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For example, an electrical network may have a transmission network of 110 kV/33 kV star/star transformers, with 33 kV/11 kV delta/star for the high voltage distribution network. If a transformation is required directly between the 110 kV/11 kV network an option is to use a 110 kV/11 kV star/delta transformer.
Rural electrification systems tend to use higher distribution voltages because of the longer distances covered by distribution lines (see Rural Electrification Administration). 7.2, 12.47, 25, and 34.5 kV distribution is common in the United States; 11 kV and 33 kV are common in the UK, Australia and New Zealand; 11 kV and 22 kV are common in ...
For example, a 100 miles (160 km) span at 765 kV carrying 1000 MW of power can have losses of 0.5% to 1.1%. A 345 kV line carrying the same load across the same distance has losses of 4.2%. [25] For a given amount of power, a higher voltage reduces the current and thus the resistive losses.
An example would be a distribution transformer with a delta primary, running on three 11 kV phases with no neutral or earth required, and a star (or wye) secondary providing a 3-phase supply at 415 V, with the domestic voltage of 240 available between each phase and the earthed (grounded) neutral point.
For example, in the United States, the most common voltage is 12.47 kV, with a line-to-ground voltage of 7.2 kV. [7] It has a 7.2 kV phase-to-neutral voltage, exactly 30 times the 240 V on the split-phase secondary side.
As an example of how per-unit is used, consider a three-phase power transmission system that deals with powers of the order of 500 MW and uses a nominal voltage of 138 kV for transmission. We arbitrarily select S b a s e = 500 M V A {\displaystyle S_{\mathrm {base} }=500\,\mathrm {MVA} } , and use the nominal voltage 138 kV as the base voltage ...
The solutions to the long line transmission equations include incident and reflected portions of the voltage and current: = + + = / + / When the line is terminated with its characteristic impedance, the reflected portions of these equations are reduced to 0 and the solutions to the voltage and current along the transmission line are wholly ...
Now V line drop = IZ line is nonzero, so the voltages and the sending and receiving ends of the transmission line are not equal. The current I can be found by solving Ohm’s law using a combined line and load impedance: I = V S Z l i n e + Z l o a d {\textstyle I={\frac {V_{S}}{Z_{line}+Z_{load}}}} .