电机与拖动基础第二版刘启新第三章部分答案

第三章 习题作业参考答案 4

．SN=750=N1IN1=N2IN2

IN1=

IN2=

5（1）此时UAX=220V，Uax=110V，由于加于绕组两端的电压不变，所以Φm不变，磁势不变。 设原边匝数为N1，

′+N1I0=N1I0I′0=I0

（2）UAX=220V，Uax=110V，Φm不变，磁势不变。

′−N1I0=N1I0′=2I0I0

8．

空载变比k=Zm=k2

=

14.43 •

••

•

•

•

•

•

=12.37A

=41.24A

N1•

I′02

P0

Rm=k

2

2

Xm==2211.92Ω

Zsh=

shpsh

=

Rsh=

Ish

=

2

35

2

Xsh==7.13Ω

Rsh75oC=Rsh

'R1=R2=

α+75234.5+75

=2.04×=2.39Ω α+θ234.5+30

1

Ro=1.195Ω 2sh75C

1

=k×

2

=7.42Ω

=2.04Ω

案

=182.35Ω 2



网

=k=2219.42Ω w

ww

.k

hd

a

N1•

′I02

w.

co

m

'

Xσ1=Xσ2=

11

Xsh=×7.13=3.56Ω 22

9．（1）

3450′=R1+kR2=0.435+Rsh=R1+R2×0.00194=0.8715Ω

230

2

′2=Xσ1+kXσ2=2.96+k2×0.0137=6.0425Ω Xsh=Xσ1+Xσ

2

Zsh==6.105Ω

2

（3） ∆u%=β

IN1(Rshcosϕ2+Xshsinϕ2)

UN1

×100%

10.（1）

k=

==25网

3

IN1==11.55A

'ZL=k2ZL=(

600+j300)Ω

•

I1=

•

UN1==8.6∠−26.58oΩ

Z

I2=kI1=215A

U2=2ZL=399.7V（2）cosϕ1=cos(26.58o

)=0.894

P1=1I1cosϕ1=10000×8.6×0.894=133.167kW （3）

0.48cosϕ2=cosϕL=costg−1=0.894（滞后）

0.96∆u=

100%=100%

2

课

后

I1=8.6A

案

'

=600.15+j300.35=671.11∠26.58oΩZ=Zsh+ZL

w

ww

.k

hd

a

80×103

IN1==23.188A

3450

23.188×0.8715

=0.586% cosϕ2=1时，∆u%=

3450

23.188×(0.8715×0.8+6.0425×0.6)

=2.905%cosϕ2=0.8(滞后)，∆u%=

3450

cosϕ2=0.8(超前)，∆u%=−1.968%

w.

co

m

=

=0.043%

β=

8.6

=

0.745 11.55

P2===99.9% P11

η=

11.（

1

）kN2

=8.66

Rsh75o

C=

Zsh=

sh

I

2

sh

=

538.86

2

=0.064Ω

I1=

=

=386.8∠−32.36oA I1

=386.8A

I2

1=

5801.8A

U2=

后

ZL=

ZL=0.1+j0.06=0.1166∠30.96oΩ （2）

β=

I1386.8==0.718 IN1538.86

cosϕ2=cos30.96o=0.858 p0+β2pshNη=1−2

βSNcosϕ2+p0+βpshN

答•

=390.6V

网

'

Z=Zsh+ZL=7.564+j4.793Ω=8.955∠32.36o

Ω



×100% 

3

w

Xsh=0.293Ω

ww

=0.3Ω .k

hda

pshN

56×103

w.

