Digital Secondary Control Architecture for Aircraft Application

Transcription

1 Digital Secondary Control Architecture for Aircraft Application PhD researchers: Henri C. Belan Cristiano C. Locateli Advisors (SAAB): Birgitta Lantto (LiU): Petter Krus (UFSC): Victor J. De Negri

2 Introduction AIRCRAFT FLIGHT CONTROL SYSTEM Moog Inc. (21) Wang, Lijian (212) Improve Efficiency Lantto, Birgitta (214) Maintenance Reliability Performance Weight 2

3 Current i sv 1V2 1V1 2V2 2V1 Bypass Valves External forces: Maximum Range: ± 7 kn; mass: 5 kg. Supply pressure: Maximum pressure: 35 bar; Middle pressure: free to select; Minimum pressure: 2 bar; Tank pressure: 7,5 bar. M 1P1 1C1 M 2P1 2C1 Cylinder: Minimum Area: 1x1-4 m 2 ; Maximum Area: 2% more; Viscous friction: 25 Ns/m; Static friction: 1 N; Kinematic friction: 7 N. 3

4 Digital Pumps in Closed Circuit Concepts Design with relatively few changes Digital Concept Funtional Unit Classification A 1 A 1 A 1 A 1 Conventional Hydraulic Circuit Digital Fluid Power 1V2 1V1 Secundary Conversion 2V2 2V1 Digital Cylinders Digital Motors i sv Limitation and Control A 2 A 2 A 1 A 1 Secondary Control Digital Flow Control Unit - DFCU Switching Hydraulic 1V2 2V2 Digital Hydraulic Transformer - DHT i sv 1V1 2V1 Primary Conversion Storage and Conditioning M Digital Hydraulic Power Management - DHPMS Digital Pumps 4

5 Concepts Design with relatively few changes A 1 A 1 A 1 A 1 1V2 2V2 Digital Concept Secondary Control A C A A A B A D p A p B p c p D 1V 1PA 1V 1PB 1V 1PC 1V 1PD i sv 1V1 2V1 p s1 1V 2PA 1V 2PB 1V 2PC 1V 2PD p s2 1V 3PA 1V 3PB 1V 3PC 1V 3PD 1V2 A 2 A 2 A 1 A 1 2V2 Tank Lines p s3 1V1 2V1 i sv 5

6 Force [kn] Sizing of the cylinder Forces Distribution Areas Rate A A = 27x, A B = 9x, A C = 3x, A D = 1x. Pressures Rate p S1 = 2.5 MPa; p S2 = 1.5 MPa; p S3 =.5 MPa n chambers n forces = ( n pressure n forces = 3 4 = 81 A = (27, 9, 1, 3) x 1-4 [m 2 ] p = (2.5, 1.5,.5) [MPa] Number of Discrete Forces 6

7 Force [kn] Force [kn] Sizing of the cylinder 27:9:3:1 5:3:2: A = (27, 9, 1, 3) x 1-4 [m 2 ] p = (2.5, 1.5,.5) [MPa] 15 1 A = (47.3, 32.5, 1, 2.7) x1-4 [m 2 ] p = (2.5, 1.5,.5) [MPa] Force [kn] Force [kn] Number of Discrete Forces Number of Discrete Forces :9:3:1 A = (12.7, 1.9, 1.3, 1.) x 1-4 [m 2 ] p = (2.5, 1.5,.5) [MPa] Number of Discrete Forces 2% :4:2:1 A = (1.8, 1.4, 1.2, 1.) x 1-4 [m 2 ] p = (2.5, 1.5,.5) [MPa] Number of Discrete Forces 7

8 Mean Difference [kn] Step 1 Areas combination Inputs: Pressures: p s1 = 2. MPa ; p s2 = 13.5 MPa e p s3 =.75 MPa; Minimum area value: 1. x1-4 m 2 ; Maximum area value: 12.5 x 1-4 m 2 ; Increment in area value to search activity: 1x1-5 m 2 ; Magnitude for consider one force different of the other: 1 N MEAN Number of Samples (x1 3 ) Standard Deviation [kn] STANDARD DEVIATION Number of Samples (x1 3 ) Filters: Minimum number of unique discrete forces: 8; Maximum mean: 1.3 kn; Maximum standard deviation: 1.5 kn; Maximum : 2.5 kn, for the 5 central forces; Minimum value for the difference of areas: -1x1-4 m 2 ; Maximum value for the difference of areas: 1.5x1-4 m 2. Maximum Absolute Difference [kn] central points ABSOLUTE DIFFERENCE Number of Samples (x1 3 ) (AA AB + AC - AD) (x1-4 ) [m 2 ] DIFFERENCE OF AREAS Number of Samples (x1 3 ) 8

