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Linköping University
IEI
Fluid and Mechatronic Systems
EXAMINATION
TMHP02
2010-02-06
Page 1(6)
SAMPLE of EXAMS: TMHP02, 2010-02-06
EXAMINATION IN
FLUID POWER SYSTEMS, (TMHP02)
Date:
Allowed educational aids: Tables: Standard Mathematical Tables or similar
Handbooks: Physics Handbook
Formularies:
Hydraulics and pneumatics, LiTH/IEI
Pocket calculator
Score:
Maximal score on each question is 10.
To get the mark 3, you will need 20 points
To get the mark 4, you will need 30 points
To get the mark 5, you will need 40 points
Solutions:
You will find the solutions of this examination on the notice board
in the A building, entrance 17, C-corridor to the right.
Results:
Results will be announced on the notice board.
GOOD LUCK!
Karl-Erik Rydberg,
Professor, (tel 28 11 87)
Examination
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Fluid and Mechanical Engineering Systems
1.
2 (6)
TMHP 02
2009-03-17
a) Metering orifice in a 3-port constant flow valve
The figure below shows a symbol circuit for a 3-ports constant flow valve. The pressure
compensator spring gives a constant difference pressure, Δpk = 2,0 MPa.
As
)(
pp
qp
qL
pL
Dpk
pT = 0
Describe qualitatively how the constant flow valve adjust the pump pressure (pp) to the
load pressure (pL). Design the orifice area As for the load flow qL = 1,5⋅10-3 m3/s when
the flow coefficient is Cq = 0,67 and the oil density ρ = 890 kg/m3.
(3p)
b) Lift- and sink-velocity for a pump controlled cylinder
A variable hydraulic machine is used for control of lift- (v+) and sink-velocity (v-) of a
vertical placed cylinder loaded by the force F. The displacement of the machine is D =
60·10-6 m3/rev and the speed is, np = 1500 rpm (constant). The displacement setting
range for the machine is -1 ≤ ε ≤ 1, which means that the load flow can be controlled in
two directions. At max displacement setting (-1 and +1) the volumetric efficiency of the
machine is, ηv = 0,94. The cylinder piston area is Ap = 5,0·10-3 m2.
Calculate the max lift- and sink-velocity (v+max, v-max) for the cylinder.
(3p)
c) Parallel control of hydraulic cylinders
In a hydraulic system is used three (3) mechanical connected cylinders as flow deviders
for parallel control of three (3) vertical arranged cylinders, which are loaded by the
forces F1 = 70 kN, F2 = 60 kN and F3 = 70 kN. All cylinders have equal piston areas,
Ap = 5,0⋅10-3 m2.
F1
p1 Ap
Ap
Ap
F2
x1
qs
ps
p2
Ap
Ap
Ap
F3
x2
p3
Ap
x3
Ap
Ap
Calculate the supply pressure (ps) required to drive the three loads when all cylinders
have friction losses, represented by the efficiency ηc = 0,95 (for each cylinder).
(4p)
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Fluid and Mechanical Engineering Systems
2.
Examination
3 (6)
TMHP 02
2009-03-17
a) Starting torque and volumetric efficiency for a hydraulic motor
A hydraulic motor with fixed displacement, Dm = 160 cm3/rev has at a pressure
difference of Δpm = 35 MPa, speed nm = 0 rev/min and oil viscosity η = 0,032 Ns/m2 a
hydro-mechanical efficiency of ηhmm = 0,76.
Calculate the motor starting torque and its volumetric efficiency in the given operation
point.
(3p)
b) Controller for a variable displacement pump
The figure illustrates a controller for a variable hydraulic pump. The controller consists
of a pressure controlled control valve and a variable orifice (a).
Which type of controller did the figure shown? Which is the value of the pump
displacement setting (εp) when the pump is started up (pp = pL)?
Describe in diagram the pump flow qL versus pump speed np.
(3p)
c) Pressure relief valve in a constant flow system
The figure shows a constant flow pump with a pressure relief valve and its
characteristics. The constant pump flow is qp = 50 litre/min. The pressure relief valve
has the reference pressure pref = 16 MPa and at the pressure 20 MPa its flow capacity is
qTBmax = 50 litre/min.
Show in a diagram how the pump pressure pp varies according to the load flow in the
range 0 ≤ qL ≤ 50 litre/min.
A new pump is installed with the flow qp = 65 litre/min, but the same pressure relief
valve is used.
Which is now the highest pump pressure (pp) occurred in the system?
(4p)
Examination
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IEI
Fluid and Mechanical Engineering Systems
3.
4 (6)
TMHP 02
2009-03-17
a) Load sensing hydraulic systems
F
xv
qp pp
DpLS
vL
DpLS
qp pp
qL pL Ap
Pin
xv
F
vL
qL pL Ap
Pin
The figure shows one load sensing hydraulic system with a constant flow pump and
one system with a variable load sensing pump. The two systems have the same load, F
= 45 kN. Cylinder piston area is Ap = 2,8.10-3 m2 and the pressure on the piston rod side
as well as friction forces can be neglected. The load flow to each cylinder is qL = 55
litre/min.
In the constant flow system is qp = 70 litre/min and the load sensing shunt-valve gives a
pump pressure 2,0 MPa higher than the load pressure pL. In the system with the variable
pump the pump pressure is 2,5 MPa higher than pL and the maximum pump flow is
qpmax = 70 litre/min.
Calculate the hydraulic power losses in the valves for the two systems respectively and
show in a flow/pressure-diagram, how the losses are distributed in the two systems.
(4p)
b) Valve characteristics
Describe in diagram the characteristics of the control valves (qL versus xv) in the
systems above. Is the valve characteristics load pressure (pL) dependent or not?
(2p)
c) Design of hydraulic accumulator
In a constant pressure system a pressure controlled variable pump and an accumulator
are used to deliver a load flow (qL) under 10 s, see diagram below. Max pump flow is
qpmax = 1,0 liter/s and its pressure setting is 300 bar. The accumulator has a preloading
pressure of p0 = 135 bar and the working pressure can vary from p1 = 150 bar to p2 =
300 bar respectively. The polytropic exponent is n = 1,7.
Calculate the ackumulator volume (V0) required to deliver the load flow (qL).
(4p)
Examination
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IEI
Fluid and Mechanical Engineering Systems
4.
5 (6)
TMHP 02
2009-03-17
Hydrostatic vehicle transmission
The figure below shows a hydrostatic vehicle transmission with some of the help
components needed to control the pressure levels in the closed circuit. The main pump
and motor are both variable machines. The Pump displacement is Dp and for the motor
Dm = 160 cm3/rev. Min displacement setting of the motor is εmmin = 0,20. The pressure
relief valves for the main circuit have an opening pressure of 42 MPa (valve 2) and 2,5
MPa (valve 3) respectively. Max engine speed is npmax = 1800 rpm.
p1
ep
M
2
np
1
3
em
nm
Mm
Last
p2
a) Max motor torque, Mmmax
Calculate the max torque Mmmax as the motor can produce if the hydro-mechanical
efficiency is ηhmm = 0,90.
(3p)
b) Transmission speed ratio and pump displacement to reach nmmax
Derive an expression for the speed ratio nm/np including losses of pump and motor and
calculate the required pump displacement (Dp) to reach, nmmax = 3500 rpm when the
total volumetric efficiency of the transmission is ηvt = 0,88.
(4p)
c) Design of booster pump
The figure above shows how the booster pump (1) is connected to the main pump.
Determine the booster pump deplacement (Dmp) so its flow will be approximately 20%
of the max flow from the main pump.
