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## Solutions Manual

One of the simplest curvesis P = P ea(t−t )where a is the average per unit growth rate, P is the demand in year t, and Pis the given demand at yeart . 94 Peak PowerDemand GW 84 86 88 90 92 96 98 100 Year .

### 1.1 The demand estimation is the starting point for planning the future electric

One of the simplest curvesis P = P ea(t−t )where a is the average per unit growth rate, P is the demand in year t, and Pis the given demand at yeart . 94 Peak PowerDemand GW 84 86 88 90 92 96 98 100 Year .

### 10 The annual load of a substation is given in the following table. During each 1.3

Annual System LoadInterval – Month Load – MW January 8 February 6 March 4 April 2 May 6 June 12 July 16 August 14 September 10 October 4 November 6 December 8 The following commands data = [ 0 1 8 1 2 6 2 3 4 3 4 2 4 5 6 5 6 12 6 . 4 6 8 10 12 time, month .

### 2.1. Modify the program in Example 2.1 such that the following quantities can be

entered by the user: The peak amplitude , and the phase angle of the sinusoidal supplyV θ v(t) =m v V cos(ωt + θ ). The impedance magnitude Z, and its phase angle γ of the load.m v The program should produce plots for i(t), v(t), p(t), p (t) and p (t), similar tor x Example 2.1.

### 6 The result for the inductive load

Z = 1.25 60 Ω is Enter voltage peak amplitude Vm = 100Enter voltage phase angle in degree thatav = 0 Enter magnitude of the load impedance Z = 1.25Enter load phase angle in degree gama = 60 (a) Estimate from the plotsP = 2000Q = 3464 ◦ 6(b) For the inductive load Z = 1.25 60 Ω, the rms values of voltage and current are◦ 6100 ◦ 6 VV = = 70.71 ◦Ω will result in (a) Estimate from the plots . 60◦ = 56.57 6−60 ◦A Using (2.7) and (2.9), we have P = (70.71)(56.57) cos(60) = 2000 W Q = (70.71)(56.57) sin(60) = 3464 VarRunning the above program for the capacitive load Z = 2.0 6−30 .

### 6 Similarly, for Z = 2.5 Ω, we get

◦) p(t) = 800 + 1000 cos(754t − 36.87 ◦ ◦= 800 + 1000 cos 36.87 cos 754t + sin 36.87 sin 754t = 800 + 800 cos 754t + 600 sin 754t= 800[1 + cos 2(377)t] + 600 sin 2(377)t p(t) is in the same form as (2.5), thus P = 600 W, and Q = 600, Var, or ◦ 6 VAS = 800 + j600 = 1000 36.87 1 ∗(b) Using , we have S = V Im m 2 1◦ ◦ 6 61000 36.87 = 200 Im Im . 6−36.87 10 6 ◦ V I =200 Z L= ◦) A (c) ◦A Therefore, the instantaneous current isi(t) = 10cos(377t − 36.87 6−36.87 .

R X . .

### 2.4. An inductive load consisting of R and X in parallel feeding from a 2400-V rms supply absorbs 288 kW at a lagging power factor of 0.8. Determine R and X

θ = cos−1 36.87◦ = 360 6 36.87◦kVA = 288 kW + j216 kvar R = |V | 2 P= (2400) 2288 × 10 3= 20 Ω X = |V | 2 Q= (2400) 2216 × 10 3= 26.667 Ω source. The two loads draw a total real power of 400 kW at a power factor of 0.8 lagging. One of the loads draws 120 kW at a power factor of 0.96 leading. Find thecomplex power of the other load. 0.8 = 36.87◦ 0.8 The total complex load is S =400 0.8 6 36.87◦ = 500 6 36.87◦kVA = 400 kW + j300 kvar The 120 kW load complex power is S =120 0.96 6−16.26 ◦= 125 6−16.26 ◦kVA = 120 kW − j35 kvar 6 The complex power is S =288 .. .. .. . .. .. . .. .. . .. .. . .. .. . .. .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . .. .. . .. .. . .. .. . .. .. .. . .. .. . . .. . . .. . .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . .. . . . . .. . . . . . .. . . . . . . . . . . . . . . . . . . . . . .. . . . .. . . . . . .. . . . . . . . . . . . . . . . . . . . . . .. . . . . .. . . . . .. . . . . . . . . . . . . . . . . . . . . . .. . .. . . . .. .. .. . .. .. . .. .. . .. .. . .. .. .. . .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. .. . .. .. . .. .. . .. .. .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. .. . .. .. . .. .. . .. .. .. . .. .. . . . .. .. . .. .. . .. .. .. . .. .. . .. .. . .. .. . .. .. . . . . . . . FIGURE 6 An inductive load, with R and X in parallel. . . . . .. . .. . .. . . .. . . . . . . . . . . . . . .. .. . .. . . .. . . . . . . . . . . . . . . .. .. . .. . . .. . . . . . . . . . . . . . .. . .. . .. . .. . . . ... .. . .. .. . .. .. . .. .. . .. .. . . .. . .. . .. .. . .. . .. .. . . . . .. .. . . . . V I R X

### 2.5. Two loads connected in parallel are supplied from a single-phase 240-V rms

6 16.26◦ 6 (8.4 + j11.2) = 1200 + j350 = 1250 36.87◦ 6 6 ◦ 36.87 A Thus, the supply voltage isV = 1200 6 ◦= 25 224000 = 60 Ω 1200 36.87◦ 6 , the current is I =30000 From S = V I∗ 218000 = 80 Ω 2 Q= (1200) X = |V | . Therefore, the second load complex power is S (a) Find the values of R and X.(b) Determine the supply voltage V .

### 2.7. Two impedances, Z

1= 0.8 + j5.6 Ω and Z . Find the kvar and the ca- pacitance in .

