric power business. George Westinghouse promoted AC,
while Edison promoted DC. So Westinghouse "saw the light," while Edison
didn't, notwithstanding Edison's most famous invention. I'm sure some
historian will set me straight, if I got this wrong.
"Stephen-I-am" <ste...@theboulets.net> wrote in message
news:1152301746.5...@m73g2000cwd.googlegroups.com...
John Fields
<JKolstad7...@yahoo.com> wrote:
>"Stephen-I-am" <ste...@theboulets.net> wrote in message
>news:1152301746.5...@m73g2000cwd.googlegroups.com...
>> How do you get less loss in the same transmission line with higher
>> voltage? Wouldn't the loss scale as the square of the voltage?
>
>It's scales with the square of the voltage *dropped across the transmission
>line*, which is (hopefully) quite small compared to the voltage dropped across
>the load. To avoid this confusion, people normally talk about "I-squared R"
>losses in transmission lines, since the same current flows through the
>transmission line as through the load.
---
That's only true if there's no transformer or a 1:1 transformer
between the line and the load.
Luhan
Tank you.
Luhan
Bill Beaty
Abstract Dissonance wrote:
> An analogy is like a freeway. If you just have a few cars going really fast
> then its usually not a problem... but what happens if you have a lot of cars
> going slow?
Wrong, since the number of electrons in a wire doesn't change. Wires
are like pipes which are always full of water, and the "water" can
flow at different speeds, but the number of water molecules in the pipe
remains the same.
TRANSMISSION LINE ANALOGY:
An electric transmission line is like a long circular drive belt
wrapped around a pair of drive wheels. But this drive belt has
high friction. (Imagine that the belt is rubbing on the ground.)
If we spin the drive wheel fast, the belt heats up from all the
rubbing, and lots of energy is wasted.
To solve the problem, we can put gear boxes at either end,
and gear the belt down to low speed (but at higher belt tension.)
Now we can transmit energy at the same rate along the belt
as before, but the belt moves very slowly and with immense
pressure.
Voltage is like pressure difference. Current is like speed.
And power transformers are like gearboxes.
(((((((((((((((((( ( ( ( ( (O) ) ) ) ) )))))))))))))))))))
William J. Beaty SCIENCE HOBBYIST website
bi...@eskimo.com http://amasci.com
EE/programmer/sci-exhibits amateur science, hobby projects, sci fair
Seattle, WA 425-222-5066 unusual phenomena, tesla coils, weird sci
Stephen-I-am
> On 7 Jul 2006 12:49:06 -0700, "Stephen-I-am"
> <ste...@theboulets.net> wrote:
>
> >Slightly off topic, but something I was wondering about.
> >
> >How do you get less loss in the same transmission line with higher
> >voltage? Wouldn't the loss scale as the square of the voltage?
> >
> >Please set me straight. ;)
Thanks for the nice diagram John. I think that the fault in my
reasoning was to assume that a higher transmission voltage meant a
correspondingly higher line current. After all, if you increase the
voltage into a fixed resistor, the current goes up.
So ... why doesn't the current go up? Is the impedance seen at the
generator lower for a higher voltage, due to a different transformer?
Stephen
Tim Williams
> So ... why doesn't the current go up? Is the impedance seen at the
> generator lower for a higher voltage, due to a different transformer?
Ya.
It's convienient to think in terms of load power. Then, current is whatever
the power is divided by the voltage you're running at. I mean, as the power
company, your customers demand some amount of power, day in, day out, for
the most part.
Tim
--
Deep Fryer: a very philosophical monk.
Website: http://webpages.charter.net/dawill/tmoranwms
John Larkin
>That's spelled "losses."
>
I think he actually meant "losers."
John
Rich Grise, Plainclothes Hippie
Too bad Mr. Rockefeller didn't get behind Mr. (Dr?) Tesla - we'd have
had wireless power transmission a century ago. )-;
Thanks!
Rich
Abstract Dissonance
"Bill Beaty" <bi...@eskimo.com> wrote in message
news:1152308890....@b28g2000cwb.googlegroups.com...
>
> Abstract Dissonance wrote:
>
>> An analogy is like a freeway. If you just have a few cars going really
>> fast
>> then its usually not a problem... but what happens if you have a lot of
>> cars
>> going slow?
>
>
>
> Wrong, since the number of electrons in a wire doesn't change. Wires
> are like pipes which are always full of water, and the "water" can
> flow at different speeds, but the number of water molecules in the pipe
>
> remains the same.
>
huh. Do you know what current is? Current is the measure of the number of
electrons passing through a cross section of wire per second. 1 Columb of
current per second is 1 amp.
So the number of electrons in a wire does change depending w.r.t to current.
Wires are not like pipes which are always full. Even if they were you can
compress water and hence change the density => changing the number of
molecules of water in the pipe.
With your (wrong) logic, if I filled a pipe full of water on the moon then
I'd have EXACTLY the same amount of water molecules in a pip that was filled
under the atlantic ocean. This isn't true because the densities are
different.
The fact is that you can sent a few electrons or many electrons in a wire
and heat is generated from friction generated when electrons collide with
atoms in the the wire. This reduces the energy of the electron by an amount
gained from the atom. It increases the vibration energy of the atom and on a
global scale we sense this as heat which is felt from radition given off by
the atoms as they stop vibrating.
John O'Flaherty
Bill Beaty wrote:
{snipped}
> Voltage is like pressure difference. Current is like speed.
> And power transformers are like gearboxes.
Now I understand transformer oil.
--
john
Joel Kolstad
"John Fields" <jfi...@austininstruments.com> wrote in message
news:tlkta2h1gqtb9g1b7...@4ax.com...
