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ADDIS ABABA UNIVERSITYADDIS ABABA INSTITUTE OF TECHNOLOGY
DEPARTEMENT OF ELECTRICAL AND COMPUTER ENGINEERING
Course Title: Introduction to Electrical Machine
ASSIGNMENT - 01
Submitted to Instructor: Dawit Habtu
Name: ID No.:
March 15, 2013
1, GivenWooden ring Ac = 300mm2
Lc = 200mm
N = 800 turns
Required
i. Filed strength HC when I = 2A assume μr =1ii. Magnetic flux density BC
iii. Current reqiered I when B=0.02wb/m2
Solution
i. HC= NI/ LC by Ampere's law F=NI=HC* LC
= 800 turns * 2A200mm
= 1600/0.2 = 8,000 wb/m
ii. BC= HC*μ μ=μ0*μr = 4π*10-7*1 = 4 π* 10 -7 = 8,000 wb/m*4π*10-7
= 32π*10-4
= 10.053mwb/m 2
iii. F=NI=HC* LC HC = B/μ =0.02wb/m 2 = 15.915kwb/m
4π*10-7
:. I = HC* LC
N = 15.915kwb/m* 0.2m
800 turns = 0.3978A
2, Given
Flux øc = 0.38mWb
Iron-ring diameter LC = 58cm=0.58m
Cross-sectional area Ac =3cm2= 0.0003m
Required
Ampere-turn NI
Solution
BC = øc/Ac = 0.38mWb = 1.27 wb/m2
0.0003m
If BC =1.27 wb/m2 from the table we can get μr
1.4-1.27 = 1000-x x= 0.13*500 + 1000 =13251.4-1.2 1000-1500 0.2
.: μr =1325
F=NI=HC* LC = BC * LC μ
F=NI = 1.27 wb/m 2 * 0.58m 1325
= 5.559At
B (wb/m2)
0.5 1.0 1.2 1.4
μr 2500 2000 1500 1000
3, Given
Horse-shoe magnet LC = 45.7cm=0.457m
Cross-sectional area Ac =6.45cm2= 0.000645mN = 500 turns each limb connected in series
Mass= 68kg
Negligible Relactance
μr = 700
Required
Current I =?
Solution
Force =m*g = 68kg*9.8m/s2 = 666.4NF=NI
I= F/2N each limb connected in series = 666.4 N 2* 500 turns = 0.6664 A