C H A P T E R 5 H y d ro sta tic T ra n sm issio n s (H y ... · H y d ro sta tic T ra n sm issio n s (H y d ro sta tic D riv es) H y d rostatic tran sm ission is th e n am e given

CHAPTER 5Hydrostatic Transmissions (Hydrostatic Drives)

�Hydrostatic transmission� is the name given to a hydraulic system consisting of apositive displacement pump and a positive displacement motor. A wider de�nitionof hydrostatic transmissions includes hydraulic systems consisting of several pumpsand/or motors and even systems in which the output devices are actuators.

The purpose of hydrostatic transmission is to convert mechanical power into hy-draulic power then to transmit and convert it back into mechanical output power ina form which matches speed and torque demands of a driven mechanism or machine.

In the previous chapter we discussed two main methods of speed control, namely:

² Resistive (valve speed control) - where the control of �ow, thus the speed of outputunits, is achieved by �ow control valves. The pump and motor are �xed (constant)displacement units.

² Volumetric (pump/motor control) - where speed is controlled by varying the dis-placement parameters of a pump, motor or both.

Valve speed control of hydrostatic transmissions, which has associated signi�cantpower losses, has limited application in the control of rotary, especially high power,hydrostatic transmissions. Its main application is in drives employing actuators asoutput units, in which continuous control of velocity is usually not required. Asvalve speed control was discussed in some detail in Section 4.2, we will discuss onlyvolumetric control (pump/motor control) of rotary hydrostatic transmissions.

In hydrostatic transmissions control of the direction of rotation of a motor its speedof rotation is accomplished by adjusting displacement of pumps, motors or both. Wemay also control the direction of rotation of a motor, and to some extent its speed,by a directional control valve interposed between a pump and a motor.

Hydrostatic transmissions may operate in either open or closed circuits. An opencircuit, shown in �g. 129, consists of a variable displacement pump 1, a variabledisplacement motor 2 and a directional control valve 3 which controls the directionof �uid �ow and thus direction of rotation of the motor. The speed of the motor iscontrolled either by changing displacements of rotary units. The return �ow fromthe motor is directed by a directional control valve to the reservoir.

A closed circuit consists of a variable displacement pump 1 and a variable or �xeddisplacement motor 2, the return �ow from the motor is supplied to the inlet port ofthe pump. In this system change of the direction of rotation of motor 2 is obtainedby using an over the centre pump 1 capable of reversing the �ow direction, �g.130. Control of speed of the motor is accomplished by changing displacement of the

193

194 Hydrostatic Transmissions (Hydrostatic Drives)

ωp

Tp

Qp pp

1

2

3Qm

pm

Tm

ωm

Fig. 129. Open circuit hydrostatic transmission

pump or the motor or both. �Cross-line� relief valves 6 and 7 protect the systemagainst excessive pressure. The volumetric losses in the system (internal and externalleakages) are replenished by an additional charge pump 3, which is protected by a lowpressure relief valve 8, or by an accumulator. The replenishing �ow is directed, viacheck valves 4 or 5, to this branch of the system which, depending on the directionof �ow, is operating at low pressure.

Tp Tm

ωmωp

1 3

5

27641

1

23

4

Fig. 130. Closed circuit hydrostatic transmission

Transformation of power transmitted by a hydrostatic transmission is described bytwo relations; dynamic ratio or kinematic ratio. Dynamic ratio id de�nes how thetransmission changes torque:

id =T2T1

(5.1)

Characteristics of an ideal hydrostatic transmission 195

where:T1; T2 - are correspondingly input and output torques.

Kinematic ratio ik de�nes how the transmission transforms input to output speed:

ik =n2n1=!2!1

(5.2)

where:!1; n1 - angular or rotation speed of input shaft!2; n2 - angular or rotation speed of input shaft

As the e¢ciency of transmission is a ratio of output power P2 on the motor shaft toinput power P1 delivered to the pump shaft, then:

´ =P2P1=T2!2T1!1

= idik (5.3)

thus, the e¢ciency of the transmission can be expressed as a product of dynamicand kinematic ratios.

