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Huhtala 2010
Hydrostatic drives
- dimensioning
Huhtala 2010
Pump controlled hydrostatic system; efficiency of the
components
p
pT
m
mT
LT
mmC
ppC
Q Q
lpQlmQpp mp
pp
p
tp
p
pp
pp
ppp
hmp
pp
vp
lppp
T
pQ
T
pC
T
pC
C
Q
QCQ
m
mmtp
mm
m
mmm
mmhmp
mmvm
lmmm
pQ
T
pC
T
pC
T
Q
C
QCQ
pC mC
Huhtala 2006
Volumetric and hydromechanical losses at pumps
and motors
Flow losses is compounded of
• Slip or leakages in narrow clearances
• Dependent on pressure
• Leakage normally laminar flow
• Oil compressibility losses
• Dependent on oil bulk modulus (oil B≈1500 MPa)
• Effective volumetric displacement smaller than geometric
Torque losses
• Based on friction in narrow lubrication films between moving parts
• Different friction types
• Viscous friction
– Direct proportional to speed
• Coulombin friction (dry friction)
– Direct proportional to pressure
– At the biggest on startingConstant friction
• Constant friction
– e.g. sealing friction
dt
pd
B
VV
0
pKQ
Huhtala 2006
Flow types
Laminar:
Flow rate is linearly dependent on pressure. Flow rate
is dependent on oil viscosity, which is changing as a
function of temperature. , K is dependent on
flow area shape and lenght.
Turbulent:
Flow rate is nonlinearly dependent on pressure. Oil
viscosity is not effecting to flow rate.
or
pKQ
pKQ
p
ACQ q
2
Huhtala 2006
p
pT
m
mT
LT
mmCppC
Q
pkQ plp pkQ mlm
p
pC mC
dt
pd
BC
Vkk
C
p
C
C
pkCdt
pd
B
VpkCQ
m
mp
mm
p
pm
mmmppp
0
0
Pump controlled hydrostatic system
Huhtala 2006
Displacement control of pump and motor
m m
mCpCmaxpC maxmC
maxmC
minmC
constC
const
kC
m
p
pm
max
1
p
p
m
ppm
C
const
CC
Huhtala 2006
Displacement
control of pump
and motor
p
m
1
2
3
2
1
max
p
p
C
C
2
1
max
m
m
C
C
1max
m
m
C
C
maxmax
,m
m
p
p
C
C
C
C11/2
3
1
max
m
m
C
C
1/3
Huhtala 2006
Load types
Constant torque or force
Constant power
Inertia force
Inertia force and viscous friction
FF
TT
m
m
FvFP
PTP
mmm
mmm
dt
dvmF
dt
dJT
mm
mm
mm
m
mm
m
fvdt
dvmF
fdt
dJT
Huhtala 2006
The angular speed of hydrostatic motor when
Constant torque or force
Constant power
Inertia force
Inertia force and viscous
friction
mp
hmmmm
p
pm kkC
T
C
C
2
tmm
mp
tmmm
p
pmQ
P
dt
d
BC
Vkk
QC
P
C
C
0
dt
dJ
BC
V
dt
dkk
C
J
C
Cm
hmmm
mmp
hmmmm
p
pm
2
2
0
2
dt
d
BC
fV
C
kkf
dt
dJ
BC
V
dt
dkk
C
J
C
Cm
hmmm
m
hmmm
mpm
hmmm
mmp
hmmmm
p
pm
2
0
2
2
2
0
2
Huhtala 2006
Example 1
Hydrostatic drive consists of variable displacement pump and constantdisplacement motor
Qmax = 500 ml/s
Volumetric displacement of the motor Vm = 2πCm = 25 ml/r
System maximum pressure pmax = 70 bar
Efficiecies are 100%
Define
• Maximum power
• Maximum rotational speed
• Maximum torque at motor axle
If constant output power is 2 kW, so define
• Minimum rotational speed
• Torque, when motor is running at full speed
• If the torque is constant at lower rotational speeds than minimum speed, sodefine the rotational speed when 20% of the system maximum power is reached
Huhtala 2006
Example 1
Nm
NmpCT
V
Qn
W
WQpP
mm
m
m
87.27
10702
1025
Max torque
r/s 20r/s25
500
speed rotationalMax
3500
105001070
PowerMax
56
max
maxmax
65
maxmaxmax
r/s) 11.43( r/s 4
87.272700
87.27T
W7003500W20%
reached ispower max of 20% hein which t speed Rotational
93.15202
2000
20002
Torque
r/s 11.43r/s287.27
2000
speed Rotational
20002
(2kW)Power output const When
mmax
max
min
minmax
n
NmnW
Nm
NmNmT
WnT
n
WnT
m
mm
m
mm
Huhtala 2006
Hydrostatic drive
Dieselmoottori
mhpp
vpp
gp
T
n
V
,
,
mhmm
vmm
gm
T
n
V
,
,
Q
pG
i
v
F
,f
r
rr nT ,
gmF aa
P, n
Huhtala 2006
Traction force diagram
F [N]
v [km/h]
F (max)
F (min)
v (max)
Piste 1
Piste 2
Corner power
F(max), v(max)
Huhtala 2006
Example for calculating corner power and
conversion ratio
• Starting values
• Diesel engine power is 19.7 kW and torque is 67 Nm
• Maximum needed speed is 35 km/h and maximum traction force is
10000 N
Huhtala 2006
Example continues
F [kN]
V [km/h]
Corner power
(Nurkka teho)
Moottorin
teho
1
2
3
4
PCP
KWWvFP vaadvaadCP 2.973600
3500010000
3.785.08.07.19
2.97
gearhstD
CP
P
PR
Corner power
Conversion ratio
Huhtala 2006
Example continues; evaluation of conversion
ratio
General guideline is
When conversion ratio is R<3 in hydrostatic drives, that leadsto variable displacementpump and constant motor
When conversion ratio is R>3 both units need to be variable.
