đức anh đề cương máy điện

đề cương máy điện 2 bằng tiếng anh cho các kĩ sư clc outline machine 2 dành cho các bạn sinh viên nghành điện do nhóm sv trường đại học điện lực tự mày mò nghiên cứu,đề cương tuyển của đh điện lực dành cho sv khôi clc ôn tập thi kết thúc học phần đề cương chỉ gồm các bài lý thuyết được bọn mình tự làm bằng tiếng anh và có hình ảnh minh họa

1.a) What are the requirements of dc machine's armature windings? b) Define the winding pitches, collector's segment and pole pitch? c) Draw a figure of the winding pitches in case of simplex lap winding? 2: present the main and linkage flux in DC machine? : Present the distributon of magnetic field on the armature surface when the brush is on the geometrical neutral axis and the main magnetic field is not considered : What is the armature magnetic field? Present the cause of spark in the armature winding of DC machine? : Present the method of improving the reversing direction rectification in DC machineby using the auxiliary magnetic pole ? Establish the magnetomotive force equations of the transformer? 7) What are the reasons and conditions for establishing a replacement circuit of a transformer? Draw the equivalent circuit of transformer and explain the parameters? draw an open circuit experiment diagram and introduce the pitch to determine the magnetizing circuit parameter of the transformer Draw a short - circuited experiment diagram and introduce the pitches to determine the primary and secondary circuit impedance of the transformer? 10: write the basic set of equations of an asynchronous machine in the case of a rotating rotor Draw the vector graph ? 11) write the basic set of equations of an asynchronous machine in the case of a rotating rotor? Draw the vector graph in motor mode and generator mode 12 Establish the equivalent circuit of induction machines in case of stationary rotor ? 13 : Draw the equivalent circuit of induction machines in the case of rotating rotors Draw a vector diagram in motor mode 14 : What is an amature reaction of synchronous machine? Which does it depend on? Which are the properties of amature reaction when working with capacitive load? 15: establish the equivalent voltage equation of syn.machine? in case of unsaturated magnetic circuit, draw the vector diagram of synchronous generator salient pole? What’s comments from the vector graph? 16 A) Establish the active power-angle equation of synchronous machine, B) draw curve of this power-angle characteristic? 1.a) What are the requirements of dc machine's armature windings? b) Define the winding pitches, collector's segment and pole pitch? c) Draw a figure of the winding pitches in case of simplex lap winding? a) Requirements for winding: - generate the necessary electromotive force such as: a passant current which produce the required torque, not give the overheating - ensure for a good changement of direction - have a simple, sure are safe structure b) +) Pitches of winding -The first winding pitch y1: is the distance between the first working-edge and the second working -edge of one winding element - Second winding pitch y2: is the distance between the second working-edge of the fist winding element and the fist working-edge of the second wingding element -The mix winding pitch: is the distance between two fist working-edge of two adjacent elements -The collector pitch yg=: is the distance between to winding-edge of one winding element Y1= Y=YG=1 Y2=Y1-YG c) Uniform winding element Level winding element 2: present the main and linkage flux in DC machine? The main flux � pass through the airgap between induced part and pole within one pole step �� is scattered flux: The magnetic flux linking the loop separately to each winding is called the flux magnetic flux, we have the stator magnetic flux, the only stator loops with the stator winding, the rotor flux magnetic floss loops with the rotor winding only Dissipation flux is characterized by dissipation resistance 3 : Present the distributon of magnetic field on the armature surface when the brush is on the