AC Load Flow Analysis

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1.0

1.0 Scope

Scope of

of a

a Load

Load Flow

Flow Analysis

Analysis

A Load Flow analysis is used to evaluate the running steady-state condition of a system, under different A Load Flow analysis is used to evaluate the running steady-state condition of a system, under different load, supply and configuration conditions. The idea behind the load flow algorithm is to determine the load, supply and configuration conditions. The idea behind the load flow algorithm is to determine the following parameters, given that the load consumption at all busses is known:

following parameters, given that the load consumption at all busses is known: a.

a. Power Power flow flow in in each each branch branch (Lines (Lines & & Transformers).Transformers). b.

b. Power Power consumption consumption at at each each source source of of generation.generation. c.

c. Voltage Voltage magnitude magnitude and and electrical electrical angle angle at at each each bus.bus. d.

d. System System Losses.Losses.

Once this data is known, corrective actions can be taken in order to ensure that every load is supplied Once this data is known, corrective actions can be taken in order to ensure that every load is supplied with power that is within the acceptable limits of operation. Such actions can include the calculation of: with power that is within the acceptable limits of operation. Such actions can include the calculation of: a.

a. Transformer Transformer tap tap settings.settings. b.

b. Reactive Reactive power power compensation.compensation. c.

c. Generator Generator set set points.points. d.

d. Emergency Emergency power power requirements.requirements.

A properly built load flow model of the system, lays a solid foundation for other important studies such A properly built load flow model of the system, lays a solid foundation for other important studies such as short circuit, motor starting, harmonic and transient stability. The load flow model constitutes the as short circuit, motor starting, harmonic and transient stability. The load flow model constitutes the basic network information and the initial

basic network information and the initial steady-state condition which these studies rely upon.steady-state condition which these studies rely upon.

2.0

2.0 Load

Load Models

Models

Loads in general can be modeled in 4 different ways: Loads in general can be modeled in 4 different ways: a.

a. Constant Constant Power Power Load Load (kVA)(kVA) b.

b. Constant Constant Current Current (Amps)(Amps) c.

c. Constant Constant Impedance Impedance (Ohms)(Ohms) d.

d. Functional Functional LoadLoad

Notes: Notes: ………. ………. ………. ………. ………. ………. ………. ………. ………. ………. ………. ………. ………. ………. Page LF.1 Page LF.1

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2.1 2.1 Constant

Constant Power

Power Load

Load (kVA)

(kVA)

This type of load is treated as constant demand of power. The load flow algorithm will attempt This type of load is treated as constant demand of power. The load flow algorithm will attempt convergence under the premise that these types of loads always draw the kVA specified by the user. convergence under the premise that these types of loads always draw the kVA specified by the user. Caution should be exercised when interpreting the convergence results in the presence constant kVA Caution should be exercised when interpreting the convergence results in the presence constant kVA loads. To illustrate this, let’s assume that a given load is rated 1.2kVA at 120 Volts; let’s further assume loads. To illustrate this, let’s assume that a given load is rated 1.2kVA at 120 Volts; let’s further assume that the system experiences excessive losses (say 20% voltage drop on the load’s supply branch). that the system experiences excessive losses (say 20% voltage drop on the load’s supply branch). Under these circumstances the only way in which the system can sustain the load at the prescribed kVA Under these circumstances the only way in which the system can sustain the load at the prescribed kVA is by supplying more current in order to compensate for the voltage drop. More current means more is by supplying more current in order to compensate for the voltage drop. More current means more voltage drop and this cycle will lead to a non-convergence situation. The analysis is basically telling us voltage drop and this cycle will lead to a non-convergence situation. The analysis is basically telling us that the system, as it has been designed, cannot supply 1.2 kVA to the load due to excessive losses. At that the system, as it has been designed, cannot supply 1.2 kVA to the load due to excessive losses. At this point one can switch the definition of the load from constant load to constant impedance and this point one can switch the definition of the load from constant load to constant impedance and determine exactly how much power can be transferred to the load. This issue will be illustrated with an determine exactly how much power can be transferred to the load. This issue will be illustrated with an example in further sections of this manual.

example in further sections of this manual.