Calculate the hydraulic output power Ph,mp, from the booster pump.
(3p)
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Fluid and Mechanical Engineering Systems
5.
Examination
6 (6)
TMHP 02
2009-03-17
a) Supply pressure for max return velocity of a pneumatic cylinder
The figure below shows a double acting pneumatic cylinder, controlled by a 3-port
valve. The cylinder, assumed as loss free, has a piston area Ap = 1,96.10-3 m2 and on the
piston rod side the ring-area is Ap/2. The valve as well as the piston rod side of the
cylinder is supplied with constant pressure ps. For negative stroking (see figure) there
are two outlet orifices with the conductance C1 = C2 = 1,5⋅10-8 m3/(Pa s). The critical
pressure ratio is b1 = b2 = 0,40. Surrounding pressure is pa= 100 kPa and the air
temperature is 293 K.
Calculate the supply pressure, ps so the cylinder return velocity can reach its maximum
value, v-max.
(4p)
b) 3-stage piston compressor for pressurized air production
The figure shows in principle a 3-stage compressor with equal pressure ratio for each
stage, with the value p2/p1 = p3/p2 = p4/p3 = 4,5. The volumetric efficiency per stage is
ηv1 = 0,84.
Calculate the compressure outlet pressure (p4) and the total volumetric efficiency, ηvtot.
(3p)
c) Sequence control of pneumatic cylinders
Two double-acting pneumatic cylinders (A and B) will be controlled in sequence as
shown in the displacement-time-diagram below. The cylinders are supplied by pressurecontrolled 4/2-valves and the end positions of the pistons are indicated by mechanically
controlled 3/2-valves. Symbols for the valves are shown below.
Draw a sequence circuit according to the given displacement-time-diagram. A
start/stop-valve shall be included, to run one cycle and then stop.
(3p)
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Fluid och Mekanisk Systemteknik
1 (5)
Exam solutions
TMHP 02
2009-03-17
EXAM SOLUTIONS
a) Metering orifice in a 3-port constant flow valve
Data: Δpk = 2,5 MPa, qL = 1,5⋅10-3 m3/s, Cq = 0,67 and ρ = 890 kg/m3.
As
)(
pp
qp
qL
pL
Dpk
pT = 0
2
Calculate the orifice area (As): Flow equation, q L = C q As
ρ
(p
p
− pL )
The compensator valve adjust the pump pressure to the load pressure so, pp = pL + Δpk.
qL
0,0015
That gives As =
⇒ As =
= 3,34⋅10-5 m2.
6
C q 2Δp k / ρ
0,67 2 ⋅ 2,0 ⋅ 10 / 890
(3p)
b) Lift- and sink-velocity for a pump controlled cylinder
D = 60·10-6 m3/rev, np = 1500 rpm, -1 ≤ ε ≤ 1, ηv = 0,94 and Ap = 5,0·10-3 m2.
+
Lift velocity: q p = ε ⋅ D ⋅ n p ⋅ηv = Ap ⋅ v + , vmax
= 1,0 ⋅ D ⋅ n p ⋅ηv / Ap = 0,28 m/s.
−
Sink velocity: ε ⋅ D ⋅ n p / ηv = Ap ⋅ v − , vmax
= −1,0 ⋅ D ⋅ n p / ηv / Ap = -0,32 m/s.
(3p)
c) Parallel control of hydraulic cylinders
F1
p1 Ap
Ap
F2
x1
qs
ps
Ap
p2
Ap
F3
x2
Ap
Ap
p3
Ap
x3
Ap
Ap
Data: F1 = 70 kN, F2 = 60 kN, F3 = 70 kN, Ap = 5,0⋅10-3 m2 and ηc = 0,95.
Calculate the required supply pressure (ps).
p1 Ap + p2 Ap + p3 Ap
F1 + F2 + F3
F1, 2,3
⇒ ps =
.
and p1, 2,3 =
Force balance: 3ps Ap =
3ηc2 Ap
Apη c
ηc
Numerically: ps =
(70 + 60 + 70) ⋅103
3 ⋅ 0,952 ⋅ 5 ⋅10−3
= 14,8 MPa.
(4p)
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Fluid och Mekanisk Systemteknik
2.
Exam solutions
TMHP 02
2009-03-17
2 (5)
a) Starting torque and volumetric efficiency for a hydraulic motor
Data: Dm = 160 cm3/rev. Δpm = 35 MPa, nm = 0 rev/min and η = 0,032 Ns/m2 → ηhmm = 0,76.
Starting torque and volumetric efficiency at nm = 0:
D
Motor torque at start: εm = 1 gives M ut = m Δpmη hmm
2π
−6
160 ⋅ 10
35 ⋅ 106 ⋅ 0,76 = 677 Nm.
Numerically: M ut =
2π
1
Volumetric efficiency at nm = 0: η vm =
. nm = 0 gives ηvm = 0.
Δpm
1 + Cv
n mη
(3p)
b) Controller for a variable displacement pump
The figure shows a constant flow controller.
qL versus np.
At start of the pump the pump pressure is pp = pL and the spring in the setting piston
will move the displacement setting to its max value, which is εp = 1,0.
(3p)
c) Pressure relief valve in a constant flow system
Data: qp = 50 litre/min, pö = 16 MPa.
Diagram for pp versus qL in the flow range 0 ≤ qL ≤ 50 litre/min
Highest pump pressure (pp) at qp = 65 liter/min.
Max opening of the pressure relief valve gives the flow eq: qTB = constant ⋅ ΔpTB ,
2
which gives p p 65
2
⎛ 65 ⎞
⎛ 65 ⎞
= p p 50 ⎜ ⎟ = 20⎜ ⎟ = 34 MPa when qL = 0.
⎝ 50 ⎠
⎝ 50 ⎠
(4p)
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Fluid och Mekanisk Systemteknik
3.
3 (5)
Exam solutions
TMHP 02
2009-03-17
a) Load sensing hydraulic systems
vL
F
xv
DpLS
qp pp
qp pp
qL pL Ap
qL pL Ap
DpLS
vL
F
xv
Pin
Pin
Data: F = 45 kN, Ap = 2,8.10-3 m2,qL = 55 litre/min.
Constant flow system with shunt-valve: qp = 70 litre/min, ΔpLS = 2,0 MPa.
Variable load sensing pump: qpmax = 70 litre/min, ΔpLS = 2,5 MPa.
Calculate the hydraulic power losses in the two valve systems: Pf = q p p p − q L p L .
Hydraulic load power: PhL = q L p L = q L F / A p , where pL = F / Ap = 16,1 MPa .
Pump pressure: p p = p L + Δp LS gives p p , CF = 18,1 MPa resp. p p ,VP = 18,6 MPa .
CF-system: qp = 70 litre/min → PfCF
Variable pump: qp = qL → PfVP =
Losses in power diagram:
70 ⋅ 18,1 ⋅ 106 55 ⋅ 16,1 ⋅ 106
=
−
= 6,4 kW
60000
60000
55 ⋅ 18,6 ⋅ 106 55 ⋅ 16,1 ⋅ 106
−
= 2,3 kW
60000
60000
p
p
pp
pL
pp
pL
Effektförluster med
fast pump
Last
qL
Effektförluster med
variabel pump
Last
qp
q
qL
q
(4p)
b) Valve characteristics
LS-system: In the two systems are pp – pL = ΔpLS, which gives q L = C q ⋅ w ⋅ x v 2 Δp LS .
ρ
In the LS-systems the valve flow is pressure comp., and qL = const.·xv, independent of pL.