For a transmission in which the above ratios are variable, it is important to de�nethe transmission range, i.e. a permissible range of dynamic ratio id and a range ofkinematic ratio ik. In practice, permissible ranges of id and ik ratios are de�nedusing criteria that over full range of these transmission ratios the transmission willoperate with e¢ciency above ´ = 0:8.

5.1 Characteristics of an ideal hydrostatic transmission

Analysis of operating parameters of a hydrostatic transmission, when the transmis-sion works under varying load and various control strategies are employed, is di¢cultdue to a large number of variable parameters and their interaction. However, if weassume that the transmission operates without losses (´ = 1), then using basic rela-tions which describe the operation of a pump and a motor we may investigate withrelative ease the e¤ect of various parameters on the performance of the transmission.

If we assume that a hydraulic system operates without losses, basic parameters of thesystem can be calculated using equations given in chapter 2. Pressure di¤erential¢ptm across the motor, under assumption of no pump/motor or line losses andsteady-state operating conditions, is the result of load torque Tload:

¢ptm =2¼Tload"mqm

(5.4)

or

¢ptm =Tload"mVÁm

(5.5)


where:qm; VÁm - stroke and unit displacements of the motor

"m - displacement parameter

In a steady-state, torque Ttm is equal to load torque Tload and as we have ignoredline losses then ¢ptp = ¢ptm and thus pump torque Ttp is equal to:

Ttp ="pqp2¼

¢ptp (5.6)

or

Ttp = "pVÁp¢ptp (5.7)

and using eq. (5.4) (or eq. (5.5)) and ¢ptp = ¢ptm, Ttm = Tload the relation betweenpump and motor torques is expressed by:

Ttp ="pVÁp"mVÁm

Ttm (5.8)

Pump delivery �ow Qtp, under assumption of no leakages, is described by equation:

Qtp = "pqpnp (5.9)

or

Qtp = "pVÁp!p (5.10)

this �ow will be delivered to the motor, thus Qtm = Qtp. As the theoretical �owdemand of the motor is described by equation:

Qtm = "mqmnm = "mVÁm!m (5.11)

then using eq. (5.9) (or eq. (5.10)) the rotation speed of the motor is de�ned by therelation:

nm ="pqp"mqm

np (5.12)

or

!m ="pV'p"mV'm

!p (5.13)

We may di¤erentiate between the following types of controls for hydrostatic trans-missions:

² Constant torque, variable power

² Constant power, variable torque

² Variable power, variable torque


ωp ωm

-1

1

+ωmmax

−ωmmax

ω

εp

a.

b.

Fig. 131. Hydrostatic transmission with a �xed displacement motor: a. graphicsymbol, b. speed characteristic of the motor

5.1.1 Constant torque, variable power ("p varying, "m = constant)

In this type of control of a hydrostatic transmission, a variable displacement pumpsupplies �uid to a �xed displacement motor. The speed of the motor is controlled bychanging the displacement of the pump. In most cases displacement parameter "pvaries between ¡1 � "p � 1 but motor displacement parameter "m = 1. The angularspeed of the motor when all parameters, with the exception of "m, are constant isde�ned as:

!m = K"p where K = const: (5.14)

and is thus a linear function of "p, �g. 131b.

5.1.2 Constant power, variable torque ("p constant, "m variable)

In this type of transmission the pump has a �xed displacement, i.e. "p = 1 butdisplacement of a hydraulic motor varies in range ¡1 � "m � 1 and thus the angularspeed of the motor, assuming that all parameters with the exception of "m areconstant, can be expressed as:

!m =K

"mwhere K = const: (5.15)

thus the speed of the motor is a hyperbolic function of "m, as shown in �g. 132b.

