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ECE 421/599 FALL 2014 - MIDTERM EXAM Problem 1 (20...

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1 ECE 421/599 FALL 2014 - MIDTERM EXAM Problem 1 (20 points): A load Z L =10+j5 is supported by a 100V rms, 60-Hz source through a line with impedance Z T =j5. a. Find the current I, voltage V L , real power, reactive power and power factor of load Z L b. Find the capacitance of the shunt capacitor making overall unity power factor (PF=1) on the load size. What is the new voltage V L on the load side? c. Respectively for the conditions with and without the capacitor, draw two phasor diagrams about V, V L , I, and the voltage of Z T . a. Z=Z T +Z L =10+j10 (1) I=V/Z=100/(10+j10)=5-j5A (2) V L =IZ L =75-j25=79.057-18.435 o V (2) S L =P L +jQ L =V L I*=(75-j25)(5+j5)=500+j250 VA P L =500W (1) Q L =250var (1) |S L |=559.0W PF=P L /|S L |=0.894 (1) b. If the load with capacitor has a terminal voltage =V L Q L +Q C =0 Q C =-QL=-250Mvar (1) X C =|V L | 2 /Q C =-25 (1) C=-1/(X C )= -1/(377X C )=106.1F (1) Z Lnew =Z L //jX C =12.5 (1) I new =V/(Z T +Z Lnew )=6.90-j2.76=7.43-21.8 o A (1) V Lnew =Z Lnew xI new = 86.20-j34.48 =92.85-21.80 o V (2) c. V ZT =Z T xI=j5 (5-j5)=25+j25 V (1) V ZTnew = Z T xI new = j5x(6.90-j2.76)=13.8+j34.5V (1) Phasor diagrams: (2 each)
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Page 1: ECE 421/599 FALL 2014 - MIDTERM EXAM Problem 1 (20 …web.eecs.utk.edu/.../Exam1_2014_Final_Solution.pdf · 1 ECE 421/599 FALL 2014 - MIDTERM EXAM Problem 1 (20 points): A load Z

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ECE 421/599 FALL 2014 - MIDTERM EXAM

Problem 1 (20 points): A load ZL=10+j5 is supported by a 100V rms, 60-Hz source

through a line with impedance ZT=j5.

a. Find the current I, voltage VL, real power, reactive power and power factor of load ZL

b. Find the capacitance of the shunt capacitor making overall unity power factor (PF=1)

on the load size. What is the new voltage VL on the load side?

c. Respectively for the conditions with and without the capacitor, draw two phasor

diagrams about V, VL, I, and the voltage of ZT.

a.

Z=ZT+ZL=10+j10 (1)

I=V/Z=100/(10+j10)=5-j5A (2)

VL=IZL=75-j25=79.057-18.435o V (2)

SL=PL+jQL=VLI*=(75-j25)(5+j5)=500+j250

VA

PL=500W (1)

QL=250var (1)

|SL|=559.0W

PF=PL/|SL|=0.894 (1)

b.

If the load with capacitor has a terminal voltage

=VL

QL+QC=0 QC=-QL=-250Mvar (1)

XC=|VL|2/QC=-25 (1)

C=-1/(XC)= -1/(377XC)=106.1F (1)

ZLnew=ZL//jXC=12.5 (1)

Inew=V/(ZT+ZLnew)=6.90-j2.76=7.43-21.8 o A (1)

VLnew=ZLnewxInew= 86.20-j34.48

=92.85-21.80o V (2)

c.

VZT=ZTxI=j5 (5-j5)=25+j25 V (1)

VZTnew= ZTxInew= j5x(6.90-j2.76)=13.8+j34.5V (1)

Phasor diagrams: (2 each)

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Problem 2 (25 points): Short answers

a. True or false for each statement? Briefly explain why. (5 points)

(1 each; 0.5 for T/F, and 0.5 for reason)

i. The top-3 energy resources of electricity generation for the overall US power grid are coal, nature

gas and hydro. False. Coal, Nature Gas and Nuclear

ii. A “bulk electric system” defined by NERC generally means the portion of system operated at

voltage of 69kV or higher. False. 100kV instead of 69kV

iii. The generation resources utilizing steam turbines are such as combined-cycle power plants,

parabolic dish concentrators, pressurized water reactors, geothermal power plants and tidal power

plants. False. Exclude tidal power.

iv. Hydroelectric power plants usually adopt round rotor generators for high speed operation

False. Salient-pole generators

v. Reactive power sources include capacitors, generators and regulating transformers

False. Transformers are not.

b. What are reliability concerns in integration of wind generation into a power grid? Give two concerns.