(2p)
c) Design of hydraulic accumulator
Data: qpmax = 1,0 litre/s, tc = 10 s, p0 = 135 bar, p1 = 150 bar, p2 = 300 bar and n = 1,7.
q − q pm tc ⎞
Accumulator working volume ΔV? Diagram and pump gives ΔV = (qLm − q pm )⋅ ⎛⎜ Lm
⋅ ⎟⎟ .
⎜ q
2⎠
Lm
⎝
ΔV = (5,0 − 1,0) ⋅ (4,0 / 5,0 ⋅ 10 / 2) = 16 litre.
p
150
ΔV 1
16
p0
135
Accumulator volume: V0 =
= 53 litre.
1 / n gives V0 =
1 / 1, 7
1 − ( p1 / p2 )
1 − (150 / 300)
(4p)
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Fluid och Mekanisk Systemteknik
4.
4 (5)
Exam solutions
TMHP 02
2009-03-17
Hydrostatic vehicle transmission
p1
ep
M
2
np
em
3
nm
1
Mm
Last
p2
3
Data: Dp = ?, Dm = 160 cm /rev, εmmin = 0,20, pref2 = 42 MPa, pref3 = 2,5 MPa, npmax = 1800
rpm, nmmax = 3800 rpm and ηvt = 0,88.
a) Max motor torque, Mmmax
ε m max Dm
Δp m maxη hmm . Max pressure difference is given by the
2π
adjustment of the two pressure relief valves (2 and 3), Δpmmax = 42 – 2,5 =39,5 MPa.
εmmax = 1,0 and ηhmm = 0,90 gives:
Torque equation: M m max =
M m max =
1,0 ⋅ 160 ⋅ 10 −6
39,5 ⋅ 106 ⋅ 0,90 = 905 Nm
2π
(3p)
b) Transmission speed ratio and pump displacement to reach nmmax
Flow equation: ε p D p n pη vp = ε m Dm n m
Pump displacement: D p =
1
η vm
, gives the speed ratio
nm ε p ⋅ D p
=
⋅ ηvp ⋅ ηvm .
n p ε m ⋅ Dm
ε m min ⋅ Dm ⋅ nm max
ε p max ⋅ n p ⋅ηvp ⋅ηvm
ηvp ⋅ηvm = ηvt = 0,88 gives pump displacement, D p =
3
0,2 ⋅ 160 ⋅ 3800
= 77 cm /rev.
1,0 ⋅ 1800 ⋅ 0,88
(4p)
c) Design of booster pump
The booster pump and the main pump is driven at the same speed np, which gives max
flow for the two pumps as, qbp = Dbp n p maxηvbp and q p max = ε p max D p n p maxηvp .
The displacement of the booster pump shall be sized to give qbp ≈ 0,2qpmax.
Assuming ηvbp ≈ ηvp gives Dbp ≈ 0,2 ⋅ D p = 0,2 ⋅ 77 → Dbp = 15,4 cm3/varv.
Hydraulic power of the booster pump, Ph,bp? Ph, bp = Dbp n pηvbp Δpbp
Δpbp = 2,5 MPa and ηvbp = 0,95 gives Ph , mp = 15,4 ⋅ 10 − 6 ⋅
1800
⋅ 0,95 ⋅ 2,5 ⋅ 106 = 1,1 kW.
60
(3p)
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Fluid och Mekanisk Systemteknik
5.
Exam solutions
TMHP 02
2009-03-17
5 (5)
a) Supply pressure for max return velocity of a pneumatic cylinder
Data: Ap = 1,96.10-3 m2, Ap/2, C1 = C2 = 1,5⋅10-8 m3/(Pa s), b1 = b2 = 0,40, pa = 100 kPa, Ts = 293 K.
Valve flow (2 orifices): q = C12 ⋅ p ⋅ ω12 (NTP). Max velocity is reached at ω12 =1,0,
which is valid when pa / p ≤ b12 , or the cylinder pressure p ≥ pa / b12 (1)
Force balance: p ⋅ Ap = ps
Ap
2
+ pa
Ap
2
⇒ ps = 2 p − pa (2)
Calculation of C12, b12 and p:
1/ 3
1
1
1
= 3 + 3 , C1 = C2 = C gives C12 = (C 3 / 2) = 1,19.10-8 m3/Pas and b1 = b2 = b
3
C12 C1 C 2
1− b
→ b12 = 1 − C122 ⋅ 2 2 = 0,244. p ≥ pa / b12 → p >= 410 kPa gives sonic flow.
C
To reach max velocity v-max the supply pressure must be ps = 2p-pa = 720 kPa.
(4p)
b) 3-stage piston compressor for pressurized air production
p2/p1 = p3/p2 = p4/p3 = 4,5 and ηv1 = 0,84.
Calculate the compressor outlet pressure (p4): p4 =
p23
.
p12
p1 = 1 bar and p2 = 4,5 bar gives p4 = 4,53 = 91 bar.
Total volumetric efficiency (ηvtot): ηvtot = ηv1 ⋅ηv 2 ⋅ηv 3 , gives ηvtot = 0,843 = 0,59.
(3p)
c) Sequence control of pneumatic cylinders, A and B
Given sequence
Circuit diagram
(3p)
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Fluid and Mechanical Engineering Systems
1.
Examination
2 (6)
TMHP 02
2008-03-15
a) Serial connected orifices in a hydraulic system
The figure shows a hydraulic line with three serial connected orifices, all with equal
area As. Inlet pressure to the first orifice is constant, p1 = 120 bar. The pressure after the
third orifice is pT = 0.
Calculate the pressures p2 and p3.
(3p)
b) Efficiency for a valve controlled system with an open-center valve
The figure illustrate a hydraulic motor, which shaft speed (nm) is controlled by an opencenter valve. The valve is supplied by a constant flow pump (qp = constant). To rotate
the motor shaft a load pressure pL = 10 MPa (independent of speed) is required and the
pump pressure is then pp = 11 MPa. Pump and motor can be assumed as loss free.
Calculate the hydraulic efficiency η h =
qL ⋅ p L
at the following motor speeds:
qp ⋅ pp
nm = 0, nm = 0,5 nmmax and nm = nmmax.
(3p)
c) Design of a hydraulic cylinder for limited suspension at loading
The figure shows a cylinder, which is loaded by a force disturbance ΔF = 50kN. When
the cylinder piston is moved the length L = 0,50 m from end position the suspension
according to the force disturbance shall not be bigger than Δx =2,0 mm. The bulk
modulus for the volume V is βe = 1000 MPa.
Calculate the piston area, Ap required to fulfil the piston suspension limit.
(4p)
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Fluid and Mechanical Engineering Systems
2.
Examination
3 (6)
TMHP 02
2008-03-15
a) Volumetric efficiency for a hydraulic pump versus shaft speed
A speed controlled hydraulic pump with fixed displacement has at a pressure difference
of Δpp = 32 MPa and the speed np = 25 rev/s a volumetric efficiency of ηvp = 0,92. The
dynamic viscosity of the oil is η = 0,032 Ns/m2.
Calculate the lowest speed np0 required for the pump to start to deliver an effective
output flow at the pressure difference 32 MPa.
(3p)
b) Controller for a variable displacement pump
The figure illustrates a controller for a variable hydraulic pump.
Which type of controller did the figure shown? Which is the value of the pump
displacement setting (εp) when the pump is started up (pp < pref)?
Describe in diagram how the controller characteristics will be changed according to
adjustment of the reference pressure (pref).
(3p)
c) Valve characteristics
Below is illustrated two different valve concepts used for pressure and flow control in
hydraulic systems.
Which are the names of the two valves in the figure and which principal function do
they have?