### 2.8. Two single-phase ideal voltage sources are connected by a line of impedance of

V 6 ◦ 120.7 + j2.4 Ω 585 16.26◦ 6 ◦0.7 + j2.4 = 42 + j56 = 70 6 500 . ◦= 28000 − j21000 VA S ◦= −24570 + j32760 VA 6−53.13 ◦) = 40950 6−53.13 6 ◦)(−70 21= (585 2 I ∗ 21= V 6−36.87 53.13◦ ◦) = 35000 6−53.13 )(70 16.26◦ 6 12= (500 1 I ∗ 12= V A S .

### 2.10. A balanced three-phase source with the following instantaneous phase volt-

plot(wt, pa, wt, pb, wt, pc, wt, p), grid xlabel(’Radian’)disp(’(b)’) V = 2500/sqrt(2);gama = acos(0.8); Z = 250*(cos(gama)+j*sin(gama));I = V/Z; P = 3*V*abs(I)*0.8 . Van = 4157√ 3 = 2400 V V I = 1767.776 ◦ (b) V =2500 √ .

### 2.11. A 4157-V rms three-phase supply is applied to a balanced Y-connected three-

With V as reference, the phase voltages are:an ◦ ◦ ◦ 6 6 6 V V V V = 2400V = 2400 V = 2400an bn an −120 −240 (a) The phasor currents are:◦ 6 V 2400 an◦ 6 AI = = = 50a −36.87◦ 48 36.87◦ 6 V 2400 bn−120 ◦ 6 AI = = = 50b −156.87◦ 48 6 Z 36.87◦ 6 V cn 2400−240 ◦ 6 AI = = = 50c −276.87◦ 48 36.87 (b) The total complex power is∗ ◦ ◦ ◦ 6 Z 6 6 6kVA S = 3V I = (3)(2400 )(50 36.87 ) = 360 36.87an a = 288 kW + j216 KVAR

### 2.12. Repeat Problem 2.11 with the same three-phase impedances arranged in a ∆ connection. Take as reference

Vab 4157 V = = 2400 Van √ 3 With as reference, the phase voltages are: Vab ◦ 6 V 4157 ab◦ 6 AI = = = 86.6ab −36.87◦ 48 36.87 √ √◦ ◦ ◦ ◦ 6 6 6 6 AI a = 3 I = ( 3 )(86.6 = 150ab −30 −30 −36.87 −66.87 For positive phase sequence, current in other lines are◦ ◦ 6 6 A AI = 150 , and I c = 150 53.13b −186.87 (b) The total complex power is∗ ◦ ◦ ◦ 6 6 6 S = 3V I = (3)(4157 )(86.6 36.87 ) = 1080 36.87 kVA abab = 864 kW + j648 kvar

### 2.13. A balanced delta connected load of

S = 3Van 6 ◦ V V 2 ❜ ❜a n . (a) I a =120 6 ◦6 + j8 = 12 6−53.13 ◦A (b) The total complex power is I∗ a V = (3)(120 6 ◦)(12 6 53.13◦ = 4320 6 53.13◦ VA = 2592 W + j3456 Var(c) V 2= 120 6 ◦− (1 + j2)(12 6−53.13 ◦= 93.72 6−2.93 1= 120 1 + j2 Ω 5 Ωj6 Ω .

### 0.6 S = 6 kW + j0 kvar

3= 0 kW − j16 kvar (a) The total complex power is ◦

### 6 A at unity power factor

The supply current is 18000◦ I = = 50 , (3)(120)(b) With the capacitor switched off, the total power is ◦ 6kVA S = 18 + j16 = 24.08 41.63 624083 −41.63 ◦ 6 AI = = 66.89 −41.63 ◦ 6(3)(120 ) ◦The power factor is cos 41.63 = 0.747 lagging.

### 2.15. Three loads are connected in parallel across a 12.47 kV three-phase supply

(a) Find the total complex power, power factor, and the supply current.(b) A Y-connected capacitor bank is connected in parallel with the loads. Find the total kvar and the capacitance per phase in µF to improve the overall power factor to 0.8 lagging.

### 3 The supply current is

2 2 L|V | (12.47 × 1000) X = =c = −j740.48 Ω∗ S j210000c 6 10 C = = 3.58µF (2π)(60)(740.48) . 4 + j3 = 40 ◦A The total complex power is .

## CHAPTER 3 PROBLEMS A three-phase, 318.75-kVA, 2300-V alternator has an armature resistance of 3.1

0.35 Ω/phase and a synchronous reactance of 1.2 Ω/phase. Determine the no-loadline-to-line generated voltage and the voltage regulation at (a) Full-load kVA, 0.8 power factor lagging, and rated voltage.

### 6 S 318750

−36.87◦ 6 AI = = = 80a −36.87∗ 3V (3)(1327.9)φ ◦ ◦ 6 6 VE = 1327.9 + (0.35 + j1.2)(80 ) = 1409.2 2.44φ −36.87 The magnitude of the no-load generated voltage is√ E = 3 1409.2 = 2440.8 V, and the voltage regulation is LL2440.8 − 2300 V. 318.75 kVA, 0.6 power factor leading, S = 318750 −53.13∗ ◦ 53.13◦ 6 S 318750 6 AI = = = 80 53.13a ∗ 3V (3)(1327.9φ ◦ ◦ 6 3.61 Vφ 6 E = 1327.9 + (0.35 + j1.2)(80 53.13 ) = 1270.4 The magnitude of the no-load generated voltage is √E = 3 1270.4 = 2220.4 V, and the voltage regulation is LL2200.4 − 2300 V.

### 69.3 V = = 40 kV

(a) The generator is delivering rated power at 0.8 power factor lagging at the rated terminal voltage to an infinite bus bar. Determine the magnitude of the generatedemf per phase and the power angle δ.(b) If the generated emf is 36 kV per phase, what is the maximum three-phase power that the generator can deliver before losing its synchronism?(c) The generator is delivering 48 MW to the bus bar at the rated voltage with its field current adjusted for a generated emf of 46 kV per phase.

