EXAMINATION Hydraulic Servo Systems, TMHP51/TEN1

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1 EXAMINATION Page 1(7) EXAMINATION Hydraulic Servo Systems, TMHP51/TEN1 Date: , 08:00 Room: TR4 Number of questions: 5 Responsible teacher: Magnus Sethson, (magnus.sethson@liu.se) Phone number during exam: , Teacher visit at: 09:00, 11:00 Allowed tables: Standard Mathematical Tables (or similar) Allowed handbooks: Tefyma (Ingelstam, Rönngren, Sjöberg), Beta (Råde, Westergren) Allowed formularies: Formula Book for Hydraulics and Pneumatics (LiTH/IEI), Mekanisk Värmeteori och Strömningslära Allowed electronics: Pocket Calculator (No smartphones or computers allowed!) Course administrator: Score: Solution: Results: Langauge: Rita Enquist ( , rita.enquist@liu.se) Maximum score on each question is 10 points, resulting in an examination maximum of 50 points. To get the mark 3 you will need 20 points, to get the mark 4 you will need 30 points and to get the mark 5 you will need 40 points. The solution will be uploaded on the course site. Results will be announced on at the latest. You are entitled to answer in either english or swedish. Please notice that the questions are not put in an increasing order of di culty. Plan your work through the exam. I hope you had a nice summer and are able to put that energy into this exam. Good Luck! Magnus Sethson Senior Lecturer August 27, 2013 On the front page and on all following pages the student must write: AID-number, TMHP51/TEN1, YYMMDD, page number

2 EXAMINATION Page 2(7) 1 Cavitation in Orifices [A, 4 points] A cavitation test is carried out using a 200bar pressure setting on the supply system. The result is seen in figure 1. Visual cavitation is observed at a downstream pressure of 73.4bar when the flow is 6.8l/min. The upstream pressure is increased to 270bar by the supply system. At what pressure can now cavitation be assumed? What is the flow then? 8 Cavitation (73.4bar,6.8l/min) Cavitation Test 6 Flow [l/min] Downstream Pressure [bar] Figure 1: The result of a cavitation test. [B, 2 points] What maximum supply pressure can be assumed when at maximum flow of 10l/min in figure 1? [C, 4 points] Suppose the arrangement of orifices is changed so that another identical orifice is added in series with the first. At a supply pressure of 300bar, what cavitation downstream pressure after the two orifices can be expected?

3 EXAMINATION Page 3(7) 2 Servo Valves Characteristics [A, 2 points] Servo valve bandwidth is one of the main characteristics for the hydraulic servo system s response. In figure 2 is the data sheet characteristics of an MOOG G valve shown. The valve is a two-stage valve without hydraulic pilot-stage. What is the bandwidth (in Hz) for a ±40% amplitude response when only considering a maximum 3dB drop in amplitude ratio? 3 0 Frequency response of 15gpm servovalve MOOG G ±40% ±100% 150 Amp. Ratio [db] Phase Lag [deg] Frequency [Hz] Figure 2: The frequency repsonse for a 15 gallon-per-minute servo-valve MOOG G [B, 2 points] Using figure 2, what is the bandwidth (in Hz) for a ±100% amplitude response when considering both a maximum 6dB drop in amplitude ratio and a maximum allowed phase lag of 45. [C, 4 points] A servo valve like in figure 2 is going to be used in position control of a system. The resonant frequency defined by the system sti ness k = 16MN/m and total mass of m = 32kg sets the hydraulic frequency. Also, in general a valve phase lag of 60 can be tolerated in such a system. And at the same time an amplitude margin of at least 6dB is required. The lowest of these three frequencies will set the practical application of the system. What is it? Assume motions of moderate amplitude (±40%) [D, 2 points] The servo-valve in figure 2 is going to be used in a test system for generating a sinusoidal flow through the test object. The requirements are ±12gpm (gallon-per-minute) with a frequency of 30Hz. The generated flow will lag behind the reference signal feed into the servo valve amplifier. How long time in milli-seconds will the lag be? (Such a measure can be used for feed-forward control in critical applications.)

