电力系统分析部分习题答案.docx
10.5 + 720,7部分习题答案第一章习题答案习题1.1T2： 10.5/12 l/242kVT4： 10/0.4kVTl： 10.5/0.4/3.3kVT3： 220/1 IkV习题12(1) 34+jl8kVA,即发出有功34kW,感性无功18kVai(2) 1.1/42.9 kV习题131299-j2250(kVA)习题14(1)各阻抗消耗：Sl=484-j484 (VA)S2=968+jl452 (VA)S3=1452+j2904 (VA)(2)电源发出：Sel=4805-j579 (VA)Se2=-1901+j4451 (VA)(3)总发出的有功：2904W 无功： 3872 (Var)可验证有功无功均平衡(或者近似平衡，与计算误差有关)。第二章习题答案习题2.1(1) x = 0.207欧/公里;r = 0105欧/公里;b = 2x2.73xl()-6西门/公里;J2.73xlOM(3) Qc = 6.6Mvar习题22Z = 14.175+ J75.27V1=y 2.43 x IO-414.175 +775271j2.43x107习题2.3(1)若线路长度为9km, R=18.9欧姆，X=0.9欧姆，P =2.34%；若线路长度为 4km, R=8.4 欧姆，X=0.4 欧姆，P = 1.04%(2) X/R=0.0476习题2.4Y(1) Z = 13.13tjl51；y= j0.922*10-313.13+jl5110O 2川.922 * 1。7 s j巾/0.922 * 10 sOOV(2) Z = 11.92+)144.1 ；y=川.94279*1()711.92+J144. 1川.94279* 10% 口LI j0.94279*10-3sY(3) Z = 12.02+jl43.87;=，0.93*1。712. 02+J143.87n/0-93*10% 口习题2.5归算到高压侧Xr=40.33：= 7.03*10-5； G, = 7.11*10" ； & = 2.44 归算到低压侧XT = 4.9408 ；%=5.74*10，Gr = 5.80* 10-5 ； & =0.2988。习题2.63华4-匕56H1 486甲 1.09 /11.582+)146.71337.5+ jAOOQKVA或(6.97-J82.64)*10% 习题2.7J0. 2731 O-0 2J2.73 J -j3习题2.8“相位超前。习题2.9末端电压大于首端电压。习题2.100.36习题2.116%习题2.12(1) 48.9, 0.368；(2) 0.105, 0.105；习题2.13(1) 4=0.98, xT. =0.1376,*=0.122, x,. =0.232；(2) % =0.733, xrp=0.103, xTr=OA10, =0.191；习题2.14有名值:1.025 欧，3.819kVo标么值：0.807, 1.05 (基值取 VB=6.3kV, SB=31.25MVA)第三章习题答案习题3.1109.21KV习题3.2首端功率：166.4 +j71.95(MW)低压边电压：122.44丫)效率 96.94%习题3.3ASr = U V.AU= -2.862-J12.4Sh =-22.96+ /11.43 S . =127.04+ /111.43 SA =52.95+ )28.57 bcJa-bJA-cJS= 177.04+ J141.43习j3.4(1) SAl = 21+ j2AMVA ； SA2 = 29+ j15.6MVA； S21 = 9 + j7.6MVA；Sl2=-9-jl.6MVAo(2) q = 105.61KV,= 106.56KVoA-(3) k = L031。习题3.5-0.935j4.262-0.481 + J2.404-0.454 +j 1.8910-0.481 + J2.4041.069-J4.728-0.588 +j2.3530-0.454 +j 1.891-0.588 +J2.3531.042-J7.543j3.33300j3.333-J3.333习题3.6Y33=1.042-j8.2430Y34=Y43=j3.667习题3.8jlZ _ KZ”(Zo + ZJK(Zio + Z?) + Z:o7 _ 11K Z”(Z0 + Z1?)，7 i “一短K；Z+4)+ Z“ 23 Jz =k；+ Nio)+ Z2Z _ _J_ z =ko7o13 h 12 &(k；(j + &)+z”)习题3.9A 与=4c +4e24 "4e4f2U2 = U1(小年)0只一2u：000000000000000000e,颂V颂习题3.10(1)为平衡节点；为PV节点。799jioJ1079.9(3)习题3.11(1)AP20-10 块、-20 - If2 Af2迭代用功率方程A = P；P _p5)= -0.4 一 U$Uj (G> cos % + % sin &)4 = " 一月(U,5) = -03- 4X4(6 , cosq, + Bsin处) 片i32 =。/一2(U,6) = 0.15-q4(G笛 sin处一% cos处) 尸1(2)迭代用修正方程 15.555610 -“=10-18.3333 田=-18.3333只/a u.& =-B*/q jLa6乌/4=回"(3)经过二次迭彳t后,(7. =1Z-2.9733, 73 =1.02 3 3Z-2.45040 一J章习题答案习题4.1(1)耳=75MW, P,=60MW； * (2) f=49. 