ECE 476 – Power System Analysis Fall 2013Homework 4

Due Date: Tuesday October 1, 2013

Problem 1. A 500-km, 500-kV, 60-Hz uncompensated three-phase line has a positive-sequence series impedancez = 0.03 + j0.35 Ω/km and a positive-sequence shunt admittance y = j4.4× 10−6 S/km. Calculate:

(a) Zc

Zc =

√

z

y=

√

0.03 + j0.35

j4.4× 10−6=

√79837∠− 4.899 = 282.6∠− 2.45 = 282.3− j12.08 Ω

(b) (γd)

γd =√zyd = 500

√

(0.03 + j0.35)(j4.4× 10−6) = 500√

1.546× 10−6∠175.1

= 0.6216∠87.55 = 0.02657 + j0.6210 per unit

(c) The exact ABCD parameters for this line.

eγd = e0.02657ej0.6210 = 1.027∠0.6210 rad = 0.8352 + j0.5975

e−γd = e−0.02657e−j0.6210 = 0.9738∠− 0.6210 rad = 0.7920− j0.5666

cosh γd =eγd + e−γd

2=

1.027∠0.6210+ 0.9738∠− 0.6210

2= 0.8137∠1.088

sinh γd =eγd − e−γd

2=

1.027∠0.6210− 0.9738∠− 0.6210

2= 0.5825∠87.87

A = D = cosh γd = 0.8137∠1.088 per unit

B = Zc sinh γd = (282.6∠− 2.45)(0.5825∠87.87) = 164.6∠85.42 Ω

C =1

Zc

sinh γd =0.5825∠87.87

282.6∠− 2.45= 2.061× 10−3

∠90.32 S

Problem 2. A 320-km 500-kV, 60-Hz three-phase uncompensated line has a positive-sequence series reactancex = 0.34 Ω/km and a positive-sequence shunt admittance y = 4.5× 10−6 S/km. Neglecting losses, calculate:

(a) Its characteristic impedance Zc.

Zc =

√

z

y=

√

j0.34

j4.5× 10−6= 274.9 Ω

(b) The value of γd.

γd =√zyd = 320

√

(j0.34)(j4.5× 10−6) = j0.3958 per unit

1

(c) The exact ABCD parameters for this line.

eγd = ej0.3958 = 1∠22.68 = 0.9227 + j0.3855

e−γd = e−j0.3958 = 1∠− 22.68 = 0.9227− j0.3855

cosh γd =eγd + e−γd

2=

0.9227 + j0.3855 + 0.9227− j0.3855

2= 0.9227

sinh γd =eγd − e−γd

2=

0.9227 + j0.3855− 0.9227 + j0.3855

2= j0.3855

A = D = cosh γd = 0.9227 per unit

B = Zc sinh γd = (274.9)(j0.3855) = j105.97 Ω

C =1

Zc

sinh γd =j0.3855

274.9= j1.402× 10−3 S

(d) The surge impedance loading in MW.

SIL =V 2rated,ll

Zc

=5002

274.9= 909.4 MW, 3φ

Problem 3. The per-phase impedance of a short three-phase transmission line is 0.5∠53.15 Ω. The three-phaseload at the receiving end is 900 kW at 0.8 p.f. lagging. If the line-to-line sending-end voltage is 3.3 kV, determine:

(a) The receiving-end line-to-line voltage in kV.

SL =300000

0.8∠ cos−1 0.8 = 300000+ j225000 VA, and VS =

3300√3∠0 = 1905∠0 V

Let the receiving-end line-to-neutral voltage be VR∠θ. Then,

SL = (VR∠θ)I∗ = (VR∠θ)

(

VS − VR∠θ

Z

)∗

SLZ∗ = 1.905VR∠θ − V 2

R

(300000 + j225000)(0.5∠53.15)∗ = 1905VR∠θ − V 2R

179982− j52562.5 = 1905VR cos θ + j1905VR sin θ − V 2R.

Equating the real and imaginary parts, we obtain

179982+ V 2R = 1905VR cos θ

−52562.5 = 1905VR sin θ.

Taking the square of each and summing, we get

(179982+ V 2R)

2 + (−52562.5)2 = 19052V 2R cos2 θ + 19052V 2

R sin2 θ

3.239352× 1010 + 359964V 2R + V 4

R + 2.76282× 109 = 3629025V 2R

V 4R − 3269061V 2

R + 3.515634× 1010 = 0

V 2R =

3269061±√

32690612 − 4(3.515634× 1010)

2

V 2R = 3258271 or 10790

Therefore,VR = ±1805 V or ± 103.9 V.

Eliminating the negative voltages and the smaller root, we get VR = 1.805 kV. Solving for θ from an earlierequation, we get θ = −0.8759. The line-to-line receiving-end voltage is

√3(1.805) = 3.126 kV.

(b) The line current.

