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Bipolar induction torque in a circular disc Morton, Ralph Mackenzie 1933

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u IBRARY I I C A T . MO. k^L^m^M^Lk^ S • ~ i BIPOLAR IliDUCTIGF TORQUE . IE A CIRCULAR DISC Ralph IvIaeKenzie Morton. A Thesis submitted f o r the Degree of MASTER OF APPLIED SCIENCE i n the. Department of ELEC TRICAL E1GI1EERI1G The U n i v e r s i t y of B r i t i s h Columbia. JU?RIL 1.933 TABLE OF CGUTEITTS Page I n t r o d u c t i o n - 1 I n v e s t i g a t i o n of Eddy Current Path 2 C a l c u l a t i o n ' o f Torque 7 Appendix Index 13 B i p o l a r Induct Ion Torque i n a C i r c u l a r D i s c . I n order to compute the torque of an i n d u c t i o n instrument I t i s necessary to know the magnitude and d i r e c t i o n of the c u r r e n t s induced i n the moving element. I n developing the theory of the torque due to b i p o l a r i n d u c t i o n i n a f l a t c i r c u l a r d i s c , i t i s u s u a l l y assumed th a t the eddy c u r r e n t s f l o w around the i n d u c i n g magnet i n a s e r i e s of c o n c e n t r i c r i n g s . While t h i s assumption i s v a l i d , i n the case where the centre of the magnet c o i n c i d e s w i t h the centre of the d i s c , i t I s not so when the magnet i s l o c a t e d e c c e n t r i c a l l y w i t h respect to the d i s c a x i s ; and torque formulas based on such an assumption give r e s u l t s c o n s i d e r a b l y i n e r r o r , the e r r o r i n c r e a s i n g as the magnets approach the edge of the d i s c . Drysdale and J o i l e y s t a t e that the t h e o r e t i c a l torque i s about 22 0% of the measured torque, the r a t i o v a r y i n g w i t h the distance between the magnets and the d i s c edge. Drysdale and J o l l e y , E l e c t r i c a l Measuring Instruments, Y o l * 2, P. 168. Ernest Benn, L t d . , London. The object of t h i s paper i s to i n v e s t i g a t e the path of the eddy cu r r e n t s induced i n a f l a t c i r c u l a r d i s c , and t o develop a trip que formula based on the a c t u a l currenl d i s t r i b u t i o n . I n v e s t i g a t i o n of Eddy Current Path. Consider an a l t e r n a t i n g current f i e l d normal to an i n f i n i t e sheet of conducting m a t e r i a l as i n F i g . 1. The r e s i s t a n c e of an annular r i n g of mean r a d i u s R and width dR i s given by t where <5~ i s the s p e c i f i c r e s i s t a n c e of the m a t e r i a l and t, the t h i c k n e s s of the sheet. 'The conductance of the r i n g Is given by 2T1 <S~~ -7? Since the <f^/7?.y~. induced i n a l l r i n g s i s the same, the current f l o w i n g i n a r i n g of outer radius Rg and inner r a d i u s R ( i s 2'77<T ye • Tf, Mext consider a p a i r of equal a l t e r n a t i n g current f i e l d s o f opposite p o l a r i t y normal to an i n f i n i t e conducting sheet as shown i n F i g . 2 . Let P move i n the d i r e c t i o n of the r e s u l t a n t e l e c t r i c f i e l d , and l e t F he on the locus of P. Since PP i s the locus of P, the component of the e l e c t r i c f i e l d normal to PI? must be zero. That i s to say there can be no. r e s u l t a n t current c r o s s i n g the locus PP. The current c r o s s i n g PF induced by the pole at JL I o . = E t l o g R i The curren t c r o s s i n g FF induced by the pole at B ! = _ E_t_ l o g £2 ' j?77(f ^ IT The r e s u l t a n t current I a + 1^ must be zero l o g RL L O • % . .0 f_ m , • Next f i n d the equation of the loc u s o f P. Refer to F i g . 3• Take the o r i g i n at 0. 2 > ,2 2 2 2 2 R/ = (s +x; *y = s +2sx+x +y 2 g 2 p 2 2 Rg = (s-x) +y = a —2sx+x +y _ 2 _ 2 But [ V / R *• 2 ? , 2 o 2 Z, Z, 2 2 2 E (s -2sx+x +y ; = n (s +2sx+x +y ) o r 2 2 T 2 2. 