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Examples of Calculations - Control of Hydraulic Systems

Problem 4.1 Cylinder drive system

Calculate maximum forces during the operation of a cylinder, piston speeds and �owthrough the directional control valve for a system shown in �g. 77. Ignore frictionforces in the cylinder and forces due to back pressure in a return line. Assume nolosses in the cylinder and the system.

Qp

po

A1

A2

F

I O II

Fig. 77. Cylinder control system

Answer: When the directional control valve is in position marked 0 then thepiston is at rest. When the directional control valve is in position II the piston willmove to the left and the maximum piston force is a function of system pressure p0 -set by a relief valve, thus:

Fmax(2) = p0A2

The maximum force during movement of the piston to the right, when the directionalcontrol valve is in its position I, is:

Fmax(1) = p0(A1 ¡A2)

Piston speed when extending is:

v2 =QpA2

Cylinder counterbalancing system 121

and

v1 =Q

A1 ¡A2

when piston is retracting. If we assume ratio of areas:

A1A2

= 2 then A1 ¡A2 = A2

and speed v1 = v2 and also Fmax(2) = Fmax(1). Thus in this system the speed of thepiston in either direction is the same as is the force if supply pressure is constant. Atconstant pump delivery �ow Qp, the maximum �ow through the directional controlvalve is when the valve is in position I, i.e. when the piston is moving to the right.In this case the �ow through the directional control valve is equal to a sum of pump�ow Qp and the �ow from the annulus volume of the cylinder

Qmax = Qp +Q2

where:Q2 - the �ow from the annulus volume of the cylinder, Q2 = v1A2

At velocity v1 equal to:

v1 =Qp

A1 ¡A2

the maximum �ow through the directional control valve is:

Qmax = Qp

µ1 +

A2A1 ¡A2

¶= Qp

A1A1 ¡A2

and when the ratioA1A2

= 2, then:

QmaxA1A2

= 2Qp answer!

Problem 4.2 Cylinder counterbalancing system

A single acting, hydraulic cylinder is loaded by a large inertial force F , �g. 78.Design a system to prevent uncontrolled descent of the piston in the case of linefailure between the directional control valve and the cylinder.

122 Examples of Calculations - Control of Hydraulic Systems

II 0 II

D

F

p0

Fig. 78. Gravity loaded hydraulic system (Solution a)

Answer - Solution a. In neutral position 0 of the directional control valve,the pump and the pilot line of a pilot operated check valve are connected to thereservoir. The pilot operated check valve, mounted directly on the cylinder, shuts theconnection between the cylinder and the directional control valve and thus preventspiston movement. When the directional control valve is moved to position II, thepilot pressure will lift the check valve o¤ its seat and the piston will start to descenddue to the gravity load on the piston. The pump �ow is directed to the reservoirthrough the pressure relief valve. The descent speed of piston will depend on the �owrate through the directional control valve to the reservoir which in turn is a functionof the sum of pressure drops across the check valve and the directional control valve.The power loss in this circuit is large as during piston descent the full �ow of thepump is discharged to the reservoir through the relief valve at the pressure set onthe relief valve.

The pressure necessary to lift the piston, ignoring hydro-mechanical losses, is:

p =4F

¼D2

where D is piston diameter.

Answer - Solution b. Lower power loss can be obtained by incorporatingin the circuit a relief valve which is set to a pressure necessary to open the pilotoperated check valve (the pilot pressure is usually 4-5 times lower than the pressureacting on the check valve). A uni-directional �ow control valve is also added to thecircuit, it allows control of the speed of the cylinder during descent. A check valve,

Forces - cylinder drive system 123

i 00 iiii i

Solution cSolution b

F F

Fig. 79. Gravity loaded hydraulic system (solutions b&c)

mounted in parallel to the �ow control valve, allows free �ow into the cylinder, �g.79b. In case of a failure of the line between the directional control valve and thepilot operated check valve, when the directional control valve in positions 0 or I,the check valve will close and prevent uncontrolled descent of the piston. When thedirectional control valve is in position II the failure of the line will not stop descentof the piston, however the piston will then travel at the speed set by a �ow controlvalve.

Answer - Solution c. An e¤ective method of protecting a system from theconsequences of failure of the line is to mount the pilot operated check valve directlyon the cylinder, �g. 79c. This circuit provides operating safety and also has theadvantage of controlling descent provided by solution b. In this solution the cylinderwill always descend at the speed controlled by the �ow control valve.

Problem 4.3 Forces - cylinder drive system

A double acting, single rod, hydraulic cylinder is controlled by a system shown in�g. 80. Calculate forces on the piston rod if the pressure settings pn1 and pn2 of therelief valves are known. Ignore losses in the system.

Answer: The system is equipped with a relief valve in the supply line and tworelief valve-check valve assemblies between the cylinder and the directional control


valve. When the directional control valve is set in position 0, the piston is held instationary position by the check valves. The up motion of the piston is obtained bysetting the directional control valve to position II, then the cylinder force, ignoringfriction losses, is:

F1 = p3A1 ¡ pn2A2 answer!

where: p3 is pump pressure.

Setting the directional control valve to position I will move the cylinder in a down-ward direction, then the cylinder force is:

F2 = p3A2 ¡ pn1A1 answer!

These relations show that, to obtain the same required force, supply pressure p3has to be higher than it would be in a circuit which does not have cylinder mountedrelief valves. If the external load acts only in one direction, only one relief valve-checkvalve assembly may be used.

0

II

I

pn2

pn1

F1

A2

A1

p3

F1

Fig. 80. Cylinder control system (Problem 4.3)

Problem 4.4 Hydraulic lock

For a hydraulic system shown in �g. 81, in which a double acting, single rod cylinderis subjected to gravity force calculate pilot pressure ps controlling the operation ofa pilot operated check valve when external force F is known. Ignore friction andhydrodynamic-forces in the pilot operated check valve and hydro-mechanical lossesin the cylinder.

Hydraulic lock 125

Fp2

F

B

B1ps

p2'

p1

I 0 II

p2'

B

B1

ps=p1pT

a0

Fs

a1

pT'

pT'

A1

A2

Fig. 81. Hydraulic control system (Problem 4.4)

Answer: When the directional control valve is set in neutral position 0 thecylinder is stationary. Placing the valve into position II will move the cylinderupward against the load. During the down motion of the piston, when the directionalcontrol valve is in position I, the pilot operated check valve must be lifted o¤ itsseat by pilot pressure ps = p1, which is set by a relief valve. Piston area is A1 andannulus area of the cylinder is A2: For this system to operate correctly, pump supply�ow Qp must be higher than the �ow demand of the cylinder, thus v1A1 � Qp. Fromthe balance of forces acting on the piston, pressure p2 is:

p2 =F

A2+ p1

A2A1

(a)

and also:

p2 = p0

2 +¢pd (b)

where ¢pd is pressure drop across the �ow control valve. The pilot operated checkvalve, which is lifted by pilot pressure ps will open when the following condition issatis�ed:

psa1 = p1a1 > p0

2a0 + Fs


and pilot pressure ps must meet condition:

ps >p02a0 + Fs

a1(c)

where:Fs - spring forcea0 - area of the poppeta1 - area of the pilot piston

We can combine the �rst two equations, eq. (a) and eq. (b), and rearrange to obtainexpression for p02:

p02 =F

A2+ p1

A1A2

¡¢pd

thus, after substituting for p02 in eq. (c) the required pilot pressure is:

ps = p1 >

F

A2¡¢pd +

F sa0

a1a0¡A1A2

answer!

and also, as p1 > 0, ratio of the pilot piston area to the valve seat area should bea1a0>A1A2. Required setting of pressure di¤erential ¢pd on the �ow control valve can

be obtained from:

¢pd =F

A2+ p1

A1A2

¡ p02

Problem 4.5 Cylinder decompression system

In some hydraulic systems when the cylinder reaches its end position the elasticenergy stored in the system (due to compliance of the �uid or load) may cause a�rebound� which may disturb the return movement of the piston. In such systemsto obtain a smooth reversal of the cylinder we must automatically release the storedelastic energy when the movement of the piston is reversed. An example of such ahydraulic system is shown in �g. 82.

Analyse operation of this system and:

² Describe its operation assuming that pressure ph set on sequence valves 3 and 5is 1=5 of maximum pressure due to load and that pressure required to operateback-pressure check valve 8 is lower than ph. Assume that the system is suppliedby a constant displacement pump, e¢ciency of the cylinder is ´c = 1, reservoir

Cylinder decompression system 127

12

3

456 7

8

O x1 h x

p1 p2

pp

A B

P T

ph

1.2 ph

F=F(x)

I IIO

Line z

Line b

Fig. 82. Decompression circuit (Problem 4.5)

pressure is p0 = 0; ignore hydraulic friction losses in hydraulic lines and checkvalves.

² Determine, on the basis of force-stroke characteristics shown in �g. 83, changesin the following operating variables in the function of piston stroke x:

� pressure p1 at the entry to the cylinder (piston head side of the cylinder)

� pressure p2 at the exit from the cylinder (piston rod side of the cylinder

� supply pressure pp� speed of piston v.

² Derive equation for the ori�ce area of �ow control valve 4 which will assure thatthe maximum speed of the piston during the return stroke is:

v2 �Q

A2

Answer - Description of operation: During piston motion to the right (direc-tional control valve in position I) the �uid �ows with minimum resistance throughcheck valve 1 to the piston head end of the cylinder. Fluid from the annulus side ofthe cylinder is returned to the reservoir, however there are two possible �ow pathsdepending on the cylinder load. In the �rst case, when pressure p1 in the cylinder


0

1<n<5

F0

h

x

x1

F

x

-F0

Fig. 83. Force F versus stroke x

head is lower than pressure ph set on sequence valve 3 (p1 < ph), the valve will beclosed and �uid from the cylinder will return to the reservoir via check valve 2, thedirectional control valve (path B-T) and check valve 8.

In the second case, when p1 > ph, �uid from the cylinder will return to the reservoirvia check valve 2 and sequence valve 3.

When the piston reaches its end position, the load (and therefore pressure p1) in-creases, see force-stroke characteristics, and the elastic energy is stored in the �uid.The increase in pressure p1 causes valve 3 to open (pressure ph is according to ourspeci�cation equal to 1/5 of the maximum pressure). When subsequently the direc-tional control valve is switched to a neutral position 0 then the piston is stopped,however, the piston rod is still under load.

