2
\$\begingroup\$

The voltage is stepped up at the beginning of transmission lines and step down again at the end of the latter. This is, of course, to reduce the current passing through the line and thus reducing the heat losses. \$P\$ loss \$= I^2 \cdot R\$. but also this equation can be rewritten as \$P\$ loss \$= V^2 / R\$. So, if we raised the voltage difference across the whole transmission line we also will increase the losses, according to the second equation. I know this happens to be false. But, what is exactly wrong with that way of thinking?!

\$\endgroup\$
7
\$\begingroup\$

schematic

simulate this circuit – Schematic created using CircuitLab

Figure 1. A comparison between power transmitted (a) directly and (b) via step-up and step-down transformers on the same pair of wires as (a).

  • In Figure 1a the power loss per cable is given by \$ P = I^2R = 1^2\cdot 1 = 1\,\mathrm{W} \$ (assuming they're in phase).
  • In Figure 1b the power loss per cable is given by \$ P = I^2R = 0.1^2\cdot 1 = 0.01\,\mathrm{W} \$ (again, assuming they're in phase).

By using a 1:10 step-up and 10:1 step-down line losses are reduced by a factor of 100.

In general, line losses will be reduced to \$ \frac {1}{n^2} \$ where n is the transformer ratio.

\$\endgroup\$
  • 1
    \$\begingroup\$ The current feeding a transformer can not be in phase. It’s an inductance. One should specify conditions where the (reactive) load created by the transformer can be neglected. \$\endgroup\$ – Incnis Mrsi Sep 2 '16 at 9:08
  • 2
    \$\begingroup\$ It most certainly can be in phase. It has to be for any real power to be transmitted. In my job I apply power-factor correction to ensure that the voltage and current drawn from the utility are as close to being in phase as I can. \$\endgroup\$ – Transistor Sep 2 '16 at 9:54
  • 2
    \$\begingroup\$ In this case magnetizing current is a detail which has really to be kept off. It has nothing to do with the basic idea of "reducing losses rising voltage" and makes just pointless confusion. Once the basics is clear one can add one by one more details as required/needed. \$\endgroup\$ – carloc Sep 2 '16 at 10:12
  • \$\begingroup\$ @carloc: Ī̲’d agree with your point in general, but Transistor presented “damnable” (Harold P. Brown) transformers. With coils and magnetic core. \$\endgroup\$ – Incnis Mrsi Sep 2 '16 at 10:17
4
\$\begingroup\$

1- the losses are equal to V^2/R not V^2*R.

2- V is the voltage drop across the the transmission line not the actual (stepped up) voltage of the line and it's a very small value comparing to the actual voltage of the line.

3- The voltage drop across the line depends on the resistance of the line and the current passing through it V=I*R.

4- When the voltage is stepped up , the current required to deliver a constant power decreses (P=V*I).so, the drop across the line decreses and then the losses also decrease.

\$\endgroup\$
2
\$\begingroup\$

You're using the wrong V. The V you should be using is the V of V = I*R where I is the current through the conductor and R is the resistance of the conductor. With higher voltage comes less current. With less current comes less voltage dropped along the length of the conductor. With less voltage dropped comes less power dissipated.

\$\endgroup\$
1
\$\begingroup\$

For transmission loss with a fixed termination voltage, the loss would be;

ΔP = ΔV²/R

then by increasing input by x%, the incremental Power ΔPi loss becomes

ΔP = ΔV(1+x)²/R assuming a fixed R

But in North America, Line voltage is regulated to within 5% for HV Transmission and 5% for Distribution and drop cables.

This can be done in substations using active tap switching and D.T.'s using either fixed or active tap changers.

But effectively it means the load regulation is 10% which implies the Zout of the network is 10% of the total load impedance in the network.

For every x% increase in line voltage drop! there would be a 2x% rise in power loss.

\$\endgroup\$
  • \$\begingroup\$ @IncnisMrsi: % is "percent" which means "parts per hundred". It's a convenient way to express small fractions. \$\endgroup\$ – Dave Tweed Sep 2 '16 at 12:38
  • \$\begingroup\$ Normally power industry uses a ratio of per unit or p.u. to normalize ratios. I prefer this method but, decided to show as an incremental delta of x% \$\endgroup\$ – Sunnyskyguy EE75 Sep 2 '16 at 12:57
  • \$\begingroup\$ OK. Make your variable and related math to use either p.u. or % consistently. \$\endgroup\$ – Incnis Mrsi Sep 2 '16 at 13:00
  • \$\begingroup\$ @IncnisMrsi: If you're going to be this nit-picky, you need to be a lot more explicit about the point(s) you're trying to make. This passive-aggressive approach is starting to irritate people and create a lot of extra work for the moderation team. I'd recommend taking a break from SE for several hours. \$\endgroup\$ – Dave Tweed Sep 2 '16 at 13:15
  • \$\begingroup\$ @Dave Tweed: Ī̲’m trying to say that, while x’s numerical value is expressed in percent (1/100), then the value in p.u. shall be x/100 and “(1+x)” is an evidently bad math. \$\endgroup\$ – Incnis Mrsi Sep 2 '16 at 13:19
1
\$\begingroup\$

This question resonated with me because I also tripped on it when I first studied circuits. Plain and simple, "what is wrong exactly" in that argument is that the correct voltage in the formula for the line losses is not \$V\$ (understood as the voltage to ground), but the voltage difference across the line (which is usually much lower compared to \$V\$).

The formula \$P=V I\$ can be very misleading. It would be better to always write it as \$P=\Delta V ·I\$, where \$\Delta V\$ is the voltage drop across the device or section of the circuit for which you want to calculate \$P\$.