5.1.3 Variable power, variable torque ("p and "m variable)

Transmissions in which displacements of both pump and motor may take any value,�g. 133, can be controlled sequentially or simultaneously. In simultaneously con-trolled transmissions both "p and "m are varied according to some program, usually


-1

1

+ωmmax

−ωmmax

ω

a.

b.

ωp ωm

Fig. 132. Hydrostatic transmission with a �xed displacement pump: a. graphicsymbol, b. speed characteristic of the motor

ωp ωm

Fig. 133. Hydrostatic transmission with a variable displacement pump and a variabledisplacement motor - graphic symbol

pump displacement parameter "p is controlled �rst and then motor displacement pa-rameter "m. The sequential control of a transmission is shown in �g. 134, and toillustrate its operation we will discuss three control cases:

² The motor is initially set at its maximum displacement ("m = 1) and the pumpat its minimum displacement ("p = 0). The straight line OA represents change ofmotor speed !m with an increase of parameter "p from zero to "p = 1. When thepump is at its maximum displacement (point A), further control of motor speedis achieved by reducing motor displacement parameter "m. Curve AG shows thechange of motor speed with variation of displacement parameter "m.

² The motor is set at "m = 0:5 (i.e. at half displacement) and the pump is set atzero displacement, "p = 0. The motor speed as a function of pump displacement,when it increases from "p = 0 to "p = 1, is represented by straight line OE.When the pump reaches it maximum displacement "p = 1, i.e. its maximum�ow, further variation of motor speed !m is obtained by decreasing the motordisplacement, by curve FG.

² When the motor is set at maximum displacement ("m = 1) and the pump dis-placement increases from "p = 0 to "p = 0:5, then straight line OB represents thechange of motor speed. In this case reduction of motor displacement "m and thecorresponding increase of motor speed is represented by curve CD.


εp =1/2

εm =1

εm =1/3

εm =1/2

εm =1/4

0 1/3 1/21/4 1

5

4

3

2

1

ωp /ωp=ιk

ωp, ωm

B C

A

E

E2

E1

F

F2

F1

GD

Fig. 134. Sequential control of an ideal hydrostatic drive

If we assume that a hydrostatic transmission should operate in the range for whichits e¢ciency is above ´ = 0:8 the sequential control of the transmission is preferableas it allows obtaining the widest control range, kinematic ratio ik and dynamic ratioid is 6¥ 7. When only the pump displacement is controlled, the maximum range ofdynamic ratio is id = 4:5 and the range of kinematic ratio is id = 4:0.

To derive operational parameters of an ideal transmission (no losses) we use equa-tions for pump input power Pp and motor output torque Tm:

Pp = "pVÁp!p¢p (5.16)

Tm = "mVÁm¢p (5.17)

Equation (5.17) shows that for a given motor, output torque depends on displacementparameter "m and pressure di¤erential across the motor ¢p. Thus when the motordisplacement is set, i.e. "m = const:, the motor output torque is proportional topressure di¤erential ¢p across the motor, �g. 135a. The straight lines for variousvalues of "m = const: pass through the origin. Plotted in logarithmic coordinatesthese lines become parallel to each other, �g. 135b.

When we reduce parameter "m but maintain a constant torque (Tm = const:) thenpressure di¤erential ¢p across the motor will increase, arrow 1 in �g. 135b. On the


other hand, we may increase the output torque at a set pressure di¤erential (¢p =const:) by increasing displacement parameter "m, arrow 2 in �g 135b.

∆p

Tmε

m=1.

0

∆pmax ∆p∆pmax

log Tm

εm=1.0

εm=const

εm=const

a. b.

2

1

Fig. 135. Characteristics of a variable displacement motor Tm = f(¢p; "m). a.Linear co-ordinates, b. Logarithmic co-ordinates.