> On Fri, 7 Jul 2006 14:12:46 -0700, "Joel Kolstad"
> <JKolstad7...@yahoo.com> wrote:
>> To avoid this confusion, people normally talk about "I-squared R"
>>losses in transmission lines, since the same current flows through the
>>transmission line as through the load.
> That's only true if there's no transformer or a 1:1 transformer
> between the line and the load.
As far power into the lin goes, the loads can just be reflected back through
whatever transformers exist so that sooner or later the *effective* load
current equals the input current.
I purposely simplified the discussion since the OP didn't appear that
experienced. Additional simplifications included ignoring the phase of
everything that's going on...
Jamie
has a lot to do with the size of the wire being used.
higher voltages can carry a load over greater distances
using smaller gauge wire. when you get to your location you
then use a step down xformer to get the desired amp and voltage
service you need.
when voltages are lower larger gauge conductors are needed.
--
Real Programmers Do things like this.
http://webpages.charter.net/jamie_5
John O'Flaherty
Abstract Dissonance wrote:
> "Bill Beaty" <bi...@eskimo.com> wrote in message
> news:1152308890....@b28g2000cwb.googlegroups.com...
> >
> > Abstract Dissonance wrote:
> >
> >> An analogy is like a freeway. If you just have a few cars going really
> >> fast
> >> then its usually not a problem... but what happens if you have a lot of
> >> cars
> >> going slow?
> >
> >
> >
> > Wrong, since the number of electrons in a wire doesn't change. Wires
> > are like pipes which are always full of water, and the "water" can
> > flow at different speeds, but the number of water molecules in the pipe
> >
> > remains the same.
> >
>
> huh. Do you know what current is? Current is the measure of the number of
> electrons passing through a cross section of wire per second. 1 Columb of
> current per second is 1 amp.
>
> So the number of electrons in a wire does change depending w.r.t to current.
The number of electrons in a wire depends on its voltage w.r.t. another
wire. The number in the pair of wires stays the same, but the negative
wire will have a slight excess of electrons over the positive one. This
will depend on the voltage between the wires and their spacing, and
will be independent of the current through them. If you have a pair of
wires at 1 cm apart, charged to 100V, they will have an imbalance of
electrons of about 34.7 billion per meter. For 12 ga. wire, the number
of electrons in the wire pair is about 1.6E25 per meter, so the
imbalance is really microscopic: 1 part in 5E14. This is true for a
charged pair of wires, whether there is zero current through them or
many amps. There'll be a change in number of electrons due to a current
that causes a change in voltage, but only in the tiny amount needed to
charge the wires' capacitance to the new voltage.
--
john
Tim Williams
news:1152316778.2...@s13g2000cwa.googlegroups.com...
>> Voltage is like pressure difference. Current is like speed.
>> And power transformers are like gearboxes.
>
> Now I understand transformer oil.
LOL! But what about ferro-resonant transformers? I don't see the ATF!!
Al
"Stephen-I-am" <ste...@theboulets.net> wrote:
One reason is that the loss in the lines is I time R squared. So, with a
higher voltage, the current is lower for the same power transmission.
R= resistance in ohms
I = current in amperes
Al
Joe Soap
B59B28.114...@news.verizon.net:
Oh dear. Power = I x I x R!
--
Joe Soap.
JUNK is stuff that you keep for 20 years,
then throw away a week before you need it.
John Fields
<ste...@theboulets.net> wrote:
>John Fields wrote:
>> On 7 Jul 2006 12:49:06 -0700, "Stephen-I-am"
>> <ste...@theboulets.net> wrote:
>>
>> >Slightly off topic, but something I was wondering about.
>> >
>> >How do you get less loss in the same transmission line with higher
>> >voltage? Wouldn't the loss scale as the square of the voltage?
>> >
>> >Please set me straight. ;)
>
>Thanks for the nice diagram John. I think that the fault in my
>reasoning was to assume that a higher transmission voltage meant a
>correspondingly higher line current. After all, if you increase the
>voltage into a fixed resistor, the current goes up.
---
True, but the generator is a constant-voltage source which supplies
whatever current the load needs through the transmission lines and
transformers in the system.
---
>So ... why doesn't the current go up?
---
It does, but it's load dependent.
---
>Is the impedance seen at the
>generator lower for a higher voltage, due to a different transformer?
---
No. Since the generator is a voltage source, its impedance should
approximate zero ohms, whatever the load.
Realistically, though, the generator will exhibit some finite
impedance which will limit its ability to deliver current into the
load, which will be the transmission line, all the transformers, and
all the loads on the transformers.
Rich Grise
> "John O'Flaherty" <quia...@yahoo.com> wrote in message
> news:1152316778.2...@s13g2000cwa.googlegroups.com...
>>> Voltage is like pressure difference. Current is like speed.
>>> And power transformers are like gearboxes.
>>
>> Now I understand transformer oil.
>
> LOL! But what about ferro-resonant transformers? I don't see the ATF!!
>
OK, I got the oil gag, but this one escapes me. Can you 'splain?
Thnaks,
Rich
John O'Flaherty
I think he means ferro-resonant ones are self-regulating, hence
automatic transmission fluid.
--
john
Robert Latest
Abstract Dissonance <Abstract....@hotmail.com> wrote
in Msg. <12ats00...@corp.supernews.com>
> Wires are not like pipes which are always full. Even if they were you can
> compress water and hence change the density => changing the number of
> molecules of water in the pipe.
Wrong on two counts. Conductors are always full of electrons --
independently of the current through them, and liquid water doesn't
appreciately change its volume under preesure.
robert
Rich Grise
Ah, got it.
Thanks!
Rich
Mark Fortune
> Wrong on two counts. Conductors are always full of electrons --
incidently, so are insulators, well except maybe hard vacuum and plasmas