INtroductor solution manunual to accomapy power system analysis and design 5th

Power triangle for the load: = Real power loss in the line is zero... The equivalent circuit referred to the low-voltage side is shown below: 3.17 a Neglecting the exciting current of th

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2.1 (a) A1= ∠ ° =5 30 5 cos30[ ° + jsin 30° =] 4.33+ j 2.5

(b) Z = +3 j8− j4= +3 j4= ∠5 53.1° Ω

(c) I =(100 0∠ °) (5 53.1∠ ° = ∠ −) 20 53.1 A°

6 3

The circuit transformed to phasor domain is shown below:

ωω

terminal of the generator

P VI

Q VI

ωω

P VI

(b) υ(t), i(t), and p(t) are plotted below: (See next page)

(c) The instantaneous power has an average value of 3.46 W, and the frequency is twice that

of the voltage or current

sin 2sin 2

Instantaneous power supplied (to sustain the changing energy in the magnetic field) has a

(c) Z R j L 1

j C

ωω

which are the reactive powers into L and C, respectively

Thus p t( )=P(1 cos 2+ ωt)−Q Lsin 2ωt−Q Csin 2ωt ←

2

2 2

2 2

2 2

2 2

2 2

2 2

0.1S10

L L L

P pf

=

=

Power triangle for the load:

=

Real power loss in the line is zero

1.867730.84 10

240

10.38385.547 10

V

∠ °

At the new pf of 0.8 lagging, P TOTAL of 60kW results in the new reactive power Q′ , such

∴The required capacitor’s kVAR is Q C =80−45=35 kVAR ←

C C

1 2 MAX

V V P

X

2.32 4 Mvar minimizes the real power line losses, while 4.5 Mvar minimizes the MVA power flow

into the feeder

2.36 Note that there are two buses plus the reference bus and one line for this problem After

converting the voltage sources in Fig 2.23 to current sources, the equivalent source

The rest is left as an exercise to the student

2.37 After converting impedance values in Figure 2.29 to admittance values, the bus admittance

2.38 The admittance diagram for the system is shown below:

V V

V V V V V

1 3

( / 3) 120.15 36.8730 24.02 66.87 A

an a

V I

circuit is shown below:

Total impedance viewed from the input terminals is

V I Z

(b) Phase voltage at load terminals V2 =120 0∠ ° − +(2 j4 5 0)( ∠ ° )

2.44 (a) The per-phase equivalent circuit for the problem is shown below:

With phase voltage V2 as reference,

( )

* 3

* 2

R S I V

LL

S I V

Using voltage division: / 3

S I

2 3

LL

P I

2.49 (a) Let Z Y = Z A= Z B= Z C for a balanced Y-load

Total apparent power dissipated in all three phases in the load

2 2

t V

1.5sin10 V2

out

t t

5.5 10

24.2227.3

3.11 Turns Ratio= =a N N1 2=66 /11.5=5.74

With high-voltage side designated as 1, and L-V side as 2,

1

(b)

2 2

j N

Note: Real and reactive losses, 0.87 kW and 1.95 kVARS, absorbed by the feeder and

transformer, are the same in all cases Highest efficiency occurs for unity P.F

(EFF=Pout Ps/ × 100=(50 / 50.87 ×100) =98.29%)

( )

2 2

2 2

The equivalent circuit referred to the low-voltage side is shown below:

3.17 (a) Neglecting the exciting current of the transformer, the equivalent circuit of the

transformer, referred to the high-voltage (primary) side is shown below:

The rated (full) load current, Ref to HV-side, is given by

3.18

2 1

V Z

(c) P spu + jQ spu =V spu I spu* =(1.0375 1.1505∠ °)(1.0 36.87∠ ° )

460

115460

10.5820,000

3

115V115

0.661320,000

20,000

173.9 A115

base

base

base

V Z I

T pu T pu Line Load pu

V I

L pu

10 40

2.778 70 pu3.6

( )( )

1 1

Y pu pu

V I

3.23 Select a common base of 100MVA and 22kV (not 33kV printed wrongly in the text) on the

generator side; Base voltage at bus 1 is 22kV; this fixes the voltage bases for the remaining buses in accordance with the transformer turns ratios Using Eq 3.3.11, per-unit reactances

on the selected base are given by

10.45The load impedance in OHMS is

LL L

L

V Z

The per-unit equivalent circuit is shown below:

