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Direct current machine
- 1. Direct current machine The purpose of this seminar to represent shunt Dc machine and Permanent magnet Dc machine as (state space equations ) and show how can represent time domain block diagram and find roots to their
- 3. The field and armature voltage equation and the relationship between torque and rotor speed may be written as Field voltage equation VF = Rf if + d𝝀f/dt = Rf if +d/dt (Lff if + Lfa 𝒊𝒂 ) ----Lfa dia/dt = 0 , d/dt = p Vf = Rf if + Lff p if = ( Rf +p Lff ) if , if = Vf / (Rf +p Lff ) ------ Lff/ Rf =𝝉f if =( Vf/rf / (1+ 𝝉f)--------------(1) Armature voltage equation Va = ra ia + d𝝀a/dt = ra ia + d/dt ( Laa ia + Laf if ) ---- dif/dt =0 , d/dt = p Va = ra ia + p Laa ia + dLaf /dt if ------ Laf = -L cos 𝜽r , LAF =L sin 𝜽r Va = ra ia + P laa ia + 𝝎r LAF if ------ Laa/ra = 𝝉𝒂 ia = 1/ra (Va- 𝝎r LAF if ) / 1+𝒑 𝝉𝒂---------------------(2) Torque equation Te = J d 𝝎r /dt + Bm 𝝎r +TL = ( Jp+Bm) 𝝎r +TL 𝝎r = (Te-TL)/(Jp+Bm)--------------------------------------(3)
- 4. time domain block diagram of a shunt connected DC machine according to equations (1 ,2 and 3) in previous slide
- 5. the state space equation for DC machine can be readily achieved by straightforward manipulation of the field and armature voltage equations as shown in matrix below :- • Ssssssssssssssssssssssssss • Where LFF and LAA are the self inductance of the field