电力系统分析第三版(于永源杨绮雯著)我国电力出版社课后答案解析

1-2、电力系统接线图分为哪两种？有什么区别？1-3、对电力系统运行的基本要求是什么？答：各部分电压等级之所以不同，是因三相功率S和线电压U、线电流I之间的关系为 5%W

WP8760pdtPTT+(8760+(8760−5000)P(MW)100302000 5000 8760t(h)解：A=2000×100+(5000−2000)×60 年 0(T

A=492800=492（

故障相得对地电压变为线电压，即为原来的3倍。

3Ia=3Ia。即故障时单相接地电1-10、消弧线圈的工作原理是什么？电力系统一般采用哪种补偿方式？为什么？

∑I

＞∑I

＜∑I

P4答：1)短线路，是指长度不超过100km的架空线路，线路电压为35KV及以下时，电纳这种线路的等值电路有T型和π型等值电路，如图二。

PU2KNS2

XU

KN

100SI

P

，励磁电纳B=I

%S

0U2

发电机电抗的百分数XG%实际上是在XG上通过额定电流时产生的电压降与额定电X

(%)=

3INXG×100U2-13、什么是有名制？什么是标幺制？标幺制有什么特点？基准值如何选取？1是按变压器实际变比计算，2是按平均额定电压之2-15、什么叫变压器的额定变比，实际变比，平均额定电压之比？在归算中如何应用？K

242(1+5%)24.2.5K2(1−2)121=1.82K2(1−3)

.5

=5.71K

110(1−2.5%)

=9.75K

35(1−5%)

=5.04K

=3.17

U=220×(1+5%)=231KVU=110×(1+5%)U=35×(1+5%)=36.75KVU=10×(1+5%)=10.5KVUl5=6×(1+5%)=6.3KVU=3×(1+5%)=3.15KVρ解:r1=

.5

=0.2625(Ωkm)sr=15.2=Dm =4000×4000×2×4000=5039.7(mm)XD

=0.1445lg+0.0157=0.1445lg5039.7+0.0157=0.423(Ω/km) b58Dmr

×1

−65839.7

×1

−6=2.69×10g=0.1RX

=rl=0.2625×60=15.75(Ω)=xl=0.423×60=25.38(Ω)×10

=bl=2.69

−6×60=×B1=

0.807×10G0.算外径为19mm,三相导线的不等边三角形排列,线间距离D12=9m,D23=8.5m,D31=UUN(%)Ur

ρ31.5ns×185

=0.085(Ω/km)r

×400=61.64(mm)D=3900×8500×=7756.46(mm)6100X

=0.1445lg

Dm+

0157

0157

=0.311(Ω/km)

r

n.64b

7.5×10mr

−6

58.64

×1

−6=3.61×10

−6(s/km)b

1.805×10

−6(s/km)g0.2-19.三相双绕组升压变压器的型号为SFL-40500/110,额定容量为40500KVA,额定电压为121/10.5KV,P

=234.4kw,Uk(%)=

I0(%)=2.315,求该变压器的参数,并作等值电路。解:`R

=P

234.4×(10.5×10 3×103)2(40500×103)2

=1.57×10−2(Ω)X=

100S

11×(10.5×=0.3(Ω)103)2100×40500×103G

PU2

×10

.6

(10.5×103)2=8.49×10−4(S)

I 2.315× =8.5×10−3(S)(%)S==0N40500×10 100×(10.5×103)22-20.三相三绕组降压变压器的型号为SFPSL-120000/220,额定容量为UUK(2−3)(%)=7.96,PK(1−

=601KW,PK(1−

=182.5KW,

P

K(1−

2)(%)=14.85,U

K(1−

3)(%)=28.25,0=135KW,I0(%)=0.663,求该变压器的参数,并作等值电路。

PK(1−

=601(KW)P'K(1−3)P'

=4P(1−3)=4P

=730(KW)=530(KW)(2−3)(2−3)P=(P

+P

P'

)=400.5(KW)P(P

K(1−

K(1−P

P'

K(2−3))=200.5(KW)P(P'

(1−2)P'(2−

K(1−3)−P)=329.5(KW)

K(1−(2−3)

K(1−2)R1

K1

N

400.5

3

3 ×(220×10)

=1.346(Ω)T1N

(12×104×103)2RN1

PK2

U2

200.5

3

3 ×(220×10)

=0.674(Ω)T2

2N

(12×10

4

×103)2R1

PUK3

N

3

3 (22010)

=1.107(Ω)T3N

(12×104×103)2II：UK(1−2)(%)=14.85,UK(1−3)(%)=28.25,UK(2−3)(%)=7.96UK1UU

(%)=1(%)=1(%)=1

K(1−K(1−K(1−

(%)+U(%)+U(%)U

(%)−K(1−U(%)−U(2−3)K(2(%)−−3)U

(2−K(1−K(1−

(%))=17.57(%))=−2.72(%))=10.68X

U

(%)U2 100S

17.57×(220=70.866(Ω)×103)2100×12×104×103X=UK2(%)U=

−2.72×(220×10

32

=−10.971(Ω)X

U100S(%)U2K3 100S

100×12×10410.68×(220×=43.076(Ω)103)2100×12×104×103G

P

135×10

=2.79×10−6(S)TU2 (220×103)2IV：B

I=0

0.663×12×10×10

=1.644×10−5(S) ×(220×103)2 220/121/38.5KV，PK(1−

=417KW,P(1−3)

=318.5KW,PK(2=314KW;−3)

U

K(1−

2)(%)=8.98,UK(1−

3)(%)=16.65,UK(2−3)(%)=10.85;P

=57.7KW,I0(%)=0.712.求该变压器的参数,解:I:P

=417(KW)(1−2)K(1−(2−

=4P(1−3)(2−3)

