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Line Model and PerformancesSyed Mohd Firdaus bin Syed Mohd Fuzi

EC11098

University Malaysia PahangAbstract

Transmission lines are representedby an equivalent model with appropriatecircuit parameters on a "per-phase" basis.The terminal voltages are expressed fromone line to neutral, the current for one phaseand, thus, the three-phase system is reducedto an equivalent single-phase system. Themodel used to calculate voltages, currents,and power flows depends on the length of

the line. In this chapter the circuitparameters and voltage and current relationsare first developed for "short" and "medium"lines. Problems relating to the regulation andlosses of lines and their operation underconditions of fixed terminal voltages arethen considered.

Introduction

The transmission lines are categorized asthree types-

1) Short transmission linethe line lengthis up to 80 km.

2) Medium transmission line the linelength is between 80km to 160 km.

3) Long transmission line the linelength is more than 160 km.

Whatever may be the category oftransmission line, the main aim is to transmitpower from one end to another. Like otherelectrical system, the transmission network

also will have some power loss and voltagedrop during transmitting power fromsending end to receiving end.

A short line is defined as a line in whichthe effect of capacitive current can beneglected. Figure 1 represents a single phaseline and Figure 2 represents the associatedphasor diagram where R and X are the

resistance and inductive reactance of theloop circuit respectively.

IS represents the sending end currentwhich will be equal in magnitude to the loadcurrent IR. VR represents the receiving endvoltage whilst VS represents the sending endvoltage which is given by the phasor sum ofthe receiving end voltage and the voltagedrop due to the line impedance. The phasordiagram is constructed by drawing areference phasor to represent the load

current IR. If the load is operating at alagging power factor the phasor representingthe receiving end voltage VR is drawn to asuitable scale leading the current phasor byan angle R (the cosine of which representsthe load power factor).

By scribing arcs representing the sendingend voltage VS, and the line voltage drop Vd

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(drawn to the same scale as the receivingend voltage) the phasor diagram can becompleted. The voltage drop due to the lineresistance is drawn in phase (i.e. parallelto) with the currentIR. The voltage drop dueto the line reactance is drawn leading thecurrent IR by 90, and hence the effectivevalues of line resistance and reactance canbe estimated for a particular value of loadcurrent. Having determined the lineparameters for a particular load current,calculation of the voltage drops may becarried out to confirm the experimentalresults.

Using the results obtained the lineregulation can be determined. This isdefined as the percentage rise in voltage at

the receiving end when full load is thrownoff, the sending end voltage remaining at aconstant value. It is therefore given by:

Transmission efficiency may also bedetermined from:

For medium line, certain lengths of linethe capacitive charging current cannot beneglected, and its effect must be taken intoconsideration. Two methods have beenevolved to determine the characteristics ofsuch lines.

The Nominal T Method

In this method the total line capacitanceis placed (i.e. lumped) at the midpoint of theline, and half the total line resistance and

inductive reactance placed at each end of theline (Figure 3).

The phasor diagram, as shown in Figure4, is constructed using load currentIR as thereferenced phasor. If the load is operating ata lagging power factor the phasorrepresenting the receiving end voltage VR isdrawn leading the current phasor by anangle R (the cosine of which represents theload power factor). The phasor representingthe voltage drop at the receiving end of theline due to line resistance (IR R/2) is drawnin phase (i.e. parallel to) with the current IR.The phasor representing the voltage drop atthe receiving end of the line due to lineinductive reactance (IR X/2) is drawnleading the current IR by 90. Bycompleting the figure formed by the threevoltage phasors the voltage across the line

capacitance (VC) can be determined.

