Vba = VbO - VaO

Vba = 100/-15 - 120;45 = 96.59- j25.88 - (84.85+ j84.85)= 11.74- jllO.i3 = 111.35;-83.95 V

Solution:

R

~.254' jl~.83 n

(a) Elements in series:

Solution:

100LJEZ = 8 ~ = ~.;)z:mr:: Sl

~ .f3:"~n Xt,= ~ n101 gj).... (Q 2~.n,

(b) Elements in parallel:I ' F' n 1 ., o~'n'+-\ ..04-'f~-;_-=- fiZ. 'I-.. ~ l:A. ':JP Y = Z = 9-:68~ Q.04 jQ.~92~ J , . S

N.....l,e: R.4-~X~B R ~=~ XC= jll~4=~'n '(-=-G+~Of' ~ ) , Ifl4->J\..- ~ 2..qJt,. (;v-\ 'j-.c. .....~ ""-0.\).~ 1.3 In a single-phase circuit Va = 120;45V and Vb = 100;-15 V with respect

to a reference node o. Find Voo in polar form.

(c) Phasor expressions in polar and rectangular form:tf..

V = 100(,(f'" = 100+ jO Va..",.~:f./~ I = 8~ = 4 - j6.93 Ar +30JThe circuit is ina-bt2(iveas I ~ V.7' let~ ..s"

1.2 If the circuit of Prob. 1.1 consists of a purely resistive and a purely reactiveelement, find R and X, (a) if the elements are in series and (b) if the elementsare in parallel. (2 i 'f...

III = 11.31 =y2 8AIVI = 141.4 = 100Vv'2

(b) rms values:

Imax = 11.31AVmax = 141.4V(a) Maximum values:

Solution:

1.1 If 1,' = 141.4sin(:..Jt + ~~t)V and i = 1l.31 Q(wt - ~O) A. find for each (a)the maximum value, (b) the rms value and (c) the phasor expression in polarand rectangular form if voltage is the iI~eference:'-Is the circuit inductive orcapacitive?

1

Chap 1Problem Solutions

1.6 A single-phase inductive load draws 10 MW at 0.6 power factor lagging. Drawthe power triangle and determine the reactive power of a capacitor to be con-nected in parallel with the load to raise the power factor to 0.85.

P 2880 WQ = 4988 - 1250 = 3738 var

p.f. = cos(tan-1~~~~) =0.61

Solution:

1.5 If a capacitor is connected in parallel with the circuit of Prob. 1.4 and if thiscapacitor supplies 1250 var, find the P and Q supplied by the 240-V source,and find the resultant power factor.

13.33

Solution:

10 . ( -106) 13.330.6 Sin cos . =cos-1O.85

6.2= 31.79

10 tan 31.79 = 6.2 varQc = - (13.33 - 6.2)

= -7.13 Mvar

1.4 A single-phase ac voltage of 240 V is applied to a series circuit whose impedanceis 10/60 D. Find R, X, P, Q and the power factor of the circuit.

2

Solution:

R = lOcos 60 = 5.0 nX = 10sin 600 = 8.66 nI = 240LJE = 24 _600 A10;60 ~p = (24)2 X 5 = 2880 WQ = (24)2 x 8.66 = 4988 var

p.f. ( _14988) = 0.50= cos tan 2880

or cos (tan-1 ~) = 0.50

Machine 1 generates 268 W, delivers 1000 var

100+ J'O - (86.6+ J'50)1 = 5 = 2.68 - j10 = 10.35;-75 A

El1* = 100 (2.68 + j10) = 268 + j1000E21* = (86.6 + j50) (2.68+ j10) = -268 + j1OO0

Solution:

1.9 Repeat Problem 1.8 if Z = 5 + jO O.

Machine 1 generates 1000 W, receives 268 var

Machine 2 absorbs 1000 W, receives 268 var

Capacitor in the line supplies (10.35)2 x 5 = 536 var

-j5= 100 (10 - j2.68) = 1000 - j268= (86.6 + j50) (10 - j2.68) = 1000 + j268

10.35;15" A100 + jO - (86.6+ j50)

= 10 + j2.68 =1 =Solution:

1.8 If the impendance between machines 1 and 2 of Example 1.1 is Z = 0 - j5 0determine (a) whether each machine is generating or consuming. power,(b) whether each machine is receiving or supplying positive reactor power andthe amount, and (c) the value of P and Q absorbed by the impedance.

A capacitor will cause the drop in current in the line because the lagging component of currentdrawn by the motor will be partially offset by the leading current drawn by the capacitor.The current drawn by the motor, however, will be unchanged if the terminal voltage remainsconstant. So the motor efficiency will remain the same. Loss in the line supplying the motorwill be less due to the lower line current. If the line to the motor from the supply bus is long,the voltage drop in the line will be reduced and this may be desirable.

Solution:

1.7 A single-phase induction motor is operating at a very light load during a largepart of every day and draws 10 A from the supply. A device is proposed to"increase the efficiency" of the motor. During a demonstration the device isplaced in parallel with the unloaded motor and the current drawn from thesupply drops to 8 A. When two of the devices are placed in parallel the currentdrops to 6 A. What simple device will cause this drop in current? Discuss theadvantages of the device. Is the efficiencyof the motor increased by the device?(Recall that an induction motor draws lagging current).

3

..................................... ------

(a)a- 1(b) 1- a2 + a(c) a2+a+j

1.12 Evaluate the following expressions in polar form:

Machine 1 absorbs 1200 W and 78 varMachine 2 delivers 1200 Wand 801 var801 - 78 = 723 var absorbed by line

In Example 2.1 the line received 536 var, half from each source. Raising IE21 caused someincrease in power transfer and some increase in Q supplied to the line, but the significant factis that raising IE21 caused that source to supply not only all the Q absorbed by the line butalso 78 var delivered to the IE11 source.

= -1200 - j801

j5 j5= 100 (-12 - jO.78) = -1200 - j78= (103.92 + j60) (-12 - jO.78) = -1247 - j720 - j81 + 46.8

= -12 + jO.78-3.92 - j60=100 - (103.92 + j60)I =

Solution:

1.11 SolveExample 1.1 if E; = 100LJE_V and E2 = 120;30 V. Compare the resultswith Example 1.1 and form some conclusions about the effect of variation ofthe magnitude of E2 in this circuit.

~I).t":::' Ean~a = -120; 2100 x 10;-600P = 1039 W deliveredQ = -600 var delivered

A voltage source Ean = -120! 210 V and the current through the source is~given by Ina = 10;60 A. Find the values of P and Q and state whether thesource is delivering or receiving each. .-,...'=1C'-it

'_IM \ \ ""_'E" '" ~Solution: .........'~ -1200L..lli!r_ ~ 12001-30' / J--

P;n ~ fetV' I~~-'1 r1_/ r- S'))\) e,(+600 var absorbed by source, since Ina defines positive current from n to a and Ean definespoint a at higher potential than n when ean is positive.)

Both machines are generators.

Machine 2 generates 268 W, receives 1000 var

Resistance in the line absorbs (10.35)2 x 5 = 536 W

1.10

4

A /_ a "2T ..'_ 7() ~ ItZ 4 t t_::.!:. (0 _(()l5t'~

~( -.-LV\.

-'\.~'

~

,.A--'

VCn - 24-~ I -~0u

b

Solution:

416240 Vy3 =

Ven = 240;_60 V

len =240;_600

= 24;-90 A10/300

1.14 a a~d~-phase system the Y-connected impedances are 10; 30 n.If Vbc = 416; ~ specify len in polar form. c

la Van 120; 2100 121225 A c= Z = =10/-150i, Vbn 120; 90 12/1050 A= Z = =10;-150Ie Ven 120;-300 12;-15 A= Z = =C- 10;-150

b

Solution:.6 Van = 120; 2100 V Vab = 208; 240 V

Vbn = 120; 90 V Vbe = 208; 1200 VVen = 120/-30 V Vea = 208L..iE V

1.13 Three identical impedances of 10/-15 n are Y-connected to balanced three-phase line voltages of 208 V. Specify all the line and phase voltages and thecurrents as phasors in polar form with Vca as reference for a phase sequence ofabc.

Solution:(a) a-I = -0.5 + jO.866- 1 = 1.732; 150e(b) 1-a2+a = 1-(-0.5-jO.866)-0.5+jO.866 = 1+j1.732 = 2.00;60"

(c) a2 + a + j = -0.5 - jO.866 - 0.5+ jO.866 + j1 = -1 + j1 = 1.414; 135~(d)ja+a2 = 11210+1;240 = -0.866-jO.5-0.5-jO.866 = -1.366-j1.366

= 1.9321225

(d) ja + a2

5

15 x 746III = v'3 = 20.39 A3 x 440 x 0.9 x 0.8P = .J3 x 440 x 20.39 x 0.8 = 12,431 W drawn from lineQ = .J3 x 440 x 20.39 x 0.6 = 9,324 var drawn from line

Solution:

1.16 Determine the current drawn from a three-phase 440-V line by a three-phase15-hp motor operating at full load, 90% efficiency and 80% power factor lagging.Find the values of P and Q drawn from the line.

e

.~

kc = 99.6 Vnk = 57.5 V

meter reading = 99.6 - 57.5 = 42.1 V

Sequence a-c-b

115nk = "2" = 57.5 Vkc = 115sin60o = 99.6 V

meter reading = 57.5 + 99.6 = 157.1 V

Sequence a-b-cSolution:

1.15 The terminals of a three-phase supply are labeled a, band c. Between anypair a voltmeter measures 115 V. A resistor of 100 ~ and a capacitor of 100 S1at the frequency of the supply are connected in series from a to b with theresistor connected to a. The point of connection of the elements to each otheris labeled n. Determine graphically the voltmeter reading between c and n ifphase sequence is abc and if phase sequence is acb.

6

S1 = 250 + j250

Letting S1 and S2 represent the load and capacitor bank, respectively,

Solution:

1.19 A three-phase load draws 250 kW at a power factor of 0.707 lagging from a440-V line. In parallel with this load is a three-phase capacitor bank whichdraws 60 kVA. Find the total current and resultant power factor.

Vt = 8.06 x 3.49 = 28.13 V to neutralLine-to-line V2 = V3 x 28.13 = 48.72 V

Letting V; equal voltage at the load, line-to-line voltage:

Z = 2+ j5 +3.41 + jO.73 = 5.41 + j5.73 = 7.88! 46.65 n1l0/V3III = 7.88 = 8.06 A from supply

Current drawn at supply:

40 + j30 13 - j6 700 + j5013+ j6 x 13 - j6'- 2053.41 + jO.732 = 3.49! 12.1 n

=5(8+j6)5+ 8+j6

Solution:Convert D. to equivalent Y having 15/3 = 5 n/phase

.1.18 A balanced-A load consisting of pure resistances of 15 n per phase is in par-allel with a balanced-Y load having phase impedances of 8 + j6 n. Identicalimpedances of 2+ j5 n are in each of the three lines connecting the combinedloads to a 110-V three-phase supply. Find the current drawn from the supplyand line voltage at the combined loads.

254 + jO + (0.3 + j1.0)(16.31 - j12.23) = 271.1 + j12.64Line-to-line voltage IVI = va 1271.1+ j12.641 = 470 V

When the reference is voltage to neutral of the motor at the terminal where I is calculated,or 440/ va = 254L.Q:.V, the supply bus voltage to neutral is

I = 20.39 (0.8 - jO.6) = 16.31 - j12.23 ASolution:

1.17 If the impendance of each of the three lines connecting the motor of Prob. 1.16to a bus is 0.3 + j1.0 n, find the line-to-line voltage at the bus which supplies440 V at the motor.

7

Solution:Assume line-to-line voltage IVI is constant. Then constant current magnitude III meansconstant 151where 151= y3 IVIII*I X 10-6 MVA.

III = 114.14+ j6.851 x lobo = 41.2 AJ3 x 220

1.21 A coal mining "drag line" machine in an open-pit mine consumes 0.92 MVAat 0.8 power factor lagging when it digs coal, and it generates (delivers tothe electric system) 0.10 MVA at 0.5 power factor leading when the loadedshovel swings away from the pit wall. At the end of the "dig" period, thechange in supply current magnitude can cause tripping of a protective relaywhich is constructed of solid-state circuitry. Therefore it is desired to minimizethe change in current magnitude. Consider the placement of capacitors atthe machine terminals and find the amount of capacitive correction (in kvar)to eliminate the change in steady-state current magnitude. The machine isenergized from a 36.5 kV, three-phase supply. Start the solution by letting Q bethe total three-phase megavars of the capacitors connected across the machineterminals, and write an expression for the magnitude of the line current drawnby the machine in terms of Q for both the digging and generating operations.

With capacitors:

III = 20,000 = 52.5 Ay3 x 220

Without capacitors:

B = cos-10.9 = 25.8414.14 tan 25.84 = 6.85

14.14 - 6.85 = 7.29 kvar

Solution:From the figure,

1.20 A three-phase motor draws 20 kVAat 0.707 power factor lagging from a 220-Vsource. Determine the kilovolt ampere rating of capacitors to make the combinedpower factor 0.90 lagging, and determine the line current before and after thecapacitors are added.