=1256(KW)==P==P==PUP=(P

+P

P')=217.5(KW)P(P

K(1−

(1−3)P

P'

K(2−3))=199.5(KW)P(P'

K(1−

(2−P

K(1−3)−P)=1056.5(KW)

K(1−(2−

K(1−2)R

U

32=217.5×10×(220×10)(12×104×103)2

=0.73(Ω)R

U2K2N1 32=199.5×10×(220×10)(12×104×103)2

=0.67(Ω)R

32=1056.5×10×(220×10)(12×104×103)2

=3.55(Ω)==10.29×(220×10)%%UN21=−1.31×(220×10==)2UU' K(2−3)(%)=2UK(1−

2)(%)=8.98UK'(1−3)(%)=2U

(%)=33.3(%)=21.7U(%)=K1(U K(1−(%)=1UK2 K(1− U(%)=1

K(1−(2−(%)+U

(%)− U −3)(%)− U (1−3)(%)−U

(%))=10.29(%))=−1.31(%)=23.01

K(1−(2−3)

K(1−2)X

U%U2

32=41.503(Ω)100×12×104×103X=T

U

−5.284(Ω)

100S

100×12×104×103X

U 100S

23.01×(220×103)2=92.807(Ω)100×12×104×103G

P

.7×103

=1.19×10−6(S)

U2

(220×103)2B

I=0(%)S100U2

0.712×124×1=1.765×10−5×10 100×(220×103)2

(S)

=125MW,cos=0.85,U=13.8KV,=1.867,=0.257,=0.18,试计算该发电机的直N N Xd Xd X'' 轴同步电抗Xd,暂态电抗Xd,直轴次暂态电抗Xd的有名值.U2

解:Z

SU

P

=1.295(Ω)X=

=1.867×1.295=2.418(Ω)11∆S=−jUPPiR+Q

Z

=0.257

=0.257×1.295=0.333(Ω)

''=0.18Z

=0.18×1.295=0.233(Ω)U

=6KV,=500A,电抗器电抗百分数XL(%)=4.I

X

4×6

=0.277(Ω)1003×500P2+Q2∆S∆P

+j∆Q

+jXL)B∆S

=−jU2B2C2 ?∆U

XU

=PiX−QiRU都要乘3倍。而线电压是相电压的3倍,将ΔU和δU前的3变成3×

3，其中一个3与ΔU、δU相乘则使相电压的ΔU、δU变成了线电压的ΔU、δU，另一个3则和等式右

U2).或dU.

−U

U

−U

×100.

−U

−U1N

UU

−

2N

U

UU

−

×100.电压调整:线路末端空载与负载时电压的数值差(U20−U2).以百分数表示,电压调整ΔU

UU

−U

×100.输电效率η%=P2×100.P++PO+jIO%SmmZ∑∑S==.+(− 8.07×10)=(30+j15)+(−j106)=30+j10.466(MVA) jUL答：∆∆S

S=

∆S

=PK+jUK%S

I%S

S

Z

ρ解:r1=

.5=0.2625(Ω/km)120sD=5m=5000(mm),r=15.2=7.6(mm)m X

=0.1445lg

Dm+0.0157=0.1445lg

00

+0.0157=0.423(Ω/km) r b

7Dmr

×10−658

=2.69×10−6

(s/km),g

R=1rl=1×0.2625×150=19.687(Ω)X=

21 xl=

×0.423×150=31.725(Ω)L 212

=2bl=2×2.69

×10−6×150=8.07×10−4(s)G0.

B S

S

==S2+∆S∆∆ =∆Pj∆QS+ L=

P'2+Q U2

(R

jX)=L

+10.466221062

×(19.687+

j31.725)=1.769+j2.85(MVA)

'.'

=(30+j10.466)+(1.769+j2.85)=31.769+j13.316(MVA)dU

=∆U2+

P'R+Q'X=L 2U+jL U

=30×19.687+10.466×31.725

+j

30×31.725−10.466×19.687=8.7+j7.03(kv)U

=U

+dU

=106+8.7+j7.03=114.7+j7.03。=114.9∠3.5（kv)SjU

=S1+(−1

B×−4L)=(31.769+j13.316)+(−j114.92×8.0710)2 =31.769+j7.99(MVA)设U

=106∠0（kv),∆U

=7.03(kv),则U=114.9∠3.5(kv)三相导线几何平均距离为7.5m,已知其始端输出的功率为120+j50MVA,始端的电压为240KV.求末端电压及功率,并作出电压向量图.ρ解:r1=

.5=0.105(Ω/Km)300sD

=7.5m=7500(mm)r=24.29=x

D=0.1445lgm+0.0157=0.1445lg7500+0.0157=0.419(/km)12.1lglgDRRL=r==+Q1′SS′=S1′=(120+j65.6384)−(6.82+j27.22)=113.18+j38.418(MVA)b

58r

−6

5800.1

−62.715×10−6(S/km)1l=0.105×200=21(Ω)X

=xl=0.419×200=83.8(Ω)

=

bl=2.715×10−6×200=5.43×10−4(S)G

Ṡ′−(−=ṠjU2

B

5.43−4)=(120+j50)−(−j2402×) 2 =120+j65.6384(MVA)∆S

(RU2

jX

)=

265.638422402

×(21+j83.8)=6.82+j27.22(MVA)̇̇−∆Ṡ

∆U

jδU

P′R+=1 Q′X U

+j

P

U

−Q′R =120×21+65.6384×83.8

+j

120×83.8−65.6384×21=33.419+j36.157(MVA)U̇=U−dU̇

=240−33.419−

j36.157=206.58−

j36.157=209.72∠Ṡ=Ṡ′−(−jU25.43×(−j209.722×

BL)=(113.18+j38.418)−10−4) 2 lglgD==S2+∆SSS=S2+=113.18+j50.36(MVA)