If we consider eq. (5.16) we can see that pump input power Pp is proportional topump parameter "p and pressure di¤erence ¢p. Thus plots of Pp = const. in ¢p; "pcoordinates are hyperbolas �g. 136a., and in logarithmic coordinates these curvesare straight lines, �g. 136b.

log ∆p

∆pmax

2

1

∆p

∆pmax

Pp=const

Pp=const

εp εp1.0 1.00 0

a. b.

Fig. 136. Characteristics of a variable displacement pump Pp = f(¢p; "p). a. linearco-ordinates, b. logarithmic co-ordinates.

However, when we maintain parameter "p = const:, then pump input power Pp willincrease with an increase in pressure di¤erential ¢p, arrow 1 in �g. 136b. On theother hand, if pressure di¤erential¢p is maintained constant then the required pump

Volumetric and hydro-mechanical losses 201

input power Pp will increase with an increase of parameter "p , i.e. with an increaseof the pump �ow Qp, as shown by arrow 2 in �g. 136b.

To �nd pressure di¤erential ¢p for an ideal hydrostatic transmission, we assumethat ¢p = ¢p12 = ¢p34 and use the equation for motor output torque:

¢p =Tm!m"pVÁp!p

(5.18)

From the above equation we can see that for a transmission operating at a setpressure di¤erential (¢p = const:) and a set value of the motor output torque (Tm =const:), the pressure di¤erential¢p is proportional to the angular speed of motor !m.Thus at constant pump delivery, in !m, ¢p coordinates, plots of the output torqueare straight lines crossing the origin, �g. 137a. For other displacement settings ofthe pump, straight lines Tm = const: are rotated in relation to the previous lines byan angle corresponding to the change of parameter "p.

At a constant setting of pump displacement parameter "p and a constant value oftorque Tm, the increase of the angular speed of motor !m, obtained by reducingmotor displacement "m, is accompanied by the increased pressure di¤erence ¢pacross the motor, arrow 1 - �g. 137b. On the other hand, increasing the angularspeed !m, at constant values of ¢p and "p, causes a decrease of the torque whichcan be developed by the motor, arrow 2 - �g. 137b.

a. b.

Tm = const

Tm = co

nst

log ∆p

∆pmax

2

1

∆p

ω m0 0

εp'

εp''

∆pmax

ω m

Fig. 137. Characteristics of hydrostatic transmission ¢p = f(!m,Tm,"m). Lines Tm= const (at ¢p = const). a. linear co-ordinates, b. logarithmic co-ordinates.

It must be stated, however, that the operational parameters of a real transmissiondi¤er from the parameters derived for an ideal transmission. This is specially thecase when the system is operating with the pump set at a small displacement, largemotor speed or when the system operates at either very high or very low pressures.


5.2 Volumetric and hydro-mechanical losses

5.2.1 Volumetric losses

Hydrostatic transmissions su¤er volumetric losses Qv which occur both in pump andmotor units. Volumetric losses (discussed in Chapter 2) are for a given speed mainlya function of load, i.e. pressure ¢p, in a system:

Qv = Kv¢p (5.19)

and are expressed in terms of the volumetric e¢ciencies of pump ´p and motor ´m,which were de�ned by equations:

´vp =QpQtp

=Qp

"p!pVÁp(5.20)

and

´vm =QtmQm

="m!mVÁmQm

(5.21)

The above equations show that displacement settings "p and "m have in�uence one¢ciencies of both the pump and the motor. In addition, some volumetric losses alsooccur in control mechanisms of the hydrostatic units. These losses are taken intoconsideration when calculating transmission e¢ciency only if the control elementsare supplied from the main circuit of the transmission, e.g. displacement or pressurecontrols, by de�ning e¢ciency ´vz of the control system. The volumetric e¢ciencyof the transmission lines for correctly manufactured systems is ´vl = 1:0 i.e. there isno line leakage. Comparing actual demand of motor Qm with pump delivery Qp weobtain expression for the angular speed of the motor:

!m ="pVÁp"mVÁm

!p´vp´vm´vz (5.22)

We can see that motor speed is lower than the speed of an ideal transmission operat-ing without volumetric losses. Kinematic ratio ik for a transmission with volumetriclosses is equal to:

ik =!m!p

="pVÁp´vm´vz´vp

"mVÁm(5.23)

ik =´vm´vz´vp

i(5.24)


where i is transmission ratio equal to:

i ="mVÁm"pVÁp

(5.25)

Thus the kinematic ratio of the transmission will diminish when volumetric losses ofthe hydrostatic units increase.