2.2 2.2 Constant

Constant Current

Current Load

Load (Amps)

(Amps)

The load (P+jQ) is treated as constant current load. The current is calculated by using the following The load (P+jQ) is treated as constant current load. The current is calculated by using the following equation: equation: ∗ ∗

−

=

V

jQ

P

II

P

P = = Load Load Active Active Power Power (kW)(kW) Q

Q = = Load Load Reactive Reactive Power Power (kVAR)(kVAR) V

V = = Load Load Rated Rated VoltageVoltage I

I = = Load Load equivalent equivalent constant constant currentcurrent

How does this work? How does this work?

When a load of rated “P” (kW), “Q” (kVAR) and “V” (Voltage) is entered, the program calculates the When a load of rated “P” (kW), “Q” (kVAR) and “V” (Voltage) is entered, the program calculates the equivalent current “I” and defines the load as a constant draw of “I” (Amps).

equivalent current “I” and defines the load as a constant draw of “I” (Amps).

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2.1 2.1 Constant

Constant Power

Power Load

Load (kVA)

(kVA)

This type of load is treated as constant demand of power. The load flow algorithm will attempt This type of load is treated as constant demand of power. The load flow algorithm will attempt convergence under the premise that these types of loads always draw the kVA specified by the user. convergence under the premise that these types of loads always draw the kVA specified by the user. Caution should be exercised when interpreting the convergence results in the presence constant kVA Caution should be exercised when interpreting the convergence results in the presence constant kVA loads. To illustrate this, let’s assume that a given load is rated 1.2kVA at 120 Volts; let’s further assume loads. To illustrate this, let’s assume that a given load is rated 1.2kVA at 120 Volts; let’s further assume that the system experiences excessive losses (say 20% voltage drop on the load’s supply branch). that the system experiences excessive losses (say 20% voltage drop on the load’s supply branch). Under these circumstances the only way in which the system can sustain the load at the prescribed kVA Under these circumstances the only way in which the system can sustain the load at the prescribed kVA is by supplying more current in order to compensate for the voltage drop. More current means more is by supplying more current in order to compensate for the voltage drop. More current means more voltage drop and this cycle will lead to a non-convergence situation. The analysis is basically telling us voltage drop and this cycle will lead to a non-convergence situation. The analysis is basically telling us that the system, as it has been designed, cannot supply 1.2 kVA to the load due to excessive losses. At that the system, as it has been designed, cannot supply 1.2 kVA to the load due to excessive losses. At this point one can switch the definition of the load from constant load to constant impedance and this point one can switch the definition of the load from constant load to constant impedance and determine exactly how much power can be transferred to the load. This issue will be illustrated with an determine exactly how much power can be transferred to the load. This issue will be illustrated with an example in further sections of this manual.

example in further sections of this manual.

2.2 2.2 Constant

Constant Current

Current Load

Load (Amps)

(Amps)

The load (P+jQ) is treated as constant current load. The current is calculated by using the following The load (P+jQ) is treated as constant current load. The current is calculated by using the following equation: equation: ∗ ∗

−

=

V

jQ

P

II

P

V = = Load Load Rated Rated VoltageVoltage I

I = = Load Load equivalent equivalent constant constant currentcurrent

When a load of rated “P” (kW), “Q” (kVAR) and “V” (Voltage) is entered, the program calculates the When a load of rated “P” (kW), “Q” (kVAR) and “V” (Voltage) is entered, the program calculates the equivalent current “I” and defines the load as a constant draw of “I” (Amps).

equivalent current “I” and defines the load as a constant draw of “I” (Amps).

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2.3 2.3 Constant

Constant Impedance

Impedance Load

Load (Ohms)

(Ohms)

The load (P+jQ) is treated as constant impedance load. The impedance is calculated by using the The load (P+jQ) is treated as constant impedance load. The impedance is calculated by using the following equation: following equation:

jQ

P

V

Z

2 2

−

=

P

V = = Load Load Rated Rated VoltageVoltage Z

Z = = Load Load equivalent equivalent constant constant ImpedanceImpedance

When a load of a rated “P” (kW), “Q” (kVAR) and “V” (Voltage) is entered, the program calculates the When a load of a rated “P” (kW), “Q” (kVAR) and “V” (Voltage) is entered, the program calculates the equivalent impedance “Z” and defines the load as a constant shunt impedance of “Z” (Ohms).

equivalent impedance “Z” and defines the load as a constant shunt impedance of “Z” (Ohms).