8

52 = o -j60where 51 + 52 = 250+ jl90 = 314,: 37.23'" kW

III 314.000 412.0 A= v'3 x 440 =p.L = cos 37.23c = 0.796 lag

Solution: \l tb~2f'-"".x ~ 1~ G:) G~)'~~ poe unit

1.24 Draw the single-phase equivalent circuit for the motor (an emf in series with in-ductive reactance labeled Zm) and its connection to the voltage supply describedin Probs. 1.16 and 1.17. Show on the diagram the per-unit values of the lineimpedance and the voltage at the motor terminals on a base of 20 kVA, 440V.Then using per-unit values find the supply voltage in per unit and convert theper-unit value of the supply voltage to volts.

1.23 The generator of Prob. 1.22 is in a circuit for which the bases are specified as100 MVA, 20 kV. Starting with the per-unit value given in Prob. 1.22, find theper-unit value of reactance of the generator windings on the specified base.

= (22)2 = 0.968 n500

X = 1.1 x 0.968 = 1.065 nBaseZ

Solution:

1.22 A generator (which may be represented by an emf in series with an inductivereactance) is rated 500 MVA, 22 kV. Its Y-connected windings have a reactanceof 1.1 per unit. Find the ohmic value of the reactance of the windings.

0.847 - 1.104Q +Q2 = 0.01 - 0.1732Q +Q20.937Q = 0.837

Q = 0.899 Mvar or 899 kvar

and equating ISI2 for the dig and swing periods, we have

lSI 1-0.1 /_60 - JQI = 1-0.05 + jO.0866 - JQIISI2 = (-0.05)2 + (0.0866 - Q)2 = 0.0025 + 0.0075 - 0.1732Q +Q2

= 0.01 - 0.1732Q +Q2

Swing period:

lSI = 10.92 (0.8 + jO.6) - JQI= 10.736 + jO.552 - JQI

ISI2 = 0.542 + 0.305 - 1.104Q + Q2 = 0.847":" 1.104Q + Q2

Dig period:

9

Yd = -j8.0 Ye = -j5.014 = 0.68(-135

Yc = -j4.013 = 1.0(-90

Yb= -j4.0Yg = -jO.8

Ya= -jO.8Yf = -j2.5

1.26 The values for the parameters of Fig. 1.23 are given in per unit as follows:

@-Yc-Yt,

(Ya+ Yb + Y.,)o

-Yd

(Yb + Yd + Ye)-Yb-Ye

The Ybus form is

-V1Yd +V2 (Yt, + Yd + Ye) - V3Yb -lt4Ye = 0-V1Y, - \l2Ye+ V4 (Yg+Yf +Ye) = 14

bus (2)bus

Rearranging equations for bus (2) and bus @ yields

Solution:

(V2 - 113) Yt, + (V2 - VI) Yd + (\12 -lt4) Ye = 0lt4Yg + (V4 - Vl) Yf + (V4 - V2)Ye = 14

bus (2)bus@

1.25 Write the two nodal admittance equations, similar to Eqs. (1.57) and (1.58), forthe voltages at nodes ~ and @ of the circuit of Fig. 1.23. Then arrange thenodal admittance equations for all four independent nodes of Fig. 1.23 into theYbus form of Eq. (1.61).

V = 1.0 + 0.777 (0.8 - jO.6) (0.031+ jO.1033)= l.0+0.777 x 0.1079/36.43= 1.0 + 0.0674 + jO.0498 = 1.0686; 2.97 per unit

iVLLI = 1.0686 x 440 = 470 V

Voltage calculations:

Base Z = (0.44)2 X 1000 = 9.68 per unit20

R = 06.3 = 0.031 per unit9. 81.0 3X = - = 0.10 3 perunit9.68

Base 1 = 20,000 = 26.24 AJ3 x 44020.39

I = 26.24 = 0.777 per unit

Per-unit base calculations:

Solution:

10

Nl Vl 1.2 X 103 10= =N2 V2 120Therefore, N2 Nl 800 80= = =10 10

(a)

Solution:

2.1 A single-phase transformer rated 7.2 kVA, 1.2kV/120 V has a primary wind-ing of 800 turns. Determine (a) the turns ratio and the number of turns inthe secondary winding, (b) the currents carried by the two windings when thetransformer delivers its rated kVA at rated voltages. Hence, verify Eq. (2.7).

Chapter 2 Problem Solutions

V = Ybus-lI

[

jO.7187 jO.6688 jO.6307 jo.6194] [ 0 ]jO.6688 jO.7045 jO.6242 jO.6258 0

= jO.6307 jO.7045 jO.6840 jO.5660 1.0; -90jO.6194 jO.6258 jO.5660 jO.6840 0.68 1-135"

[

0.9285 - jO.2978 ] [ 0.9750;-]7.78 ]= 0.9251 - jO.3OO9 = 0.9728;-18.02

0.9562 - jO.2721 0.9941 / -15.890.8949 - jO.3289 0.9534/-20.18

jO.6688 jO.6307 jo.6194]jO.7045 jO.6242 jO.6258jO.7045 jO.6840 jO.5660jO.6258 jO.5660 jO.6840

= Ybus-lI

[

jO.7187jO.6688= Zbus = jO.6307jO.6194

(~](~]

where Y bus -1

YbusV = IYbus-lybusV

j2.5] [Vl] [ 0 ]j5.0 V2 0jO 113 = 1.01-90

-j8.3 V4 0.68;-135

j4.0j40

-j8.8jO

j8.0-jI7.0

j4.0j5.0

Compute voltages:

[

-jI4.5j8.0j4.0j2.5

Solution:Using the Y bus solution of Problem 1.25, substitute the given admittance values:

Substituting these values in the equations determined in Prob. 1.25, computethe voltages at the nodes of Fig. 1.23. Numerically determine the correspondingZbus matrix.

11

9/9/2003

Page 1 of 1

Printed for Bruce Mork

Dr. Mork

Looks like we are off to a good start, good to be thinking about thesedetails and doing some review/refreshing.

Thanks for the comments on the Ch.1 problems, I will try to go thruthese and then issue any corrections that may be required.

4) Careful with conjugates: remember that the conjugate of a complex numberhas the same magnitude, but the sign of its angle is changed. Forexample, if I = 10L3QoA, then 1*= 10/-30 A. Thus, negating a complexnumber is not the same as taking its conjugate.

3) To correctly calculate complex power consumed by (or flowing in to) acircuit element, Sin = VI* = P + jQ, where V and I havereference polarity/direction according to passive sign convention.

2) The negative sign associated with I is most likely dueto how I is defined on the circuit, i.e. the assumed referencedirection of current flow that is marked on the circuit.

1) As I mentioned in the first lecture when we discussedEuler's identity, it is standard practice to define phasorsaccording to the cosine (real) component and this is termed"cosine reference." Therefore, when converting from time domainto phasor domain, we must first convert all sinusoidal functions toequivalent cos functions. By sketching out a sine and a cosinefunction, it becomes clear that a sine is just a cosine that hasbeen delayed by 90. Therefore, sin(wt) = cos(wt -90) orcos(wt) = sin(wt + 90).

Some comments that could help with the Ch.1 review problems:

Glad to see this kind of exchange on the e-mail list,that is what I have been hoping for.

To: ee5200-I@mtu.eduSubject: sign convention, conjugate, cosine reference

EE 5200

9/9/2003

Page 1 of 1

Printed for Bruce Mark

Dr. Mork

The author's solution is therefore correct. From this we might agreethat it is indeed important it is to know the difference between activeand passive sign convention... Working at MS level, we need tounderstand the concepts and details. These seemingly simpleproblems bring that out.

Sin = (Ean)(lan)* = (Ean)(-Ina)* = Pin + jOin.===============================================Passive sign convention (load convention) calculated S consumed:

Sout = (Ean)(lna)* = (Ean)(-Ian)* = Pout + jQout=================================================Active sign convention (generator convention) calculates S produced:

The meaning of S=VI* really hinges on whether active or passive sign conventionis used. Use of double subscripts makes it much easier to explain andto understand. The general equation S = VI* is clearly being implementedhere using ACTIVE sign convention, the subscripts clearly tell us.

11.10: I also assumed S=Ean(-lan)* and got a different answer than thesolutions.

In prob. 1.2, the calculation method is correct, just update the valuesaccording to the solution of problem 1.1.

Yes, this is correct, we use cosine as the common basis/reference forexpressing all phasor angles. Peak values must be divided by sqrt(2)to get RMS values.

1.1, 1.2: v(t) is given as v(t)=141.4sin(wt+30) and i(t) is given asi(t)=11.31cos(wt-30). However, don't you need to put v(t) in terms ofcosine, which becomes v(t)=141.4cos(wt+30-90)? Using v(t) as the reference,this give V=100@Oand 1=8@30. Current is now leading the voltage and thecircuit is capacitive. Does this sound right?

Let's take a look at the first couple of items in question:

To: ee5200-I@mtu.eduSubject: Solutions, Probs 1.1 and 1.10

EE 5200

9/912003

Page 1 of 2

Printed for Bruce Mork

A philosophical observation:Studying at the MS level, our goal is not only to learn more advanced "stuff"but to also improve our understanding of the fundamentals and concepts.Actually, encountering these errors and confusions in these review problemsmay have taught us more than if the solutions had all been totally correct --we had to stop and question what's going on, go back to basic concepts,

If you rearrange these terms, you end up with what is given in Eqn 1.56.To the author's credit, he suggests this two-step approach in the 2nd paragraphon p. 30.

/ (kvbase.new=z / MVAbase,new)

9/9/2003Printed far Bruce Mark

Dr Mork

See you all in class tomorrow morning, we will go through some moreper unit things.

Any more points of uncertainty or possible errors? Please go ahead andstart the discussion here, hopefully this is helpful.

and figure it out.

Page 2 af2EE 5200

9/9/2003Printed for Bruce Mork

There are some devices, in cyclic loading applications,that increase overall motor efficiency by reducing the sourcevoltage to the motor when the mechanical load on the motoris removed/reduced, and then restores full voltage when themotor is loaded down again. Not restoring full voltage, oroperating a loaded induction motor at reduced voltagewill draw excessive current, resulting in a very low efficiencyand extreme 11\2 R heating of the armature windings, thus burningit out. One basic type of motor protection is thus to trip the motoroff line if the voltage is too low and/or the current is too high,and/or if the winding temperature gets too high.

Dr. Mork

This is essentially a power factor correction situation, nointernal changes have been made to the motor, it is stilloperating in the same way and with the same efficiency.Since efficiency is related only to real power P, the Qthat is produced by the caps has no effect on motorefficiency.

Adding shunt capacitors (shunt compensation) reduces theinductive component of the current being drawn from themains, i.e. flowing down the line, thus reducing the netcurrent flowing in the line. This reduces the 11\2 Rline losses. The current flowing into the motor, however,is unchanged (assuming the capacitor placement has notchanged the terminal voltage).

The author's rationalization seems to be sound:

At 12:02 PM 9/3/2003 -0500, you wrote:

11.7: I believe that they drew the correct conclusion about the efficiencybut for the wrong reason. Anybody care to comment?

To: ee5200-I@mtu.eduSubject: Chapter one problem 1.7

Page 1 of 1EE 5200

(a)

Solution:

2.1 A single-phase transformer rated 7.2 kVA, 1.2kV/120 V has a primary wind-ing of 800 turns. Determine (a) the turns ratio and the number of turns inthe secondary winding, (b) the currents carried by the two windings when thetransformer delivers its rated kVA at rated voltages. Hence, verify Eq. (2.7).

Chapter 2 Problem Solutions

r V = Ybus -II

[

jO,7187 jO.6688 jO,6307 jo.6194] [ ]jO,6688 jO,7045 jO.6242 jO.6258 jO.6307 jO,7045 jO.6840 jO.5660 1.0/-90jO,6194 jO.6258 jO.5660 jO.6840 0.68/-135"

[

0.9285-jo.2978] [0.9750/-17,78]0.9251 - jO,3009 0.9728;-18.02

= =0.9562 - jO.2721 0.9941 / -15.890.8949 - jO,3289 0.9534/-20,18

[~l~[~]

Y bus-lybusVjO,66.88 jO,6307 jo.6194]jO.7045 jO.6242 jO.6258jO,7045 jO,6840 jO.5660jO,6258 jO.5660 jO,6840

= Ybus -II

[

jO,7187jO.6688

= Zbus = jO.6307jO,6194

where Y bus -1

YbusV = I

Compute voltages:

1.2 X 103= = 10120

80080= =10

Nl VI=

N2 V2

Therefore, N2 Nl= -10

j2.5] [ VI] [ 0 ]j5,0 V2 _ 0jO V3 - 1.0/-90

-j8.3 V4 0,68/-135

j4,0j4.0

-j8,8jO

j8.0-j17.0

j4,0j5.0

[

-j14,5j8,0j4,0j2,5

Solution:Using the Y bus solution of Problem 1,25, substitute the given admittance values:

Substituting these values in the equations determined in Prob. 1.25, computethe voltages at the nodes of Fig. 1.23. Numerically determine the correspondingZbus matrix.