The line charging current IC due to linecapacitance is calculated from:

IC = 2 f C VC (3)

and drawn at an angle of 90 to thephasor representing the voltage VC, it is thentransposed to the end of the current phasor

IR to enable the phasor representing thesending end current IS to be drawn. Thephasor representing the voltage drop at thesending end of the line due to line resistance(IS.R/2) is drawn in phase with the currentIS. The phasor representing the voltage dropat the sending end of the line due to lineinductive resistance (IS.X/2) is drawnleading the currentIS by 90. By completing

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the figure formed by the three voltagephasors the sending end voltage VS can bedetermined. The sending end power factor isgiven by the cosine of the angle S.

The Nominal Method

In this method the total line resistanceand inductive reactance is placed (i.e.lumped) at the midpoint of the line, and halfthe total line capacitance placed at each endof the line (Figure 5).

The phasor diagram (Figure 6) isconstructed using load current IR asreference phasor. If the load is operating at alagging power factor the phasor representingthe receiving end voltage VR is drawnleading the current phasor by an angle (thecosine of which represents the load powerfactor). The capacitor at the receiving end of

the lineIC1 is calculated from:IC1 = f C1 VR (4)

The phasor representing this current isdrawn leading the receiving end voltage by90. By completing the figure formed by thecurrent phasors, the current through the lineresistance and reactance can be determined,and hence the voltage drops due to the lineresistance (IRR) and line reactance (IRX).The capacitor current at the sending end ofthe lineIC2 is calculated from:

IC2 = f C2 VS (5)

The phasor representing this current isdrawn leading the sending end voltage VSby 90. By completing the Figure formed bythe current phasorsIC1 andIC2 the sendingend currentIS may be determined.

The sending end power factor is given bythe cosine of the angle S.

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Experiment Details

EXPERIMENT 1 :

SHORT LINE REPRESENTATION

PROCEDURE

a. Resistive Load

1. The circuit is connected as shownin Appendix A, but only 100%resistive load (L1) is selected

2. CB1 (at the secondary of thesupply transformer) and CB2 (atthe 'send' end of the single phaseline) are shut. The power is

connected to the line with acurrent of roughly 5 Ampere.

3. The analogue voltmeter is used tomeasure the voltage drop (Vd)across the line impedance. Themultifunction meters is used tomeasure the sending andreceiving voltages, currents andpower. These meters can also be

used to show power factor, but forgood practice, equation 6 is usedto calculate the power factor foreach end of the line. The sendingend voltage is remained fairlyconstant - determined by localmains and cables.

4. The results are recorded in Table1.

b. Resistive and Inductive Load

1. Procedure a is repeated, but a50% inductive load is used inparallel with the resistive load

(shown by dotted lines of theconnection diagram in AppendixA).

EXPERIMENT 2 :

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SINGLE PHASE LINE - MEDIUM ORLONG LINE INVESTIGATION(NOMINAL T METHOD)

PROCEDURE

a. 10F Shunt

1. The circuit is connected as shownin Appendix B. A 100% resistiveand 50% inductive load areselected.

2. CB1 (at the secondary of thesupply transformer) and CB2 (atthe 'send' end of the single phaseline) are shut. The power isconnected to the line.

3. The right-hand analoguevoltmeter is used to measure thevoltage drop (Vd1) across the twoinductances at the sending end,and the voltage drop (Vd2) acrossthe two inductances at the

receiving end. In Appendix B, theright-hand dotted line shows wireto move to make this connection.The multifunction meters is usedto measure the sending andreceiving voltages, currents andpower. These meters can also beused to show power factor, but forgood practice, equation 6 is usedto calculate the power factor foreach end of the line. The sendingend voltage is remained fairlyconstant - determined by localmains and cables.


b. 20F Shunt

1. Procedure a is repeated but the 20F line capacitor is connected (shown by the dotted lines of theconnection in Appendix B).

EXPERIMENT 3 :

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SINGLE PHASE LINE - MEDIUM ORLONG LINE INVESTIGATION(NOMINAL PI METHOD)

PROCEDURE

a. 5F + 5F

1. The circuit is connected as shownin Appendix C. A 100% resistiveand 50% inductive load areselected.

2. CB1 (at the secondary of thesupply transformer) and CB2 (atthe 'send' end of the single phaseline) are shut. The power isconnected to the line. CB4 isclosed to connect the multi-function meter in the lower panelcorrectly and complete the line.