2.4 2.4 Functional

Functional Load

Load

A functional load is a load that is composed of any combination of constant load, constant current, A functional load is a load that is composed of any combination of constant load, constant current, and/or constant impedance. For example, in a single phase system, the total load at 120 Volts may be and/or constant impedance. For example, in a single phase system, the total load at 120 Volts may be (12 kW+j0 kVAR) broken down as follows:

(12 kW+j0 kVAR) broken down as follows: 30%

30% as as constant constant power power load: load: 3.6 3.6 kW kW – – 3.6 3.6 kWkW 60%

60% as as constant constant current: current: 7.2 7.2 kW kW – – 60.0 60.0 AmpsAmps 10%

10% as as constant constant impedance: impedance: 1.2 1.2 kW kW – – 12.0 12.0 ohmsohms

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3.0

3.0 Generator

Generator Models

Models

The Load Flow study algorithm, specifies each bus based on; Active/Reactive power consumption (Load The Load Flow study algorithm, specifies each bus based on; Active/Reactive power consumption (Load or PQ Bus), Voltage magnitude (Generation or PV Bus), and Voltage reference (Swing Bus). Supply or PQ Bus), Voltage magnitude (Generation or PV Bus), and Voltage reference (Swing Bus). Supply busses that contain synchronous generators may also be specified as PQ busses depending on busses that contain synchronous generators may also be specified as PQ busses depending on application, or re-designation from PV busses if the prescribed limits of reactive power generation are application, or re-designation from PV busses if the prescribed limits of reactive power generation are reached during the Load Flow Analysis.

reached during the Load Flow Analysis.

The Swing bus is the only bus at which the power is not specified. Net power flows cannot be fixed in The Swing bus is the only bus at which the power is not specified. Net power flows cannot be fixed in advance at every generating bus, since the network power losses are not known until the study has advance at every generating bus, since the network power losses are not known until the study has been completed. The Swing bus is therefore defined based solely on voltage and angle, leaving the been completed. The Swing bus is therefore defined based solely on voltage and angle, leaving the rest of the variables (kW & kVAR, including system losses) free to adjust (SWING) to the requirements rest of the variables (kW & kVAR, including system losses) free to adjust (SWING) to the requirements of the network. In other words, Active and Reactive Power are the unknown variables of the Swing of the network. In other words, Active and Reactive Power are the unknown variables of the Swing bus.

bus.

In light of the above, a Load Flow Analysis is not mathematically feasible (by computer, by hand, or In light of the above, a Load Flow Analysis is not mathematically feasible (by computer, by hand, or otherwise) without the presence of a Swing bus or busses in the network. The Swing bus designation is otherwise) without the presence of a Swing bus or busses in the network. The Swing bus designation is bestowed upon the stiffest busses in

bestowed upon the stiffest busses in the network, traditionally Power Company or the network, traditionally Power Company or Utility busses.Utility busses. The table below summarizes the critical variables for each type of bus:

The table below summarizes the critical variables for each type of bus:

Bus Type

K

n

ow

n

V

a

r

i

a

b

l

es

U

n

k

no

w

n

V

a

r

ia

b

l

e

s

Swing

V = 1.0 PU

_δ_δ

= 0 deg

P & Q

Load

P

&

Q

V &

δδ

PQ

Generator

P

&

Q

V &

δδ

PV

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4.0 Conducting a Balanced Load Flow Analysis (File: EDM7.axd)

Step 2.

Run a Load Flow Connectivity Error Check”, and correct any problems as required.

Step 1.

Proceed to open the file EDM7.axd”

Notes: ………. ………. ………. ………. ………. ………. ………. Page LF.5

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Generator Utility Supply

Scenario Name Switch

Position Table _{1. Co-Gen.} _{2. Normal} _{3. Emergency}

BKR-1 Closed Closed Open

BKR-2 Closed Open Closed

Notes: ………. ………. ………. ………. ………. ………. ………. Page LF.6

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Notes: ………. ………. ………. ………. ………. ………. ………. Page LF.7

Generator Model

Generator Load Flow set-points. Generator Type. Generator Full Rating in kVA.

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Notes: ………. ………. ………. ………. ………. ………. ………. Page LF.8

Utility Model

m e

The positive sequence impedance is passed onto the Load Flow model. When X” is selected, the positive

sequence SC impedance, is included in the Load Flow model for the Swing bus. This allows the progra to calculate the voltage drop at th “non-ideal” Swing-Bus.