Switching the directional control valve into position II causes the piston to retract.As sequence valve 3 is still in its open position it prevents build up of pressure in lineb thus sequence valve 5 remains shut. Discharge of the �uid through �ow controlvalve 4 causes a gradual movement of the piston and reduction of the pressure inthe piston head volume of the cylinder. When pressure p1 drops to the level ofpressure ph, valve 3 closes and the pressure in line b increases. Pressure in lineb controls sequence valve 5, which opens and connects the piston head volume ofthe cylinder to the reservoir thus releasing all the energy stored in the �uid duringcylinder stopping. As pressure in line b increases further and reaches the value ofapprox. 1:2¢ph, sequence valve 6 opens and allows �uid �ow to the annulus volumeof the cylinder. The purpose of valve 6 is to maintain pressure that will keep valve5 open and valve 3 closed independently of the load on the piston.


During the decompression phase, it is necessary to prevent cylinder cavitation byallowing �uid to enter the annulus volume of the cylinder. This is achieved byallowing �uid �ow through line z to the annulus volume via check valve 7, withvalve 8 providing the necessary back-pressure.

0

nF0/A

hx1

F/A

x

p1

Fig. 84. Plot of pressure p1 vs. stroke x

Piston head pressure p1 = p1(x): When an external force, retracting the piston,is exerted on the cylinder rod (F > 0), the value of pressure p1 is a function of theload:

p1 =F

A1answer!

where A1 is the area of the piston.

In the other case, when an external force is extending the piston (F < 0), tensionforce pressure p1 is equal to pressure in the reservoir p1 = 0. Plot of pressure p1 asa function of piston stroke is shown in �g. 84.

As we assumed that 1 < n < 5, then pressure p1 =F0A1

is larger than pressure ph set

on valves 3 and 5, and:

F0A1

> ph

Annulus pressure p2 = p2(x): When pressure p1 > ph valve 3 is open, therefore,annulus pressure is equal to pressure in the reservoir p2 = 0. This occurs duringout-stroke of the piston and during return stroke up to the moment when there isa jump change in force from +F0 to ¡F0. When the piston is subjected to tension


force F0, pressure p2 is de�ned by the relation:

p2 =F0A2

where A2 is the annulus area of the cylinder.

Plot of pressure p2 as a function of piston stroke x is shown in �g. 85.

0

hx1 x

F0/A2

p2

Fig. 85. Plot of pressure p2 vs. stroke x

Pump pressure pp = pp(x): To extend the piston the directional valve is shiftedto position I. Then pump pressure pp is equal to pressure p1:

p1 = pp =F0A1

answer!

During the return stroke, valve in position II, pump pressure is equal to zero untilsuch time as valve 3 closes, and then it is equal to the sum of pressure p2 and thepressure di¤erence across sequence valve 6:

pp =F0A2

+ 1:2ph answer!

Plot of pressure pp as a function of piston stroke x is shown in �g. 86

Piston velocity v = v(x): Piston velocity v1 during the out-stroke is constantand depends on pump delivery Qp:

v1 =QpA1

answer!

During the return stroke, velocity of the piston until the moment when valve 3 closesand valve 5 opens, is equal to:

v02 =QdA1

answer!


where Qd is �ow through restrictor valve 4.

0

nF0/A1

hx1

F0/A1

x

pp

F0/A2+1.2ph

Fig. 86. Pump pressure pp vs. stroke x

Flow rate through restrictor 4 can be calculated from equation:

Qd = Kvad

r2p1½

where:Kv - �ow coe¢cientad - area of the ori�ce½ - �uid densityp1 - load pressure

Substitute expression for p1:

p1 =F

A1

and rearrange, then:

v02 = Kvad

s2

½A31

pF

Thus velocity of the piston is dependent on restrictor �ow area ad. After valve 3closes and valve 5 opens (this will occur when force +F0 changes to ¡F0) the velocityof the piston during the return stroke is:

v002 =QpA2


Plot of piston velocity as a function of piston stroke is shown in �g. 87. Velocityduring working stroke is taken to be positive. Various values of parameter ad areshown by dotted lines. This plot was drawn using �ow area ad of the ori�ce forwhich steady maximum velocity during the return stroke is equal to v002 .

hx1

v

v1

-v''2

x

ad1ad1

ad2>ad1

ad3>ad2

0

Fig. 87. Plot of piston velocity v vs. stroke x

Maximum ori�ce �ow area: From the condition v02 = v00

2 for x = h we obtain:

QpA2

= Kvadmax

s2

½A31

pnF 0

thus:

admax =QpA1KvA2

r½A12nF 0

answer!

Problem 4.6 Speed control of two cylinders in parallel

A system of cylinders is shown in �g. 88. System parameters are:

² pump stroke displacement qp = 2£ 106m3rev¡1

² pump rotation speed np = 1500 rpm

² piston areas A1 = A2 = A, A = 25£ 10¡4m2

² force - cylinder 1 F1 = 1000 N

² force - cylinder 2 F2 = 2000 N

The relief valve is set at pressure p0 = 2:5 MPa, ignore losses in the system.

Calculate speed of pistons v1; v

2and delivery pressure p for the following cases:

Speed control of two cylinders in parallel 133

1. a. All throttling valves are fully open (no �ow resistance)

b. Valves S2 and S3 are fully open, and pressure drop on valve S1 is:

¢pS1 = k1QS1 Pa k1 = 1£ 1010 Nsm¡5

c. Valve S3 fully open, and pressure drops on valves S1 and S2 are:

¢pS1 = k2QS1 Pa k2 = 5£ 1010 Nsm¡5¢pS2 = k3Q

2S2Pa k3 = 1£ 1015 Ns2m¡8

d. Valves have following pressure drops:

¢pS1 = k2QS1 Pa k2 = 5£ 1010 Nsm¡5¢pS3 = k3Q

2S2Pa k3 = 1£ 1015 Ns2m¡8

¢pS3 = k2QS3 Pa k2 = 5£ 1010 Nsm¡5

∆pS1

∆pS2

F1

p0

p

p0

qp

np

A2

A1

S3

∆pS3

F1

S2

S1

Cylinder 1

Cylinder 2

Fig. 88. Hydraulic control system (Problem 4.6)

Solution of case a. When restrictor valves are fully open, we can calculate thesupply pressure to both cylinders from equations:

p1 =F1A=

1000

25 £ 10¡4= 400000Pa = 0:4 MPa answer!

p2 =F2A=

2000

25 £ 10¡4= 800000Pa = 0:8 MPa answer!

And as p1 < p2, cylinder 1 will extend �rst with velocity:


v1 =QpA=npqp60A

=1500£ 2£ 10¡6

60£ 25£ 10¡4= 2£ 10¡2 ms¡1 answer!

When the piston of cylinder 1 reaches its end position, the piston in cylinder 2 willextend with the velocity, �g. 89:

v2 = v1 =QpA= 2£ 10¡2 ms¡1 answer!

0

0.02

v [m/s]

t [s]

v1 v2

Fig. 89. Velocity of pistons versus time (case a)

Solution of case b. Valves S2 and S3 are fully open. The pump delivery �ow is:

Qp =npqp60

=1500£ 2£ 10¡6 £ 1

60= 50£ 10¡6 m3s¡1

Pressures in cylinders 1 and 2 are:

p1 =F1A+ k1QS1

p2 =F2A=

2000

25 £ 10¡4= 8£ 105 Pa = 0:8 MPa answer!

If the pressure set on the relief valve is p0 > p1 and p0 > p2, then we may considerthe following cases:

² For condition p1 < p2, cylinder 2 is at rest and cylinder 1 will start moving, thenQS1 = Qp:

p1max =F1A+ k1Qp =

1000

25£ 10¡4+ 1010 £ 5£ 10¡5 = 9£ 105 Pa = 0:9 MPa

thus p1max > p2 and this case is not possible.

² Both cylinders move simultaneously when p1 = p2:


F1A+ k1QS1 =

F2A

thus:

QS1 =F2 ¡ F1Ak1

=2000¡ 1000

25£ 10¡4 £ 1010= 40£ 10¡6 m3s¡1

If:

Qp = QS1 +QS2

then:

QS2 = Qp ¡QS1

Initial velocities of the pistons are:

v1 =Q1A=40 £ 10¡6

25£ 10¡4= 1:6£ 10¡2 ms¡1

v2 =QS2A

=10 £ 10¡6

25£ 10¡4= 0:4£ 10¡2 ms¡1 answer!

Piston 1 will reach its end position in time:

t1 =h

v1

where h is piston stroke.

Piston 1 will reach its end position earlier than piston 2 as v1 > v2. After time t1piston 2 moves with increased velocity as pump full �ow is directed to cylinder 2,�g. 90:

v02 =QpA=50£ 10¡6

25£ 10¡4= 2£ 10¡2 ms¡1 answer!

Piston 2 will reach its end position after time:

t2 = t1 +h¡ v2t1v02

= t1(1 +v1 ¡ v2v2

)

Pump delivery pressure p = p2 = 0:8MPa < p0


0.020

0.016

0.010

0.004

t1 t2 t [s]0

v [m/s]

t [s]

v1 v2'

v2

Fig. 90. Velocity of pistons versus time (Case b)

Solution for case c. Valve S3 is fully open, and pressure drops on valves S1and S2 are:

¢pS1 = k2QS1 k2 = 5£ 1010 Nsm¡5¢pS2 = k3Q

2S2

k3 = 1£ 1015 Ns2m¡8

Supply pressure to cylinders:

p1 =F

A+ k2QS1

p2 =F

A+ k3Q

2S2

Supply �ow rates to cylinders:

QS1 = Av1QS2 = Av2

(a)

If the relief valve is closed, p0 ¸ p1 and p0 = p2, then:

Qp = QS1 +QS2 = A(v1 + v2)

thus:

v1 + v2 =QpA=50£ 10¡6

25£ 10¡4= 2£ 10¡2 ms¡1 (b)

We may now consider again, as we did in case b, that p1 = p2 so both pistons moveat the same time:

F

A+ k2QS1 =

F

A+ k3Q

2S2


If we substitute eq. (a) and the values of coe¢cients then:

1000

25£ 10¡4+ 5£ 1010 £ 25£ 10¡4v1 =

2000

25£ 10¡4+ 1015 £

¡25£ 10¡4

¢2 £ v2

The equation above allows calculation of either velocity v1 or v2. For example, byusing eq. (b) and transforming the above equation we obtain:

v22 + 0:02v2 ¡ 3:36£ 10¡4 = 0

thus:

v2 = ¡0:01 + 0:0209 = 0:0109 = 1:09£ 10¡2 ms¡1 answer!

and using again eq. (b) we calculate velocity v1:

v1 = 2£ 10¡2 ¡ 1:09£ 10¡2 = 0:91£ 10¡2 ms¡1 answer!