and armature volotge respectively. • P is the short-hand notation for the operator 𝒅 𝒅𝒕 . • 𝝎𝒓 is the rotor speed . • LAF is the mutual inductance between the field and the rotating armature coils. • Vf = rf if + 𝒅𝒊𝒇 𝒅𝒕 LFF ------------------------------------------- equ(4) • Va= 𝝎𝒓LAF if +ra ia + 𝒅𝒊𝒂 𝒅𝒕 LAA -----------------------------equ(5) • Torque equation is Te = J 𝒅𝝎𝒓 𝒅𝒕 +B 𝝎𝒓+ TL-----------------equ(6)
- 6. rearrange equations (4,5 and 6) by derived yield • 𝒅𝒊𝒇 𝒅𝒕 =- 𝑹𝒇 𝑳𝑭𝑭 if+ 𝟏 𝑳𝑭𝑭 Vf -----------------------equ(7) • 𝒅𝒊𝒂 𝒅𝒕 =- 𝒓𝒂 𝑳𝑨𝑨 ia - 𝑳𝑨𝑭 𝑳𝑨𝑨 if 𝝎𝒓+ 𝟏 𝑳𝑨𝑨 Va --------equ(8) • 𝒅𝝎𝒓 𝒅𝒕 = - 𝑩𝒎 𝑱 𝝎𝒓 + 𝑳𝑨𝑭 𝑱 if ia - 𝟏 𝑱 TL --------equ(9) where Te= LAF if ia • Now write state space in matrix
- 7. Permanent –magnet DC machine • The equations that describe the operation of a permanent - magnet DC machine are identical to those of a shunt- connected dc machine with the field current constant .for the permanent- magnet dc machine (LAF if )replaced by (kv)which is a constant determined by the strength of the magnet, the reluctance of the iron and the number of turns of the armature winding. The time domain block diagram may be developed for the permanent -magnet machine by using equations ( 2 and 3) with( kv ) substituted for (LAF if) • ia = 1/ra (Va- 𝝎r kv) / 1+𝒑 𝝉𝒂--------------------------(10) • 𝝎r = (Te-TL)/(Jp+Bm)--------------------------------------(11)