Calculated SC Impedance values. Short Circuit Levels.

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4.1 Running the Load Flow & the Load Flow Options Screen

Notes: ………. ………. ………. ………. ………. Step 2.

Select the scenario to be analyzed.

Step 1.

Invoke the Load Flow program by selecting either of these commands. Step 6. Select “OK”. Refer to steps 6a & 6b in the next page. Step 5.

Select the rest of the options.

Step 4.

Verify the “Scenario” to be studied (Normal). Step 3.

Select the calculation method.

………. ……….

Page LF.9

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Notes: ………. ………. ………. ………. ………. Step 6a. Fine-tune the advanced options as needed. Step 7.

Here the program is letting the user know that there is a dead island caused by the “Normal” scenario configuration. This scenario calls for BKR-2 to be opened, which isolates the generator from the system. Select “Yes” to initiate the

analysis. Step 6b._{Select “OK”.}

………. ……….

Page LF.10

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4.2 Viewing the Load Flow Analysis Results

Step 2.

To view the results, select the desired output format from the “Results” menu.

Select “TD LoadFlow Format 1”

Step 3.

Select the desired report sections and options to be displayed.

Step 4. Select “OK”.

Step 1.

Verify that the analysis has converged. Notes: ………. ………. ………. ………. ………. ………. ………. Page LF.11

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Notes: ………. ………. ………. ………. ………. ………. ………. Page LF.12

Partial view of the output results for scenario 2 “Normal Power”.

Exercise

Using the same procedure explained in this section, conduct load flow analyses for the cases 1 & 3 (Co-Gen, and Emergency).

Compare the results of the 3 scenarios. Do the results make sense? Why?

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4.3 Convergence & Constant kVA Loads (File EDM8.axd)

Step 2.

Model this load as a “Constant kVA” load.

Step 1.

Open the file named “EDM8.axd”. Notes: ………. ………. ………. ………. ………. ………. ………. Page LF.13

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Step 3.

Run the load flow analysis, and verify that in this case it does NOT converge.

Step 4.

View the report. The results, which in cases of

non-convergence ARE NOT TO BE RELIED UPON, seem to indicate that the losses are too great. The system is not able to supply 750 kVA to the load.

Notes: ………. ………. ………. ………. ………. ………. ………. Page LF.14

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4.4 Convergence & Constant Impedance Loads (File EDM8.axd)

Step 1.

Open the file named “EDM8.axd”.

Step 2.

Model this load as a “Constant Impedance” load.

Step 3.

Run the load flow analysis, and verify that in this case it DOES converge. Notes: ………. ………. ………. ………. ………. ………. ………. Page LF.15

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Step 4.

View the report. The results confirm that due to the high losses in the transformer and the line, the system is only able to supply 484.5 kVA to the load.

Voltage available at the load: 385.85 Volts

kVA

5 .

484

255

412 kVA

=

2

+

2

=

Voltage drop at the load: 19.62%.

Notes: ………. ………. ………. ………. ………. ………. ………. Page LF.16

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4.5 Voltage Control Techniques (File EDM7.axd)

The EDSA Object Oriented Load Flow analysis program can assist in correcting voltage levels by means of the any of the following techniques:

- Transformer tap settings.

- Static reactive power adjustment (Capacitors & Inductors). - PV Generator reactive power adjustment (Exciter Control). - A combination of any the above techniques.

The first step is to prepare the network as follows:

- Select the transformer(s) with adjustable taps that will be used as voltage control transformers. - Select the bus or busses at which static reactive compensation is to be considered.

- Select the voltage control generators by defining them as PV units. PQ generators cannot regulate voltage.

Step 1.

Open the file named “EDM7.axd”. Notes: ………. ………. ………. ………. ………. ………. ………. Page LF.17

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Step 2.

Define the settings for all the potential voltage control means desired. Notes: ………. ………. ………. ………. ………. ………. ………. Page LF.18

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Step 3.

Following the guidelines explained in previous sections of this manual, run a Load Flow Analysis using scenario No.1 (Co-Generation).

Step 4.