As v1 < v2, piston 1 will reach its end position before piston 2. Calculation ofvelocity v01 when v2 = 0 is carried in the same way as in the case b. Pump pressurep is calculated from equation:

p = p1 =F

A+ k2QS1 =

F

A+ k2v1A

p =1000

25£ 10¡4+ 5£ 1010 £ 0:91£ 10¡2 £ 25£ 10¡4 = 1:54MPa answer!

and as p0 = 2:5 MPa then p < p0 so the assumption that the relief valve is closed istrue.

Solution of case d. Pressure drops across restrictor valves were de�ned previ-ously by the equations:

¢pS1 = k2QS1 k2 = 5£ 1010 Nsm¡5¢pS3 = k3Q

2S2

k3 = 1£ 1015 Ns2m¡8¢pS3 = k2QS3 k2 = 5£ 1010 Nsm¡5

If we initially assume that the relief valve is closed, then �ow through restrictor S3is QS1 = Qp and the pressure drop on this valve is:

¢pS3 = k2Qp = 5£ 1010 £ 50£ 10¡6 = 2:5MPa

However, as additional pressure drops also occur on valves S1, S2 and in the cylindersthen it obvious that the above assumption is not correct. Thus, the relief is operating,and the system is working at pressure p0 = 2:5MPa set by the relief valve. Under


such condition velocities of the pistons can be calculated from equations:

p1 = p0 =F1A+ k2Av1 + k2A(v1 + v2) (c)

p2 = p0 =F2A+ k3(Av2)

2 + k2A(v1 + v2) (d)

Substituting valves data into (c) and (d) we obtain:

2:5£ 106 =1000

25£ 10¡4+ 5£ 1010 £ 25£ 10¡4v1 + 5£ 1010 £ 25£ 10¡4 £ (v1 + v2)

2:5£ 106 =2000

25£ 10¡4+ 1015 £ (25£ 10¡4v2)2 + 5£ 1010 £ 25£ 10¡4 £ (v1 + v2)

which after rearrangement becomes:

250v1 + 125v2 ¡ 2:1 = 0 (e)

6250v22 + 125(v1 + v2)¡ 1:7 = 0 (f)

Expression for velocity v1 is obtained using eq. (e):

v1 =2:1¡ 125v2

250

which after substitution in eq. (f) yields equation for velocity v2:

6250v22 + 125

µ2:1¡ 125v2

250+ v2

¶¡ 1:7 = 0

or:

v22 + 0:01v2 ¡ 1:04£ 10¡4 = 0

Solution of this equation gives value of velocity v2:

v2 = 6:3578£ 10¡3 ms¡1 answer!

and �nally velocity v1 is

v1 = 5:22£ 10¡3 ms¡1 answer!

The velocity of piston 2 is larger than velocity of piston 1, so piston 2 will reach itsend position at time t:

Articulated platform - cylinders in series 139

t =h

v2

and subsequently the velocity of piston 1 will change to v01 which can be calculatedfrom equation:

p0 =F1A+ k2Av

0

1 + k2Av0

1

Velocity v01 is, �g. 91:

v0

1 =p0 ¡

F1A

2k2Aanswer!

v0

1 =2:5£ 106 ¡

1000

25£ 10¡42£ 5£ 1010 £ 25£ 10¡4

= 0:84£ 10¡2 ms¡1

Problem 4.7 Articulated platform - cylinders in series

A hydraulic system operating an articulated platform, shown in �g. 92 and �g. 93,has the following data:

² piston areas of cylinder 1 and 2, A11 = 50 cm2, A21 = 30 cm

2

² annulus areas of cylinder 1 and 2; A12 = 30 cm2; A22 = 10 cm

2

² forces acting on pistons 1 and 2; F1 = 60 kN; F2 = 50 kN

² hydro-mechanical e¢ciencies of cylinders 1 and 2; ´hm1 = 0:96, ´hm2= 0:94:

t [s]0

0.0084

v [m/s]

v1

v2

v1'

0.0100

0.0057

Fig. 91. Velocity of pistons versus time (Case d)


Assume volumetric e¢ciencies of cylinders ´v1 = ´v2= 0 and calculate pressuresp1and p2.

Answer: The connections between cylinders are shown in �g. 93. Force balancefor cylinder 1:

F1´hm1

= p1A11 ¡ p2A12

and for cylinder 2:

F2´hm2

= p2A21 ¡ p1A22

Substituting data for cylinder 1:

60£ 103

0:96= p1 £ 50£ 10¡4 ¡ p2 £ 30£ 10¡4

and for cylinder 2:

50£ 103

0:94= p2 £ 30£ 10¡4 ¡ p1 £ 10£ 10¡4

we may calculate pressures p1and p2:

1

2

I

0

II

G

Fig. 92. Articulated platform (Problem 4.7)

Two cylinders in series - position error 141

p1 = 28:93MPa answer!

p2 = 27:38MPa answer!

p1

p2

F1 F2

A11 A22

A12A21

12

Fig. 93. Cylinder connections

Problem 4.8 Two cylinders in series - position error

A hydraulic system with two cylinders connected in a series is shown in �g. 94. Thefollowing data is available:

² piston areas for cylinder 1 and 2; A11 = 20 cm2; A21 = 15 cm

2

² annulus area for cylinder 1 and 2; A12 = 15 cm2; A22 = 10 cm

2

² length of stroke is s = 50 cm for both cylinders

² forces F1 = F2 = 15:0 kN.

² The volumetric losses in cylinders are de�ned by the following equation:

´vc =1

1 +K¢p

v

(a)

where:¢p - cylinder pressure di¤erentialv - piston velocityK - constant; K = 10¡1 cmMPa¡1s¡1

² pump has stroke displacement qp = 15 cm3rev¡1

² pump rotation speed np = 2800 rpm:

² pump volumetric e¢ciency ´vp = 1.

Calculate position error e between pistons when one of the pistons reaches its endposition. Also determine to which cylinder we must connect in parallel a restrictorvalve to eliminate this position error. The calculation should be carried out assuming


A12 A22

A11 A21

Q2

Qvc1 Qvc2

F1 F2

p3

Q1

p1

p2

Fig. 94. Hydraulic system (Problem 4.8)

of no losses with exception of the volumetric losses due to leakages in cylinders.Ignore pressure losses in directional control valve.

Answer: Pressure in the cylinder 2 is calculated from equation, p3 = 0 :

F2 = p2A21

p2 =F2A21

=15000

15£ 10¡4= 10 MPa

and the pressure in cylinder 1 from equation:

F1 = p1A11 ¡ p2A12p1 =F1 + p2A12

A11=15£ 103 + 10£ 106 £ 15£ 10¡4

20£ 10¡4= 15 MPa

Pump delivery �ow rate Qp, assuming pump volumetric e¢ciency ´vp = 1, is:

Qp = npqp =2800

60£ 15 = 700 cm3s¡1

The volumetric e¢ciency of cylinder 1:

´vc1 =v1A11Qp

where v1 is piston velocity in cylinder 1. Thus equating this equation with the eq.(a) we obtain:

Two cylinders in series - position error 143

1

1 +K¢p

v1

=v1A11Qp

thus:

v1v1 +K¢p

=v1A11Qpv1

v1 =QpA11

¡K¢p

The pressure di¤erential ¢p in cylinder 1 is:

¢p = ¢pc1 = p1 ¡ p2 = 5MPa

therefore:

v1 =700

20¡ 5£ 10¡1 = 34:5 cms¡1

Volumetric losses Qvc1, due to internal leakage in cylinder 1, are:

Qvc1 = Qp ¡ v1A11 = 7000¡ 34:5£ 20 = 10 cm3s¡1

Return �ow Q2 from cylinder 1 is described by equation:

Q2 = v1A12 +Qvc1 = 34:5£ 15 + 10 = 527:5 cm3s¡1

Similar calculations for cylinder 2 allows us to �nd piston velocity v2: Again vol-umetric e¢ciency of the cylinder is a ratio of the demand �ow of the cylinder tosupply �ow Q2, thus:

´vc2 =v2A1Q2

and

1

1 +K¢p

v2

=v2A21Q2

v2 =Q2A21

¡K¢p

where ¢p is pressure di¤erential in cylinder 2 which, and if we ignore pressure lossesin the return line, it is equal to:


¢p = ¢pc2 = p2 = 10 MPa

Finally, the velocity of piston 2 is:

v2 =527:5

15¡ 1 = 34:2 cms¡1

Volumetric losses Qvc2 due to internal leakage in cylinder 2 are:

Qvc2 = Q2 ¡ v2A21 = 527:5¡ 34:2£ 15 = 15 cm3s¡1

As velocity of the piston in cylinder 1 is v1 = 34:5 cms¡1 and velocity of the piston

in cylinder 2 is v2 = 34:2 cms¡1, thus the piston in cylinder 1 will reach its endposition sooner than the piston in cylinder 2. Position error e will be:

e = s1¡s2 = s¡v2t1 = s¡v2s

v2= s

µ1¡

v2v1

¶= 50£(1¡

34:2

34:5) = 0:435 cm answer!

where:s - maximum stroke of pistons

s1;s2 - strokes of pistons in cylinders 1 and 2t1 - stroke time of the piston in cylinder 2

To eliminate the position error we may install a restrictor valve, in parallel withcylinder 1, which will allow some �uid to bypass cylinder 1. Such a valve shouldallow by-pass �ow Qv = Qvc2¡Qvc1 = 5 cm3s¡1 at pressure di¤erence ¢p = 5 MPa:

Problem 4.9 Two cylinders in parallel - lifting system

A hydraulic lifting system is shown in �g. 95. Following data is available:

² mass M = 5000 kg

² pump stroke displacement qp = 50 cm3rev¡1


² pistons �areas are 25 cm2, pistons are cushioned at the end of their stroke, howeverthis should be ignored in calculations.

² hydro-mechanical and volumetric e¢ciencies of the pump are ´hmp = 0:96; ´vp =0:94

² hydro-mechanical and volumetric e¢ciencies of cylinders are ´hmc = 0:93, ´vc =0:98

Calculate time required to lift the load to height h = 1:0m :

Answer: Actual pump delivery �ow Qp is:

Qp = ´vpnpqp = 0:94£ 1420£ 50 = 66740 cm3min¡1

Two cylinders in parallel - lifting system 145

Mass M

A11

A21

Qc Qc

qp

np p0

Fig. 95. Hydraulic lifting system (Problem 4.9)

and the �ow into each cylinder is equal to:

Qc =Qp2= 33370 cm3min¡1

To calculate pressure in cylinders we use the equation for cylinder hydro-mechanicale¢ciency ´hmc:

´hmc =FeFtc

=FepA

where:Fc - actual cylinder forceFtc - theoretical cylinder force

therefore:

p =Fe

A´hmc=

Mg

2£ ´hmcA=

5000£ 9:812£ 0:93£ 25£ 10¡4

= 10:5 MPa

Volumetric e¢ciency of cylinder is:


´vc =vcA

Qc

thus:

vc =´vcQcA

=0:98£ 3337025£ 60

= 21:8 ms¡1

and, �nally, lifting time t is equal to:

t =h

vc=100

21:8= 4:6 s answer!