5.2.2 Hydro-mechanical losses

Hydro-mechanical losses are de�ned by hydro-mechanical e¢ciencies of the individ-ual elements of the transmission. The e¢ciency of pump ´hmp and the e¢ciency ofmotor ´hmm were de�ned as:

´hmp =TtpTp

and ´hmm =TmTtm

(5.26)

Additional hydro-mechanical losses are due to pressure losses ¢pc in the controlelements of the transmission and pressure losses ¢pl due to friction in the deliveryand return lines of the transmission.

Hydraulic e¢ciency ´hl of the hydraulic line is de�ned as a ratio of pressure di¤erence(p3 ¡ p4) at the motor ports 3 and 4 to pressure di¤erence (p2 ¡ p1) at the pumpports 1 and 2, �g. 130:

´hl =p3 ¡ p4p2 ¡ p1

=¢p34¢p21

(5.27)

where pressure p3includes pressure loss ¢pl in the hydraulic line and losses ¢pc in

the control elements:

p3 = p2 ¡¢pl ¡¢pc (5.28)

Thus, the hydraulic e¢ciency of the line is:

´hl =p2 ¡¢pl ¡¢pc ¡ p4

p2 ¡ p1(5.29)

Pressure loss ¢pl is a function of the �ow rate Q and �uid viscosity ¹, thus ¢pl =f(Q;¹).

Dynamic ratio id of the transmission can now be de�ned as:

id =TmTp

=Ttm´hmm´hmp

Ttp(5.30)

substituting for Tm and Tp, we obtain:

id ="mVÁm¢p34"pVÁp¢p21

´hmp´hmm (5.31)


or:

id = i´hmp´hmm´hl (5.32)

Thus the hydro-mechanical e¢ciencies of the pump and motor, control elements andhydraulic lines decrease the dynamic ratio of a transmission.

5.2.3 E¤ect of load

In general, steady-state analysis of hydraulic systems is carried out under the as-sumption of incompressibility of working �uid, which is re�ected in an instant andexact reaction of a system to parametric changes. The real hydraulic �uid is, how-ever, non-homogeneous - it contains free air and a mixture of liquid and air, thus itis necessary to consider the e¤ects of �uid compressibility on system performance.This is especially important in the case of systems which must provide exact loadpositioning and fast response to control signals.

12.0

10.0

8.0

6.0

4.0

2.0

0 15.010.05.01.0

14.0

p [MPa]

B [MPa]

m = (Vp/V)100%

1 - m = 0.001

2 - m = 0.01

3 - m = 0.1

4 - m = 0.3

5 - m = 0.5

6 - m = 1.0

Fig. 138. Dependence of bulk modulus on amount of air and operating pressure. m- air content (%) , Vp - air volume, V -liquid volume.

Compressibility of �uid is characterized by its bulk modulus. Bulk modulus dependson many factors, e.g. the type of hydraulic �uid, temperature, pressure and degreeof aeration. The e¤ect of aeration on the value of bulk modulus can be quite drastic.Fig. 138 shows how bulk modulus varies with pressure and the content of air in


a typical hydraulic �uid. Bulk modulus B of the �uid allows the determination ofsti¤ness coe¢cient C of a hydraulic system.