11

....,.....,.~.Ii"

(b)

Z' = (Z~yZl = (~~yz, = 100 x 2.4/36.9 n2= 240/ 36.9 n = 192 + j144 n

(c)

1111 1V11 1.2 x 103 A 5 A=IZ21

= =240IS}I = IVII 1111 = 1.2 x 103 x 5 VA = 6 kVA

Z2V2 V2 1V212

= = S;/V2 =t, S;(120)2 n= 6 x lOS/ -36.9

= 2.4/36.9 n = (1.92 + j1.44) n

S2 = IS21 Li = 6 x 103(36.9 VA

h = (~r(a)

Solution:

2.2 The transformer of Prob. 2.1 is delivering 6 kVA at its rated voltages and0.8 power factor lagging. (a) Determine the impedance Z2 connected across itssecondary terminals. (b) What is the value of this impedance referred to theprimary side (i.e. Z~)? (c) Using the value of Z~ obtained in part (b), determinethe magnitude of the primary current and the kVA supplied by the source.

Right-hand side of Eq, (2.7):

Left-hand side of Eq. (2.7)

= 60 AIhlrated = 7.2 X 103120h 6h = 60 = 0.1N2 1- = - = 0.1N} 10

= Right-hand side of Eq. (2.7)

Left-hand side of Eq. (2.7):

Ihlrated =7.2 x 103 =

IVdrated lIt Ira ted = 1V21rated Ih Irated1.2 x 103 Ih Irated = 120 Ih Irated7.2 x 1031.2 X 103 = 6 A

Srated =

(b)

12

(a) Write the impedance form [Eq. (2.24)] of the system equations(b) Write the admittance form [Eq. (2.26)] of the system equations(c) Determine the primary voltage VI and the primary current h when the

secondary is(i) open circuited and has the induced voltage V2 = 100LlL.v(ii) short circuited and carries the current 12 = 2 (90 A

2.4 For the pair of mutually coupled coils shown in Fig. 2.4, consider that LII =1.9 H, L12 = L21 = 0.9 H, L22 = 0.5 H and Tl = T2 = 0 il. The system isoperated at 60 Hz.

El = V27r X 60 x 1000 x 1.5 x 100 x 10-4 V = 4.0 kV(c) With given values,

(c)

(b)

(t) = B(t) A = BmAsin(27rft)

Solution:(a)

(a) the instantaneous flux (t) in terms of Em, f, A and t.,(b) the instantaneous induced-voltage el (t), according to Eq. (2.1).(c) Hence show that the rms magnitude of the induced voltage of the primary

. is given by lEI I = v27i j N1BmA.(d) If A = 100 em", f = 60 Hz, Em = 1.5 T and N, = 1000 turns, compute

IEll

2.3 With reference to Fig. 2.2, consider that the flux density inside the center-leg ofthe transformer core, as a function of time t., is E (t) = Em sin (27ijt) where Emis the peak value of the sinusoidal flux density and f is the operating frequencyin Hz. If the flux density is uniformly distributed over the cross-sectional areaA m2 of the center-leg, determine

F"fI 13

Leakage reactances: Xl = 37.7 n

j 75.4 .0j 37.7.0

LlI = L1 1 - aLz 1 = 1.9 - 2 x 0.9 H = 0.1 HLZI = L2z-L1z/a = 0.5-2 x 0.9/2 H = 0.05 H

a2 L21 = 4 x 0.05 H = 0.2 HLm = aLz 1 = 2 x 0.9 H = 1.8 Hw 12071"rad/sec

Solution:

[ 2/lxJO) = 1O-z x 1/~90 [ -~:~6~]V1hence V1 = 117.30L..!L. V

t, = 1.11/-90 A

2.5 For the pair of mutually coupled coils shown in Fig. 2.4, develop an equivalent- Tnetwork in the form of Fig. 2.5. Use the parameter values given in Prob. 2.4and assume that the turns ratio a equals 2. What are the values of the leakagereactances of the windings and the magnetizing susceptance of the coupled coils?

(ii)

[ lOci';0] = 100/ 90 [ ~:A~~] 11hence 11 = 0.295/-900 A

V1 = 211.llL..Q: V

(c) (i)

[ }~] = _j1O-2 [~j~~r:~~)-1 [ ~ J= -j10-2 [-?:?6~ -1:~gg][ ~ ]

(b) From Eq. (2.25),

Solution:

(a) From Eq. (2.22) and (2.23),

[ ~] = jw [ f~~ f~;][}~] = j1207r [6: g:~][ f~J= j102 [~j~~r:~~~][ i~]

14

'v

(b)

R2 c: Rda2 1.8/100 n 0.Dl8 n= =X2 l::. Xda2 = 2.2/100 n 0.022 n=

ZI

~II ~ jx,

lv' lv,(c)

Zl (1.8 + j2.2) n12,FL IS2/V2!I-B = 7200 -36.90 A = 60/-36.9 A120 1h,FL =

h,FL = 6.0(-36.9 AaaV2,FL = 1200 VV1,FL = aV2,FL + h,FLZl

Solution:(a) With turns ratio a = 1.2 x lOS/120 = 10,

Rl = rl + a2r2 = 0.8 + 100 x 0.01 n = 1.8 nXl = Xl + a2X2 = 1.2+ 100 x 0.01 n = 2.2 n

(a) the combined winding resistance and leakage reactance referred to the pri-mary side, as shown in Fig. 2.8,

(b) the values of the combined parameters referred to the secondary side(c) the voltage regulation of the transformer when it is delivering 7.5 kVA to

a load at 120V and 0.8 power factor lagging.

2.6 A single-phase transformer rated 1.2 kV/120 V, 7.2 kVA has the following wind-ing parameters: Tl = 0.8 0, Xl = 1.2 0, TZ = 0.01 Oand X2 = 0.01 O. Deter-mine

I 75.4 nx2 =75.4 nX2 = - = 18.85 n4

Magnetizing susceptance: Bm 1 1= -- = SwLm 12071"x 1.8= 1.474 x 10-3 S

p:,

15

Voltage VI = 20 V; Current II = 6.0 A; Power WI = 36 W

Short-Circuit Test (Secondary Shorted)

Voltage V2 = 120Vi Current 12= 1.2 A; Power W2 = 40 W

Open-Circuit Test (Primary Open)

2.8 A single-phase transformer rated 1.2 kV/120 V, 7.2 kVA yields the followingtest results:

Low voltage: T

High voltage: T0.774= -2- = 0.387 n

x = 3.85 x 0.387 = 1.49 n

(220)2= 0.387 x - = 0.097 n440

x = 1.49(~~r= 0.373 n

For equal loss in high- and low-voltage windings,

0.774 = 3.85J3.082 - 0.7742 = 2.98 n

= 22.73 A (low voltage)50002205000

= 11.36 A (high Voltage)440~ = 0.774 n11.36235 .

11.36 = 3.08 n (R, Z, X high-voltage)X 2.98

=R

Solution:

Rated I ==

R =Z =X =

2.7 A single-phase transformer is rated 440/220 V, 5.0 kVA. When the low-voltageside is short circuited and 35 V is applied to the high-voltage side, rated currentflows in the windings and the power input is 100 W. Find the resistance andreactance of the high- and low-voltage windings if the power loss and ratio ofreactance to resistance is the same in both windings.

= 1200+ 6.0 /_36.9 (1.8 + j2.2) V = 1216.57 I 0.1ge VIV2.FLI = 120 V1\t2.nl - Vl.FL/a = 121.66 V

% Regulation = (121.66 - 120) /120 = 1.38 %

16

\j= 98.81 %TJ = 5400 + 40 + 25

2.9 A single-phase transformer rated 1.2 kV/120 V, 7.2 kVA has primary-referredparameters RI = TI+a2r2 = 1.0 f2 and Xl = Xl +a2X2 = 4.0 f2. At rated voltageits core loss may be assumed to be 40 W for all values of the load current.

Core 1055 at V2Winding loass at 12 =

.

120 V = 40 W50 A = 11212R' = 502 x 0.01 W = 25 W

= 6.0 kVA at 0.9 p.I. = 6 x 103 x 0.9 W = 5400 W5400

Power output at 52

50 A

(c) When S2 = 6.0 kVA and V2 = 120 V,

12 = 6 X 103 A =120

R' = R/a2 = 0.01 n X' = X/a2 = 0.0318 nG~ = 2.78 X 10-3 S B;" = 9.606 X 10-3 S

(b)

R = Wdlf = 36/6.02 n = 1.0 nIZI = VI!h = 20/6.0 n = 3.33 nX = )IZI2 - R2 = 3.18 n

From the short-circuit test,

Gc = G~/a2 = 2.78 x 10-5 SBm = B'm/ a2 = 9.606 x 10-5 S

Therefore,

= )IY~12 _ G~2 = 9.606 X 10-3 Sa = 1.2 x 103/120 = 10

B'm

(a) From open-circuit test,

G~ = W2/V22 = 40/1202 S = 2.78 X 10-3 SIY~1 = h/V2 = 1.2/120 S = 0.01 S

Solution:

(a) the parameters RI = TI + a2T2: Xl = Xl + a2X2' C; and Em referred to theprimary side, Fig. 2.7

(b) the values of the above parameters referred to the secondary side(c) the efficiency of the transformer when it delivers 6 kVA at 120 V and

0.9 power factor.

Determine

-17

1121 = ~ A = 63.245 AWinding loss at 1121 = 40 W

Therefore,

11*12 R P.2 a2 = core

(b) Load current at which TJ is maximum is given by

% Regulation

hFL .= aV2 FL + -- (Rl + ;X1), a'60= 120 x 1OLQ:.+ 10/36.9 n.o + j4.0) V = 1190.6L.LL

= 119.06 - 120 = -0 78 crt120 . 70

V1,FL

(ii) cos s = 0.8, leading 0 = 36.9TJ = 98.698% because it does not depend on whether 0 is leading or lagging.

120

12,FL .aV2 FL + -- (Rl + ;X1), a120 x lOd+ ~ (-36.9 (1.0+ j4.0) V = 1219.3/0.73120 V 1"2,FLI = 1V1,FLI/a = 121.93 V121.93 -120 = 1.61 %

V1,FL =V1,FL =

!V2,FLI =% Regulation =

(a) (i) cosO = 0.8, lagging 0 = -36.9V2 = 120~ V

12 = 7122~0;-36.9 = 60/ -36.9 ATotal loses = 40 + 602 X 1.0 W = 76 W

100Output power = 7.2 x 103 x 0.8 W = 5760 W

5760TJ = 5760 + 76 = 98.698 %

Solution:

(a) Determine the efficiency and regulation of the transformer when it delivers7.2 kVA at lI2 = 120V and power factor of (i) 0.8 lagging, (ii) 0.8 leading.

(b) For a given load voltage and power factor it can be shown that the efficiencyof a transformer attains its maximum value at the kVA load level whichmakes the 12R winding losses equal to the core loss. Using this result,determine the maximum efficiencyof the above transformer at rated voltageand 0.8 power factor, and the kVA load level at which it occurs.

18

.......... J

\;r-:,Circuit B: 1.22 x 10610 X 103 n = 144 n

Circuit C: 120210 x 103 n = 1.44 n

(b) Impedance bases

Transforrner B-C

Transformer A-B Primary: 50029.6 x 103 x jO.05 = j1.302 n

Secondary:1.52 x 1069.6 X 103

x jO.05 = j11.719 n

Primary:1.22 X 106 n7.2 x 103 x jO.04 = j8.0

Secondary: 1202 n7.2 x 103 x jO.04 = jO.08IVl2 1202VlLi = 6 x 103 I cos-1 0.8 = 2.4/36,9 nLoad:

(a) Ohmic impedances

Solution:

(a) Determine the value of the load impedance in ohms and the actual ohmicimpedances of the two transformers referred to both their primary andsecondary sides.

(b) Choosing 1.2 kV as the voltage base for circuit B and 10 kVA as thesystemwide kVA base, express all system impedances in per unit.

(c) What value of sending-end voltage corresponds to the given loading condi-tions?

Transformer A-B: 500 V/1.5 kV, 9.6 kVA, leakage reactance = 5%Transformer B-C: 1.2 kV/120 V, 7.2 kVA, leakage reactance = 4%Line B: series impedance = (0.5 + j3.0) nLoad C: 120 V, 6 kVA at 0.8 power factor lagging

2.10 A single-phase system similar to that shown in Fig. 2.10 has two transformersA-B and B-C connected by a line B feeding a load at the receiving end C. Theratings and parameter values of the components are '

6071.57 + 40 + 40= 120 x 63.245 VA = 7.589 kVACorresponding kVA level

= 98.700 %TJma:: =

= 120 x 63.245 x 0.8 W = 6071.57 W6071.57

Output

19

R = 3 x 2380 = 7140 nIf the .t.-connected resistors are reconnected in Y, then the resistance to neutral will be threetimes as great and

Solution:

R =

Ihnel = 8,000 = 33.47 A.J3 x 138138,000/ v'3 = 2380 n

33.47

2.12 Solve Prob. 2.11 if the same resistances are reconnected in Y.

Solution:

2.11 A balanced ..6.-connected resistive load of 8000 kW is connected to the low-voltage, ..6.-connected side of a Y-..6.transformer rated 10,000 kVA, 138/13.8 kV.Find the load resistance in ohms in each phase as measured from line to neutralon the high-voltage side of the transformer. Neglect transformer impedance andassume rated voltage is applied to the transformer primary.

