- No category
DC Machine Modeling: Back EMF, Torque, and Simulink Simulation
advertisement
Modeling of DC Machines Dr.Tun Lin Naing Back EMF • A compact form of back emf equation is given by • ππ = πΎππ ππ [V] (1) • Where, • ππ = induced emf or back emf of armature circuit [V] • πΎ = constant of armature circuit [V.sec/Wb.rad] • ππ = field flux [Wb] • ππ = rotor speed in [rad/sec] • In practical, it will very difficult to measure the exact value of field flux. Back EMF • If field flux is kept constant, the back emf equation can be rewriting as • ππ = πΎππ ππ = πΎπ ππ [V] (2) • Where, • πΎπ = back emf constant [V.sec/rad] • Back emf constant in V/rpm is = πΎπ π.π ππ 2π πππ 1 πππ 2π π.πππ π π × × = πΎπ = πΎπ πππ 1 πππ£ 60 π ππ 60 πππ£ 30 πππ • In this case, back emf is directly proportional to motor speed and πΎπ is become a proportional constant. • We can use equation (2) for permanent magnet DC machines and separately excited DC machine with rated constant field current. Back EMF • Equation (1) can also be written in terms of field current as • ππ = πππ ππ [V] (3) • Where, • π = mutual inductance between armature and field circuit [H] • ππ = field current [A] • We can use equation (3) for shunt or series connected DC machines and separately excited DC machines with field control. Electromagnetic Torque • Air-gap power can be expressed as • ππ = ππ ππ = ππ ππ π π ππ • ππ = π π (4) • Where, • ππ = air-gap power [W] • ππ = armature current • Substituting the Equation (2) in (4) gives πΎπ ππ ππ • ππ = = πΎπ ππ ππ (5) Electromagnetic Torque • Therefore, motor torque constant is equal to back emf constant and its unit is V.s/rad for constant flux machine. • ππ = πΎπ ππ • Where, • πΎπ = torque constant [N.m/A] (6) Armature Circuit π π πΏπ + π£π π − + π£πΏπ − • By applying KVL to armature circuit, • −ππ + π£π π + π£πΏπ + ππ = 0 • ππ −ππ + π π ππ + πΏπ π + ππ = 0 ππ‘ • πΏπ πππ = ππ − π π ππ − ππ ππ‘ (7) + π£π − + ππ − π π =Armature Resistance (Ω) πΏπ =Armature Inductance (H) Mechanical Equation • Mechanical dynamic equation of motor load system can be written as • ππ = • π½ πππ ππ‘ πππ π½π + π½πΏ + π΅ππ + ππΏ ππ‘ = ππ − π΅ππ − ππΏ (8) • Where, • π½π = moment of inertia of rotor [kg.m2/sec2] • π½πΏ = moment of inertia of load [kg.m2/sec2] • π½ = moment of inertia of rotor and load [kg.m2/sec2] • π΅ = viscous friction coefficient [N.m/(rad/sec)] • ππΏ =load torque [N.m] Preparing for Simulink Simulation PMDC Machine • Rewriting Equation (7) and (8) • ππ = 1 πΏπ π£π − π π ππ − ππ ππ‘ • ππ = πΎπ ππ • ππ = πΎπ ππ 1 • ππ = π½ ππ − π΅ππ − ππΏ ππ‘ Parameters for Simulation PMDC Machine Sr.No Description Symbol Value Unit 1 Rated armature voltage ππ 6 V 2 No load speed Ωπ 351.6 rad/sec 3 No load armature current πΌπ 0.15 A 4 Rated load torque ππΏ 0.00353 N.m 5 Armature resistance π π 7 Ω 6 Armature inductance πΏπ 120 mH 7 Back emf constant πΎπ 1.41 x 10−2 H 8 Moment of