4 EXAMINATION Page 4(7) 3 Hydraulic Servo Systems [A, 4 points] A servo valve is used for producing a pressure response according to the diagram in figure 3. The dynamics may be approximated by a first order system with an initial delay. What is the system bandwidth,! s in Hz for both the 50bar and 120bar response? Assume 1ms delay for the 50bar response and a 2ms delay for the 120bar Response 120bar 50bar Pressure Response [bar] Time [s] Figure 3: The diagram shows two step responses from a pressure control application. The reference pressures are set to 50bar and 120bar.

5 EXAMINATION Page 5(7) [B, 2 points] Figure 4 show schematically a valve controlled position servo. The system is quite old. At centre position of the piston the chamber volumes are equal, V 1 = V 2 = Vt 2 =0.46litre. Thecylinder piston area, A p is 15.9cm 2. The moving mass m t is 42kg. The system has been tested with the piston in centred position using step responses. It has been seen from these tests that the damping is quite good, h =0.31. New oil where used during the tests with a bulk modulus of e = 1GPa. From this, calculate the hydraulic resonance frequency! h and the servo valve e ective leakage coe cient K ce.! h = s h = K ce 2A p 2 ea p m t s 1 V V 2 em t 1 V V 2 (1) (2) Figure 4: A classical servo system with a symetric cylinder, a controller and a servo-valve. [C, 2 points] Most of the time the system in figure 4 is operated in an o -centre position, V 1 =0.25V t and V 2 =0.75V t. What are the dynamic system characteristics ( h,! h )then? [D, 2 points] After some time it is discovered that the system characteristics is changed due to the rather old components. Investigations find a problem with supply system tank. Its poor design adds quite a lot of air into the system. As a consequence the sti ness of the oil is reduced by 20%. In the o -centre position described above, what are the air-a ected dynamic system characteristics ( h,! h ) now? Assume oil density unchanged.

6 EXAMINATION Page 6(7) 4 System Models [A, 6 points] In figure 5 is a servo system seen with its symmetrical cylinder, feedback and servo-valve. Draw the fully reduced block-diagram of the closed-loop system. Indicate the feedback controller and the electromagnet as separate blocks. Also indicate the intermediate signals i v and x v along with the input and output signals X r, X p and F L. Assume the hydraulic damping and frequency are represented by h and! h. The oil is represented by e and the valve by K ce and K q. V t = V 1 + V 2 Figure 5: A servo system with a symetrical cylinder, feedback loop and a servo valve. The feedback loop is represented by a position transducer with a transducer gain K f, a controller K sa and a electromagnet K ix. The damping is represented by B p. The external load force is F L. [B, 4 points] K ix represents the amplifier stage of the electric power stage for the electromagnet. The force applied by the electromagnet is opposed by the flow-forces inside the valve. For larger velocities of the piston, ẋ p, these forces can be significant. Compare two cases, one where the valve is operating close to zero valve position, (x v 0), to another case where the piston is moving at maximum speed, ẋ p =0.6m/s. What flow force can be expected in this later case? Assume C q =0.67, = 69, maximum x v =0.4mm and the diameter of the valve is 6mm. The supply pressure P s is 210bar and during this maximum speed motion a pressure di erence across the servo-valve s outlet ports (marked A and B) of 60bar have been measured.

7 EXAMINATION Page 7(7) 5 Feedback Systems [A, 10 points] Draw a reduced and simplified block diagram for the closed-loop system in figure 6. Also, calculate the hydraulic characteristics! h and h for the given parameter values. Finally, indicate in a sketch the principle look of the open-loop system Bode-diagram and indicate! h and the amplitude margin A m by value. There is a substantial internal leakage in the cylinder, C ip = The external damping B p can be neglected. V 1 = V 2 =9.0l, A p = 65cm 2,m t = 180kg, K ce = ,K q = 0.28, e =1.03GP a. The mechanical linkage forms a regulator G reg that implements the control law x v =0.4(x ref x p ). Figure 6: A servo system with mechanical linkage as feedback system and a symetric cylinder. Input to the system is an externally applied motion to the x ref point.