Show in diagram the characteristics (pressure versus flow) for the two valves.
For the left valve show it’s working range (p/q-diagram) and how the flow forces
influence the control characteristic.
(4p)
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Fluid and Mechanical Engineering Systems
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Examination
4 (6)
TMHP 02
2008-03-15
a) Constant pressure system and load sensing system
Constant pressure system with variable pump
Load sensing system with variable pump
The figure shows a hydraulic system of constant pressure type with a pressure
compensated closed-center valve and a load sensing system with variable pump and a
closed-center valve. Both systems have to operate a lifting load, F = 45 kN and the load
flow to each cylinder is qL = 55 litre/min. The cylinders, which are equal with the piston
area Ap = 2,8.10-3 m2 can be assumed loss free. The pressure relief valves are closed
during operation.
In the constant pressure system the pump pressure is pp = 25 MPa and the maximum
pump flow is qpmax = 70 litre/min.
In the load sensing system the max pump flow is qpmax = 70 litre/min and the pump
controller creates a pump pressure which is 2,0 MPa higher than the load pressure pL.
Calculate the pump input power, Pin for the two systems at the given loading case and
when the pump overall efficiency is ηpt = 0,88.
Losses according to the return flow from the cylinders can be neglected.
(4p)
b) Control valve characteristics
Describe in diagram the characteristics of the control valves (qL versus xv) in the
systems above. Is the valve characteristics load pressure (pL) dependent or not?
(2p)
c) Hydraulic accumulator in a constant pressure system
In a constant pressure system (see figure above) a hydraulic accumulator with the total
volume, V0 = 20 liter shall be installed. The acc. pre-loading pressure p0 = 135 bar and
the working pressures are p1 = 150 bar and p2 = 250 bar respectively. The accumulator
cycle time is about 20 s and it has been designed for an effective polytrophic exponent
of n = 1, 6 (adequate for a surrounding temperature of 20o C).
Assume that the system shall be used out-doors at a temp of -30o C, which gives an
effective polytrophic exponent of n = 2,0.
Which accumulator volume (V0) is now required if the accumulator has to work with
the same volume diff. (ΔV) and the same pressure range as in the first design case?
(4p)
Examination
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Fluid and Mechanical Engineering Systems
4.
5 (6)
TMHP 02
2008-03-15
Hydrostatic vehicle transmission with a mechanical gear-box in series
em
ep
Dieselengine
Pin
np
Dp
Dp
nm
Dm
Mm
UL
nd
Md
Last
The figure shows a hydrostatic transmission equipped with a variable pump and a
variable motor, which is mechanically connected to a gear-box. Pump and motor
displacement are Dp and Dm respectively. The lowest motor setting is εmmin = 0,2.
Pump- and motor settings are controlled in sequence. Max pump speed is npmax = 2100
rpm. Transmission max pressure difference is Δpmax = 42 MPa. The mechanical gear
ratio between the hydraulic motor shaft and the driving shaft is UL = nm/nd = 65.
The vehicle driving shaft has the following performance requirements:
Required driving torque at start is Mdmax = 95 kNm. At max vehicle velocity the
driving shaft speed is ndmax = 70 rpm.
a) Dimensioning of pump- och motor-displacement
Calculate the pump displacement, Dp and the motor displacement Dm, which will fulfil
the performance requirements.
The volumetric efficiencies for pump and motors are ηvp = 0,9 and ηvm = 0,94 respectively and the hydro-mechanical motor efficiency is ηhmm = 0,88 (start efficiency).
(6p)
b) Max theoretical power, Pinmax for the transmission and it’s TR-value
Calculate (loss free) max input power, Pinmax, which can be transferred by the pump
when Pinmax is reached at the pump displacement setting εpN = 0,85.
⎛
⎞
n
Calculate loss free the transmission TR-value, ⎜⎜ TR = d max ⎟⎟ , which is the speed range
ndN ⎠
⎝
where the transmission can transfer max input power.
(4p)
Examination
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Fluid and Mechanical Engineering Systems
5.
6 (6)
TMHP 02
2008-03-15
a) Lowering velocity for a valve controlled single-acting pneumatic cylinder
The figure explains a pneumatic cylinder, with the lowering velocity (vs) controlled by
two orifices in serial connection. The cylinder, which is assumed as loss free, is loaded
by the force F = 950 N. The cylinder piston areas are A1 = 1,96.10-3 m2 and A2 =
1,47.10-3 m2. The supply pressure for the control valve is ps = 800 kPa. The conductance for the valve orifices are C1 = C2 = 2,0.10-8 m3/Pas and their critical pressure
ratios are b1 = b2 = 0,35.
vs
F
A2
pa
ps
pa
)(
b1, C1
b2,
C2
p1
A1
Calculate the piston velocity, vs when the reference pressure is p0 = 100 kPa and the
pressurised air temperature is 293 K.
(5p)
b) Volumetric efficiency for a piston compressor
The figure shows the volumetric efficiency versus pressure ratio for a single-stage
piston compressor when the relative “dead-volume” is ε = 0,06 and ε = 0,12.
Which is the approximate value of the volumetric efficiency, ηv when ε = 0,08 and the
pressure ratio p2/p1 = 8? How can the compressor efficiency ηv increases?
(2p)
c) Sequence control of pneumatic cylinders
Two double-acting pneumatic cylinders (A and B) will be controlled in sequence as
shown in the displacement-time-diagram below. The cylinders are supplied by pressurecontrolled 4/2-valves and the end positions of the pistons are indicated by mechanically
controlled 3/2-valves. Symbols for the valves are shown below.
Draw a sequence circuit according to the given disp-time-diagram. A start/stop-valve
shall be included. Observe that the stroke A+ has reduced velocity.
(3p)
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1 (5)
Solutions for exam
TMHP 02
2008-03-15
EXAM SOLUTIONS
1.
a) Series connection of orifices in a hydraulic system
p1 = 120 bar and pT = 0
Calculate the pressures p2 and p3.
Flow equations: q1 = C q As
2
ρ
( p1 − p 2 ) = C q As
2
ρ
( p2 − p3 ) = C q As
2
ρ
( p 3 − 0) .
Equal orifices gives, p1 − p2 = p2 − p3 = p3 → p2 = p1 − p3 , p3 = p2 / 2 , and
p2 = p1 / 1,5 = 120 / 1,5 = 80 bar ; p3 = p2 / 2 = 80 / 2 = 40 bar.
(3p)
b) Efficiency for a valve controlled system with an open-center valve
qp = constant, pL = 10 MPa (independent of speed) and pp = 11 MPa. Pump and motor
can be assumed as loss free.
q ⋅p
Calculate η h = L L at the motor speeds nm = 0, nm = 0,5 nmmax and nm = nmmax.
qp ⋅ pp
nm = 0 → η h 0 =
0,5 ⋅ q p ⋅ p L
0 ⋅ pL
= 0 ; nm = 0,5·nmmax → η h 0,5 =
= 0,45 ;
qp ⋅ pp
qp ⋅ pp
nm = nmmax → ηh1 =
q p ⋅ pL
= 0,91
qp ⋅ pp
(3p)
c) Design of a hydraulic cylinder for limited suspension at loading
ΔF = 50kN, L = 0,50 m, βe = 1000 MPa och Δxmax =2,0 mm.
Calculate the piston area, Ap for Δxmax =2,0 mm. Hydraulic spring constant for the
2
β e Ap
ΔF β e Ap
=
= [V = Ap L] =
[N / m] , which gives Ap min = ΔF ⋅ L .
cylinder:
ΔX max ⋅ β e
ΔX
V
L
Numerically: Ap min =
50 ⋅ 103 ⋅ 0,5
= 0,0125 m2.