2 . r 2 mc+n'1 I m n y * fx - m 2 *n , 2 1 2 m 2n 2 I , 2(m-n)J (m-n) 2 ^ _ 4 - This i s the equation of a f a m i l y of c i r c l e s of r a d i u s ...fti*1... having- t h e i r centres at - m-n 2 2 m -MI ^bi-n) y = 0 Equation ( l ) describes the path of the eddy cur r e n t s induced i n a t h i n i n f i n i t e sheet by a p a i r of equal a l t e r n a t i n g current poles l80° out of phase w i t h each other, A s e r i e s of these c i r c l e s together w i t h a sketch of the p h y s i c a l arrangement of the poles i s shown i n F i g . 4 and 5 . In order to describe the flow c i r c l e s i t i s con venient to,know the r e l a t i o n between s = the h a l f distance between the pole centres. R = the radius of any current path., •3, - the distance between the centre of the c i r c l e of r a d i u s R and the centre of the in d u c i n g pole. This r e l a t i o n i s obtained as f o l l o w s : s _ m^+n^ _ m+n . 2(m-n) 2 = n 2 . m-n s = m+n .2' m ~ 2s - X n = R - e .". e = . ( R - e ) ' 2s - R + <S - R + ,2 2 2 R - 2 e s - € = ; . 0 S = R 2 ~GC (2) 2 e =/s'2 + R 2 - S : . (3) I t .should be noted from (2) that 8. i s the parameter, clef i n i n g ; an e n t i r e family,,of current; flow c i r c l e s , , and t h a t i f the radius of one c i r c l e ;i's 'known"'together w i t h the distance between i t s centre and the centre of the pole, then S can be computed; and by s u b s t i t u t i n g t h i s .value - of S i n (j)» the value of R corresponding' to any value of can'be found, thus d e f i n i n g the e n t i r e f a m i l y of fl o w . l i n e s ' f o r . ' t h a t p a r t i c u l a r case, A s e r i e s of fl o w l i n e s drawn by the above method i s shown i n F i g . 6. Consider an a l t e r n a t i n g ; current pole l o c a t e d e c c e n t r i c a l l y w i t h respect to a conducting- d i s c . Let. R' = ra d i u s of d i s c t - •'•.••<£? - distance between centre of pole and centre of d i s c . Then the boundary of the di s c de f i n e s . one - f l o w l i n e , and by means of equations .(2) and (3) the', flow l i n e s w i t h i n the d i s c can be computed. • I n order to o b t a i n some.experimental check on the preceding work, a shallow c i r c u l a r trough of mercury was set up i n conjunction w i t h an al t e r n a t i n g , current magnet as shown i n F i g . 7* Two pointed e l e c t r o d e s made contact w i t h the mercury, and were connected to a v i b r a t i o n galvanometer tuned to the supply frequency, p r o v i s i o n being made f o r moving these contacts and p l o t t i n g t h e i r p o s i t i o n r e l a t i v e to the d i s c . The e q u i p o t e n t i a l l i n e s i n the mercury d i s c were found and p l o t t e d by means of t h i s apparatus. The method gave good r e s u l t s , because a very s l i g h t d e v i a t i o n o f one electrode from the e q u i p o t e n t i a l l i n e p a ssing through the second, e l e c t r o d e caused a lar g e d e f l e c t i o n of the galvano meter. A p l o t of the e q u i p o t e n t i a l l i n e s to f u l l s c a l e i s shown i n F i g . 6. The equation of the e q u i p o t e n t i a l l o c i i s the orthogonal t r a j e c t o r y of the o r i g i n a l f l o w l i n e equation ( l ) and i s gi v e n by . a , _ > ,2 , / 2 2 3c + i y +• o) = v s + b This defines a f a m i l y of c i r c l e s having t h e i r centres on the y- a x i s and passing through the centre of each of the p o l e s . By means o f equations (2) and (3) the y-axis was lo c a t e d f o r the value of R and used i n the mercury set up, and the s e r i e s of e q u i p o t e n t i a l and flow c i r c l e s was drawn. The e q u i p o t e n t i a l c i r c l e s described by the above method were found t o correspond very w e l l w i t h the e x p e r i -mental r e s u l t s , and c o n s t i t u t e d a check on the preceding theory. I n F i g . 