If both units are variable the total conversion range is divided to both units in identical ratio. Rp is hydraulic pump and Rm is hydraulic motor conversion ratio.
So
7.23.7 RRR mp
Huhtala 2006
Example continues
In our case it should be used variable displacement units, because the conversion ratio is 7.3.
If we are using variable displacement pump and constant motor, the pump size will be increasing. That is because the wholeconversion range will be executed by means of the pump
The new conversion ranges are now
1 ja 3.7 mp RR
Huhtala 2006
Defining the size of hydraulic components
The pump volumetric displacement can be calculated as
mgearhstp
vaadvaadp
Rnp
vFV
max
•It can be seen from the equation that the demands (sppedand traction force) is effecting together with pressure to the size of the pump. When pressure level is increasing the size of the pump and motor is decreasing.
Huhtala 2006
Example continues…
The size of the hydraulic motor is defined as
r
ivn
where
np
vFV
gearvaad
m
gearmhmm
vaadvaadm
max
maxmax
Huhtala 2006
Example continues …
Ja
ko
va
ihd
e
KardaaniKardaaniTasauspyörästöTasauspyörästö
Diesel
Variable pump/
Constant motor
Variable pump/
variable motor
Δpmax (bar) 350 400 350 400
Vp (cm3/r)/ (nearest standard
size)
66/71 57/56 26/25 23/25
Vm (cm3/r )/ (nearest standard
size)
48/46 42/42 48/46 42/44
According to the equations the size
of the components will be as
defined in the table. The gear ratio
between hydraulic motor and
wheel is igear = 18.
Huhtala 2006
Traction force curves
0 5 10 15 20 25 30 35 40 45 [km/h]0
100
200
300
400
500
600
700
800
900
1000
[daN]1
2Max.drive resistance (mass max.), gradient = 10.0%
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1.0
Hyd
rau
lic efficie
ncy
Velocity v
Tra
ctiv
e e
ffort
FG
es
Drive Diagram
Proj. resp. No. Date 07.12.2004 File VPVM.FA4
FADI 4.1IHA/Huhtala
0 5 10 15 20 25 30 35 [km/h]0
100
200
300
400
500
600
700
800
900
1000
1100
1200
[daN]1
2
3Max.driv e resistance (mass max.), gradient = 10.0%
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1.0
Hyd
rau
lic efficie
ncy
Velocity v
Tra
ctive
effo
rt F
Ge
s
Drive Diagram
Proj. resp. No. Date 07.12.2004 File VPFM.FA4
FADI 4.1IHA/Huhtala
Variable pump /
constant motor
Both variable
Huhtala 2006
Example continues …
Diesel
When using low speed motors the size of the motors are Vm =
160 cm3/r (4 motors). The pump size is almost the same as
previous case.
Huhtala 2006
Traction force when using variable pump
and radial pistons motors (low speed
motors )
0 10 20 30 40 [km/h]0
50
100
150
200
250
300
350
400
450
500
550
600
650
700
750
800
850
900
[daN] 1
2
Max.drive resistance (mass max.), gradient = 10.0%
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1.0
Hyd
rau
lic e
fficie
ncy
Velocity v
Tra
ctive
effo
rt F
Ge
s
Drive Diagram
Proj. resp. No. Date 07.12.2004 File VPFM.FA4
FADI 4.1IHA/Huhtala