2

=113.18+j50.36(MVA)；U̇

=209.72∠−10°(KV)。ρ解:r1=

.5=0.332(Ω/sDm=5m=5000(mm),r=13.7=x=0.1445lg5000+0.0157=0.425(Ω/km)b

58r

85−6

580085

−6=2.647×10

−6gRX

=rl=0.332×80=26.56(Ω)=xl=0.429×80=34.32(Ω)

bl=2.647×10−6

×80=2.12×10−4(s)G=0 (-jU2

BL)=(15+j10)+(−j1102×

2.12×10−4)=15+j8.717(MVA)S

+QU2

(RL+

)=L15

8.7172

×(26.56+j34.32)=10.66+j0.854(MVA)

'.'

=(15+j8.717)+(10.66+j0.854)=15.66+j9.57(MVA) S1=S1+(−jU

B2

)=(15.66+

j9.57)+(−j116×

2.12×−104)=15.66+j8.14(MVA)2P'R+Q'XX

–Q'R

15.66×26.56+9.57×34.32dU

U

+jδU

L1U

L+j

==U2+dUUU2(RL++j15.66×34.32−9.57×26.56=6.417+j2.442(KV)116U

=U

−dU

=116−6.417−j2.442=109.583−

j2.442(KV)3-11.220kv单回架空电力线路,长度为220km,电力线路每公里的参数为r=0.108Ω/km,x =2.66×10

=0.42Ω/km,b

s/km,线路空载运行,当线路末端电压

=rl=0.108×220=23.76(KV)X

=xl=0.42×220=92.4(Ω)

bl=2.66

×10−6×220=5.852×10−4(s)因为线路空载运行,所以S

'=

B

)=−j2052×

=−j12.3(MVA)SSS=LP

'2 +Q

)=×(23.76+j92.4)=0.085+j0.33(MVA)S

=S'+∆S

=−j12.3+0.085+j0.33=0.085−j11.97(MVA)dU

=∆U2+

P'+Q'X L2U

XL+j

–Q'RL

−12.3×.4

12.3×23.76+j=−5.544+j1.4256(KV) 2=205−5.544+j1.4256=199.456+j1.4256(KV) 5.852×=S'+(−jU2+1.42562)×

L)=(0.085−

j11.97)+[−j(119.4562

10−4S

=0.085 U1= SS =S3+∆∆S3-12.有一台三绕组变压器,其归算至高压侧的等效电路如图所示,Z

=2.47+j65Ω,Z

=2.47−j115Ω,Z=2.47+j37.8Ω,S

=5+j4MVA,S

=8+j6MVA,当变压器变比为110/38.5(1+5%)/6.6kv时,试计算高、中压侧的实际电压..'. U=U3×

0(KV∆.S

P2+Q2 (RU

jX

)=

×(2.47+j37.8)=0.0247+j0.378(MvA)'2

=(8+j6)+(0.0247+j0.378)=8.0247+j6.378(MVA)dU

U

j

PR+QX

jPX−QRU∆

=

3+=

δ

U

+

3T3

U

=8×2.47+6×37.8

8×37.8−6×2.47+j

=2.4656+j2.87589(KV)U'

U

+.dU

=100+j2.4656+j2.8757=102.4656+j2.8758(KV)U

,S

∆.S

P2+Q2 (

jX

)=

×(2.47−j1.5)=0.0084−j0.005(MVA)U =S2+∆ST2=(5+j4)+(0.0084−j0.005)=5.0084+j3.995(MVA)

P'X

P'X

–Q'RdU

=∆U2+

T2

+j

T22

=5.0084×2.47− j5.0084×(−1.5)−3.995× 3.995×1.5+102.46562+2.8758

2.47=0.062−102.46562

+2.8758

U.=U.'−.dU 102.4+

=

j3.04(KV)DDm1= +S'

=13.033+j10.373(MVA)11

∆

P+Q''

(R

+

13.033

×(2.47+j65)=0.0652+j1.716(MvA)

ST1'

102.46562+2.87582S.'=.S''.+∆S

=(13.033+j10.373)+(0.0652+j1.716)=13.0982+j12.089(MVA)dU1=∆U1+

P''XT1 U'

j

–Q''RT1 U'

13.033×2.47+10.373×652.5

+j

13.033×65−10.373×2.472.5

=6.89+j8(KV)U.

=U

+.dU

=109.3556+j10.8758(KV)U1=109.9(KV)UU

=U×

=102.445(KV)38.5(1+5%)=102.445×

38.5×1.05

=37.65(KV) 110 3-13.某电力线路导线为LGJ-185,长度为100km,导线计算外径为19mm,线路末端负荷为r

ρ.5=0.17(Ω/km)185R

=rl=0.17×100=17(Ω)34000×4000×2=5.04×103(mm)×4000DDm2=35500×5500×2×5500=6.93×103(mm)lglgD==−j110×SS =S2+∆S=(90+j20)−j1.68=90+j18.32(MVA)r= Dx=0.1445lgm1+0.0157=0.1445lg5.04×10 r

+0.0157=0.41(Ω/km)xD

=0.1445lg+0.0157=0.1445lg6.93×10

+0.0157=0.43(Ω/km) r X

=xl=0.41×100=41(Ω)X

=x

l=0.43×100=43(Ω)b58D×10−6r

=2.78×10−6(s/km)b58m2×10−6r

=2.65×10−6(s/km)BL1=bl=2.78×10−6×100=2.78×10−4(s)BL2=bl=2.65×10−6×100=2.65×10−4(s)1∆.S