- 8. time domain block diagram of a permanent- magnet DC machine according to equations (10 and 11) in previous slide
- 9. To write state space equations used same equations (8 and 9) with (kv) substituted for (LAF if) and (Te) substituted (kv ia) • 𝒅𝒊𝒂 𝒅𝒕 =- 𝒓𝒂 𝑳𝑨𝑨 ia - 𝒌𝒗 𝑳𝑨𝑨 𝝎𝒓+ 𝟏 𝑳𝑨𝑨 Va ---------equ(12) • 𝒅𝝎𝒓 𝒅𝒕 = - 𝑩𝒎 𝑱 𝝎𝒓 + 𝒌𝒗 𝑱 ia - 𝟏 𝑱 TL ------------equ(13) • Now write state space in matrix PX = [ A ] X + [ B ] U
- 10. Transfer function for permanent- magnet DC machine to find the roots this block in time domain we will convert it to laplace and compare it with main equation to find the roots
- 11. Block diagram of a permanent- magnet dc machine for a step change in applied voltage , therefore TL will be equal to zero G(s) = 𝟏 𝒓𝒂 𝒌𝒗 (𝟏+𝝉𝒂𝒔)(𝑩𝒎+𝑱𝒔) , 𝑯 𝒔 = 𝒌𝒗 ∆𝜔𝑟(𝑠) ∆𝑉𝑎(𝑠) = 𝐺(𝑠) 1 + 𝐺 𝑠 𝐻 𝑠
- 12. ∆𝝎𝒓(𝒔) ∆𝑽𝒂(𝒔) = 𝟏 𝒓𝒂 𝒌𝒗 (𝟏 + 𝝉𝒂𝒔)(𝑩𝒎 + 𝑱𝒔) 1 + 𝟏 𝒓𝒂 𝒌𝒗 𝟐 (𝟏 + 𝝉𝒂𝒔)(𝑩𝒎 + 𝑱𝒔) ∆𝝎𝒓(𝒔) ∆𝑽𝒂(𝒔) = 1 𝑘𝑣 𝜏𝑎 𝜏𝑚 𝑠2+ 1 𝜏𝑎 + 𝐵𝑚 𝐽 𝑠+ 1 𝜏𝑎 ( 1 𝜏𝑚 + 𝐵𝑚 𝐽 ) −−−−−− −equ(14) ∆𝒊𝒂(𝒔) ∆𝑽𝒂(𝒔) = ∆𝝎𝒓(𝒔) ∆𝑽𝒂(𝒔) ∗ ∆𝒊𝒂(𝒔) ∆𝝎𝒓(𝒔) ∆𝒊𝒂(𝒔) ∆𝝎𝒓(𝒔) = 𝑩𝒎 + 𝑱𝒔 𝒌𝒗 ∆𝒊𝒂(𝒔) ∆𝑽𝒂(𝒔) = 𝟏 𝒓𝒂 𝒌𝒗 (𝟏 + 𝝉𝒂𝒔)(𝑩𝒎 + 𝑱𝒔) 1 + 𝟏 𝒓𝒂 𝒌𝒗 𝟐 (𝟏 + 𝝉𝒂𝒔)(𝑩𝒎 + 𝑱𝒔) ∗ 𝐵𝑚 + 𝐽𝑠 𝑘𝑣 ∆𝑖𝑎(𝑠) ∆𝑉𝑎(𝑠) = 1 𝜏𝑎 𝑟𝑎 𝑠 + 𝐵𝑚 𝐽 𝑠2 + 1 𝜏𝑎 + 𝐵𝑚 𝐽 𝑠 + 1 𝜏𝑎 ( 1 𝜏𝑚 + 𝐵𝑚 𝐽 ) −−−−−−−−−−−−− −equ(15) 𝜏𝑚 = 𝐽 𝑟𝑎 𝒌𝒗 𝟐
- 13. Block diagram of a permanent- magnet dc machine for a step change in torque load , therefore Va will be equal to zero G(s) = 𝟏 (𝑩𝒎+𝑱𝒔) , 𝑯 𝒔 = −𝟏 𝒓𝒂 𝒌𝒗 𝟐 (𝟏+𝝉𝒂𝒔) ∆𝜔𝑟(𝑠) ∆𝑇𝐿(𝑠) = −𝐺(𝑠) 1 − 𝐺 𝑠 𝐻 𝑠