### 17.32 V = = 10 kV

φ √ 3◦ 6(a) For kVA. 24000 kVA, 0.8 power factor lagging, S = 24000 36.87∗ ◦

### 6 S 24000

−36.87◦ 6 I = = = 800 A a−36.87 ∗ 3V (3)(10)φ ◦ −3 ◦ 6 6 E = 10 + (j5)(800 = 12.806 14.47 kV φ−36.87 ) × 10 (b) (3)(13.4)(10) 3|E||V |P = = = 80.4 MWmax X 5s ◦ (c) At maximum power transfer , and solving for the armature current from δ = 90, we have E = V + jX Is a ◦ ◦ 6 613400 90 − 10000◦ 6 AI a = = 3344 36.73 j5

### 3.4. A 34.64-kV, 60-MVA, three-phase salient-pole synchronous generator has a

(a) Form the phasor diagram shown in Figure 17, we have |V | sin δ = Xq = Xq . (a) Referring to the phasor diagram of a salient-pole generator shown in Figure 17, show that the power angleδ is given by δ = tan .

φ √ 3◦ 6(a) For

### 60 MVA, 0.8 power factor lagging, S =

(a) Referring the secondary impedance to the primary side, the transformer equiv- alent impedance referred to the high voltage-side isZe1 = 0.2 + j0.45 +µ 2400 240¶ 2(0.002 + j0.0045) = 0.4 + j0.9 Ω . = 2453.933 − 24002400 × 100 = 2.247% (c) At full-load 0.8 power factor leading S = 150 6−36.87 I V∗ V V 1.44◦ 6 ◦) = 2387.004 6 1= 2400 + (0.4 + j0.9)(62.5 ◦A 2′ 6 ◦2400 = 62.5 6 = 150000 V∗ = S∗ = 150000 = S∗ .

### 3.6. A 60-kVA, 4800/2400-V single-phase transformer gave the following test

to primary = 12.500 A at -36.87 degrees Primary no-load current = 1.462 A at -53.13 degreesPrimary input current = 13.910 A at -38.56 degrees Primary input voltage = 5848.290 V at . 7.37 degrees Voltage regulation = 21.839 percentTransformer efficiency = 85.974 percent Maximum efficiency is 86.057%occurs at 53.33kVA with 0.8pf 50 .

### 3.7. A two-winding transformer rated at 9-kVA, 120/90-V, 60-HZ has a core loss of 200 W and a full-load copper loss of 500 W

(a) With windings carrying rated currents, the auto transformer rating is S = (120)(175)(10−3 (21000)(0.8) (21000)(0.8) + 200 + 500× 100 = 96 % . The load complex power is S = 18000 6 36.87◦kVA I 1= 18000 6−36.87 ◦(3)(7.2) = 833.333 6−36.87 ◦A .

### 18 MVA, 0.8 power factor lagging at 4.16 kV. Determine the line-to-line voltage at the high-voltage terminals of the transformer

Ze1 = 0.12 + j0.82 V 1 V ′ 2= 7200 V I . V 1= 7.2 6 ◦ 6−36.87 ◦)(10 −3) = 7.7054 6 3.62◦kV Therefore, the magnitude of the primary line-to-line supply voltage is V .

### 3.9. A 400-MVA, 240-kV/24-KV, three-phase Y-

The primary is supplied from a feeder with an impedance of . V′ Vs = 1.2 + j6 = 0.6 + j1.2 Ze1 Zf .

### 3.11. A three-phase, Y-connected, 75-MVA, 27-kV synchronous generator has a

Using rated MVA and voltage as basevalues, determine the per unit reactance. Then refer this per unit value to a 100- MVA, 30-kV base.

### 9.72 The generator reactance on a 100-MVA, 30-kV base is

µ ¶ µ ¶ 2100 27 X = 0.926 = 1.0 pupunew 75 30

### 3.12. A 40-MVA, 20-kV/400-kV, single-phase transformer has the following series

impedances: Z = 0.9 + j1.8 Ω and Z = 128 + j288 Ω 1 2 Using the transformer rating as base, determine the per unit impedance of the trans-former from the ohmic value referred to the low-voltage side. Compute the per unit impedance using the ohmic value referred to the high-voltage side.

### 20 Z = 0.9 + j1.8 + (128 + j288) = 1.22 + j2.52 Ω

= 488 + j10084000 = 0.122 + j0.252 pu 40 = 4000 ΩZpu2 2 Z B 2 = (400) 2(0.9 + j1.8) + (128 + j288) = 488 + j1008 Ω The high-voltage base impedance is 20¶ 2: . 1: X = 0.16 90¶ µ = 0.30 pu The load impedance in ohms isZ L 120 400¶ The per unit line reactance is Line: X =µ 2100 = 400 Ω Z BL = (200) 2= 0.081 pu The base impedance for the transmission line is 20¶ 18 µ100 ) 2: X = 0.09 = 0.25 pu G 80¶ µ100 2: X = 0.20 = 0.20 pu T 80¶ µ100 = (VL−L 2 S ∗L(3φ) 90¶ .

## B5 B6

22 Voltage base for the tertiary side of is T 4µ ¶ 4 V = 110 = 4 kVBT 110 The per unit impedances on a 100 MVA base are:µ ¶ 100

## G: X = 0.24 = 0.30 pu

80µ ¶ 100 :T

### 1 X = 0.0 = 0.20 pu

However, the base voltage at bus 4 for the motor is 22 kV, thereforeµ ¶ µ ¶ 2100 20 M : X = 0.225 = 0.27 pu 68.85 22 Impedance bases for lines 1 and 2 are 2(220) Z B2 = = 484 Ω 100 2(110) Z B5 = = 121 Ω 100 . 40¶ 40¶ 100 = 0.120µ = 0.18 pu ZST 40¶ 100 = 0.072µ = 0.24 pu ZP T 100 = (4) = 0.096µ The three-winding impedances on a 100 MVA base are ZP S = −j1.60.16 = −j10 pu Therefore, the load impedance in per unit is ZL(pu) 2100 = 0.16 Ω = (4) The base impedance for the load is ZBT 2j10 = −j1.6 Ω .

