I III IIIIII I — .— — __’___ — — ,— ___——'__—;. — — I I III I IU'IN _mhm SOME. PROBLEMS IN THE DESIGN AND OPERATION OF ELECTRICAL MACHINERY AND THEIR SOLUTION TIIGQIS for the Degree of E‘ E. George T. Smith I936 VIII: .; III 3‘ x .IfiI‘ I' '- .)."v-""l. :'. k‘fIQ -< . n {:4 3;- ,\ _ 2“?" O “‘13: ijla‘. 3:“ 3f. "33; I” . ‘4‘ o y:"I‘- l ‘I I “II“ R‘rj-‘I' . d-K‘k. . Q; :n‘\' I. “III . . I f ‘I-‘ .' V , J‘Ix‘ \Q}I ;"t‘*.IOV ’hI“- g. h 4%R :I;|I 7‘ fi‘n" .I L... ,‘l .I ‘_ r IL,“ “ . . ;‘. I. 2‘ - 7%; )Igzt? \;" .‘.'.\1.fl.‘-)‘f IPI‘ZJII A J V “W” I W" .=-'- 96”: ' a» 24‘ far“; we: #33“ 31' av ' V C . \". .V- . 0.!,.‘¢_' "A. ‘."~- '5. o OiIIII :6, 'fl!’ $1.5 !.':I“ ”I ‘.II " H: u; «3’. “fl”! ”I‘Q-xg‘. sibk II‘ II J"- f.‘(I/ I .IA‘I.‘I‘ I ' ”1.“! " I.‘ “ J I.’ ":IUI. n J ’1' } ‘2‘“:‘7: ‘5’ far... 5‘“ V ’ .1 :10“ fig ‘ '?_ ‘5‘ “| . q n ‘ u " ‘. . .A~ '.' . ..I~I I '1" t .- u k, . VS” ( .- 2&9. : : . ~ -_ A. o Ii'xI'I-‘I (-o‘ E‘ ‘I ,3. Afi-I‘ 'V N)! I}. ‘ 'r ,l ’VI‘I‘S kvg ., . ,‘ . ‘I..' > . 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V ’ I I ‘ v' . o r '. .7: ‘, I: J p 5?." 1;: {Ar 9% '1' i 5 s‘ 2“ ._ .‘L‘ 3., .. . m . ‘5! . in “f 5: .- '3:- F c P O SOME PROBLEMS IN THE DESIGN AND OPERATION OF ELECTRICAL MACHINERY AND THEIR SOLUTION C V I A THESIS PRESENTED TO THE FACULTY OF MICHIGAN STATE COLLEGE OF AGRICULTURE AND APPLIED SCIENCE BY \,d9fi GEORGE TQISMITH h A CANDIDATE FOR THE PROFESSIONAL DEGREE OF ELECTRICAL ENGINEER JUNE, 1956 THESIS Part I DERIVATION OF A GENERAL "EMF-FLUX" EQUATION FOR ELECTRICAL MACHINES In part one of this thesis I wish to show how a very general:form of the equation, showing the:relation between the magnetic flux in.maxwells and the generated volts in electrical machines can be derived for use in practical design work; and also to show how,by simple and logical changes, the general equation can be adap- ted to several different types of electrical machines. All the usual types of rotating electrical machines have alternate north and south magnetic poles distributed evenly around their air gaps. The width of the field of influence of all the poles is usually equal, and the ave- rage magnetic strength of the north poles is equal to the average magnetic strength of the south poles but, of course, opposite in polarity. The distance between two successive magnetic poles is called the pole pitch. This pole pitch can be expressed in electrical degrees, the distance between the center of one magnetic pole and the center of the next succeeding magnetic pole being 180 electrical degrees, and the distance between.any point on a magnetic pole of given polarity and a correspond- ing point on the next succeeding magnetic pole of the same polarity being equal to 360 electrical degrees. (On a 2 pole machine a mechanical degree equals one electrical degree; on a 4 pole machine 1 mechanical de- gree equals 2 electrical degrees; onm.6 pole machine 1 mechanical degree equals 5 electrical degree§ etc.) I (\rn‘fiyg‘ ' (in: ) j-‘)’\).}r’;a (2) The pole pitch may also be measured as a circum- ferential distance in inches if it is indicated what circumference is intended; that is to say, whether on the surface of the rotor, the inside surface of the stationary member or stator or the mean circumference in the air gap. Inches circumferential distance will usually be used here except when the term pole pitch is used in a general way. 5 The useful flux per pole of an electrical machine is that magnetic flux which leaving that member of an electrical machine where the magneto-motive force exists, would pass entirely within a one turn full pitch coil directly on the surface of the other member of the machine. A full pitch coil, for any cylindrical rotor type of machine, is a coil whose parallel sides lie just one pole pitch apart. Now to derive the general "voltage-flux" equation first consider a single turn full pitch coil rotated at a uniform rate in the magnetic field of a.maohine as shown in diagram.l. (The diagram is a general repre- sentation of a two pole machine and a similar diagram could be made for any even number of poles.) See blue print, next page. The space between the two large circles in diagram 1 represents the airgap of the machine. The two small circles a-l and b-l represent the two sides of a one turn full pitch coil. The magnetic field is represented vertically in the iron parts of the machine and radially in the air gap. In the position of the (5) coil shown at avl, b-l, the plane of the coil is in the axis of the magnetic field that is parallel to the direction of the field. It is plain that for1shis posi- tion, namely with the sides of the one turn full pitch coil at the centers of the magnetic poles, there is zero magnetic flux passing through the coil. Now if either the magnetic field or the coil are rotated so that the relative position of the coilvvith respect to the field is as shown at a-2 and b-2 there will be magnetic flux passing through the coil, that is the amount of magnetic flux within the one turn full pitch coil has changed. Therefore a voltage hasleen generated in the coil. The direction of the voltage, of course, is such that if a current were allowed to flow in response to this voltage, the magnetomotive force of this current would tend to oppose the change of flux taking place. Now if the flux density in the air gap is uniform between the positions a-l, b-l and a-2, b-2 it is evident that the total change of flux hithin the coil will be proportional to the amount of movement between these positions. Hence, if the rotation is.uniform the rate of change, and therefore the voltage generated, will be proportional to the density ofizhe magnetic field. That is to say for uniform rotation the voltage generated is proportional to the density of the magnetic field through which the sides of the coil are passing. This is so not because the sides of the .coil are "cutting lines of magnetic force", but because the rate at which the magnetic flux within the coil is (4) changing is proportional to the density of the magnetic field through which the sides of the coil are passing. Electrical engineers speak of "field forms" by which is meant the way the magnetic field of an electri- cal machine varies across the circumferential space occupied by a magnetic pole, or a curve plotted to show the variation of the magnetic density over this space occupied by a magnetic pole. The density of the mag- netic field is never constant over a whole pole pitch in rotating electrical machines, that is tossay the "field form" is never rectangular, Neither is the field ever a perfect triangle or a perfect trapezoid. In some specially constructed synchronous machines the no load field form may have the shape of the sine curve. In polyphase induction machines the average field form set up by the currents in the primary wind- ings during a cycle comes so close to a "sine field" that no error great enough to effect the practical design of induction machines is made by assuming that all the constants of a "sine field form" apply to polyphase induction machines. In the usual direct current machine or salient pole synchronous machine the density of the no load field may be constant or nearly so over the central part of a pole piece but it decreases, first slowly then.more rapidly as we pass from the center of the pole piece towards the edges or "tips" of the pole piece, and the magnetic density drops off rapidly beyond the pole tips and passes through zero to the reverse polarity midway (5) between the main poles. However, regardless of the "field form" it is still true that the voltage generated in the coil in diagram.l, by uniform rotation, is at every instant proportional to the magnetic density of the field through which the sides of the coil are passing at that instant. The logic of this statement will be apparent if a smaller and smaller movement is consi- dered at a time until an infinitely small element of the motion is considered; for the magnetic density of any field form can be considered as constant for an infinitely small element of the rotation, which elementary portion of the rotation will take place in infinitely short time so that the rate of change of flux through the coil and the voltage which it generates in the coil are at each instant proportional to the density of the magnetic field through which the sides of the coil are passing at the given instant. From the above reasoning it is evident that with any single full pitch coil in a rotating electrical machine the voltage wave will have the same shape as the shape of the'field form. (More is to be said about field form later.) Then if the field form is sine shaped the voltage wave generated by a relative rotation of'such.a.field and a full pitch coil will be a sine wave. and since the resultant of a number of equally spaced sine curves is also a sine curve, it.