Modeling, Identification and Control, Vol. 31, No. 2, 2010, pp. 67–77, ISSN 1890–1328
Control of a Buoyancy-Based Pilot Underwater Lifting Body
Finn Haugen
Telemark University College, Kjoelnes Ring 56, N-3918 Porsgrunn, Norway. E-mail: fi[email protected]
Abstract
This paper is about position control of a specific small-scale pilot underwater lifting body where the lifting force stems from buoyancy adjusted with an air pocket in the lifting body. A mathematical model is developed to get a basis for a simulator which is used for testing and for designing the control system, including tuning controller parameters. A number of different position controller solutions were tried both on a simulator and on the physical system. Successful control on both the simulator and the physical system was obtained with cascade control based on feedback from measured position and height of the air pocket in the lifting body. The primary and the secondary controllers of the cascade control system were tuned using Skogestad’s model-based PID tuning rules. Feedforward from estimated load force was implemented in combination with the cascade control system, giving a substantial improvement of the position control system, both with varying position reference and varying disturbance (load mass).
Keywords: Underwater, buoyancy, air lift, position control, cascade control, feedforward control, distur- bance, estimation, Skogestad model-based controller tuning
1 Introduction ancy. Several control functions were tried both on a simulator and on the physical system. The only Underwater lifting operations is a common task in e.g. method which worked well on the physical system was the offshore oil and gas production industry. This pa- standard cascade control with positional control as the per is about control of a small-scale physical pilot un- primary (master) control loop and air mass in the lift- derwater lifting system.1 The lifting principle is ad- ing body as the secondary (slave) control loop. The justment of the buoyancy by controlling the amount of selected control strategy has certain similarities with air in the air pocket of the lifting body. Buoyancy can the control structure described in a US patent by Ot- provide a large lifting force with little energy, but it terblad and Dovertie(1985) where the inner loop is requires a control system. A position control system is based on a measurement of the lifting force.2 The sim- designed and implemented to keep the body with load ulator and the control system are implemented in a at a reference position. A mathematical model is de- LabVIEW program running with cycle time 0.02 sec veloped to get a basis for a simulator which is used for on a PC. testing and for designing the control system, including The paper is organized as follows: The system is de- tuning controller parameters. scribed in Section2. A mathematical model is derived Fossen(1994) and others, describe position control of in Section3. This model is the basis of a simulator of underwater vehicles, but there is not much research re- the system, and it is also used for design and tuning ported about stabilizing underwater bodies using buoy- of the controllers and the observer (estimator). Con-
1This project was initiated and funded by the company Miko 2Actually, in the patent the secondary loop is based on the Marine, Oslo, Norway. time-derivative of the force measurement.
doi:10.4173/mic.2010.2.3 c 2010 Norwegian Society of Automatic Control Modeling, Identification and Control trol system design, including state observer design, is described in Section4. Experimental results are pre- sented in Section5. Conclusions are given in Section 6.
2 System Description
Fig.1 shows schematically the lifting body, including the position control system. Fig.2 (left) shows a photo of the lifting body inside a water tank used for testing the position control system. Fig.2 (right) shows the actuator which is a pneumatic control valve which ad- justs the air flow into the air pocket of the lifting body. The air flows out of the air pocket via a valve with Figure 2: Left: Water tank with lifting body. Right: fixed (manually adjustable) opening. Pneumatic control valve (Samson, Type 3241, Series 250) for controlling the air flow
PC with to the cylinder. LabVIEW u [V] in this sense, but to design and implement a proper Air Ps [bar] Fin [kg/s] position control system for the selected configuration, supply patm [bar] Surface assuming that the principal results are transferable to a different configuration. Fout [kg/s] La [m] y [m] Air Sensor La [m] Air 3 Mathematical Modeling mcyl [kg] (Ultrasound) pocket
ma [kg] pa [bar] La [m] 3.1 Variables and parameters Va [m3] rho a [kg/m3] Variables and parameters with values are given below. y [m] Water Sensor mb [kg] • y [m] is position of lift body. Position is zero at the y [m] Vb [m3] (Pressure) water surface, and positive direction is downwards.
• flift [N] is buoyancy lifting force. Load ml [kg]
• fh [N] is hydrodynamic drag or damping (friction) force acting on the lifting body from the water. Figure 1: The lifting body with load.
• fg [N] is gravity force on the system. The lifting body position y is measured with a pres- sure sensor. The position is calculated from the pres- • fd [N] is any independent environmental force in sure measurement, since the position is proportional to addition to the forces defined above. the hydrostatic pressure. The height of the air pocket in the lifting body is measured with an ultrasound level • mtot [kg] is total (resulting) mass to be lifted by sensor.3 the lifting body. It is certainly an open question whether the con- figuration shown in Fig.1 represents optimal design. • mb [kg] is mass of the ballast water. Alternative configurations are controlling the outlow, • m [kg] is mass of air in air pocket and controlling both inflow and outflow simultaneously air with split-range control. However, it was not a purpose • madd = 2.0 [kg] is added (virtual) mass related of the present project to find the optimal configuration to the forced motion of water as the lifting body 3It was important that the cap of the lifting body where the moves. sensor was mounted was isolated (we used foam plastic), to avoid disturbing sound reflections. • mcyl = 6.0 [kg] is mass of the body.
