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?!
Electronic – The Relation of power heat losses with V & I in transmission lines
heathigh voltagepowerpower-engineeringstep-down
Related Solutions
Placing two or more transmission lines (transformers, cables, ...) in parallel is a common way to obtain increased capacity and redundancy against single failures. It also lets you take one line out of service for maintenance, while the second line carries the load.
Mathematically, you would obtain the same results by combining the parallel transmission lines, but you would lose the ability to simulate a scenario where only one of the two transmission lines is in use. Simulating contingency scenarios ("what happens if one line has a fault?") is one of the main reasons we do power systems analysis, so I would tend to model the two transmission lines as two separate elements.
In many cases, if you encounter zero reactance or zero impedance, this often represents bus ties or breakers. However, for the IEEE 14-bus test system, that is not the case.
As you can see from the numbers, the reactance is quite high. Obviously a breaker or a short cable can not have zero resistance and a high reactance. Reactors used for reactive power compensation are usually modeled at the buses and not at the branches. It could, as Li-aung suggests be a series reactor used for limiting fault currents. However, it seems to be a bit large for that purpose. Also, I believe these are usually used between breakers and loads, not between different buses in a meshed grid.
Although it's not a physically correct representation, these numbers represent transformers. Transformers do of course have resistance, but compared to the reactance this is close to negligible (at least when it comes to power flow).
From To R X B
Bus Bus pu pu pu
4 7 0.0 0.20912 0.0
4 9 0.0 0.55618 0.0
5 6 0.0 0.25202 0.0
As you can see from the single line diagram below, the branches with zero resistance are transformers.
A common problem in power flow calculations is when impedance values are set to zero. This is because the current through a branch in general is given by the voltage difference divided by the impedance. When the impedance is zero, this gives an infinite current, thus an unfeasible power flow solution. It is common practice to set the impedance values to a very small value in such cases (for instance R = 0, X = 0.0001 pu). In AC systems, zero resistance is often not a problem, but zero reactance is, as this will cause the decoupled load flow, and the DC load flow algorithms to fail.
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Best Answer
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).
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.