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Electronic & System Essay Sample

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Electronic & System Essay Sample

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

Full load rotor frequency =Ns – Nr/Ns*Full load frequency

                                         = 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

                                                                                Full load current= 20Kw/415 =48.2A

                          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

                            230               230Ω                                          Full load IL=17.64Kw/230

  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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