435 Hz, =117. 375W； P： =82. 625W：习题4.2PGl = 45.45(MW)(1) Pn = 500(A/W),136.36(A/W), C = Cl + C2 + Ci = 54044.65(/h)%=318.18(MW)f &=136.36(MW)(2) Pn = lOOO(MW), P(.2 = 409.09(MlV), C = + Cz + C3 = 95409(/h)al454.55(A/W),& = 318.18(MW)(3) Pn = 2000(A/lV), PG2 = 954.55(MW), C = + G + C3 = 182227(/h)& = 127.21(MW)习题4.3& = 50(MW)(1) E)=500(MW), (%？=133.33(MW), C = Q + C2 + C3 = 54046(/h) o % = 316.67(MW)七= 136.36(MW)(2) 2=1000(MW), PG2 = 409.09(A/W), C = + C2 + C3 = 95409(/h)% = 454.55(MW)& = 380(MW)(3) Pn = 2QQQ(MW),=800(例W), C = + G + C3 = 182490(/h)&=820(MW)第五章习题答案（略）第六章习题答案习题6.1习题6.5（略）习题6.6（a）系统C变电站开关的额定断开容量S = 1000MVA （提示：下图中K1 处短路容量）有名值 =3.922 25073*115= 4.922KA(b)有名值= 3.873=50 = 4.861KA3115(c)有名值/尸=6.0292506*115=7.561 KA习题6.7K点短路容量为：Sq = 737.3ME4第七章答案部分:习题7.1U an，Q = - Ug习题7.2（略）习题7.3对应零序网络为:习题7.4(1) X, =0时-J2.179i =!=_L_al j(Xj + XT)+ j(X2 + XT)/(Xr + 3Xp) J0.4588"=-0.208乙=J0.4534一(X+3Xp)X、+ Xr + Xr+3X”4o=-<i- iai = J2.179-7O.4534=J1.7256。=晨 +也+ a 心=3.449/131.4。/， = I”。+ a J + aial = 3.449N486有名值:=2.179* =2.179* 二 =0.421KAal b 斤 115/ =0.4534*/ =0.4534* U -D38.5V3*115=0.0876KA/1O =1.7256*/ =1.7256* mvn38.573*115= 0.333必。=02Q 5/. =/ =3.449*7 = 3.449*-= = 0.667 KAb c b V3*115-J1.677(2) X,=46Q时j(X； + XQ+ J(X2 + XT)/(Xr+ 3Xp) ;0.596-(Xr + 3XJ 5iM/ = J0.954X, + Xr + Xr+3XaO=乙 + 3 + /心=4=(。+。3, + &/心=2-523/154.4,= L0 + a& +a%=2.523/25.6。有名值:IQ 5/fl = 1.667*7. = L667* 二 =0.324K4al"3115/ =0.954* =0.954*38.56*115= 0.184K4IQ 5/0 =0.723*/ = 0.723 * = 0.140/C4""4*115I.=o38 5/ =/ = 2.523*/w =2.523*-j=- = 0.488KAh c R 0*115X上没有流过正、负序电流，因为在正负序网中正、负序电流对称，a,b,c 三相之和为0, X,上只能流过3倍零序电流。但X大小对正、负序电流有一定 影响，X,增加使正序及相电流变小，而使负序电流与零序电流之比增大。习题7.5 1 Q0/, =1,112* J =0.5Q2KA “*230I go/ =0.343* 二 =0.155K>V3*230i on/“o = 0.769 * 二 =0.347KA 73*230过T1变压器中性点电流：/n= 0.3740过T2变压器II绕组中性点电流：/厂 =0.374K4过T2变压器I绕组中性点电流：/7,， = 0.511KA过T3变压器中性点电流：q=0.531必 习题7.6有名值/ = 1.6324*=0.764KA 的*37习题7.7负序网J 而门XT, L零序网JXti /Mo 必匕ka21aor1ora11Z-12O01Z1203-1/32/3-1/3由零序网可得:几。大X + Xl)X= -J2.667：, L = " = L=-j2.667 且(=3乙=一#在正序网中Urj(X“ + X + XQ(Xh + X；)可得Xr=X_= 0.0167J，J 一由正序网可得山j(Xcl + XTl + XLl)= -J2.0,可得% = iaJ=-j667由负序网可得可得i-c=T0667Ti 侧由系统C提供的三相故障电流大小为:=-J2.0+ ;2.0| = 0a2lr.-a =2.0a2-a = 3.464 a ic 一 a，i门=3.464T2侧由发电机提供的三相故障电流大小为: 乙_乙卜卜加.667+ J0.667| = 0 a2/ . -a = 0.667a2 -a = 1.155a /门-orl=1.155 A 1A -第八章习题答案（略）