Longer explanation:

The correct argument for higher voltages producing lower transmission losses goes as follows. Take a simple circuit consisting of an ideal DC generator with voltage \$V_0\$, a transmission line having resistance \$R\$, and a load having resistance \$R_{load}\$. Let us label the voltage at the load as \$V_{load}\$. The circuit is readily solved: $$ I = \frac{V_0}{R + R_{load}} $$

The voltage drop across the line is therefore: $$ (V_0 - V_{load}) = I R $$

So that the power losses on the transmission line are: $$ P_{loss} = (V_0 - V_{load}) I = I^2 R $$ Note that the formula here is not \$P_{loss}=V_0 I\$; actually, \$V_0 I\$ is the total power (line+load power) given by the generator. The power consumed by the load is: $$ P_{load} = V_{load} I = (V_0 - IR)I = V_0 I - I^2 R $$ That is, the load takes all the power delivered by the generator, minus the transmission losses.

Now, to complete the argument, it is necessary to point out that in a power system the load resistance \$R_{load}\$ is not really constant. Rather, it is the power demand \$P_{load}\$ that remains roughly constant against changes of voltage \$V_{load}\$. Then it is easy to see by looking at the formulas above that by increasing \$V_0\$ (using transformers), we reduce \$I\$, and therefore \$P_{loss}\$ is reduced with respect to \$P_{load}\$. If you want to follow through the whole mathematical solution, you need to solve the so-called power-flow equation, which is not overly difficult to do (it's a second-degree equation on the voltage).

\$\endgroup\$
  • \$\begingroup\$ best answer. yea , I have mistaken the two voltages as you've mentioned . \$\endgroup\$ – Fadi Sep 3 '16 at 20:04
0
\$\begingroup\$

On "What's wrong with thinking that way". For a given Power output you need, think about "few packets (Q) with high energy (low I, high V)", Versus "many packets with low energy (high I, low V). Now think you have to move those packets through a long wire with non-null resistance.

That's why home power supply in Europe had been turning towards a higher voltage delivering model as technology allowed to do it safely.

\$\endgroup\$
  • 2
    \$\begingroup\$ (1) We don't think of AC power as "packets". It's continuous current. (2) European voltages haven't changed little in 100 years other than some convergence to 230 V (where some used 220 V). This was for standardisation rather than technological improvement. \$\endgroup\$ – Transistor Sep 1 '16 at 21:38
  • \$\begingroup\$ AC power, stands for power delivered via "Alternating Current". The packets of Energy are the electrons, measured in Coulombs (Q). \$\endgroup\$ – user2807874 Sep 2 '16 at 8:17
  • \$\begingroup\$ Electrons and their charge are irrelevant to the question posed. That’s a plain electromagnetism + the Ohm’s law. No electronics. \$\endgroup\$ – Incnis Mrsi Sep 2 '16 at 9:24
  • \$\begingroup\$ Electrons and their "voltage to the ground" (not their charge), are at the core of this question. As some people has greatly explained, the bigger the 'V to ground', the smaller the 'I' to deliver the same ammount of power, and the smaller the 'I', the smaller the loss of energy within the wire (V drop on the wire). \$\endgroup\$ – user2807874 Sep 2 '16 at 16:52
  • \$\begingroup\$ By the way, "Voltage to the ground" is a measurement akin to "potential ENERGY" in mechanical engineering. \$\endgroup\$ – user2807874 Sep 3 '16 at 0:13
0
\$\begingroup\$

your question is good and challenging, but it has a simple answer, in a power transmission lines the sending power is Ps=VsIs (S for abbreviation of sending), therefore because we wanna have a maximum power without any changing, Ps consider to be constant, so if we increase the voltage (Vs) with transformer, the current (Is) will decrease, then in a transmission line, voltage drop is DeltaV=ZIs , therefore power losses with equation Ploss=((deltav)^2)/Z will be decrease. I think this question can be a question for alot of students, and i like it.

\$\endgroup\$
-1
\$\begingroup\$

In the current (amps) equation, R relates to series resistance of the wire. In the case of volts, R relates to leakage resistance from the wire to ground. Two different resistances entirely.

Having said that if the leakage losses are X then doubling the voltage makes the losses 4X so, there does come a point when the two losses may become equal but, there would be sparks flying before that happened.

\$\endgroup\$
-1
\$\begingroup\$

Original question stems from confusion about where exactly the Ohm’s law should be applied and what namely does “\$R\$” refer to.

There are, roughly, three sources of power losses related to transmission, that are determined by:

  • current in lines,
  • transmission voltage,
  • conversion (in a broad sense, AC/AC included).

Losses determined by current are mostly (but not exclusively) due to wire resistance. The Ohm’s law gives \$I^2\cdot R_{\mathrm{wire}}\$, right. But \$I\cdot R_{\mathrm{wire}}\$ has nothing to do with the transmission voltage – it’s the voltage drop in wires.

Losses due to transmission voltage (such as currents leaking from/between wires) may play a noticeable rôle in cables, but can be neglected in overhead power lines until the isolation breaks down.

There are also losses directly and indirectly inflicted by conversion. Even a simple transformer has inherent power losses due to its magnetic core. Moreover, a transformer uses an out-of-phase (reactive) AC current that put an additional term to \$I\$ of the line feeding it (hence increasing ohmic losses to some extent). So, unlike the model presented in Transistor’s answer, transformers do not act for free. But, for sufficiently long distances (where \$R_{\mathrm{wire}}\$ is of significant value) and heavy loads stepping the transmission voltage up pays off. Decrease in ohmic losses can be much larger than increase in all other ones.

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.