### 1 T

Hence the base on its HV side is V B1 = 20µ 200 20¶ = 200 kV This fixes the base on the HV side ofT 2at V B2 = 200 kV, and on its LV side at V B m = 200µ 20 200¶ = 20 kV . 6 ◦pu 6 ◦= 0.90 18 = Vm 36.87◦pu and the motor terminal voltage is 6 100 = 0.45 36.87◦ 6 45 = (b) The motor complex power in per unit is Sm .

V 1 . .

## CHAPTER 4 PROBLEMS A solid cylindrical aluminum conductor 25 km long has an area of 336,400 4.1

2.8 × 10 (a)√ −3d = 336400 = 580 mil = (580)(10 )(2.54) = 1.4732 cm 2πd 2 A = = 1.704564 cm 4 3ℓ 25 × 10 −8R = ρ = 4.106 Ω 1= 2.8 × 10 −4A 1.704564 × 10 (b)µ ¶ µ ¶ 228 + 50 278 R = R = 4.106 = 4.6 Ω 2 1228 + 20 248 A transmission-line cable consists of 12 identical strands of aluminum, each 4.2. 2.8 × 10◦ Find the C AC resistance per Km of the cable.

### 4.3. A three-phase transmission line is designed to deliver 190.5-MVA at 220-

The total transmission line loss is not to exceed 2.5 percent of the rated line MVA. If the resistivity of the conductor material is−8 Ω-m, determine the required conductor diameter and the conductor 2.84 × 10size in circular mils.

### 2.5 P = (190.5) = 4.7625 MW

What is the value of the inductance per conductor?r . Find the geometric mean radius of a conductor in terms of the radiusr of an individual strand for .

### 4.6. One circuit of a single-phase transmission line is composed of three solid 0.5- cm radius wires. The return circuit is composed of two solid 2.5-cm radius wires

The arrangement of conductors is as shown in Figure 35. Applying the concept of theGM D and GM R, find the inductance of the complete line in millihenry per kilometer.

### 6 D = (20)(25)(15)(20)(10)(15) = 1.802 m

mq 1 1 1 9− − − 4 4 × 0.005 × 5 × 10)(e × 0.005 × 5 × 5)(e × 0.005 × 5 × 10) = 0.5366 mr ³ ´ 4 D = (eS X 1 2 4− S Y× 0.025 × 5 Therefore

### 4 D = e = 0.312 m

16.802 L = 0.2 ln = 0.6888 mH/Km X0.5366 and 16.802 L = 0.2 ln = 0.79725 mH/KmY 0.312 The loop inductance isL = L + L = 0.6888 + 0.79725 = 1.48605 mH/Km X Y

### 4.7. A three-phase, 60-Hz transposed transmission line has a flat horizontal con-

The GM R of each conductor is 1.515 cm.(a) Determine the inductance per phase per kilometer of the line.(b) This line is to be replaced by a two-conductor bundle with 8-m spacing mea- sured from the center of the bundles as shown in Figure 38. If the line inductance per phase is to be 77 percent of the inductance in part (a), what would be the GM R of each newconductor in the bundle?a c b ✛ ✲✛ ✲ ♠ ♠ ♠D = 8 m D = 8 m 12 23✛ ✲ D = 16 m 13 FIGURE 37 Conductor layout for Problem 4.8 (a).

### 3 GM D = (8)(8)(16) = 10.0794 m

10.0794 Db Sor Db S (b) L 12= 8 m D . 23= 8 m D 13= 16 m ab c .

### 4.9. A three-phase transposed line is composed of one ACSR 1,431,000 cmil, 47/7

✲ ✛ ❢ ❢ ❢✛ ✲✛ ✲ ✛ ✲ . 23= 14 m D 13= 28 m ab c .

### 4.10. A single-circuit three-phase transmission transposed line is composed of four

(14)(14)(28) = 17.63889 m and From (4.53) and (4.90), we haveGM R L = 1.09 ✲ ✲ ✛ = 0.889 mH/Km L = 0.2 ln GM D= 0.2 ln 17.63889 3= 21.8 cm and from (4.58) and (4.92), we get (3.5103/2)(45) 4q C= 1.09 3= 20.66 cm GM R (1.4173)(45) 4q 3q . 23= 14 m D 13 = 28 ma bc .

## GM RC 0.218

The function gmd is used to verify the results. The following commands [GMD, GMRL, GMRC] = gmdL = 0.2*log(GMD/GMRL); %mH/Km C = 0.0556/log(GMD/GMRC); %microF/Kmresult in GMD = 17.63889 m GMRL = 0.20673 m GMRC = 0.21808 m L =0.8893 C =0.0127 A double circuit three-phase transposed transmission line is composed of two

### 4.11. ACSR 2,167,000 cmil, 72/7 Kiwi conductor per phase with vertical configuration

The conductors have a diameter of 4.4069 cm and a GM Rof 1.7374 cm. The circuit arrangement is , a b c 1 1 1 .

### 2 Find these values when the circuit arrangement is , . Use function

a b c a b c 1 1 1 2 2 2[GMD, GMRL, GMRC] =gmd , (4.58) and (4.92) in MATLAB to verify your re-sults. For the , configuration, the following distances are computed in a 1 b 1 c 1 c 2 b 2 a 2a c 1 2✛ ✲ ❡ ❡ ❡ ❡ S = 16 ma 1 a 2✻ H = 10 m 12✛ ❄ ✲ ❡ ❡ ❡ ❡ S = 24 m b b b b 1 1 2 2✻ H = 9 m 23❄ ✛ ✲ ❡ ❡ ❡ ❡ S c c = 17 m 1 2c a 1 2 FIGURE 42 Conductor layout for Problem 4.11.

