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