华科大辜承林主编《电机学》课后习题答案

P型:设U2''。。37?60000。?202200(V)?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。(1A5)(V)∴U1?21254 I1?54.6

用简化等效电路:

(A)U1?21254(V)(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。

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

'43A ∴U1??212542. I1?I1??53.(V)699。(见上题)∴U1?3U1??36812

?1?2.699。?(-36.87。)=40.82。 cos?1?0.76) (滞后

U1N?10kV,U2N?400V,3.46 一台三相电力变压器,额定数据为:SN?1000kVA,

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 k?U1N?U2N??100.433?25 Uk??4003A( )?230.95(V) Ik??57.74K?K?PKPk??U?3.867(kW)Z??230.9557.74?4(?) kI32?386757.742?1.16(?) xk?ZK?R2?42?1.162?3.83(?)

KRk?PK?2Ik?(2)方法一:

??I2I2NU1N?I1N? I2N? I1N?3SN3U2N?10003?0.41152?1443.42(A)∴??1443.42?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??U2N∴U2?3?4003?230.947(V)

?(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?''。1155U?U20 则I2I2???46.2?36.87 ?k?25?46.2(A)设21000U1N??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)

∴U2?

3.47 一台三相电力变压器的名牌数据为:SN?20000kVA,U1N/U2N?110/10.5kV,

??Yd11接法,f?50Hz,Zk?0.105,I0?0.65%,P0?23.7kW,

?'U2?k?224.9 ∴U2?3?225?389.7(V)

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?10SN?323.35(A) U1N??10003?5773.67(V)

3U2N??6.3kV I2N??3?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??UKU3?5503?317.55(V)

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

'Zm?U2N?I0??10 ?6.34.73?1331.92(?)23'Rm?P0?I0?'''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(?)

Zm2** Rm?0.916?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:

Zk?Rk?Uk?Ik?Pk?I2k?*0.982?0.055 Zk?17.856?317.55323.35?0.982(?)*0.0575973.33?17.856?0.0032 ?323.352?0.057 Rk*Xk?Z*2k*2?Rk?0.0552?0.00322?0.0549

*Rk*xk**(2) ??1 ?U?0?Rkcos?2?Xksin?2 ∴tg?2????0.00320.0549??0.05829

。∴?2??3.3359 ∴是容性负载 cos?2?0.998(超前)

**?U(3)??1 d??Rsin??Xk2kcos?2?0 d?2∴tg?2?*Xk*Rk。??86.66?0.0549?17.156 (感性)cos?2?0.0583(滞后) 20.0032?Umax?0.0032?0.0583?0.0549?sin86.66?0.055

(4) ?m?P0PkN2(即P0??mPkN时效率最大)= 2240?35973.3?3=0.06124

?max?1?

2P0??mPkN?mSNcos?2?P0??mPkN6720?0.06124?179205623.22?17319.680 ?0.06124?5600?1??99.6330?10?0.9?6720?0.06124?179202981.16?10?5623.223.49 有一台三相变压器,SN?5600kVA,U1N/U2N?10/6.3kV,Yd11联接组。变

压器的空载及短路试验数据为: 试验名称 空载 短路 线电压 U1/V 6300 550 线电流 I1/A 三相功率P/W 7.4 323.3 6800 18000 备注 电压加在低压侧 电压加在高压

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