Produce a report that includes the “Bus Voltage” results and a voltage violation summary for all busses outside a ±10% voltage tolerance. The screen-capture shown here is a partial view of the report showing the voltage violation report

Notes: ………. ………. ………. ………. ………. ………. ………. Page LF.19

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4.5.1 Voltage Control / Transformer Taps

Step 1.

Select “Voltage Control”.

Step 3.

Select “Transformer Tap Adjustments” as a means of control. Press “OK”.

Step 2.

Complete the “Voltage Control Limits” section as indicated here. Notes: ………. ………. ………. ………. ………. ………. ………. Page LF.20

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Step 6.

Select “No’ for now. This option will be explained later. Step 5.

Review the results and select “Done” to exit. Notes: ………. ………. ………. ………. ………. ………. ………. Page LF.21

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4.5.2 Voltage Control / Capacitors-Inductors

Step 1.

Step 3.

Select “Capacitor/Inductor...” as a means of control. Press “OK”

Step 2.

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Step 6.

Select “No’ for now. This option will be explained later. Step 5.

Review the results and select “Done” to exit.

Required amount of reactive compensation, in per unit of base kVA. To obtain actual kVA, multiply by the base kVA (10,000 kVA).

Notes: ………. ………. ………. ………. ………. ………. ………. Page LF.23

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4.5.3 Voltage Control / Transformer Taps & Generator Settings

Step 1.

Step 3.

Select “Transformer…” and “Generator…” as combined means of control. Press “OK”

Step 2.

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Step 6.

Select “Yes’ to test the results.

Step 5.

Review the results and select “Done” to exit.

Required amount of additional generator reactive compensation, in per unit of base kVA. To obtain actual kVA, multiply by the base kVA (10,000 kVA).

Step 7. Select OK to acknowledge. Notes: ………. ………. ………. ………. ………. ………. ………. Page LF.25

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Step 8.

Select “Yes” and verify convergence.

Step 9.

The solution proves to be effective as indicated here.

Notes: ………. ………. ………. ………. ………. ………. ………. Page LF.26

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Step 10. Select “Exit”.

Step 11. Select “Yes”.

Step 12.

Save the file under a different name, if you wish to avoid over-writing the original one. Press “Save”.

Conduct a vol age control analysis for case 1 (Co-Generation), using ONLY Generator Reactive Power control as means of correcting the voltage drops in the network. Use ±7% as the acceptable tolerance. Is the calculation successful? Why?

t Exercise Notes: ………. ………. ………. ………. ………. ………. ………. Page LF.27

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4.6 Load Flow Transformer Sizing (File EDM7.axd)

Step 1.

Run a Load Flow analysis on the subject file.

Step 2.

Select “Xfr Sizing”.

Step 3.

Enter the “Design Margin” in %, and select “OK”.

Notes: ………. ………. ………. ………. ………. ………. ………. Page LF.28

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4.7 Back Annotation of Load Flow Results (File EDM7.axd)

Step 1.

Click here to select the “Back Annotation” tool.

Step 2. Select “ON”.

Step 3.

Select the desired components from the “AC Load Flow” section.

Step 4. Select “OK”. Notes: ………. ………. ………. ………. ………. ………. ………. Page LF.29

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Load Flow back annotation. Notes: ………. ………. ………. ………. ………. ………. ………. Page LF.30

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5.0 Unbalanced Load Flow Analysis (File EDM9.axd)

Note:

These three static loads will be entered as unbalanced loads.

Step 2.

Step 1.

Proceed to open the file “EDM9.axd” Notes: ………. ………. ………. ………. ………. ………. ………. Page LF.31

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5.1 Preparing the Master File Editor

The steps described in this section, will enable the “Unbalanced Load Flow” tab at every load in the single line diagram. The user can enter, as required, the percentile distribution of the load across the phases. These % figures are based on the full load condition entered in the “Load Flow” tab. The initial default assumed by the program is a balanced state as indicated here.

Step 3.

Check the “View Unbalanced Load Flow Fields” option. Step 2.

Select the “Visibility” tab. Step 1.

Click here to invoke the “Master File Editor”.

Notes: ………. ………. ………. ………. ………. ………. ………. Page LF.32

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5.2 Preparing the Unbalanced Loads

Step 1.

Double click on each of the static loads fed from bus J#1066, and edit the Unbalanced Load Flow conditions as indicated in steps 2 & 3.