- 14. ∆𝝎𝒓(𝒔) ∆𝑻𝑳(𝒔) = −𝟏 (𝑩𝒎 + 𝑱𝒔) 1 + 𝟏 𝒓𝒂 𝒌𝒗 𝟐 (𝟏 + 𝝉𝒂𝒔)(𝑩𝒎 + 𝑱𝒔) 𝜏𝑚 = 𝐽 𝑟𝑎 𝒌𝒗 𝟐 ∆𝝎𝒓(𝒔) ∆𝑻𝑳(𝒔) = −1 𝐽 𝑠 + 1 𝜏𝑎 𝑠2 + 1 𝜏𝑎 𝑠 + 1 𝜏𝑎 𝜏𝑚 −−−−−− −equ(16) ∆𝒊𝒂(𝒔) ∆𝑻𝑳(𝒔) = ∆𝝎𝒓(𝒔) ∆𝑻𝑳(𝒔) ∗ ∆𝒊𝒂(𝒔) ∆𝝎𝒓(𝒔) ∆𝒊𝒂(𝒔) ∆𝝎𝒓(𝒔) = − 𝟏 𝒓𝒂 𝒌𝒗 𝟏 + 𝝉𝒂 𝒔 ∆𝒊𝒂(𝒔) ∆𝑻𝑳(𝒔) = −𝟏 (𝑩𝒎 + 𝑱𝒔) 1 + 𝟏 𝒓𝒂 𝒌𝒗 𝟐 (𝟏 + 𝝉𝒂𝒔)(𝑩𝒎 + 𝑱𝒔) ∗ − 1 𝑟𝑎 𝑘𝑣 1 + 𝜏𝑎 𝑠 ∆𝑖𝑎(𝑠) ∆𝑉𝑎(𝑠) = 1 𝜏𝑎 𝜏𝑚 𝑘𝑣 𝑠2 + 1 𝜏𝑎 𝑠 + 1 𝜏𝑚 𝜏𝑎 −−−−−−−−−−−−− −equ(17)
- 15. Exmaple :- For a permanent DC motor ra = 7Ω ,, LAA= 0.012 H , J= 1.036*𝟏𝟎−𝟔 , Bm= 6.03 *𝟏𝟎−𝟔 N.m.s , kv = 0.0141, the armature voltage stepped fro zero to (6 v) and the torque load TL = 0 . • Solution from the block diagram below ∆𝐓𝐋 𝐬 = 𝟎 we will use equations ( 14 and 15 ) to find ∆𝝎𝒓(𝒔) 𝒂𝒏𝒅 ∆ 𝒊𝒂(𝒔) and then ia and 𝝎𝒓 ,
- 16. ∆ωr(s) ∆Va(s) = 1 kv τa τm s2+ 1 τa + Bm J s+ 1 τa ( 1 τm + Bm J ) −−−−−−−−−−−−−−−− −equ(14) ∆ia(s) ∆Va(s) = 1 τa ra s + Bm J s2 + 1 τa + Bm J s + 1 τa ( 1 τm + Bm J ) −−−−−−−− −equ(15) • 𝜏𝑎= 𝐿𝐴𝐴 𝑟𝑎 = 0.012 7 = 0.017 𝑠𝑒𝑐 , 𝜏𝑚 = 𝐽 𝑟𝑎 𝑘𝑣2 = 1.06 ∗ 10−6 ∗ 7 0.01412 = 0.037 𝑠𝑒𝑐 • compare denominator (characteristic equation ) with main second order equation to find damping ratio and ( 𝜔𝑛)and then roots • 𝑆2 + 2 ℰ 𝜔𝑛 𝑆 + 𝜔𝑛2-------------------------------- equ(1) 𝜔𝑛2 = 1 𝜏𝑎 1 𝜏𝑚 + 1 𝐽 = 1 0.017 1 0.037 + 6.03 ∗10−6 1.06 ∗10−6 = 1925 𝑠𝑒𝑐−2 , 𝜔𝑛 = 43.9 𝑟𝑎𝑑 𝑠𝑒𝑐 2ℰ𝜔𝑛 𝑆 = 1 𝜏𝑎 + 𝐵𝑚 𝐽 𝑆 , 2 ℰ ∗ 43.9 = 1 0.017 + 6.03 ∗10−6 1.06 ∗10−6 ; ℰ = 0.735 , Equation (1) become 𝑆2 + 64.533 𝑆 + 1925 So b1,b2= −𝑏± 𝑏2−4𝑎𝑐 2𝑎 = 𝑥 = −64.533± 64.5332−4∗1925 2 = 32 ∓𝑗29.8 𝑏1, 𝑏2 = (𝛾 ± 𝑗𝛽) ∆𝜔𝑟 𝑠 = ∆𝑣𝑎 𝑠 𝑘𝑣 𝜏𝑎 𝜏𝑚 𝑆 𝑆+𝑏1 𝑆+𝑏2 , 6 0.0141∗0.017∗0.037 ( 𝐴 𝑆 + 𝐵 𝑆+𝑏1 + 𝐶 𝑠+𝑏2 )
