

Voltage and Efficiency
Name: Jerry B.
Status: Educator
Age: 20s
Location: N/A
Country: N/A
Date: May 2002
Question:
If all else is equal, (as it never is), is higher voltage
more efficient than lower voltage? Power is lost according to R/I
squared , I think, so higher voltage will lose less in
transmission. Other than that, is a higher voltage motor more efficient?
Replies:
Power is lost according to I*I*R, the current squared times the
resistance it passes through. So its best to minimize the current. Since
power is (simplified a bit) voltage * current, then if you wish to keep
the power transfer the same, and drop the current, you must raise the
voltage. Hence the high voltage lines which cross the country  high
voltage, low current, lower transmission losses.
The speed of an electricity really means the speed at which a wave of
electric potential goes down a transmission line. This is fairly close
to the speed of light (off by a constant factor, like 2/3, which has to
do with the geometry of how the wires are strung, and is pretty similar
for a lot of different wires, do not get caught up in this!)
You do not need to know any SR or GR for this statement, after all the
starting point for SR is that the speed of light is a constant (in free
space  again it slows down in various media). All this is covered in
"classical" electromagnetic field theory, worked out in the 1880's.
Steve Ross
If you transmit power at a rate of P watts, using a voltage V and a
current I, then P = IV. If the transmission line has a resistance R,
the power wasted heating the wires is IIR (I squared R losses).
Plugging in the first equation gives waster power = (PPR)/(VV).
So to transmit power P over a line with resistance R, the power lost
get reduced as the voltage is increased by 1/(VV), so as you say, less
power is lost in transmission at higher voltages, which, of course,
why long distance transmission lines are operated at very high
voltages, often in excess of 1 million volts.
Motors are more complicated, since magnetic fields are proportional to
the current, so high currents (or many turns of wire) are necessary
The speed of electricity does not change (much) as the voltage is
increased. When you close a light switch, an electric field is set up
through the wire and the light bulb at nearly the speed of light.
However, the electrons whose motion constitutes the electric current
move only very slowly in the direction the electric field pushes them,
only about 0.0001 m/s! Electrons are mostly moving randomly inside
the wire, similarly to gas molecules. Their random motion is at
speeds average some 1.0E6 m/s (= 1,000,000 m/s).
What is S R and G R?
Feel free to send in your questions at any time! I hope you find my
response helpful. Good luck with your education group!
Best, Dick Plano
Jerry,
I apologize in advance if my explanation sounds over simplified. But we
try to write our responses such that they can be understood by as large an
audience as possible.
Transmitting electrical power is done much more efficiently at higher
voltages due to (I^2*R) losses. If you recall, Power = I^2*R. Lets
compare two cases, A and B for transmitting power to a small residential
community, at the rate of say, 1 MW. In other words, the generating
station needs to supply whatever power it takes such that 1 MW arrives at
the end user (compensating for I^2*R losses).
Case A: (1,000 Volt @ 1,000 Amperes) > Power = 1 MW (1 million watts)
Case B: (10,000 Volts @ 100 Amperes) > Power = 1 MW (1 million watts)
R = electrical resistance = 0.0001 Ohms / foot.
L = length of transmission line = 1 mile = 5280 feet.
Rt = total resistance = 0.528 Ohm
P(loss, CASE A): = (1,000 amperes)^2 * 5.28 Ohm = 0.528 MW
P(loss, CASE B): = (100 amperes)^2 * 5.28 Ohm = 0.00528 MW
Since Power loss is proportional to I^2, stepping up the voltage by a
factor of 10 will reduce power losses by 10^2.
Psupply required (CASE A): 1.52800 MW (not reasonable)
Psupply required (CASE B): 1.00528 MW (more reasonable)
** MOTOR EFFICIENCY **
I dont have enough experience to answer your motor question, but I will
guess. I would be willing to bet that a 120 V motor would be less
efficient than a 240 V motor. I might be wrong though.
** ELECTRON VELOCITY **
I like your thinking. I believe this is where the analogy fails. To the
best of my knowledge, voltage has very little role in electron propagation
velocity. Velocity, however, is very dependent upon the
type/configuration/thickness of the insulating dielectric. For a very
rough estimate I believe electricity (the field) travels in the range of
about 0.66 to 0.9 times the speed of light in a conductor.
Darin Wagner
Your first question about the efficiency of transmission lines is
outside my expertise. The bottom line is that electricity (alternating
current) is transmitted at high voltage, so there must be some economies in
doing so. I suspect the answer is more complicated than Ohm's Law.
Your second inquiry about the "speed of electric current" is also more
complicated than it appears at first glance, because the "speed of the
electric current" does not have a single definition. My search for an answer
led me to the website: http://www.amasci.com/miscon/speed.html which
discusses your question in some detail. Rather than paraphrasing that
discussion I think going to that site would be more helpful to you.
Both of your questions are very good  the answers, however, are not as
simple as the questions!!!
Vince Calder
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Update: June 2012

