LIU/IEI/FluMeS
2007-10-26
Exercises for TMHP51
Collection of Exercises
for the course
TMHP51 – Hydraulic Servo Systems
LIU / IEI / FluMeS
2007-10-26
1 (20)
LIU/IEI/FluMeS
2007-10-26
Exercises for TMHP51
2 (20)
Chapter 2 - Orifices
Example 2.1
In a hydraulic system the pressure p1=5 MPa is to be reduced by orifices to
p2=0,2 MPa at the flow q=0,333⋅10-3 m3/s.
Calculate the number of orifices needed to avoid cavitation in any place.
Calculate also the diameters for the different orifices. The orifices are
assumed as sharp edged with the flow coefficient Cq=0,81. Density of the oil
is 880 kg/m3.
Example 2.3
The figure shows a flapper-nozzle system with the supply pressure p1.
Describe schematically in a diagram how the pressure ratio p2/p1 varies
according to the displacement x.
Investigate at which pressure ratio the sensitivity,
dp2
will reach its
dx
maximum value. Assume turbulent flow through the orifice.
The flapper-nozzle is designed for x0 <
d
and the orifice area is A = π ⋅ d ⋅ x .
16
LIU/IEI/FluMeS
2007-10-26
Exercises for TMHP51
3 (20)
Chapter 3 – Flow forces
Example 3.1
The figure schematically shows the working of a constant flow valve. For
the flow through the variable orifice the following relation ship can be
established:
q = As C q
2( p 2 − p 3 )
ρ
.
In a working point the orifice configuration gives that AsCq=9,6⋅10-6.
Data:
d1
=
20 mm
p1
=
20⋅106 N/m2
Ff
=
600 N
d2
=
40 mm
3
ρ
=
880 kg/m .
Determine
a)
the magnitude of the constant flow, if no consideration is taken to the
flow forces acting on the valve spool.
b)
the magnitude of the constant flow, if the action of the flow forces
acting on the valve spool are concidered. As the magnitude of the
flow forces has to be calculated the flow calculated in a) can be used.
Determine the flow at p3 = 2 MPa, p3 = 8 MPa and p3 = 15 MPa.
LIU/IEI/FluMeS
2007-10-26
Exercises for TMHP51
4 (20)
Example 3.3
The figure illustrates an open-center valve with different area gradients for
the different orifice locations.
Calculate the total flow force for the case when the valve spool displacement, xv = l = 4 mm (l = L2 - L1), F=0 and A2/A1=1,0.
Data:
Cq
d
=
=
0,7
0,02 mm
qs
ρ
=
=
2,0⋅10-3 m/s
870 kg/m3.
LIU/IEI/FluMeS
2007-10-26
Exercises for TMHP51
5 (20)
Chapter 5 – Component dynamics
Example 5.10
A valve controlled cylinder with lift load and without meter-out-restriction to tank is
loaded with a mass Mt and extern load FL. Its viscous friction is Bp. The valve has zero
over and under lap. The cylinder is supplied with a constant pressure ps.
a) Derive the linearised and Laplace transformed equations needed for describing the
dynamic behaviour of the system. Derive also the transfer function of the system
sΔX p (s )
, with the mass-spring system
b) Compare the open system transfer function,
ΔX v (s )
below and identify k0, b0 and m0.
FL (s)
. Investigate what will
X p (s)
happen if Bp = Cip = Cp = 0, i.e. the total damping of the system is zero. Assume that
BpKce<<Ap2.
c) Determine the stiffness of the system in 5.10 a, S(s) =
LIU/IEI/FluMeS
2007-10-26
Exercises for TMHP51
6 (20)
d) Increase the damping with a leakage orifice, KsL. How will the leakage paths KsL, the
flow between the two cylinder volumes, Cep and the flow between the cylinder and
the free air, Cip, affect the steady state characteristics of the system?
e) Increase the damping with a pressure feedback to in this case the hydraulic controlled
manoeuvre valve. Assume BpKce<<Ap2 and that the feedback is stable. Neglect the
mass and the friction of the spool. How will the damping affect the stiffness? Is it
possible to use the benefits but avoiding the drawbacks from pressure feedback?
How?