geometrical neutral axis and the main magnetic field is not considered - If N is the total number conductor of the winding , the number ampere conductor per unit length equival A= N iu πD iu = ( A/cm) With i 2a is the current in conductor D is the outer diameter of the armature (cm) A is the number of branched part - If the circuit is symetrical with the midpoint of two brushes, according to the law of total current, the magnetomotive force of armature part at a point x is Fux = A.2.x - (A/pair of pole) If bypassing the magnetic impedance of magnetic poles of the terminals and rotors, Fux distance δ is the magnetic force required to force the path through the two is Fuwx = H uwx 2δ The word bromide in the opening will be - Bux = µ0 H ux = µ0 with µ0 H ux Fux A = µ0 x 2δ δ is magnetic field magnitude at the base of a segment x is permeability of air : What is the armature magnetic field? Present the cause of spark in the armature winding of DC machine? • • - In order to produced magnetic field of armature part, we introduc DC current in to the armature part by the brush as normal working condition The cause of spark Mecanical cause + Collector is not consistant with the central axis + Rotating part is not well balenced + The surface of collector is not flat + Force on brush is inappropriate - Electromagnetic force + The reactant of electromotive force does not remove completely the reversing electromotive force + The current’s distribation on the contact surface is not uniform : Present the method of improving the reversing direction rectification in DC machineby using the auxiliary magnetic pole ? 1) 2) The fandaneted way to impove the reversing dỉrection is to create an extural magnetic field also knows as a reversig magnetic field at neutral zone by placing auxiliary pole between the main pole The magnetomotive force ( emf ) of auxiliary pole Ft must be : a) Direction ,inverse with horizonal emf of induced part Fuq b) Magnitude have to neutrualize the effect of Fuq and also create the auxiliary field to genenate the reversing emf epk Therefore ,the auxiliary pole in the generator must be set to have same polarity as the main pole with the egles of the winding elememt at the auxiliary pole appwatching inverseinthe motor mode d) Due to the reactant emf epk=Iu and edc is proportial with Bdc ,so Bdc have to be ratio with Iu e) Therefore , auxiliary pole winding must be connected part winding and Iu is changed only is a zone that auxiliary polr circuit is not saturated The air gap between auxiliary pole and induced part =( 1,5 - ) time the main pole and induced part gap The width of auxiliary pole =( 0,4 – 0,8 ) time the width of reversing director zone The auxiliary pole is only located on machine with P > 0,3 KW ,munber of auxiliary pole = number of main pole c) 3) 4) Establish the magnetomotive force equations of the transformer? With load : I1W1+I2W2 Open circuit : I0W1 Because the applied voltage in primary winding U1=Udm=const  m = const So, the balance magnetomotive force equations is obtained as follow: I1W1+I2W2 = I0W1 Complex form W1+2W2 = 0W1   Where =0+(-2 ) =0 + (-2’) ’= - When the transformers works with load the total magnemotive force (i1.w1+i2.w2) of the primary and secondary winding generated the main magnetic flux in transformers - When the transformers works in the opens circuit case: the secondary winding is opened the current in primary winding is i0, the magnemotive force i0.w1 produces the main magnetic flux in transformers 7) What are the reasons and conditions for establishing a replacement circuit of a transformer? Draw the equivalent circuit of transformer and explain the parameters? Reason: In order to simplify the transformation of transformer and power system, we replace the electric circuit and magnetic circuit of transformer to an equivalent circuit This consist of resistance and reactance called an repalacement circuit of transformer Condition: In the process of conversion It makes sure that the physical and energy process have not to change, such as transmission power, machine