3.24 (a) The per-unit voltage at bus 4, taken as reference, is

0.665 36.87Current drawn by the motor is

0.95 0

m m

S I V j

The impedance diagram in PU is shown below:

3.26 Base impedance on the low-voltage, 3.81kV-side is

Per-unit value should be the same as 6.2 pu

6648.4

3.27 Transformer reactance on its own base is

3.32 (a)

For positive sequence, V H1 leads V M1 by 90°, and V H1 lags V X1 by 90°

For negative sequence, V H2 lags V M2 by 90°, and V H2 leads V X2 by 90°

(b)

For positive sequence V H1 leads V X1 by 90° and V X1 is in phase with V M1 For

Δ – zig/zag transformer can be used to obtain the advantages of a Δ − transformer Y

without phase shift

(c)

For positive sequence, V H1 lags V X1 by 23.4°

For negative sequence, V H2 leads V X2 by 23.4°

3.33

3.34 (a)

(b)

(b)

6 3

and transformer ac absorbs 12.5 Mvars The total reactive power delivered by the

open-Δ transformer to the resistive load is therefore zero

3.38 (a) The single-line diagram and the per-phase equivalent circuit, with all parameters in per

unit, are given below:

Current supplied to the load is

= ∠ − ° +

Terminal voltage of the generator is 23 1.0374× =23.86 kV

The real power supplied by the generator is

*

ReV I t a = 1.0374 2.4 cos× −26.02° +55.84° =2.16 pu

Per-unit positive-sequence reactance diagram

S base = 100 MVA

V base H= 500 kV in transmission-line Zones

V base X= 20 kV in motor/generator Zones

1500

0.1pu1000

3

18

201500

3 18 0.81000

28.87 kA

20 360.14

0.1 0.1

20.7454 21.88 pu

base X

H X

I

I I

1.506 kA;

base X

H X

I

I I

and selecting a base of 20 kV in the generator circuit,

The voltage base in the transmission line is 230 kV and the voltage base in the motor circuit

176.3

2 1

3.46 The motors together draw 180 MW, or 180 0.6 pu

With phase-a voltage at the motor terminals as reference,

13.20.9565 0 pu13.8

The magnitude of the voltage at the generator terminals is

Note that the transformer phase shifts have been neglected here

shown below:

230

0 kV,3

a n

a n a j

S

S I

V I

Three-phase complex power supplied by the generator is

diagram is drawn below:

Generator current magnitude = I1 = 1659.1 A ←

Three-phase complex power delivered to the load is

On the common base of 100 MVA for the entire system,

2 1

2 2

2 1

2 2

0.05 per unit

=

(b)

Per unit positive sequence

(c)

Per unit positive sequence

3.53 With a base of 15 MVA and 66 kV in the primary circuit, the base for secondary circuit is

15 MVA and 13.2 kV, and the base for tertiary circuit is 15 MVA and 2.3 kV,

21

3.54 The constant voltage source is represented by a generator having no internal impedance On

a base of 5 MVA, 2.3 kV in the tertiary, the resistance of the load is 1.0 pu Expressed on a

5

On a base of 15 MVA, 13.2 kV, the reactance of the motor is

15

7.5

X′′ = × =

The impedance diagram is given below:

Note that the phase shift that occurs

between the Y-connected primary and

the Δ-connected tertiary has been

10 kVA is transformed by magnetic induction

100 kVA is transformed electrically

At full-load current, winding losses

479.5 316.256 A

H H

V V

The autotransformer connection is shown below:

With windings carrying rated currents, the autotransformer rating is

LOSS

The total autotransformer loss is same as the two-winding transformer, since the windings are subjected to the same rated voltages and currents as the two-winding transformer

3.59

Load Load Load

(a) No regulating transformer, C=1.0

Using current division:

2 1

Using the admittance parameters from Example 4.14(a)

The voltage magnitude regulating transformer increases the reactive power delivered by

line L-2 43.970 (from 0.2222 to 0.3198) with a relatively small change in the real power delivered by line L2

22 21

The phase-angle regulating transformer increases the real power delivered by line L2

31.8% (from 0.3849 to 0.5074) with a relatively small change in the reactive power

delivered by line L2

3.60 Losses are minimum at 0 degrees = 16.606 MW (there can be a +/- 0.1 variation in this

value values of the power flow solution tolerance)