9 Number of Unique Forces Step 1 - Result 5 4 p = (2., 13.5,.75) [MPa] Force [kn] Number of Discrete Forces High pressure 2. MPa High pressure 14. MPa High pressure 8. MPa Middle Pressure [MPa] A A A B A C A D Areas: A A = 12.1; A B = 1.6; A C = 1. and A D = 1.1 (x1-4 [m 2 ]) 9

10 Typical force levels Surface Takeoff/Landing Ferry Flight Dogfight/Turb. Flight Military Aircraft Pitch 2% 1% 6% - 1% Row 2% 1% 6% - 1% Yaw 5% 5% 6% - 1% Civil Aircraft Pitch 4% 2% 6% - 1% Row 4% 2% 6% - 1% Yaw 1% 1% 6% - 1% 5% 3.5 kn: (Rudder in takeoff movement) 1% 7 kn: (Aileron in ferry flight movement) 2% 4 kn: (Aileron in takeoff movement) 4% 28 kn: (Elevator in takeoff movement) 6% to 1% (42kN to 7 kn) 1

11 Mean Difference [kn] Step 2 Operation points Inputs: Areas: A A = 12.1; A B = 1.6; A C = 1. e A D = 1.1 (x1-4 [m 2 ]); Maximum force value: 3.5 kn; Minimum force value: -2. kn; Maximum working pressure: 35 MPa; Minimum working pressure: 2. MPa; Pressure of reservoir:.75 MPa; Increment in pressure value to search activity:.5 MPa MEAN 18 N Number of Samples (x1 3 ) Standard Deviation [kn] STANDARD DEVIATION 2 N Number of Samples (x1 3 ) Maximum Absolute Difference [kn] ABSOLUTE DIFFERENCE 6 N Number of Samples (x1 3 ) Number of Unique Forces NUMBER OF FORCES Number of Samples (x1 3 ) Filters: Maximum mean: 18 N; Maximum standard deviation: 2 N; Maximum absolute difference: 6 N; Minimum number of unique discrete forces:

12 Force [kn] Step 2 Result A = (12.1, 1.6, 1., 1.1) x 1-4 [m 2 ] p = (3.5, 2.5,.75) [MPa] Number of Discrete Forces Force [kn] [35., 26.,.75] [2., 13.5,.75] [13.5, 45.,.75] [ 8.5, 6.5,.75] [ 3.5, 2.5,.75] [ p s1, p s2, p s3 ] MPa 1% 4% 2% 1% 5% Number of Discrete Forces 12

13 Simulation results 13

14 Energy Loss [W] [mm] Energy efficiency i sv i sv 1V2 1V1 1V2 1V A A 1 A 1 A 1 A 1 Current 1 HybridConcept 2 HybridConcept DigitalConcept A 2 2V2 2V1 A 1 A 1 2V2 2V1 Tank Lines p s1 p s2 1 % 71 1V 2PD % 67 % 19 % p s V 1PA 1V 2PA 1V 3PA Time [s] A C A A A B A D p A p B p c p D 1V 1PB 1V 2PB 1V 1PC 1V 2PC 1V 1PD 1V 3PB 1V 3PC 1V 3PD 14

15 Conclusions It is possible to control a system using a combination of areas and pressures that does not provide an equidistant distribution for the discrete forces values. The digital technical efficiency was proven superior to the current system considered and hybrid proposals from an energy savings point of view However, more research, is needed, especially in relation to control and fault tolerance, to allow the technique to be used in aircraft One suggestion to address this challenge is to evaluate the possibility of using a smaller control surface to control the small variations of load and use the digital actuator for greater demands situations. 15

16 Failure cases A C A A A B A D p A p B p c p D 1V 1PA 1V 1PB 1V 1PC 1V 1PD Rigth p s1 1V 2PA 1V 2PB 1V 2PC 1V 2PD p s2 1V 3PA 1V 3PB 1V 3PC 1V 3PD Tank Lines p s3 If a valve has one closed failure, the system would still be able to generate 54 discrete force values (3 3 x2 1 ), for each combination of pressure. If a valve has one open failure, which can be considered a failure in an actuator chamber, the system will operate with 27 discrete forces (3 3 ). If one pressure line fails, the number of discrete force will be reduced to 16 (2 4 ). 16

17 PhD researcher: Henri Carlo Belan, Cristiano C. Locateli, Advisor (SAAB): Birgitta Lantto, (LiU): Prof. Petter Krus, (UFSC): Prof. Victor J. De Negri,