The sti¤ness of a hydraulic system is expressed as a ratio of load change ¢F (or¢T ) to the corresponding position change ¢x (or ¢®) of the loaded element:

C =¢F

¢x(5.33)

or

C =¢T

¢®(5.34)

High hydraulic sti¤ness is very important in systems driving machine tools, robotsetc. The sti¤ness of a system is a function of the changes in �uid volume in the systemwhich are due to the variation of system pressure. The change in �uid volume dueto a change in pressure is de�ned by the relation:

V = V0

µ1¡

p

Ba

¶(5.35)

where:V0; V - �uid volume at atmospheric pressure and at pressure pBa - apparent bulk modulus of the �uid

Apparent bulk modulus Ba combines ideal bulk modulus B with the e¤ects of aer-ation and compliance of hydraulic lines and system elements (e.g. actuators). Ininitial design calculations we may assume the value of apparent bulk modulus of amineral oil to be Ba = 1200 MPa. The compressibility of �uid and the complianceof the system (mainly of transmission lines) change the e¤ective �ow rate in thesystem. Thus, the e¤ective �ow rate in a hydrostatic system can be expressed byequation:

"p!pVÁp ¡Kvp¢p¡V0Ba

d¢p

dt= "m!mVÁm +Kvm¢p = Q (5.36)

where:Kvp - coe¢cients of volumetric loss in pumpKvm - coe¢cients of volumetric loss in motor¢p - load pressure, ¢p = f(Tload)

This is a basic equation for a hydrostatic transmission, it includes the e¤ects ofthe compressibility of �uid and the volumetric losses in the pump and motor. Thisequation is true, regardless of the type of transmission and the type of loading,as long as the volumetric losses in the transmission line can be ignored. To allowanalysis of the e¤ect of load on the speed of the motor the above equation can bewritten in the form:


!m ="pVÁp"mVÁm

!p ¡(Kvp +Kvm)

"mVÁm¢p¡

V0Ba"mVÁm

d¢p

dt(5.37)

5.2.4 Constant torque loading

We will analyse the characteristics of a transmission which is controlled by variationof pump �ow delivery and which is subjected to constant torque Tm = Tload = const.

When we vary "p and maintain "m = const:, the pressure in the system is:

Tm = Ttm´hmm = "mVÁm¢p34´hmm (5.38)

¢p34 =Tm

´hmmVÁm"m(5.39)

If we assume that hydro-mechanical e¢ciency ´hmm of the motor is independent ofmotor speed !m, then pressure ¢p is constant:

d¢p

dt= 0 (5.40)

Thus the last term in the equation for !m; eq. (5.37), is equal zero and substitutingeq. (5.40) into eq. ((5.37)) we obtain:

!m ="pVÁp"mVÁm

!p ¡(Kvp +Kvm)Tm"2mV

2Ám´hmm

(5.41)

Looking at this equation we can see that as torque Tm increases the speed of themotor decreases due to volumetric losses in the pump and the motor. Fig. 139shows the e¤ect of volumetric losses on the speed of the motor when displacementparameter "p is varied, a represents system volumetric loss when !p = 0 and b is adecrease of motor speed due to volumetric losses.

5.2.5 Inertial Loading

The following two cases are considered:

Case a. Inertial loading - compressibility of �uid and viscous friction are ignored

Case b. Inertial loading - compressibility of �uid is taken into account

Case c. Inertial loading - compressibility of �uid and viscous friction are included

The input in each case is a step change in pump displacement parameter "p: Pumpangular speed !p is assumed to be constant. We assume that the load on the motoris in the form:

Tload = Tm = Jd!mdt

(5.42)


and using the de�nition for hydro-mechanical e¢ciency of the motor we obtain:

¢p34 =Tm

´hmm"mVÁm=

J

´hmm"mVÁm

d!mdt

(5.43)

when we introduce this relation into the equation for !m, eq. (5.37), the followingequation for motor speed is obtained:

!m =!pVÁp"mVÁm

"p ¡J(Kvp +Kvm)

"2mV2Ám´hmm

d!mdt

¡V0J

Ba"2mV2Ám´hmm

d2!mdt2

(5.44)