500 3VS,base = 3 X 1.2 X 10 = 400 V1.5 X 10

Therefore, the required sending-end voltage is

Vs = 400 X 1.0642 = 425.69 V

VR = 120 V = 1.0 per unit1.667 ! 36.9 + (0.0104 + jO.15778)

Vs = 1.0 x 1.667136.9 = 1.0642 per unitThe sending-end voltage base is

1.667 L36.9 p,u.-.......--(0.QJ04 + j 0.15778) p.u.

R.E.

(c) Sending-end voltage calculations

Transformer A-B:.11.719

jO.08138 per unit)l44 =Transformer B-C:

8jO.0556 per unit) 144 =

Line B:(1.5 + j3.0) 0.0104 + jO.0208 per unit=144

Load: 2.4 36.90 = 1.667! 36.9 per unit1.44Lii!Llt:.

Per unit impedances on new bases:

20

(b) kVAsupplied = 220 X 26.24 X 10-3 = 5.772 kVA(c) The load must be reduced to (5.0/5.772) x 100 = 86.6% or 4.33 kW for each transformer.(d) The current and voltage in each of the remaining two transformers are not in phase.

Output of each transformer before the reduction in load is,

51 VabI~ = 220 10c x 26.24/ 30 = 5000 + j2886 VA52 = VcbI~ = 220/60 x 26.24/270 = 5000 - j2886 VA

Ia = 10,000 -300 = 26.24;-30 AJ3 x220~t, 26.24/2100 AIe = 26.24; 90 A

(a) Vab and Vbe remain the same after removing the third transformer, so Vea is also thesame and we have a three-phase supply, and these voltages are: Vab= 220/0 V, Vbe =220; 2400 Vand Vea = 220; 120 V. Then, Van = 127/-30 V, Vbn = 127/210 VandVen = 127190 V. The line currents are

aSolution:

2.13 Three transformers, each rated 5 kVA, 220 Von the secondary side, are conected6-6 and have been supplying a balanced 15 kW purely resistive load at 220 V.A change is made which reduces the load to 10 k\\T, still purely resistive andbalanced. Someone suggests that, with two-thirds of the load, one transformercan be removed and the system can be operated open-A. Balanced three-phasevoltages will still be supplied to the load since two ofthe line voltages (and thusalso the third) will be unchanged.To investigate further the suggestion

(a) Find each of the line currents (magnitude and angle) with the 10 kW loadand the transformer between a and c removed. (Assume Vab = 220&. V,sequence abc.)

(b) Find the kilovoltamperes supplied by each of the remaining transformers.(c) What restriction must be placed on the load for open-A operation with

these transformers?(d) Think about why the individual transformer kilovoltampere values include

a Q component when the load is purely resistive.

21r;,fr

2.15 A three-phase transformer rated 5 MVA, 115/13.2 kV has per-phase seriesimpedance of (0.007+ jO.075) per unit. The transformer is connected to a shortdistribution line which can be represented by a series impedance per phase of(0.02+ jO.10) per unit on a base of 10 MVA, 13.2 iv. The line supplies a bal-anced three-phase load rated 4 MVA, 13.2 kV, with lagging power factor 0.85.

(a) Draw an equivalent circuit of the system indicating all impedances in perunit. Choose 10 MVA, 13.2 kVA as the base at the load.

(b) With the voltage at the primary side of the transformer held constant at115 kV, the load at the receiving end of the line is disconnected. Find thevoltage regulation at the load.

Load Z

BaseV = 20.5 kV(20.5)2

= -- = 4.20 n1002.B1 0= 4.20 (36.B7 = 0.669/36.87 per unit

BaseZ

At the load,

(22.5)2-'--~ / cos-1 O.B = 2.B1 / 36.B7c n (low-voltage side)1BO

Load Z

(b)

Solution:

(a) Each single-phase transformer is rated 200/3 = 66.7MVA. Voltage rating is (345/v3) /20.5or 199.2/20.5 kV.

2.14 A transformer rated 200 MVA, 345Y/20.5~ kV connects a balanced load rated180MVA, 22.5 kV, 0.8 power factor lag to a transmission line. Determine

(a) the rating of each of three single-phase transformers which when properlyconnected will be equivalent to the above three-phase transformer and

( b ) the complex impedance of the load in per unit in the impedance diagramif the base in the transmission line is 100 MVA, 345 kV.

Sl = 4333 + j2500 VAS2 = 4333 - j2500 VA

Note that Q is equal in magnitude but opposite in sign. There is no Q output from theopen delta. After the load reduction,

22

v= 0.375/-35.97 per unitVR,FL = 0.375;-35.97 x 2.S; 31.8 = 0.937;-4.17 per unitVR,NL = Vs = 1.0

1 - 0.937V.R. = 93 x 100 = 6.72 %O. 7

2.16 Three identical single-phase transformers, each rated 1.2 kV/120 V, i.2 kVAand having a leakage reactance of 0.05 per unit, are connected together toform a three-phase bank. A balanced Y-connected load of 5 n per phase isconnected across the secondary of the bank. Determine the Y-equivalent per-phase impedance (in ohms and in per unit) seen from the primary side whenthe transformer bank is connected (a) Y-Y, (b) Y-A, (c) A-Y and (d) A-A.Use Table 2.1.

2.668/ 35.97=0.014+ 0.02+ 2.125+j(0.150 + 0.10 + 1.317)1.01.0

I =

(b) Voltage regulation calculations

(values are in per unit)

jl.317Vs

2.125Va

1

Transformer Z = 1_0(0.007+ jO.075) = 0.014+ jO.150 per unit0

Vs = 1.0 per unitLine Z = 0.02+ jO.l0 per unit

Load IZI (13.2? x 1(}00 = 43.56 D= 3400/0.85Base Z at load

(13.2? 17.42 D= =10Load Z 43.56 -108- 2.50/31.8= 17.42/cOS .oo =

= 2.125+ j1.317 per unit

115 IN ~ ;:a.~_13_.2_kV ---l~ Load

(a) Base voltages are shown on the single-line diagram.

Solution:

23

2.17 Figure 2.17a shows a three-phase generator supplying a load through a three-phase transformer rated 12 kVA/600 V Y, 600 kVA. The transformer has per-phase leakage reactance of 10%. The line-to-line voltage and the line current at

z.:

12 tV 1120V

~xmA

3.33 n

l.2fjkV 1l20V

10 n

(d) L!.-L!. connection:

IVLLI = 1200 VIViti = 120 V

R'r = 5 C200) 2 = 500 nx 120XI = 3.33 n from part (c)Z' = (500 + j3.33) nL

= 66.67 n7.2 x 3 x 1030.05 per unit = 66.67 x 0.05 n =(166.67 + j3.33) n

= 166.67 n5003

1200 V120v3 V

5 x ( 1200 )2 =120V3

12002

fVLLI =IViIi =R'r =

Zb

XI =ZL =

(c) b,.-Y connection:

1500 n

(b) Y-~ connection:

fVLLI = 1200 x Vi vIVzIi = 120 V

(1200V3)2RL' 5= x 120 =

XI = 10 n from part (a)ZL (1500 + jlO) n

IVLLIIVII I

= 1.2 X 103 x v3 V= 120v3 V

R'r = 5 x (1200~)2 = 500 n120y3

(1.2.13/ x 106 = 200 n7.2 x 103 x 3

XI = 0.05 per unit = 200 x 0.05 n =ZL = (500 + jl0) n

(a) Y- Y connection:

Solution:

24

Line voltage at the loadLine current at the load

v3WLI = 571 V= Ihl = 400 A= V II = 329.65/-33.34 n

L L 400/ -66.900.824/33.6 n

Load irr:pedance - ZL

329.65;-33.34 VVL = Vl.Ja = 6.593;-3.34 kV20/30

(12 X 103)2. 600x 103 x 0.1 = 24.0 n11.9 ._10 kV = 6.87 kVy3-II. = 20;-36.9 AILa* = 20 x 20/-36.9 - 30 A = 400/ -66.9 AV _ 'X I = 6.87 0 _ (24.0~ x 20/-36.9) kVs J I S L![_ lOOO6.593;-3.34 kV

Xl =

Let Vs =

Then, Is =i, =V'L

=

Current ratio = _!_ = 0.05/30a*

12 X 103 300 20( 300600 ~ =a =Voltage ratio =

(a)

a: 1

Solution:

(a) Determine the line current and the line-to-line voltage at the load, and theper-phase (equivalent-Y) impedance of the load.

(b) using the line-to-neutral voltage VA at the transformer primary as refer-ence, draw complete per-phase phasor diagrams of all voltages and currents.Show the correct phase relations between primary and secondary quanti-ties.

(c) Compute the real and reactive power supplied by the generator and con-sumed by the load.

the generator terminals are 11.9 kV and 20 A, respectively. The power factorseen by the generator is 0.8 lagging and the phase sequence of supply is ABC.

25

(a) Draw a diagram of the transformer connections showing the polarity markson the windings and directions chosen as positive for current in each wind-ing so that the currents will be in phase.

2.19 A single-phase transformer rated 30 kVA, 1200/120 V is connected as an auto-transformer to supply 1320V from a 1200 V bus.

(c) Same results as in Problem 2.17.

ref

'GEJd30IL' = Is = 20A

VL= 329.65V

..~~===30='==:J:I~::::==:re~f~.I 3.34" Ys = 6.87 rv..VL =6.593 leV

(b)

h = ILa = 400;-36.9 + 30 A = 400/-6.9 AVL = Vila = 329.65;-3.34 + 30 V = 329.65/26.7 V

l/a = 0.05/-30a = 20/-30

Solution:

2.18 Solve Prob. 2.17 with phase sequence ACB.

(a) Final answers remain the same except for the following intermediate results:

395.6/33.56 kVA3 x 329~65/-33.34 x 400L...QJE kVA =1000= 329.7 kW + j218.7 kvar

3Vs1s = 3 x 6.87L..!L. x 20( 36.9 kVA = 412.2/36.9 kVA= 329.8 kW + j247.3 kvar

(c) Pg + jQg from the generator is 3Vs1s, where

IL =400 A

refref "2(_'_3_3_4' ....:.VS= 6.87 kV30' ..VL' = 6.593 iv

VL= 329.65 V

26

(b)

Note that, once we consider loss, we no longer have an ideal transformer; and both windingresistance and reactance as well as magnetizing current and core loss must be considered.The applied voltage and input current will be greater than the values shown to achieve ratedoutput, in which case the equivalent circuit corresponding to Fig. 2.7 would be used.

Pout = 250.x 1320 = 330,000 W Pin = 330,928 W1] = 330,000 X 100 = 9970/( Rated kVA = 330,000330,928 . 0

Loss remains the same in the autotransformer because current in the windings and voltageacross the windings are unchanged. For the autotransformer,

Pin = 30, 928 WPout = 30,000 WLoss = 928 W

Connected for 1200/120-V operation (regular transformer),

= 25 Arated IHv =

= 250 A

30,0001200

30,000120rated It =

1320V

250A~Solution:

+

( b) Mark on the diagram the values of rated current in the windings and atthe input and output.

(c) Determine the rated kilovoltamperes of the unit as an autotransformer.(d) If the efficiency of the transformer connected for 1200/120 V operation

at rated load unity power factor is 97%, determine its efficiency as anautotransformer with rated current in the windings and operating at ratedvoltage to supply a load at unity power factor.

27

(a) Thru XI the current is II = ~ x 1.0(-300 = 0.577- jO.333 and thru X2 the current ish= i x 1.0;-300 = 0.289 - jO.167. Into bus thru Xl>

P + jQ = Vb1i = 0.577+ jO.333 per unit

ImcCSolution:

2.21 Two buses 0 and are connected to each other through impedances Xl = 0.1and X2 = 0.2 per unit in parallel. Bus b is a load bus supplying a current1 = 1.0/-30 per unit. The per-unit bus voltage Vb is1.0LlL.. Find P and Qinto bus b through each of the parallel branches (a) in the circuit described, (b)if a regulating transformer is connected at bus b in the line of higher reactanceto give a boost of 3% in voltage magnitude toward the load (a = 1.03), and(c) if the regulating transformer advances the phase 2 (a = jn'/90). Use thecirculating-current method for parts (b) and (c), and assume that v~is adjustedfor each part of the problem so that Vb remains constant. Figure 2.26 is thesingle-line diagram showing buses a and b of the system with the regulatingtransformer in place. Neglect the impedance of the transformer.