£ollows that the resultant voltage wave generated by a group of equally spaced (6) full pitch coils will also be a sine voltage wave. However, most electrical machines have coils which are less than full pitch, that is, the two parallel sides of the coil lie a little less than oné pole pitch apart. It will be found that the voltage generated at any instant in such a coil is prOportional to the sum.of the two magnetic densities through which the two sides of the coil are passing at the given instant. The easiest way to understand that this is so is to imagine the coil as consisting of two coils each having one imaginary side of infinitely small dimensions passing back through the center of the shaft or axis of rotation. In this case these two imaginary halves of the coil are out of the magnetic field and it is evident that the voltage generated in each half of the coil by uniform rotation will be proportional to the density of the magnetic field through which the outer or real sides of the coil are passing, at each instant, and if the two imaginary halves of the coil were properly connected in series the total voltage at each instant would be proportional to the sum.of the magnetic densities through which the two real sides of the coil were passing at the given instant. To carry this reasoning still farther it is evi- dent that the two voltages generated in the two imaginary halves of the "short pitched" coil will be out of phase by the amount by which the short pitched coil falls short of being a full pitch coil. So that the voltage generated ina coil which is less than full pitch can (7) be considered as the vector sum of two voltage waves which are out of phase by the same amount which the coil falls short of being a full pitch coil. The amplitudesof these two voltage waves generated by the two sides of the coil are, of course, one half the amplitude of the voltage wave whichxnould be generated by the complete turns of a full pitch coil having the same number of turns. Since a full pitchczoil could also be separated into two imaginary halves in the same manner as the above mentioned coil, which was less than.full pitch, it is evident that the voltage of a full pitch.coil can be considered as the sum of the two equal voltage waves generated by the two sides of the coil. These considerations show why it is sometimes more convenient for a designer of electrical machines to consider the voltage generated in each side, or conductor, of a single turn of a coil rather than to consider the voltage generated in a complete turn. The above reasoning also shows that all that is necessary to obtain the voltage wave generated in a group of coils (such as a phase group of coils) in any winding, is to plot a voltage wave each ordinate of which is proportional to the sum of the magnetic densities through which each conductor of the group of coils is passing at the instant which that ordinate of the voltage wave is to represent. This is the simplest and most direct way of obtaining the voltage wave generated in any given winding when acted on by the field forms found in commercial electrical machines and it (8) is the one used by designing engineers in originally determining their design constants for the various types of windings and for various "field forms." Before proceeding with the derivation it is best to choose symbols for and tociefine the various factors which will enter into the "EMF-flux" equation; hence let: El== The volts to be generated per phase or per winding. ¢= The useful flux per pole in maxwells. Cp‘; The ratio of theaverage magnetic density to the maximum magnetic density of the magnetic field.form. Cwu: Ratio of the effective volts per conductor of the given winding, to the maximum volts which would be generated by a single conductor ro- tated in the same field. Number of effective series conductors per phase or perwninding. t: H u f == Line or primary frequency in cycles per second. Sin a1=Sine of one half the pitch angle of the g coils in electrical degrees, that is, the sine of one half the angle spanned by the coils in electrical degrees, one pole pitch being 180 electrical degrees. R.P.S;Revolutions per second of relative motion between the winding being considered and the magnetic field. P = The number of poles. T1 fl Number of effective series turns per phase or per winding: Then N1 =3 2 X Tl Tc.= Turns in one coil No a Number of effective series conductors in one coil = 2 Tc Equations will be identifiedtay numbers in a circle following the equations thus (:>, (:) etc.) (9) Before proceeding with the derivation of the EMF-flux equation it should be remembered that the voltage generated in all the usual types ofealectrical machines, even in the direct current machine, is alter- nating in its nature and that the voltage of the usual D.C. machine is uniform and constant in direction only because the voltage of each coil is rectified by the commutator and the instantaneous voltage of all coils between positive and negative brushes is summed up by the con- nection to the commutator andlarushes. So that the voltage in a single full pitch coil of any of the usual types of electrical machines can be considered as alternating in its nature and passing through one cycle when the coil or field moves over a space of two pole pitches. Now all of the useful flux from a magnet pole passes through a full pitch coil When its sides are exactly midway between the magnetic poles, that is when the center of the coil is exactly in line with the center of a magnetic pole. Ninety electrical degrees later when the sides of the coil are exactly on the radial line passing through the center of the magnetic poles, that is when the center of the coil is exactly midway between magnetic poles the flux passing through the coil will be zero. This movement corresponds to one half a pole' ‘pitch or one fourth a cycle, that is four times that Inuch movement is required for one cycle. Therefore the average voltage generated in“a single full pitch coil “— (10) will Be: x .—x Ave. volts per coil= Ml— Q.) But since fic= 2Tc3 Tc 3 (:) Ave. volts per coil.— QA’N2/iX/Vg @ This, for a single full pitch coil, will be the average of a voltage wave which has the same shape as the field form for, as previously explained, the voltage wave shape generated by a single:full pitch coil is the same shape as the field form. Diagram 2 below shows the"field form" or distribu- tion of magnetic density in the airggap over one pole pitch of annual type of electric machine such as a sal- ient pole synchronous machine. See blue print, next page. . As previously defined, the field form factor or Cp factor of this field form.will be the ratio of its average density to its maximum density. For a practi- cal field form which is neither a sine curve or a straight sided geometrical figure this Cp constant is usually found by taking the ratio of the average of a large number of ordinates to the maximum ordinate. The CD factor forishis field form has been so determined and is equal to .67. Since as previously explained the voltage wave generated by a single full pitch coil when rotated rela- tive to such a magnetic field form will have the same shape as the field form it is evident that tOfpt the ‘maximum.value of such a vBItage wave the average value o 0 v ‘ 0' (ll) of thexroltage wave should be divided by the field form factor CD; then since fox—W @ Maximum.volts for a si gle full pitch coil 1.: 5"ng A/g_ @ / - " ()9 Now having this formula for the maximum volts fix 2 Ave. volts per coil: generated in a single full pitch coil all that is nec- essary to obtain the effective or "root mean square" value of the voltage generated in this full pitch coil is to multiply the above formula by the CW factor. 