U20II?146.1V
3.39 有一台单相变压器,U1N/U2N?220/110V当在高压侧加220V电压时,空载电流
为I0,主磁通为?0。今将X、a端联在一起,Ax端加330V电压,此时空载电流和主磁通为多少?若将X、x端联接在一起,在Aa端加110V电压,则空载电流和主磁通又为多少?
解:(1)设高压绕组为N1匝,低压绕组为N2匝
则
N1N2?220110?2
U1N原来主磁通: ?0?4.44fN1?2204.44fN1
现在匝数为N1?N1?1.5N1(Z,a端连在一起) ∴??''03304.44f1.5N1' ∴?0??0'3301.51?220?1 ∴主磁通没变,∴励磁磁势
'I01.5F0?F0而I0?1.5N1?I0?N1 ∴I0??1223I0
12(2)若将Z,x连在一起,则匝数为:N1?现在的主磁通?0?'UAm4.44f12N1?N1
N1?'110?24.44fN1??m 不变
12'∴励磁磁势不变F0?F0而F0?I0?'N1 ∴I0'?1N1?I0?N1 2F0?I0N1 ∴I0?2I0
'
kVA,高、低压侧额定电压3.40 有一台单相变压器,额定容量SN?5000U1N/U2N?35/6.6kV。铁柱有效截面积为1120cm,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。
2 解:?m?BmA?1.45?1120?10N1?U1N4.44f?m?4?0.1624(Wb)
?35?104.44?50?0.16243?971(匝)
3N2?k?U2N4.44f?m?6.6?104.44?50?0.1624?183(匝)
U1NU2N?356.6?5.303
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
(1100?1100)110?110k1??10
?2.273(A) I2N?SNU2NI1N?SNU1N?50001100?1100?5000110?110?22.73(A)
2:两高压绕组串联,两低压绕组并联 k2?U1U2?(2200110?20
5000110A( )I2N? I1N?2.273?45.45(A)
3:两高压绕组并联,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:ZRm?P12I1m?UI?1100.5?220(?) P1?I12Rm
Zm?Rm?216.3(?)
22?100.52?40(?) Xm?UI1无铁心时:Zm0? Rm0?P1I02?110100?1.1(?) I ?1 ?100001002?1(?)
2 Xm?1.1 ( )?1?0.458?3?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,
一二次侧绕组的电阻和漏电抗的数值为:R1?4.32?;R2?0.0063?;f?50Hz。
X1??8.9?;X2??0.013?,试求:
E4.44fN?1104.44?50?1000?4.955?10Wb
?4 (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 x2??kx2??26.1?0.013?8.8557? ∴RK?R1?R2?4.32?4.287?8.607?
''22xk?x1??x2??8.9?8.8557?17.457?
ZK?RK?XK?22228.607?17.457?19.467?
(2)折算到低压测:
R1?''R1k2?4.3226.12?0.0063? x1??'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?'2'20.0126?0.0261?0.029?
22(3)阻抗机值: Zb?U1NI1N?U1NSNU1N?6000?6000100?103?360?
*4.287360R1?*4.32360?0.012 R2??0.01191 x1??*8.9360?0.0247
x2??Zk?**8.8557360?0.0246 Rk?*8.607360?0.0239 xk?*17.457360?0.0485
19.467360?0.05408
??ZI??(8.607?j17.457)?16.667?143.33?j290.95 (4) Ukk1N I1N? Uk?*SNU1N?Uk??1006?16.667A
290.956000143.?3j36000也可以,但麻烦。
****∵Uk?Zk ∴Uk?5.400 Ukr?Rk?2.3900 Ukx?Xk?4.8500
***(5) ?U??(Rkcos?2?Xksin?2) ∵是满载 ∴??1
**?2?(a)cos?2?1 sin0 ?U?0.0239?1?2.3900
(b) cos?2?0.8(滞后) sin?2?0.6 ?U?1?(0.023?90?.80 0.0?485?0.60)4.822(c) cos?2?0.8(超前) sin?2??0.6 ?U?1?(0.023?90?.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?876260?3.37 k2?R2k'2?3.37?0.15?1.7035(?)
2x2??3.37?0.964?10.948(?)
I?1 2R1I?0RmX1?R2X2?I?2?U2ZLI?1 R1X1?R2X2?I?2?U2ZL?U1?U1Xm
P型等效电路 简化等效电路
??kU?0?3.37?60000?202200(V) P型:设U2?U20则U22'。'。。。?'?I2?I2R?1803.37?36.87?53.41?36.87?42.728?j32.046
。。??U??(R?R'?jx?jx')I?' U1211?22?2 =20200?(2.19?1.7035?j15.4?j10.948)?53.41?36.87 =20220?(3.8935+j26.348)?53.41-36.87
=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?j12600??U112661.8584.33。?1.6786?81.63?0.2443?j1.6607
。??I??I??0.2443?j1.6607?42.728?j32.046?42.9723?j33.7067?54.615?38.12。I1021A5)(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?U2'ZL?U1
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 设U22?'312?36.87。3k?53.41?36.87
3U1??36812(V) I1?I1??53.43A
。。?'?212542.699(见上题)∴U1?∴U1?
百度搜索“77cn”或“免费范文网”即可找到本站免费阅读全部范文。收藏本站方便下次阅读,免费范文网,提供经典小说综合文库华科大辜承林主编《电机学》课后习题答案(4)在线全文阅读。
相关推荐: