Validación de la Propuesta del Manual de Usabilidad y Accesibilidad



 +

2

(3.63)

and similarly, the corresponding reactive power at the sending end, the maximum input vars, is given by

Q V

S= ZS²sinθ (3.64)

As can be observed, both Equations 3.63 and 3.64 are independent of V_R voltage, and Equation 3.64 has a positive sign this time.

3.9.3 P^erCent v^oltage r^egUlation

The voltage regulation of the line is defined by the increase in voltage when full load is removed, that is,

Percentage of voltage regulation= − V V ×

V

S R

R

100 (3.65)

or

Percentage of voltage regulation= V ^NL −V ^FL V

R R

R

, ,

,FFL

×100 (3.66)

where

│VS│ = magnitude of sending-end phase (line-to-neutral) voltage at no load

│VR│ = magnitude of receiving-end phase (line-to-neutral) voltage at full load

│VR,NL│ = magnitude of receiving-end voltage at no load

│VR,FL│ = magnitude of receiving-end voltage at full load with constant |VS| Therefore, if the load is connected at the receiving end of the line,

│VS│ = │VR,NL│ and

│VR│ = │VR,FL│

An approximate expression for percentage of voltage regulation is

Percentage of voltage regulation≅I R ±X

R cosΦR sinΦRR

VR







×100 (3.67)

EXAMPLE 3.3

A three-phase, 60-Hz overhead short transmission line has a line-to-line voltage of 23 kV at the receiving end, a total impedance of 2.48 + j6.57 Ω/phase, and a load of 9 MW with a receiving-end lagging power factor of 0.85.

(a) Calculate the line-to-neutral and line-to-line voltages at the sending end.

(b) Calculate the load angle.

Solution

METHOD 1. USING COMPLEX ALGEBRA:

(a) The line-to-neutral reference voltage is

V V

R R

( ) ( )

, .

L N L L

V

− = −

= × ∠ ° = ∠ °

3 23 10 0

3³ 13 294 8 0

The line current is

I= ×

× × × × −

= −

9 10

3 23 10 0 85 0 85 0 527 265 8 0 85

6

3 . ( . . )

. ( .

j j00 527

225 83 140 08 . )

. .

= − j A

Therefore,

IZ= − +

= ∠ −

(225.93 140.08)(2.48 6.57) 265.8 31.8°

j j

( )(77.02 69.32°

1866.8 37.52° V

∠

= ∠

)

Thus, the line-to-neutral voltage at the sending end is

V_{S(L N)} V_{R(L N)} IZ 14,803 4.4° V

− = − +

= ∠

The line-to-line voltage at the sending end is

V_{S(L L)} 3V_{S(L N)}

25,640 4.4 25,640 4.4 V

− = −

= ∠ ° + ° =30 ∠3 °

(b) The load angle is 4.4°.

METHOD 2. USING THE CURRENT AS THE REFERENCE PHASOR:

(a)

V_R cos θR + IR = 13,279.06 × 0.85 + 265.8 × 2.48 = 11,946 V V_R sin θR + IX = 13,294.8 × 0.527 + 266.1 × 6.57 = 8744 V

then

V_S(L−N) = (11,946.39² + 8744²)^1/2 = 14,803 V/phase V_S(L−L) = 25,640 V

(b)

θS θR δ= tan1 8744

11,946 36.2°

= + 





=

−

δ = θS – θR = 36.2 – 31.8 = 4.4°

METHOD 3. USING THE RECEIVING-END VOLTAGE AS THE REFERENCE PHASOR:

(a)

V_S(L−N) = [(V_R + IR cos θR + IX sin θR)² + (IX cos θR + IR sin θR)²]^1/2

IR cos θR = 265.8 × 2.48 × 0.85 = 560.3 IR sin θR = 265.8 × 2.48 × 0.527 = 347.4 IX cos θR = 265.8 × 6.57 × 0.85 = 1484.4 IX sin θR = 265.8 × 6.57 × 0.527 = 920.3 Therefore,

V_{S(L N)−} =[(13,279 560.3 920.3)+ + ²+(1484.4 347.4) ]− ^211/2

2 2 1/2

[14,759.7 1137 ] 14,803 V

= +

=

V_{S(L L)−} = 3V_{S(L L)−} =25,640 V

(b)

δ = 





= ° tan− 1137

14,759.7

1 4 4.

METHOD 4. USING POWER RELATIONSHIPS:

The power loss in the line is

P_loss I R

2 6

3 265.8 2.48 10 0.526 MW

=

= × × × ⁻ =

3²

The total input power to the line is

P_T= +P P

= + =

loss

9 0.526 9.526 MW

The var loss in the line is

Q_loss I X

2 6

3 265.8 6.57 10 1.393 Mvar lagging

=

= × × × ⁻ =

3²

The total megavar input to the line is

Q P

T R Q

R

= +

= × + =

sin cos

θ θ ^loss 9 0.526

0.85 1.393 6.973 Mvvar lagging

The total megavolt-ampere input to the line is

S_T=

(

P_T +Q_T

)

= =

2 21 2/

2 2 1/2

9.526 +6.973 11.81 MVA

( )

(a)

V S

S TI

(L L)

11.81 106

3 265.8 25,640 V

− =

= ×

× =

3

V V

S S

(L N) L L

3 14,803 V

− = ⁽−⁾ =

(b)

cos ,

. .