losses,etc… The Equivalent Circuit of Transformer: draw an open circuit experiment diagram and introduce the pitch to determine the magnetizing circuit parameter of the transformer open circuit experiment open circuit in secondary winding A * * W I0 U source -the V1 U20 primary winding is applied a voltage U1 = U1dm and the secondary winding is open I0 % = x 100% R0 = =R1 + Rm ≈ Rm Z0 = ≈ Zm X0= Z02 – R02) ≈ Xm I0 is the open circuit current X0 is the open circuit reactance Z0 is the open circuit inpedance R0 is the open circuit resistance Rm is magnetic resistance which prensent the main magnetic flux Xm is magnetic reactance which prensent the main magnetic flux Zm is magnetic inpedance which prensent the main magnetic flux Draw a short - circuited experiment diagram and introduce the pitches to determine the primary and secondary circuit impedance of the transformer? - The secondary winding is shorted Determine: Un, In, Pn Zn= - The primary winding is applied a low voltage for obtaining I1n= I1dm rn=r1+r2’= Pn/In2 - Based on meters , we determine : Un , In , Pn xn= => So , we calculate : r1=r2’= rn/2 ; x1=x2’=xn/2 10: write the basic set of equations of an asynchronous machine in the case of a rotating rotor Draw the vector graph ? U1= -E1+I1(r1+jx1)= -E1+I1* Z1 0=E’2 –I’2*(r’2/s +jx’2)=E’2 –I’2(x’2+jx’2+([1-s]/s)r’2 E’2 =E1 I1+I’2=I0 -E1=I0 (rm+jxm) =I0 *Zm Where U1, E1, I1: voltage,emf and current of stator winding r1, x1, Z1: resistance, reactance and impedance of satator winding E’2, I’2:emf and current of rotor winding which area converted to stator winding r’2, x’2: resistance, rectane of rotor which are conveted to stator winding s: coefficient of slide rm: resitance of magnetic circuit which is characteristic for iron losses xm:reactance of magnectic circuit which presents the mutual inductance between stator and rotor I0:current in magnectic circuit which produces the mmf F0 The vector graph 11) write the basic set of equations of an asynchronous machine in the case of a rotating rotor? Draw the vector graph in motor mode and generator mode Emf : electromotive force Mmf :magnetomotive force U1=-E1+I1(r1+jx1)=- E1+ I1Z1 0=E2'- I2'(r'2+jx2')= E2'- I2'(r'2+jx2'+ r2') E2'= E1 I1+ I2'= I0 -E1= I0(rm+jxm)= I0 Zm Where: U1, E1, I1: voltage, emf and current of stator winding r1, x1, Z1: resistance, reactance and Impedance of stator winding E'2, I'2 : emf and current of rotor winding which are converted to stator winding r'2, x2' : resistance, reactance of rotor winding which are converted to stator winding s : coefficient of slide rm : resistance of magnetic circuit which is characteristic for iron losses xm :reactance of magnetic circuit which presents the mutal inductance between stator and rotor I0 : is current in magnetic circuit which produces the mmf F0 12 Establish the equivalent circuit of induction machines in case of stationary rotor ? - - When rotor is non-rotating : f2 = f1, so the voltage transform rotio : Ke ==  Converting rotor emf to stator emf, we have: E1= E2’ = KeE2 The equivalent emf equation in stator circuit U1 =-E1 + I1 (r1 + jx1 )= -E1 + I1Z1 = - (E1 + Eσ1 ) + I1.r1 Eσ1 : is dissipated emf induction in stator I1r : is drop voltage in stator circuit Z1=r1 + jx1 : is impedance of stator winding - Because rotor winding is shorted, the equivalent emf equation in rotor circuit is: = -E2 + I2 ( r2 + jx2 ) = -E2 + I2Z2 r2 : rotor resistance x2: dissipated reactance in rotor winding Z2: impedance of rotor winding - Similar to transformer, we have : -E1 = I0 Zm = I0 ( rm + j xm ) rm : resistance of magnetic circuit which is characteristic for iron losses Xm: reactance of magnetic circuit which produce the mmf F1 - Converting the resistance due to losses is :m2I22 r2 = m1I2΄ r2’ ‘  r2 = kiker2=k.r2 Converting X2 due to the phase between F2 and I2 const : Tan φ2 = = X2’ = x2 = k.x2 Impedance : Z2’= r2’ + jx2’ = k( r2 + jx2 )= k.Z2  - K=kike : is converted ratio of resistance, reactance and impedance Conclusion: the final equation of induction machines in case of stationary rotor U1 =-E1 + I1Z1 = -E1 + I1.