发布时间 : 星期三 文章电机学课后习题与答案更新完毕开始阅读
'40.34 R2?(Rk?R1)X0?0XK?(0.936?0.5)?40.34?3.712?0.48?'X1??X2??X0?X0?XKX0'(R2?X02)?40.34?2X40.34?3.71240.34?(0.482?40.342)?1.9(?)
Xm?X0?X1??40.34?1.9?38.44(?)
5.34 一台三相异步电动机的输入功率为10.7kW,定子铜耗为450W,铁耗为200W,转差率为s=0.029,,
试计算电动机的电磁功率、转子铜耗及总机械功率。
Pem?P1?Pcu1?PFe?10700?450?200?10050(W)电磁功率 PCU2?SPem?0.029?10050?291.45W Pmec?(1?S)Pem?10050?291.45?9758.55W速960r/min,额定负载时cos?1
5.35 一台JO2-52-6异步电动机,额定电压为380V,定子Δ联接,频率50Hz,额定功率7.5kW,额定转
?0.824,定子铜耗474W,铁耗231W,机械损耗45W,附加
损耗37.5W,试计算额定负载时,
(1)转差率;
(2)转子电流的频率; (3)转子铜耗; (4)效率;
(5)定子电流。
60f3000??1000r
minp3n?n1000?960??0.04 ∴S?1n11000 (1)
n1?(2) (3)
f2?Sf1?0.04?50?2HZ
Pmec?p2?pmec?pad=7500+45+37.5=7582.5W
Pmec?(1?S)Pem
p206.82?103?60Tem?em??1317N.M?12??1500P7582.5PCU2?SPem?S?mec?0.04??316W
1?S1?0.04
(4)
?PCU2?PFe?Pmec?Pad=474+316+213+45+37.5=1103.5W
PP27500?=2???87.17%
P1P2??P7500?1103.5CU?P?P?3U1I1cos?1?3?380I1?0.824
7500?1103.5?I1??15.86A
3?380?0.8245.36 一台4极中型异步电动机,PN?200kW,UN?380V,定子Δ联接,定子额定电流
IN?385A,频率50Hz,定子铜耗pCu2?5.12kW,转子铜耗pCu2?2.85kW,铁耗pFe?3.8kW,机械损耗pmec?0.98kW,附加损耗pad?3kW,R1?0.0345?,
???0.195?;起动时,??0.022?,X2Xm?5.9?。正常运行时X1??0.202?,R2???0.11?。试??0.0715?,X2由于磁路饱和与趋肤效应的影响,X1??0.1375?,R2(5)P1求:
(1)额定负载下的转速、电磁转矩和效率;
(2)最大转矩倍数(即过载能力)和起动转矩倍数。 解:(1)Pmec PCU2
?p2?pmec?pad=200+0.98+3=203.98kw
?SPem Pmec?(1?S)Pem
41
?pcu2S?pmec1?S 即
2.85S?203.981?S
S=0.01378
n?(1?S)n1?(1?0.01378)?1500?1479rmin
p2.85Pem?cu2??206.82kw
S0.01378p206.82?103?60Tem?em??1317N.M
?12??1500PP2200?=2???92.7%
P1P2??P200?5.12?2.85?3.8?0.98?3(2)Tmaxm1u121 ?2?12(R1?R12?(X1??X2??))603?3802???3186N.m
222??15002?(0.0345?0.0345?(0.202?0.195))PN200?103?60TN???1291.97N.m
?2??1479T3186km?max??2.466
TN1292Tst??m1PU12R22???(X1??X2??)2?2?f1??R1?R2??1?0.04?3??0.34?20.52?10287.72W0.04
?3?2?3802?0.07152??50???0.0345?0.0715???0.1375?0.11????22?2721.5N.m
kst?