Rated kVA= 270,000, but see the note which accompanies the solution of Problem 2.19.

r.: = 250 x 1080 = 270,000 WPin = 270,928 W

TJ = 270,000 x 00 = 99.7 %270,928 1

As in Prob. 2.19, Loss= 928 W. As an autotransformer,

250 A__.Solution:

2.20 Solve Prob. 2.19 if the transformer is to supply 1080 V from a 1200 V bus.

'i~............cC"

~..:.,

128

a-I ~V = -j0.425(jO.08 + jO.12) = 0.0850a = 1.085 turns ratio

jO.8 + jO.12 = -j0.425Iob,circ =~V

To eliminate vars to bus thru Xl, we need in the X2 branch-j0.425

= 2.279- j0.4251.034+ jO.1823 - 1.0jO.081.05L1.!E- 1.0

lAb = =jO.08

In reactance Xl,Solution:

2.22 Two reactances Xl = 0.08 and X2 = 0.12 per unit are in parallel between twobuses 0 and in a power system. If Va = 1.05/10 and Vb= 1.0L.JL. per unit,what should be the turns ratio of the regulating transformer to be inserted inseries with X2 at bus so that no vars flow into bus from the branch whosereactance is Xl? Use the circulating-current method, and neglect the reactanceof the regulating transformer. P and Q of the load and Vb remain constant.

P + jQ = VbI'" = 0.461+ jO.335 per unitand into bus @ thru X2,

P + jQ = VbI'" = 0.405+ jO.165 per unitNote: Compare P and Q found in parts (b) and (c) with part (a).

Into bus @ thru x.,

LOa. - 1.0 = 0.9994+ jO.0349 - 1.0 = -0.0006 + jO.0349-0.0006 + jO.0349 = 0.116 '0.002

jO.3 + J0.577 - jO.333 - (0.116+ jO.002) = 0.461 - jO.3350.289 - jO.167 + 0.116+ jO.002 = 0.405 - jO.165

~V(c)

P + jQ = 0.289+ jO.267 per unitand into bus @ thru X 2,

P + jQ = 0.577+ jO.233 per unitInto bus @ thru Xl,

0.289+ jO.167 per unitand into bus @ thru X2,

P+jQ = VbI:i

0.03 '0(b) ~V = 0.03; Icirc = jO.3 = -J .1

h = 0.577 - jO.333 - (-jO.l) 0.577 - jO.23312 = 0.289 - jO.167 + (-jO.l) = 0.289 - jO.267

29

Of:

W~Iqold S!l{~ lOJ A.rOP'eJS!+"8SS! pOtP~UI'lUaim:::>-~uq'ern:::>.I!:::>=u ap!s a~'e'llOA-.M0l atn UOA){ (;1 'YAW S JO as'eq 'e =nSlaUI10JSUB1'l aq~ p'eopaAo lOU H!.Mq:::>rq.M.P"BOI{'e~Ol =n JO saladu.relloA"B~dUIlUTIUIpcBlUaql pUB p-eoI I"B~O~YAW S I'eu~po dq~ .IoJ .rauucjsuan q:::>'edJO and-lTIO aladUI'Bl{OA'B~dlUatn puy 'd'B'l A){ S11 alp uo SU!"BUIalqO!q.M.'z .rauirojstranoi parsdtnoo lawloJsUBl'l Wql JO ap!s a~B~IoA-MoI aq'l praaoa a~'elloA U! ~sooq%9 B aA!-2 0+ A){ TTl ~B 'las alB 1 laUI.IoJsUBl'l uo SdB'l aq'l JI papBollaAo S!.rauuojstran leH.maU lBq'l os pa'l!UlH aq lSTIUIPBO{{'elO'l aq~ qJ!q.M oa s~ladUIB'lIoA-'8~aUI alp pUB 'l~W.IOJSUBl'l qJ'ea JO andino ~lladwB~IoAB~aUI atn 'lawloJsUBl'lqOBa q~nO.Iq'l anrn lad U! iuarrno aq'l JO apn'l!~BW aq'l PU!d 'l!UTIlad LOoO = Xq'l!1t\' VAf~ SI parer S! G .ratnrojsrran pUB ''l!un lad 600 = X q'l!.M VAl'\I 0(;parar S! 1 l~UlIOJSUB-\L o~u~~BI lOpBJ laMod 80 re A){ (;1 'YAW SEJO PBoIB J(Iddns oa F~IIB.red U! CllBlado A~ \7(;1/ AII pa'lBl qJ"Ba SlaUI.IO]SUB1'lo..t'\.~ EZ'Z

= l1:sl= 11s1= 1:X= tx

VAW (;LI91"0 + SLS1"O= S x SLSYoO

VAW SOLIEE91"0 + SLS1"O= cE x

- 91"0anm lad E91"0 = (SI/9) x WoOatun lad SLS1"O = (O(;/S) x 600

S6Z00{ - E6EoO = (90f - soo)_g_ = 1:1 (;LISooor - LOvoO = (90f - soo)2L = II SOLI

1:>"PI-1:III:>"PI+ III

YAW 6nI = SE x 01 x vtvOO,,0(;

= l1:sl

= PsiVAW vOl)(; = S~ x OOIx SsoOv~voO = IssI"or - sssol =ESSoO= ILIvoOr - Lovool =

(;II"O{- = (SLsro + EE91"O)r =9~OoO

0(;0(;01 =

VAW ~Ovt = (zot/ss) oa peo] a:mpall

3.3 A three-phase round-rotor synchronous generator has negligible armature re-sistance and a synchronous reactance Xd of 1.65 per unit. The machine isconnected directly to an infinite bus of voltage l.OLlL per unit. Find the in-ternal voltage E of the machine when it delivers a current of (a) 1.0; 30 per

ea' = 17962.9coswt -4.1484 x 10-3 x 1207rX ia ...axsinwt= 17962.9 coswt - 1.5639 ia m.x sin wt

eo 'max = J 17962.92 + (1.5639iCl .... J2 = 29869.1 VHence, 2.amx = 15259.4 A

t, iClm ... /..;2 = 10.79 kAP3 = .j3x 22 x 10.79 x 1 MW = 411.2 MW

If Va = 17962.9 coswt, then ia = ia ..... coswt and

VLL = 22 kVV"max = (..;2/J3) x 22000 V = 17962.9 V

Given:

45855e,,'m.... = 3838 x 2500 V = 29869.1 V

Solution:Using the values in the solution of Example 3.1,

3.2 The three-phase synchronousgenerator described in Example 3.1 is operatedat 3600 rpm and suppliesa unity power factor load. If the terminal voltage ofthe machine is 22 kVand the fieldcurrent is 2500A, determine the line currentand the total power consumptionof the load.

PSO = 24

P2S 25

Pso and P25 must be even integral numbers lowest value where P60 = 2.4P2S Thus,

=

=2 x 60 x 60 2 x 60 x 25

speed =

P60 60

Solution:Let P = number of poles:

3.1 Determine the highest speed at which two generators mounted on the sameshaft can be driven so that the frequency of one generator is 60 Hz and thefrequency of the other is 25 Hz. Howmany poles does each machine have?

Chapter 3 Problem Solutions

31

Va= l.Op.u.

1.65p.u.

v; l.Op.u.

8 = -30EiL.. = 1.0LK + 1.65/ 60

= 2.32/38 per unit

(c)

B = 0Eid = 1.0LK + 1.65/90

= 1.93/58.8 per unit

(b)

B = 300Eid = 1.O~ + 1.65/ 120

= 1.44/ 830 per unit

(a)

Ei~ = VaLK + Jail. x.: 900= 1.OL.Jr+ 1.0~ X 1.65/900= 1.0LK + 1.65; 900 + B

Solution:

unit, (b) 1.0LJE per unit and (c) 1.0(-300 per unit to the infinite bus. Drawphasor diagrams depicting the operation of the machine in each case.

32

Ei = 11.95 kV(b) Open-circuit voltage:

(c) Short-circuit voltage:

E, 1.195 .Zd = 1.653 per unit = 0.7242 per unit = 2090.7 A

EiL.. = V L.Q:+ faZdL Q + (J= 1.0L..Q:+ 0.693 x 1.6530( 112.37 per unit= 1.195i 61.83 = 11.95;61.83 kV

Va; 1.0p.u.

Zd = (0.1 + j1.65) per unit = 1.653i 86.53 ~ ZdLQV

101.0 per unit= 10 per unit =

t, = 50x103 A = 2886.75 AV3 x 10fa

2000 .0.693 per unit2886.75 per unit =

B = cos-10.9 = 25.8

(a) Choosing Vb = 10 kV and MVAb= 50 MVA:

Solution:

(a ) Determine the internal voltage E, and the power angle 0 of the machine.Draw a phasor diagram depicting its operation.

(b) What is the open-circuit voltage of the machine at the same level of exci-tation?

(c) What is the steady-state short-circuit current at the same level of excita-tion? Neglect all saturation effects.

3.4 A three-phase round-rotor synchronous generator, rated 10 kV, 50 MVA hasarmature resistance R of 0.1 per unit and synchronous reactance Xd of 1.65 perunit. The machine operates on a 10 kV infinite bus delivering 2000 A at 0.9power factor leading.

33

1.337 ; 22.8 per unit

New Ia = 0.75 x 0.4818 per unit = 0.3614 per unit90 - () = 60.27Eill = VaLJt + IaXdL 90 - ()

= O.9375LQ: + 0.3614 x 1.65;60.27= 21.4;22.8 kV L-L

(b)

Base I =Therefore, t, =

S ==

Thus, P =Q =

Va = 15/16 per unit = 0.9375 per unit24

EiI_~. = 16.127.4 per unit = 1.5;27.4 per unitEiil. - VaLQ: = IaXd/90 - B

1.5/27.4 - 0.9375LQ: = Ia x 1.65; 90 - ()Ia! 90 - () = 0.4818;60.27 per unit

Ia/-() = 0.4818;-29.73 per unit200 x 103 kA = 7.217 kAJ3 x 160.4818 x 7.217 kA = 3.477 kA0.9375 x 0.4818 per unit = 0.4517 per unit90.34 MVA90.34 cos29.73 MW = 78.45 MW9O.34sin29.73 Mvar = 44.80 Mvar

(a) Using 16 kV, 200 MVA base;

Solution:

(a) Determine the line current and the three-phase real and reactive powerbeing delivered to the system.

( b ) If the mechanical power input and the field current of the generator arenow changed so that the line current of the machine is reduced by 25% atthe power factor of (a), find the new internal emf E, and the power angle6.

(c) While delivering the reduced line current of (b), the mechanical powerinput and the excitation are further adjusted so that the machine operatesat unity power factor at its terminals. Calculate the new values of E, and6.

3.5 A three-phase round-rotor synchronous generator, rated 16 kV and 200 MVA,has negligible losses and synchronous reactance of 1.65 per unit. It is operatedon an infinite bus having a voltage of 15 kV. The internal emf E, and thepower angle 6 of the machine are found to be 24 kV (line-to-line) and 27.4,respectively.

34

Val 0 + IaXd( 90 + B= 0.9375~ +~ x 1.65( 90 - 36.9 per unit0.9375= 1.6258;25.7" per unit = 26.0( 25.7 kV

e = -36.90.5 per unit Xd = 1.65 per unit= SIVa = 0.5/0.9375 per unit

0.9375 per unitVa(or Yt) =S

(a) From Prob. 3.5,

Solution:

Draw a phasor diagram for the operation of the machine in cases (a), (b) and( c).

(a) Determine the internal voltage Ei, power angle fJ and the line current ofthe machine.

(b) If the field current of the machine is reduced by 10%, while the mechanicalpower input to the machine is maintained constant, determine the newvalue of fJ and the reactive power delivered to the system.

(c) The prime mover power is next adjusted without changing the excitationso that the machine delivers zero reactive power to the system. Determinethe new power angle 6 and the real power being delivered to the system.

( d) What is the maximum reactive power that the machine can deliver if thelevel of excitation is maintained as in (b) and (c)?

3.6 The three-phase synchronous generator of Prob. 3.5 is operated on an infinitebus of voltage 15 kV and delivers 100 MVA at 0.8 power factor lagging.

3.48kA

New L; = 0.3614 per unit B = 0"EiLl. = 0.9375LJL. + 0.3614 x 1.65; 90'"

= 1.111; 32.5 per unit = 17.8; 32.5 kV

( c)

35

when the synchronous generator has non-zero armature resistance R.

p - R21;IXJ {IEil (Rcosb+Xdsinb) -.lvtIR}

IViI .Q = R2 +XJ {Xd (IEil cosb -IVi\) - R IEil sinS}

3.7 Starting with Eq. (3.31),modify Eq. (3.38) to showthat

V.=O.9375 p.u.