3. The left-hand analogue voltmeteris used to measure the voltagedrop (Vd) across the inductancesof the line. Analog ammeters is

used to measure the capacitorcurrents.. The multifunctionmeters is used to measure the linecurrents, sending and receivingvoltages, currents and power.These meters can also be used toshow power factor, but for goodpractice, equation 6 is used tocalculate the power factor foreach end of the line. The sendingend voltage is remained fairlyconstant - determined by localmains and cables.


b. 10F + 10F

1. Experiment is repeated with two10 F line capacitors (shown bythe dotted lines of the connectionin Appendix C).

Results:

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Table 1: Blank table for Experiment 1

Voltages,Currentsand Powers

100%ResistiveLoad

100%ResistiveLoad +50%InductiveLoad

SendingVoltage (Vs) 129.7 V 129.8 V

SendingCurrent (Is)

5113.8mA

5193.2mA

Sending

Power (Ws) 635.88 W

520.5

W

ReceivingVoltage(VR) 119.6 V 105.9 V

ReceivingCurrent (IR)

5128.8mA

5179.7mA

ReceivingPower (WR) 612.54 W

493.98W

Line Volt

Drop (Vd) 35 V 34 V



Voltages,Currents andPowers

10 FShunt

20 FShunt

Sending Voltage(Vs)

129.3V

129.4V

Sending Current(Is)

4079.9mA

3968.7mA

Sending Power(Ws)

373.98W

388.02W

ReceivingVoltage (VR) 87.2 V 89.5 V

Receiving

Current (IR)

4281.6

mA

4367.5

mA

ReceivingPower (WR)

336.76W

352.74W

Line Volt Drop(Vd1) 28 V 27 V

Line Volt Drop(Vd2) 29 V 30 V

CapacitorVoltage (Vc) 108 V 110 V

CapacitorCurrent (Ic) 0.34 A 0.67 A

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Voltages,Currents andPowers

5 F +5 F

10 F+ 10F

SendingVoltage (Vs)

129.4V

129.4V

SendingCurrent (Is)

3661.7mA

3537.5mA

Sending Power(Ws)

316.74W

325.22W

Sending EndCapacitorCurrent (Ic1) 0.2 A 0.4 A

ReceivingVoltage (VR) 79.4 V 80.3 V

ReceivingCurrent (IR) 3868.7mA 3915.5mA

ReceivingPower (WR)

278.04W

284.88W

Receiving EndCapacitorCurrent (Ic2) 0.15 A 0.26 A

Line Volt Drop(Vd) 68 V 68 V

Line Current

(ILine)

3.685

A

3.547

A

Discussions

Experiment 1: Short LineRepresentation

Procedure A - Resistive Load

1)

Use your results to find the powerfactor, line regulation andtransmission efficiency.

Power Factor:

pF =

Sending end :

pF =

()()= 0.9587

Receiving end :

pF =

()()= 0.9986

Line Regulation:

V.R =

x 100

V.R =

x 100 = 8.44%

Transmission Efficiency:

Efficiency =

x 100

Efficiency =

x 100 = 96.33%

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2) From the phasor diagram as shown

in Figure 2, find the effective valuesof line resistance and reactance (IRRand IRX) for your recorded loadcurrent (receive end).

cos R =

sin R =

X = VR cos R

Y = VR sin R

cos S =

IRR = VS cos S - X

IRR = VS cos S - VR cos R

IRR = (129.7 x 0.9587) - (119.6 x0.9986)

IRR = 4.91V

sin S =

IRX = VS sin S - Y

IRX = VS sin S - VR sin R

IRX = (129.7 x sin 16.52) - (119.6 xsin 3.03)

IRX = 30.56V

3) Use your results to calculate theactual value of resistance andinductance and compare it with thevalues written next to the inductorsof the single phase line.