SL = VRI∗

300000 + j225000 = (1805∠− 0.8759)I∗

I = 207.8∠− 37.75 A

(c) The phasor diagram with the line current I, as reference.

I

VS

VR

Problem 4. To maintain a safe “margin” of stability, system designers have decided that the power angle θ12 :=θ1− θ2, where θ1 is the phase angle of the sending-end voltage and θ2 is the phase angle of the receiving-end voltage,cannot be greater that 45. We wish to transmit 500 MW though a 300-mile line and need to pick a transmission-line voltage level. Consider 138−, 345−, and 765−kV lines. Which voltage level(s) would be suitable? As a firstapproximation, assume that the voltage magnitudes on sending and receiving ends are equal, i.e., V1 = V2 and thelines are lossless, i.e., γ = jβ, with β = 0.002 rad/mi.

The real power delivered to the receiving end for a lossless line is

P =VRVS

X ′sin(θ1 − θ2).

Next, we determine X ′ for a lossless line as follows:

X ′ = ωLl

(

sinβl

βl

)

= 2π60(300)L

(

sin(0.002 · 300)0.002 · 300

)

= 106432.6L.

We are given typical values for C for the three voltage levels, so we can solve for L using β = ω√LC.

138 kV: C138 = 8.84× 10−12 F/m = 1.422× 10−8 F/mi =⇒ L138 = 1.979× 10−3 H/mi =⇒ X ′

138 = 210.63 Ω

345 kV: C345 = 11.59× 10−12 F/m = 1.865× 10−8 F/mi =⇒ L345 = 1.509× 10−3 H/mi =⇒ X ′

345 = 160.6 Ω

765 kV: C765 = 12.78× 10−12 F/m = 2.056× 10−8 F/mi =⇒ L765 = 1.369× 10−3 H/mi =⇒ X ′

765 = 145.7 Ω

Assuming VR = VS = V , we get

P =V 2

X ′sin(θ1 − θ2).

For the case of 138 kV line,

sin(θ1 − θ2) =PX ′

138

V 2138

=(500× 106)(210.63)

1380002= 5.53,

which does not have a solution. For the case of 345 kV line,

sin(θ1 − θ2) =PX ′

345

V 2345

=(500× 106)(160.6)

3450002= 0.6746 =⇒ θ1 − θ2 = 42.43 < 45,

which satisfies the safe margin of stability. For the case of 765 kV line,

sin(θ1 − θ2) =PX ′

765

V 2765

=(500× 106)(145.7)

7650002= 0.1245 =⇒ θ1 − θ2 = 7.15 < 45,

which is well under the safe margin of stability. Therefore, both the 345 kV and 765 kV line are suitable to transmit500 MW through the line.

Problem 5. Given a transmission line described by a total series impedance Z = zd = 20+ j80 Ω and a total shuntadmittance Y = yd = j5× 10−4 Ω.

(a) Find its characteristic impedance Zc, γd, eγd, sinh γd, and cosh γd.

Zc =

√

z

y=

√

zd

yd=

√

20 + j80

j5× 10−4=

√164924.225∠− 14.036 = 406.11∠7.02 Ω

γd =√

zdyd =√

(20 + j80)(j5× 10−4) =√0.04123∠165.96 = 0.2031∠82.98 = 0.02482 + j0.2016 per unit

eγd = e0.02482ej0.2016 = 1.025∠0.2016 rad = 1.0042 + j0.2052

e−γd = e−0.02482e−j0.2016 = 0.9755∠− 0.2016 rad = 0.9557− j0.1953

cosh γd =eγd + e−γd

2=

1.0042 + j0.2052 + 0.9557− j0.1953

2= 0.9799∠0.2885

sinh γd =eγd − e−γd

2=

1.0042 + j0.2052− 0.9557 + j0.1953

2= 0.2017∠83.10

(b) Suppose that the line is terminated in its characteristic impedance Zc. Find the efficiency of the transmissionline in this case, i.e., find η = −P21/P12, where P21 is the active power flowing from the receiving end to thesending end of the line, and P12 is the active power flowing from the sending end to the receiving end of the line.

V1 = V2 cosh γd+ I2Zc sinh γd

I1 =V2

Zc

sinh γd+ I2 cosh γd

If the line is terminated in Zc, then V2 = ZcI2 and the above equations become

V1 = V2 cosh γd+ V2 sinh γd = V2 (cosh γd+ sinh γd) = V2eγd

I1 = I2 sinh γd+ I2 cosh γd = I2 (cosh γd+ sinh γd) = I2eγd.

Thus,V2 = V1e

−γd and I2 = I1e−γd.

With γ = α+ jβ, the complex power from receiving end to sending end can be written as

S21 = −V2I∗

2 = −V1e−γd

(

I1e−γd

)∗

= −V1e−αde−jβdI∗1 e

−αdejβd = −S12e−2αd.

Since α is real, the efficiency of the line is

η =−P21

P12= e−2αd = e−2(0.02482) = 0.952