0,002 ⋅ 1000 ⋅ 106
(4p)
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2.
Solutions for exam
TMHP 02
2008-03-15
2 (5)
a) Volumetric efficiency for a hydraulic pump versus shaft speed
Data: Δpp = 32 MPa, np = 25 rev/s and η = 0,032 Ns/m2 gives ηvp = 0,92.
Calculate the lowest speed np0 required to deliver an effective output flow at 32 MPa:
nη
Δp
η vp = 1 − Cv p ⇒ Cv = p (1 − η vp ) = 2,0 ⋅ 10 −9 . At np0 the efficiency is ηvp = 0, and
n pη
Δp p
n p 0 = Cv
Δp p
η
Numerically: n p 0 = Cv
Δp p
η
= 2,0 ⋅ 10 −9
32 ⋅ 106
= 2,0 rev/s.
0,032
(3p)
b) Controller for a variable displacement pump
Constant pressure controller.
Influence from a change in pref.
At start the pump pressure is pp < pref and the spring in the stroking cylinder (and the
control pressure) moves the displacement setting to its maximum value, εp = 1,0.
(3p)
c) Valve characteristics
The left valve is a direct controlled pressure relief valve, which control the pressure p1
before the valve.
The right valve is a 2-port constant flow valve, which control the load flow, qL after the
valve.
Valve characteristics:
(4p)
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3.
Solutions for exam
TMHP 02
2008-03-15
3 (5)
a) Constant pressure system and load sensing system
Constant pressure system with variable pump
Load sensing system with variable pump
Data: F = 45 kN, qL = 55 litre/min, Ap = 2,8.10-3 m2. CP-system: pp = 25 MPa and qpmax
= 70 litre/min. LS-system: qpmax = 70 litre/min and pp = 2,0 MPa + pL.
Calculate the pump input power, Pin ηpt = 0,88
CP-system: PinCP =
LS-sys.: PinLS
qp pp
ηtp
[
]
= q p = qL =
qL p p
ηtp
. PinCP
55 ⋅ 25 106
=
⋅
= 26 kW.
0,88 60000
⎛F
⎞
⎛ 45 ⋅ 10 3
⎞
+ ΔpLS ⎟⎟
qL ⎜⎜
+ 2 ⋅ 10 6 ⎟⎟
55 ⋅ ⎜⎜
−3
Ap
q ( p + ΔpLS )
⎝ 2,8 ⋅ 10
⎠ = 18,8 kW.
⎠. P =
= p L
= ⎝
inLS
0,88 ⋅ 60000
ηtp
ηtp
(4p)
a) Control valve characteristics
CP-system: Pressure compensated valve gives: qL = Cq ⋅ w ⋅ xv 2 Δpc where Δpc is a
ρ
constant pressure drop controlled by the compensator.
LS-system: The LS-pump gives pp – pL = ΔpLS, and q L = C q ⋅ w ⋅ x v 2 Δp LS .
ρ
In both the CP- and the LS-system the valve flow is pressure compensated, which gives
qL = constant·xv, independent of pL.
(2p)
c) Hydraulic accumulator in a constant pressure system
Data: V0 = 20 litre, p0 = 135 bar, p1 = 150 bar, p2 = 250 bar, tc = 20 s and n = 1,6.
p
ΔV 1
1/ n
p0
p0 ⎡ ⎛ p1 ⎞ ⎤
Accumulator volume: V0 =
⎢1 − ⎜ ⎟ ⎥ = 4,92 litre.
1 / n and ΔV = V0
p1 ⎢ ⎜⎝ p2 ⎟⎠ ⎥
1 − ( p1 / p2 )
⎣
⎦
p
150
ΔV 1
4,92
p0
135
=
V0 for n = 2,0: Numerically: V0 =
= 24,3 litre.
1/ n
1/ 2,0
1 − ( p1 / p2 )
1 − (150 / 250 )
(4p)
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4.
4 (5)
Solutions for exam
TMHP 02
2008-03-15
Hydrostatic vehicle transmission with a mechanical gear-box in series
em
ep
Pin
Dieselengine
Dp
np
Dp
nm
Dm
Mm
UL
nd
Md
Last
Dp = ?, Dm = ?, εmmin = 0,2. Pump and motor displ. settings are controlled in sequence.
npmax = 2100 rpm, Δpmax = 42 MPa, UL = nm/nd = 65, Mdmax = 95 kNm, ndmax = 70 rpm,
ηvp = 0,9, ηvm = 0,94 och ηhmm = 0,88
a) Dimensioning of pump- och motor-displacement
Motor displacement, Dm?
ε
D
M
Max motor torque: M m max = m max m Δp max ⋅η hmm = d max , gives with εmmax = 1,
2π
UL
2π ⋅ M d max
2π ⋅ 95 ⋅ 103
Dm =
, Dm =
= 250 cm3/rev.
6
U L ⋅ Δp max ⋅η hmm
65 ⋅ 42 ⋅ 10 ⋅ 0,88
Pump displacement, Dp? q ep = ε p D p n pη vp and qem = ε m Dm nm / ηvm . With nm=UL·nd,
the speed ratio is,
U L ⋅ nd ε p D p
=
ηvpη vm . εp = 1,0, εm = εmmin and nd =ndmax gives the
np
ε m Dm
pump displacement: D p =
Numerically: D p =
U L ⋅ nd max ⋅ ε m min ⋅ Dm
.
n p ⋅ηvp ⋅η vm
65 ⋅ 70 ⋅ 0,2 ⋅ 250
= 128 cm3/rev.
2100 ⋅ 0,9 ⋅ 0,94
(6p)
b) Max theoretical power, Pinmax for the transmission and it’s TR-value
Calculate, loss free, the max input power, Pinmax that can be transferred by the pump at
the displacement setting εpN = 0,85.
Pinmax at εpN = 0,85?
Pin max = ε pN D p n p max Δp max → Numerically:
Pin max = 0,85 ⋅ 128 ⋅ 10 − 6 ⋅
2100
⋅ 42 ⋅ 10 6 = 160 kW.
60
⎛
⎞
n
Calculate loss free the transmission TR-value (Theoretical Range), ⎜⎜ TR = d max ⎟⎟ .
ndN ⎠
⎝
ε p max ⋅ D p 1
⋅
nd max ε m min ⋅ Dm U L
1
1
→ Numerically: TR =
= 5,88.
=
TR =
=
ε pN ⋅ D p 1 ε pN ⋅ ε m min
ndN
0,85 ⋅ 0,2
⋅
ε m max ⋅ Dm U L
(4p)
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5.
Solutions for exam
TMHP 02
2008-03-15
5 (5)
a) Lowering velocity for a valve controlled single-acting pneumatic cylinder
vs
F
A2
pa
ps
pa
)(
b1, C1
b2,
C2
p1
A1
Data: A1 = 1,96.10-3 m2, A2 = 1,47.10-3 m2, F = 950 N, ps = 800 kPa, C1 = C2 = 2,0.10-8 m3/Pas,
b1 = b2 = 0,35, p0 = pa = 100 kPa och T = 293 K (Kt = 1,0).
Calculate the piston velocity, vs: Valve flow and cylinder flow at Kt = 1 gives,
C ⋅ω ⋅ p
p
q = C12 ⋅ p1 ⋅ ω12 (NTP), q = A1 ⋅ v s ⋅ 1 (NTP) → Lowering velocity: v s = 12 12 0 .