866 previous definition of Cw° We have then: ‘Effective volts for one full pitch coil jéXJZIQ‘IAQ"Cu' (:) /o.'x Cr For a single full pitch coil Cw will be the same as the ratio of the root mean square value of the field form.to the maximum value of the field form. It is evi- dent that this must be so since the voltage wave for such a coil has the same shape as the field form itself. If there are several equally spaced full pitch coils connected together in series, as for example the phase group for one pole of a polyphase I. C. winding, the voltage wave generated in each individual coil will have the same shape as the field form, but the resultant voltage generated in the whole group of several coils in series will be the vector sum of several such voltage waves one for each coil and displaced in phase by a number of electrical degrees equal to the spacing of the coils from each other. (12) It is evident that the Cw factor for a windingxf several equally spaced coils will not be the same as for a single full pitch coil and will not be equal to the ratio of the "root mean square" value of the field form to the maximum value of the field form. The method of working out the voltage waveggenerated in a phase group of four equally spaced coils connected in series as well as the method of working out the Cw factor for such a winding when acted on by the field form shown in diagram 2 will now be given. Taking a case of equally spaced coils with 12 coils in a pole pitch and considering the voltage generated in 4 of these coils in series (which would be like a polar phase group of a three phase winding), the voltage wave shape generated in the group of coils when acted on by the field form shown in diagram 2 may be determined as follows: The voltage wave that is generated in each individual coil has the same shapeas the field form and since the four coils are in series and spaced one twelveth of a pole pitch or 15, electrical apart, the resultant voltage wave will be the sum of four voltage waves of the same shape as the field form and which are out of phase with each other by 15 electrical degrees. This is shown in diagram 3 below. 866 blue print, next page. NOW to find the Cu factor forishese four equally spaced coils in this particular field form, this resul- tant voltage wave is simply spaced off equally into a considerable number of division, say 48 divisions, _ - 1.3-: 1L ,_ ” (15) the ordinate of the curve is read at the center of each division, these ordinates are squared and‘then the average of the squared values found. The square root of the average of“the squared values is found, which is evidently the "root mean square" or effectivexralue of this voltage wave. Then dividing this effective value of the voltage wave by the number of effective conductors in series in the whole group of coils, the effective volts per conductor of the winding is obtained. Then the ratio of this value of effective volts per conductor of the winding to the maximum volts which would be generated in one conductor alone will be,by definition,the Cw factor forthese:four equally spaced coils covering one third of a pole pitchvvhen acted on by this field form. ((In the graphical method of solution shown in diagram 3 the ordinates of the voltage wave are only relative values and do not representzictual volts; also for the purpose of*working out this Cw factor it is Just as well to assume one turn coils, that is one conductor in each coil side. When this is done the value of Cw is the same as the ratio of the root mean square of the resultant voltage divided by the number of coils to the maximum of one of the separate voltage waves generated in each coil, the coils being full pitch coils. The Cw factor for this particular field form, (diagram 1) has been.worked out in just this way for a winding of equally spaced coils covering one third of a pole pitch and this Cw\factor is equal to .710. (14) Now if we call the number of coils per pole in any one winding the "polar group of coils" or "polar phase . group of coils" and let: Eg“ the effective or root mean square voltage generated in one polar group of coils and, Ng:: the number of conductors in series in one polar group of coils It is evident that by working out the CW factor in the manner previously described for the particular "polar group of coils", that is foreavenly spaced coils covering the percentage of the pole pitch which the given "polar group of coils" covers, that the formula for E3 follows directly: rom the.formula for effective volts for one full pitch coil by simply substituting Ng for No and using the correct value of Cw: that is: Effective volts for one full pitch coil ¢x fo At/QLC'V /Zfl; (2p (:) and ‘£?=r 95X'2qut/fiéx‘ch’ I ’2ghr é; (:) In the first case Cw is of course for a single full pitch coil in the given field form and in the second case CW must be the correct one for a winding of evenly spaced coils covering the percentage of the pole pitch covered by this particular "polar group of coils." That is the later equation above becomes general for a polar group of full pitch coils as long as the proper value of Cw is used. (15) Now if E1 = the effective or root mean square volts for the winding or per phase winding as the case may be, and N1 = the effective series conductors per winding or per phase winding: it is evident that for a full pitchwinding all we need to do to get the total effective voltage for all the polar groups which are in series in a winding or one phase winding is to substitue these values of E1 and N1 instead of Eg and N in the above formula 8 for E3 and we have 1E3: igjxikf;94{3r<7uf I [Q7’X'tao (:> This is the general equation for“the effective A. C. Voltage generated in a winding or phase winging except that it is still for a winding having full pitch coils only. It has been demonstrated many times that When a winding is made up of equally spaced coils each of which spansless than a pole pitch, that is of coils which are less than full pitch, the resultant effective or root mean square voltage decreases directly as the sine of one half the pitch angle of "span" of the coils expressed in electrical defrees. This is true for any number of coils covering any fractional part of one pole pitch and for any ordinary field form which is found in electrical machines. (16) Now if o(=:the pitch angle of the coils in elec- trical degrees, a full pitch coil spana ning 180 electrical degrees and, Sine-g- : the pitch factor or "chord factor" of the 2' winding, which for a full pitch coil = 1. Since the effective voltage varies as the sine of one half the pitch angle of the coils of a winding the general formula for the i. C. voltage generated in a winding becomes: E1: ¢x2fx&/i)(/:wa4— S/fiQ—‘t Usually El is known from the circuit with Which the machine is to operate, and fly is assumed by the designer to suit the rating of the machine which he is designing. 