华科大辜承林主编《电机学》课后习题答案

解:(1)设高压绕组为N1匝,低压绕组为N2匝

N1N2?220110?2

U1N220?0?4.44?fN14.44fN1

原来主磁通:

现在匝数为N1?N1∴?'0?1.5N1(Z,a端连在一起)

?'0?0?3304.44f1.5N1 ∴

?3301.51220?1 ∴主磁通没变,∴励磁磁势

I0''?1.5N1?I0?N1 ∴I0F0'?F0而I0 ?1.5?23I0(2)若将Z,x连在一起,则匝数为:N1?12N1?现在的主磁通?0'U12N1

110?2?4.44Am?4.44fN1??m 不变 f1N21∴励磁磁势不变F0?F0而F0'''1?I0?2N1 ∴I0'?12N1?I0?N1

'?2I0 F0?I0N1 ∴I0

3.40 有一台单相变压器,额定容量SN?5000kVA,高、低压侧额定电压

U1N/U2N?35/6.6kV。铁柱有效截面积为1120cm2,铁柱中磁通密度的最大值Bm?1.45T,试求高、低压绕组的匝数及该变压器的变比。

解:?mU?BmA?1.45?1120?10?4?0.1624(Wb)

335?10 N1?4.441fN?m?4.44?50?0.1624?971(匝)?10N2?4.442fN?m?4.446.6?50?0.1624?183(匝)

35Nk?U1??5.303 6.62NUU3

3.41 有一台单相变压器,额定容量为5kVA,高、低压侧均由两个绕组组成,一次侧每

个绕组的额定电压为U1N?1100V,二次侧每个绕组的额定电压为U2N?110V,

用这个变压器进行不同的联接,问可得几种不同的变比?每种联接时的一、二次侧额定电流为多少?

共有4种:

1:两高压绕组串联,两低压绕组串联

k1?(1100?1100)110?110?10

I1N?U1NN?50001100?1100?2.273(A) I2N?US?5000110?110?22.73(A)

N2NS2:两高压绕组串联,两低压绕组并联

00k2?U1U2?(22110?20

A( )I2N?5000110?45.45(A) I1N?2.2733:两高压绕组并联,两低压绕组串联

k3?1100220?5 I1N?50001100?4.545(A) I2N?22.73(A)

4:两高压绕组并联,两低压绕组并联

k4?1100110?10,I1N?4.545(A) I2N?45.45A( )

3.42 将一台1000匝的铁心线圈接到110V、50Hz的交流电源上,由安培表和瓦特表的读

数得知I1?0.5A、P1?10W,把铁心取出后电流和功率就变为100A和10Kw。设不计漏磁,试求:

(1)两种情况下的参数、等效电路;

(2)两种情况下的最大值。 (1)有铁心时:Zm1102?UI?0.5?220(?) P1?I1Rm

22Rm?P1I12?100.52?40(?) Xm?Zm?Rm?216.3(?)

无铁心时:Zm0110?UI1?100?1.1(?)

Rm0?P1I2?100001002?1(?)

0 Xm?1.12?1?0.4583(?)

I1 U1RmXm

(2) E?U?110 E?4.44fN?m ∴?m

3.43 有一台单相变压器,额定容量SN?100kVA,额定电压U1N/U2N?6000/230V,

?4110E?4.44??4.955?10Wb fN4.44?50?1000R1?4.32?;R2?0.0063?;一二次侧绕组的电阻和漏电抗的数值为:f?50Hz。

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。

(1) R2?kR2?(6000230)?0.063?4.287?

'22x2?kx?26.1?0.013?8.8557? ?2?'22k?26. 1∴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*7?0.01191x*?R1*?4.32360?0.012 R2?4.28 1?3608.9360?0.0247

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

Uk?ZkI1N?(8.607?j17.457)?16.667?143.33?j290.95

I1N?U1NN?1006?16.667A

143.?3j3290.95*Uk?60k0?06000也可以,但麻烦。

**SU∵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

0.?80) 0.0?485?0.60(b) cos?2?0.8(滞后) ?U?1?(0.023?94.822(c) cos?2?0.8(超前) sin?2??0.6

9 ?U?1?(0.023?0.?80.0?485?0.?60)0

0.968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

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