3.61 Minimum occurs at a tap of 1.0 = 16.604 (there can be a +/- 0.1 variation in this value

values of the power flow solution tolerance)

The per-unit positive-sequence network is:

3.63 A radial line with tap-changing transformers at both ends is shown below:

1

V′ and V2′ are the supply phase voltage and the load phase voltage, respectively, referred to

are the tap settings in per unit The impedance Z includes the line impedance plus the

referred impedances of the sending end and the receiving end transformers to the

neglecting the phase shift between V S and V R as an approximation, and noting that V S=t V S 1′

and V R =t V R 2′, it can be shown that

where Pφ and Qφ are the load real and reactive powers per phase and it is assumed that t t S R = 1

closed, only a very small fraction of that current goes through the load impedance, because

it is much larger than the transformer impedance; so the superposition principle can be

3.65 Same procedure as in PR 3.49 is followed

eq

eq a

S

D L

D

4.24 Reduces the electric field strength at the

conductor surfaces and hence reduces corona, power loss, communication

interference, and audible noise

1.61433.576 10

Pl R A

75+228.10.0807 /km

per conductor (at 75% current capacity)

For 4 conductors per phase:

4.6 Total transmission line loss 2.5( )

Pl A R

4.9 (a) Lint =0.05mH/ km Per Conductor

The total inductance decreases 4.1% (increases 5%) at the conductor diameter increases

4.13 (a) The total line inductance is given by

r D r

Doubling the separation between the conductors causes only about a 13% rise in inductance

7 3

4.15 For each of six outer conductors:

4

(decrease 1.6%) as the phase spacing increases 10% (decreases 10%)

4.21 (a) From Table A.4:

2 3

7 1

4.23 (a) The geometric mean radius of each phase is calculated as

( )( ) ( )

AB

BC

AC

D D D

with I a as reference, instantaneous flux linkage is

4.8ln0.25 /12

(b) ( 12)

11 1

3.2ln0.25 /12

4.37 (a) Eq (4.9.15): capacitance to neutral 2 F/m

ππε

4.40 (a) D eq =38.8 8.8 17.6× × =11.084 m

12 1

11.084ln

9.069ln

C1, Y1, and I chg decrease 1.5% (increase 1.7%)

As the phase spacing increases 10% (decreases 10%)

LL

C

D D

ππε

12

11 1

Sc D C

Q

ππ

For the larger, 1351 kcmil conductors (smallel, 700 kcmil conductors)

4.43 (a) For drake, Table A.4 lists the outside diameter as 1.108 in

Total three-phase reactive power supplied by the capacitance is given by

15.12 m and 0.0164 m

eq

12 23 31 9

C

H D

Neglecting Earth effect, (8.854 10 12)

20ln0.06677

H H

( ) 1 ( )3

0.693ft

N GMR=rN A −  = × × 

T E

4.51 (a) From Problem 4.30,

the conductor diameter increases 25 % (decreases 25%) The ground level electric field

Chapter 5

Transmission-Lines: Steady-State Operation

ANSWERS TO MULTIPLE-CHOICE TYPE QUESTIONS

base L L

S I

(ii) For nominal T-circuit

A0.8

3126 V3.126 kV

(b) Sending-End 3-Phase Power = P S=3 1905.3 207.76 0.79( )( )

5.11 (a) The series impedance per phase Z = +(r j L lω )

P P

φη

φ

R S

P P

φ φ

5.12 (a) The nominal π circuit is shown below:

The line admittance for 100 mi is

3 3

185.5 10100

C

Y X

sinh 0.034862 sin 0.39848 radians

φ φ

At the sending end

Line to line voltage magnitude 1.1102 215

16

(b) Refer to Table 5.1 of the text

X X A

0.4973 87.55134.8 85.3

l

l

γγ

5.24 Z′ = =B 98.25 86.69∠ °Ω =5.673+j98.09Ω

6 2

F

l l

( )

6 3

99.68

274.91.319 10 S

S R

V V P

5.30 0 ( )

1 1

0

/2

eq SL C

eq SC

L z

Z

D

Ln D

μππε

9 0

3771

1036

velocity of

propagation

/1

3 4

πω

×

5.31 (a) For a lossless line, β ω= LC =2π( )60 0.97 0.0115 10× × −9

0.001259 rad/km

=

30.97 10

60.0115 10

Z j

From the practical line loadability,

For a lossless line, the maximum power that can be transmitted under steady-state