1"or this reason, it is more convenient for the designer to solve this general A. C. voltage equa- tion for the useful flux per pole,¢ , instead of the voltage per winding or per phase E1. When so transposed the equation 12:3): [0’ x-a- - 2 x/V, xfx 57/599 5-,, (9) This is the general equation for useful flux per pole which will apply to all the usual types of A. C. machines,'both induction and synchronous. In this general "flux" equation E1 and f are deter- mined from the given characteristics of the electric circuit with which the machine is to operate. N1 and sinegg are determined by the designer to suit the rating ofishe machine he is designing. The usual method of determin- ing Cw has already been described and it is a long pro- 0933. (1'7) Thevvriter has, however, developed a very quick method of determining Cw and the ratio Cp/Cw directly from CD. For’all the field forms found in ordinary electrical machines this short method.has been found to give results which are quite as accurate as the longer method. This quick method will be described in the following discussion of field forms and field form factors. Uiagram 4 shows the curves of the values of the Cw factor for various geometrically shaped field forms, a sinusoidal field form and one taken:from the design of an actual alternator. These Cw factors were all worked out by the long method previously described and are plotted against the percentage of one pole pitch covered by a winding of equally spaced coils. The value of Cp which applies to each field form is given along the curve which applies to that particular field form. See blue print, next page. In diagram 4 the calculated points for the*various field forms are shown on the curves by small circles or squares around these points. It should be noted that the.curves when drawn through these calculated points are smooth and consistent and that for increasing values of C the value of Cw increases for any given part P of the pole pitch covered by the winding. Note also 110w closely the curve of Cw for thezictual alternator ifield.form follows theczurve of Cw for the 1/3 flat 130;; field. It was consistency of this data as well as the C 1/}? 1/4" .9557 9/7542 MM fame; o a .0 . h I -— o I 0 I l oao-Q—o.- I H--. 5 . . t ; 1 O I I O A i 9. o . I o - 4.- -----t . I o .’ ao< 0 0/ 02] . or /‘0 ‘ JDART OF A POLE 529/ 727/ cat/£72203 éa/ mag/:fié 7cm 00/45 or 7725‘ MwM¢ a? (3‘71. /_,. 131/ twmw/V; Jar/V (18) results of many actual designs which led thevvriter to devise the following curves forcietermining the ratio of Cp/Dw directly from theiralue of Cp as calculated from the field form. This set of curves oft:he1?atio Cp/Cw plotted against Cp is shown on curve sheet 5 and is obtained by reading across the curves of curve sheet 4 on various vertical ordinates. Any given ver- tical ordinate on curve sheet 4, or course, corresponds to aggiven distribution of the winding; foreaxample the .333 ordinate represents the distribution of winding corresponding to the usual type of three phase winding where the polar phase group of coils is distributed over 1/3 of a pole pitch. Ihis grouping of the coils is often called "60 degree grouping" since the group of coils covers 60 electrical degrees. Having curve sheet 5 fortietermining theI'atio Cp/Cw directly from Cp, the designer is in a position to use theggeneral equations (8) and (9) as soon as the shape of the field form is determined and its Cp factor is figured. The usual procedure is to assume a value of useful flux per pole ¢ which will be very near what will be required for theggiven size and rating of the machine, then to solve the general equation (9) for the effective series conductors per winding or per phase, N1; then choose an actual number of series conductors N1 as near to this required value as possible considering the number of slots etc. The general equation (9) is then used again to determine the final value of useful r-. 1 i ;, ‘I 5 . ’, _--‘_.. I. 0 o . .-. .— . .. . ..1 §‘ . . 0 "V ., . . -1. . - .-. ._.- ~ . . i . L1. . . . .. o o -4.._. o o i . . _. .. . - . . -—-_+.- t , . 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C II I l —..-- - O O . a I b 1 l. U . . A .. .. . . .. .-.- .- ... . .. .. . _ . . ‘A- n _ . . ._. .. ., r .. . . . . . .. . ,-.-.. l. . _, o. . l -A- l‘.. .. .. ‘ o . n i . . . . , _. . - .. . . .._ it I..- . ‘. , f _. .-. ~~—.- , -.. -.. V 7" o _ . . . P. . c... t ... . -- . t » —-—- ... .. I n ‘ §.. . . . a.. .. -, o ., , O n v I O. .» . .. . . .. - .. .. , . . . -. , . ..... .-.- ‘ I ¢.. . - .. - - ... .,-. .- . .. . - V ~ . . . . . . .. o ‘3 . o. v 1 - -_4-. D . . O . . i . v w. . .4 I - O A ,. O . . O ,.. . - 3 . _ .1. no.) - . ‘_. .. . .- . . . a .o . .' . -. . . . . a. L.-- 4‘ .. . . . . l 1 y - .. .o- 1. . ~ .. on ,. . 9 . . ¢ - . . . . . M- -. . . . . . . _ .. O . v - . - _ _.-A . ,- .. O .’ - O -4 t C . .. l . O r- . . o . .. . (19) flux per pole,¢. If the winding isto be less than full pitch the quantity sinegir will be less than unity and thisITactor must also be taken into account in making these solutions. The general formulas (8) and (9) are in the proper form for use in designing the usual forms of:3alient pole synchronous machine. ”or the polyphase induction motor the field form constans from curve sheet 4 fore: sine shape field can be used, since the average shape of the field form set up by the windings of these machines comes very close to a sine field. Substituting the value CP==.636 for a sine field in equation (8) and (9) for the polyphase induction motor there results: __ ¢x2fXM xC‘wa/liid 5." ' /0? x .636 or ¢_ E,’ x /0-{ .636 ' "ZR M xiv/fig“ '7»;- ® These equations (10) and (11) can be used in this form for the polyphase induction motor reading the value of Cw on the curve for the sine field on curve sheet 4 and for 60°, 900, or 1200 winding distribution depending on what kind of a polyphase winding is being used. However, these equations can be further simplified by introducing what is usually called theciistribution factor. The distribution factog is usually represented.by Cd and is equal to the ratio of the effective volts (20) for aggiven number oftsurns used in theciistributed winding to the effective volts in a concentrated winding of the same number of turns. This is evidently the same thing as the ration of Cw for the distributed windingt:o Cw for a concentrated or single coil winding in the same field, and since the value of cw for a concentrated winding in a sine field is equal to .707 (see curve sheet (4) there results for the sine field form: 0 CW = ah .707 or Cw .707 x Ga GE? Dubstituting this value of Cw in equations a it becomes: For the polyphase induction motor: 3 ._ x2 M’xC' i n70? .® 1" II. .x .636 It will be noticed that the iggg in this formula is equal to 1.11, which is commonly called the "form factor of the sine field? It is evident that this should be the form factor since .707 is equal to the ratio of the effec- tive volts to the maximum volts for the case of a concen- trated winding in a sine field, while .636 is the ratio of the average volts to the maximum volts with the same winding in a sine field. Combining the numerical factor 2 with the numerical factor 1.11 the equation for the polyphase induction motor becomes: __¢x2-22.fx4,/x C/xJ/lt-‘g_ . 3? ° = ’4‘”? e 2.22:: fox 01x J/MEQEL (21) “ow in the polyphase induction motor since g6 is the total useful flux per pole it is evident that it is the maximum value of flux wlich would pass through a full pitch coil at the instant when the axis or center line of that coil coincided with the center of a pole of the rotating magnetic field. For this reason it can be compared directly with the maximum value of the useful flux of a transformer, and since in the usual type of transformer the winding is always concentrated and there is no such thing as a winding less than a full pitch winding; the values of Cd and Sineg can both be said to equal one in this case. Hence for the transformer the equations become: Ez7¢xfifixfx7i / /0§' and ¢= Eat/0., 7W2: 77xf Which is the usual form of the "EMF-flux? equation for the transformer. The voltage ratio or transformer ratio between the secondary and the primary windings of a polyphase induc- tion motor can be easily derived from a consideration of equation (10), which is an expression for the voltage per primary winding in terms of flux per pole, conductors per winding, etc. The voltage ratio or transformation ratio of a polyphase induction motor is usually called.7r: and is equal to the ratio of the standstill open circuit secon- (22) dary voltage