A1
p0
Calculation of C12 and ω12:
1/ 3
1
1
1
= 3 + 3 , C1 = C2 = C ger C12 = (C 3 / 2) = 1,59.10-8 m3/Pas and b1 = b2 = b →
3
C12 C1 C 2
1− b
b12 = 1 − C122 ⋅ 2 2 = 0,178. pa / p1 ≤ b12 → p1 >= 561 kPa gives critical flow.
C
F
950
+ pa =
+ 100 ⋅ 10 3 = 585 kPa.
Forces: F = ( p1 − pa ) ⋅ A1 → p1 =
−3
A1
1,96 ⋅ 10
The outlet orifices gives critical flow, means that ω12 = 1.0, which gives the lowering
C ⋅ p 1,59 ⋅ 10 −8 ⋅ 100 ⋅ 103
velocity as: v s = 12 0 =
= 0,81 m/s.
A1
1,96 ⋅ 10 −3
(5p)
b) Volumetric efficiency for a piston compressor
The diagram gives for ε = 0,08 and the pressure ratio p2/p1 = 8 the efficiency ηv = 0,63.
The best way to improve the volumetric efficiency (ηv) is to split up the compression in
several stages in series connection, with equal pressure ratio for each stage.
(2p)
c) Sequence control of pneumatic cylinders, A and B
Displacement-time-diagram
Symbol circuit
(3p)
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Hydraulic system with open-center valve
The figure shows a hydraulic system with a fixed pump, Dp = 98⋅10-6 m3/rev, driven at
constant speed np = 1500 rev/min, a pressure relief valve with opening pressure pref =
20 MPa (characteristics shown in the figure) and a control valve with open-center and
blocked load ports. The open-center orifice area is a linear function of the valve spool
displacement (xv), Aoc = woc(xvmax – xv) where the area gradient is woc = 0,022 m and
xvmax = 0,005 m. The flow coeff is Cq = 0,67 and the oil density ρ = 890 kg/m3.
np
Pin
Dp
qp
pp
xv
pref
pT = 0
pp [MPa]
)(
1.
2 (6)
EXAMINATION
TMHP 02
2007-03-17
Blocked
load ports
Aoc
26
20
0
0
Pressure relief valve
qPR [litre/min]
qp
a) Pump pressure versus control valve displacement
Calculate and show in a diagram how the pump pressure (pp) varies according to the
valve displacement (xv) in the range 0 ≤ xv ≤ xvmax when the pressure relief valve is
assumed to have ideal characteristics and all leakage flow is neglected.
(Calculate the pump pressure for at least 3 different values of xv.)
(5p)
b) Maximum input pump power, Pinmax
Calculate the input pump power at maximum loading, Pinmax, with respect to the pressure relief valve characteristics and when the hydro-mechanical efficiency of the pump
is ηhmp = 0,90.
(2p)
c) Flow forces in the open-center valve
Show with equations in which operation point, for the system above, the flow forces
from the open-center orifice will have its maximum value, Fsmax.
Assume that the pump and pressure relief valve have ideal characteristics.
(3p)
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2.
EXAMINATION
TMHP 02
2007-03-17
3 (6)
a) Telescopic cylinder for constant velocity
The figure shows a telescopic cylinder loaded by the force F. The biggest piston area is,
A1 = 0,025 m2. Relations for other areas are, A11 = A2 and A21 = A3. Cylinder losses can
be neglected.
Calculate the pressure p1 when the cylinder is loaded by the force F = 80 kN.
When the cylinder is loaded by the force F the total cylinder suspension is measured to
12 mm. Calculate approximately the suspension for the biggest piston (A1) relatively to
the cylinder bottom housing.
(4p)
b) Controller for a variable displacement pump
The figure shows a controller for a variable hydraulic pump.
Describe how the pump pressure (pp) is controlled according to the load pressure (pL)
and show in a diagram the load flow (qL) versus the pump speed (np) for different
adjustments of the load orifice area (a).
(3p)
b) Volumetric efficiency for a hydraulic motor
A hydraulic motor with fixed displacement Dm = 80 cm3/rev is supplied with a constant
flow qem = 1,35⋅10-3 m3/s. At a pressure difference of Δpm = 32 MPa the motor shaft
speed is measured to nm = 16 rev/s.
Calculate the volumetric efficiency of the motor (ηvm) in this operation point.
(3p)
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a) Constant flow system and load sensing system
Constant flow system with open-center valve
F
vL
xv
qp
Load sensing system with variable pump
vL
F
xv
DpLS
qp pp
pp
qL pL Ap
q L pL A p
Pin
Pin
The figure shows a hydraulic system of constant flow type with an open-center valve
and a load sensing system with variable pump. Both systems have to operate a lifting
load, F = 40 kN and the load flow to each cylinder is qL = 55 litre/min. The two
cylinders are equal with the piston area Ap = 2,8.10-3 m2 and its efficiency according to
friction and pressure forces on the piston rod side is ηc = 0,85. The pressure relief valves
are closed during operation.
In the constant flow system the pump pressure pp, is 1,0 MPa higher than the load
pressure pL and the constant pump flow is qp = 70 litre/min.
In the load sensing system the max pump flow is qpmax = 70 litre/min and the pump
controller creates a pump pressure which is 2,5 MPa higher than the load pressure pL.
The overall efficiency of the pumps is ηtp = 0,88 and the input pump power is Pin.
Calculate the system efficiency: η s =
qL pL
for the two systems.
Pin
(6p)
b) Accumulator system for motor driving
A hydraulic system with an accumulator, a constant flow and a motor is illustrated in
the figure below. The figure also shows the motor torque versus speed. The motor
displacement is Dm = 100.10-6 m3/rev and the load troque is Mm. The total accumulator
volume is V0 and its polytropic exponent is n = 1,65. Max accumulator pressure is p2 =
35 MPa and the pre-loading pressure is p0 = 0,9·p1. When the motor is running at
maximum speed (see diagram) the pressure losses from accumulator to the motor shaft
(CF-valve and motor losses) is 5,0 MPa. The motor leakage flow is neglectable.
Ack.
pö
pT = 0
250
KF-ventil
p L pm
Dm
nm M m
Mm [Nm]
ps
)(
3.
4 (6)
EXAMINATION
TMHP 02
2007-03-17
0
nm [rpm]
1000
Calculate the accumulator volume V0, required for motor driving at maximum speed during a
time period of 10 s when the accumulator is assumed to be fully loaded at the time t = 0 s.
(4p)
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4.
5 (6)
EXAMINATION
TMHP 02
2007-03-17
Hydrostatic vehicle transmission with two hydraulic motors
em
ep
Dieselengine
Pin
np
Dm1
em
Dp
Dm2
Um=1
nm
UL
Um=1
nd
Md
Last
The figure shows a hydrostatic transmission equipped with a variable pump and two
variable motors, which are hydraulically connected in parallel and the motor shafts are
mechanically connected to one shaft. The motor displacements are equal (Dm1 = Dm2)
and its displacement settings are equally controlled from 1,0 to the lowest setting, εmmin
= 0,2. Pump and motor settings are controlled in sequence. Max pump speed is npmax =
2200 rpm. Transmission max pressure difference is Δpmax = 42 MPa. The mechanical
gearing between the motor shaft and the driving shaft is UL = nm/nd = 60.
The vehicle driving shaft has the following performance requirements:
Required driving torque at start is Mdmax = 108 kNm. At max vehicle velocity the
driving shaft speed is ndmax = 80 rpm. Max driving power is Pdmax = 120 kW.
a) Design of pump and motor displacement
Calculate the required pump displacement, Dp and the motor displacement Dm1 and Dm2
(Dm1 = Dm2), which will fulfil the performance requirements.