Problem 4.10 Pump torque - open circuit hydrostatic drive

Select the sizes of a pump and a motor for the system shown in �g. 96. Determinethe required input torque when the relief valve pressure p0 is set to p0 = 21 MPa,and pump rotation speed is np = 2£103 rpm. The hydraulic motor should haverotation speed nm = 5 revs

¡1 at load torque Tm = 100 Nm. Ignore volumetric andhydraulic losses.

poTp

np

Tm

nm

Fig. 96. Open circuit hydrostatic drive (Problem 4.10)

Answer: The following equations apply for a system without losses:

Tp = VÁp¢pp Qp = VÁp!pTm = VÁm¢pm Qm = VÁm!m

To minimize the size of the pump and the motor, it is necessary to choose systemoperating pressure as high as possible. Thus we will choose an operating pressureonly slightly lower than the maximum pressure set on the relief valve. Ignoringlosses, we assume:

¢pp = ¢pm = 20MPa < p0

System pressure - open circuit hydrostatic drive 147

Unit displacement VÁm of a motor is calculated from:

VÁm =Tm¢pp

=100

20£ 106= 5£ 10¡6 m3rad¡1 answer!

and using value of VÁm, we can calculate the �ow demand of the motor:

Qm = VÁm!m = 5:0£ 10¡6 £ 2£ ¼ £ 5 = 157£ 10¡6 m3s¡1

If the relief valve is closed then:

Qp = Qm thus Qp = VÁm!p

and unit displacement of the pump is equal to:

VÁp =Qm!p

=157£ 10¡6 £ 602000£ 2£ ¼

= 0:75£ 10¡6 m3rad¡1 answer!

Required pump input torque Tp is:

Tp = VÁp¢pp = 0:75£ 10¡6 £ 20£ 106 = 15:0 Nm answer!

Problem 4.11 System pressure - open circuit hydrostatic drive

The system shown in �g. 97 has the following data:

² rotation speed of electric motor ne = 1430 rpm

² torque of hydraulic motor Tm = 0:07nm Nm, where rotation speed of hydraulicmotor nm rpm

² stroke displacements of pump and motor qp = qm = 35 cm3rev¡1

² volumetric e¢ciency of pump ´vp = 0:94

² volumetric e¢ciency of motor ´vm = 0:95

² overall e¢ciency of pump ´p = 0:88

² overall e¢ciency of motor ´m = 0:90

² relief valve is set at p0 = 20 MPa

Using above data and ignoring volumetric and hydraulic losses in the lines calculatethe rotation speed of the motor nm, hydraulic motor output power Pm and pressureat the motor inlet port p3.

Answer: Pump delivery �ow:

´vp =QpQtp

=Qpnpqp


ne

po

p3

Load

nm

Pm

Qrv


Qp = ´vpnpqp (a)

When the relief valve is closed and the e¢ciency of the delivery line is ´v1 = 1 (nolosses), then:

Qm = Qp (b)

The rotation speed of the motor is calculated from the equation for motor volumetrice¢ciency:

´vm =QtmQm

=nmqmQm

and

nm =´vmQmqm

answer!

From eq. (a) and eq. (b) we obtain, qp = qm:

nm = ´vm´vpnpqpqm

= 0:95£ 0:94£ 1430 = 1277 rpm

According to the problem statement, the torque of the motor is:

Tm = 0:07nm = 1277£ 0:07 = 89:4 Nm

and the motor output power is:

Pm = Tm!m = Tm2¼nm60

= 89:3£2£ ¼ £ 1277

60= 11:9 kW answer!

We can calculate pressure di¤erential across the motor using equation for hydro-mechanical e¢ciency of the motor:

Motor e¤ective power - open circuit hydrostatic drive 149

´hmm =TmTtm

=Tm2¼

¢pmqm

therefore:

¢pm =Tm2¼

´hmmqm

and as:

´hmm =´m´vm

then:

¢pm =Tm2¼´vm´mqm

=89:3£ 2£ ¼ £ 0:9535£ 10¡6 £ 0:90

= 16:92 MPa

We assume that the return line pressure p4 = 0, thus:

p3 = ¢pm = 16:92 MPa answer!

this pressure is lower than the setting of the relief valve p0 = 20 MPa, therefore theassumption that Qm = Qp was correct.

Problem 4.12 Motor e¤ective power - open circuit hydrostatic drive

For the system analysed in Problem 4.11, �g. 97, calculate rotation speed of themotor nm, actual motor power Pm and motor input pressure p3 when the relief valveis set to p0 = 15MPa. Torque of the motor was de�ned by equation:

Tm = 0:07nm Nm

where nm was rotation speed of the motor in rpm. Other system data are:

² pump and motor displacements; qp = qm = 35 cm3rev¡1

² volumetric e¢ciency of motor; ´vm = 0:95

² overall e¢ciency of motor; ´m = 0:90

Answer: In Problem 4.11 system pressure was calculated to be 16:9 MPa. Ifwe now assume that the relief valve is set at p0 = 15 MPa and that it has an ideal�ow-pressure characteristics, the pressure in the system will be limited to 15 MPa.Equations for hydro-mechanical e¢ciency ´hmm of the motor and motor torque Tmare:

´hmm =TmTtm

=2¼Tm¢pmqm


Tm = 0:07nm Nm

which when combined yield:

´hmm = 0:07nm2¼

¢pmqm(a)

The overall e¢ciency of the motor is equal to:

´m = ´vm´hmm

then after substituting in the above eq. (a) and rearranging we obtain:

´m´vm

= 0:072¼nm¢pmqm

where ¢pm = p3. Thus:

nm =´m¢pmqm0:07´vm2¼

nm =0:9£ 150£ 105 £ 35£ 10¡6

0:07£ 0:95£ 2£ ¼= 1130 rpm answer!

then torque of the motor is:

Tm = 0:07nm = 0:07£ 1130 = 79:1 Nm answer!

and the output power of the motor:

Pm = Tm!m = 79:1£2£ ¼ £ 1130

60= 9:4 kW answer!

Problem 4.13 Analysis of open circuit hydrostatic drive

Open circuit hydrostatic drive is shown in �g. 98. Input data is as follows:

² rotation speed of electric motor ne = 1430 rpm

² torque of hydraulic motor Tm = 0:07nm Nm, where rotation speed of hydraulicmotor nm rpm

² stroke displacements of pump and motor qp = qm = 35 cm3rev¡1

² volumetric e¢ciency of pump ´vp = 0:94

² volumetric e¢ciency of motor ´vm = 0:95

² overall e¢ciency of pump ´p = 0:88

² overall e¢ciency of motor ´m = 0:9

² volumetric and hydraulic losses in the lines are ignored.

Analysis of open circuit hydrostatic drive 151

The valve has the following characteristics:

Qrv =

½0 for p3 � p0

K(ps ¡ p0) for p3 > p0

where:

K - �ow coe¤cient, K = 40£ 10¡11 m3MPa¡1=2s¡1p - system pressurep0 relief valve set pressure, p0 = 13 MPa

Using above data and ignoring volumetric and hydraulic losses in the lines calculatethe rotation speed of the motor nm, hydraulic motor output power Pm and pressureat the motor inlet port p3.

ne

po

p3

Load

nm

Pm

Qrv


Answer: Pump delivery �ow is:

Qp = ´vpnpqp =0:94£ 1430£ 35£ 10¡6

60= 784:1£ 10¡6 m3s¡1 (a)

The maximum motor pressure is dependent on the characteristics of the relief valve.Remembering that we may ignore losses in delivery and return lines:

¢pm = p0 = p3

Pressure p = p3 at which system will operate after relief valve opens is calculatedusing the equation for �ow through the relief valve:

Qrv = K(p3 ¡ p0)

p3 =QrvK

+ p0 (b)

We use the expression for hydro-mechanical e¢ciency of the motor ´hmm and calcu-late pressure di¤erential ¢pm:

´hmm =TmTtm

=Tm

¢pmVÁm


then as Tm = 0:07nm then pressure di¤erential across the motor is:

¢pm =0:07nm2£ ¼´hmmqm

(c)

Thus, as ¢pm = p3, we can equate eq. (b) with eq. (c):

0:14¼nm´hmmqm

=QrvK

+ p0

and obtain the expression for �ow through the relief valve:

Qrv =0:14¼nmK

´hmmqm¡Kp0 (d)

The equation for volumetric e¢ciency of a motor is:

´vm =QtmQm

=!mVÁmQm

from which we obtain the equation for motor demand �ow:

Qm =nmqm £ 10¡6

60´vm(e)

Pump delivery Qp is equal to the sum of �ows through the relief valve Qrv and themotor Qm, thus we can calculate rotation speed of the motor:

Qp = Qrv +Qm

taking into consideration equations (d) and (e) we obtain:

Qp =0:14¼nmK

´hmmqm¡Kp0 +

nmqm60£ 106´vm

where:

´hmm =´m´vm

=0:9

0:95= 0:947

Rearranging and using (a) we get expression for motor speed:

nm =Qp +Kp0

0:14¼K

´hmmqm+qm £ 10¡6

60£ ´vm

=

Motor speed - open circuit hydrostatic drive 153

nm =784:1£ 10¡6 + 40£ 10¡11 £ 13£ 106

0:14£ ¼ £ 40£ 10¡11

0:947£ 35£ 10¡6+35£ 10¡6

60£ 0:95

= 1010 rpm answer!

The motor output torque is:

Tm = nm0:07 = 1010£ 0:07 = 70:7 Nm

and the motor output power is equal to:

Pm = T! =70:6£ 2£ ¼ £ 1010

60= 7:47 kW answer!

Finally pressure p3; which in the absence of losses is equal to pressure di¤erentialacross the motor, is calculated from equation:

p3 = ¢pm =Tm´vm2¼

´mqm=70:6£ 0:95£ 2£ ¼0:90£ 35£ 10¡6

= 13:4 MPa (answer!)