per phase winding to the primary voltage per phase winding. If we let E2= this open circuit secondary voltage per phase winding then 7.; a If we indicate otherssecondary quantities by a sub 2 and primary quantities by a sub 1 and remember that at standstill the frequency in the secondary of the motor is the same as the primary frequency, that is the rotating magnetic field is then acting on the secondary at synchronous speed, it is evident from equation (10) that: E" Qx2§XAgXCgiflligf , 2' /0. x .536 @ and that: fit 72.52.: Jae" C'zu’fli . ® ' d‘ 1:. /GIX1C%7"JZ“Fji; This is the usual form of the transformation ratio or voltage ratio for the polyphase slip-ring rotor induction motor in which N2 would be the effective series conductors in a secondary phase winding,cqb the pitch angle of theesecondary coils in electrical degrees, and ng the cW factor for the secondary wind- ing, which is usually equal to .675 since the secon- daries of slip ring motors usually have three phase windings with each phase winding distributed over one third of a pole pitch, (see curve sheet 4). The quanti- ties in the denominator of this equation are primary winding quantities as previously explained and Cwl is taken from curve sheet 4 for the sine field form and for .333, .5 or .666 part of a pole pitch covered by (25) the phase winding, depending on whether the primary winding is the usual three phase, two phase, or a "consequent pole" connected three phase winding. Once the value ofTis determined the primary volts per phase multiplied by?p gives the open circuit standstill secondary volts per phase. For the polyphase induction motor with a squirrel cage secondary or rotor, H2==l since each rotor bar is considered as a separate phase and hence has but one conductor. For the squirrel cage rotor Cw2==.707 since each bar is a concentrated winding acted on by a sine field. (See curve sheet 4.) For a squirrel cage rotor in which the slots and bars are parallel to the shaft, that is for a rotor which is not "spiralled" sine %>= 1. Then if we let 7c'equal the transformation ratio for a motor with a squirrel cage rotor we have, for a motor with an unspiralled rotor 7.: 1x.707X1 _.._. '707 c chwxj/A’Egl chnxwfi'g: @ ) If the squirrel cage rotor is spiralled the voltage generated at one end of a rotor bar is out of phase with that generated at the other end by an amount equal to the angle of the spiral expressed in electrical degrees, just as the voltage generated in one side of a "short pitch" coil is out of phase with the voltage generated in the other side of the coil by a phase angle equal to the number of electrical degrees by which the coil falls short of being awfull pitch coil. (24) Hence if m=:the number of electrical degrees which the squirrel cafe rotor is spiralled. Then for the squirrel cage rotor «f /70'/’7 @ or [flflfia-Afl __3 g: - and SAIZ';Z .xwwr‘} :1 (#5 Using this value instead of l in equation (Iv—mu 7._ .7o7x5/A/5 .(/ 2._2_ ‘ MA C'Wz xswdg @ and if the primary volts per phase are multiplied by this value of Zthe total effective volts generated in a rotor bar at standstill are obtained. For the case of the direct current machine it is evident that the voltage generated between positive and negative brushes is equal to the sum of the instantaneous voltages being generated in all the coils between positive and negative brushes. This is of course the same thing as the average voltage per coil times the number of coils in series between the positive and negative brushes. Referring back to formula the x x average volts per coil " Q 2/12!” @ in which assstated before {6 is the useful flux per pole, No is the series conductors per coil andfis the frequency in the armature, the voltage in the separate coils of a D.C. armature being reilly an alternating voltage and being rectified by the commutator to give the "D. C." voltage outside the armature. ‘1’ I (25) Now if Ed H'the D. C. voltage between positive and nega- tive brushes. the total coils in the armature Sc n the number of parallel circuits into which Sc coils are divided b the connection to the commutator and brus es. Then the number of coils in series between positive and negative brushes 2—23. Now since the voltage between positive and negative brushes is equal to the average volts per coil multiplied by the number of coils in series between positive and negative brushesvve will S get this voltage by multiplying formula (:) byfig , then :Lhixz w .4; Nos 5 /0' I‘m, But n c-'=-'the number of series conductors between positive and negative brushes which should be called H1 for the D. C. machine, hence: N Substituting this value of N1 in equation it becomes: $44—$3ij 9 But since R. P. M.='4€:y: where P is the number of poles, - RIM P f" [22. Substituting this value of f in equation it becomes __ 1 NV 3241 E;- /o x (a. Which is the standard "EMF-flux" equation for D. C. 663 ® machines, Eibeing the generated "D. C." voltage between positive and negative brushes. (54) Part II DETEAMINArlOE OF THE ACTUAL CURRENT IN THE ROTOR BAdS AND IN THE END hINeS OF POLYPHASE MOTORS AND ALSO THE PRIMARY EQUIVALENT OF THE hESISTANCE OF THE BOTOh BARS AND END RINGS OF SUCH SQJIRBEL CAGE WINDINGS In order to determine the actual value of current in the oars of a squirrel cage rotor we must first derive an eXpression for this current in terms of the primary equivalent of the secondary current. Taxing first a general case of a motor having any number, Phg of phasesia the secondary; while the primary has a number of phases ph, also for this general case let: E1 =: the primary volts per phase E2 == the normal secondary volts per phase (that is the open circuit standstill secondary volts per phase.) 12 ‘= the current in each secondary phase winding. 1 12 1: the primary equivalent, in each primary phase winding, of the secondary current, 12. How the primary equivalent 1% of the secondary current 12 is the vector difference between the total current per primary phase at the given load and the excitating or no load current per primary phase. Hence this primary equi- valent of the secondary current is easily found. It is evident that "KVA' represented by this primary equivalent of the secondary current together with the primary voltage and the primary number of phases is equal to the "KVA" in the secondary circuit; that is to the "KVA" represented by the secondary current, the normal (35) secondary voltage and the secondary number of phases. Hence: ElngxPhl=szszrfiz But if 5r is the ratio of the normal secondary voltage to the primary voltage that is if 7: £1 4"? (See part I of this thesis) Then '57? 3 in ® Substituting this value for E2 in equation (53) we get ElxléxPhlf- 7' 1311.121th 9 Inviding both sides of this equation by El IéxPhl =7’ zszPhg 6 Then =4. as. 1 12 TXth 112 (‘9 This is the general equation for the secondary current in the secondary of a polyphase induction motor, and gives the secondary current per phase in terms of the voltage trannformation ratio, the number of primary and secondary phases and the primary equivalent of the secondary current. Equation (37) is used as it is for the slip ring motor which has a phase wound rotor or seenndary. For the squirrel cage rotor motor each rotor slot or bar is considered as a "phase" (36) and hence the current per phase is, in this case, the current per rotor bar, and the transformation ratio is 7; where 7c; .70 7 x 5/1: (”2””) /4‘.x C3g.x,5AME‘EE% as eXplained and derived in part 1 of this thesis. Then is we let: Ig :: the effective current in a bar, or per slot, of a squirrel cage winding corresponding to a given primary current, SB :: the number of rotor slots (which takes the place of. ma in the case of a squirrel cage rotor.) Substituting these squirrel cage quantities in the general formula (37) we get _...