1.0

b

a

ωm

εm

Actual c

haract

eristic

Idea

l cha

ract

eristic

Fig. 139. E¤ect of volumetric losses on the speed of the motor when the displacementparameter "p is varied

Case a. As in this analysis compressibility of the �uid is ignored (Ba =1), theabove equation takes the following form:

J(Kvp +Kvm)

"2mV2Ám´hmm

d!mdt

+ !m =!pVÁp"mVÁm

"p (5.45)

We can de�ne time constant ¿ :

¿ =J(Kvp +Kvm)

"2mV2Ám´hmm

(5.46)

We can observe that if a system has large inertial load J and large volumetric lossescharacterised by coe¢cients Kvp and Kvm this will result in large time constant ¿and thus in slow response of the system.


Constant k:

k =!pVÁp"mVÁm

(5.47)

and we can now rewrite eq. (5.45) as:

¿d!mdt

+ !m = k"p (5.48)

which is a �rst order di¤erential equation. With step change in pump displacement

parameter ep and with initial values of !m(0) = 0 andd!mdt

= 0 the above equation

after transformation into Laplace domain becomes:

!m(s) =kep

s (¿s+ 1)=

kep=¿

s (s+ 1=¿)(5.49)

which after expansion into partial fractions and transformation back into time do-main yields expression for time response of the motor to step change in e.g. "p):

!m(t) = !o

h1¡ e¡t=¿

i(5.50)

where !o is a steady-state speed of the motor:

!o =!pVÁp"mVÁm

"p (5.51)

Case b. We will use again eq. (5.44) and this time compressibility of the �uid isincluded in the analysis. Eq. (5.44) after some rearranging becomes:

d2!mdt2

+Ba(Kvp +Kvm)

V0

d!mdt

+Ba"

2mV

2Áp´hmm

JV0!m =

Ba"2mV

2Ám´hmm

JV0

!pVÁp"mVÁm

"p

(5.52)we notice that this equation is a second order di¤erential equation and thus it canbe written as follows:

¢¢

!m +2»!n¢

!m +!2n!m = k"p (5.53)

where:!n - natural frequency of the system³ - damping ratio

Comparing eq. (5.52) with eq. (5.53) we �nd that natural frequency !n of thesystem is equal to:


!n = "mVÁm

r´hmmBaJV0

(5.54)

and the damping ratio ³ is expressed by:

³ =Ba(Kvp +Kvm)

2V0

1

!n=Ba(Kvp +Kvm)

2V0"mVÁm

sJV0

´hmmBa(5.55)

which after rearranging becomes:

³ =J (Kvp +Kvm)

"2mV2Ám´hmm

"mVÁm2

r´hmmBaJV0

(5.56)

and �nally, constant k is equal to:

k =Ba"

2mV

2Ám´hmm

JV0

!pVÁp"mVÁm

The equation for damping ratio ³ shows that its value will increase as volumetriclosses Kvp and Kvm in transmission units and load moment of inertia J increase.Thus amplitudes of damped oscillations of the motor speed will be reduced duringthe transient state. The frequency of damped oscillations is expressed by:

!d = !n

q1¡ ³2 (5.57)

The solution of equation (5.52) is in the form:

!m = !0

�1¡ e¡³!nt

½cos(!dt) +

³!n!d

sin(!dt)

¾¸(5.58)

and it represents time response of speed !m of the motor. Steady-state speed of themotor !0 is equal to:

!0 =k"p!2n

=

´hmmBa"2mV

2Ám

JV0

VÁp"mVÁm

µ"mVÁm

r´hmmBaJV0

¶2 =!pVÁp"mVÁm

(5.59)

Case c. Solving a more general loading case, i.e. assuming that the load on themotor includes viscous friction, then load equation is:

Tload = Tm = Jd!mdt

+ f!m (5.60)


where f is viscous friction coe¢cient. Pressure di¤erential across the motor is thenrepresented by:

