Power Loss Equation
nahuatl...@gmail.com
physics and need all the help I can get.
In the equation for electrical power through a line, P = V*I. By Ohm's
Law, V = I*R, so P = (I*R)*R = I^2*R. So it makes sense that to reduce
loss, we'd want to make the current to be as small as possible while
delivering the same amount of power, by increasing the voltage.
But we can also rewrite the equation in terms of voltage, such as the
following: P = V*I = V*(V/R) = V^2/R. So it seems power loss also
increases with voltage, as one can guess from P = V*I as the V and I
are symmetrical!
What's going on? I suspect I have conflated terms here, where there
are different meanings of voltages and powers in the equations. As you
suspect, I was taught and learned physics by plugging in numbers into
formulas, and it's only now I am relearning the subject to actually
understand the _physical_ phenomena.
nahuatl...@gmail.com
bert
You are confused between the total power delivered,
which is roughly V*I when V is the line voltage,
and the power lost in transmission, which is
(dV)*I when dV is the voltage drop from one end
of the transmission line to the other.
A high supply line voltage (e.g. 440,000 volts)
delivers the required power at a low current
which minimises the transmission loss.
--
OPPT
See discussion of line loss here:
<http://www.bsharp.org/physics/stuff/xmission.html>
Also note the reason for choosing AC versus DC.
gjed...@gmail.com
For the *same* current, power loss *does* increase for higher voltage,
but this issue is about the V/I ratio for a *given* power. It's
generally better (though its strictly more complicated than this) to
have a bigger V (which since P=I*V allows smaller I for the same
power).
Randy Poe
My $.02: When you make the substitution V = I*R or I = V/R in
reference to the power line, then I is the current through the
transmission line, R is the resistance of the transmission
line, and V is the voltage drop ACROSS THE TRANSMISSION
LINE. It is not the voltage of the source, or the voltage
across the thing plugged into the line (the "load").
So it's a mistake to think that a higher voltage source means
a higher value for this particular V. On the contrary, when
you use a higher voltage source and have less current
through the lines, you'll have a lower voltage drop on
the lines and the voltage delivered to the load will be closer
to that at the source.
Ex: Suppose the line resistance is 1 Ohm, the source
voltage is 10 kV and the current drawn is 1 Amp (delivering
~1 kW of power at the output). Then the voltage drop on
the line is 1 V and the voltage delivered is actually 9999 V.
The power loss in the line is (1 V)^2/1 Ohm = 1 Watt.
Now suppose the source voltage is 20 kV. The current
drawn is only 0.5 Amp to deliver that same 1 kW.
The voltage drop on the line is 0.5 V and the voltage
delivered is 19999.5 V. The power loss in the line is
(0.5 V)^2/ 1 Ohm = 0.25 Watt.
- Randy
Temp
Previous replies have likely answered your question adequately.
JIC things are still not clear, here are some more references that
explain the situation in a simple way.
<http://www.newton.dep.anl.gov/askasci/phy00/phy00347.htm>
<http://www.eejitsguides.com/environment/elec-transmission.html>
(and the Next page)
<http://www.eejitsguides.com/environment/elec-voltage.html>
Mohan Pawar
news:1191942309.7...@22g2000hsm.googlegroups.com...
Consider we want to supply power Ph to a home-equipment. Further,
assume that the electric power comes from power
station say many miles away by transmission lines with resistance
Rt. Therefore power required by home equipment is
given by
Ph = I*V, where subscript h shows variables related to home. V and
I have no subscripts because they are same for
home and power station since we didn't use transformers and if
charge didn't accumulate or drain in a power line then
constant current would flow through it from power station to home.
Now, if P_loss = the power loss due to heating of transmission lines
then
P_loss = Rt*I^2 = Rt*Ph^2/V^2
Now, one can see that to minimize P_loss, the practically useful
equations is P_loss= Rt*Ph^2/V^2 because it has two
constraint variables Rt and Ph that cannot be changed. To reduce
Rt, there is cost to pay for making transmission lines
thick. Ph is the power required to be delivered that cannot be
changed. Hence only alternative is to maximize V and that
is exactly what the step up transformers do. They bring high voltage
to city and then step it down as needed for domestic
or industrial use.
--
Best regards
Mohan Pawar
MP Classes LLC
www.mpClasses.com
US Central Time: 1:13 PM 10/9/2007
CWatters
<nahuatl...@gmail.com> wrote in message
news:1191942176....@o3g2000hsb.googlegroups.com...
> I apologize if this seems like a basic equation, but I am relearning
> physics and need all the help I can get.
>
> In the equation for electrical power through a line, P = V*I. By Ohm's
> Law, V = I*R, so P = (I*R)*R = I^2*R. So it makes sense that to reduce
> loss, we'd want to make the current to be as small as possible while
> delivering the same amount of power, by increasing the voltage.
>
> But we can also rewrite the equation in terms of voltage, such as the
> following: P = V*I = V*(V/R) = V^2/R. So it seems power loss also
> increases with voltage, as one can guess from P = V*I as the V and I
> are symmetrical!
no that last bit is wrong...
Power loss is actually ((Vstart - Vend)^2)/R
where Vstart is the voltage at the start of the power line and Vend is the
voltage at the end.
jmor...@idirect.com
To give the same answer as others have given...
Both equations are correct, but the missing point is: when the
current is flowing, how much of the voltage drop is across the LINE
and how much is across the LOAD?
Randy Yates
For one thing you're oversimplifying. The model you're implying (in
simple DC terms) is a transmission line resistance in series with a
load resistance, but you didn't (at least not explicitly) mention
these.
The thing you've missed is as follows. Let's use "w" for "wire" (line)
and "l" for "load." So we have Pw, Vw, Rw, Pl, Vl, Rl, but only one I
since Iw = Il (the wire is in series with the load).
Yes, decreasing the current through the line decreases the amount of
power dropped across the line. Pw = I^2 * Rw.
But in order to maintain the same power delivered to the load, Rl must
increase. Pl = I2^2 * Rl2 = I^2 * Rl, or Rl2 = I^2 * Rl / I2^2, where
I2 is the lowered current and Rl2 is the increased load resistance.
Now what happened to the voltage across the line, Vw2? In the first
scenario, the current is I and the voltage across the line is
Vw = I * Rw.
In the second scenario the current is I2 and the voltage across
the line is
Vw2 = I2 * Rw.
Since I2 < I, then Vw2 < Vw, and the overall power through the
wire is reduced.
--
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