£. [22; I [g—Zx‘gxzz‘ the equation gives the actual current in the rotor bar (or per rotor slot) for any given load current and corresponding value of I; in the primary. Now having the actual effective value of the current Ib in the rotor bar it is easy to gigure the (IZR) loss in the bar part of the rotor winding. All that is necessary to do this is to figure the resistance of the rotor bars as if they were all connected together in one long bar and to multiply this resistance by (Ib)2. Now that we have an expression for the actual effective (37) current in the rotor bars, the next step is to find an expression for the current in the rotor and rings; which it will be comparatively easy to do. As explained in part I of this thesis, the average "field form" set up by the currents in the primary winding of a polyohase motor is very close to a sine shaped field form. Therefore the voltage wave generated in each rotor bar of a squirrel cage rotor of such a motor, will be approximately a s.ne voltage wave. This means that the current in each rotor bar will vary appnoximately according to the sine law. Also, since the rotor slots are evenly Spaced the sine current waves of the different bars will be Out of phase with each other by a phase angle corres- ponding to the spacing of the rotor slots in electrical degrees. rhis being the case the instantaneous values of the currents in the bars covering a pole pitdh vary as the sins of the angle of the displacement of each bar in . electrical degrees from the position of zero bar current. This is shown diagramatically in diagram 6 below. Since the voltages and currents in the rotor bars are in ooposite directions over adjacent pole pitches of the rotor because adjacent poles of the magnetic field are of Opposite polarity, it is quite evident that the currentls flowing into the and ring from half the bars under a oole of the magnetic field will Join together and flow one way (38) See blue print,next page. in the and ring toward the adjacent pole of the magnetic field. While the bar currents flowing into the and ring from the other half of the magnetic pole under considers- .S‘I/fo' ('t/Xlé‘Wf/Wfff/VIZVG 7775' VAR/A 71.7, t 1).: t 27+ 4 51/7 I” 77/5 floral? 3M5 ‘Zb £7071:a ‘ 20¢wa 0F * 307mm 0 O O O O O O O O O O O O 3 M/aarszcnoy 2 mo (WE’RE/VT MA x/M/M (”RE/VT Zero coma/r j 703/ Y/OA/FOP BARS 7’05/770/1/ FOR BARS Porxrxomrozvmm N“ y m 4 rfi q 4 n a p. q q q r’i DIAGRAM 0F PORT/av . Effie 7'0}? h : b. L, I ”IS AMP V E E/VD R/Me's I J ‘JI N, g I, 1 _‘3 l i } I + + + POI/V for raw 0F 7’0//V7' 0F MAX/HUM cwmwr ZERO war/yr WIN UM C‘l/R/VFA/‘T //V 7W! 17”; M T/IE fi/A/G m Th’é' R/A/ 9 fl/AGRAM 6. (59) tion will join together and flow the other way in the ring toward the adjacent pole on the other side of magnetic pole being considered. From these considerations it is also evident that the maximum current in the ring will occur at a point in the ring which is midway between a north and south magnetic pole at that particular instant, and that the current will be zero in the end ring at points which are in axial alignment with the center of a magnetic pole of the field. This is shown diagramatically in diagram 6. Hence the maximum current in the and ring is equal to the average current in a rotor bar times the number of bars in one half ef a pole pitch. Also since as explained before the instantaneous currents in the rotor bars vary according to the sine law over a pole pitch, it is evident that the current at different pointsqlong the and ring, at any instant, will vary according to the sine law, and that the instan- taneous current at any point in the end ring will be approximately proportional to the sine of the displacement of the given point in electrical degrees from the point of xere current in the rirg. (The larger the number of rotor bars or slots per pole the more nearly this is true.) To put these statements into the form of an equation let 13 = the maximum current in a bar (40) lb = the effective current in a bar 13 = the maximum current in one and ring. I = the effective current in one end ring the number of rotor slots or bars no to N P = the number of poles for which the primary is wound or connected. Then first we have since the currents vary according to the sine law: a. I], = .7071; and Ir = .7011: and AVE. CURRENT 1N THE BAR = .63613 Then: In: 1 :52 x.63613 . 2 P But from equation (39) above In :1 48-7 and IR: V- Ir .707 6 069 substituting these values of IE and IR in equation (41) we get @ 45-, = 82 Zolzfi x Ib ' 2? .707 or multiplying through by .707 _. 32 Ir - 'Ef" .6561b which is an equation for the Effective current in the end rings in terms of the effective current in the rotor bar. Now having the effective value of current in the end rings all that is necessary to figure the correspondizg (IZR) loss in the end ring part of the squirrel cage winding is to figure the resistance of the and rings at both ends of (41% the rotor as if both ends were stretched out into one long piece of metal of the same cross section as the end rings have, and to multiply E;iS resistance by (Ir)2’ The Ib in equation (44) will of course be the one corresponding to the desired value of primary load current and its component I; which is the primary equivalent of the secondary current for the given load. The expressions for the effective current in the rotor bars and and rings and the method suggested for figuring the (IZR) loss in the rotor bars and the end rings provide one method for figuring this loss in the squirrel cage rotor. However it is often more convenient for the designer to have e "primary equivalent" winding which will be expressed in the same terms as resistance in the primiry phase winding and which he can treat in his calculations as if this primary equivalent of the secondary resistance were actually in the primary phase. Then if we let 3% == the primary equivalent, per primary phase winding, of the secondary resistance It is evident that l 2 g ._ Phl x (12) x R - IE x (Res. of all Bars)‘t I? 1 (Res. of Rings) The best way to use this equation to solve for l 2 equation (38) becomes Ib =fxg‘: (For the case when 1%:- l) R is to assume that I::‘ l for in this case Ib from (42) and II. from equation (44) becomes 3 Ir 2: 7%- x '656 x Ib (Where Ib is the value) (from equation (46) ) (when 12 " l ) Then for Ié: 1 equation (45) becomes 1 2 2 R (I ) 1 Res. of all Bars (I ) x(Res of Rings) . 2 f5] Th1 Ib and I: being in this case figured from equation (46) and (47); that is being the values corresponding to 1 12: 1e Now all the designer has to do to get (123) 1088 in the squirrel cage rotor is to figure the primary equivaleng I1 of the secondary current for each lodd for which he 2 wishes to maxe a calculation and use this to figure the rotor (123) loss from the following equation: . 2 l 2 1 Secondary (I R) lose=(Iz) x I’h1 x Re Using B; as figures from equation (48) and using the number of primary phases Phl since R% as derived in equation.(48) is the primary equivalent, per primary phase winding, of the resistance of the secondary squirrel cage winding. (43) Part III DYNAMIC BRAKING OF SYNCHRONOUS MOTORS AND THE PnEDETERMINATION OF THE CORRECT VALUE OF THE EXTERNAL RESISTANCE PER PHASE TO GIVE THE MAXIMUM BRAKING TORQUE AT THE BEGINNING OF THE BRAKING PERIOD. The connections necessary in order to use dynamic brahing for stopping a synchronous motor are as follows: The motor being in operation driving its load and having its normal or fiull load field excitation, the three armature terminals (a three phase machine being considered) must be disconnected from the three phase A.C. supply lines and then a star connected three phase resistance of the correct value must be immediately connected to the three phase armture terminals. These changes of connections being made with suitable switching apparatus, either man- ually or electromagneticly operated. The field excitation of the synchronous machine being next at its normal value during all the braxing period. These connections are shown diagramatically in diagram 7 below. It is very evident that as soon as the synchronous motor is disconnected from the A.C. supply line and con- nected to the dynamic braxing resistance, as shown in diagram 7, the synchronous machine ceases to Operate: as a motor and begins to Operate as a generator; being driven by the stored energy in its own rotor and in the machine which the, synchronous motor was driving. )~ 7m? 299155 AC. 601% ,V A/Mé‘ \2; Tfl/f’ii 2954:; 57m ewe/[c7252 ’72” 3’5"” jXIVC'fl/Pa/oflj Mom/T RES'IJW‘E FOR Exam? 0F Prim £071ch WW/crm/(we m? .