### 3 GM D = (15.1057)(15.1057)(17.7049) = 15.9267

From (4.51) and (4.66), we haveq pb D = D (0.017374)(0.45) = 0.08842s s × d =q √br = (0.044069/2)(0.45) = 0.09957 r × d = From (4.56) and (4.93), we havep p D = (0.08842)(25.1644) = 1.4916 D = (0.08842)(24.0000) = 1.4567 pD = (0.08842)(25.1644) = 1.4916 SCandp pr = (0.09957)(25.1644) = 1.5829 r = (0.09957)(24.0000) = 1.5458 A Bpr = (0.09957)(25.1644) = 1.5829 Cq 3 GM R = (1.4916)(1.4567)(1.4916) = 1.4799Lq 3 GM R = (1.5829)(1.5458)(1.5829) = 1.5705C Therefore, the inductance and capacitance per phase are GM D 15.9267L = 0.2 ln = 0.2 ln = 0.4752 mH/Km GM RL 1.4799and 0.0556 0.0556 C = = = 0.0240 µF/KmGM D 15.9267ln ln

## GM RC 1.5705

The following commands [GMD, GMRL, GMRC] = gmd L = 0.2*log(GMD/GMRL); %mH/KmC = 0.0556/log(GMD/GMRC); %microF/Km result in Circuit Arrangements Enter row vector [H12, H23] = [10 9] Cond. (enter 1 for single cond.) = 2 Bundle spacing in cm = 45GMD = 15.92670 m GMRL = 1.47993 m GMRC = 1.57052 mL = 0.4752C = 0.0240 For the , configuration, the following distances are computed in me- a b c a b c 1 1 1 2 2 2 ter.

### 2 From (4.54), we have

q 4 D AB = (10.7703)(22.3607)(22.3607)(10.7703) = 15.5187q 4 D = (9.6566)(22.3886)(22.3886)(9.6566) = 14.7036BCq 4 D = (19.0065)(25.1644)(25.1644)(19.0065) = 21.8698CA The equivalent geometric mean distance isq

### 3 GM D = (15.5187)(14.706)(21.8698) = 17.0887

From (4.51) and (4.66), we haveq pb D = D (0.017374)(0.45) = 0.08842s s × d =q √br = (0.044069/2)(0.45) = 0.09957 r × d = From (4.56) and (4.93), we havep p D = (0.08842)(16) = 1.1894 D = (0.08842)(24.0000) = 1.4567 pD = (0.08842)(17) = 1.2260 SCandp pr = (0.09957)(16) = 1.2622 r = (0.09957)(24.0000) = 1.5458 A Bp (0.09957)(17) = 1.3010 r C =q 3 GM R = (1.1894)(1.4567)(1.2260) = 1.2855Lq 3 GM R C = (1.2622)(1.5458)(1.3010 = 1.3641Therefore, the inductance and capacitance per phase are GM D 17.0887 L = 0.2 ln = 0.2 ln = 0.5174 mH/Km GM RL 1.2855and 0.0556 0.0556 C = = = 0.02199 µF/KmGM D 17.0887ln ln GM RC 1.3641 The function gmd is used in the same way as before to verify the results.

### 4.12. The conductors of a double-circuit three-phase transmission line are placed

√ D = D = D = D = D = D = 3Da 1 c 1 a 1 b 2 b 1 a 2 b 1 c 2 c 1 b 2 c 2 a 2q 4 D = D = D = D D D DAB BC CA a b a b a b a b 1 1 1 2 2 1 2 2q 1 4= D D = 3 Da 1 b 1 a 1 b 2 ❦ ❦ ❦ ❦❦ ❦ ✛ ✲ . a .

### 4.14. A three-phase, 60-Hz, untransposed transmission line runs in parallel with a

telephone line for 20 km. The power line carries a balanced three-phase rms current◦ ◦ ◦ 6 6 6of A, A, and A I = 320I = 320 I = 320a b c −120 −240 The line configuration is as shown in Figure 45.

❧❧ ❧ ❧ . ❧ ❧ .

### 5.1. A 69-kV, three-phase short transmission line is 16 km long. The line has a per

Determine the sending endvoltage, voltage regulation, the sending end power, and the transmission efficiency when the line delivers(a) 70 MVA, 0.8 lagging power factor at 64 kV.(b) 120 MW, unity power factor at 64 kV. Selectingoption 1 and entering data for part (b), will result in Line performance for specified receiving end quantities ◦Vr = 64 kV (L-L) at 0 Pr = 120 MW Qr = 0 Mvar ◦Ir = 1082.53 A at 0 PFr = 1 ◦Vs = 69.0096 kV (L-L) at 10.9639 ◦Is = 1082.53 A at 0 PFs = 0.981747 Ps = 127.031 MW Qs = 24.069 MvarPL = 7.031 MW QL = 24.069 Mvar Percent Voltage Regulation = 7.82754Transmission line efficiency = 94.4649 mance of Problem 5.1.

### 6 Z = 2 + j7 = 7.28 74.054

−6z = 0.05 + j0.45 Ω per Km and a per phase shunt admittance of y = j3.4 × 10 siemens per km. Find the sending end voltage and cur-rent, voltage regulation, the sending end power and the transmission efficiency when the line delivers(b) 200 MVA, 0.8 lagging power factor at 220 kV.(c) 306 MW, unity power factor at 220 kV.

### 4 The receiving end voltage per phase is

Selectingoption 1 and entering data for part (b), will result in Line performance for specified receiving end quantities ◦Vr = 220 kV (L-L) at 0 Pr = 306 MW Qr = 0 Mvar ◦Ir = 803.042 A at 0 PFr = 1 ◦Vs = 230.029 kV (L-L) at 12.6033 ◦Is = 799.862 A at 2.5008 PFs = 0.984495 Ps = 313.742 MW Qs = 55.900 MvarPL = 7.742 MW QL = 55.900 Mvar Percent Voltage Regulation = 5.0732Transmission line efficiency = 97.532 mance of Problem 5.3. (a) Determine the total Mvar and the capacitance per phase of the Y-connected ca- pacitors when the sending end voltage is 220 kV.

### 5.5. A three-phase, 345-kV, 60-Hz transposed line is composed of two ACSR 1,113,000, 45/7 Bluejay conductors per phase with flat horizontal spacing of 11 m

The resistance of each conductor in the bundle is 0.0538 Ω perkm and the line conductance is negligible. Using the nom- inal π model, determine the ABCD constant of the line.