LIU/IEI/FluMeS
2007-10-26
Exercises for TMHP51
9 (20)
Chapter 6 – Servo systems
Example 6.1
A hydraulic position servo system has the following data:
Servo valve
Cylinder and load
Cq
=
0,61
Ap
=
4,40⋅10-3 m2
w
=
1,00⋅10-2
L
=
0,15 m
-12
5
=
2,1⋅10 m /Ns
Vt
=
0,72⋅10-3 m3
Kc0
xvmax =
1,00⋅10-3 m
Bpmax =
3500 Ns/m
Ctp
=
0 m5/Ns
Mtmin =
3,0 kg
System parameters
ps
=
21,0 MPa
ρ
=
855 kg/m3
βe
=
1,2⋅109 N/m2
b
=
a
a) Determine the stationary prestanda limits of the servo, i.e. draw the piston velocity as
a function of Fl.
b) What is the power loss in the valve as Fl=0 and piston velocity=0?
c) Draw the Bode-diagram and determine the amplitude margin as Mt=Mtmin.
d) Calculate the static stiffness.
LIU/IEI/FluMeS
2007-10-26
Exercises for TMHP51
10 (20)
Example 6.2
The position servo from problem 6.1 is to be used with the mass Mt=100 kg.
a) How will the increased mass affect resonance frequency, ωh, damping δh and
amplitude margin Am?
Implement a laminar restrictor between the load ports, so that the amplitude margin is
10 dB.
b) Determine the stationary performance limits, i.e. draw vp as a function of FL. What is
the maximal load that can be controlled by the servo? What is the power loss at this
maximum load?
c) What is the magnitude of the power loss at FL=0? If we use a underlapped valve with
the same properties around xv=0 instead of the valve plus the orifice from a), what
would the power loss be at FL=0?
d) Determine the stationary closed loop stiffness, S(s), with the load port restrictor and
with the underlapped valve respectively.
LIU/IEI/FluMeS
2007-10-26
Exercises for TMHP51
12 (20)
Example 6.4
The servo from problem 6.1 is provided with a dynamic load pressure feedback, see
figure below:
a) Dimension the pressure feedback (orifice constant Kcd, spring rate kd and piston area
Ad, so that the stability margin will be 10 dB as the mass is Mt = 100kg.
b) How are the stationary performance limits for the servo affected by the feedback?
c) Determine the power loss for the valve when FL = 0.
d) Calculate the stationary closed loop stiffness for the servo.
LIU/IEI/FluMeS
2007-10-26
Exercises for TMHP51
Answers to exercises
Example 2.1
4 orifices are required. The orifice diameters shall be selected as:
d1 = 2,56 mm, d2 = 3,15 mm, d3 = 3,86 mm, d4 = 4,68 mm.
Example 2.3
Pressure ratio:
p2
1
=
p1 C ⋅ x 2 + 1
Max sensitivity
dp2
dx
for
max
p2 3
=
p1 4
Example 3.1
a) Constant flow capacity (no flow forces): qv = 3,16⋅10-4 m3/s.
b)
q2MPa = 3,136⋅10-4 m3/s
q8MPa = 3,141⋅10-4 m3/s
q15MPa = 3,149⋅10-4 m3/s
Example 3.3
Total flow force: Fst = 14,2 N
Example 5.10
-
Example 5.12
-
Example 5.13
-
Example 6.1
•
a)
x p = 0,217 ⋅ 1 − 1,08 ⋅ 10 −5 ⋅ FL
b)
Pf = 926 W
c)
Am = 20,7 dB
d)
ΔFL
= 1,0⋅109 N/m
ΔX p
18 (20)
LIU/IEI/FluMeS
2007-10-26
Exercises for TMHP51
Example 6.2
a)
Ct = 4,43⋅10-11 m5/Ns
b)
FL = 58 kN, Pf = 8 kW
c)
Pf = 20,4 kW
ΔFL
= 4,5⋅107 N/m
ΔX p
d)
Example 6.3
a)
Ad/kd= 4,6⋅10-11 m3/N
b)
The steady state performance will not be affected
c)
Pf = 926 W
ΔFL
= 4,5⋅107 N/m
ΔX p
d)
Example 6.4
a)
-
b)
The steady state performance will not be affected
c)
Pf = 926 W
d)
ΔFL
= 1,0⋅109 N/m
ΔX p
Example 6.9
Dp = 100 cm3/rev., Dm = 35 cm3/rev.
QPrv = 2,5 litre/s (Pressure relief valve), QSv = 1,5 litre/s (Servo valve),
Example 6.12
a)
b)
c)
Kes = 1,76⋅10-3 m/Vs, Kv = 17 1/s (Kq0 = 0,62 m2/s)
ΔTL
⋅
Δθ m
⇒∞
ε0 = 0 rad/s (if Kt = 1,0)
19 (20)