θs T T

P

=S =9 526=

11 81 0 807lagging

Therefore,

θ δ

s=

= − =

36.2°

36.2° 31.8° 4.4°

METHOD 5. TREATING THE THREE-PHASE LINE AS A SINGLE-PHASE LINE AND HAVING V_S AND V_R REPRESENT LINE-TO-LINE VOLTAGES, NOT LINE-TO-NEUTRAL VOLTAGES:

(a) The power delivered is 4.5 MW

I_line= × A

× × =

4 5 106

23 103 0 85. 230 18

. .

R_loop = 2 × 2.48 = 4.96 Ω

X_loop = 2 × 6.57 = 13.14 Ω

V_R cos θR = 23 × 10³ × 0.85 = 19,550 V VR sin θR = 23 × 103 × 0.527 = 12,121 V

IR = 230.18 × 4.96 = 1141.7 V IX = 230.18 × 13.14 = 3024.6 V

Therefore,

550 1141.7 12,121 3024.6 20,691.7 15

Calculate percentage of voltage regulation for the values given in Example 3.3 (a) Using Equation 3.65

(b) Using Equation 3.66

Solution

(a) Using Equation 3.67,

Percentage of voltage regulation

14

(b) Using Equation 3.67,

Percentage of voltage regulation≅ ×I R ±X

R ( cosθR sinn )

3.9.4 rePreSentationof mUtUal imPedanCeof Short lineS

Figure 3.16a shows a circuit of two lines, x and y, that have self-impedances of Z_xx and Z_yy and a mutual impedance of Z_zy. Its equivalent circuit is shown in Figure 3.16b. Sometimes, it may be required to preserve the electrical identity of the two lines, as shown in Figure 3.17. The mutual impedance Z_xy can be in either line and transferred to the other by means of a transformer that has a 1:1 turns ratio. This technique is also applicable for three-phase lines.

EXAMPLE 3.5

Assume that the mutual impedance between two parallel feeders is 0.09 + j0.3 Ω/mi per phase.

The self-impedances of the feeders are 0.604∠50.4° and 0.567∠52.9° Ω/mi per phase, respec-tively. Represent the mutual impedance between the feeders as shown in Figure 3.16b.

Solution

Z_xy = 0.09 + j0.3 Ω

Z_xx = 0.604∠50.4° = 3.85 + j0.465 Ω Z_yy = 0.567∠52.9° = 0.342 + j0.452 Ω

1 1

3

3 2

2

4 4

(a) (b)

Zxx

Z_yy Zyy – Z_xy

Z_xx – Z_xy

Zxy

Z_xy

FIGURE 3.16 Representation of mutual impedance between two circuits.

1

3

2

4 Z_yy – Z_xy

Zxx – Zxy

Zxy

1 : 1

FIGURE 3.17 Representation of mutual impedance between two circuits by means of 1:1 transformer.

Therefore,

Z_xx − Z_xy = 0.295 + j0.165 Ω Z_yy − Z_xy = 0.252 + j0.152 Ω Hence, the resulting equivalent circuit is shown in Figure 3.18.

3.10 MEDIUM-LENGTH TRANSMISSION LINES (UP TO 150 MI OR 240 KM) As the line length and voltage increase, the use of the formulas developed for the short transmission lines give inaccurate results. Thus, the effect of the current leaking through the capacitance must be taken into account for a better approximation. Thus, the shunt admittance is “lumped” at a few points along the line and represented by forming either a T or a π network, as shown in Figures 3.19 and 3.20. In the figures,

Z = zl For the T circuit shown in Figure 3.19,

V I Z I Z V

FIGURE 3.18 Resultant equivalent circuit for Example 3.5.

C G

FIGURE 3.19 Nominal T circuit.

or

Alternatively, neglecting conductance so that I_C = I_Y

FIGURE 3.20 Nominal π circuit.

