《电机学》(华中科大出版社,辜承林,第二版)课后答案

发布时间 : 星期六 文章《电机学》(华中科大出版社,辜承林,第二版)课后答案更新完毕开始阅读

(2) E?U?110 E?4.44fN?m

110E∴?m?4.44fN?4.44?50?1000?4.955?10Wb

3.43 有一台单相变压器,额定容量SN?100kVA,额定电压

U1N/U2N?600/023V0,f?50Hz。一二次侧绕组的电阻和漏电抗的数值

?4为:R1?4.32?;R2?0.0063?;X1??8.9?;X2??0.013?,试求: (1)折算到高压侧的短路电阻Rk、短路电抗Xk及短路阻抗Zk;

?、短路电抗Xk?及短路阻抗Zk? (2)折算到低压侧的短路电阻Rk (3)将上面的参数用标么值表示;

(4)计算变压器的阻抗电压及各分量;

(5)求满载及cos?2?1、cos?2?0.8(滞后)及cos?2?0.8(超前)等三

种情况下的?U,并对结果进行讨论。

'22)?0.063?4.287? k?26. 1 (1) R2?kR2?(6000230'22 x2??kx2??26.1?0.013?8.8557?

∴RK'?R1?R2?4.32?4.287?8.607?

xk?x1??x2??8.9?8.8557?17.457?

22ZK?RK?XK?8.6072?17.4572?19.467?

(2)折算到低压测:

R14.32 x1??R1'?k2?26.12?0.0063?'x1?k2?286..912?0.013?1

∴Rk?R1?R2?0.0063?0.0063?0.0126?

'xk?x1'??x2??0.0131?0.013?0.0261?

''Zk'?Rk'?Xk'?0.01262?0.02612?0.029?

22(3)阻抗机值:

1NZb?UI1N?U1NSNU1N?6000?6000?360? 100?103*87?0.01191x*?R1*?4.32360?0.012 R2?4.2360 1?8.9360?0.0247

***8.85578.60717.457x2??0.0246R??0.0239x? ?kk360360360?0.0485

* Zk?19.467360?0.05408(4) Uk?ZkI1N?(8.607?j17.457)?16.667?143.33?j290.95 I1N?U1NN?1006?16.667A

*?60k0? Uk0U**∵Uk?Zk ∴U?5.4S143?.3j3290.956000**00kkr也可以,但麻烦。

**?Xk*?4.8500 U?Rk?2.3900 Ukx**(5) ?U??(Rkcos?2?Xksin?2) ∵是满载 ∴??1

?1 sin?2? 0?U?0.0239?1?2.3900

(b) cos?2?0.8(滞后) sin?2?0.6

(a)cos?2 ?U?1?(0.023?90.?806 )0.0?485?0.04.822(c) cos?2?0.8(超前) sin?2??0.6

0)?80.0?485?0.?60.968 ?U?1?(0.023?90.0

说明:电阻性和感性负载电压变化率是正的,即负载电压低于空载电压,容性负

载可能是负

载电压高于空载电压。

3.44 有一台单相变压器,已知:R1?2.19?,X1??15.4?,R2?0.15?,

X2??0.964?,Rm?1250?,Xm?12600?,N1?876匝,N2?260匝; 当cos?2?0.8(滞后)时,二次侧电流I2?180A,U2?6000V,试求:

?及I?,并将结果进行 (1)用近似“?”型等效电路和简化等效电路求U11比较;

(2)画出折算后的相量图和“T”型等效电路。

(1) k?I1 N1N2'22k?Rk?3.37?0.15?1.7035(?) ?876?3.37 22260'2x2?3.37?0.964?10.948(?) ?R1I0RmX1?R2X2?I2U2ZLI1 R1X1?R2X2?I2U2ZLU1U1Xm

P型等效电路 简化等效电路

'。37?60000?202200(V)P型:设U2?U20则U2?kU20?3.

'。。。1802I2'?IR?387?53.41?36.87。?42.728?j32.046 .37?36.''U1?U2?(R1?R2'?jx1??jx2)I ?2。 =20200?(2.19?1.7035?j15.4?j10.948)?53.41?36.87。

(3.8935+j26.348)?53.41-36.87。 =20220? =20220?1422.5244.72。?20220?1010.78?j1000.95 =20220?1422.5244.72。?20220?1010.78?j1000.95 =21230.78+j1000.95=21254.362.699(V)

I0?U1Zm?21254.362.6991250?j12600U1。?12661.856786?81.63?0.2443?j1.6607 。?1.84.33I1?I0?I2?0.2443?j1.6607?42.728?j32.046?42.9723?j33.7067?54.615?38.12。(V)(A)∴U1?21254 I1?54.615

用简化等效电路: U1?21254(V)(A)(不变) I1?I2?53.41

比较结果发现,电压不变,电流相差2.2%,但用简化等效电路求简单 。

I1 R1X1'?I0'R2'X2?'I2'ZLU1U2

T型等效电路

3.45 一台三相变压器,Yd11接法,R1?2.19?,X1??15.4?,R2?0.15?,X2??0.964?,变比k?3.37。忽略励磁电流,当带有cos?2?0.8(滞后)的负载时,U2?6000V,I2?312A,求U1、I1、cos?1。