Assume that the volumetric efficiencies for pump and motors are ηvp = 0,9 and ηvm =
0,94 respectively and the hydro-mechanical efficiency of the motor is ηhmm = 0,85 (start
efficiency).
(6p)
b) Max input effekt Pinmax and pump displacement setting
Calculate maximum input pump power, Pinmax when the transmission overall efficiency
is assumed to have a value of ηtt = 0,78.
Calculate loss free the pump displacement setting, εpN required to transfer the
calculated input power Pinmax to the same amount of hydraulic power.
(Use the calculated pump displacement from task a. If you miss a value of Dp assume one, which gives
you a reasonable value of εpN.)
Make a judgement from your calculated εpN if the size of the pump is reasonable
according to the maximum power it has to transfer.
(4p)
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a) Piston velocity for mechanical serieal connection of pneumatic cylinders
The figure shows two pneumatic cylinders in mechanical serial connection, controlled
by four individual 3-port valves. The cylinders, which can be assumed as loss free, have
equal piston areas, A1 = 1,96.10-3 m2 and A2 = 1,47.10-3 m2. The valves are supplied
with the pressure ps = 700 kPa. The conductance for the inlet orifices during positive
stroke is C1 = C2 = 1,5.10-8 m3/Pas and for the outlet orifices C3 = 1,0.10-8 m3/Pas and
C4 = 1,2.10-8 m3/Pas. The critical pressure ratios are b1 = b2 = b3 = b4 = 0,25.
A2
A2
A1
p1
b1,
C1 ) (
ps
A1
p3
p2
b2,
C2 ) (
ps
vmax
p4
)(
b3,
C3
)(
b4,
C4
Calculate the maximum piston velocity, vmax when the reference pressure is p0 = 100
kPa and the pressurised air temperature is 293 K.
(4p)
b) Pneumatic system efficiency
A pneumatic system and its sub-efficiencies are shown in the figure. The air expansion/
compression efficiency, he/c is shown in a diagram. System pressure ratio is p2/p1 = 8
and the volumetric compressor efficiency is ηvk = 0,75 and the load/actuator efficiency
is ηL = 0,80.
F
Pk
nk
p1
p2
he/c
hvk
v
pa
)(
5.
6 (6)
EXAMINATION
TMHP 02
2007-03-17
p App
pL2
hL
Calculate the system overall efficiency, ηtot and describe how this efficiency will be
influenced if the pressure ratio is reduced.
(3p)
c) Sequence control of pneumatic cylinders
Two double-acting pneumatic cylinders (A and B) will be controlled in sequence as
shown in the displacement-time-diagram below. The cylinders are supplied by
pressure-controlled 4/2-valves and the end positions of the pistons are indicated by
mechanically controlled 3/2-valves. Symbols for the valves are shown below.
A
B
1
0
1
4/2-ventil
3/2-ventil
0
Draw a sequence circuit according to the given disp-time-diagram. A start/stop-valve
shall be included in the circuit. Observe that the stroke A- has reduced velocity.
(3p)
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SOLUTIONS
TMHP 02
2007-03-17
SOLUTIONS FOR EXAM, TMHP02
Hydraulic system with open-center valve
Pin
Dp
qp
pp
pp [MPa]
xv
)(
np
pref
Aoc
pT = 0
-6
Blockerade
lastportar
26
20
0
Tryckbegränsningsventil
qTB [liter/min]
0
qp
3
Dp = 98⋅10 m /rev, np = 1500 rev/min, pref = 20 MPa, Aoc = woc(xvmax – xv), woc = 0,022
m, xvmax = 0,005 m, Cq = 0,67 and ρ = 890 kg/m3.
a) Pump pressure versus control valve displacement
Pump and open-center valve gives the flow, q p = D p n p = C q woc ( xv max − xv )
ρ ⎛⎜
2
ρ
p p (pT=0),
2
2
⎞
⎞
890 ⎛
0,00245
⎟
and p p = ⋅
. Pump pressure: p p =
⎟ .
⋅⎜
2 ⎜⎝ C q woc ( x v max − xv ) ⎟⎠
2 ⎜⎝ 0,01474 ⋅ ( x v max − xv ) ⎟⎠
xv = 0 gives pp = 0,49 MPa and xv = 4,2 mm ger pp = 19,2 MPa. The diagram for pp
versus xv is shown in the diagram below
Dp n p
20
pp [MPa]
1.
0,5
0
xv [mm]
4,2
5
(5p)
b) Maximum input pump power, Pinmax
Pressure relief valve characteristics and pump efficiency, ηhmp = 0,90, gives:
Pump input power: Pin = M in ⋅ ω p = D p
diagram) and np = 1500 rpm → Pin max
pp
η hmp
⋅ n p . ppmax = 26 MPa (according to
26 ⋅ 106 1500
= 98 ⋅ 10
⋅
= 71 kW
0,90
60
−6
(2p)
c) Flow forces in the open-center valve
Flow force from the open-center: Fs = 2C q ⋅ woc ⋅ (x v max − xv ) ⋅ p p ⋅ cos δ (1)
Flow eq. for the open-center: qoc = C q woc ( x v max − xv )
Ekv (1) and (2) gives: Fs = 2 ⋅
qoc
2 pp / ρ
2
ρ
p p (2).
⋅ p p ⋅ cos δ . Fs max = 2 ρ ⋅ qoc max ⋅ p p max ⋅ cos δ
Max flow forces appears when the product of flow through the open-center (qocmax = qp)
and pressure drop (pp) have reach it’s max value, which is true just when the pressure
relief valve start to open.
(3p)
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2.
SOLUTIONS
TMHP 02
2007-03-17
2 (5)
a) Telescopic cylinder for constant velocity
Data: A1 = 0,025 m2, A11 = A2 and A21 = A3. The cylinder is loss free.
Calculate the pressure p1 when the cylinder is loaded by the force F = 80 kN.
Force equations for the cylinder: p3 A3 = F , p2 A2 = p3 ( A3 + A21 ) = 2 p3 A3 and
4F
p1 A1 = p2 ( A2 + A11 ) = 2 p2 A2 gives p1 A1 = 4 p3 A3 = 4 F ⇒ p1 =
A1
Numerically: p1 =
4 ⋅ 80 ⋅ 103
= 12,8 MPa.
0,025
Cylinder suspension: Total cylinder suspension is ΔxL = 12 mm. The relative
suspension for each piston will be the same, which gives Δx1 + Δx2 + Δx3 = Δx L and the
Δx
suspension of piston A1 will be as, Δx1 = L = 4,0 mm.
3
(4p)
b) Controller for a variable displacement pump
The pump pressure (pp) is controlled according to the load pressure (pL) by the pump
controller valve. Pressure balance for the valve is, pp = pL + Δpc, pp will follow pL.
a ökar
Characteristics:
qL
0
np
(3p)
c) Volumetric efficiency for a hydraulic motor
Data: Dm = 80 cm3/rev, qem = 1,35⋅10-3 m3/s, Δpm = 35 MPa gives nm = 16 rev/s.
D n
D n
Formula Book gives with εm = 1,0, q em = m m ⇒ η vm = m m
η vm
qem
Numerically: η vm =
80 ⋅ 10 −6 ⋅ 16
= 0,95
1,35 ⋅ 10 −3
(3p)
LINKÖPINGS TEKNISKA HÖGSKOLA
IEI
Fluid och Mekanisk Systemteknik
a) Constant flow system and load sensing system
Constant flow system with open-center valve
F
vL
xv
qp
Load sensing system with variable pump
vL
F
xv
DpLS
qp
pp
pp
qL pL Ap
qL pL Ap
Pin
Pin
Data: F = 40 kN, Ap = 2,8.10-3 m2,qL = 55 litre/min, ηc = 0,85 och ηtp = 0,88.