(2.5 points for one and 5 points for any two)

1) Congestion issues due to long distance transmission between wind generation and load centers.

2) Mismatch in supply-time and demand-time curves

3) Errors in wind forecast

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c. Which statements about the per-unit system are correct? (Total 6 points)

i. The circuit laws are still valid in per-unit systems.

Correct (1)

ii. A minimum of four base quantities, i.e. SB, VB, IB and ZB, have to be given for a per-unit system

Incorrect (1). Giving any two is the minimum (1)

iii. A bus with voltage at 0.8pu must mean that its actual voltage is as low as 80% of the nominal value

Incorrect (1). Only when the nominal value is chosen as the base (1)

iv. The per-unit values of impedance, voltage and current of a transformer are the same regardless of

whether they are referred to the primary or the secondary side

Correct. (1)

d. For a 3-phase transmission line with line-to-line voltages Vab , Vbc, and Vca, phase voltages Van , Vbn, and

Vcn, and currents Ia, Ib and Ic, which of the following formulas can calculate its 3-phase complex power?

(Total 4 points)

i. *

3a n a

V I Yes

ii. *

3b n b

V I Yes

iii. *

3 ( 1 3 0 )a b a

V I No

iv. *

3 ( 1 9 0 )a b c

V I Yes

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e. A 3-phase transposed transmission line has 2 bundled identical conductors

per phase placed at a distance of d. Which of the following will necessarily

increase its inductance? Briefly explain why. (total 5; 0.5+0.5 each)

i. Moving the left line to the left while fixing the other two

Yes. GMD L

ii. Moving the middle line to the left while fixing the other two

No. L if D12 becomes too small

iii. Moving the middle line upward while fixing the other two

Yes. GMD L

iv. Increasing d between the two bundled conductors of each phase

No. GMR L

v. Add a 3rd conductor which is identical to the existing two conductors

such that each phase has symmetric 3 bundled conductors.

No. GMR L

23

2 3

1 / 61 / 3 2 / 3 1 / 6

3

1 / 2 1 / 2 1 / 6

2

1

s s

s

s s s

G M R D d G M R D d

G M R D d d d

G M R D d D D

3

1 2 2 3 1 30 .2 ln 0 .2 ln m H /k m

L L

DL

G M R

D

G R

DG

M

M D

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This question is for ECE599 students only (5 points):

f. A salient-pole synchronous machine has internal emf E, Ra0, and Xd=2Xq. It is operated as a

synchronous motor load. Draw phasor diagrams with E, V, Id, Iq, jXdId and jXqIq to respectively

illustrate two conditions having |E|=|V| and |E|<|V|. Does the machine supply or absorb reactive

power for each condition?

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TAKE-HOME PROBLEMS (Due by today midnight 12:00AM)

Problem 3 (15 points): As shown in the figure, a three phase bundled line has two conductors per phase. D=10m and d=0.1m.

Each conductor has radius r=0.02m. The line spacing is shown in the figure, which is measured from the center of the bundle.

Consider transposition,

a. Calculate the GMR of each conductor, and GMD, GMRL

and GMRC of the line.

b. Determine the inductance and capacitance per phase per

kilometer

GMR=r’=re-1/4=0.02e-1/4=0.0156m (2)

GMD=(2DD)1/3=(2010)1/3=12.60m (3)

GMRL=(r’d)1/2=(0.01560.1)1/2=0.0395 (3)

GMRC=(rd)1/2=(0.02.1)1/2=0.0447 (3)

L=0.2ln(GMD/GMRL)=0.2ln(12.6/0.0395)=1.153mH/km (2)

C=0.0556/[ln(GMD/GMRC)]=.0556/[ln(12.6/0.0447)]=0.00986F/km (2)

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Problem 4 (20 points): A 3-phase line has an impedance of ZT=j3 and feeds two

parallel, balanced 3-phase loads respectively in and Y connections. Z1= j30 and

Z2=5. The line is energized at the sending end from a 3-phase balanced supply with

line-to-line voltage |VL|=346.41V.

a. Determine 3-phase real and reactive powers drawn from the supply, and the

power factors on the supply and load sides.

b. Determine the total real and reactive powers absorbed by the 3-phase line and

each 3-phase load

c. Take the phase voltage Van on the load side as the reference, i.e. Van=|Van|0o. Determine VAB, IA, Vab, and Van. Draw a

phasor diagram about those phasors together with the voltage of ZT in phase a.

a.