¢p34 =Tm

´hmm"mVÁm=

J

"mVÁm´hmm

d!mdt

+f

"mVÁm´hmm!m (5.61)

and its derivative by:

d¢p34dt

=J

"mVÁm´hmm

d2!mdt2

+f

"mVÁm´hmm

d!mdt

(5.62)

We substitute the above equations into eq. (5.37) and obtain:

!m =!pVÁp"mVÁm

"p ¡(Kvp +Kvm)

"mVÁm

µJ

"mVÁm´hmm

d!mdt

+f

"mVÁm´hmm!m

¶¡

¡V0

Ba"mVÁm

µJ

"mVÁm´hmm

d2!m(t)

dt2+

f

"mVÁm´hmm

d!m(t)

dt

¶(5.63)

which after rearranging becomes:

k1d2!mdt2

+ k2d!mdt

+ k3!m = k4ep (5.64)

where constant k1:

k1=V0J

Ba"2mV2Ám´hmm

(5.65)

constant k2:

k2=

(Kvp +Kvm)J +V0f

Ba´hmm"

2mV

2Ám

(5.66)

constant k3:

k3 = 1 +(Kvp +Kvm) f

"2mV2Ám´hmm

(5.67)

and �nally coe¢cient k4 is equal to:

k4 =!pVÁp"mVÁm

(5.68)


As eq. (5.64) represents again a second order di¤erential equation, we can againwrite it as:

¢¢

!m(t) +2»!n¢

!m(t) +!2n!m(t) = kep(t) (5.69)

then natural frequency !n is equal to:

!n =

rk3k1=

2

66664

Ã

1:0 +(Kvp +Kvm)f

´hmm"2mV

2Ám

!

V0Ba"2mV

2Ám

J

´hmm

3

77775

1=2

=

=

vuut³"2mV

2Ám´hmm + fKvp + fKvm

´Ba

V0J(5.70)

damping ³ is expressed by equation:

³ =k2pk1

2k1pk3= :5

k2pk1k3

=

0:5

0

BB@

(Kvp +Kvm)J +V0f

Ba"2mV

2Ám´hmm

1

CCA

vuutÃ

V0J

Ba"2mV2Ám´hmm

!Ã

1:0 +(Kvp +Kvm) f

"2mV2Ám´hmm

!

=1

2

JBaKvp + JBaKvm + V0fr³"2mV

2Ám´hmm + fKvp + fKvm

´BaV0J

(5.71)

and k is expressed by:

k =k4k1=

!pVÁp"mVÁmV0J

Ba"2mV2Ám´hmm

=!pVÁp"mVÁmBa´hmm

V0J(5.72)

Finally, time response of motor speed !m to step input ep is described by equation:

!m = !0

�1¡ e¡³!nt

½cos(!dt) +

³!n!d

sin(!dt)

¾¸(5.73)

where steady-state speed of the motor !0 is equal to:


!0 =k"p!2n

=

!pVÁp"mVÁmBa´hmm

V0J

("2mV 2

Ám´hmm+fKvp+fKvm)Ba

V0J

=

!pVÁp"mVÁm

1 +(Kvp +Kvm) f

"2mV2Ám´hmm

"p (5.74)

and damped frequency as before is calculated using:

!d = !n

q1¡ ³2 (5.75)

The values of !d, !n and ³ are de�ned by modi�ed expressions which re�ect the e¤ectof friction coe¢cient f . Time responses of motor speed in a hydraulic transmissionsubjected to various loading conditions are shown in �g. 140.

t

ωm

inertia loading

inertia load and fluid compressibility

resistive load

π/ωd 3π/ωd

Fig. 140. Time responses of motor speed in a hydraulic transmission

Further, a more realistic analysis of hydrostatic transmissions and their dynamics,taking into consideration non-linearities in the system, requires application of digitalsimulation techniques.

546

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