Single line diagram

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Single line diagram

Transcript of Single line diagram

Page 1: Single line diagram

Single line diagram

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Representation of essentials of a system in a most simplified form.

Main components of a power plantGeneration plantTransmission lineDistribution lineIndustrial, commercial and residential load

Nowadays all of these are in the form of three phase

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A one-line diagram or single-line diagram (SLD) is a simplified notation for representing a three-phase power system.

The one-line diagram has its largest application in power flow studies.

Electrical elements such as circuit breakers, transformers, capacitors, bus bars, and conductors are shown by standardized schematic symbols.

Instead of representing each of three phases with a separate line or terminal, only one conductor is represented.

It is a form of block diagram graphically depicting the paths for power flow between entities of the system.

Elements on the diagram do not represent the physical size or location of the electrical equipment, but it is a common convention to organize the diagram with the same left-to-right, top-to-bottom sequence as the switchgear or other apparatus represented.

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Impedance diagram and reactance diagramSynchronous generator are replaced by:Transformer is replaced by:Transmission lines are replaced by:

In many power system studies synchronous generator resistance, resistance of transformer windings, resistance of transmission lines, line charging and magnetizing circuits of transformer are neglected.

The impedance diagram then become reactance diagram

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Example 8.2A three phase synchronous generator delivers 10 MVA at a

voltage of 10.5 kV. The line impedance is 5 ohm. Determine the voltage drop in the line in per unit and in volts. Use the reference base as 10 MVA at 11 kV.

S3φ= 10 MVA

(Sb)3φ = 12MVA

Vl=10.5kV

Vlb= 11kV

Vpu=(Vl)pu= Vl/Vlb=10.5/11

Spu=S3φ/(Sb)3φ=10/12

Ipu=Spu/Vpu=10/12*11/10.5= 0.8730158

Zb= Vlb2/ (Sb)3φ =112 /12

Zpu= ZΩ/Zb=5*12/ 112 =0.4958677

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Voltage drop in line per phase:ΔVpu= Zpu* Ipu= 0.4958677 *0.8730158=0.4329pu

ΔVpu=Vactual/Vb

ΔVactual= ΔVpu*Vb=0.4329*11kV/root(3)=2749.36 V