=−

B

.2.78×10−4 =−j1.68(MVA)

∆S=−

B

.2.65×10−4=−j220× =−j6.4(MVA) ==S2+∆ SC2

=(90+j20)−j6.4=90+j13.6(MVA)==(11.85+j28.58)−==P−−Q−−Q算负荷为算负荷为S2=90+j40MVA,SdUdU1%=dUdU2%=%%=dU1%−dU2%=33.68%−8.7%=25%∆S1

P'2

+Q'221(RL+jXL1)=

+.322

×(17+j41)=11.85+j28.58(MVA)∆S1

P'2

+Q'222(R

+jX

+.6

×(17+j43)=2.9+j97.36(MVA)∆.S=.∆S.−∆S1

(2.9+j7.36)=8.95+j21.22(MVA)

jδUR

QXX

LP+j21

U

'R21L

90×17+18.32×41

90×41−18.32×17+j

=20.74+j30.7(kv)dU

=∆U+

=22RU

QXX

P+j22

U

'R22L =90×17+13.6×43

90×43−13.6×17+j

=9.6+j16.54(kv)

dU1×100%

.74

30.72

×100%=33.68%110

dU2×100%

9.6

×100%=8.7%3-14.对图所示环式等值网络进行潮流计算,图中各线路的阻抗为Z

=10+j17.32Ω,Z

=20+j34.6Ω,Z

=25+j43.3Ω,Z

=10+j17.3Ω,各

=50+

=40+j15MVA,且U1=235KV.Z

=10+j17.32≈10+17.3Z

=20+j34.6=(10+17.3)×2Z

=25+j43.3=(10+17.3)×2.5SS2×(2+2.5+1)+S3×(2.5+1)+S++Zl1)+S3(Z++Zl1)+SSS4×(2.5+2+1)+S3×(2+1)+S==109.2+j52.3−(90(90+j40)=19.2+j12.3(MvA)SS=1SSS1+SS2+S3+=90+j40+50+j30+40+j15=180+j85(MVA)Z

=10+j17.3则S1

S2(Z*+Z

l

Z

).+(Z+Z*

Z*

+Z*+Z

1+2+2.5+1 (90+j40)×5.5+(50+j30)×3.5+(40+j15)=109.2+j52.3(MvA)S'1Z

S4(Z +Z

Z+Z

Zl1 1+2+2.5+1 (40+j15)×5.5+(50+j30)×3+(90+j40)=70.77+j32.7(MvA)−23S

70.77+

−S

=

j32.7−(40+

j15)30.77+

j17.7()S'

=109.2+j52.3+70.77+j32.7=179.97+j85(MVA) S +=S2+3+SUU3=URRL3+QSS=23+=='S+=109.4+j52.67+3+j5.27=122.4+j57.94(MVA)−−dU2=225.95−j5.82−(3.656+j1.85)=222.294−j7.67(kv)=++jP∆.S

P223

+Q(RU2

jX

)=

212.322×(20+j34.6)=0.2+j0.37(MVA) S

=19.2+j12.3+0.2+j0.37=19.4+j12.67(MVA) =S2+

S

=90+j40+19.4+j12.67=109.4+j52.67(MVA)∆S1S

P'2 2(Rl1+ l1.'.

)=

.67

×(10+j17.32)=3+j5.27(MVA)d

U

j

PR

+QX

j

PX

−QR

δ

U

1L1

U

1L1

112.4×10+57.94×17.32

+j

112.4×17.32−57.94×10

=9.05+j5.82(KV)U

=U

−dU

=235−(9.05+j5.82)=225.95−

j5.82(kv)=226∠−1.5.（KV

PX

P“

–Q”RdU

=∆U2+

U

j

U

19.4×20+1267×34.6

+j

19.4×34.6−12.67×20

=3.656+j1.85(KV) 222.426∠−2.

P

U

X

L

X−QU

R =30.77×25+17.7×43.32.42

30.77×43.3−17.7×25+j2.426

=6.9+j4(kV)==U3+dU==25KW,UK(%)=7.5;T 3=222.426+6.9+j4=229.326+j4(KV)SP

=2MVA,PK=24KVA,UK(%)=6.5XXT1=UK(%)UXXT2=UK(%)USST1=的变比为K=KT1 T211KV

=35/

时,每台变压器的总功率为多少?当K

=34.125/11KV,K

R=

×(35320.48(= Ω(8×106)2

100S

7.5×(35×103)2100×8×106

=11.48(Ω)R

32=×(35×10)(2×106)2

35(Ω

100S

6.5×(35×103)2100×2×106

=39.8(Ω)Z=0.48+j11.48(Ω),Z

=7.35+j39.8(Ω)

(8.5+j5.3)(7.35−j39.8)(0.48− j11.48)+(7.35−

=6.48+j4.34(MVA)

(8.5+j5.3)(0.48−(0.48− (7.35− j39.8)

=2.00+j0.96(MVA)II:如右图,设U2=35KV则U=35×11×34.125=34.125(KV) 3511−−∆U=U

−U

=0.875(KV)

∆U

0.875×35(0.48− j11.48)+(7.35

=0.092−j0.58(MVA)ZZ1=16j120Ω,Z2=33+j89Ω,ZSSA=

=(6.48+j4.34)+(0.092−j0.58)=6.572+j3.76(MVA)

=(2.00+j0.96)−(0.092−

j0.58)=1.91+j1.54(MVA)