__1, FIFO,SW- 3 C5”; ARM/477ml: 73425- W 29? 7%.? colt/Mic 7/0449 fm 1))”ch flak/A”? éf' A jVA/(l/PO/VOZ/fi MAM/ME p/A MAM Z 9 7'; 2 -/2 —3( (44) At the very beginning of this braking period the frequency of this synchronous machine (now acting as a generator) will be approximately equal to the normal line value since the speed of the machine has not yet decreased at that time. The voltage of the machine Just before the dynamic brazing resistance is connected to its armature terminals will be the no load voltage of the machine acting as a generator and Operating at full speed and with a field excitation equal to the full load field current of the machine as a synchronous motor. It is e7ident that this voltage may be 20 to 35% higher than the A.C. supply line voltage, espeically if the synchronous motor has been Opera- ting at a leading power factor, and so has an over-excited field. The amount of braking torque which will be exerted on the synchronous machine to slow it down in speed will depend on the amount of electrical energy delivered to the resis- tance by the machine acting as a generator. This ammunt of electrical energy will depend on the total generated voltage, “or no load voltage, generated in the armature of the syn- chronous machine for the given field current; the synchronous impedance per phase in the armature of the synchronous michine and the resistance per phase in the star connected dynamic braking resistance which is connected to the armature terminals. (48) The total voltage generated intthe armature of the syn- chronous machine during this dynamic braking process can be considered as made up of two components, one component being the synchronous impedance drop in the armature of the machine and the other component the resistance drOp, which latter is mostly used up in forcing the current through the star connected resistance connected to the armature terminals, since the resistance of the armature is relatively very low. These two components of the total generated voltage are at right angles to each other in phase relatien, or at least very nearly so. The data which it is necessary to have, either from test or from design calculations, in order to be able to work out the required resistance for dynamic braking of a synchronous motor, is enough data to plot the following curves; The no load saturation curve, usually plotted in percent of normal voltage as ordinates against field current in percent of the no load full voltage value as abscissa; the zero power factor full load current saturation curve, and the 100% power factor full load current saturation curve. These curves for'a 600.32, 450 RPM, 3 phase, 60 cycle, 2200 volt, 80% leading power factor synchronous motor are shown on curve chart 8. On this curve chart 100% field current is taken as that I 1..- 5 as -.<-——. ___)Ln_ $73- 2v!— (46) field current which will give full voltage as a generator on no load. Since this particular machnne is rated on the basis of 80% power factor at full load current, the full load field current will be that field current which will give full voltage, or 100% voltage, when the machine is Operating with full load armature current and 80% power factor, as represented by the 100% voltage point on the full load 80% power factor saturation curve, that is by point Z on curve chart 8. This is found to be 185% field current, and this is the field current which is assumed to be flowing in the field circuit during the period of dynamic braking. This valueof field current is indicated In curve chart 8 by a vertical line marked full load.field current. The total generated voltage in the armature at the beginning of the braking period will be the voltage value shown on the no load saturation curve on this vertical line corresponding to 185% field current. This total generated voltage is found to be 131.5% of full voltage for this particular machine. See point A on curve chart 8. This machine is a star connected machine, as most 'alternators and synchronous motors are, and is wound for a 2200 line so that the normal voltage per phase is 2200%'wfii or 1270 VOltBe (47) Hence the total generated voltage per phase at the beginning of the braking period, being 131.5% of normal, will be equal to 1.315 x 1270 = 1670 volts. As previously mentioned this total generated voltage is made up of two components, when dynamic braking by means of external resistance is used. One of these components is the synchronous impedance drop in the armature and the other component is the resistance drop or IR drOp in the external resistance. Since the synchronous impedance of the armature acts lime reactance, these two components are approximately at right angles to each other in phase relation. The problem now is to choose the value of the external resistance per phase so that the maximum possible amount of electrical energy will be delivered to the external resis- tance, for when this is done the maximum amount of braking torque will be exerted to stop the synchronous machine. It has been found that this electircal energy has a maximum value when the external resistance per phase is made equal to the synchronous impedance per phase of the armature of the synchronous machine. Since this synchronous impddance acts lire a reactance this is similar to the case of an electric circuit having a fixed reactsnce and variable resis- tance, and being acted on by an alternating current voltage. (48) In this case the amount of electrical energy in the electric circuit becomes a maximum when the resistance is madeaqual to the fixed reactance. This latter fact can be demonstrated as follows: In a circuit acted on by an A.C. voltage and containing a given reactance. X, and some variable value of resistance, R, the impedance Z of the circuit will be Z a 1KIE ‘k 32 and with an impressed A.C. voltage, E, the current,I, will be I“ ;X!'* Bg But the electrical power expended in such a circuit is always equal to 123 or WATTS = 123 and from equation (51) above I2 = Re 1X2 f” R2 hence, mere .. E2 ) xR 1:31-33 or 2 war-rs = E 3 1211-35 Taking the first derivative of this expression with e e e 0 respect to the variable B and equating this first derivative to zero to solve for the value of B which will give the (49) maximum value of watts we get: (x2+ 32) 193-11223) x as ._. o (X2+R ) or (x2 + s2) E2 = 232mg then dividing both sides by E2 x2+s = 232 01‘ 32=x2 from which 381, for the condition of maximum power in watts. And it has been found by trial that the braking torque at the beginning of the braking period, by this method of braking, is a maximum when the external resistance per phase is made equal to the synchronous impedance of the armature per phase. flow, referring again to curve chart 8, since the no load voltage of 1670 volts corresponding to full load field current, as shown at po1nt A, is to be divided into two equal components which are approximately at right angles to each other in phase, one of these components being the syn- chronous impedance drOp in the armature and the other the external (IR) drop per phase; both of these components will be equal to .707 x 1670 = 1180 volts. That is: SYN. IMO! DROP -‘-"- IR DROP=.707 x 1670 31180 volts That is to say, the external resistance per phase must be 0 9 9 9 (so) so chosen that the (IR) drOp per phase in the external resistance will be 1180 volts and the synchronous impedance drOp in the armature of the synchronous machine per phase will be 1180 volts with the current which will flow when the external resistance is connected. But since the synchronous impedance of the armature depends on the current being supplied by the synchronous machine, the current is an unknown quantity and the best