Problem 4.14 Motor speed - open circuit hydrostatic drive

A hydraulic pump which supplies a hydraulic motor, �g. 99, has rotation speed nm =15 rev s¡1. What is the rotation speed of a motor which has stroke displacementqm = 15qp (where qp is pump stroke displacement). Volumetric e¢ciency of thepump is ´vp = 0:9, and the motor internal leakage �ow Qvm = 2Qvp.

qp qmQvp Qvm

QpQm

np nm

Fig. 99. Hydraulic System (Problem 4.14)

Answer: The theoretical delivery �ow of a pump and the theoretical �ow demandof a motor are calculated from relations:

Qtm = nmqm

Qtp = npqp


Volumetric e¢ciencies of a pump and a motor are de�ned by expressions:

´vp =QpQtp

=Qtp ¡QvpQtp

= 1¡QvpQtp

and (note that Qvm = 2Qvp)

´vm =QtmQm

=Qtm

Qtm +Qvm=

1

1 +QvmQtm

=1

1 +2QvpQtm

(a)

where:Qp;Qm - actual pump supply and motor demand �ows

Qtp;Qtm - theoretical pump supplu and motor demand �owsQvp; Qvm - leakage �ows of pump and motor

As the pump is directly connected to the motor thus Qp = Qm and:

´vpnpqp =nmqm´vm

(b)

or:

1¡ ´vp =QvpQtp

therefore the leakage �ow of the pump is expressed by:

Qvp = (1¡ ´vp)Qtp (c)

The volumetric e¢ciency of the motor, using relations (c) in (a) is:

´vm =1

1 + 2(1¡ ´vp)QtpQtm

which we can substitute in eq. (b):

´vpnpqp =nmqmQtm + 2(1¡ ´vp)Qtp

nmqm

as the stroke displacement of motor is qm = 15qp then after simpli�cation we obtain:

´vpnp = 15£ nm + 2(1¡ ´vp)np

so, �nally, we can calculate motor rotation speed nm:

Parallel drive of two hydraulic motors 155

nm =3´vp ¡ 215

np =3£ 0:9¡ 2

15£ 15 = 0:7 revs¡1 answer!

Problem 4.15 Parallel drive of two hydraulic motors

In a hydraulic system shown in �g. 100, a single �xed displacement pump supplies�uid to two motors connected in parallel.

np

qp

Tm1 nm1

Tm2 nm2

qm1

qm2

Fig. 100. Drive of two motors in parallel (Problem 4.15)

Following data is available:

² rotation speed of the pump np = 50 revs¡1

² torque loading of hydraulic motors Tm1 = 173 Nm and Tm2 = n2m2k

² stroke displacement of the pump qp = 40:8£10¡6 m3rev¡1

² stroke displacements of motors qm1 = 60 £ 10¡6 m3rev¡1 and qm2 = 50 £10¡6 m3rev¡1

² hydro-mechanical e¢ciency of the motors; ´hmm = 0:90, constant over wholeoperating range

² volumetric e¢ciency of the pump; ´vp = 0:98, constant over whole operatingrange

Ignore volumetric and hydraulic losses in hydraulic lines and calculate rotationspeeds of motors.

Answer: Torques developed by the motors are:

Tm1 =¢pmqm12¼

´hmm


Tm2 =¢pmqm22¼

´hmm

Using the �rst of the above equations we calculate pressure drop in the motors:

¢pm =2¼Tm1qm1´hmm

=2£ ¼ £ 173

60£ 10¡6 £ 0:9= 20£ 106 Pa

then the torque of the second motor is:

Tm2 =¢pmqm22¼

´hmm =20£ 106 £ 50£ 10¡6

2£ ¼£ 0:9 = 143:2 Nm

Rotation speed nm2 of the second motor is calculated from equation:

Tm2 = n2m2k

and is equal to:

nm2 =

rTm2k

=p143:2 for k = 1

nm2 = 11:97 revs¡1 answer!

If we ignore leakages in the motor and hydraulic lines �ow demand Qm2 of motor 2is:

Qm2 = nm2qm2 = 11:97£ 50£ 10¡6 = 598:5 cm3s¡1

and as �ow delivery Qp of the pump is:

Qp = npqp´vp = 50£ 40:8£ 10¡6 £ 0:98 = 2000 cm3s¡1

then demand �ow Qm1 of motor 1 is:

Qm1 = Qp ¡Qm2 = 2000¡ 598:5 = 1401:5 cm3s¡1

and �nally rotation speed nm1 of motor 1, keeping in mind that we assumed volu-metric e¢ciency ´vm = 1; is calculated from equation:

nm1 =Qm1qm1

=1401:5£ 10¡6

60£ 10¡6= 23:4 revs¡1 answer!

Pump �ow - meter-in control of a motor 157

Problem 4.16 Pump �ow - meter-in control of a motor

Calculate �ow rate Qp of a pump driving a hydraulic motor in a circuit shown in�g. 101. The required output power of the motor is P = 25 kW. Calculate also thepump delivery pressure p and �ow coe¢cient Krv of the relief valve which has thefollowing �ow characteristics:

Qrv =

½0 for ps � p0

Krv(ps ¡ p0)pp for ps > p0

where p0 is relief valve pressure setting.

∆pd

Qp

po

Qd=Qmpp

∆pm

np nm

P

Fig. 101. Meter-in control of hydraulic motor (Problem 4.16)

The pump operating rotation speed np = 18:0 revs¡1 and its volumetric e¢ciency is

´vp = 0:95. Flow through the restrictor is governed by the following equation:

Qd = Kadp¢pd (a)

where:¢pd - pressure drop across the restrictorad - �ow area of the restrictor, ad = 3:0 cm

2

K - constant, K = 5:0 MPa¡1=2ms¡1

The motor has the following parameters:

² stroke displacement qm = 80:0 cm3rev¡1

² rotation speed nm = 14 revs¡1

² volumetric e¢ciency ´vm = 0:93

² overall e¢ciency ´m = 0:68

Calculations should be carried out assuming that actual �ow demand Qm of themotor is equal to 72% of pump �ow rate (Qm = 0:72Qp). Ignore losses in deliveryline.


Answer: We calculate pressure drop in the motor using equation:

P = Qm¢pm´m = qmnm¢pm´hmm

then

¢pm =P

qmnm´hmm

and as

´hmm =´m´vm

=0:68

0:93= 0:73

therefore

¢pm =25£ 103

80£ 10¡6£14£ 0:73= 30:58 MPa

Pressure p is calculated using equation:

p = ¢pm +¢pd

We use the equation for �ow through the restrictor valve to calculate pressure drop¢pd:

¢pd =

µQd

Kad

¶2

and as

Qd = Qm =qmnm

´vm=80£ 10¡6 £ 14

0:93= 1:2 Ls¡1

therefore

¢pd =

µ1:2£ 10¡3

5:0£ 3£ 10¡4

¶2= 0:64 MPa

and pressure p is:

p = 30:58 + 0:64 = 31:22 MPa answer!

Flow delivery of the pump is:

Qp =Qm

0:72

Qp = qpnp´vp =Qm

0:72

Pump �ow - meter-in control of a motor 159

thus required stroke displacement of the pump:

qp =Qm

0:72np´vp=

1:2£ 10¡3

0:72£ 18£ 0:95= 0:097 Lrev¡1

Coe¢cient Krv is calculated using relation:

Qb = Krv(p¡ p0)pp

where �ow Qb through the relief valve is:

Qb = Qp ¡Qm

as the actual �ow delivery of the pump is equal to:

Qp = qpnp´vp = 0:097£ 18£ 0:95 = 1:66 Ls¡1 answer!

then �ow through the relief valve is equal to:

Qb = 1:66¡ 1:2 = 0:46 Ls¡1

Pressure p0 at which relief valve will open is calculated using equation:

p0 = ¢pm1 +¢pd1

where ¢pmp1 and ¢pd1 are pressure di¤erentials when Qp = Qm, thus:

¢pm1 =P

Qm´hmm

¢pm1 =25£ 103

1:66£ 10¡3 £ 0:73= 20:63 MPa

Pressure drop ¢pd1on the restrictor valve when Qp = Qm is equal to:

¢pd1 =

µ1:66£ 10¡3

5£ 10¡3 £ 3£ 10¡4

¶2= 1:2 MPa

so, pressure p0 set on the relief valve should be equal to:

p0 = 20:63 + 1:2 = 21:83 MPa

Coe¢cient Krv can be now calculated:

Krv =Qb

(p¡ p0)pp


Krv =0:46

(31:22¡ 21:83)p31:22

= 8:76£ 10¡3 Ls¡1MPa¡3=2 answer!

Problem 4.17 Motor speed - meter-in control of a motor

For the hydrostatic drive shown in �g. 102 calculate �ow Qd through the restrictorvalve and a maximum rotation speed nm of the motor.

Qp

po

Qd=Qmpp

∆pm

np nm

T

∆pd

Fig. 102. Meter -in control of hydraulic motor (Problem 4.17)

Use the following data:

² pump stroke displacement qp = 0:12 Lrev¡1

² pump rotation speed np = 20 revs¡1

² pump volumetric e¢ciency ´vp = 0:95

² motor stroke displacement qm = 0:16 Lrev¡1

² motor volumetric e¢ciency ´vm = 0:95

² motor hydro - mechanical e¢ciency ´hmm = 0:80

² motor load torque T = 60 Nm

² pressure setting of the relief valve p0 = 7 MPa

Flow through the restrictor valve is de�ned by equation:

Qd = Kadp¢pd

where:¢pd - pressure drop on the valvead - valve �ow area, ad = 2 cm

2

K - �ow constant, 4:25 MPa¡1=2ms¡1

Answer: Hydraulic motor power Pm is following equations:

Pm = T!m

Pm = Qtm¢pm´hmm = qmnm¢pm´hmm

Motor speed - pressure compensated �ow control valve 161

and as:

nm =!m

2¼

therefore:

qm!m¢pm´hmm2¼

= T!m

and pressure drop ¢pm across the motor is equal to:

¢pm =2¼T

qm´hmm=

2£ ¼ £ 600:16£ 10¡3 £ 0:8

= 2:95 MPa

As the system is working at maximum pressure thus the relief must be open and thepressure drop ¢pd on the restrictor valve is:

¢pd = p0 ¡¢pm = 7:0¡ 2:95 = 4:05 MPa

Flow through the restrictor valve is, according to the problem statement, de�ned byequation:

Qd = Kadp¢pd = 4:25£ 2£ 10¡4 £

p4:05 = 1:71 Ls¡1 answer!

The maximum pump delivery �ow is equal to:

Qp = qpnp´vp = 0:12£ 20£ 0:95 = 2:28 Ls¡1

The maximum rotation speed nm of the motor is calculated using Qd and the knownvalue of the volumetric e¢ciency of the motor:

Qd =Qtm

´vm=qmnm

´vm

where Qtm is the theoretical �ow demand of the motor. Finally, the maximumrotation speed of the motor is equal to:

nm =Qd

qm´vm =

1:71

0:16£ 0:95 = 10:1 revs¡1 answer!

Problem 4.18 Motor speed - pressure compensated �ow control valve

For the system which is shown in �g. 103 calculate the maximum torque developedby a hydraulic motor and the overall e¢ciency of the system when the restrictorvalve is fully open. Use the following data:

² pump stroke displacement qp = 0:18 Lrev¡1


² pump rotation speed np = 16 revs¡1


² pump overall e¢ciency ´p = 0:75

² the relief valve is set at p0 = 7 MPa: The relief valve is closed when the restrictorvalve is fully open

² the pressure di¤erential valve has pressure di¤erential ¢pc = 0:7 MPa

² pressure drop through the fully open restrictor valve is ¢pd = 0:3 MPa

² motor stroke displacement qm = 0:14 Lrev¡1

² motor volumetric e¢ciency ´vm = 0:93

² motor hydro-mechanical e¢ciency ´hmm = 0:85

Qp

po

pp

∆pm

np nm

∆pd∆pc

qp qm

Fig. 103. Speed control of hydraulic motor (Problem 4.18)

Answer: The pressure drop on the motor is:

¢pm = p0 ¡¢pc ¡¢pd = 7:0¡ 0:7¡ 0:3 = 6:0 MPa

and the maximum motor torque may be calculated using equation:

Tmmax=qm¢pm´hmm

2¼=0:14£ 10¡3 £ 6:0£ 106 £ 0:85

2¼= 113:6 Nm answer!

Overall e¢ciency of the system is a product of overall e¢ciencies of the pump, motorand the �ow control valve, thus:

´ = ´p´m´fv

The pump overall e¢ciency is given as:

´p = 0:75

and the overall e¢ciency of the motor is de�ned by equation:

Speed control of cylinder - pressure compensated controls 163

´m = ´vm´hmm = 0:85£ 0:9 = 0:76

As the overall e¢ciency of the �ow control valve (i.e. sequence and restrictor valves)is de�ned by equation:

´fv =¢pmp0

=p0 ¡¢pc ¡¢pd

p0=6:0£ 106

7:0£ 106= 0:86

thus, the overall e¢ciency of the system is:

´ = 0:75£ 0:76£ 0:86 = 0:49 (answer!)

Problem 4.19 Speed control of cylinder - pressure compensated controls

The hydraulic system shown in �g. 104 consists of a �xed displacement pump, arelief valve, three directional control valves (a, b and c), restrictor type �ow controlvalve (d, e) and a hydraulic cylinder. The following data is available:


² pump unit displacement V'p = 20 mLrad¡1

² cylinder piston diameter D = 60mm

² cylinder rod diameter - d = 25mm

² cylinder force F = 10 kN

² force losses in cylinder Ffr = 100 N

² pump hydro-mechanical e¢ciency ´hmp = 0:80


² the relief valve is set at p0 = 15:0 MPa

Flows through the valves, when pa; pb and pc units are MPa, are:

for valve (a):

Qa = 3:0ppa Ls

¡1 (a)

for valve (b):

Qb = 2:5ppb Ls

¡1 (b)

for valve (c):

Qc = 4:5ppc Ls

¡1 (c)


Constant pressure di¤erential ¢pd = 0:4 MPa across restrictor (d) is maintainedby valve (e). The �ow control valve assures constant velocity of the piston, vc =600 mms¡1.

Calculate pump �ow rate Qp, pump input power Pp, pressure drop on �ow controlvalve ¢pr, and the overall e¢ciency of the system ´sys. Calculate also system e¢-ciency when the restrictor-type, �ow control valve is replaced by a by-pass type �owcontrol valve.

Qp

p3

pp

np

Q1

F

p4 Q2

φD

φdb

c

d

e

a

p0=15 MPa

Fig. 104. Cylinder speed control (Problem 4.19)

Answer: Flow rate Q1 into cylinder piston volume is:

Q1 =¼D2

4vc =

¼ £ (0:06)2

4£ 0:6 = 1:7 Ls¡1

and �ow rate Q2 into annulus volume is:

Q2 =¼(D2 ¡ d2)

4vc =

¼ £ (0:0602 ¡ 0:0252)4

£ 0:6 = 1:4 Ls¡1

Speed control of cylinder - pressure compensated controls 165

The actual �ow delivery of the pump is calculated using the pump theoretical �owrate Qtp and its volumetric e¢ciency ´vp:

Qtp = VÁp2¼np60

= 20£ 10¡6 £2£ ¼ £ 103

60= 2:09 Ls¡1

Qp = Qtp´vp = 2:09£ 0:92 = 1:93 Ls¡1 answer!

As pump �ow is larger than �ow into the cylinder piston volume Qp > Q1 then therelief valve must be open and the input power to the pump is then:

Pp =p0Qtp

´hmp=15£ 106 £ 2:09£ 10¡3

0:8= 39:2 kW answer!

The e¤ective power of the cylinder is calculated using equation:

Pc = Fvc = 10£ 103 £ 0:6 = 6 kW


´sys =Pc

Pp=

6

39:2= 0:15 answer!

Pressure loss in directional control valve (b), eq. (b), due to �ow from the annulusvolume of the cylinder into the reservoir is:

¢pb2 =

µQ2

2:5

¶2=

µ1:4

2:5

¶2= 0:31 MPa

and the pressure loss due to �ow into the cylinder:

¢pb1 =

µQ1

2:5

¶2= 0:46 MPa

Using eq. (a) we can calculate pressure loss in directional control valve (a):

¢pa =

µQ1

3

¶2=

µ1:7

3

¶2= 0:32 MPa

Cylinder pressure p3 is calculated from the following equation:

p3¼D2

4= F + Ffr +¢pb2

¼(D2 ¡ d2)4

thus,


p3 =4(F + Ffr)

¼D2+¢pb2

µ1¡

d2

D2

¶

p3 =4£ (10£ 103 + 100)¼ (60 £ 10¡3)2

+ 0:31£ 106Ã

1¡¡25 £ 10¡3

¢2

(60 £ 10¡3)2

!

= 3:83 MPa

Finally, pressure drop on the pressure di¤erential valve (e) is:

¢pr = p0 ¡¢pa ¡¢pd ¡¢pb1 ¡ p3¢pr = 15¡ 0:32¡ 0:4¡ 0:46¡ 3:83 = 10 MPa answer!

Pressure drop ¢pr is the inherent loss associated with a chosen control strategy.

c d

e

To valve a

To valve b

Fig. 105. By-pass �ow control (Problem 4.19)

If we replace the restrictor type �ow control valve with a by-pass �ow control valve,�g. 105, then pump pressure is:

pp = ¢pa +¢pd +¢pb1 + p3

where:

¢pa =

µQp

3

¶2=

µ1:93

3

¶2= 0:41 MPa

as the relief valve is closed because the pump pressure is lower than the relief valvesetting. The pump pressure is:

Hydraulic lift system 167

pp = 0:41 + 0:4 + 0:46 + 3:83 = 5:1 MPa

The power input to the pump is equal to:

Pp =pQtp

´hmp=5:1£ 106 £ 2:09£ 10¡3

0:8= 13:32 kW answer!


´sys =Pc

Pp=

6

13:32= 0:45 answer!

The e¢ciency of this system in which a by-pass �ow control valve is used is thusthree times higher than the e¢ciency of the system with a restrictor type �ow controlvalve.

Problem 4.20 Hydraulic lift system

A hydraulic system operating a lift is shown in �g. 106. Calculate power P necessaryto drive the pump and overall e¢ciency of system ´sys. System parameters are asfollows:

² load F = 13:4 kN

² dead weight of the lift F0 = 450N² lifting velocity v = 0:6 mms¡1

² pump delivery �ow Qp = 1:5 Ls¡1

² pump overall e¢ciency ´p = 0:85

² cylinder piston area A2 = 0:5A1² piston diameter D = 100mm

² e¢ciency of cylinder ´c = 1

The �ow rate through directional control valve (e) in directions P-A, B-T, and indirections P-B and A-T is de�ned by equation (¢p in MPa):

Q = K1p¢p

where K1 is �ow coe¢cient, K1 = 440 LpmMPa¡1=2

The �ow rate through valves (d) and (g) is de�ned by:

Q = K2p¢p

where K2 is �ow coe¢cient, K2 = 570 LpmMPa¡1=2 and ¢p is in MPa.


c

d

0.35 MPa

b

a

h

e

f

Fc

F+F0 v

A2

A1

g17 MPa

A

T

P

B

Fc

p3

p4

Fig. 106. Hydraulic lift system (Problem 4.20)

Answer: The power necessary to drive the system is calculated using equation:

P = Pp = Qp¢p12´p

(a)

In the case when directional control valve (d) is in a closed position, the �ow pathfrom the pump to the cylinder is through by-pass �ow control valve (c) and direc-tional control valve (e) (path P-A). If we ignore pressure losses in hydraulic lines,the pressure di¤erence acting on the cylinder piston is:

¢p12 = ¢pc +¢pP¡A + p3

To evaluate pressure losses in individual control elements we must �rst calculate �owQ2 to the cylinder and return �ow Q1:

Q1 = A1 £ v2

and as piston area is equal to:

Hydraulic lift system 169

A1 =¼D2

4=¼ £ 0:1002

4= 7:854£ 10¡3m2

and velocity of the piston is:

v2 =v

i=0:6

2= 0:3 ms¡1

where i is overall ratio of pulley system, therefore:

Q1 = 7854£ 10¡6 £ 0:3 = 2:36£ 10¡3 = 2:36£ 10¡3 m3s¡1 = 141:4 Lpm

At the same time the return �ow from the cylinder is:

Q2 = A2 £ v2

where area A2 is equal to:

A2 = 0:5£A1 = 0:5£ 7:854£ 10¡3 = 3:927£ 10¡3m2

thus return �ow Q2 is

Q2 = 3:927£ 10¡3 £ 0:3 = 1:18£ 10¡3 m3s¡1 = 70:7 Lpm

Pressure drop ¢pP¡A is equal to:

¢pP¡A =

µQ2

K1

¶2=

µ70:7

440

¶2= 0:026 MPa

Assuming pulleys e¢ciency ´pl = 1, cylinder pressure p3 is:

A2 £ p3 = Fc + p4 £A1

where:

Fc = i£ (F + F0) = 2£ (13400 + 450) = 27700 N

Pressure in the return line is calculated from the relation:

p4 = ¢pg +¢pB¡T =

µQ1

K2

¶2+

µQr

K1

¶2

p4 =

µ141:4

570

¶2+

µ141:4

440

¶2= 0:062 + 0:103 = 0:165 MPa

thus:


p3 =Fc+p4A1A2

=27700 + 0:165£ 106£7854£ 10¡6

3927 £ 10¡6= 7:38 MPa

and pressure ¢p12 in pump delivery line is equal to:

¢p12 = 0:35 + 0:026 + 7:384 = 7:76 MPa

Using eq. (a) we can calculate the required input power:

P =1:5£ 10¡3 £ 7:76£ 10¡6

0:85= 13:7 kW answer!

