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# Electronic & System

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Question 1

Part i

Ns =120f/p

Full load slip = Ns-Nr/Ns                                                            = 120*50/4

Ns is synchronous speed.                                                            = 1,500 rev/min

Nr is the rotor speed.                                                             Nr = 1,453rev/min

Therefore, the slip is equal to; 1,500-1453/1,500=0.031

Part ii

= 0.031*50

= 1.55Hz

Part iii

From the No load test:         power

3*VL*IL*cosф

Cosф=3.46*10e3

3*415*18

=74.48

Io=18/_74.48 = 4.813+j17.34

Equalizing voltage: E1=Vph-Vab

Vab=IL Zab

Thus; 415/31/2*-226=13.6v/

13.6=64.7 * Zab

Zab=13.6v/64.7A

=.21Ω

CosФ=37000/3*415*64.7

=0.8

Zr=K*2 (R2/S+jX2)     1.12(R1/.031+jX2)

But also R2=R1*K2   =0.16X1.12    = 0.1936

Zr=Zt*K2     =0.211.12=0.254

jXm=(.254-0.1936) 1/2=0.164Ω

Also W=R1I

0.24=182XRc,=.0.001/0.31=0.23

JXm= 0.68Kw/182=2.04X104/Zm2

=0.395

Now;                     =0.395/0.031=12.74Ω

R1                               Jm1                                     Jm2

0.16                  .13j

415                                        Rc         JXm                           R1/s

0.023         12.7Ω           0.193/0.031

Q 2

Part i

The approximate equivalent circuit

R1               JXi                             jX2

IL         I22

Io                 1.496w          4.51 w                        4.51w

Rc                     jxm                                                                                  R22/s

2.2w            190w                                                                                0.05

Slip= (NS-Nr)/Ns           Ns=120*50/6=1000rev/min

Nr= (1000-967)/1000

=0.033

Thus     X1=X2=watts

R2=R1 =1.496W

Part ii

Thus IL=Io+I 2

13.1A=Il   Io=2.2w+j190w     but W=VI

Current at Rc=7.5/2.2=3.4w   Ir=3.4/415=8.19A

Also at jXm =7.5Kw/190w=39.47

Im =39.47/415=.95A

13.1A=Io +I22

Io= (8.192 +.952)1/23  =8.24/__6.61

I22=13.1-8.24   =4.86A

Air gap power=3*IL2*R2/s        =3*4.862* 1.496/0.033 =3212.26w

Copper rotor losses =slip*air gap power

=3212.26w*.033= 106.004w

Mec. losses     = 3212.26w (1-0.033)       =   311490.26w

Output power =311490.26w-100=31049.25w

Full load output torque=31049.25w*(1-0.033)/ (4∏*50/6) =306.61N/m

Part iii

Current at rated voltage=7.5Kw/32*415*0.906 = 11.51A

Q3

Part i                                                                                           data

205A

Torque=962N/m

Output=20Kw

Ratio =1:0.7

Rated values

415/50HZ/8pole                                                                     Io=205A

T1=962N/m

Ф =100%

V2= 0.7*415 =290.5

Ea2/Ea1=N2/N1 =I1/I2*Φ1/Φ2                          Therefore=205A/48.2A=.7/1=962/T1

T2=0.7*48.2*967/05*1 =159.15N/m

Q4

Part i

IL

I                   IA                            Data

P=15kw

0.3Ω                     Efficiency =0.85

N1=600rev/min

I.o=2.5A

230V                                                                       At Shunt field current,

230Ω                                                             230/230=1A

Armature current

Ra=0.2Ω   =2.5-1=1.5A

E1=Vt-IaRa+VF

Vs= (230-1.5*0.2) + (1.5*0.3)

Ea1= (230-0.45) =229.55V

IL

If                          Ia             input power =15Kw/0.85                                                                                                                         =17.64Kw

1.                      =76.72A

RR                Armature current =76.7-1

=75.72A

Ea1=230-75.72*0.2

=214.65V

Ea1/EA2=N1/N2

Ra=0.2Ω 229.55/214.65=600/N2

N2=600*214.65/229.55

= 561.07N/m

REFERENCE

Theraja and AK Theraja., (2002.)Textbook of electrical technology: (vol-2) AC and DC machines / BL23rd rev. edn. New Delhi: S. Chand.pp

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