- 17. Now we find (A,B,C ) 𝑨 = lim 𝒔→𝟎 𝑺 𝑺 (𝑺+𝒃𝟏)(𝑺+𝒃𝟐) 𝒔=𝟎 = 𝟏 (𝑺+𝒃𝟏)(𝑺+𝒃𝟐) 𝑺=𝟎 = 𝟏 𝒃𝟏 𝒃𝟐 B= lim 𝒔→−𝒃𝟏 𝑺+𝒃𝟏 𝑺 (𝑺+𝒃𝟏)(𝑺+𝒃𝟐) 𝒔=−𝒃𝟏 = 𝟏 𝑺 (𝑺+𝒃𝟐) 𝒔=−𝒃𝟏 = 𝟏 𝒃𝟏 𝒃𝟏−𝒃𝟐 𝑪 = lim 𝒔→−𝒃𝟐 𝑺+𝒃𝟐 𝑺 (𝑺+𝒃𝟏)(𝑺+𝒃𝟐) 𝒔=−𝒃𝟐 = 𝟏 𝑺 (𝑺+𝒃𝟏) 𝒔=−𝒃𝟐 = 𝟏 𝒃𝟐 𝒃𝟐−𝒃𝟏 ∆𝝎𝒓 𝒔 = 𝟔𝟕𝟔𝟓𝟐𝟏( 𝑨 𝒔 + 𝑩 𝒔+𝒃𝟏 + 𝑪 𝒔+𝒃𝟐 ) = 𝟔𝟕𝟔𝟓𝟐𝟏 𝟏 𝒃𝟏 𝒃𝟐 𝟏 𝒔 + 𝟏 𝒃𝟏 𝒃𝟏−𝒃𝟐 𝟏 𝒔+𝒃𝟏 +
- 18. ∆ia(s) = ∆𝑣𝑎(𝑠) τa ra s + Bm J s(s2 + 1 τa + Bm J s + 1 τa ( 1 τm + Bm J )) −−−−−−−− −equ(15) • Let a= 𝐵𝑚 𝐽 = 5.7 𝑠𝑒𝑐−1 • ∆𝑖𝑎 𝑠 = 6 0.017 ∗7 ( S+a s (s+b1)(s+b2) ) = 50.42 𝐷 𝑠 + 𝐸 𝑠+𝑏1 + 𝐹 𝑠+𝑏2 • 𝑫 = lim 𝒔→𝟎 𝒔 (𝒔+𝒂) 𝑺 (𝑺+𝒃𝟏)(𝑺+𝒃𝟐) 𝒔=𝟎 = (𝒔+𝒂) (𝑺+𝒃𝟐) (𝑺+𝒃𝟏) 𝒔=𝟎 = 𝒂 𝒃𝟏 𝒃𝟐 • 𝑬 = lim 𝒔→−𝒃𝟏 (𝑺+𝒃𝟏)(𝑺+𝒂) 𝑺 (𝑺+𝒃𝟏)(𝑺+𝒃𝟐) 𝒔=−𝒃𝟏 = 𝑺+𝒂 𝑺 (𝑺+𝒃𝟐) 𝒔=−𝒃𝟏 = 𝒂−𝒃𝟏 𝒃𝟏 𝒃𝟏−𝒃𝟐 • 𝑪 = lim 𝒔→−𝒃𝟐 (𝑺+𝒃𝟐)(𝑺+𝒂) 𝑺 (𝑺+𝒃𝟏)(𝑺+𝒃𝟐) 𝒔=−𝒃𝟐 = 𝑺+𝒂 𝑺 (𝑺+𝒃𝟏) 𝒔=−𝒃𝟐 = 𝒂−𝒃𝟐 𝒃𝟐 𝒃𝟐−𝒃𝟏 • ∆𝒊𝒂 𝒔 = 𝟓𝟎. 𝟒𝟐 𝒂 𝒃𝟏 𝒃𝟐 𝟏 𝒔 + 𝒂−𝒃𝟏 𝒃𝟏 𝒃𝟏−𝒃𝟐 𝟏 𝒔+𝒃𝟏 + 𝒂−𝒃𝟐 𝒃𝟐 𝒃𝟐−𝒃𝟏 𝟏 𝒔+𝒃𝟐 𝒍𝒂𝒑𝒍𝒂𝒄𝒆 𝒊𝒏𝒗𝒆𝒓𝒔𝒆 • ∆𝒊𝒂 𝒕 = 𝟓𝟎. 𝟒𝟐( 𝒂 𝒃𝟏𝒃𝟐 + 𝒂−𝒃𝟏 𝒃𝟏 𝒃𝟏−𝒃𝟐 𝒆−𝒃𝟏𝒕 + 𝒂−𝒃𝟐 𝒃𝟐 𝒃𝟐−𝒃𝟏 𝒆−𝒃𝟐𝒕) • - complex algebra making use euler's identity a couple of times • ∆𝒊𝒂 𝒕 𝟓𝟎. 𝟒𝟐 𝒂 𝜸 𝟐 +𝜷 𝟐 [−𝒆−𝜸𝒕(cos 𝜷𝒕 − 𝜸 𝟐+𝜷 𝟐+𝒂𝜸 𝒂𝜷 sin 𝜷𝒕) ] • ∆𝒊𝒂 𝒕 = 𝟎. 𝟏𝟒𝟗[𝟏 − 𝒆−𝟑𝟐.𝟑 𝒕 (cos 𝟐𝟗. 𝟖𝒕 − 𝟏𝟎. 𝟑 sin 𝟐𝟗. 𝟖𝒕)] • ia= 𝒊𝒂 𝟎 + ∆𝒊𝒂 𝒕 −−−−−−−−−− −𝒊𝒂 𝟎 = 𝟎 • So ia = 𝟎. 𝟏𝟒𝟗[𝟏 − 𝒆−𝟑𝟐.𝟑 𝒕 (cos 𝟐𝟗. 𝟖𝒕 − 𝟏𝟎. 𝟑 sin 𝟐𝟗. 𝟖𝒕)]
- 19. References • Paul c. Krause .analysis electric machines and drive system. USA: wiley interscience,2002,pp.76-101