Finally, the overall e¢ciency of the system is equal to:

´sys =Fcv2

P

´sys =27700£ 0:313700

= 0:61 answer!

Problem 4.21 Flow-pressure characteristics of combined hydraulicelements

Flow-pressure characteristics of two hydraulic elements A and B are shown in �g.107. Determine �ow-pressure characteristics for the assembly of these elements in-series and in-parallel.

∆p

B

A

Q

Fig. 107. Flow-pressure characteristics of hydraulic elements A and B (Problem4.21)

Answer: In the case of a parallel connection of the elements a total �ow §Qis the sum of �ows through the individual elements. However the pressure drop, if

Flow-pressure characteristics of pumping unit 171

we ignore losses in connecting lines, is the same in each element. We obtain the�ow-pressure characteristics of the assembly by adding �ow rates through elementA and element B at constant pressure drop ¢pA = ¢pB = ¢p, and repeating theprocedure for various values of ¢p.

A

B

QA

QB

Qtotal

A

B

QA

A+B

QA

QB

∆p

Qtotal

∆pA=∆pB

B

A

Q

∆pA

∆pB

∆p

Qtotal

∆ptotal

∆pB

∆pB

Q

A+B

B

A

A B

∆ptotal

∆pB∆pA

∆ptotal

Fig. 108. Flow-pressure characteristic curve. a.- parallel connection; b.- seriesconnection. Series connection

In a series connection of the elements the �ow through each element is the same,however the total pressure drop is the sum of pressure drops on the individual el-ements. We obtain the �ow-pressure characteristic curve by summing the pressuredrops ¢pA and ¢pB on elements A and B at a constant �ow Q and repeating theprocess for various values of �ow Q. The resultant characteristic for parallel andseries connections are shown in �g. 108.

Problem 4.22 Flow-pressure characteristics of pumping unit

Given the �ow-pressure characteristics of a pump and a relief valve, �g. 109, deter-mine �ow-pressure characteristics of the pumping unit which includes these elements.

Answer: In a pumping unit a pump and a relief valve are connected in parallel.We obtain the unit�s characteristics by assuming that the system �ow is equal to:

Q = Qp ¡Qrv


p

Qp

p

p0

Qrv

Fig. 109. Flow-pressure characteristics. Left - relief valve, right - pump

where:Qp - �ow rate delivered by the pumpQrv - �ow through the relief valve to reservoir

For p � p0 (the pressure at which the relief valve opens) �ow through the relief valveis Qrv = 0 and thus Q = Qp, and in this pressure range the characteristic curve ofthe unit is identical to the characteristic curve of the pump.

p

p0

px

Qrv

p2

B(Qx,px)

A(Q1,p0)

D(Qp,px)

Qrv

Qx

p

p0

p2

Qrv Q

Qrv

C

Fig. 110. Flow-pressure characteristics of pumping unit (pump - relief valve)

For p > p0, relief valve �ow Qrv > 0 and the characteristic curve of the unit isfound using two points A(Q1; p0) and B(Qx; px). Point A(Q1; p0) de�nes the �owand pressure at which the unit starts working with the relief valve open, whereaspoint B(Qx; px) is obtained by subtracting relief �ow Qrv at pressure px from pump�ow Qp at the same pressure. Point C in which characteristic curve cuts y-axisdetermines pressure p2 at which full �ow of the pump is returned to the reservoirvia the relief valve, �g. 110.

Graphical analysis of hydraulic system 173

If the pump was not protected by the relief valve then system �ow would followpump characteristics, e.g. point D(Qp; px) shows that pump would deliver its full�ow (less leakage) at pressure px: One outcome of this would be overloading of theprime mover supplying power to the pump.

Problem 4.23 Graphical analysis of hydraulic system

A hydraulic circuit shown in �g. 111 is assembled from components having �ow-pressure characteristic curves also shown in �g. 111 (note that the purpose of thiscircuit is only to illustrate the method - it has no practical application!). Determineoperating characteristics of the system using the graphical method.

Qp

Qrv

np

p

Qd

p p p

Qp QdQrv

Fig. 111. A hydraulic system and characteristic curves of the system components

Answer: Subtracting relief �ow Qrv from pump �ow Qp (see Problem 4.22) weobtain the characteristic curve of the pumping unit, �g. 112, represented by lines(1,5). As the �ow from the pumping unit �ows through the restrictor valve, andthe input pressure to the restrictor is equal to the delivery pressure, we obtain pointB in which the characteristic curve of the restrictor intersects the pumping unit�scharacteristic (1, 5). This point de�nes system pressure p and �ow through therestrictor valve Qd.

Point A on pump characteristic curve corresponding to pressure p determines pump�ow Qp. The �ow through the relief valve is thus equal to:

Qrv = Qp ¡Qd


QpQd Q

p

p

p0

5

3 1

A(Qp,p)

Qrv

B(Qd,p)

Fig. 112. System characteristic curve (�rst approach)

This problem may be also solved using a di¤erent approach - by summing the �owsthrough relief valve Qrv and the �ow through restrictor valve Qd at the same pressurewe obtain characteristic (3,4)1 shown in 113.

Qp QQd

p

p

p0

3

4

A(Qp,p)

Qrv

B(Qd,p)

1

Fig. 113. System characteristic curve (second approach)

Point A - an intersection of (3,4) with characteristic (1) of the pump determinessystem pressure p and pump output �ow. Flow Qd through the restrictor valve isdetermined by point B which lies on restrictor characteristic curve (3) and has y-axis coordinate of p. We obtain relief valve �ow Qrv by subtracting �ow through therestrictor Qd from pump �ow Qp.

1 A characteristic curve for two elements in parallel - see problem 4.21

Principle of motor control - graphical method 175

Problem 4.24 Principle of motor control - graphical method

For a hydraulic system shown in �g. 114 demonstrate graphically the principle ofmotor control. The operation of the individual elements is de�ned by their static�ow-pressure characteristics. Ignore system losses and assume hydro-mechanicale¢ciency of motor ´hmm = 1.

Qp

Qrv

np

p

nm

Tm

p3

Qd

p4

Fig. 114. Meter-in control of hydraulic motor (Problem 4.24)

Answer: The �ow-pressure characteristics, lines (1,2) in �g. 115, for the pump-relief valve system are determined in the same way as for Problem 4.22. The hy-draulic pressure in the system, as we are ignoring pressure losses in the lines, is onlydependent on load and the pressure drop on the restrictor valve.

Pressure drop on the hydraulic motor is obtained from equation:

¢pm = p3 ¡ p4 =2¼Tmqm

and if the motor load is constant, then the actual pressure drop on the motor is alsoconstant and is equal to p3 (assuming no pressure loss in the return line of the motor)- this is illustrated by line (3) parallel to Q-axis. Pressure drop on the restrictorvalve, assuming a turbulent �ow, is de�ned by equation:

¢pd =K2

a2dQ2d

where:ad - restrictor �ow areaK - �ow coe¢cientQd - �ow through the restrictor valve

Thus pressure in the delivery line upstream of the restrictor valve is equal to thesum of pressure drop ¢pd on the restrictor valve and load pressure p3. Curves(40),(400) and (4000) represent �ow-pressure characteristics of the restrictor valve forvarious �ow areas ad. The intersection of curve (4) with characteristics (1,2) of the


p

p2

p1

Qtm'' Qtm''Qm''

p3

ad'''

ad'

ad''

4'

4''4'''

3

Qm'

Qm'''Qtm'''

5''' 5'

5''

Q

2

1

A

C

B

3

1α

Fig. 115. Static characteristics of the motor control system

pumping unit determines operating parameters of the system. Point A allows us to�nd �ow Q0m to the motor when the �ow area of the restrictor is a

0

d. In a similar waypoints B and C determine Q00m and Q

000

m for other settings of ad, where a000

d < a00

d < a0

d.Rotation speed nm of the motor is de�ned by equation:

nm =Qtm

qm(a)

and Qtm is de�ned by:

Qtm = Qm ¡Kvmp3 (b)

where Kvm is motor leakage coe¢cient.

This relation, for various displacements of the motor, is represented by line (5)(tan(®) = Kvm). Thus the theoretical �ow demand of the motor is de�ned by apoint of intersection of the Q-axis by line (5). We may select, at large load, asuitable opening ad of the restrictor valve and vary the theoretical �ow demand ofthe motor, i.e. its rotation speed.

The motor will have a maximum rotation speed when the restrictor valve is fullyopen, ad = admax. Whereas, when the restrictor valve is fully closed, ad = 0, themotor will be stationary and all �ow from the pump will be directed to the reservoirthrough the relief valve.

By-pass control of hydraulic motor - graphical method 177

Variations in settings of the restrictor valve have small e¤ect on the variation of motordemand when characteristic curve (4) of the restrictor cuts the pump characteristiccurve (line 1, pointA). The correct control range for this type of control correspondsto those settings of ad for which characteristic curve (4) cuts characteristic curve ofthe power unit (line 2) - i.e. for ad < a

00

d.

We should note that in actual systems the load on the motor will usually vary, thusthe pressure drop ¢pm will also vary and characteristic curve (4) will change itsposition relatively to the characteristic curve of the pumping unit (lines 1,2). Thusfor a �xed setting of the restrictor valve the point of intersection of characteristics,which determine the operating parameters of the drive, may also change.

Problem 4.25 By-pass control of hydraulic motor - graphical method

A hydraulic system which uses a by-pass �ow control is shown in �g. 116. Staticcharacteristics of the system elements are known. Using these characteristics demon-strate graphically principles of by-pass control of motor speed and determine thesystem operating parameters. Ignore pressure loses in delivery line.

