电机学I作业(3)

2020-03-26 20:37

I1*?rk*?xk*I2*RL*U1*2-9:

?rm*xm*Im*??U2*

解:(1)空载时,U0?UN,则U0*?1,计算激磁阻抗: zm*?11??50 I0*0.02rm*?p0*133125000??2.66 22I0*0.0222xm*?zm*?rm*?49.93

短路时,Ik?IN,则Ik*?1,计算短路阻抗:

zk*?uk*?0.105

rk*?pkN*?600?0.0048

12500022xk*?zk*?rk*?0.1049

??1?0?,则I??1??36.87? (2)设U2*2*???U??I??r?jx???1?0??1??36.87??0.0048?j0.1049?U1*2*2*k*k*           ?1.0699??175.7?

??I??I??1.0699??175.7???1??36.87??1.015?142.26? I1*m*2*2.66?j49.93(3)?U??U1*?1??100%??1.0699?1??100%?6.99% ??P2U2*I2*cos?21?1?cos36.87????99.27% P1U1*I1*cos?11.0699?1.015?cos??175.7?142.26??实用公式:

?U??(rk*cos?2?xk*sin?2)?100%   ?1?0.0048?0.8?0.1049?0.6??100%?6.678%

- 10 -

?=βSNcos?21?125000?0.8??99.27% 22βSNcos?2?p0??PKN1?125000?0.8?133?1?600(4)当?????

p0133??0.471时,有最大效率: PkN600?max=β?SNcos?20.471?125000?0.8??99.44"??βSNcos?2?p0??PKN0.471?125000?0.8?133?0.471?600

2-11:设有一台50kVA,50Hz,6300/400V,Yy连接的三相铁芯式变压器。空载电流I0?0.075IN,空载损耗p0?350W,短路电压uk??0.055,短路损耗

pKN?1300W。(最好用标么值,并与下面方法比较)

(1)试求该变压器在空载时的参数r0和x0,以及短路参数rk、xk,所有参数均归算到高压侧,作出该变压器的近似等效电路;

(2)试求该变压器在供给额定电流且cos?2?0.8滞后时的电压变化率及效率 解 (1)

IN?SN50KVA??4.58A 3U1N3?6300VZ0?Zm?r0?U1U1N/3??10584? I00.075INP0350??988? r0*?1.24 3I023?(0.075?1.58)22x0?Z0?r02?105842?9882?10534? x0*?13.2

Uk?0.055?6300/3?200V

Ik?I1N?4.58A

Zk?Uk200??43.67 Ik4.58rk?Pk1300??20.66 rk*?0.026 3IN23?4.582- 11 -

xk?Zk2?rk2?43.672?20.662?38.47? xk*?0.0485 (2)

?U%?I1Nrkcos?2?I1Nxksin?2?100

U1?4.58?20.66?0.8?4.58?38.47?0.6?100?4.99%

6300/3??

?SNcos?250000?0.8??96%

?SNcos?2??2PKN?P050000?0.8?1300?350- 12 -

第3章 三相变压器及运行

p54:3-1

B(X)b(x)A(Z)(a)c(y)C(Y)Dd0

Dy11

B(X)bxyzA(Z)(a)cC(Y)Dy11

Bc(y)XYZA(a,z)Cb(x)Yd7

Yy10

- 13 -

BbxyzcXYZCA(a)Yy10

p54:3-2

BbxyzcBXYZCcbA(a)XYZCYd9

A(a)Yy10B(X)

A(Z)(a)C(Y)b(x)Dd4

c(y)p54:3-4解: (1)ukI*?zkI*?zkI250323??0.0568 23z1NI6300500?10- 14 -


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