68 Haugen, “Control of a buoyancy-based pilot underwater lifting body”
• mcyllift = 2.0 [kg] is mass of the water correspond- 3.2 Mathematical modelling ing to volume taken up by the lifting body in the A mathematical model describing the motion of the water (causing buoyance force). lifting body was developed from the following modeling • ml = 5.0 [kg] (default value) is mass of load at- principles: tached to the lifting body. • Equation of motion of the lifting body with load 3 • Vb [m ] is volume of ballast. • Mass balance of the air in the air pocket of the • V [m3] is volume of lifting body. cyl lifting body 3 • Va [m ] is volume of air pocket in lifting body. Modeling details are given in the following sections. 2 • Acyl [m ] is cross-sectional area of lifting body. 2 • Ap [m ] is a cross-sectional area parameter used 3.2.1 Equation of motion to calculate f . d Applying Newton’s Second Law gives • Lcyl = 0.6 [m] is length of lifting body. mtoty¨ = −flift − fh + fg + fd (1) • d = 0.2 [m] is lifting body diameter.
mtot in eq. (1) is described in detail below. • La [m] is height of air pocket in lifting body. 3 • ρa [kg/m ] is density of air in air pocket (at the mtot = mb + madd + mcyl + ml (2) present depth). where 3 • ρaatm = 1.2 [kg/m ] is density of air at the surface (atmospheric pressure). mb = ρwVb (3) 3 • ρw = 1025 [kg/m ] is water density. Vb = Vcyl − Va (4) m V = a (5) • Fin [kg/s] is air mass flow into lifting body from a ρa the compressor, through the control valve. m is given by the mass balance of the air in the lifting • F [kg/s] is air mass flow out from lifting body a out body, cf. eq. (20) below. Now, eq. (1) becomes through outlet valve. • K = 0.004 [N/(m/s)2] is hydrodynamic drag or ma h mtot = ρw Vcyl − + madd + mcyl + ml (6) damping (friction) force. ρa
• Kv = 250 [kg/s] is valve constant of inlet (control) where valve. Vcyl = AcylLcyl (7)
• Kvout is valve constant of outlet (manual) valve. In eq. (6), ρa is a function of depth y given by the Gas Law: • patm [Pa] is atmospheric air pressure. p(y) ρa(y) = (8) • ps [Pa] is the pressure of the air out from air com- RT pressor (air supply). R is the specific gas constant, and T is the temperature. Assuming constant T , eq. (8) gives • ∆pin [Pa] is pressure drop across air inlet valve. ρ (y) ρ (0) ρ • ∆pout [Pa] is pressure drop across air outlet valve. a = a ≡ aatm (9) pa(y) pa(0) patm • ∆patm = 101000 [Pa] is atmospheric pressure. Here, • u [V or %] is valve control signal. pa(y) = patm + ρwg (La + y) (10) • g = 9.81 [m/s2] is gravity constant. Now eq. (9) gives
The parameters Kh and Kvout were adjusted manu- ally in a simulator until the simulated position showed pa(y) ρa(y) = ρaatm (11) the similar response as the measured position. The patm parameter madd was given an assumably reasonable ρwg (La + y) = ρaatm 1 + (12) value. patm
69 Modeling, Identification and Control
Here we will make the assumption that ∆pin is the pressure drop across the valve:
y La (13) ∆pin = ps − pa (22)
which will hold when the lifting body is at relatively Typically, large depths. This implies ps pa (23) Hence, ρwgy r ρ (y) ≈ ρ 1 + (14) ps a aatm F ≈ K f (u) (24) patm in v v1 patm which is used in eq. (6). Each of the terms at the right fv1 (u) is the valve flow characteristic function normal- side of eq. (1) are now described: In eq. (1), the lifting ized with values between 0 and 1. The valve used in force flift is given by this project has an equal percentage valve characteris- tic function. However, since we have an air flow meter flift = ρwVag (15) installed in the rig, we have decided to model the valve with experimental relation between control signal and where flow. Fig.3 shows the air flow Fin [kg/s] as a function of the valve control signal u [%]. The supply pressure Va = AcylLa (16) was ps = 1 bar. The circles are experimental data, and πd2 the lines are just linear interpolations. Acyl = (17) −3 4 x 10 2 In eq. (1), 1.8 dy dy fh = Kh (18) dt dt 1.6
1.4 where Kh is adjusted manually during experiments. In eq. (1), 1.2