### 3 GM D = (11)(11)(22) = 1.859 m

qGM R = (45)(1.268) = 7.5538 cm Lq GM R = (45)(3.195/2) = 8.47865 cmC 13.859 L = 0.2 = 1.0424 mH/Km−2 7.5538 × 10 0.0556C = = 0.010909 µF/Km 13.859ln −2 8.47865×100.0538 −3Z = ( )(150) = 4.035 + j58.947 2−6 )(150) = j0.0006169 Y = j(2π60 × 0.9109 × 10The ABCD constants of the nominal π model are µ ¶ µ ¶ZY (4.035 + j38.947)(j0.0006169) A = 1 + = 1 + 2 2 = 0.98182 + j0.0012447B = Z = 4.035 + j58.947 µ ¶ZY C = Y 1 + = j0.00061137 4 Using lineperf and option 5, result in Type of parameters for input SelectParameters per unit length r ( Ω), g (Siemens), L (mH), C (µF) 1 Complex z and y per unit length r + j*x ( Ω), g + j*b (Siemens) 2 Nominal π or Eq. π model 3 A, B, C, D constants 4 Conductor configuration and dimension 5 To quit Select number of menu→ 5 When the line configuration and conductor specifications are entered, the following results are obtainedGMD = 13.85913 m GMRL = 0.07554 m GMRC = 0.08479 mL = 1.04241 mH/Km C = 0.0109105 micro F/Km Short line modelZ = 4.035 + j 58.947 ohms Y = 0 + j 0.000616976 Siemens ·¸ 0.98182 + j0.0012447 4.035 + j58.947 ABCD= −3.8399e − 007 + j0.00061137 0.98182 + j0.0012447

### 5.6. The ABCD constants of a three-phase, 345-kV transmission line are

A = D = 0.98182 + j0.0012447 B = 4.035 + j58.947C = j0.00061137 The line delivers 400 MVA at 0.8 lagging power factor at 345 kV. Determine the receiving end quantities, voltage regulation, and the transmission efficiencywhen the line is sending 407 MW, 7.833 Mvar at 350 kV.

### 5.8. Then, solve each Problem using hand calculations

The line is 400 Km long, and for the purpose of this problem, a losslessline is assumed.(a) Determine the transmission line surge impedance Z , phase constant β, wave-c length λ, the surge impedance loading SIL, and the ABCD constant. Determine the sending end quantities and voltage regulation.(c) Determine the receiving end quantities when 1920 MW and 600 Mvar are being transmitted at 765 kV at the sending end.(d) The line is terminated in a purely resistive load.

### 3 GM D = (14)(14)(28) = 17.689 m

q 4 3 GM R = 1.09 (45) (1.439) = 20.75 cm Lq 4 3 GM R = 1.09 (45) (3.625/2) = 21.98 cm C17.6389 L = 0.2 = 0.88853 mH/Km −220.75 × 10 0.0556 C = = 0.01268 µF/Km17.6389ln −2 −9β = ω = 0.001265 Radian/Km LC = 2π × 60 0.88853 × 0.01268 × 10◦ βℓ = (0.001265 × 400)(180/π) = 29 2π 2πλ = = = 4967 Km β 0.001265s s −3L 0.88853 × 10 Z = = = 264.7 Ωc −6 C 0.01268 × 10 2 2(KV ) (765) LratedSIL = = = 2210.89 MW Z 264.7c The ABCD constants of the line are◦ A = cos βℓ = cos29 = 0.8746◦ B = jZ sin βℓ = j264.7 sin 29 = j128.33c 1 1◦ C = j sin βℓ = j sin 29 = j0.0018315 Z 264.7c D = A (b) The complex power at the receiving end is◦ 6 S = 2000 36.87 = 1600 MW + j1200 Mvar R(3φ)◦ 6735 ◦ 6kV V = = 424.352 R √

### 3 The current per phase is given by

∗ ◦S 6 R(3φ) 2000000− 36.87 ◦ 6 I = = = 1571.02 A R− 36.87 ∗ ◦ 6 3 VR 3 × 424.352 The sending end voltage is◦ 6 V = AV + BI = (0.8746)(424.352 ) + (j128.33) S R R◦ ◦ −36 6kV ) = 517.86 18.147 (1571.02 × 10 −36.87 The sending end line-to-line voltage magnitude is√ S|V S(L−L) | = 3 |V | = 896.96 kV 6 I = CV + DI = (j0.0018315)(424352 ) + (0.8746) S R R◦ ◦ 6 6 A(1571.02 ) = 1100.23 −36.87 −2.46 The sending end power is∗ ◦ −3 6 I 6 S = 3V 2.46 S(3φ) SS = 3 × 517.86 18.147 × 1100.23 × 10 = 1600 MW + j601.59 Mvar◦ = 1709.3

### 20.6 Voltage regulation is

896.96− 735 0.8746Percent V R = × 100 = 39.53%735 (c) The complex power at the sending end is ◦ 6 MVAS = 1920 MW + j600 Mvar = 2011.566 17.354 S(3φ) ◦ 6765 ◦ 6 V = = 441.673 kV S√

### 6 S(3φ) 20115

∗ ◦S − 17.354◦ 6 AI = = = 1518.14 S− 17.354 ∗ ◦ 6 3 V 3 × 441.673S The receiving end voltage is◦ 6 V = DV = (0.8746)(441.673 R S S− BI ) − (j128.33) ◦ ◦ −36 6) = 377.2 kV (1518.14 × 10 −17.354 −29.537 The receiving end line-to-line voltage magnitude is √S |V R(L−L) | = 3 |V | = 653.33 kV The receiving end current is◦ 6 I + AI ) + (0.8746) R S S= −CV = (−j0.0018315)(441673 ◦ ◦ 6 6 A(1518.14 ) = 1748.73 −17.354 −43.55 6 I 6 S = 3V R 43.55 R(3φ) R = 3 × 377.2 − 29.537 × 1748.73 × 10= 1920 MW + j479.2 Mvar ◦ 6= 1978.86 14.013 MVA Voltage regulation is 765− 653.33 0.8746Percent V R = × 100 = 33.88%653.33 (d) ◦ 6735 ◦ 6kV V = = 424.352 R√