R =

R =

= 0.96

X =

X =

= j5.96

X = jwL

X = J2fL

L =

L =

L =

()= 18.97mH

Compare to original L = 11.46mH x2 = 22.92mH

4) Verify and validate the value ofsending voltage and current by usingthe equation that you have learned inthe class in terms of ABCDparameters for the short transmissionline.

VS = VR + ZIR

Z = R + X

Z = 0.96+j5.96

VS = 119.6 + (0.96 +j5.96)(5128.8m)

VS = 128.22

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Power Factor:

pF =

Sending end :

pF =

()()= 0.7722

Receiving end :

pF =

()()= 0.9006

Line Regulation:

V.R =

x 100

V.R =

x 100 = 25.57%


Efficiency =

x 100

Efficiency =

x 100 = 94.90%

2) From the phasor diagram as shown

in Figure 2, find the effective values

of line resistance and reactance (IRRand IRX) for your recorded loadcurrent (receive end).

cos R =

sin R =

X = VR cos R

Y = VR sin R

cos S =

IRR = VS cos S - X

IRR = VS cos S - VR cos R

IRR = (129.8 x 0.7722) - (105.9 x0.9006)

IRR = 4.86V

sin S =

IRX = VS sin S - Y

IRX = VS sin S - VR sin R

IRX = (129.8 x sin 39.45) - (105.9 x

sin 25.76)IRX = 36.45 V

3) Use your results to calculate theactual value of resistance andinductance and compare it with thevalues written next to the inductorsof the single phase line.

R =

R =

= 0.94

X =

X =

= j7.04

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X = jwL

X = j2fL

L =

L =

L =

()= 22.41mH


4) Verify and validate the value of

sending voltage and current by usingthe equation that you have learned in

the class in terms of ABCDparameters for the short transmissionline.

VS = VR + ZIR

Z = R + X

Z = 0.94 +j7.04

VS = 105.9 + (0.94 +j7.04)(5179.7m)

VS = 116.62

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cos R =

sin R =

X = VR cos R Y = VR sin R

Apply KVL to find the equation ofVc

VC = IR (

) + VR

IR (

) = VC - VR

From phasor diagram we get,VR = X + jY

VR = VR cos R + jVR sin R

IR (

) = VC - (VR cos R + jVR

sin R)

IR (

) = VC - VR cos R - jVR sin

R)

IR (

) = IR( + j

)

IR (

) = VC - VR cos R

IR (

) = 108 - (87.2)(0.9020)

IR (

) = 29.35V

IR (

) = jVR sin R

IR (

) = j(87.2)(sin 25.58)

IR (

) = j37.65

3) Use your results to calculate the

actual value of resistance andinductance and compare it with thevalues written next to the inductorsof the single phase line.

=

(

)

=

= 6.85R = 13.7

=

(

)

=

= j8.79

X = j17.58

= jwL

= j2fL

L =

L =

()= 27.98mH


4) Verify and validate the value ofsending voltage and current by usingthe equation that you have learned inthe class in terms of ABCDparameters for the mediumtransmission line.

Y = g + jXc

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Y = 0 + j(2 x 50 x 10)

Y = j3.14mS

Z = R +jXL

Z = 13.7+j17.58

A = D = 1 +

A = D = 1 +()( )

A = D = 0.97 + j0.02

B = Z(

)

B =(

)(()( )

)

B = 13.32 + j17.48

C = Y

C = j3.14m

VS = AVR + BIRVS = (0.97 + j0.02)(87.2) + (13.32 +j17.48)(4281.6m)

VS = 161 < 28.4 V

IS = CVR + DIR

IS = (j3.14m)(87.2) + (0.97 +j0.02)(4281.6m)

IS = 4.15 A

Procedure B - 20F Shunt

1) Use your results to find the power factor,

line regulation and transmissionefficiency.