inertia π½ 1.06 x 10−6 Kg-m2 9 Friction coefficient π΅ 6.04 x 10−6 N.m/(rad/sec) State-space Model • Rewriting Equation (7) and (8) gives πππ 1 • = π − π π ππ − πΎπ ππ ππ‘ πΏπ π ππ 1 • π = πΎπ ππ − π΅ππ − ππΏ ππ‘ π½ • In matrix form • πππ ππ‘ πππ ππ‘ π πΏπ πΎπ π½ − π = πΎ πΏπ π΅ − π½ − π (9) 1 πΏπ ππ + ππ 0 • Compact form of Equation (10) is • π₯ = π΄π₯ + π΅π’ 0 π£ π 1 ππΏ π½ (10) Roots of Matrix ‘A’ from state-space equation • Matrix A is • π΄= π πΏπ πΎπ π½ − π πΎ πΏπ π΅ − π½ − π • Eigenvalues of matrix A • πππ‘ π΄ − ππΌ = 0 • π΄ − ππΌ = • π΄ − ππΌ = π πΏπ πΎπ π½ π − π πΏπ πΎπ π½ − π πΎ πΏπ 1 − π π΅ 0 − π½ πΎ − π πΏπ π − π΅ 0 − π½ − π 0 1 0 =0 π Roots of Matrix ‘A’ from state-space equation π π − −π πΏπ πΎπ π½ • πππ‘ π΄ − ππΌ = πππ‘ • π π − −π πΏπ π΅ − −π π½ π π΅ • π2 + π + πΏπ π½ − πΎπ − πΏπ π΅ − −π π½ πΎπ 2 − π½πΏπ =0 =0 π΅π π + πΎπ 2 π+ =0 π½πΏπ • The roots of the above equation gives • π1 , π2 = π π΅ − πΏ π+ π½ ± π π π π΅ 2 π΅π π + πΎπ 2 + −4 πΏπ π½ π½πΏπ 2 (11) Roots of Matrix ‘A’ from state-space equation • If we neglect the viscous friction coefficient, we can let π΅ = 0. Therefore, Equation (11) will become • π1 , π2 = π π 2 πΎπ 2 −4 π½πΏ πΏπ π π − πΏπ ± π 2 • Next, letting electromechanical time constant ππ = π½π π πΏπ 2 and armature electrical time constant ππ = πΎπ • π1 , π2 = 1 − ± ππ 1 2 4 − ππ ππ ππ 2 π π 1 1 =− ± 2ππ ππ 1 ππ − 4 ππ Roots of Matrix ‘A’ from state-space equation • Finally, • π1 , π2 = − • π1 , π2 = 1 1 ± 2ππ 2ππ 1 2ππ 1− 4ππ ππ −1 ± 1 − 4ππ ππ (12) • It can be observe that the motor is stable on open-loop operation because its roots will always have a negative real parts. • Question: Show that a DC motor is stable on open-loop operation. Discuss on Eigenvalues • Eigenvalues of open-loop controlled DC motor is • π1 , π2 = 1 2ππ −1 ± 1 − 4ππ ππ • If ππ ≥ 4ππ , the roots of π1 πππ π2 are negative real numbers. • If ππ ≤ 4ππ , the roots of π1 πππ π2 are negative complex numbers. Block Diagram of DC Motor • From Equation (7) • πΏπ πππ + π π ππ = ππ −ππ ππ‘ • Using Laplace transform • π πΏπ πΌπ π + π π πΌπ (π ) = ππ −πΈπ (π ) • πΌπ π π −πΈ (π ) = π π π π +π πΏπ (13) ππ + − πΈπ 1 π π + π πΏπ πΌπ Block Diagram of DC Motor • From Equation (8) πππ •π½ + π΅ππ = ππ − ππΏ ππ‘ • Using Laplace transform • π π½Ωπ (π ) + π΅Ωπ (π ) = ππ − ππΏ ππ − ππΏ • Ωπ (π ) = π΅+π π½ (14) ππ + − ππΏ 1 π΅ + π π½ Ωπ Block Diagram for Armature Controlled DC Motor • By Combining Equation (13) and (14), we can draw the block diagram of a DC motor with constant field flux as follow. ππΏ ππ + − πΈπ 1 π π + π πΏπ πΌπ πΎπ πΎπ ππ − + 1 π΅ + π π½ Ωπ Transfer Functions ππ + − πΈπ 1 π π + π πΏπ πΌπ πΎπ ππ 1 π΅ + π π½ πΎπ ο§ Transfer function between motor speed and terminal voltage is πΎπ 1 Ωπ (π ) πΊ πΎπ π π + π πΏπ π΅ + π π½ = = = πΎπ πΎπ ππ (π ) 1 + πΊπ» 1 + π π + π πΏπ π΅ + π π½ + πΎπ2 π π + π πΏπ π΅ + π π½ Ωπ (π ) = ππ (π ) πΎπ /π½πΏπ (π΅πΏπ + π½π π ) π΅π π + πΎπ2 2 