τ答:电能损耗=I2Rt=+QRt,为某个时段t内的电能损耗.PU2PT=Wττ

∆W∆P,+ +j120Ω,Z4 3=48=60+j152Ω~,S1=170+j40MVA,S~2=50+j30MVA~,S3=40+j15MVA.计算:网络的自然功率分布;网络的

)+S

+ZZ

+Z+Z

(Z*

+Z*).+S

Z

=168.86+j62.39(MVA)==S23+S++R2+R++R2+R==(R1+()RSSAO+SS)+S=2

+Z3Z

+Z

(Z*

+Z*).+S

Z

=91.29+22.38(MVA)SA+S =260.15j84.77(MVA),S1+S2+S3+=260+

j85(MVA)S.12=.SA−.S1=(168.86+j62.39)−(170+j40)=−1.14+j22.39(MVA)

.−S

12=(50+j30)−(−1.14+j22.39)=51.14+j7.61(MVA)=(51.14+j7.61)+(40+j15)=91.14+22.61(MVA)

S

(R+

RR

(R

+R)+SR3=(170+j40)(33+48+60)+(50+j30)(48+60)+(40+j15)×6016+33+48+60=202.26+j62.23(MVA)

=S

(R+

RR

(R

R+R)+SR1=(40+j15)(33+48+16)+(50+j30)(16+33)+(170+j40)×1616+33+48+60=57.64+j22.7(MVA)

=

=(17.74+j7.77)+(40+j15)=57.74+j22.77(MVA)S

S

S

S′

III:∆

P

U

U

+U()R3+U()N

R=10.705+0.34+2.65+10.93=24.625(KW)∆P(=A⋅O)2R+(U 12⋅O)2U

R+(23⋅O)2R+(U

S′U

)2R

=14.8+1.046+0.37+4.776=20.992(KW)∆W=∆P∆P

−=(24.625−20.992)×8760=3.633×8760=31825.08(J)

=0.8.电力线路装设有两台SFZ-7500/35

型变压器并联运

S

=7500KVA,U

35/11KV,=75KW,UK(%)=7.5,=9.6KW,I0(%)=0.8.两台变压器全年投入P )ρ解:r1=

.5=0,45(Ω/KM)70sr=11.4=D

=3.5m=3500(mm)x

D=0.1445lgm+0.015700=0.1445lg +0.0157=0.4186(Ω/KM)5.7b

7Dmr

10−658

=2.72×10−6

=1rl=1×0.45×20=4.5(Ω)21 X=1xl=1×0.4186×20=4.186(Ω)L 212==INNU∆∆QKT=I∆∆QKT=UB=2bl=1×2.72×10−6×20=1.088×10−4(S)L R 32=PKU=75×10×(35=1.63(Ω)×10) (7.5×106)2XT=

U

(%)UK

75×(35×=12.25(Ω)=103)2100×7.5×106G

P

×103

=7.84×10−6(S)

U2

(35×103)2B

100×(35×103)2

=4.9×10−5

(S)∆P

∆P

=9.6KW=0.0096(MW) =0.8×=0.06(MVar)(%)S

7.5∆PZT

=∆PK

=0.075(MW)K =7.5×=0.5625(MVar)(%)S

I:cosϕ=0.8a.S

ϕ

=10∠arccos0.8=12.5∠36.870.8=10+j7.5(MvA)∆S=∆Ṡ+∆ṠTj0.06)+2[(

=2(0.0096+

.5)27.5

×0.075

.5)27.5

=0.0192+j0.12+0.104+j0.78=0.1232+j0.9(MvA)PP2A+QPP2A+Q==SA+∆S

=10+j7.5+0.1232+j0.9=10.1232+j8.4(MVA)∆

(R

.1232+jX)=

8.42

×(4.5+j4.186)=0.636+j0.59(MVA)

U2

352̇̇S=S +∆S

=10.1232+j8.4+0.636+j0.59=0.7592+j1.49(MVA)∆S

=∆Ṡ

=0.1232+j0.9+0.636+j0.59=0.7592+j1.49(MVA)

S=10×0.5∠arccos0.8=6.25∠36.87�=5+∆S

=∆Ṡ+∆ṠKT j0.06)+2[(

=2(0.0096+

6

×0.075

)27.5×

)27.5×=0.0192+j0.12+0.026+j0.19=5.0452+j4.06(MVA)

=5+j3.75+0.0452+j0.31=5.0452+j4.06(MVA)∆

(R

jX

0452

4.062

×(4.5+j4.186)=0.154+j0.143(MVA)

U2

352S

=5.0452+j4.06+0.154+j0.143=5.1992+j4.203(MVA)∆

=∆Ṡ

=0.0452+j0.31+0.154+j0.143=0.1992+j0.453(MVA)Ṡ=10×0.25∠arccosϕ=10×0.25∠arccos0.8=2.5+j1.875(MvA)

ϕ==P∆S=∆Ṡ+∆ṠTj0.06)+2[(

=2(0.0096+

25)2

×0.075

25)27.5

=0.0192+j0.12+0.0065+j0.0488=0.0257+j0.1688(MVA)

=2.5+j1.875+0.0257+j0.1688=2.5257+j2.044(MVA)∆Ṡ(R +QA

+

)=

044352

(4.5

j4.186)=0.039+j0.036(MVA)tt1+∆PS=S +∆S

=2.5257+j2.044+0.039+j0.036=2.5647+j2.08(MVA)∆

=∆Ṡ

=0.0257+j0.1688+0.039+j0.036=0.0647+j0.205(MVA)∆P=0.7592(MVA),∆P (MVAP

∆P×100=

592.7592

×100=7.06P∆P

×100=

1992

×100=3.83P1992∆P3×100=P

0.039×100=1.522.5647∆W=∆P

t22

+∆P

t

=0.7592×2000+0.1992×3000+0.039×3760=1518.4+597.6+146.64=2262640(KWh)II.cos=0.7a.Ṡ=10∠arccos0.7= =10+j10.2(MVA)∠57