Hence,

for a nominal T circuit, the general circuit parameter matrix, or transfer matrix, becomes

For the π circuit shown in Figure 3.20,

V_S=V_R× Y I+ _R Z V_R

By substituting Equation 3.78 into Equation 3.79,

I_S=  + YZ V_R ZI_R Y Y V_R I_R

Alternatively, neglecting conductance,

I = I_C2 + I_R where

I_C2 1Y V_R

= 2 × yields

I= 1Y V× +I

2 ^{R R} (3.81)

Also,

V_S = V_R + IZ (3.82)

By substituting Equation 3.81 into Equation 3.82,

V_S=V_R+ Y V× _R+I_R Z







1 2 or

V YZ V Z I

A

B

S= + R R





 + × 1 1

2  (3.83)

and

I_C1= 1Y V× _S

2 (3.84)

By substituting Equation 3.83 into Equation 3.48,

I_C1 Y 1 1YZ V_R Y ZI_R 2

1

= × + 2





 + ×

1

2 (3.85)

and since

I_S = I + I_C1 (3.86)

by substituting Equation 3.81 into Equation 3.86,

I_S= YV_R+I_R+ Y + YZ V_R YZI_R





 + 1

2

1

2 1 1

2 1

2

or

for a nominal π circuit, the general circuit parameter matrix becomes

A B

As can be proved easily by using a delta–wye transformation, the nominal T and nominal π circuits are not equivalent to each other. This result is to be expected since two different approxima-tions are made to the actual circuit, neither of which is absolutely correct. More accurate results can

be obtained by splitting the line into several segments, each given by its nominal T or nominal π circuits and cascading the resulting segments.

Here, the power loss in the line is given as

Ploss = I²R (3.95)

which varies approximately as the square of the through-line current. The reactive powers absorbed and supplied by the line are given as

QL = Q_absorbed = I²XL (3.96)

and

QC = Qsupplied = V²b (3.97)

respectively. The QL varies approximately as the square of the through line current, whereas the QC

varies approximately as the square of the mean line voltage. The result is that increasing transmis-sion voltages decrease the reactive power absorbed by the line for heavy loads and increase the reactive power supplied by the line for light loads.

The percentage of voltage regulation for the medium-length transmission lines is given by Stevenson [3] as

Percentage of voltage regulation

LN FL

= V −

A V

V

S

R

R ,

,

,,FL

×100 (3.98)

where

|V_S,LN| = magnitude of sending-end phase (line-to-neutral) voltage

|V_S,FL| = magnitude of receiving-end phase (line-to-neutral) voltage at full load with constant |V_S|

|A| = magnitude of line constant A

EXAMPLE 3.6

A three-phase 138-kV transmission line is connected to a 49-MW load at a 0.85 lagging power factor. The line constants of the 52-mi-long line are Z = 95∠78° Ω and Y = 0.001∠90° S. Using nominal T circuit representation, calculate the

(a) A, B, C, and D constants of the line (b) Sending-end voltage

(c) Sending-end current (d) Sending-end power factor

(e) Efficiency of transmission

Solution

V_{R(L N)} 138 kV

3 79,624.3 V

− = =

Using the receiving-end voltage as the reference, V_R(L−N) = 79,624.3∠0° V The receiving-end current is

IR= ×

(a) The A, B, C, and D constants for the nominal T circuit representation are

A= + YZ 0.9535 j0.00999

0.9536 0.6°

91,377 17,028.8 92,951.2 10.6° V

∠ −

= +j = ∠

or

VS(L−L) = 160,996.2∠40.6° V (c) The sending-end current is

IS=0.001 90° 79,674.8 0° 0.9536 0.6° 241.18∠ × ∠ + ∠ × ∠ −31..8°

196.72 39.5 200.64 11.3° A

= −j = ∠ −

(d) The sending-end power factor is

θs = 10.6° + 11.3° = 21.9°

cos Φs = 0.928

(e) The efficiency of transmission is

η =

= ×

= × ×

output input

3

3 100

138 10³ V I V I

R R R

S S S

cos cos Φ Φ

2241 18 0 85 160 996 2 200 64 0 928 100 94 38

. .

, . . .

. %

×

× × ×

=

EXAMPLE 3.7

Repeat Example 3.6 using the nominal π circuit representation.

Solution

(a) The A, B, C, and D constants for the nominal π circuit representation are

A= + YZ

= ∠

1 1 2 0.9536 0.6°

B = Z = 95∠78° Ω

C Y= + Y Z

= ∠ + ∠ ∠

= − 1 4

2

0.001 90° 1

4(0.001 90°)2(95 78°)

4.93379 10× ⁻⁶+j9.7677 10× ⁻⁴=0.001 90.3° S∠

D= + YZ A=

Therefore,

V_{S(L N)−} =0 9536 0 6. ∠ . ° ×79 674 8 0, . ∠ ° + ∠ ° ×95 78 241 46. ∠ −331 8 (c) The sending-end current is

I_S=0 001 90 3. ∠ . ° ×79 674 3 0, . ∠ ° +0 9536 0 6. ∠ . ° ×241 18. ∠ −331 8

(d) The sending-end power factor is

θS = 10.7° + 11.37° = 22.07°

and

cos θS = 0.927

(e) The efficiency of transmission is

η

The discrepancy between these results and the results of Example 4.4 is due to the fact that the nominal T and nominal π circuits of a medium-length line are not equivalent to each other. In fact, neither the nominal T nor the nominal π equivalent circuit exactly represents the actual line because the line is not uniformly distributed. However, it is possible to find the equivalent circuit of a long transmission line and to represent the line accurately.

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