Tst2721.5??2.11 TN12925.37 一台三相8极异步电动机的数据为:
PN?200kW,UN?380V,
f?50Hz,
nN?722r/min,过载能力kM?2.13。试求:
(1)产生最大电磁转矩时的转差率; (2)s=0.02时的电磁转矩。 (1) Sm推倒如下:
?SN(km?k2m?1
(即TemTN21??TmaxSN?SmkmSmSN2?TN时)
?2kmSNSm??N2?0 求一元二次方程即可 n1?n750?722??0.03733 SN?n1750 SmSm?0.03733?(2.13?2.132?1)?0.1497
(2)
42
pN200?103?60??2646.6N.m TN??2??722Tmax?kmTN?2.13?2646.6?5637.2N.m Tem2 ?TmaxSm?SSSmTem22???0.2625 0.14970.025637.27.485?0.1336?0.020.1497Tem?5637.2?0.2625?1480N.m
5.38 一台三相4 极异步电动机额定功率为28kW,UN?380V,?N?90%,cos??0.88,
定子为三角形联接。在额定电压下直接起动时,起动电流为额定电流的6倍,试求用Y-Δ起动时,起动电流是多少?
解:
28?103 IN???53.72A
3UNcos?N?3?380?0.85?0.9PN直接起动时的起动电流:
Ist?6IN?6?53.72?322.3A
用Y-△起动时:
Ist??Ist?107.4A 35.39 一台三相绕线转子异步电动机,PN比ke?155kW,IN?294A,2p?4,UN?380V,Y
???0.06?,?1?1,电动势及电流的变??0.012?,X1??X2联接。其参数为R1?R2?ki?1.2。现要求把起动电流限制为3倍额定电流,试计算应在转子回路每相中接入多大?3IN?3?294?882A
的起动电阻?这时的起动转矩为多少?
解: Ist 起动时阻抗:
Est?UN380??0.249? 3Ist3?882 Est?(R1?R2??Rst?)2?(X1??X2?)2?Rst??0.218?0.012?2?0.194?
∴每相接入的起动电阻为:
?R1?R2??Rst??0.2492?0.062?0.218?
Rst?Rst?0.194??0.1347?] 2keki1.2Tst?m1pu12R2??Rst?2??22?f1?(R1?R2??Rst?)?(X1??X2??)???
43
?
3?3059.7N.m
222?3.14?50??0.218?0.12??0.02?,额定负载时nN?1480r/min,
3?2?38022??0.218?0.012?5.40 一台4极绕线型异步电动机,50Hz,转子每相电阻R2若负载转矩不变,要求把转速降到1100r/min,问应在转子每相串入多大的电阻?
解: n160f60?50??1500r
minp2n?n1500?1480SN?1N??0.01333
n11500n?n1500?1100S?1??0.2667
n11500?R2R2?R??SNS
∵负载转矩不变 ∴电磁转矩不变
?S?0.2667???1?R2?(?1)?0.02?0.38?
0.01333?SN?U1N?380V,5.41一台三相4极异步电动机,定子Y接法,cos?N?0.83(滞后),R1?0.35?,
?N?20.5A,??0.34?,sN?0.04,R2机械损耗和附加损耗之和为288W,设 I1N?I2 R?试求:
(1)额定运行时输出功率、电磁功率和输入功率;
(2)额定运行时的电磁转矩和输出转矩。 (1)2P=4, n1minn1?nN1500?1426??0.0493 (2)SN?n11500 f2?sf1?0.0493?50?2.47HZ
与5.31一样
(4)Pem
?1500r
?spcu2
pcu2?3I2?2R2??3?5.822?2.82?286W
p286?pem?cu2??5812.6W
s0.0493p5812.6?60Tem?em??37N.m
?12??15005.42 一台三相4极绕线式异步电动机,f1?50Hz,转子每相电阻R2?0.015?,额定运行时转
子相电流为200A,转速nN?1475r/min,试求:
(1)额定电磁转矩;
0?0SN?1515001475?0.01667
0.015?3?2002?0.01667?10800(W)
2P?mIem22R2STem?Pem?110800 ?2??1500?688(N?m) (2)在转子回路串入电阻将转速降至1120r/min,求所串入的电阻值(保持额定电磁转矩不变);
?1120SN?1500?0.2533 150044