=2(0.0096+j0.06)+

.28)27.5

×0.075+j(

28)27.5

=10+j10.2+0.1552+j1.12=10.1552+j11.32(MVA)∆

P

+Q

(R

+jX)15

+11.32

×(5+

̇̇S=S +∆S

=10.1552+j11.32+0.85+j0.79=11+j12.11(MVA)∆

=∆Ṡ+∆Ṡ=0.1552+j1.12+0.85+j0.79=1+j1.91(MVA)T b.Ṡ=10×0.5∠arccos0.7==5+j5.1(MVA)∆S=∆Ṡ+∆ṠTj0.06)+2[(

=2(0.0096+

×0.075+j(

)27.5×

)27.5×=0.0192+j0.12+0.034+j0.25=0.0532+j5.47(MVA)

=5+j5.1+0.0532+j0.37=5.0532+j5.47(MVA)

P2+Q

(R+

5.05322+jX1)5.472=

× j4.186)=(4.5+0.2+

j0.189(MVA)A UN ̇̇S=S +∆S

=5.0532+j5.47+0.2+j0.189=5.2532+j5.659(MVA)∆

=∆Ṡ

=0.0532+j0.37+0.2+j0.189=0.2532+j0.559(MVA)c.Ṡ=10×0.25∠arccos0.7=3.57∠45.57�=2.5+j2.55(MVA)0.7∆S=∆Ṡ+∆ṠTj0.06)+2[(

=2(0.0096+

3

×0.075+j(

3

)27.5×

)27.5×=0.0192+j0.12+0.0085+j0.064=0.0277+j0.184(MVA)̇SS

=

+ =2.5j2.55+0.0277+j0.184=j2.734(MVA)∆Ṡ+ 2.527+

P2+Q

(R+

2.5272+jX1)2.7342=

× j4.186)=(4.5+0.05+

j0.047(MVA)A UN ̇̇S=S +∆S

=2.527+j2.734+0.05+j0.047=2.577+j2.78(MVA)∆

=∆Ṡ

=0.0277+j0.184+0.05+j0.047=0.0777+j0.23(MVA)∆P

=1(MW),=0.2532(MW),∆P

=0.0777(MW)tt1+∆P∆P1×100=

111

×100=9.1P∆P

×100=

2532

×100=4.82P2532∆P

×100=

0777

×100=3P77∆W=∆P

t22

+∆P

t

=1×2000

+0.2532×3000+0.0777×3760=2000+759.6+292.152=3051752(KWh)III:cos=0.9 ∠arccos0.9=11∠25.84

=9.9+

=2(0.0096+j0.06)+

×0.075+j(

7.5×7.5×=0.0192+j0.12+0.08+=0.09+j0.725 Ṡ=+̇A S

=9.9+j4.79+0.0992+j0.725=10+j5.515(MVA)

P2+Q

(R+

102+jX1)5.5152=

× j4.186)=(4.5+0.48+

j0.44(MVA) A UN ̇̇S=S +∆S

=10+j5.515+0.48+j0.44=10.48+j5.955(MVA)∆

=0.0992+j0.725+0.48+j0.44=0.5792+∆

j1.165(MVA)Ṡ=4.9+j2.4(MVA)

10×0.5∠arccos0.9=5.5∠25.84

∆S=∆Ṡ+∆ṠTj0.06)+2[(

=2(0.0096+

×0.075+j(

)27.5×

)27.5×=0.0192+j0.12+0.02+j0.15=0.0392+j0.27(MVA)

=4.9+j2.4+0.0392+j0.27=4.94+j2.67(MVA)=

P2+Q

(R+

4.942+jX1)2.672=

× j4.186)=(4.5+0.116+

j0.108(MVA)A UN ̇̇S=S +∆S

=2.49+j1.358+0.029+j0.0275=2.52+j1.385(MVA)∆

=∆Ṡ

=0.024+j0.1578+0.029+j0.0275=0.0532+j0.1853(MVA)

Ṡ=10×0.25∠arccos0.9=

=2.47+j1.2(MVA)∆S=∆Ṡ+∆Ṡ̇Tj0.06)+2[(

=2(0.0096+

×0.075+j(

2

)27.5×

)27.5×=0.0192+j0.12+0.005+j0.0378=0.0242+j0.1578(MVA)

=2.49+j1.2+0.0242+j0.1578=2.49+j1.358(MVA)∆ =PS ̇(R

+Q

jX

)=

358

×(4.5+j4.186)=0.029+j0.0275(MVA)̇̇S=S +∆S

=2.49+j1.358+0.029+j0.0275=2.52+j1.385(MVA)∆

=∆Ṡ

=0.0242+j0.1578+0.029+j0.0275=0.0532+j0.1853(MVA)∆P

=0.5792(MW),=0.1552(MW),

∆P

=0.0532(MW)∆P

×100=

5792

×100=5.53P.48∆P

×100=

0.1552

×100=3.07P

5.056∆P

×100=

0.0532

×100=2.1P

2.52∆W=∆Pt11+∆Pt

+∆P

t

=0.5792×2000+0.1552×3000

+0.0532×3760=1158.4+465.6+200.032=1824032(KWh),,==∑PL+∑∆P+∑∆P,,

S和网络的有功损耗∑∆P。而且还应具有一定的备用容P

位位调节功率（负荷的调节效应）KL，即：K

∆P∆f

(MWHz)，K∆PL*=∆fL*。 ∆PK=− G(MWHz)，G ∆f

K

∆=P−∆f答:所谓的机组调差系数，是以百分数表示的机组空载运行时f0与额定条件下fN的差值，f−f即：σ%0 N×100%f=

σ%:,

O,则在O点

P

=PG,稳态运行,运行频率为f1.负荷出现增量∆PL,

+∆P

>P

=PL电力系统的单位调节功率K

=K

+KL。,++∆PGB)=K∆∆f=(∆Pabab=KA(∆P−−∆PGB)−KB(∆PKKA+Kabab可看成一电源:∆P−−∆PGA)+(∆P调节功率之和KGΣ不可能过大，KS自然不可能过大。5-10.?差调发电机组发出的功率,使频率特性向上移动.设发电机组增发∆

P,则运行点又将从点O'∆f

P

=∆P则=0,亦即实现答:如图，设A、B两系统发电机组均能参加一次调频，且均有调频厂即均可以参加二次调∆P

P

电机组二次调频增发出力，以KA、KB表示A,B两系统的单位调节功率，联络线功率为∆Pab，且设其正方向是由A系统流向B系统。对A系统∆Pab可看成一负荷:∆PLA+∆P

−∆P

∆f对B系统∆

−(∆P

∆f.

K+KA

−∆PGB)∆P

−∆P)

联络线上功率∆P==850×50−KKGA=800,KGB=0,K==KLA+K==KLB+KKKA+K==850×50−解解:I.∆PGA=∆PII.II.∆PGA=∆P5-12.A、B两系统由联络线相连如图所示,已知A系统K

=800MW/HZKLA=50MWHZ∆P

=100MWB系

K

=700MWHZK

=40MW/HZ,∆PLB=50MW.求在下列情况下频率的变化量∆f和联络线功率的变化量∆P

=0,∆P

=100,∆P

=50；K

=800,K

=700,K

=50,K

=则∆P

=∆P

−∆P

=100,∆P

=∆P

−∆P

50K

=K

+K

=850,K

=K

+K

=740.

∆f=∆PA+∆PK+KA

100850+

=0.0943(HZ)∆P

K∆PA KA+K

−K

∆P

850+740

740×100=−19.8(MW)

=0,∆P

=100,∆P

=LA=50,KLB=40.则∆P

=∆P

−∆P

=100,∆P

=∆P

−∆P

=KA GA=850,KB GB=40.∆f

∆P= A+∆P

100+50850+40

=0.168.∆P

K

∆P

−K

∆P

40×100=43.26(MW)

KK

850+40==0,KLA=50,K==KLB+K解解:I.∆PGA=∆PGB=50,∆P==KLA+KGA=850,K==KLB+KII.II.∆PGA=60,∆PIII.∆

P

=∆P

=0,∆P

=100,∆P=50;K

=K

=则∆P

=∆P

−∆P

=100,∆P

=∆P

−∆P

50K

=K

+K

=50,K

=∆f

∆P= ∆P

=100+5050+40

K

+K∆PK∆P

∆P

−

=50×50−40×100

=−16.67(MW).

K

K

50+405-13.仍按题5-12中已知条件,试计算下列情况下的频率变化量∆f和联络线上流过的功率∆P,,

=100,∆P

=K∆P

=∆P

−∆P

=50,∆P

==∆P−∆PLB

0.∆f

∆P= ∆PK

++K

50+0=0.0314(HZ)=850+740∆P

K∆PA KA+K

−K

∆P

850×0−740×50850+740

=−23.27(MW)

=0,∆P

=100,∆P

=50;KKA=KLA+KGA=850,K==KLB+KB GB=740.∆P

=∆P

−∆P

=40,∆P

=∆P

−∆P

=III.III.∆PGA=0,∆PGB=60,∆P==KLB+K==PL,则PL∗ =1,SAA=∆P−∆PGA=100,∆P∆f

∆P= ∆PK

++K

40+50=0.0566(HZ)=850+740∆PK∆PB

KK

∆PK

−

850×50−740×40850+740

=8.11(MW)

=100,∆P

=K

=K+K=850,KLA GB LA

=740.=∆P−∆PLB

=−10.∆f

∆P+= ∆PK+KA

100−850+

=0.0566(HZ)∆P

K∆PA KA+K

−K

∆PA=850×(−10)−740×100−51.89(MW)850+740

=2,主调频电厂额定容量为系统负荷的20%,当系

P

=1,f

N

=1.1(发电机组K

=2,调频厂S

=0.2P

=0.2P

=0.2,S

=

P'P'

=

G

=1.1发电机在一次调频的作用下出力增加了P'

=0.1=K

∆f∆=0.3∆∆P∆P

+ 'G∗∆P'K

∆fK

∆f=0.1+2×

=0.1+0.012=0.112由K

∆f

=0.1

=0.1=0.1∆f0.3

×50=二次调频投入后频率上升

∆

f

0.3−0.250

∆P

−∆P=K+

∆f

K

L'∗0.112−∆P

=2+

50=18.67解得：∆P

=0.0747<0.1bbRi+Qbb---用电设备端电压,Ub---用户侧电压归算至高压侧.;.?U

=(KU

P

b)/KXiKU

U

U

U 通过改变发电机机端电压UG,改变变压器变比K1,K2,改变无功功率分布或串联电容补偿来改变用电设备端电压Ub..,.U

'---低压侧实际电压.