''312?36.87?202200。 则I2 设U2??。3k?53.41?36.87。

43A ∴U1'??212542.(见上题)∴U1?3U1??36812 I1?I1??53.(V)699。?1?2.699。?(-36.87。)=40.82。 co?s1?0.(76) 滞后

3.46 一台三相电力变压器,额定数据为:SN?1000kVA,U1N?10kV,

U2N?400V,Yy接法。在高压方加电压进行短路试验,Uk?400V,Ik?57.7APk?11.6kW。试求: (1)短路参数Rk、Xk、Zk;

(2)当此变压器带I2?1155A,cos?2?0.8(滞后)负载时,低压方电压U2。 (1)求出一相的物理量及参数,在高压侧做试验,不必折算

N?k?U1?2N?U100.433?25 Uk??40034( ?230.95(V) Ik??57.7AK?K?PKPk??3?3.867(kW) Zk?U?230.9557.74?4(?) IKRk?

PK?2Ik?2?386757.742?1.16(?) xk?ZK?R2?42?1.162?3.83(?)

(2)方法一:

??I2I2NU1N?I1N? I2N? I1N?3SN3U2N?10003?0.4?1443.42(A)∴??11521443.42?0.8

Zb?SN3U1N?10003?10?57.74(A)?I1N?

*RkZb∴ Zb?1000057.74?99.99?100 ∴Rk??0.0116

x*?2?0 .6xk?Zkb?0.0383 sin∴

**?U??(Rkcos?2?Xksin?2)?0.8?(0.0116?0.8?0.0383?0.6)?0.02581U2? ∴U2??(1??U)U2N? ?U?1?U2NU2N??U2N3?4003?230.947(V)

∴U2??(1?0.02581)?230.947?225(V) ∴U2?3U2??3?225?389.7(V) 方法二:利用简化等效电路

I2?''。1155I?46.2?36.87U?U0I2???46.2(A)设 则 2?22?k251000U1N??I2(?RK+jXK)+U2? U1N??3??5773.67? 。’∴5773.67cos??j5773.67sin??46.2?36.87?473.15?U2?

=184.836.28?U2??149.16?j109.35?U2?

'5773.67cos??149.16?U2?

5773.67sin??109.35 解得:U2'??5623.5(V)

?224.9 ∴U2?3?225?389.7(V) ∴U2??

3.47 一台三相电力变压器的名牌数据为:SN?20000kVA,

??Zk?0.105,I0?0.65%,U1N/U2N?110/10.5kV,Yd11接法,f?50Hz,

'U2?kP0?23.7kW, Pk?10.4kW。试求:

(1)?型等效电路的参数,并画出?型等效电路;

(2)该变压器高压方接入110kV电网,低压方接一对称负载,每相负载

阻抗为16.37?j7.93?,求低压方电流、电压、高压方电流。 答案 与P138例3.5一样

3.48 一台三相变压器,SN?5600kVA,U1N/U2N?10/6.3kV,Yd11接法。

在低压侧加额定电压U2N作空载试验,测得,P0?6720W,I0?8.2A。。

在高压侧作短路试验,短路电流Ik?I1N,Pk?17920W,Uk?550V,

试求:

(1)用标么值表示的励磁参数和短路参数;

(2)电压变化率?U?0时负载的性质和功率因数cos?2; (3)电压变化率?U最大时的功率因数和?U值; (4)cos?2?0.9(滞后)时的最大效率。 先求出一相的物理量

I1N??SN3U1N?56003?10?323.35(A) U1N??10003?5773.67(V)

3U2N??6.3kV I2N??SN3?560033?6.313?296.3(A)

P0??67203?2240(W) I0??I0?8.23?4.73(A)

PK??179203?5973.33(W) IK??I1N??323.35(A)

Uk??UKU'Zm?'Rm?U2N?I0?P0?I0?3?5503?317.55(V)

U1N?I1N?N?k?U1?5773.676300?0.916 Zb?2N??5773.67323.35?17.856(?)

?10 ?6.34.73?1331.92(?)23'''2?Zm?Rm?1331.922?93.422?1328.64? ?22404.732?100.12(?) Xm折算到高压侧:

'Zm?k2Zm?0.9162?1331.92?1117.6(?)

Zm** Rm?0.9162?100.12?84(?) RmZm?Z??1117.6?8417.856?4.7 17.856?62.59'Xm?0.9162?1328.64?1114.8(?) Xm?1114.817.856?62.43

短路参数:

U?*0.982?17.856?0.055 ZkZk?Ikk??317.55323.35?0.982(?)Rk?Pk?I2k?*0.0575973.33?17.856?0.0032 ?323.352?0.057 Rk*2k*2?Rk?0.0552?0.00322?0.0549

*Xk?Z

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