Constant flow system: pp - pL = 1,0 MPa, qp = 70 litre/min.
Load sensing system: qpmax = 70 litre/min, pp − pL = ΔpLS = 2,5 MPa.
Calculate the system efficiency, η s =
qL pL
for the two systems.
Pin
Hydraulic load power: PhL = q L p L där p L = F /( Apη c ) = 16,8 MPa .
Pump input power: Pin =
qp pp
η tp
⇒ ηs =
qL p L
ηtp
qp pp
CF-system: qp = 70 l/min, pp = 16,8+1 = 17,8 MPa ⇒ η sCF =
LS-system: qp = qL, pp = 16,8+2,5 = 19,3 MPa ⇒ η sLS =
55 ⋅ 16,8
0,88 = 0,65.
70 ⋅ 17,8
55 ⋅ 16,8
0,88 = 0,766.
55 ⋅19,3
(6p)
b) Accumulator system for motor driving
Given data: Dm = 100.10-6 m3/rev, Mm (see diagram). Accumulator: V0, n = 1,65, p2 = 35 MPa, p0 = 0,9p1.
Pressure losses Δpf= 5,0 MPa and driving time t = 10 s.
Ack.
pö
pT = 0
250
KF-ventil
pL pm
Dm
nm Mm
Mm [Nm]
ps
)(
3.
3 (5)
SOLUTIONS
TMHP 02
2007-03-17
0
nm [rpm]
1000
Calculate the accumulator volume V0 for the given load:
ΔV ⋅ p1 / p0
Accumulator volume: V0 =
, where ΔV = qm⋅t = Dm·nm·t = 0,0167 m3.
1/ n
1 − ( p1 / p 2 )
2πM m
D
The motor pressure losses included in Δpf gives: M m = m Δp ⇒ Δp =
where
2π
Dm
min accumulator working pressure is p1 = Δp + Δpf = 15,7 + 5,0 = 20,7 MPa. Acc preloading pressure is p0 = 0,9·p1 = 18,6 MPa. Numerically the accumulator volume is,
V0 =
0,0167 ⋅ 20,7 / 18,6
= 68,2⋅10-3 m3 = 68,2 litre.
1 / 1, 65
1 − (20,7 / 35)
(4p)
LINKÖPINGS TEKNISKA HÖGSKOLA
IEI
Fluid och Mekanisk Systemteknik
4.
4 (5)
SOLUTIONS
TMHP 02
2007-03-17
Hydrostatic vehicle transmission with two hydraulic motors
em
ep
Dieselengine
Pin
Dm1
em
Dp
np
Dm2
Um=1
nm
UL
Um=1
nd
Md
Last
Given data: Dm1 = Dm2, equal motor displacement setting control from 1,0 to minimum
value, εmmin = 0,2. npmax = 2200 rpm, Δpmax = 42 MPa and UL = nm/nd = 60.
a) Design of pump and motor displacement
Motor displacement, Dm1 and Dm2?
ε
( D + Dm 2 )
M
Max total motor torque: M m max = m max m1
Δp max ⋅η hmm = d max , gives with
2π
UL
εmmax = 1, Dm1 + Dm 2 =
2π ⋅ M d max
2π ⋅ 108 ⋅ 103
, Dm1 + Dm 2 =
= 317 cm3/rev.
U L ⋅ Δp max ⋅ η hmm
60 ⋅ 42 ⋅ 106 ⋅ 0,85
Equal motor displacements gives Dm1 = Dm2 = 160 cm3/rev.
Pump displacement, Dp? q ep = ε p D p n pη vp and qem = ε m ( Dm1 + Dm 2 )nm / η vm . Using the
relation nm =UL·nd, the transmission speed ratio is
ε p Dp
U L ⋅ nd
ηvpη vm .
=
ε m ( Dm 2 + Dm 2 )
np
With εp = 1,0, εm = εmmin and nd =ndmax we have: D p =
and numerically, D p =
U L ⋅ n d max ⋅ ε m min ( Dm 2 + Dm 2 )
,
n p ⋅ηvp ⋅ηvm
60 ⋅ 80 ⋅ 0,2 ⋅ (320)
= 165 cm3/varv.
2200 ⋅ 0,9 ⋅ 0,94
(6p)
b) Max input effekt Pinmax and pump displacement setting
Max input power? Pin max =
Pd max
ηtt
, gives Pin max
120 ⋅ 103
= 154 kW.
=
0,78
Pump displacement setting for Pinmax, εpN? Pin max = ε pN D p n p max Δp max →
ε pN =
Pin max
154 ⋅ 103 ⋅ 60
gives ε pN =
= 0,61.
D p n p max Δp max
165 ⋅ 10 −6 ⋅ 2200 ⋅ 42 ⋅ 106
The calculated pump displacement setting value is relatively low, which means that the
pump can transfer higher power than Pinmax (for εp = 1,0 approx. 250 kW). The pump is
over-sized, but the transmission fulfils the performance requirements.
(4p)
LINKÖPINGS TEKNISKA HÖGSKOLA
IEI
Fluid och Mekanisk Systemteknik
a) Piston velocity for mechanical serieal connection of pneumatic cylinders
A2
A2
A1
p1
b1 ,
C1 ) (
ps
A1
p3
p2
b2 ,
C2 ) (
ps
vmax
p4
)(
b3 ,
C3
)(
b4 ,
C4
Data: A1 = 1,96.10-3 m2, A2 = 1,47.10-3 m2, ps = 700 kPa, C1 = C2 = 1,5.10-8 m3/Pas, C3 = 1,0.10-8
m3/Pas, C4 = 1,2.10-8 m3/Pas, b1 = b2 = b3 = b4 = 0,25, p0 = 100 kPa and T = 293 K (gives Kt = 1,0).
Max piston velocity: vmax = v1max + v2max (sum of the 2 piston velocities).
p
Valve flow: q = pm ⋅ C ⋅ ω (NTP), Kt = 1 and cylinder flow: q = v ⋅ Ap
(NTP).
p0
Assume that the piston velocities are determined by the valve outlet orifices (C3, C4),
and the valve feeding pressure (pm) is equal to the cylinder volume pressure (p)
C ⋅ ω ⋅ p0
respectively (p2 and p4 in the figure). One piston velocity is expressed as, v =
.
Ap
ω = 1,0 gives max velocity. According to the figure the total max velocity is,
v max =
C 3 ⋅ p 0 C 4 ⋅ p0
p
+
= (C 3 + C 4 ) 0 .
A2
A2
A2
Numerically : v max = (1,0 + 1,2) ⋅ 10 −8
100 ⋅ 103
= 1,5 m/s.
1,47 ⋅ 10 −3
(4p)
b) Pneumatic system efficiency
Data : p2/p1 = 8, ηvk = 0,75 and ηL = 0,80.
F
Pk
p1
hvk
nk
p2
he/c
v
pa
)(
5.
5 (5)
SOLUTIONS
TMHP 02
2007-03-17
p App
pL2
hL
System overall efficiency: ηtot = ηvk·ηe/c·ηL = 0,75·0,33·0,80 = 0,20.
Lower pressure ratio means that ηvk and ηe/c increases.
(3p)
b) Sequence control of pneumatic cylinders
Given displacement-time-diagram
A
B
1
Circuit solution
A
a0
a1
B
b0
)(
0
1
0
b1
START
SI
SII
a0
b0
a1
b1
(3p)