Step 3.

Enter a per-phase distribution as indicated here. Assume the same percentile values for all three static loads.

Step 2.

Select the “Unbalanced Load Flow” tab. Notes: ………. ………. ………. ………. ………. ………. ………. Page LF.33

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5.3 The Unbalanced Load Flow Analysis Interface

Step 5.

Allow the program to adjust the tolerance by selecting “Yes”. Step 4.

Select the desired options and press “OK”.

Step 3.

Select “Continue”. Step 2.

Select “Run”.

Step 1.

Select “AC 3-Phase Unbalanced”. Notes: ………. ………. ………. ………. ………. ………. ………. Page LF.34

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Partial view of the final report. To exit select “Done”

Step 6.

Verify that the calculation has converged and select “View Results”. Notes: ………. ………. ………. ………. ………. ………. ………. Page LF.35

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6.0 Load Flow Analysis Voltage / Profile Method (File EDM10.axd)

Note:

These four loads will be entered as cycling loads.

Step 2.

Step 1.

Proceed to open the file “EDM10.axd” Notes: ………. ………. ………. ………. ………. ………. ………. Page LF.36

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6.1 Preparing the Master File Editor

Notes: ………. ………. ………. ………. ………. Step 5.

Define the “Load-Cycling” time periods. Select “OK”.

Step 4.

Select the “Time Periods” tab. Step 1.

Click here to invoke the “Master File Editor”.

Step 3.

Check the “View Voltage Profile Fields” option.

Step 2.

Select the “Visibility” tab.

………. ……….

Page LF.37

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6.2 Preparing the Cycling Loads

Step 1.

Double click on each of the loads fed from bus J#1066, and edit the cycling

conditions as indicated in this page.

Notes: ………. ………. ………. ………. ………. ………. ………. Page LF.38

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6.3 The Voltage Profile Load Flow Interface

Step 1. Select “AC Voltage Profile”. Step 3.

Define the “Load Flow Options”, and select “OK”.

Step 4.

Verify convergence for all periods, and select “Done”.

Step 5.

To view the results, select “Results”. Step 2. Select “Run”. Notes: ………. ………. ………. ………. ………. ………. ………. Page LF.39

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6.4 Graphical View of the Results

Step 3.

Select “OK” to return.

Step 2.

View and edit the graphs. When done press “Exit”.

Step 1.

Select the bus to view, by clicking on it once. Notes: ………. ………. ………. ………. ………. ………. ………. Page LF.40

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7.0 Single Phase Load Flow Analysis / Building the Job-File

Step 4.

De-select the password requirement, and press “OK”.

Step 3.

Name the file, and select “Open”.

Step 2.

Select Electrical One Line Single-Phase.axd”, and press “OK”.

Step 1.

Click here to create a new job-file. Notes: ………. ………. ………. ………. ………. ………. ………. Page LF.41

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Master Job-File Editor settings.

Step 5.

Following the same guidelines outlined in the “Main User Interface” section of this manual, proceed to build the required single-phase network.

Single-Phase Element catalog. This catalog contains both, single-phase busses and branches.

Notes: ………. ………. ………. ………. ………. ………. ………. Page LF.42

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7.1 Conducting a Single-Phase Load Flow Analysis

(File: EDM6.axd)

Exercise

Step 3.

The AC Load Flow Analysis program is designed to analyze both single and three-phase systems. To continue, simply repeat the same steps explained in section 4.0 of this manual.

Step 3.

Invoke the AC Load-Flow Analysis program.

Step 2.

Run a load-flow

connectivity error check.

Step 1.

Proceed to open the single-phase file named “EDM6.axd”. Notes: ………. ………. ………. ………. ………. ………. ………. Page LF.43

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8.0 Practical Exercise (File: EDM7X.axd)

Power Factor Correction

Capacitor Bank.

Using the File “EDM7X.axd”, shown in the above picture, complete the following questionnaire:

1. Total Power in kVA demanded from the utility supply: ______________kVA 2. Power Factor at the Utility Bus: _______________% 3. Total System Losses on kW: ______________kW 4. Voltage Drop at 480 V Bus J#1066: _______________% 5. List the transformers loaded beyond 100% of their capacity: F:______ T:______

F:______ T:______ Notes: ………. ………. ………. ………. ………. ………. ………. Page LF.44