solution which the writer has found for this problem is the following graphical one. First the poLnt K is laid off on the ordinate of curve chart 8 corresponding to full load field current and at a percent voltage cOrresponding to 1180 volts, that is at 93% voltage since 1180 volts is 93% of the normal full phase voltage of 1270 vOlts. Now since the external resistance connected to the machine constitutes a 100% power factor load, it is evident that the point K is a point on a 100% P.F. load current curve for some value of current. New the synchronous impedance drop of a synchronous machine for any given load current and given field excitation, is equal to the distance between the no load saturation.curve and the zero power factor saturation curve for the given load (51) current, this distance being measured along the vertical ordinate corresponding to the given field excitation and being expressed in percent of nOrmal voltage. Hence the synchronous impedance drop for full load current and full load field excitation for this machine is equal to the dis- tance AM'on curve chart 8. Point M being the point on the zero power factor full load current saturation curve corres- ponding to full load field current. In percent of normal voltage this full load synchronous impedance drOp,AM, equals 131.5 - 88.5 or 43% which is equal to .43 x 1270 or 546. volts per phase. Now, as we have already determined, the synchronous impedance droo for maximum braking torque is to equal 1180 volts or 93% of normal voltage, see equation.(60). Next then, we proceed to lay off a distance AB on the full load field line equal to 93% , or in other words we make 13 equal to 131.5 - 93 or 80.9% of normal voltage. It is evident than that the point B is a point on the zero power factor saturation curve correSponding to the still unknown braking current at the beginning of the braking period. On curve chart 8 the base of the triangle, 00L or EDP, that is 00 or ED is equal to the percent field excita- tion required to give full load armature current on short circuit. The height of this triangle Lfl or FP is e ual to (52) the reactance drOp of the armature per phase expressed in percent of normal voltage and the distance NO or PD is equal to the demagnetizing effect of the armature current at full load current and zero powerfactor, expressed in percent of the field current at full voltage and no load. These quantities being determined either from test or from design calculations, as the case may be. As is well known to electrical engineers if this triangle 00L or EDI is moved so as to keep its vertex L or F on the no load saturation curve and its base 00 or ED parallel to the position 00, the vertex D of the triangle will follow along the zero power factor full load current saturation curve. But as stated a few paragraphs above, the point B is a pOint on the zero power factor saturation curve of a load current which is the still unknown current at the beginning of the braking period. If then, a triangle HBG is drawn similar to the triangles DOD and ED? and.having one vertex at B and being drawn of the correct size so that its vertex G Will fall on the no load saturation curve, it is evident that if this traangle is moved so as to keep its base [3 parallel to 00 and its vertex G on the no load saturation curve, its vertex B will describe the zero power factor saturation curve of a load current corresponding to the unknown braking current; in the same way that the vertex (53) D of the triangle EDP followed the zero power factor full load current saturation curve when moved in a similar manner. Furthermore, Just as the height or base or the triangles DOB and EDP are proportional to the fullload current, so the height or base of this similar triangle HBG are prepor- tional to the unknown braxing current at the beginning of the braking period. Then all we need to do to find this braxing current is to multiply the full load current by the ratio of EB : ED, for example, But full load current equals 131. amperes, and.the line HB scales 148. units long while line ED scales 83. units long. Therefore: INITIAL BR.A.C1NG coarsrr g.- x11 = $.51 x 151.: 253.5 Amps] This initial braxing current is amperes per phase and is the same in the armature winding of the synchronous machine and In each "leg" or phase of the external resis- tance, since both the armature and the external resistance are connected in star. Now since we have previously determined the voltage drOp per phase across the external resistance to be 1180 volts, and have now determined the current at the beginning of the braking period to be 233.5 amperes per phase it is easy to find the required resistance per phase to give the maximum braking torque or effort at the beginning of the ( 54) braxing period. If we call this resistance RB then for this machine; R‘ = 1130 = 5.07 oars PER PHASE . f3 2—155. 6 The watts (123) loss developed in each phase of the resistance by this braxing current of 235.5 amperes will be (255.5)2 x as and the total watts (1%) loss in all three phases at the beginning of the braking period will be 3 times this quantity. If we call these total watts W3 then for this machine; 175: 3 x (233.5)2 x 5.07=830,000. WATTS This is equivalent to 1110. HP of braking energy. Then the maximum braxing torque TB at the beginning of the braking period can be found from the usual formula giving the relation between torque, T, horsepower, HP, and speed, RPM , name 1y, r =% x 5,250. Then in this case :13D .11.1.9. x 5,250 = 12,950; (LBS. AT own room rooms) 450 The flywheel effect or W32 in pounds times the radius of gyration in feet squared of this motor is equal to 9,000; lbs. x (Pt.)2. If this same maximum braxing torque were to be exerted throughout the entire braking period from synchronous speed (55) to standstill, that is from 450 RPM to zero speed, the tide requi-ed to step the motor alone could be figured from the well known formula, 2 tgwsxarm . 307. x T (:3 which gives the relations between time in seconds, t, W32 in pounds feet squared, speed in RPM and accelerating or retarding torque, T, in pounds at one foot radius. Then in this case the time in seconds to step the motor alone would be 2 _WRxRPM__9000.x450 _. . . t - m - Wo.’ 1°25°°°nd5 @ The average speed during this braking time would be égg" 1/2 or 3.75 revolutions per second. Hence the number of revolutions required to stop would be, in this 0889 REV. TO STOP: 1; x R.P.S. = 1.02 x 3.75 = 5.825 REVOLUTlOES 0f cOurse, it must be remembered that this braking time and revolutions to stop are for the motor alone, having a WRZ:= 9000. If the WR2 of the driven machine was, say, twice as great as the motor, that is if the tOtal HR2 was equal to 27,000., the nu ber of seconds as well as the number Oi revolutions to stop would be three times as great as figured abOve. This braxing time and revolutions to stop are based on (55) the assumption that the maximum braxing torque TB is going to be exerted throughout the whole braxing period from synchronism to standstill. In order that this may be true the external resistance must be decreased during the braxing period in such a way that the watts expended in the external resistance will dedrease directly as the Speed decreases. In making such a change in the external resistance it must also be remembered that the total gen- erated voltage of the synchronous machine, with constant field current, will decrease directly with the speed. The resistance of the armature winding can.be taken into account, with sufficient accuracy, by subtracting the armature resistance per phase, which is comparatively small, from the figured external resistance to be connected per phase. This can be done with sufficiently accurate results because the component of voltage representing the (IR) drOp in the armature Winding and the component of voltage representing the (IR) drOp in the external resis- tance, both being in phase with the current are in phase with each other. ”'111'1111’111’111111111111111111111111111’1111“ 1293