6, the t h e o r e t i c a l e q u i p o t e n t i a l curves are drawn i n f u l l , w h i l e the experimental p o i n t s are shown by the small c i r c l e s . C a l c u l a t i o n of Torque, The preceding work i n d i c a t e s the path of the currents induced i n a, d i s c . The next step i s to develop torque r e l a t i o n s based on t h i s current d i s t r i b u t i o n . I n order to compute the torque produced by the flo w of curren t induced i n the d i s c by one magnet across the f i e l d of the second, magnet, i t i s necessary to know the magnitude and d i r e c t i o n of the c u r r e n t . The following data i s r e q u i r e d : Diameter of d i s c . Thickness of d i s c . S p e c i f i c r e s i s t a n c e of d i s c m a t e r i a l . Dimensions of magnets. Length of a i r gap. L o c a t i o n of poles. Magnitude and phase of f l u x i n each magnet. Supply frequency. Refer to F i g . 8. The f i r s t step i s to describe the flow l i n e c i r c l e s through g and h. Then a l l the current that flows under the pole B due to the Cmf* induced i n the d i s c by the pole A , must flow i n the crescent bounded by these two c i r c l e s . C o n s t r u c t i o n : Given. & - uv distance between centre . of d i s c and centre of pole A. R = radius; of d i s c . o Evaluate s by means of equation ( 2 ) . Draw the y-axis at a distance s from the centre of pole A. B i s e c t uh at K and draw kh normal to uh i n t e r - s e c t i n g the y-axis at b. J o i n bh and draw hb normal to hb and i n t e r s e c t i n g the x - a x i s at b . With centre b t and r a d i u s b h describe a c i r c l e . This c i r c l e i s the flow l i n e through the p o i n t h. By means of t h i s c o n s t r u c t i o n draw flow c i r c l e s through g and j , the d i a g o n a l l y opposite corner, and centre of the pole, r e s p e c t i v e l y . Let R, = r a d i u s of f l o w c i r c l e through g. Rg = r a d i u s of f l o w c i r c l e through h. & i distance between centre of R . c i r c l e and centre of pole A. d*2 = distance betT/een centre of Rg c i r c l e and centre of pole A. ' T = thickness of d i s c . 6~ •- s p e c i f i c r e s i s t a n c e microhms/cm^. / 2 2 = / S * R, ' - S ^ = / s 2 + R 2 2 _ S . A l l dimensions i n centimetres. ../Knowing the above s i x f a c t o r s i t i s p o s s i b l e to work out the r e s i s t a n c e of the r i n g . The c a l c u l a t i o n i s r a t h e r l o n g and'Is . gi v e n i n the appendix. The f i n a l , expression i s -6 Resistance ~ 277 0" -r 10 _ X : l o g • 2 1 ohms a e e, R The current, f l o w i n g i n the r i n g can now be computed. Let 0 j •. * f l u x i n magnet A, Tptf-S. - y~ - supply frequency.. .,- 'enif induced i n r i n g by (j> t •'It, - r e s i s t a n c e of r i n g surrounding magnet A.' I = current f l o w i n g i n r i n g due to E . . i t a (3 B ~ f l u x d e n s i t y o f magnet B. • Q = e l e c t r i c a l phase angle between L - mean l e n g t h of path of current under magnet B. • / o - g ^ > fk = ^ Z T l o g e ^ ^ ~ X 10 Then E '•«?•' 2 77 / <j> # R ^ . 2 77 6" / i s obtained f r o m . E i g . . a s f o l l o w s : -10* L e i . a ' , be the" centre of. the flow; c i r c l e through J . • (the centre of the pole; ..B t h r o u g h '^J draw ^ ^ i n t e r  s e c t i n g the e i ^ E R and Eg at d and s r e s p e c t i v e l y , Then L : may be taken as the e f f e c t i v e area/of the pole B divi d e d ; by: .de, V: . ;.r ( M / i ^ . : ^ ^ ; •-• The force, between the cu r r e n t induced' by /magnet A- and the : f i e l d of magnet->B> ;.ls"" " •'.' -v .' Since the f l o w l i n e s ; under the pole B are not p a r a l l e l the d i r e c t i o n of the fo r c e Var.Ies: over ^he pole . f a c e , f o r example, -the. force due to the .current element at .'the. p'oint.v- h)-; '-Would; be' .along hb' .and the; f o r c e due to . the curr e n t 'at ;.g- would, be along go',/ I t i s ' assumed that the r e s u l t a n t f o r c e a c t s along- Ja' , that; is. to say, along the normal to the flow l i n e passing under the centre;: of the pole.;' •;'.;."