π + π + π½πΏπ π½πΏπ (15) Ωπ Transfer Functions ππΏ 1 π΅ + π π½ + − ππ 1 π π + π πΏπ πΎπ Ωπ πΎπ ο§ Transfer function between motor speed and load torque is 1 Ωπ (π ) πΊ π π + π πΏπ π΅ + π π½ = = = πΎπ πΎπ ππΏ (π ) 1 + πΊπ» 1 + π π + π πΏπ π΅ + π π½ + πΎπ2 π π + π πΏπ π΅ + π π½ Ωπ (π ) = ππΏ (π ) π π + π πΏπ /π½πΏπ (π΅πΏπ + π½π π ) π΅π π + πΎπ2 2 π + π + π½πΏπ π½πΏπ (16) Steady-state Equations for Armature Controlled DC Machines • From equation (7) (neglecting voltage drops in carbon brushes), • ππ = π π πΌπ + πΈπ [V] (17) • From Equation (8), • ππ = π΅Ωπ + ππΏ [N.m] (18) • Where, • πΈπ = πΎπ Ωπ [V] (19) • ππ = πΎπ πΌπ (20) [N.m] Analysis on Steady-state Equations of Armature Controlled DC Machines • By neglecting viscous friction coefficient say π΅ = 0, Equation (18) can be rewritten as • ππ = ππΏ (21) • By substituting Equation (20) into Equation (21), • πΎπ πΌπ = ππΏ ππΏ • πΌπ = πΎπ (22) • From Equation (17), we can rewrite as • πΈπ = ππ − π π πΌπ • πΎπ Ωπ = ππ − π π πΌπ (23) Analysis on Steady-state Equations of Armature Controlled DC Machines • Substituting Equation (22) into (23) gives π πΎπ • Ωπ = π − π π π πΎπ2 πΏ (24) • We can also rewrite the Equation(17) as ππ −πΈπ ππ πΎπ • πΌπ = = − Ωπ π π π π π π (25) • According to Equation (24), motor speed will decrease if load torque increase. • If speed decrease, armature current will increase according to Equation (25). Speed-torque Characteristic of An Armature Controlled DC Machine • We can draw speed-torque characteristic of an armature controlled DC machine from Equation (24) and (25). Modeling of SEDC Machine + π£πΏπ − + ππ − − + π£π π − πΏπ + π£πΏπ − π£π π + πΏπ + π£π − π π ππ π π + π£π − ο Apply KVL to armature circuit ο Apply KVL to field circuit πππ πΏπ = π£π − π π ππ − ππ (26) ππ‘ πππ πΏπ = π£π − π π ππ ππ‘ ππ = πππ ππ , ππ = πππ ππ (27) Notes on SEDC Motor • Steady-state equations of SEDC machine are ππΌπ ππ −πΈπ ππ • πΌπ = = − Ωπ π π π π π π π π • Ωπ = π − π 2 ππΏ (speed-torque characteristic) ππΌπ ππΌπ • Without field voltage, π π π • πΌπ = π (higher armature current) • Ωπ = 0 (rotor will not start to rotate) • Therefore, make sure to supply the field circuit before supply the armature circuit. Prepare for Simulink Simulation • We can rewrite Equation (26) and (27) as 1 πΏπ 1 • ππ = πΏπ • ππ = π£π − π π ππ − ππ ππ‘ (28) π£π − π π ππ ππ‘ (29) • ππ = πππ ππ • ππ = πππ ππ • ππ = 1 π½ ππ − π΅ππ − ππΏ ππ‘ (30) (31) (32) • We can draw Simulink blocks using above model equations. These blocks can represent the separately excited DC machine. Shunt-Excited DC Machine ππ ππ‘ ππ + π£π − π π + π£π π − πΏπ + π£πΏπ − π π πΏπ + π£π π − + π£πΏπ − ο Apply KVL to armature circuit πππ πΏπ = π£π − π π ππ − ππ (33) ππ‘ ππ = πππ ππ , ππ = πππ ππ + ππ − π£π = π£π ο Apply KVL to field circuit πΏπ πππ = π£π − π π ππ ππ‘ (34) Shunt-excited DC Machine • We can rewrite Equation (33) and (34) as 1 πΏπ 1 • ππ = πΏπ • ππ = π£π − π π ππ − ππ ππ‘ (35) π£π − π π ππ ππ‘ (36) • ππ = πππ ππ • ππ = πππ ππ 1 • ππ = ππ − π΅ππ − ππΏ ππ‘ π½ (37) (38) (39) • We can draw Simulink blocks using above model equations. These blocks can represent the shuntexcited DC machine. • Speed-torque characteristic is same as SEDC machine. Parameters for SEDC and ShEDC Machine Sr.No Description Symbol Value Unit 1 Rated armature voltage ππ 240 V 2 Rated field voltage ππ 240 V 3 Rated speed Ωπππ‘ππ 127.7 rad/sec 4 Rated armature current πΌπ 16.2 A 5 Rated load torque ππΏ 29.2 N.m 6 Armature resistance π π 0.6 Ω 7 Armature inductance πΏπ 0.012 H 8 Mutual inductance π 1.8 H 9 Field resistance π π 240 Ω 10 Field inductance πΏπ 120 H 11 Moment of inertia π½ 1 Kg-m2 12 Friction coefficient π΅ 1 x 10−6 N.m/(rad/sec) π£π π − π£πΏπ − ππ π π + π£π − π π πΏπ + π£π π − + π£πΏπ − + + Series-excited DC Machine + ππ ππ = ππ − ο Apply KVL to armature circuit πππ πΏπ + πΏπ = π£π − π π + π π ππ − ππ ππ‘ ππ = πππ ππ , ππ = πππ ππ = πππ2 (40) Series-excited DC Machine • We will rewrite the Equation (40) as, πππ • πΏ = π£π − π ππ − ππ ππ‘ (41) • Where, • πΏ = πΏπ + πΏπ and π = π π + π π • Therefore, we also can prepare for simulation as • ππ = 1 πΏ π£π − π ππ − ππ ππ‘ • ππ = πππ ππ • ππ = πππ2 • ππ = 1 π½ ππ − π΅ππ − ππΏ ππ‘ (42) (43) (44) (45) Steady-state Equation of Seriesexcited DC Machine • From equation (41) (neglecting voltage drops in carbon brushes), • ππ = π πΌπ + πΈπ [V] (46) • From Equation (8), • ππ = π΅Ωπ + ππΏ [N.m] (47) • Where, • πΈπ = ππΌπ Ωπ [V] (48) • ππ = ππΌπ2 (49) [N.m] Steady-state Equation of Seriesexcited DC Machine • By neglecting viscous friction coefficient say π΅ = 0, Equation (47) can be rewritten as • ππ = ππΏ (50) • By substituting Equation (49) into Equation (50), • ππΌπ2 = ππΏ • πΌπ = ππΏ π (51) • From Equation (46), we can rewrite as • πΈπ = ππ − π πΌπ • ππΌπ Ωπ = ππ − π πΌπ (52) Steady-state Equation of Seriesexcited DC Machine • Substituting Equation (51) into (52) gives • Ωπ = ππ πππΏ − π π (53) • We can also rewrite the Equation(46) as π −πΈπ π ππΌ = π − π Ωπ π π π π ππ • πΌπ 1 + Ωπ = π π ππ • πΌπ = π +πΩπ • πΌπ = π • Note that Ωπ = ∞ with ππΏ = 0. (54) Parameters for Series-excited DC Machine Sr.No Description Symbol Value Unit 1 Rated armature voltage ππ 230 V 2 Rated speed Ωπππ‘ππ 209.52 rad/sec 3 Rated armature current πΌπ 12.88 A 4 Rated load torque ππΏ 10.675 N.m 5 Armature resistance π π 1.5 Ω 6 Armature inductance πΏπ 0.12 H 7 Mutual inductance π 0.0675 H 8 Field resistance π π 2.33 Ω 9 Field inductance πΏπ 0.03 H 10 Moment of inertia π½ 0.02365 Kg-m2 11 Friction coefficient π΅ 0.0025 N.m/(rad/sec)
0
advertisement
Download
advertisement
Add this document to collection(s)
You can add this document to your study collection(s)
Sign in Available only to authorized usersAdd this document to saved
You can add this document to your saved list
Sign in Available only to authorized users