Vt (E v.) 0.9375 ( 937-) .= - i - t = -- 1.46322 - O. ::> per urutx, 1.65= 0.2787 per unit = 59.74 Mvar

(d) For Yr, e. and x, fixed, Qmc% occurs when e = 0. Hence,

-1 Vt _ -1 ( 0.9375) _ - e = cos E. - cos 1.46322. - 00.15

PE.Yr . 0.9375 x 1.46322 . 50 150 .= -- Sill () = Sill . per unitx, 1.65

= 0.638 per unit = 127.65 MW

(c) When Q = 0,

New e. = 0.9 x 1.6258 per unit = 1.46322 per unitP = 0.5 x 0.8 per unit = 0.4 per unit6 . -1 (PXd) , . -1 ( 0.4 x 1.65 ) = 28.76sin VrEi = Sill 0.9375 x 1.46322

NewQ Vr= - (E. cose - Vt)Xd0.9375= -- (1.46322 cos 28.76 - 0.9375)1.65

= 0.196 per unit = 39.2 Mvar

(b)

36

The distance n-k on the loading capability diagram is

IEil 2.0625 .-- =. urnts = 1.1393 unitsIViI x, 1.00:>x 1.7241The angle formed by points k-n-o is 38.5. Hence, point k is marked as shown. By readingfrom the chart, Pk = 0.7 per unit and Qk = 0.31 per unit.

P + jQ = IVd21Pk + jQk] = 1.052 (0.7 + jO.31) per unit= 1.052 (0.7+ jO.31) x 635 MVA

P = 490 MW Q = 217 Mvar

IVtl 25.2 kV = 25.2 . 1.05 per unit= 24 per urnt =IEil 49.5 kV = 49.5 . 2.0625 per unit= 24 per urnt =

Solution:

3.8 The three-phase synchronous generator described in Example 3.4 is now oper-ated on a 25.2 kV infinite bus. It is found that the internal voltage magnitudeIEil = .49.5 kV and that the power angle 6 = 38.5. Using the loading capa-bility diagram of Fig. 3.14, determine graphically the real and reactive powerdelivered to the system by the machine. Verify your answers using Eqs. (3.38).

Separating real and imaginary parts,

IVillEil . IVrl2RP = (R2 + XJ) {Rc(l';6 + Xdsmc5}- (R2 + XJ)

IViI { .= R2+XJ IEil(Rcos6+Xdsmc5)-IViIR}IVillEil . IVil2Xd

Q = (R2 + X~) {Xdcos6 - Rsmc5} - (R2 +XJ)

= R21;~J {Xd (IEilc(l';6 -IViI) - R IEil sinc5}

=

1V;IIEild -IVrI25 = P + jQ == ViI~ = (R _ jXd)

IVrIIEil(cosc5-jsin6)-IVrI2(R - jXd)

IVrllEd (coso- j sin6) (R + jXd) -IVrI2 (R + jXd)(R2 + XJ)

10. IEil LJ-lVrl LJE= (R + jXd)t: = IEil /-6 -IViIDo (R - jXd)

Solution:From Eq. (3.55), V = E; + 1a(R + jXd) and, therefore,

I..~I!

I[11(

37

-cos420 ]-cos (-60) mHcos 360

Solution:

(a) From Table 3.1,

(a) Using Eq. (3.41), determine the instantaneous values of the flux linkagesAa, Ab, Ac and Af when ()d = 60.

(b) Using Park's Transformation given by Eqs. (3.42) and (3.43), determine theinstantaneous values of the flux linkages Ad, Aq and Ao, and the currents is,iq and io when ()d = 60.

(c) Verify results using Eqs. (3.45) - (3.46)

if = 4000 A

During balanced steady-state operation the field current and a-phase armaturecurrent of the machine have the respective values

i.; = 0.3771 mHMf = 31.6950 mHLff = 433.6569 mH

3.9 A three-phase salient-pole synchronous generator with negligible armature re-sistance has the following values for the inductance parameters specified inTable 3.1,i; = 2.7656 mHM; = 1.3828mH

From Eq. (3.38),

P IvtllEil . = Xd sm o1.05 x 2.0625 . (38.50) x 635 MW = 496.5 MW

= 1.7241 SIDIVtlQ = Xd (lEd cos 6 -IYtI)

= 1.05 (2.0625 cos (38.5"') - 1.05) x 635 Mvar = 218.2 Mvar1.7241

38

[' , 1] ['1 -~l'2 '2 - 7s V6p = If ~ -~ 0 = 1 12 2 T2 -721 1 1 1 1

V2 72 72 T3 7J V3

[ ::] [~ 1 -~l[ 935610] [ 975381 1761 -13.9215 = 76.0016 W~T72 -721 1 -79.6395 073 73 73P AClbc

[::] [~ 1 -fi][ 10] [ -12.2474 ]761 o -20 = 2~.2132 kA72 -721 1 )3 1073 73p iabc:

( c)

Ld = L, +M,+ iLm = 4.71405mH

With if = 4 kA and Lff = 433.6569mH,

[~n [ L., Lab i.: i., 1 [l~]= Lbo Lbb u: u,Lea t.; Lee LeILla LIb LIe Lff[ 257m5

-1.0057 -1.5713515~75][ 10 1-1.0057 2.57705 -1.57135 15.8475 -20 wi-T= -1.57135 -1.57135 3.1427 -31.6950 10

15.8475 15.8475 -31.6950 433.6569 4

[ 935610 1-13.9215 W~T= -79.63951259.2026

[

2.57705 -1.0057 -1.57135]= -1.0057 2.57705 -1.57135 mH

-1.57135 -1.57135 3.1427

Lobe, I ,c,. [i:; 1 = MI [ ~~:f;d- 120) 1LeI cos (Od - 240)

= 31.695 [~~:f.~~oo)] mH = [ ~~:::~~] mHcos (-:180) -31.695

lobe ,c,. [::] = 20000 [ :!~~~O:ciO)] A = [-;~] kAie sin (-210) 10

39

(a)

Solution:

Determine id, iq and io.

(a) Using the P- Transformation matrix of Eq. (3.42), find the direct-axis cur-rent id and the quadrature-axis current iq. What is the zero-sequence cur-rent io?

( b ) Suppose that the armature currents are

'La - V2x1000sin(Od-Oa) Aib ic = 0

'La -ib -'I.e

3.10 The armature of a three-phase salient-pole generator carries the currents

J2 x 1000 sin (Od - Oa) AJ2 x 1000sin (Od - 1200 - Oa) AV2 x 1000 sin (Od - 2400 - Oa) A

Lq = L, +M, - iLm = 3.58275 mHv1MJ = 38.8183 mH

Ad Ldid+-AMfif = 4.71405x (-12.2474)+38.8183x4 Wb-T= 97.5381 Wb-T

Aq = Lqiq = 3.58275 x 21.2132 Wb-T = 76.0016 Wb-TAO = Loio = 0 (since io = 0)AJ = AMJid + Llfif = 38.8183 x (-12.2474) + 433.6569 x 4 Wb-T

= 1259.20 Wb-T

40

3.12 The single-line diagram of an unloaded power system is shown in Fig. 3.22.Reactances of the two sections of transmission line are shown on the diagram.The generators and transformers are rated as follows:

3 3L~ + M, - '2Lm = 2.7656+ 1.3828 - '2 x 0.3771 mH = 4.71405 mH12071x 4.71405 X 10-3 n = 1.777 n

3 MJ _ 3 31.69502i,- "2 Lff = 4.71400 -"2 x 433.6569 mH = 1.2393 mH12Or. x 1.2393 x 10-3 n = 0.467 n

3 (MJLD + M'bLff - 2MfMDMr)Ld -- 2 LffLD - M:4.71405 _ ~ (316952 x 4.2898+ 3.15232 x 433.6569 - 2 x 31.6950 x 3.1523 x 37.0281) mH

2 433.6569 x 4.2898 - 37.028120.9748 mH12071x 0.9748 X 10-3 n = 0.367 n

Ld =

Xd =L~ =X' =d

L~ =

==

x~

Solution:

L,= 2.7656 mHMs = 1.3828 mRLm = 0.3771 mR

LD = 4.2898 mRMD = 3.1523 mH

Lff = 433.6569mHu,= 31.6950 mHMr = 37.0281 mR

3.11 Calculate the direct-axis synchronous reactance Xd, the direct-axis transientreactance Xd and the direct-axis subtransient reactance X:J of the 60 Hz salient-pole synchronous machine with the following parameters:

= Ii [f -f -,! 1 x loooh [ sin (O~ - Oa) 1 Av'272v'2 0[ :: 1

41

(b)

Gen 1: X" = 0.2 x: = 0.50 per unit3 rating T2 = 220/18 kV, 30 :MVA

Base in trans. line: 220 kV, 50 MVABase for Gen 2 = 18 kV

Gen 2: X" 50= 0.2 x 30 = 0.333 per unitBase for Gen 3 = 22 kV

Gen 3: X" (20r 50= 0.2 22 x 30 = 0.275 per unit

Transformer T1: X 50 0.20 per unit.01 x-25

Transformer T2: X 50= .01 x 30 = 0.167 per unitTransformer T3: X = .01 x 50 = 0.143 per unit35

jO.27S

jO.143D

EB

jO.20 jO.OS26 C jO.1033 jO.167

Solution:(a)

( a) Draw the impedance diagram with all reactances marked in per unit and'with letters to indicate points corresponding to the single-line diagram.Choose a base of 50 MVA, 13.8 kV in the circuit of Generator 1.

(b) .Suppose that the system is unloaded and that the voltage throughout thesystem is 1.0 per unit on bases chosen in part (a). If a three-phase short cir-cuit occurs from bus C to ground, find the phasor value of the short-circuitcurrent (in amperes) if each generator is represented by its subtransientreactance.

(c) Find the megavolt amperes supplied by each generator under the conditionsof part (b).

20 MVA, 13.8 kV, X:; = 0.20 per unit30 MVA, 18 kV, X:; = 0.20 per unit30 MVA, 20 kV, X:; = 0.20 per unit25 MVA, 220Y/13.8~ kV, X = 10%Single-phase units each rated 10 MVA,127/18 kV, X = 10%35 MVA, 220Y/22Y kV, X = 10%

Generator 1:Generator 2:Generator 3:Transformer T1:Transformer T2:Transformer T3:

42

3.13 The ratings of the generators, motors and transformers of Fig. 3.23 are:

20 MVA, 18 kV, X; = 20%20 MVA, 18 kV, X; = 20%30 MVA, 13.8 kV, X:; = 20%20 MVA, 138Y120Y kV, X = 10%15 MVA, 138Y113.8!::.kV, X = 10%

Generator 1:Generator 2:Synchronous motor 3:Three-phase Y-Y transformers:Three-phase Y-t::. transformers:

1.0 x 1.278 x 50 MVA = 63.9 MVA1.0 x 1.658 x 50 MVA = 82.9 MVA1.0 x 2.392 x 50 MVA = 119.6MVA

EiJl =Ei2I2 =Ei3I3 =

ISll =IS21 =IS31 =

Eil Ei2 = Ei3 = 1.0~ per unitXl = 0.50+ 0.20+ 0.0826 per unit = 0.7826 per unitX2 = 0.333+0.167+ 0.1033 per unit = 0.6033 per unitX3 = 0.143+0.275 per unit = 0.418 per unit

E 1 .II = j;ll = 0.7826/-90 per unit = 1.278/-90 perunit

h = ji22 = 0.~33 /_90 per unit = 1.658/-90 per unitEi3 = -01 /_90 per unit = 2.392;-90 per unitjX3 .418 .II +h + Is = (1.278+ 1.658+ 2.392) /_90 per unit = 5.328/-90 per unit

50 x 106 A = 131.22 AJ3 x 220 x 1035.328 x 131.22A = 699 A

Is =

If =

hase at C =

IIfl =

(c)

SIC

(b)

= 0.1033 per unit100968

2202 = 968 fl50

= 0.0826 per unitBaseZ

80968

Transmission lines:

43

OA05 per unit= 0.20 X (18)2 5020 x20 =X"Gens. 1& 2:

y-y 50= 0.1 x- = 0.250 per unit20y-~ 50= 0.1 x- = 0.333 per unit15

Transformers :

40 n line: ZBase impedance in lines 138250 = 381 n

40= 381 = 0.105 per unit20= 381 = 0.053 per unit20 n line: Z

138 kV138 kV20 kV

13.8 kV

40 n lines20 n linesGen. 1 & 2Motor 3

Base voltages are:

jO.250jO.250

(a)

Solution:

(c) Find the megavolt amperes supplied by each synchronous machine underthe conditions of part (b).

(a) Draw the impedance diagram for the power system. Mark impedances inper unit. Neglect resistance and use a base of 50 MVA, 138 kV in the 40-0line.

(b) Suppose that the system is unloaded and that the voltage throughout thesystem is 1.0 per unit on bases chosen in part (a). If a three-phase short cir-cuit occurs from bus C to ground, find the phasor value of the short-circuitcurrent (in amperes) if each generator is represented by its subtransientreactance.

44

diameter = 0.1672 x 1000 = 167.2mils/strand

Solution:

4.1 The all-aluminum conductor identified by the code word Bluebell is composedof 37 strands having a diameter of 0.1672 in. Tables of characteristicsof all-alumi conductors list an area of 1,033,500 emil for this conductor(1 emil = (1i / X 10-6 in2). Are these valuesconsistent with each other? Findthe overall of the strands in square millimeters.

