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ch3sol

# ch3sol - Chapter 3 Problems Solutions 3.1 Based on the...

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Chapter 3 Problems Solutions 3.1 Based on the cross product of [3.2] with the vectors B and I separated by 30 ° , the mag- nitude of the force is ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 sin 30 0.25 2cos 5cos 0.5 F i t B t t t ϖ ϖ = ° = l ( 29 2 1.25cos 0.625 0.625cos 2 N F t t ϖ ϖ = = + The force F acts perpendicular to a plane that combines both B and I with direction de- termined as the direction that the thumb of the right hand points when the fingers are crossed from I to B . 3.2 The H -field is directed CCW at any particular radius r measured perpendicular to the centerline of the conductor. The current enclosed by a circular path of radius r is 2 2 2 2 4 4 r r Ir I I D D π π = = By use of [3.4], the magnitude of the H -field at any distance / 2 r D is 2 2 2 4 2 2 2 / r I Ir D Ir H r r D π π π = = = After substituting known values, ( 29 2 2 4 5 6.366 r r H D D π = = 3.3 The assumption of / 0 dB dt = used in derivation of [3.10] is no longer valid; thus, ( 29 d B x dB e B v x dt dt = = + l l l ( 29 1 2 2 sin cos e B B t v x B t ϖ ϖ ϖ = + + l l The first term of e is commonly called a speed voltage . The second term is known as a transformer voltage . 3.4 The MATLAB file cckt1.m is changed by use of an editor so that 2 4 0.10 m = = l l . After execution, the screen gives 19

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3.5 The solution is identical to Example 3.3 through the point of calculating the core mmf c F . Starting at that point and neglecting air gap fringing terms in [3.17], ( 29 ( 29 ( 29 7 6 3 4 10 0.05 0.05 1.571 10 H 0.002 o g d π μ δ - - × = = = × P l The required air gap mmf is 6 0.004 2546.1 A-t 1.571 10 T g g φ - = = = × F P The total coil current follows as 2826.7 2546.1 10.75 A 500 c g I N + + = = = F F Using the coil current of Example 3.3 as the correct value, 10.75 10.54 % Error 100% 1.99% 10.54 - = × = 3.6 ( a ) The leakage permeance P l of Example 3.4 is valid; however, the air gap permeance has changed in value. By [3.17], ( 29 3 3 0.52 0.308 g o d d μ δ δ = + + + P l l ( 29 ( 29 ( 29 ( 29 7 6 0.05 0.05 4 10 0.52 0.05 0.05 0.308 0.003 1.113 10 H 0.003 π - - = × + + + = × cr B and cr H of Example 3.4 are unchanged. Only the mean length path 3 m l need be re- calculated. 3 2 0.10 0.05 0.003 0.147 m m h δ = + - = + - = l l The mmf across the leakage flux path F l and the resulting leakage flux φ l follow as 2 4 3 2 2 m m cr m g H φ = + + + F P l l l l 6 0.20 0.20 0.004 4083.5 0.147 5010.9 A-t 2 2 1.113 10 - = + + + = × F l 20
( 29 ( 29 8 5010.9 9.425 10 0.472 mWb φ - = = × = F P l l l The coil flux is then 4 0.472 4.472 mWb T φ φ φ = + = + = l ( b ) The left-hand core flux density is ( 29 ( 29 ( 29 1 0.004472 1.883 T 0.05 0.05 0.95 T cl B dSF φ = = = l Either enter Fig. 3.12 or use hm27.m to find 10,131 A-t cl H = . The coil mmf follows by 2 4 1 2 2 m m cl m NI H = + + + F l l l l 0.20 0.20 10,131 0.15 5010.9 8556.7 2 2 NI = + + + = And the coil current is 8556.7 17.11 A 500 NI I N = = = Using the value of leakage flux from Example 3.4, the percentage increase in leakage flux due to the 0.001-in increase in air gap length is ( 29 0.472 0.364 100% 29.7% 0.364 - = 3.7 Based on [3.17] and [3.18] for the given dimensions, ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 50 50 1.04 0.616 1.04 50 50 0.616 2 2 1.0404 50 50 0.52 0.308 0.52 50 50 0.308 2 2 wd w d wd w d δ δ δ δ + + + + + + = = + + + + + + or the plane opposing face air gap has a 4.04 percent larger permeance than the equal area opposing face air gap.

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ch3sol - Chapter 3 Problems Solutions 3.1 Based on the...

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