Z1Y=-j30/3=-j10 (1)

ZL=Z1Y//Z2=-j1015/(-j10+15)=4-j2 (1)

First, use VAB as the reference: VAB=346.41V

VAn lags VAB by 30o, so VAn=346.41/3-30o=200-30o V

Z=ZT+ZL= 4+j1 (1)

Ia=VAn/Z=34.87-j33.72=48.51-44.04o (1)

Ss3=3VAnIa*=28235W+j7059var=2910414.04o

Ps3=28235W (1) Qs3=7059var (1)

PFs=Ps3/|Ss3|=0.9701 lagging (1)

Van=IaZL=72.05-j204.62=216.9-70.60o V (1)

SR3=3VAnIa*=28235W-j14118var=31568-26.57o (1)

PS3=28235W (1) QS3=-14118var (1)

PFR= PR3/|SR3|=0.8944 leading (1)

b.

Sline =3|Ia|2ZT=j21176var (1)

S31= P31+jQ31=3|Van|2/Z1*=-j14118var (1)

S2= P32+jQ32=3|Van|2/Z2*=28235W (1)

c.

Vab=Van330o =285.28-j244.52=375.7-

40.6oV (1)

Using Van =216.9-70.60o as the reference:

Van= 216.9 V

Ia=48.51(-44.04+70.60)o =48.5126.56o A

(0.5)

Vab=375.730o V (0.5)

VAB=346.4170.60o V (0.5)

VTa=ZTIa=145.5116.6oV (0.5)

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Phasor diagram (3)

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Problem 5 (20 points): Two round-rotor generators, G1 and G2, are connected to support a load. The 3-phase power and line-

line ratings are given below. The stator resistances of generators are ignored

G1: 80MVA 20kV X=10%

T1: 125MVA 20/200kV X=15%

T2: 80MVA 200/20kV X=20%

G2: 75MVA 22kV X=12.5%

Line: 200kV Z=j100

Load: 200kV S=120MW+j40Mvar

a. Draw an impedance circuit diagram showing all per-unit impedances on a common base of 100MVA and 20kV on the G1

side.

b. G2 supplies 60MW+j30Mvar with a terminal voltage of 20kV at bus 4. Take that voltage as the reference. Determine

voltages at buses 1, 2 and 3 in per-unit and kV, and the internal emfs of G1 and G2 in per-unit and kV.

These two questions are for ECE599 students (10 points):

c. Under the system condition in b, draw a phasor diagram including the following quantities in per-unit:

i. The emfs of the generator and motor

ii. The voltages at buses 1, 2, 3 and 4

iii. The currents of the line, load and G2

d. Assume that the load changes at bus 3, a capacitor bank is added to bus 3 to maintain its voltage in the range of

180~200kV, and the internal emfs of G1 and G2 are constant. What is the maximum real power of the load that G1 and

G2 are able to support without violating their theoretical power transfer limits?

a.

VB1=20kV (1)

VB2=VB3=20(200/20)=200 kV (1)

VB4=200 (20/200)=20 kV (1)

ZB=(VB2)2/SB=(200)2/100=400 (1)

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XG1=0.1 (100/80) =0.125 pu (1)

XT1=0.15 (100/125) =0.12 pu (1)

XT2=0.2 (100/80) =0.25 pu (1)

XG2=0.125 (100/75) (22/20)2=0.2017

pu (1)

ZT=ZT/ZB=j0.25 pu (1)

ZL=2002/(120-j40)=300+j100 (1)

ZL=ZL/ZB=0.75+j0.25 pu (1)

Circuit diagram (1)

b.

SG2=(60+j30)/100=0.6+j0.3 pu (1)

Take V4 as the reference

V4=20/VB4=1pu (1)

IG2=SG2*/V4*=0.6-j0.3=0.67-26.6o pu (1)

V3=V4-IG2XT2=0.925-j0.15

=0.9371-9.21o pu (1)

I3=V3/ZL=1.05-0.55j=1.185 -27.65 o pu (2)

IG1= I3-IG2=0.45-0.25j

=0.515-29.05 o pu (2)

V2=V3+ZTIG1=0.9875-0.0375j

=0.9882-2.175o pu (1)

c

Phasor diagram (3)

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d.

V3=0.9~1pu

Let V3=1pu

Pmax=3|EG11/(XG1+XT1+ZT)|+ 3|EG21/(XG2+XT2)|=13.46pu=1346MW (3)


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