Question about Transmission Line Losses

[Techterp]

Hi there,

If I'm using the equation P_loss = P^2 * R / V^2 to calculation the power lost on a cable, where R = p(resistivity) * L(length) / A(cross sectional area), then how long is this power loss for exactly?

Thanks,
Sydney

 

[Windadct]

I think the equation "shakes out" to just P_{loss = I²} * R

What I do not understand is "then how long is this power loss"

 

[ElectricRay]

Where did you find this equation as Windadct said you should have:

$$P_{loss} = I^2 * R$$

Or:

$$P_{loss} =\frac {V^2} R $$

Note that this voltage is the voltage drop accros the transmission line. Than if you know R and the copper section you could calculate the length of the cable for example. There is no such thing as the length of the power loss. The losses that exists on a transmission line are depending on the physical properties of the line hence the equation you wrote:

$$R = \rho * \frac l a$$

For this reason energy companies transform the voltage up so the current that flows thru the cable will be lower (the transported power is constant in this case $S = U*I$ ) and they can use smaller (thinner cables). This is easier for practical use.

 

[donpacino]

Hi there,

If I'm using the equation P_loss = P^2 * R / V^2 to calculation the power lost on a cable, where R = p(resistivity) * L(length) / A(cross sectional area), then how long is this power loss for exactly?

Thanks,
Sydney

When I was an undergrad I was confused by this too. power is instantaneous.
I guess the actual answer to your question is however long that voltage is held over the wire!

That said this might help you:

P = J/s

power is equal to one joule (a unit of energy) per second (a unit of time)

 

[Averagesupernova]

Is there confusion concerning how LONG the transmission line is in actual length? If not i would have to simply agree with everyone who had replied so far.

 

[Windadct]

The shaking out

P_loss = (P² * R) / V₂ = ( ( V * I ) ² * R )/ V² = (V² * I² * R )/ V² = I² * R

So you can start with the Power at the source and drop out the V, and be left with current flow.

 

[ElectricRay]

The shaking out

P_loss = (P² * R) / V₂ = ( ( V * I ) ² * R )/ V² = (V² * I² * R )/ V² = I² * R

So you can start with the Power at the source and drop out the V, and be left with current flow.

Ok clear Windadct I see now where the formula from the OP comes from. Thanks!

 

[russ_watters]

Hi there,

If I'm using the equation P_loss = P^2 * R / V^2 to calculation the power lost on a cable, where R = p(resistivity) * L(length) / A(cross sectional area), then how long is this power loss for exactly?

Whatever length you plug in...unless I'm misunderstanding your question...

 

[Babadag]

If by "power loss “you meant actually "a power" it does not depend on duration. Considering
the load power flowing through conductor is constant and supplied voltage it will be constant
the power loss will be n* I^2*R [n=2 for single phase and 3 for three-phase].
If by "power loss “you meant actually "energy loss" it does depend on duration.
W=Ploss*t [or better W=integral(ploss*dt)].
The power is measured in watts[W] and energy in J[Joules]- or kWh[kilowatt-hour]for instance.
The conductor resistance depends on some factors as temperature-and if it is about a.c. current also on skin effect and proximity effect.
In a transient heating process the heat produced by conductor losses-for high voltage cable there are insulation losses also, shield losses and armature losses-will determine the conductor temperature rise and only a part of the losses will be evacuated.
The evacuated power depends on conductor temperature and will rise accordingly.
Now since the conductor temperature depends on power losses and heat evacuation in a steady state the power losses have to be equal to the evacuated power and then the conductor temperature will be steady [and maximum].The evacuated power-what it is called "cooling power” depends on conductor temperature, ambient temperature-usually air or underground-and the thermal resistance of the insulation and other sheathes and jackets, conduits and so on up to free air or earth. Also depends on other heat sources from vicinity-usually other cables.