(r1+jx1) =E΄2 - I΄2.(r΄2+jx΄2) = -E΄2 + I΄2.Z΄2 E2’ = E1 I1 + I2’ = I0 -E1 = I0 Zm = I0 ( rm + j xm ) U1 E1 I1 : voltage, emf, current of stator winding r1 , x1, z1 : resistance, reactance, impedance of stator winding E΄2 + I΄2 emf and current of rotor winding which are converted to stator winding r΄2, x΄2, Z΄2: resistance, reactance, impedance of rotor winding which are converted to stator winding rm: is resistance of magnetic circuit which is characteristic for iron losses xm: is reactance of magnetic circuit which present the mutual inductance between stator and rotor I0: is current in magnetic circuit which produce the mmf Fo The equivalent circuit When rotor is non-rotating ( stationary rotor ) and rotor winding is shorted, for limiting the current I1 & I2 in stator winding and rotor winding to nominal value, the applied voltage should be decreased → the electromotive force ( emf ) E1 decreases → the main flux in machine decrease → the magnetomotive force ( mmf ) F0 very small → ignore F1 + F2 = F0 ≈ (1) → I1 + I2 (2) With (1) & (2) we have: I1 = U1/( Z1 + Z΄2 ) = U1 / Zn Where: Zn = Z1 + Z΄2 = rn + jxn is short circuit impedance of IM 13 : Draw the equivalent circuit of induction machines in the case of rotating rotors Draw a vector diagram in motor mode 14 : What is an amature reaction of synchronous machine? Which does it depend on? Which are the properties of amature reaction when working with capacitive load? - - - When a synchronous generator's rotor is spun, a voltage EA is induced in the generator's stator windings If a load is attached to the terminals ofthe generator, a current flows But a three-phase stator current flow will produce a magnetic field of its own in the machine This stator magnetic field distorts the original rotor magnetic field, changing the resulting phase vo ltage This effect is called armature reaction because the armature (stator) current affects the magnetic field which produced it in the first place In an alternator like all other synchronous machines, the effect of armature reaction depends on the power factor i.e the phase relationship between the terminal voltage and armature current The properties of armature reaction under capacitive load is to help the main field flux and thereby increasing the electromagnetic force induced in the armature Therefore, when a synchronous generator is capacitively loaded, It’s terminal voltage will increase 15: establish the equivalent voltage equation of syn.machine? in case of unsaturated magnetic circuit, draw the vector diagram of synchronous generator salient pole? What’s comments from the vector graph? • Generator synchronous With symmetric load, the equation of phase the voltage is as follows U = E� – I(ru+jxσu) (1) Where U: terminal voltage of generator ru+jxσu : the resistance & the reactance of the armature winding E� the emf of armature winding • In case of unsaturated magnetic circuit: E� = E + Eu (2) Combining the above two equations (1)&(2), we have: U = E� +Eu– I(ru+jxσu) Eu = -jIu U = E – j I(ru+jxσu) - I.ru = E – jIxdb - I.ru Where: xdb = xu + xσu is the synchronous reactance of salient pole synchronous generator Vector diagram of synchronous generator salient pole Left – in case of inductive load Right - in case of capacitive load 16 A) Establish the active power-angle equation of synchronous machine, B) draw curve of this power-angle characteristic? A) Active power –angle characteristic -the active power-angle characteristic of synchronous machine is expressed by: P=f (); when E= const, U= const Where it the angle between to vectors of E and U Ignoring rư since it is much smaller than Xđb, Xd and X, we obtain P=m.U.I.cos a)in case of salient synchronous machine according to the simplified vector diagram with Rư=0, we have: Id= Iq= With: So P= m.U.I.cos = m.U.I.cos () = m.U.(Icos.cos = m.U.(Iq.cos + Id.sin) Finally we have: P=.sin+.(-).sin2 = PE+PU Inper unit of machine (P.U) P*=.sin+.( - ).sin2 b) in case of non-sqliend synchronous machine: P=.sin B) Characteristic curve of power angle