8 9/7/13 6:36 PM /Users/magse/exams/TM.../TMHP51_ m 1 of 2 %Solution to exam in TMHP format compact; %% disp Task 1A ; C2=(200e3 73.4e3)/200e3 P2=270e3*(1 C2) XwCq=6.8/sqrt(200e3 73.4e3) % this combines all of valve size, opening, Cq and density q=xwcq*sqrt(270e3 P2) disp Task 1B ; P1=10^2/(C2*XwCq^2) disp Task 1C ; %assume cavitation in first orifice q=xwcq*sqrt(300e3*c2) % have second orifice already cavitated then? %upstream pressure of second orifice = downstream pressure of first orifice: P2=300e3*(1 C2) % downstream pressure of second orifice according to restriction: P3=P2 q^2/xwcq^2 % NEGATIVE!!! means that cavitation takes place first in second orifice, % contrary to assumption above % for the second orifice: P2=300e3/(1+C2) P3=P2*(1 C2) %% disp Task 2A ; % from diagram: 71Hz 71 disp Task 2B ; % from diagram: 34.5Hz < 67.5Hz > 34.5Hz 34.5 disp Task 2C ; % 6dB > 79Hz, 60deg phase > 140Hz f=sqrt(16e6/32)/(2*pi) fmin = 79 disp Task 2D ; % from diagram, phase lag = 37grad tlag=1000/30*37/360 %% disp Task 3A ; f120=1/( ) f50 =1/( ) disp Task 3B ; betae=1e9; Ap=15.9e 4; mt=42; V1=0.46e 3; V2=0.46e 3; Vt=V1+V2; dh=0.31 w=sqrt((betae*ap^2)/mt*(1/v1+1/v2)) % in rad/s Kce=dh*2*Ap/sqrt(betae*mt*(1/V1+1/V2)) disp Task 3C ; V1=0.25*Vt; V2=0.75*Vt; w=sqrt((betae*ap^2)/mt*(1/v1+1/v2)) % in rad/s dh=kce/2/ap*sqrt(betae*mt*(1/v1+1/v2)) disp Task 3D ;

9 9/7/13 6:36 PM /Users/magse/exams/TM.../TMHP51_ m 2 of 2 betae=0.8e9; w=sqrt((betae*ap^2)/mt*(1/v1+1/v2)) % in rad/s dh=kce/2/ap*sqrt(betae*mt*(1/v1+1/v2)) %% disp Task 4A ; % see formula book disp Task 4B ; Cq=0.67; d=69; x=0.4e 3; dv=0.006; Ps=210e5; Pab=60e5; w=dv*pi; dp=0.5*(ps Pab); Fs=2*abs(2*Cq*w*x*(dP)*cos(d/180*pi)) % 2 orifices, inlet and outlet %% disp Task 5A ; betae=1.03e9; Ap=65e 4; mt=180; V1=9e 3; V2=9e 3; wh=sqrt((betae*ap^2)/mt*(1/v1+1/v2)) % rad/s Kce=2.2e 10+8e 11; dh=kce/2/ap*sqrt(betae*mt*(1/v1+1/v2)) Kq=0.28; Kreg=0.4; Kv=Kq/Ap*Kreg Am= 20*log10(abs(Kv/( 2*dh*wh)))

10 9/7/13 6:35 PM MATLAB Command Window 1 of 2 >> what MATLAB Code files in the current folder /Users/magse/exams/TMHP51/TMHP51_ TMHP51_ >> TMHP51_ Task 1A C2 = P2 = XwCq = q = Task 1B P1 = e+05 Task 1C q = P2 = P3 = e+04 P2 = e+05 P3 = e+04 Task 2A ans = 71 Task 2B ans = Task 2C f = fmin = 79 Task 2D tlag = Task 3A f120 = f50 = Task 3B dh = w = Kce = e 11 Task 3C w = dh = Task 3D w = dh = Task 4A Task 4B Fs = Task 5A wh = dh =

11 9/7/13 6:35 PM MATLAB Command Window 2 of Kv = Am = >>