Chapter 4 Problem Solutions

1511 = 1521 = 1.0 x 50 x 0.9606 MVA = 48.03 MVA1531 = 1.0 x 50 x 3.0 MVA = 150 MVA

Ei2 = Ei3 = 1.0~ per unitX2 = 0.405+ 0.250+0.053+ 0.333 per unit = 1.041 per unit0.333 per unit

1121 = IEill = 104.0per unit = 0.9606 per unitIXd 1. 1

IE- I 10_'_3 = _. _ per unit = 3.0 per unitIX31 0.3334.9212 per unit

50 x 106 A = 2091.8 AJ3 x 13.8 X 1034.9212 x 2091.8 A = 10.29kA

SIC

Eil =Xl =X3 =lId =

1131 =11,1 =

hase at C =11,1 =

( c)

50X" = 0.20 x 30 = 0.333 per unit

(b) If a fault occurs at C, by symmetry equal currents are input from generators 1 and 2.Moreover, no current should exist between busses A and B through the jO.l05 per unitbranch. If this branch is omitted from the circuit, the system simplifies to

jX1 II

45

Motor 3:

diameter = 0.1672 x 1000 = 167.2 mils/strandSolution:

4.1 The all-aluminum conductor identified by the code word Bluebell is composedof 37 strands each having a diameter of 0.1672 in. Tables of characteristicsof all-aluminum conductors list an area of 1,033,500 emil for this conductor(1 emil = (7i /4) X 10-6 in2). Are these values consistent with each other? Findthe overall area of the strands in square millimeters.

Chapter 4 Problem Solutions

1511 = 1521 = 1.0 x 50 x 0.9606 MVA = 48.03 MVA1531 = 1.0 x 50 x 3.0 MVA = 150 MVA

Eil = Ei2 = Ei3 = 1.0L..!t per unitXl = X2 = 0.405 + 0.250 +0.053 + 0.333 per unit = 1.041 per unitX3 = 0.333 per unit

1121 = II~:II = 1~~1 per unit = 0.9606 per unitIE.31 1.0 . 3 0 .IX31 = 0.333 per Unit = . per unit4.9212 per unit

50 x 106 A = 2091.8 A.J3 x 13.8 X 1034.9212 x 2091.8 A = 10.29 kA

Ihl =

1131 =11,1 =

Ibase at C =11,1 =

( c)

SIC

jXl

X" = 0.20 x ~~ = 0.333 per unit(b) If a fault occurs at C, by symmetry equal currents are input from generators 1 and 2.

Moreover, no current should exist between busses A and B through the jO.l05 per unitbranch. If this branch is omitted from the circuit, the system simplifies to

Motor 3:

45

................... -----------------------

Solution: From Eq. (4.10),

4.4 The energy density (that is, the energy per unit volume) at a point in a mag-netic field can be shown to be B2/2J.L where B is the flux density and J.L is thepermeability. Using this result and Eq. (4.10) show that the total magnetic fieldenergy stored within a unit length of solid circular conductor carrying currentI is given by /-LI2/167r. Neglect skin effect and thus verify Eq. (4.15).

At 75C"and corrected for stranding,

228+ 75Rdc = 1.02 x 2 20 x 0.0878 = 0.1094 O/km, 75C

28+

{0.333 X 10-2)2= 7r 4 x 37 = 3.222 x 10-4 m2

2.83 x 10-8= 3.222 X 10-4 x 1000 = 0.0878 O/krn at 20C

Area

Solution:

4.3 An all-aluminum conductor is composed of 37 strands each having a diameterof 0.333 ern. Compute the de resistance in ohms per kilometer at 75 C. Assumethat the increase in resistance due to spiraling is 2%.

228 + 50Rdc = 22 x 0.01678 x 5.28 = 0.09932 O/mile

8+20This value does not account for skin effect and so is less than the 60-Hz value.

At 50 C,

Rdc = 1.02 x 0.01645 = 0.01678 0/1000' at 20CCorrected for stranding,

= 0.0164517.0 x 10001,033,500

Solution:

4.2 Determine the de resistance in ohms per km of Bluebell at 20 C by Eq. (4.2)and the information in Prob. 4.1, and check the result against the value listed intables of 0.01678Oper 1000ft. Compute the de resistance in ohms per kilometerat 50 C and compare the result with the ac 60-Hz resistance of 0.1024 n/milisted in tables for this conductor at 50 C. Explain any difference in values.Assume that the increase in resistance due to spiraling is 2%.

condo area = (167.2)2 X 37 = 1,034,366 cmils(compared to 1,033,500 cmils)

strand diam. = 0.1672 x 2.54 x 10 = 4.24 mmcondo area = ~(4.24)2 x 37 = 5224 mm2

46

4.6 A single-phase 60-Hz overhead power line is symmetrically supported on a hor-izontal crossarm. Spacing between the centers of the conductors (say a andb) is 2.5 rn. A telephone line is also syrmnetrically supported on a horizontalcrossarm 1.8 m directly below the power line. Spacing between the centers ofthese conductors (say c and d) is 1.0 rn.

(a) Using Eq. (4.36) show that the mutual inductance per unit length betweencircuit a-b and circuit c-d is given by

4 x 10-7 1 . IDadDbc HIn V o.: m

Lint = 2 (~ X 10-7) x 1000 x 1609 = 0.161 mH/mile

0.412r' = -2- x 0.7788 = 0.1604 em

L = 4 X 10-7 In (30~~:) 1609 x 1000 = 4.85 mH/mileDue to internal flux,

Solution:

4.5 The conductor of a single-phase 60-Hz line is a solid round aluminum wirehaving a diameter of 0.412 em. The conductor spacing is 3 m. Determine theinductance of the line in millihenrys per mile. How much of the inductance isdue to internal flux linkages? Assume skin effect is negligible.

= 2 X Eint = 1.. p]2 = !!:_ Him]2 ]2 167r 87r

= 47rX 10-7 Him = .!. x 10-7 Him87r 2

Total energy per unit length is

Energy stored in the tubular element of thickness dx, unit length and radius r:

B2 B2dE .....L . dV = .....L (hx 1 . dx) J

2/)- 2/)-p2x2]2 1;__-:::--,- . - , 27rx dx J47r2r4 2/)-/)-]2x3= --dxJ47rr4

IIHu,I;1.l'IitI

47

Dac = )(1.25 - 0.5)2 + 1.82 = 1.95 mDad = }(1.25 + 0.5)2 + 1.82 = 2.51 m

(b)

Mutual Inductance

>'c-d (linkage of the loop) is given by

-7 Dbd>'d = 2 x 10 x In -D Wb-t/mad

Similarly,

>'c 2 X 10-7 x In DDbC Wb-t/rnac

Therefore,

>'c = 2 X 10-7 (Ia In D1 + t,In D1 + t,In 2;. + laIn D1 )~ ~ ~ *

since 2:-1 = 0 for the group, Eq, (4.36) remains valid.Ia = =I, = I A (and I; = I = 0)

(a) Let circuit a-b carry the current I, so that

Solution:

r- 2.5m ___,e ,---

1.8m

CD +--11m r-

where, for example, Dad denotes the distance in meters between conductorsa and d.

(b) Hence compute the mutual inductance per kilometer between the powerline and the telephone line.

(c) Find the 60 Hz voltage per kilometer induced in the telephone line whenthe power line carries 150 A.

48

Solution:

(a) Bundle: 83

4.9 Find the GMR of each of the unconventional conductors shown in Fig. 4.15 interms of the radius r of an individual strand.

4.8 Find the GMR of a three-strand conductor in terms of r of an individual strand.III,:1:1

J"

j

Solution: Given this bundle: 00GMR = \/(0.779r x 2r x 2r)3 = r.y4 x 0.779 = 1.46r

due to 10. cd7 21.5

2 x 10- I0.1n 20.5

due to Ib cd7 19 since Ib = -10.= -2 x 10- I In-a 18

due to 10. and h cd = 2 X 10-710. In 21.5 x 18 = -0.01288 x 10-710.20.5 x 19M cd -0.01288 x 10-7:H/m= =10.Vcd = wMI = 377 x 150M x 103 = 0.0728 V/km

Solution:

18m

@

2.5m ~r---

4.7 If the power line and the telephone line described in Prob. 4.6 are in the samehorizontal plane and the distance between the nearest conductors of the twolines is 18 m, use the result of Prob. 4.6(a) to find the mutual inductancebetween the power and telephone circuits. Also find the 60 Hz voltage perkilometer induced in the telephone line when 150 A flows in the power line.

(c) Vcd = wMI = 377 x 1.01 x 10-7 x 103 x 150 = 5.71 V/km

Note also that Ia and hare 1800 out of phase. So, due to Ia and h,-7 2.51

cd = 4 x 10 laIn 1.957 2.51 7M = 4 x 10- In - = 1.01 x 10- Hzrn

1.95

-7 1 2.51 } Note that flux through2 x 10 Ia n 1.95 .71251 c-ddueto IaIS

= -2 x 10- L; n--'g1. 5 opposite that due to h

due to l ged =due to Ib cd

Flux linkages with c-d:

49

4.11 Solve Example 4.2 for the case where side Y of the single-phase line is identicalto side X and the two sides are 9 m apart as shown in Fig. 4.9.

D12 = 2r D14 = 4rD13 = JDi4 - DL = 2rY3Ds = "\}(rl)7 (Df2Dr3D14D17)6 (2r)6 = ~ x "'.j(22r2 x 3 x 22r2 x 22r2 x 2r x 2r)6

= 2r {/3 (0.779) = 2.177r"W_ 10 x 12

L = 4 x 10 71n 7 x 1000 x 1609 = 4.51 mH/mi2.1 7 x 0.05

Outside conductors are counter-clockwise numbered 1 through 6. The center conductor isnumber 7. Each radius is r and the distances between conductors are:

Solution:

4.10 The distance between conductors of a single-phase line is 10 ft. Each of itsconductors is composed of six strands symmetrically placed around one centerstrand so that there are seven equal strands. The diameter of each strand is0.1 in. Show that Ds of each conductor is 2.177 times the radius of each strand.Find the inductance of the line in mR/mile.

(d) Bundle: ~

GMR = ~0.779r x 31(4 x 4 x 2 x 2 X 2Y3)3 (24 X 2Y3)3 r30= r x 2 x 31/12 X 0.7791/6 = 2.1Or

(e) Bundle: CfX)

GMR = \1(0.779)3 x 8r2 x 8r2 x 4r2 = 1.704r

50

The conductor is Rook.

Xd = 0.2361 fl0.651 - 0.2361 = 0.415 fl/mi at I-ft spacing

Solution:From Table A.3 at 7-ft spacing:

4.13 Which conductor listed in Table A.3has an inductive reactance at 7-ft. spacingof 0.651 Sl/mi?

D~ = 0.0386 ft1 ft = 2.54 X 12/100 = 0.3048 mDs = 0.3048 x 0.0386 = 0.01177 ftXL = 2 X 10-7 (In 0.O~177) x 377 x 1000 = 0.335 fl/km at 1 m spacing

Solution:From Table A.l for Rail at l-ft spacing:

4.12 Find the inductive reactance of ACSR Rail in ohms per kilometer at 1-m spac-ing.

o; = {fDadDaeDaJDbdDbeDbJDedDeeDc!Dlld = Dbe = Dc! = 9 mDlle = Dbd = DbJ = Dee = vm mDaf = Ded = J122 + 92 m = 15 mti; 19 X vTi7 X 15 X vm X 9 X v'il7 X 15 X vm X 9 m = 10.940 mD~ 0.481 (from Example 4.2) for both sides

-7 10.940 7Lx = Lv = 2 X 10 In 0.481 H/m = 6.249 x 10- HyrnL = L; + Lv = 12.497 X 10-7 H/m

sideyside x

r- 9m ~Io CD -,-

6m

0+6m

Solution:

51

The conductor is Finch.

Deq = {/1Ox 10 x 20 = 12.6 ftXL = 377 x 2 x 1O-7In~ x 1000 = 5.17 n/km0.0133D. = 0.0133/0.3048 = 0.0436 ft

Solution:

4.17 A three-phase 60-Hz line has fiat horizontal spacing. The conductors have aGMR of 0.0133 m with 10 m between adjacent conductors. Determine theinductive reactance per phase in ohms per kilometer. What is the name of thisconductor?

Deq = {/25 x 25 x 42 = 29.72 ftL 7 29.72 . 2.24 mH/mi= 2 x 10- In ----s4 x 1000 x 1609 =0.02

XL = 0.377 x 2.24 = 0.84 n/mi

Solution:

4.16 A three-phase 60-Hz transmission line has its conductors arranged in a trian-gular formation so that two of the distances between conductors are 25 ft andthe third distance is 42 ft. The conductors are ACSR Osprey. Determine theinductance and inductive reactance per phase per mile.

