Essay Title: 

Elictronic & system

March 24, 2016 | Author: | Posted in asian studies, history

Running head : Electronic System

Electronic System

Name

Student Ref

University

Course : Electrical Systems 204

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 [banner_entry_middle]

VL IL cosf

Cosf 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 .21

CosF 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 .164

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 .74

R1 Jm1 Jm2 0 .16 .13j

415 Rc JXm R1 /s

0 .023 12 .7 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

F 100 V2 0 .7 415 290 .5 Full load current 20Kw /415 48 .2A

Ea2 /Ea1 N2 /N1 I1 /I2 ?1 ?2 Therefore 205A /48 .2A .7 /1 962 /T1

T2 0 .7 48 .2 967 /05 1 159 .15N /m

Q4

Part i

IL

If IA Data

15kw ? Efficiency 0 .85 N1 600rev /min I .o 2 .5A

230V At Shunt field current

f

R

o… [banner_entry_footer]

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