3

Ish=IN1==538.86A

co

m

'ZL=k2ZL=7.5+j4.5=8.746∠30.96oΩ

18×103+0.7182×56×103=1−×100%=98.66% 33230.718×5600×10×0.858+18×10+0.718×56×10

ηmax时,βm=

==0.567 2p0

ηmax=1−×100%

βScosϕ+2pmN20

2×18×10

=1−×100%=98.70%33

0.567×5600×10×0.858+2×18×10

3

课

后

答案

网

w

ww

4

.k

hda

w.

co

m

第三章 习题作业参考答案 4

．SN=750=N1IN1=N2IN2

IN1=

IN2=

5（1）此时UAX=220V，Uax=110V，由于加于绕组两端的电压不变，所以Φm不变，磁势不变。 设原边匝数为N1，

′+N1I0=N1I0I′0=I0

（2）UAX=220V，Uax=110V，Φm不变，磁势不变。

′−N1I0=N1I0′=2I0I0

8．

空载变比k=Zm=k2

=

14.43 •

••

•

•

•

•

•

=12.37A

=41.24A

N1•

I′02

P0

Rm=k

2

2

Xm==2211.92Ω

Zsh=

shpsh

=

Rsh=

Ish

=

2

35

2

Xsh==7.13Ω

Rsh75oC=Rsh

'R1=R2=

α+75234.5+75

=2.04×=2.39Ω α+θ234.5+30

1

Ro=1.195Ω 2sh75C

1

=k×

2

=7.42Ω

=2.04Ω

案

=182.35Ω 2



网

=k=2219.42Ω w

ww

.k

hd

a

N1•

′I02

w.

co

m

'

Xσ1=Xσ2=

11

Xsh=×7.13=3.56Ω 22

9．（1）

3450′=R1+kR2=0.435+Rsh=R1+R2×0.00194=0.8715Ω

230

2

′2=Xσ1+kXσ2=2.96+k2×0.0137=6.0425Ω Xsh=Xσ1+Xσ

2

Zsh==6.105Ω

2

（3） ∆u%=β

IN1(Rshcosϕ2+Xshsinϕ2)

UN1

×100%

10.（1）

k=

==25网

3

IN1==11.55A

'ZL=k2ZL=(

600+j300)Ω

•

I1=

•

UN1==8.6∠−26.58oΩ

Z

I2=kI1=215A

U2=2ZL=399.7V（2）cosϕ1=cos(26.58o

)=0.894

P1=1I1cosϕ1=10000×8.6×0.894=133.167kW （3）

0.48cosϕ2=cosϕL=costg−1=0.894（滞后）

0.96∆u=

100%=100%

2

课

后

I1=8.6A

案

'

=600.15+j300.35=671.11∠26.58oΩZ=Zsh+ZL

w

ww

.k

hd

a

80×103

IN1==23.188A

3450

23.188×0.8715

=0.586% cosϕ2=1时，∆u%=

3450

23.188×(0.8715×0.8+6.0425×0.6)

=2.905%cosϕ2=0.8(滞后)，∆u%=

3450

cosϕ2=0.8(超前)，∆u%=−1.968%

w.

co

m

=

=0.043%

β=

8.6

=

0.745 11.55

P2===99.9% P11

η=

11.（

1

）kN2

=8.66

Rsh75o

C=

Zsh=

sh

I

2

sh

=

538.86

2

=0.064Ω

I1=

=

=386.8∠−32.36oA I1

=386.8A

I2

1=

5801.8A

U2=

后

ZL=

ZL=0.1+j0.06=0.1166∠30.96oΩ （2）

β=

I1386.8==0.718 IN1538.86

cosϕ2=cos30.96o=0.858 p0+β2pshNη=1−2

βSNcosϕ2+p0+βpshN

答•

=390.6V

网

'

Z=Zsh+ZL=7.564+j4.793Ω=8.955∠32.36o

Ω



×100% 

3

w

Xsh=0.293Ω

ww

=0.3Ω .k

hda

pshN

56×103

w.

3

Ish=IN1==538.86A

co

m

'ZL=k2ZL=7.5+j4.5=8.746∠30.96oΩ

18×103+0.7182×56×103=1−×100%=98.66% 33230.718×5600×10×0.858+18×10+0.718×56×10

ηmax时,βm=

==0.567 2p0

ηmax=1−×100%

βScosϕ+2pmN20

2×18×10

=1−×100%=98.70%33

0.567×5600×10×0.858+2×18×10

3

课

后

答案

网

w

ww

4

.k

hda

w.

co

m