Then V1=V2(coshγl+sinhγl)=V e2 γl =V e e2 αl j lβ (1)

e I

e S

e P

α

5.34 For a lossless line, Z C L

2

2 2

12

l ZY

l V

γπ

the voltage at the open receiving end is higher

than that at the sending end,for small ,for

the medium and long-line models

For maximum power, dP dQ/ = 0 :

S R MAX

R

ZV V

( )( ) ( )

( )

2 max

R LL

P I

From Table A.4, the thermal limit for 3 ACSR 1113 kcmil conductors is

3 1.11 3.33 kA/ p hase.× = The current 1.966 kA corresponding to the theoretical steady-state stability limit is well below the thermal limit of 3.33 kA

and unity power factor

Q pf

230

529100

base L base base

SIL SIL

From Eq (5.4.21) of the text,

Nominal voltage level for the transmission line is

5.43 The maximum amount of real power that can be transferred to the load at unity pf with a bus

voltage greater than 0.9 pu (688.5 kV) is 2950 MW

5.44 The maximum about of reactive power transfer that can be transferred to the load with a bus

voltage greater than 0.9 pu is 650 Mvar

988 MW

R

thermal limits are not exceeded

δδδ

×

exceeded Therefore the voltage drop limit determines loadability for this line Based on 95 per unit

δδ

5.47 (a) l=200 kM; The steady-state limit is:

At the voltage drop limit:

(b) By shifting the origin from 0′ to 0, the power diagram is shown in Fig (b) above

For a given load and a given value of |V R|, 0′ =A |V R| |V S| |B| the loci of point A will

two such circles (known as receiving-end circles) are shown below:

(c) Line 0A in the figure above is the load line whose intersection with the power circle

2

received for the two sending-end voltages

The reactive power that must be supplied at the receiving end in order to maintain constant |V R| when the sending-end voltage decreases from |V S1| to |V S2| is given by

AB, which is parallel to the reactive-power axis

Problem 5.37(a) solution, one gets

(b) For a lossless line, Z C = L C/ is purely real and γ = jβ is purely imaginary Also

Z′ is now the impedance of a pure inductance

C

V P

θβ

SIL

l

θβ

(c) For βl=0.002 radiansl =(0.1146 )l °, and θ12 = ° , 45

Applying the result of part (b), one gets

0.707sin 0.1146

SIL

P

°

(d) Thermal limit governs the short lines;

Stability limit prevails for long lines



 ←



5.53 The maximum power that can be delivered to the load is 10,250 MW

5.54 For 8800 MW at the load the load bus voltage is maintained above 720 kV even if 2 lines

are taken out of service (8850 MW may be OK since the voltage is 719.9 kV)

5.55 From Problem 5.23, the shunt admittance of the equivalent π circuit without compensation is

at full load Therefore

Y Z

compensated line are

3

compensated line series capacitors

receiving-end series capacito

Equivalent ABCD parameters of the compensated line are

3

Receiving end Sending end Uncompansated line from Pr 5.13

The equivalent series reactance of four lines with two intermediate substations and one line section out-of-service is then:

From Eq (5.4.26) with δ = °35 , V R=0.95per unit, and P=9000 MW;

L REAC

N X

N B

3 Line

5.62 See solution of Pr 5.18 for γl Z, C,coshγl, and sinh lγ

For the uncompensated line:

γγγ

the new series arm impedance is

γγ

Thus, the maximum power that can be transmitted is increased by about 300%

5.63 The shunt admittance of the entire line is

6

From Fig 5.4 of the text, for the case of ‘shunt admittance’,

The voltage regulation with the shunt reactor connected at no load is given by

0.0667124.13

−

=

which is a considerable reduction compared to 0.247 for the regulation of the

uncompensated line (see solution of Pr 5.18)

5.64 (a) From the solution of Pr 5.31,

The capacitive reactance is given by (see Eq 2.3.5 in text)

433.38567.85

L C

C

X

j S

The receiving end voltage per phase 500 0 kV 288.675 0 kV

5.65 The maximum amount of real power which can be transferred to the load at unity pf with a

bus voltage greater than 0.9 pu is 3900 MW

5.66 The maximum amount of real power which can be transferred to the load at unity pf with a

bus voltage greater than 0.9 pu is 3400 MW (3450 MW may be OK since pu voltage is 0.8985)