(精品)《电机学》课后习题答案

***8.85578.60717.457R??0.0239xx2??0.0246 k k?360?0.0485 ?360360* Zk?19.467360?0.05408(4)

&?ZI&Ukk1N?(8.607?j17.457)?16.667?143.33?j290.95

I1N?U1NN?1006?16.667A

?j290.95*kUk?6000?143.33也可以,但麻烦。 6000**S&U∵Uk?Zk ∴Uk(5)

****?5.400 Ukr?Rk?2.3900 Ukx?Xk*?4.8500

**?U??(Rkcos?2?Xksin?2) ∵是满载 ∴??1

(a)cos?2?1 sin?2?0 ?U?0.0239?1?2.3900

sin?2?0.6

(b) cos?2?0.8(滞后)

?U?1?(0.0239?0.8?0.0485?0.6)?4.82200 (c) cos?2?0.8(超前) sin?2??0.6

?U?1?(0.0239?0.8?0.0485?0.6)??0.96800

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?,并将结果进行比较; (1)用近似“?”型等效电路和简化等效电路求U11(2)画出折算后的相量图和“T”型等效电路。 (1) k?N1N2'22k?Rk?3.37?0.15?1.7035(?) ?876?3.37 22260'2x2?3.37?0.964?10.948(?) ?I1 R1I0RmX1?R2X2?I2U2ZLI1 R1X1?R2X2?I2U2ZLU1U1Xm

P型等效电路 简化等效电路

'。。&&0。?3.?kU37?60000?202200(V)?U20。则U 22P型:设U2'&。I'。1802&I???36.87?53.41?36.87?42.728?j32.046 R3.372'&?U&?(R?R'?jx?jx')I& U1211?22?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 。?1.84.33&&?I&?I&?0.2443?j1.6607?42.728?j32.046?42.9723?j33.7067?54.615?38.12。I102(A)(V)∴U1?21254 I1?54.615

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。 &?202200 则I&?312?36.87 设U22?'。'。3k?53.41?36.87。

'。&43A ∴U(见上题)∴U1?3U1??36812 I1?I1??53.(V)?212542.6991?

?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??4003?230.95(V) Ik??57.74(A)

K?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*xk?Zkb?0.0383 sin?2?0.6

**?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?''。1155&&?U0 则IUI2?2??46.2?36.87 2?k?25?46.2(A)设2&&1000&?I&U1N?2(?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)

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