Tutorial 3 Soln

7/31/2019 Tutorial 3 Soln

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Tutorial # 3

1. Given: A ferromagnetic ring, Lm = 36 cm, A = 3 cm2, N = 400 turns. I = 1.4 A, and

ϕ = 1.4 mWb.

Required: µr .

Solution :

F = N . I = ϕ .R , ,

400 x 1.4 = 1.4 x 10-3

.R ,

R =3104.1

4.1400−

x

x= 4 x 10

5A/Wb,

R = A

L

r o

m

µ µ ,

µr =745 104103104

36.0−−

x x x x x π = 2387

2. Given: A cast steel ring, Lm = 30 cm, Lg = 2 mm, A = 3 cm2, and B = 0.6 T.

Required: F .

Solution :

Bc = Bg = 0.6 T,

From the cast steel magnetization curve

Hc = 430 At/m,

F c = Hc . Lm = 430 x 0.3 = 129At,

F g = Hg . Lg =o

g B

µ x Lg ,

=7104

6.0−

xπ x 2 x 10-3 = 955At,

F = F c + F g = 129 + 955 = 1084 At

I

N

Lm



Magnetization Curves

3. Given: A sheet steel, B1 = 0.4T, B2 = 0.8 T, B3 = 1.2 T, and B4 = l.6 T.

Required: Average value of permeability (B/H) ( µ).

Solution :

For each value of B the corresponding H is found from the sheet steel

magnetization curve. The results are as follows:

B (T) H (At/m) µ (H/m)

0.4 130 3.08E-03

0.8 370 2.16E-03

1.2 720 1.67E-03

1.6 2650 6.04E-04

4. Given: A cast iron pipe as shown in figure, N = 100 turns, from the figure do = 8 cm, di = 6 cm,

and pipe length (L) = 6 cm.

Required: (a) ϕ for I = 10 A.

(b) I when using a cast steel pipe with the same ϕ as in part (a).

Solution :

(a) F = N . I = 100 x 10 = 1000 At,

A = (ro – ri) x L = (4 - 3) x 6 = 6 cm2,



dm = (do + di)/2 = (8 + 6)/2 = 7 cm,

Lm = π dm = 0.07 π m

H =π 07.0

1000.=

m L

I N = 4547 At/m,

From the cast iron magnetization curve

B = 0.72 T,

ϕ = B x A = 0.72 x 6 x 10-4 = 432 Wb.

(b) For the same flux, ϕ = 432 Wb,

B = 0.72 T,


H = 500 At/m,

F = H . Lm = 500 x 0.07 π = 110 At,

F = N . I, therefore,

I =100

110=

N

F = 1.1 A.

5. Given: A sheet steel as shown in figure, N = 200 turns, ϕ varies from 4 to 12 mWb in steps of

2 mWb.

Required: Calculate the magnetization curve and plot ϕ vs I for;

(a) The circuit without air gap.

(b) The circuit with air gap.

Solution :

(a) Lm = 18+18+18+18 = 72 cm,

A = 8 x 10 = 80 cm2

= 8 x 10-3

m,

The steps to calculate the magnetization

curve and the corresponding values for

I for each value of ϕ are as follows;

1- Calculate B, B = A

ϕ ,



2- Find the corresponding value for H from the sheet steel magnetization

curve.

3- Calculate F , F = H . Lm , and finally,

4- Calculate I, I =

N

F .

The results are as follows:

ϕϕϕϕ (mWb) B (T) H (At/m) FFFF (At) I (A)

4 0.5 180 130 0.65

6 0.75 310 223 1.12

8 1 350 252 1.26

10 1.25 750 540 2.70

12 1.5 1660 1195 5.98

(b) Lm = 18+18+18+18 -0.1 = 71.9 cm ≅ 72 cm,

A = 8 x 10 = 80 cm2

= 8 x 10-3

m,

The steps to calculate the magnetization curve and the corresponding values

for I for each value of ϕ are as follows;

1- Calculate B, B = Bc = Bg = A

ϕ ,

2- Find the corresponding value for Hc from the sheet steel magnetization

curve (this values are the same as in part (a)).

3- Calculate F c, F c = Hc . Lm (this values are the same as in part (a)),

4- Calculate F g, F g =o

g B

µ . Lg,

5- Calculate F , F = F c +F g , and finally,

6- Calculate I, I = N

F .

The results are as follows:



B (T) Hc (At/m) F cF cF cF c (At) F gF gF gF g (At) F cF cF cF c (At) I (A)

0.5 180 130 398 527 2.64

0.75 310 223 597 820 4.10

1 350 252 796 1048 5.24

1.25 750 540 995 1535 7.67

1.5 1660 1195 1194 2389 11.94

6. Given: A sheet steel core as shown in figure, N = 280 turns, ϕ = 1 mWb, core thickness is

3 cm.

Required: Coil current (I).

Solution :

A = 3 x 4 = 12 cm2

= 12 x 10-4

m2

,

Bg = A

ϕ =

4

3

1012

101−

−

x

x= 0.83 T,

Hg =o

g B

µ =

7104

83.0−

xπ = 0.661 x 10

6

At/m,

0 . 0 0 5 . 0 0 1 0 . 0 0 1 5 . 0 0

4

6

8

1 0

1 2

F l ux ( mW b )

I (A)



F g = Hg x Lg = 0.661 x 106

x 0.001 = 661 At,

Bc = Bg = 0.83 T,


Hc = 350 At/m,

F c = Hc x Lc = 350 x 0.16 = 55 At,

ϕL = ϕR = ϕ /2 = 0.5 mWb,

BL = BR = A

2 / ϕ =

4

3

1012

105.0−

−

x

x= 0.42 T,


HL = HR = 130 At/m,

F L = F R = HL x L L = 130 x 0.4 = 52 At,

F = F R + F c + F g = 52 + 55 + 661 = 768 At,

I = N

F =

280

768= 2.74 A.

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Tutorial 3 Soln