• : ; This, force :, can be r e s o l v e d .into two components, one a c t i n g through the a x i s of the d i s c , and the, .other t a n g e n t i a l t o the d i s c . The torque ,1s due to t h i s second •: . component. '.Let' ^ = angle . Ct'j.lTr.:; _ ; ; ; - = ' r a d i u s , arm :7/V~ ' •;.\'...;Then''torque.',-.^ ".'. F s i n ^ > . -11- • Since''an e q m l torque i s produced by the current I s f l o w i n g under, the f i e l d The t o t a l torque i s twice the above, and can be expressed by T = 0.2 x 2 1 1 f x / ^ B i c / f s i n & s i n (Jj 10* dyne cm. In order to check the preceding theory, an aluminum d i s c mounted on a v e r t i c a l a x i s In jewel bearings was;.'.fitted w i t h a; t o r s i o n head as shown i n F i g . 11.: The instrument was . c a r e f u l l y c a l i b r a t e d , and the torque developed by the d i s c under the i n f l u e n c e of two magnets connected to a two phase supply -was measured. The f l u x from each magnet was measured by means of search c o i l s wound on the poles and connected to an. AC potentiometer. The dimensions of the apparatus were as f o l l o w s : Diameter of d i s c 12*7 em, : .Thickness o f d i s c .IO55 cm. S p e c i f i c r e s i s t a n c e 2.83 microhms per cm. cube. . (aluminum) Dynamometer constant 18.5 dyne em. per degree. Pole dimensions 2.2 x 1.9 cm. Search c o i l 20 t u r n s . The r e s u l t s obtained w i t h t h i s apparatus are g i v e n i n the ..appendix. When ••'the.-space angle between the , poles was the c a l c u l a t e d torque v a r i e d from 63. to 937° of the measured torque, depending on the distance between the magnets and the edge of the d i s c . When the poles were spaced out to 90° the c a l c u l a t e d torque v a r i e d from 5.6.5 to 76.3fo of the measured torque. I t should be noted that a number of assumptions s i m p l i f i c a t i o n s have been made i n developing the pre ceding theory. I t i s d i f f i c u l t to f i n d the magnitude and d i r e c t i o n of the r e s u l t a n t force due to the current f l o w i n g under'the magnet, because of the v a r y i n g current d e n s i t y . The e f f e c t s of the d i s c leakage inductance on the flow contours has not been taken i n t o account. In the ease of a low r e s i s t a n c e d i s c , the d i s c reactance would decrease the magnitude and change the phase angle of the induced \ c u r r e n t s . The torque would a l s o be modified by the e f f e c t \ \ \ of f r i n g i n g i n the d r i v i n g magnets. \ v In view of the number of v a r i a b l e s and the com- \ \ p l e x i t y of the phenomena i n v o l v e d , i t i s f e l t that the t h e o r e t i c a l r e s u l t s are i n about as good agreement w i t h the observed q u a n t i t i e s as could be expected. Apparently very l i t t l e work has been done on the i s u b j e c t , and i t i s probable that f u r t h e r research would y • / r e s u l t i n the determination of an accurate expression f o r / the torque.. -13 ..APPENDIX. Table of r e s u l t s . P l o t t e d r e s u l t s . Worked, out Example. C a l c u l a t i o n of R e s i s t a n c e . Drawings. -14- Pole Angle , b5° Distance 'between edge of pole and Disc Edge. mm. Measured Torque Dyne Cm. Computed Torque Dyne Gnu % 2260 1420 63 • 6 3710 2840 .. ' 7b.5 12 4?60 4100 86.0 17 5300 4920 93*0 . -Pole '..Angle. 0 1680 950 3b.3 3080 1950 63. 