Power Factor:

pF =

Sending end :

pF =

()()= 0.7556

Receiving end :

pF = ()()= 0.9024

Line Regulation:

V.R =

x 100

V.R =

x 100 = 44.58%


Efficiency =

x 100

Efficiency =

x 100 = 90.91%

2) From the phasor diagram as shown inFigure 2, find the effective values of line

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resistance and reactance (IRR and IRX)for your recorded load current (receiveend).

cos R =

sin R =


Apply KVL to find the equation of Vc

VC = IR (

) + VR

IR (

) = VC - VR

From phasor diagram we get,

VR = X + jY


IR (

) = VC - (VR cos R + jVR sin R)

IR (

) = VC - VR cos R -jVR sin R)

IR (

) = IR(

+ j

)

IR (

) = VC - VR cos R

IR (

) = 110 - (89.5)(0.9024)

IR (

) = 29.24V

IR (

) =jVR sin R

IR (

) = j(89.5)(sin 25.52)

IR (

) = j38.56

3) Use your results to calculate the actualvalue of resistance and inductance andcompare it with thevalues written next tothe inductors of the single phase line.

=

(

)

=

= 6.69

R = 13.38

=

(

)

=

= j8.82

X = j17.64

= jwL

= j2fL

L =

L =

()= 28.07mH

Compare to original L = 11.46mH x 2 =22.92mH

4)

Verify and validate the value of sendingvoltage and current by using theequation that you have learned in theclass in terms of ABCD parameters forthe medium transmission line.

Y = g + jXc

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Y = 0 + j(2 x 50 x 20)

Y = j6.28mS

Z = R +jXL

Z = 13.38 +j17.64

A = D = 1 +

A = D = 1 +()( )

A = D = 0.94 + j0.04

B = Z(

)

B =(

)(()( )

)

B = 12.64 + j17.43

C = Y

C = j6.28m

VS = AVR + BIRVS = (0.94 + j0.04)(89.5) + (12.64 +j17.43)(4367.5m)

VS = 160.52 < 29.77 V

IS = CVR + DIR

IS = (j6.28m)(89.5) + (0.94 +j0.04)(4367.5m)

IS = 4.11 A

Experiment 3: Single Phase Line -Medium or Long Line Investigation(Nominal Pi Method)

Procedure A - 5F + 5F

1) Use your results to find the power factor,line regulation and transmissionefficiency.

Power Factor:

pF =

Sending end :

pF =

()()= 0.6685

Receiving end :

pF =

()()= 0.9052

Line Regulation:

V.R =

x 100

V.R =

x 100 = 62.97%


Efficiency =

x 100

Efficiency =

x 100 = 87.78%

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2) From the phasor diagram as shown in

Figure 2, use it to find the effectivevalues of line resistance and reactance(IRR and IRX) for your recorded loadcurrent (receive end).

cos R =

sin R =


Apply KVL to find the equation ofVS

VS = ILZ + VR

ILZ = VS - VR


VR = X + jY


ILZ= VS - (VR cos R + jVR sinR)

ILZ= VS - VR cos R -jVR sin R)

ILZ = IL(R + jX)

ILR = VS - VR cos R

ILR = 129.4 - (79.4)(0.9052)

ILR = 57.53V

jILX= jVR sin R

jILX = j(79.4)(sin 25.15)

jILX = j33.74

R =

R =

= 15.61

IRR = (3868.7m)(15.61)

IRR = 60.39V

X =

X =

= j9.16

IRX = (3868.7m)(j9.16)

IRX = 35.44V

3) Use your results to calculate the actualvalue of resistance and inductance andcompare it with the values written nextto the inductors of the single phase line.

R =

= 15.61

X =

= j9.16

X= jwL

X = j2fL

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L =

L =

()= 29.16mH

Compare to original L = 11.46mH x

2 = 22.92mH

4) Verify and validate the value of sending

voltage and current by using theequation that you have learned in theclass in terms of ABCD parameters forthe medium transmission line.