D = 12.7 ft{/D x -D x 2D = if2D = 16Solution:

4.15 A three-phase line is designed with equilateral spacing of 16 ft. It is decided tobuild the line with horizontal spacing (DI3 = 2DI2 = 2D23)' The conductorsare transposed. What should be the spacing between adjacent conductors inorder to obtain the same inductance as in the original design?

Solution:For ACSR Dove conductors, D, = 0.0314 ft. Given that D = 10 ft,

10XL = 21i X 60 x 2 x 10-7In 0.0314 x 103 n/km = 0.4346 n/km

Alternatively, from Table A.3,

Xa = 0.420 n/mi Xd = 0.2794 n/miXL = 0.420+ 0.2794 n/mi = 0.6994 n/mi

= 0.6994 x 0.6214 n/mi = 0.4346 n/mi

4.14 A three-phase line has three equilaterally spaced conductors of ACSR Dove. Ifthe conductors are 10 ft apart, determine the 60 Hz per-phase reactance of theline in n/krn.

52

4.19 A three-phase underground distribution line is operated at 23 kV. The threeconductors are insulated with 0.5 em solid black polyethylene insulation and liefiat, side by side, directly next to each other in a dirt trench. The conductoris circular in cross section and has 33 strands of aluminum. The diameter ofthe condutor is 1.46 ern. The manufacturer gives the GMR as 0.561 em andthe cross section of the conductor as 1.267 em", The thermal rating of the lineburied in normal soil whose maximum temperature is 30C is 350 A. Find thede and ac resistance at 50 C and the inductive reactance in ohms per kilometer.To decide whether to consider skin effect in calculating resistance determine thepercent skin effect at 50 C in the ACSR conductor of size nearest that of theunderground conductor. Note that the series impedance of the distribution lineis dominated by R rather than XL because of the very low inductance due tothe close spacing of the conductors.

Power transmission capability per conductor cost if resistance is neglected is IVsIIVRI/(X .A)based on our cost assumption where A is the cross-sectional area of the conductor. There-fore, the product X . A must be minimized. Assuming Deq is fixed, examining the Tableshows that in comparing any two conductors the percent difference in A is much greater thanthat of X. So, A is the controlling factor, and Partridge or Waxwing would be selected.However, resistance cannot be neglected. A conductor must be large enough in cross sec-tion that melt-down caused by 1112R loss will not occur under the most extreme operatingconditions. The reference (Aluminum Electrical Conductor Handbook) gives information onthermal effects. If reactance causes too high a voltage drop on a line, double-circuit lines orbundled conductors must be provided. The reference (Analytical Development of LoadabilityCharacteristics for EHV and UHV Transmission Lines) contains information on maximumtransmission capability of lines.

Solution:

Note to Instructor: The purpose of this problem is to stimulate thestudent's examination of Table A.3 and is worthwhile in introducingclass discussion of conductor selection.

where Vs and VR are the line-to-neutral voltages at the sending and receivingends of the line and X is the inductive reactance of the line. This relationshipwill become apparent in the study of Chap. 6. If the magnitudes of Vs and VR areheld constant and if the cost of a conductor is proportional to its cross-sectionalarea, find the conductor in Table A.3 which has the maximum power-handlingcapacity per cost of conductor at a given geometric mean spacing.

4.18 For short transmission lines if resistance is neglected the maximum power whichcan be transmitted per phase is equal to

53

1.8 m

e CD i-11m ~

I--4.76 m --I

-vD x D x 2D = ~ D = 33

D = ~ = 2.38m

Solution:

4.20 The single-phase power line of Prob. 4.6 is replaced by a three-phase line ona horizontal crossarm in the same position as that of the original single-phaseline. Spacing of the conductors of the power line is D13 = 2D12 = 2D23, andequivalent equilateral spacing is 3 m. The telephone line remains in the positiondescribed in Prob. 4.6. If the current in the power line is 150 A, find the voltageper kilometer induced in the telephone line. Discuss the phase relation of theinduced voltage with respect to the power-line current.

= ~2.46 x 2.46 x 2 x 2.46 = 3.0993.099= 3n x 1000 x 2 x 10-7 In 0.561 = 0.129 n/km

Since temperature rise would account for a factor of 1.121, skin effect is only about 0.2%. Withinsulation thichness of 0.5 ern center-to-center conductor spacing is 2 x 0.05+1.46 = 2.46 em.So,

0.3831 = 1.1230.0646 x 5.28=

Waxwing has an area of 266,800 cmils and for this conductor

Skin effect can be estimated from the values in Table A.3. The area 1.267 em? is

1.267 x (_1_)2 x ~ X 106 = 250,000 cmils2.54 1r

Rsoo, de

228 + 50= 228 + 20 = 1.121

pi 2.83 x 10-8= A = 1.267 X 10-4 = 0.223 n/km= 1.121 x 0.223 = 0.250 n/km

Note to Instructor: When assigning this problem, it may be advisableto outline part of the procedure.

Solution:

54

D~ = 0.0415 (2.54 x 12 x 10-'2) = 0.0126 mX = 2 X 10-7 X 103 x 377In 13.86 = 0.528 n/km

0.0126

Bluejay:

Deq = ..vll x 11 x 22 = 13.86 mSolution:

4.21 A 60-Hz three-phase line composed of one ACSR Bluejay conductor per phasehas fiat horizontal spacing of 11 m between adjacent conductors. Comparethe inductive reactance in ohms per kilometer per phase of this line with thatof a line using a two-conductor bundle of ACSR 26/7 conductors having thesame total cross-sectional area of aluminum as the single-conductor line and11 m spacing measured from the center of the bundles. The spacing betweenconductors in the bundle is 40 ern.

The induced voltage leads Ie by 90+ 30 = 120; that is, V is in phase with Ie

Ie - Ib = J3Ie/300

ide = 2 X 10-7 J3Ia In ~:: ;300 W1mM = 9.29 X 10-8 HimV = wM x 150 = 377 x 10-8 x 9.29 x 150 x 1000 = 5.25 V/km

Since Ib lags Ia by 1200,

Total flux linkages

due to h, ide

due to l, ide

= Dbd = \}1.8'2+ (2.38 + 0.5)2 = 3.40 m-7 3.40= 2 X 10 I; In 2.60-7 3.40

= 2 x 10 Ibln 2.60-7 3.40

= 2 x 10 (Ia - h) In 2.60

Dee

2.60 m

The center conductor of the 3-phase line causes no flux linkages with d-e since the conductoris at an equal distance from d and e.

55

s = 60 X 10-6 Cjkmqb = qc = -30 X 10-6 Cjkm

Vbc = 10-6 (60 In i - 30 In ~ - 30 In~) where r is in metersbk 2 r 210-6 x 60

= 27rx 8.85 X 10-9 = 744.5 V

Solution:'

5.1 A three-phase transmission line has flat horizontal spacing with 2 m betweenadjacent conductors. At a certain instant the charge on one of the outsideconductors is 60 j.LCIkm, and the charge on the center conductor and on theother outside conductor is -30 j.LCjkm. The radius of each condutor is 0.8 cm.Neglect the effect of the ground and find the voltage drop between the twoidentically charged conductors at the instant specified.

Chapter 5 Problem Solutions

f- 2m -+- 2m -Io 0

D. = 0.0386 ft = 0.0386 (2.54 x 12 x 10-2) = 0.0118 mDb = ~0.01l8 x 0.45 x 0.45 = 0.1337 msX = 2 X 10-7 x 103 x 37710 11.34 = 0.3348 njkm

0.1337

Rail:

Solution:Deq ~9 x 9 x 18 = 11.34 m

4.22 Calculate the inductive reactance in ohms per kilometer of a bundled 60-Hzthree-phase line having three ACSR Rail conductors per bundle with 45 embetween conductors of the bundle. The spacing between bundle centers is 9, 9and 18m.

D, = 0.0314 (2.54 x 12 x 10-2) = 0.00957 m

D~ = y'0.OO957x 0.4 = 0.0619 In13.86

X = 2 X 10-7 X 103 x 3771n 0.0619 = 0.408 njkm

Dove is the conductor for bundling:

56

qo. = 60 X 10-6 Cjkmqb = qc = -30 x 10-6 Cjkm

Vbc = 10-6 (60 In ~ - 30 In~ - 30 In~) where r is in meters21rk 2 r 2

10-6 X 60= 2 8 8~ 10 ~ = 744.5 V1rX .ox -

I- 2m -+ 2m -lo 0

Solution:"

5.1 A three-phase transmission line has flat horizontal spacing with 2 m betweenadjacent conductors. At a certain instant the charge on one of the outsideconductors is 60 j.J.C/km, and the charge on the center conductor and on theother outside conductor is -30 j.J.C/km. The radius of each condutor is 0.8 cm.Neglect the effect of the ground and find the voltage drop between the twoidentically charged conductors at the instant specified.

Chapter 5 Problem Solutions

D, = 0.0386 ft = 0.0386 (2.54 x 12 x 10-2) = 0.0118 mD~ = ~0.01l8 x 0.45 x OA5 = 0.1337 m

7 3 11.34 ....X = 2 x 10- x 10 x 377 In 0.1337 = 0.3348 u/km

Rail:

Deq = ~9 x 9 x 18 = 11.34 mSolution:

4.22 Calculate the inductive reactance in ohms per kilometer of a bundled 60-Hzthree-phase line having three ACSR Rail conductors per bundle with 45 cmbetween conductors of the bundle. The spacing between bundle centers is 9, 9and 18 ID.

D, = 0.0314 (2.54 x 12 x 10-2) = 0.00957 mD~ = JO.OO957 x OA = 0.0619 m

13.86X = 2 X 10-7 X 103 x 377 In 0.0619 = OA08 Q/km

Dove is the conductor for bundling:

56

L'.,

,

9.276 X 10-3 J.LF /km9.276 x 10-12 F[tt: =I 20n (1.196)/24

1.196 1r = -- = - ft2 12

27r x 8.85 X 10-12F/m =

For Cardinal conductors,Solution:

5.4 Using Eq. (5.23), determine the capacitance to neutral (in J.LF /km) of a three-phase line with three Cardinal ACSR conductors equilaterally spaced 20 ftapart. What is the charging current of the line (in A/km) at 60 Hz and 100kVline to line?

Solution:

Xc = 1.779 X 106 I ~ D . rni = 0.2115 MD .mi50 n 0.0268Bc

1 = 4.728 J.LS/rni= XcX' = :~ x 0.1074 MD rniaX~

60= 50 x 0.0683 MD .miXc = ~~ (0.1074+ 0.0683) MD mi = 0.2109 MD .miBe = 4.742 ~S/rni

From Eq. (5.12), at 1-ft spacing and 25 Hz,

1.779 6 1Xc = 25 x 10 In 0.00670 = 356,200 D .mi

5.3 Solve Example 5.1 for 50 Hz operation and 10 ft spacing.

Xc = 2.965 X 104 In ~ = 196,100 D .mir5In- = 6.614rr = 0.00670 ft, or 0.0805 in

= (2 x 0.0805 x 1000)2 = 25, 992 eire mils

Solution:At 5-ft spacing,

5.2 The 60-Hz capacitive reactance to neutral of a solid conductor, which is oneconductor of a single-phase line with 5 ft spacing, is 196.1 kO-mi. What valueof reactance would be specified in a table listing the capacitive reactance inohm-miles to neutral of the conductor at 1-ft spacing for 25 Hz? What is thecross-sectional area of the conductor in circular mils?

57

............................................................q

X - 3.256 X 108 = 1619 nc - 125 x 1609

5.7 (a) Derive an equation for the capacitance to neutral in farads per meter ofa single-phase line, taking into account the effect of ground. Use the samenomenclature as in the equation derived for the capacitance of a three-phase

For 125 miles,

Xc

= ~12 x 12 x 24 = 15.12 m0.0328/2 = 0.0164

= 2.862 x 109 In 15.12 = 3.256 x 108 n .m60 0.0164

r

Solution:

5.6 A three-phase 60-Hz line has fiat horizontal spacing. The conductors have anoutside diameter of 3.28 em with 12 m between conductors. Determine thecapacitive reactance to neutral in ohm-meters and the capacitive reactance ofthe line in ohms if its length is 125 mi.

Cn = 150 x 0.01336 = 2.004 JLFXc = 0.1987 X 106 = 1325 n

150 .

Solution:

Osprey diam. =o.; =Cn =

=Xc =

0.879 in~25 X 25 X 42 = 29.72 ft27rX 8.85 X 10-12 F[tt:

In ?g";~9)~~8.301 x 10-12 F [tn = 8.301 x 10-6 x 1.609 JLF 1m = 0.01336 J.LF/mi

106377 x 0.01336 = 0.1985 x 106 n.mi

From Table A.3, X~ = 0.0981. Interpolation from Table A.4 yields Xd = 0.0999+0.72(0.1011-0.0999) = 0.1006. From Table A.4, Xc = 0.1987 X 106 n .mi.For 150 miles,

5.5 A three-phase 50-Hz transmission line has its conductors arranged in a triangu-lar formation so that two of the distances between conductors are 25 ft and thethird is 42 ft. The conductors are ACSR Osprey. Determine the capacitance toneutral in microfarads per mile and the capacitive re