power distribution and utilization

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Power Distribution

and Utilization

LABORATORY MANUAL NO.12

Submitted By: Samrah Imtiaz

Submitted To: Rizwan Khan

Roll no.: 2011-EE-312 Submission Date: 7th May, 2014



LT ANALYSIS OF DISTRIBUTION NETWORK WITH CAPACITOR

INSTALLATION USING FDR-ANA

FDR-ANA:

FDR-ANA is a text based software tool which is used for the analysis of the HT and LT power systems.

The information about the power system is input to the software in a table based format, describing theinterconnections of conductors and cables. The software performs the calculations and displays theresult. The software is presently used by the Water and Power Development Authority in Pakistan to

perform the HT and LT feeder analysis. The information provided by the feeder analyzer software can

be summarized as:

Line Losses.

Transformation Losses

Total Power losses

% Power Losses

%annual Energy Losses

Transformer Loading Total length of the line

Total KVA

What is power factor?

Power factor is the relationship between working (active) power and total power consumed (apparent power). Essentially, power factor is a measurement of how effectively electrical power is being used. Thehigher the power factor, the more effectively electrical power is being used. A distribution system's

operating power is composed of two parts: Active (working) power and reactive (non-working magnetizing) power. The ACTIVE power performs the useful work - the REACTIVE power does not. It's only function

is to develop magnetic fields required by inductive devices.

Why improve low power factor?

Low power factor means poor electrical efficiency. The lower the power factor, the higher the apparent power drawn from the distribution network. When low power factor is not corrected, the utility must provide

the nonworking reactive power in addition to the working active power. This results in the use of largergenerators, transformers, bus bars, wires, and other distribution system devices that otherwise would not benecessary.

How can a capacitor improve the power factor?

A poor power factor is usually caused by coils; by inductors within electric motors. A capacitor is theopposite of a coil, and can improve the power factor by moving the current phase angle forward, moretowards the voltage. Power factor degradation from inductive loads occurs because inductors resist a change

in current, by "presenting" a higher resistance to a step (or any) change in current. As a result, current lagsvoltage, and power factor suffers. Capacitors are opposite to inductors. They resist a change in voltage, by

"presenting" a lower resistance to a change in voltage. As a result, current leads voltage.



Suppose that a capacitor is placed in parallel to an inductor. If the value of capacitance is adjusted so itexactly cancels the inductance at 60Hz, then the combination as a whole behaves purely as a resistor.

If capacitive loads were common, then they would cause problems similar to inductors.

However, adding capacitance in a case where there is also inductance serves to raise voltage, particularlysince the conductors (power lines, etc.) leading up to the load are also resistive, inductive, and partially

capacitive. By raising the voltage, power factor is improved, and the inductive loads (usually motors) causefewer losses in power lines.

Procedure

First step is to run the exe file.

The following window will appear in front of you.

Create a new file.

Enter the file name.

Select the distance in feet so type F.

Enter the data from single line diagram.

Press Pg Dn and then F to finish. Enter the value of current, voltage, load factor and power factor.

If study with growth is required than press Y otherwise press N.

Than option of installation of capacitor. Press Yes we have to analyze the system with the

installation of capacitor.

Enter the values of current and power factor etc.

Then tell the capacitor bank size that is normally of 20 kVAR.

The feeder analyzer will tell you the proposed nodes on which there is a need to install capacitors.

Enter all these nodes.

Then display result.

Press S for summary.



Now we have to improve our power factor and for that purpose we should know the size of

capacitor bank in KVARs.

We will use the formula :

KVAR = KW (tan (angle 1)-tan (angle2))

Where KW is actual power and we can calculate it by summing all the lodes present on the node nearestto the power grid

Angle1 can be calculated from the power factor that we use in previous analysis

Angle2 can be calculated from the power factor that we want to achieveAnd that is 0.9 where previous power factor is 0.85.

Again proceed in the same way as we have before but with different capacitor bank size.

The new results are:



And the summary is:

power distribution and utilization

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