### 3 The receiving end current per phase is given by

◦ R 6 V 424352 ◦ 6 A I R = = = 1604.357 Z 264.5L The complex power at the receiving end is∗ ◦ ◦ −3 6 I = 3(424.352 )(1604.357 = 2042.44 MW 6 S = 3V R(3φ) R R) × 10 The sending end voltage is ◦ 6 V = AV + BI = (0.8746)(424.352 ) + (j128.33) S R R◦ ◦ −36 6) = 424.42 29.02 kV = (1604.357 × 10 The sending end line-to-line voltage magnitude is √S |V S(L−L) | = 3 |V | = 735.12 kV The sending end current is◦ 6 I = CV + DI = (j0.0018315)(424352 ) + (0.8746) S R R◦ ◦ 6 6(1604.357 ) = 1604.04 28.98 A The sending end power is∗ ◦ −3 6 I S(3φ) SS = 3 × 424.42 29.02 × 1604.04 − 28.98 × 10 6 S = 3V = 2042.4 MW + j1.4 Mvar◦ = 2042.36 0.04 − 7350.8746 Percent V R = × 100 = 14.36%735

### 5.9. The transmission line of Problem 5.8 is energized with 765 kV at the sending end when the load at the receiving end is removed

(a) Find the receiving end voltage.(b) Determine the reactance and the Mvar of a three-phase shunt reactor to be installed at the receiving end in order to limit the no-load receiving end voltage to735 kV.(a) The sending end voltage per phase is ◦ 6765 ◦ 6kV V = = 441.673 S√ 3 When line is open I = 0 and from (5.71) the no-load receiving end voltage isRgiven by V 441.673 SV = = = 505 kV R(nl)cos βℓ 0.8746 The no-load receiving end line-to-line voltage is √ V =

### 3 V = 874.68 kV

R(nl) R(L−L)(nl)(b) For V = 735 kV, the required inductor reactance given by (5.100) is RLL ◦sin(29 ) X = (264.7) = 772.13 Ω Lsh 765◦ ) − cos(29735 The three-phase shunt reactor rating is 2 2(KV ) (735) Lrated= = 699.65 Mvar Q 3φ = X 772.13Lsh

### 5.10. The transmission line of Problem 5.8 is energized with 765 kV at the sending

end when a three-phase short-circuit occurs at the receiving end. Determine the receiving end current and the sending end current.

### 5.11. Shunt capacitors are installed at the receiving end to improve the line per- formance of Problem 5.8. The line delivers 2000 MVA, 0.8 lagging power factor

(a) The equivalent line reactance for a lossless line is given by′ ◦ X = Z sin βℓ = 264.7 sin(29 ) = 128.33 Ωc The receiving end power is−1 6 S = 2000 cos (0.8) = 1600 + j1200 MVA R(3φ)For the above operating condition, the power angle δ is obtained from (5.93) (765)(735) 1600 = sin δ128.33 ◦which results in δ = 21.418 . Using the approximate relation given by (5.94), the net reactive power at the receiving end is 2(765)(735) (735) ◦ ◦Q = cos(21.418 cos(29 ) = 397.05 Mvar R(3φ) ) −128.33 128.33 Thus, the required capacitor Mvar is S C= j397.05 − j1200 = −j802.95.

### 5.12. Series capacitors are installed at the midpoint of the line of Problem 5.8

providing 40 percent compensation. Determine the sending end quantities and thevoltage regulation when the line delivers 2000 MVA at 0.8 lagging power factor at 735 kV.

### 6 R(3φ) 2000

−36.87◦ 6 kA I = = = 1.57102R −36.87∗ ◦ 6 3V R 3 × 424.35 Thus, the sending end voltage is◦ 6 V = AV + BI S R R= 0.92476 × 424.352 + j77 × 1.57102 −36.87 ◦ 6kV = 474.968 11.756 √ and the line-to-line voltage magnitude is 3 V = 822.67 kV. TheS |V | =S(L−L)sending end current is ◦ 6 I = CV + DI = (j0.0018805)(424352 ) + (0.92476) S R R◦ ◦ 6 6 A(1571.02 ) = 1164.59 − 36.87 −3.628 The sending end power is∗ ◦ −3 6 6 S = 3V I 3.628 SS(3φ) S = 3 × 474.968 11.756 × 1164.59 × 10 = 1600 MW + j440.2 Mvar◦ 6= 1659.4 15.38 MVA Voltage regulation is822.67/0.92476 − 735 Percent V R = × 100 = 21.035%

### 5.13. Series capacitors are installed at the midpoint of the line of Problem 5.8, pro-

The receiving end power is −1 6 S = 2000 cos (0.8) = 1600 + j1200 MVA R(3φ)′ With the series reactance X = 77 Ω, and cos βℓ = A = 0.92476, the power angleδ is obtained from (5.93) (765)(735) 1600 = sin δ 77◦which results in . Using the approximate relation given by (5.94), the δ = 12.657net reactive power at the receiving end is 2(765)(735) (735) ◦Q = cos(12.657 0.92476 = 636.75 Mvar R(3φ) ) − 77 77 Thus, the required shunt capacitor Mvar is SC = j636.75 − j1200 = −j563.25.

### 5.14. The transmission line of Problem 5.8 has a per phase resistance of

Obtain the voltage profile forthe uncompensated and the compensated line.(e) Find the receiving end and the sending end current when the line is terminated in a three-phase short circuit.(f) For the line loading of part (a), determine the Mvar and the capacitance of the shunt capacitors to be installed at the receiving end to keep the receiving end volt-age at 735 kV when line is energized with 765 kV. Determine the Mvar and the capacitance of the shunt capacitors to beinstalled at the receiving end to keep the receiving end voltage at 735 kV when line is energized with 765 kV at the sending end.(i) Obtain the receiving end circle diagram.(j) Obtain the line voltage profile for a sending end voltage of 765 kV.(k) Obtain the line loadability curves when the sending end voltage is 765 kV, and the receiving end voltage is 735 kV.