12 4050 3020 • 74.3 17 4550 3440 76.5 - I d  l e s t ITo. 13. Angle between poles 6 5 0 . Distance between edge o f pole and edge of d i s c 6m.m. E.M.F. induced i n 20 t u r n search c o i l Pole A 1.220 V Pole B 1.370 ¥ Angle between E a and E D = = 8 2 . 5 0 s i n G = .991. Dynamometer readi n g = 2 0 1 ° . See page jj f o r instrument data. F i r s t a p l a n of the d i s c was drawn showing the two magnets i n t h e i r proper l o c a t i o n as i n F i g . 8. Then S was computed 2 2 • ' R D - e _ 6.35 - 4.63 S = _o - = 2 . 0 1 . 2 2 x 4.55 By means of the c o n s t r u c t i o n given on page flow c i r c l e s were described through the p o i n t s g, h, and j Measure R =. 2.92 cm. A R = 5 .50 cm. 3 e » / s 2 + R 2 - S / " 2 2 / 2.01 + 2 , ! <£> = / 2.01 -i- 2.92 - 2.01 = I . 3 3 JL I 2 2 <f= B = / 2.01 +'5.50 - 2 . 0 1 = 3 . 8 7 . Resistance of r i n g ~ -AMI X - — 5 = — _ e R T 2.303 l o g . . _ J _ A . • 1 0 ^ l H B 2 77 x 2.83 x _ I = 574' microhms. .105 l o g 1.34 - l b - Current f l o w i n g i n r i n g 1.220 - 10 l i n ~ , -±u_ x . ss 10.62 amps. 20 574 F l u x d e n s i t y i n magnet: B /3 % _ , V x - i B 2 pole area 1 «370 K _ i i L = 432 l i n e s / cm. 40 x 60 4 e 2 The f o l l o w i n g q u a n t i t i e s are measured from F i g . 8, ^ajw » ^ = 2 5 . 5 ° s i n ^  = .430 de = 2.7 em. Mean l e n g t h of path under pole = - f t 0 ^ ^ r.e— = 4 * 2 = I . 5 6 cm, 2.7 = Radius.arm = 4 .65 cm. Torque *• 0,2 x 10*62 x 452 ±'1.56-3: 4 .65 x .991 x .430 = 2850 dyne cm. Measured torque = 2 01 x 18.5 = 3710 dyne cm. Computed', torque- as per cent of measured torque - 76.5?° • -17- : C a l c u l a t i o n o f Ring. Resistance. Refer to F i g . 9 . Let Rj = r a d i u s of outer boundary of r i n g . R 0 = r a d i u s of innter.boundary of r i n g . •^y.;^" d i s t a n c e between'magnet centre and centre of c i r c l e R . • <?2 = distance betweeh magnet centre and centre of c i r c l e R •• ^'f.-'• \•>"''-di^ .'- ='•;' s p e c i f i c r e s i s t a n c e 'of d i s c . m a t e r i a l . :1 - .-thiclcness of disc'. , •.;; Then providedXthat R^ , i s only s l i g h t l y greater than Rg, the r i n g r e s i s t a n c e i s given by m I 2 2 l / 2 ' (a +R. +2aRcos 6) ' - R 0 ' ' 'This i n t e g r a l i s evaluated as f o l l o w s : T / Let I = JJA^zl f*j*Z^*Wx™*±J>* 1 T / 2 2 2 ^ 0 a -i-Rj - R +2aR( cos © I - ^ R , <-R2)R2 /* ^  d© 2 5 / ~ g 2 ~ / a. +R( -R2 -;-2aRj cos O E v a l u a t i o n of I ( Let U - Then s i n 0 = e = 77 tan ;"• © •'- • 2 :. • • 2U 1 - 0 u = 0 d0 2 du 1 + (/ S 2 ^ 1 A OT^glj2 2du 1 * U 2du / r • -2 2 ) 2 >/• 2 2 w 0 |(a-R, } - R 2 J yfc | (a+R, ) -Rg |/l+U Let U t a n f When U = 0 ^ = 0 du = a * R, 2 a - R. sec 17. 2d" (R ^ ) | (a+R, ) see f ^ > seeZ<f J<f h*>f-\M) * ~ V > z-4 sec d(tan (y ) t a n -19- 2^ = (a+R,)2 - R - _- 2-£" x U ^ R 2 ) ( a € , / 7 7 (T ^ > R 2 ) < a + R i > E v a l u a t i o n of I j = <TR2 (R( +R2) / - . 2 1 J a 2 + R,* - ^ 2 2 + 2aRy cos 0 o L e t U = tan 9^ ^ 1-C/ 2 * >, 2a o Then ' cos 0 = -—~-—5- d y ~ 1 *6> 2 1 v(J 2 I = 2 <TR2 • • s /" 2du T I 2 2 _ 2 . 2 . _ / i - t ^ 2 ^ »V I f / 20 2 7 ( ( a ^ j ^ j u ' ^ a ^ ) 2 - ^ 2 } 2 ^ R 2 ( V - R 2 ) f.ar,-V/ ( a" B« ) 2" R2 2 7? ( f R 2(R +R2) . , '. 1 « I + I • s : ' 7 7 < T(R ftR 2) ( R t ^ R 2 t a )  The above expression gives the r e s i s t a n c e of a crescent provided, that R, i s o n l y s l i g h t l y greater than R^» In order to f i n d the r e s i s t a n c e of a t h i c k crescent such as i n F i g . 1 0 , i t i s necessary to change the above expression i n t o a conductance, and i n t e g r a t e the conductances between the outer and inner r a d i i of the t h i c k crescent. This i s done as f o l l o w s : l e t c = conductance = —™- R 2 = R R } = R + dR -21-- 2 2 - = 2<f77 s e 1 - b a l o g 7/<f . e e A a b 2 77 <T Resistance — — x — — — 2.303 l o g 10 Fig 4. 

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