Y = g + jXc

Y = 0 + j(2 x 50 x 5)Y = j1.57mS

Z = R +jXLZ = 15.61 +j9.16

A = D = 1 +

A = D = 1 +()( )

A = D = 0.99 + j0.01

B = ZB = 15.61 + j9.16

C = Y(

)

C = (j1.57m)(()( )

)

C = -9.62 + j1.56m

VS = AVR + BIR

VS = (0.99 + j0.01)(79.4) + (15.61 +

j9.16)(3868.7m)

VS = 143.64 < 14.61 V

IS = CVR + DIRIS = (-9.62 + j1.56m)(79.4) + (0.99 +

j0.01)(3868.7m)

IS = 3.83 A

Procedure B10F + 10F

1) Use your results to find the power factor,

line regulation and transmissionefficiency.

Power Factor:

pF =

Sending end :

pF =

()()= 0.7105

Receiving end :

pF =

()()= 0.9061

Line Regulation:

V.R =

x 100

V.R =

x 100 = 61.15%


Efficiency =

x 100

Efficiency =

x 100 = 96.33%

2) From the phasor diagram as shown in

Figure 2, use it to find the effectivevalues of line resistance and reactance

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(IRR and IRX) for your recorded loadcurrent (receive end).

cos R =

sin R =


Apply KVL to find the equation of VS

VS = ILZ + VR

ILZ = VS - VR


VR = X + jY


ILZ= VS - (VR cos R + jVR sin R)

ILZ= VS - VR cos R -jVR sin R)

ILZ = IL(R + jX)

ILR = VS - VR cos R

ILR = 129.4 - (80.3)(0.9061)

ILR = 56.64V

jILX= jVR sin R

jILX = j(80.3)(sin 25.03)

jILX = j33.97

R =

R =

= 15.97

IRR = (3915.5m)(15.97)

IRR = 62.53V

X =

X =

= j9.58

IRX = (3915.5m)(j9.58)

IRX = 37.51V

3) Use your results to calculate the actual

value of resistance and inductance andcompare it with the values written nextto the inductors of the single phase line.

R =

= 15.97

X =

= j9.58

X= jwL

X = j2fL

L =

L =

()= 30.49mH

Compare to original L = 11.46mH x 2 =22.92mH

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4) Verify and validate the value of sendingvoltage and current by using theequation that you have learned in theclass in terms of ABCD parameters for

the medium transmission line.

Y = g + jXc

Y = 0 + j(2 x 50 x 10)

Y = j3.14mS

Z = R +jXL

Z = 15.97 +j9.58

A = D = 1 +

A = D = 1 +()( )

A = D = 0.98 + j0.03

B = Z

B = 15.97 + j9.58

C = Y(

)

C = (j3.14m)(()( )

)

C = -39.36 + j3.12m

VS = AVR + BIR

VS = (0.98 + j0.03)(80.3) + (15.97 +j9.58)(3915.5m)

VS = 146.76 < 15.78 V

IS = CVR + DIR

IS = (-39.36 + j3.12m)(80.3) + (0.98 +j0.03)(3915.5m)

IS = 3.83 A

Conclusion

From this experiment, we havelearned a lot regarding the line model andperformances in transmission line. Wehave been able to find the regulation and

transmission of a simple inductive linewithout shunt capacitance, and constructthe relevant phasor diagram. We alsohave successfully find the regulation andtransmission efficiency of a longer linewith shunt capacitance, represented bythe nominal T circuit, and construct therelevant phasor diagram. Lastly, we havebeen able to find the regulation andtransmission efficiency of a longer linewith shunt capacitance, represented bythe nominal pi circuit, and construct the

relevant phasor diagram. For theconclusion, we have successfully donethis experiment and have archived all thegoals.

References

1. Mohd. Redzuan Ahmad, Ruhaizad

Ishak. (2011) Teaching ModuleElectrical Power System

Pahang; Universiti Malaysia Pahang.

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