### 5.15. The ABCD constants of a lossless three-phase, 500-kV transmission line are

A = D = 0.86 + j0 B = 0 + j130.2C = j0.002 (a) Obtain the sending end quantities and the voltage regulation when line delivers 1000 MVA at 0.8 lagging power factor at 500 kV. As a result of this, the compensated ABCDconstants become · ¸ · ¸ · ¸ · ¸ 1 1 ′ ′A B A B jX c jX c 1 − 1 − 2 2= ′ ′C D C D 1 1 where is the total reactance of the series capacitor.

### 5.16. A three-phase 420-kV, 60-HZ transmission line is 463 km long and may be

The line is energized with 420 kV at the sending end. When the load at the receiving end is removed, the voltage at the receiving end is 700 kV, and◦ 6the per phase sending end current is 646.6 90 A.(a) Find the phase constant β in radians per Km and the surge impedance Z in Ω.c (b) Ideal reactors are to be installed at the receiving end to keepS R |V | = |V | = 420 kV when load is removed.

### 5.17. A three-phase power of 3600 MW is to be transmitted via four identical 60-

From the impedance diagram the following data is constructed for use with the function Y = ybus(Z) 1 . (a) The self admittances are Y 11= −j2 + (10 − j20) + (10 − j30) = 20 − j52 Y 22= −j1 + (10 − j20) + (16 − j32) = 26 − j53 Y 33= (2 − j1) + (10 − j30) + (16 − j32) = 28 − j63 Therefore, the bus admittance matrix is Ybus =  20 − j52 −10 + j20 −10 + j30 −10 + j20 26 − j53 −16 + j32−10 + j30 −16 + j32 28 − j63   .

### 6.4. A fourth-order polynomial equation is given by

S 1 V(1) 2 Z 12= 0.02 + j0.04 S 2= 280 MW +j60 Mvar FIGURE 51 One-line diagram for Problem 6.6.y 12 = 1 0.02 + j0.04= 10 − j29 The per unit load at bus 2 is S 2= − 280 + j60 100 = −2.8 − j0.60Starting with an initial estimate of V(0) 2= 1.0+j0.0, the voltage at bus 2 computed from (6.28) for three iterations are The line loss is SL12 . Construct a power flow diagram and show the direction of the line flows.(c) Check the power flow solution using the lfgauss and other required programs.(Refer to Example 6.9.) .

### 2 P

In the two-bus system shown in Figure 59, bus 1 is a slack bus with (1) 2 | = −0.0214 |V(2) 2 | = 0.91 + (−0.0214) = 0.8886 pu (2) 2 = −0.13 + (−0.0164) = −0.1464 radian∆|V (1) 2 = −0.0164δ Obtaining the solution of the above matrix equation, voltage at bus 2 in the second iteration is∆δ . ◦−δ 3 (0)= 1.0 + j0, and keeping |V 2 (0)= 1.05 + j0 and V V 2(b) Using Newton-Raphson method, start with the initial estimates of 3| 2)+40|V 3 +δ 2| sin(90 3 3||V 1)−20|V 3 +δ ◦−δ 1| sin(90 3||V 3= −20|V 2) Q (a) Show that the expression for the real power at bus 2 and real and reactive power at bus 3 areP 2 Also, computing the elements of the Jacobian matrix, the set of linear equations in the second iteration becomes· (2) .

### 6 The slack bus voltage is

V 1 = 1.0 20 pu, and the bus 2 voltage magnitude is |V | = (0) (0) (0)= 0.0, and δ = 1.05 pu. tap at bus nllinedata=[1 2 0.0 0.025 0.0 1 1 3 0.0 0.05 0.0 1 2 3 0.0 0.05 0.0 1]; disp(’Problem 6.12(c)’)lfybus % form the bus admittance matrix lfnewton % Power flow solution by Gauss-Seidel methodbusout % Prints the power flow solution on the screen lineflow % Computes and displays the line flow and lossesThe above statements are saved in the file ch6p12c.m.

### 6.13. For Problem 6.12:

Perform two iterations.(b)Check the power flow solution for Problem 6.12 using the decouple and other required programs. The slack bus voltage is V = 1.0 0 pu, and the bus 2 1 (0)voltage magnitude is 3|V | = 1.05 pu.

### 1.0 The new bus voltages in the first iteration are

Obtain the power flow solution by the following methods: (a) Gauss-Seidel power flow (see Example 6.9).(b) Newton-Raphson power flow (see Example 6.11).(c) Fast decoupled power flow (see Example 6.13). 3| = 0.925 + (−0.0162) = 0.9088 pu (c) The power flow program decouple is used to obtain the solution.

### 6.14. The 26-bus power system network of an electric utility company is shown in

MW Mvar 20.0 40.0 26 15.0 31.0 13 28.0 15.0 25 48.0 89.0 12 27.0 54.0 24 0.0 27.0 41.0 31.0 27.0 55.0 16 50.0 64.0 3 70.0 25.0 15 15.0 22.0 2 12.0 24.0 14 4 10.0 48.0 29.0 20 0.0 0.0 7 15.0 75.0 19 76.0 17 6 67.0 30.0 18 153.0 50.0 1 38.0 78.0 5 . 14 21 20 22 15 24 23 10 19 25 11 16 12 1 9 4 7 6 18 13 8 5 26 3 2 .

## GENERATION DATA

Bus Voltage Generation Mvar Limits No. 1 1.025 2 1.02079.0 40.0 250.0 3 1.02520.0 40.0 150.0 4 1.050 100.0 40.0 80.0 5 1.045 300.040.0 160.0 26 1.015 60.0 15.0 50.0 The Mvar of the shunt capacitors installed at substations and the transformer tap settings are given below.

## SHUNT CAPACITORS

4.0 2.0 Bus No. Mvar The line and transformer data containing the series resistance and reactance in per unit and one-half the total capacitance in per unit susceptance on a 100-MVA baseare tabulated below.

## LINE AND TRANSFORMER DATA

Bus Bus R, X, 1 Bus Bus R, X, 1

### 2 B

6 7 0.0053 0.0306 0.001016 20 0.0239 0.0585 0.000 80 2 B,No. pu pu pu No.

c . .