=U–

U=U

–PR+QXU

UU

'=

UiU

=U

UU

U

=UtImax

UimaxU'

,U tImax

UiminU',,,;⇒⇒UtI=KU.,,U

UIminU'

Q

X

(KU'

−U),这样可充分利用电容器的容量,使在满足调压要求的Q

KU'X

(KU

−U

Q

KU'X

inIm

−U

⇒K=

'U

+U'UIminImin

U'2 2UImin+U

U

选择一最靠近的标准抽头U,则K=UtI',将K代入Q式即可解出Q,这样,可保证在满tI U C .,,,6-13.有一降压变压器归算至高压侧的阻抗为2.44+j40Ω,变压器的额定电压为∆U

28*2.44+14*40

U低max=U高max−∆Umax=113−5.56=107.44(kv)Utmax=U低

U=107.44* ∆U

10*2.44+6*40

U低min=U高min−∆Umin=115−2.3=112.7(kv)Utmin=U低

U

=112.7*6.3=107.58(kv)U

maxmin

)=110.19(kv)选择110kv分接头低低max=U =107.44*6.3=U'

6.15>6U'

=U

U

=112.7*6.3=6.45(KV)<6.6∆U%

6.15−6*100%=2.5%∆U%45−

*100%=2.27%S

=110±2×2.5%

=0.8,最小负荷为7MVA,cos

163×110

×10−3

=4.93(Ω)T 20X

U(%)UK

.5×1102100×20

=63.525(Ω)S

=18∠arccos0.8=14.4+

=7∠arccos0.7=4.9+j5(MVA)∆

.4

10.82

×(4.93+j63.525)=0.13+j1.7(MVA)Tmax ∆

522×(4.93+j63.525)=0.02+j0.26(MVA)

=14.53+j12.5(MVA)S=S+∆S

=4.92+j5.26(MVA)UUtmax=U低低max=U∆U

=14.53×4.93+12.5×63.525=8.05(kv)7.5∆U

=4.92×4.93+5.26×63.525=3.33(kv)7.5U低max=U高max−∆Umax=107.5−8.05=99.45(kv)U低min=U高min−∆Umin=107.5−3.33=104.17(kv)U

UUU

99.45×=104.17

11 ×102.5%10×107.5%则公共抽头UtI=1

)=106.92(kv)

=107.25KVU'

U

99.45×107.2

=10.2(kv)<10.25KV选择110−2×2.5%分接头,U'

U'

=U

U

=104.7

4.

=11(kv)>10.75KV最大负荷时,选择110−2×2.5%

U

=104.5kvU'低max=U

UUN‘tIm

=99.45

4.

=10.5(kv)>10.25KV

2×2.5%

U

=107.25kv

=U

UUtImin

=104.17×

=10.68(kv)<10.75KV6-15.某变电所有一台降压变压器,变压器额定电压为110±2×2.5%/11kv解:U' ≥ = ≤ =10.75imax1.025U

10.25

U'

=(1−7%)×10=9.3<10.25

=(1−2%)×10=9.8<10.75则UtI=115.5kv,不满足调压要求.Uimax=

U'U

=9.3×115.5=UtIm

UUN‘i

=97.65

1.025×

imin=

U'U

=9.8×115.5=UtIm

UUN‘i

=102.9

1.075×

UtI=1

tIm

+U

)=105.05(kv)U

=104.5kvU'

=U

U

=97.65×

=10.28(kv)>10.25

U'

U

=102.9×

=10.83(kv)>10.75

则重新选择U=107.25kvU'

= =10.01(kv)<97.65×107.210.25KV

U'

=102.9

=10.55(kv)<10.75U'tIm

U'tIm6-16.水电厂通过SFL-40000/110型升压变压器与系统连接,变压器归算至高压侧阻抗为2.1+j38.5Ω,额定电压为121±2×2.5%/10.5kv,系统在最大、最小负荷时高压母线电∆UU =U + =112.09+7.74=119.83(kv)Utmax=U低

U.5=119.83× 最小负荷时,,∆UU低max=U高min+∆Umin=115.92+7.48=123.4(kv)Utmin=U低

U

3.4.5

则公共抽头U

TIm

+U

)=12.18(KV)U′=UU低 =119.83×10.5=

U

10.4>10U′=U

UU

=123.4×10.5=10.7<11

∆U%

10.4−10×100%=4%∆U%.7−11×100%=2.7%SFL−31500/110

U

=110/11KV,UK(%)=10.5.最大负荷时变电压偏移在最大,最小负荷时为二次网络额定电压的2.5%−7.5%.试根据调压的要求接并联电容器和同期调相机两种措施,确定变电所10KV母线上所需补偿设备的最小容ρ解:r1=

.50.2625(Ω/sx

D=0.1445lgm+0.0157=0.1445lg×10

+0.0157=0.423(Ω/km)15.2/2

=1rL=21

1×0.2625×70=9.19(Ω)X=

xL=

×0.423×70=14.8(Ω)X=

.5×

=20.167(Ω)2100S2×100×31.5U

=100.5(KV),U

=112(KV)KK=U′CCmax)(UUU2max+2U′U=10(1+2.5%)=10.25(KV)U′2min

=10(1+7.5%)=10.75(KV)1并联电容器U

=U

U2NU′112×10.7

=114.6(KV)选择U′则K=

=115.5KV,

UU

5.5

=10.5Q

K2U′2max

U

C X 2Cmax 10.510.25

×(10.25

100.5)=21.9(Mvar)14.8+ .1672并联调相机U′2+2U′22max 2min

U

=10.25×100.5+2×10.75×11210.252+2×10.752

=10.227U

=KU

=10.227×11=112.5(KV)UK=UtI

2.75

=10.25

Q

U

10.252×10.25100.5

X

2