Answer: Flow from the pumping unit (pump-relief valve) is divided into a �ow

Qrv

p

Qd

np

nm

QmQp p3

p4

Tm

Fig. 116. By-pass control of hydraulic motor (Problem 4.25)

through the by-pass restrictor valve Qd to the reservoir and a �ow Qm supplying themotor. In contrast to series �ow control, the relief valve acts here as a safety valveand is closed during normal operation of the system. By changing Qd = F (ad), atconstant Qp, we vary �ow Qm to the motor and therefore rotation speed nm:

Qm = Qp ¡Qd (a)

Flow Qd through the restrictor valve is de�ned by equation:

Qd = adKpp3 ¡ p4 (b)

where: ad �ow area of the restrictor and K is �ow coe¢cient.


p

p2

p1

p3

Qtm'Qtm'' Qm'' Qm'

Q

5'5''

6'6''

ad'

ad''

3

1

4

2

A

A'

A''

Fig. 117. Static characteristic curve of by-pass control of hydraulic motor

The �ow Qd for a given setting of the valve (area ad) thus depends on pressuredi¤erence ¢p34, i.e. the load of the motor. By varying the opening of the valve wecan change the speed of the motor, at the same time load variation at the set valueof ad will also a¤ect the motor speed.

The �ow-pressure characteristics (lines 1,2) de�ne operating limits of the pumpingunit (pump-relief valve) and line 3 represents the load characteristics of the motorpm = p3, �g. 117. Curve 4 represents the actual characteristics of the motor whenhydro-mechanical e¢ciency is taken into consideration.

By combining the characteristics of the pumping unit and the restrictor valve (bysubtracting �ow rates Qd from Qp at various values of system pressure) we obtaincurves 5� and 5�, which represent characteristics of the pumping unit combined withthe restrictor valve, corresponding to di¤erent �ow areas ad of the restrictor valve.

Point A which is the intersection of curve 5�with the load curve of the motor de�nespressure in delivery line and �ow rate Qm supplied to the motor at valve openinga0d. Similarly, point A� de�nes the motor operational parameters for a

00

d the areaa0d < a00d. Lines 6� and 6� determine the theoretical �ow demands Q

0

tm and Q00tm ofthe motor (see Problem 3.22).

We can see that by changing the opening of the restrictor valve we change the theo-retical �ow demand of the motor and thus its rotation speed according to equation:

nm =Qtm

qm

The motor will reach its maximum speed when the restrictor is fully closed ad = 0(point A is then placed on the characteristic curve of the pumping unit), whereasby increasing the valve opening we may reduce rotation speed to zero.

Analysis of hydraulic winch drive - graphical method 179

Problem 4.26 Analysis of hydraulic winch drive - graphical method

A hydraulic winch consists of an electrically driven pump, a relief valve, two restrictorvalves and a hydraulic motor driving a drum, �g. 118. The static �ow-pressurecharacteristic curves for hydraulic elements (a), (b), (c), (d) and (e) are shown in�g. 119 for the chosen load and selected settings of the restrictor valves. Determinegraphically the actual �ow demand of the motor and its rotation speed if motor unitdisplacement VÁm = 0:2 Lrad

¡1and volumetric e¢ciency ´vm = 1.

Qrv

pQd1

Qd2Qp

np nm

Qm

Mg

Qp

e

pp

d

a

b c

Fig. 118. Hydraulic winch

Answer: Graph I shows the pressure-�ow characteristic of the pump (line a)with superimposed characteristics of the relief valve (dotted line); these two linesde�ne the operating range of the pumping unit. Graph II shows the relief valvecharacteristics. Graph III represents the �ow-pressure characteristics of restrictorvalve (c), whereas graphVI shows the �ow-pressure characteristics of restrictor valve(d). Graph IV shows motor characteristic. Flow in the delivery line of the pumpingunit is equal to �ow through valve (c). Thus, pressure downstream of restrictor valve(c) is obtained by subtracting, for various values of delivery �ow, pressure drop onvalve (c) from the pump delivery pressure. Plot of the pressure downstream of therestrictor as a function of �ow Qd1 through restrictor (c) is shown in graph V -curve f. This pressure is also equal to the pressure drop on by-pass restrictor valve(d). Subtracting, at the same pressure, from �ow Qd1 �ow Qd2 through valve (d) weobtain the characteristic curve of the system - curve (g). The point of intersectionof curve (g) with characteristic curve of motor (e) de�nes the operating parametersof the motor, i.e. pressure pm = 4:3 MPa and actual �ow demand:

Qm = 0:2 Ls¡1 (answer!)

The rotation speed of the motor is de�ned by equation:

nm =Qtm

qmnm=Qm´vm2¼VÁm

and substituting the data:


nm =0:2£ 0:922£ ¼ £ 0:2

£ = 0:14 revs¡1 = 8:6 rpm (answer!)

1.00.0

10.0

0.0

Qrv

a

Qp [ L/s]

pp [MPa]

Pump

pm=4.3

1.00.0

10.0

0.0

Qm=0.2

Qd2

g

f

e

a

b

Q [ L/s]

p [MPa]

∆pd1

1.00.0

10.0

0.0

c

∆pd1

Qd1 [ L/s]

∆pd1 [ MPa]

Valve 1

1.00.0

10.0

0.0

Qrv b

Qrv [L/s]

pp [MPa]

Relief valve

1.00.0

10.0

0.0

dQd2

Qd1 [L/s]

∆pd2 [MPa]

Valve 2

1.00.0

10.0

0.0

e

Qm [ L/s]

pm [MPa]

Motor

I

VIV

IVIII

II

Fig. 119. Hydraulic winch characteristics (Problem 4.26)

Problem 4.27 Speed of hydraulic motor - analytical and graphicalapproach

Calculate rotation speed nm of the motor and overall e¢ciency ´sys of the systemshown in �g. 120. System parameters are as follows:

Speed of hydraulic motor - analytical and graphical approach 181

² pump rotational speed np = 40 revs¡1

² pump displacement qp = 40 cm3rev¡1


² pump volumetric e¢ciency ´mp = 0:94

² motor displacement qm = 80 cm3rev¡1

² pump volumetric e¢ciency ´vm = 0:97

² pump volumetric e¢ciency ´hmm = 0:96

Load on hydraulic motor is de�ned by equation:

Tload = T0 +Knm

where:T0 - constant torque, T0 = 20 Nmnm - motor speedK - constant, K = 6:0 Nmrev¡1

p0

p

QpQrv

np Tload

nm

Fig. 120. Drive of a hydraulic motor (Problem 4.27)

The relief valve characteristics are:

Qrv =

½0 for p � p0

Krv(p¡ p0) for p > p0

where:Qrv - relief valve �owp - system pressurep0 - relief valve set pressure, p0 = 10 MPaKrv - constant, Krv = 400 cm

3MPa¡1s¡1

Ignore leakages and pressure losses in the lines. Solve the problem using the analyt-ical and graphical approach.

Answer (analytical solution): Pump delivery �ow:

Qp = npqp´vp = 48£ 40£ 0:95 = 1824 cm3s¡1

When the relief valve is closed then Qp = Qm, and the motor speed is:


nm =Qp

qm´vm =

1824

80£ 0:97 = 22:12 revs¡1

and as the load torque is de�ned the by equation:

Tload = T0 +Knm = 20 + 6£ 22:116 = 152:7 Nm

then pressure drop across the motor is equal to:

¢pm =Tload2¼

qm´hmm=

152:7£ 2£ ¼80£ 10¡6 £ 0:96

= 12:49 MPa

As ¢pm > p0, the relief must open and Qp = Qrv+Qm, where Qm is motor demand�ow rate de�ned by equation::

Qm =nmqm

´vm

thus, for p > 10 MPa:

npqp´vp = Krv

µ2¼(T0 +Knm)

qm´hmm¡p0

¶+nmqm

´vm

and �nally rearranging the last equation and substituting the data we obtain:

nm = 18:55 revs¡1 answer!

We again use the equation for load torque to calculate the pressure drop on themotor at rotation speed nm = 18:55 revs

¡1:

¢pm = p3 =(20 + 6£ 18:55)£ 2¼80£ 10¡6£0:96

= 10:74 MPa

The output power of the motor is:

Pm = Tloadnm2¼ = (20 + 6£ 18:55)£ 18:55£ 2¼ = 15:3 kW

and the pump input power is equal to:

Pp =qpp3np

´hmm=40£ 10¡6£10:74£ 106£48

0:94= 21:93 kW

thus, overall e¢ciency ´sys of the system is:

´sys =Pm

Pp=15:3

21:93= 0:697 answer!

Speed of hydraulic motor - analytical and graphical approach 183

3

1

2

4

A

B

C

E

p [MPa]

Q [L/s]

15.0

2.001.821.541.00

1.63

5.0

10.0

10.8

12.5

0

D

Fig. 121. Graphical solution of Problem 4.27

Answer (graphical solution): Pump delivery �ow rate Qp, at a constant volu-metric e¢ciency, is represented by line (1), �g. 121. The �ow-pressure characteristicof the relief valve is a straight line de�ned by equation:

Qrv = Krv(p¡ p0)

Line (2) is plotted using coordinates p = 10 MPa, Qrv = 0, and p = 12:5MPa,Qrv = 1:0 Ls

¡1 (point C).

The pumping unit (pump-relief valve) characteristics are represented by half-lines(1) and (3) according to equation Q = Qp¡Qrv, intersection pointA has coordinatesp = 10 MPa and Q = 1:824 Ls¡1:

Motor characteristic curve (line 4) is de�ned by equation:

Tload = T0 +Knm

and because:

¢pm =Tload2¼

qm´hmm


thus, for nm = 0 , the motor �ow is Qm = 0 and the motor characteristic line passesthrough point D which has the following p-axis coordinate:

¢pm =2¼T0qm´hmm

= 1:63 MPa

The second point on the motor characteristic line (point E) may be de�ned usinganalytical solution, where for Qm = 1:824 Ls

¡1 we found that the pressure drop onthe motor was pm = 12:49 MPa. Intersection point B of line (4) with the character-istic of the pumping unit (pump-relief valve) de�nes system operating parameters.Coordinates of point B are:

Qm = 1:540 Ls¡1 = 1540 cm3s¡1

pm = 10 MPa

thus rotation speed of the motor is, as in the analytical solution, equal to:

nm =Qm´vmqm

=1540£ 0:97

80= 18:6 revs¡1 answer!

Problem 4.28 Pump - accumulator system

Hydraulic cylinder shown in �g. 122 has stroke volume Vc = 0:6 L. The duration ofa working stroke is t1 = 2 s, intervals between subsequent working strokes vary butare not smaller than t2 = 0:5 s. Calculate the size of the pump and the accumulator,to assure maximum utilisation of hydraulic energy. The minimum cylinder pressureduring working stroke is p3 = 9 MPa and the relief valve is set at 12 MPa.

Qp p0

p3

FQ [L/s]

t1 t2

Q

Qp

t [s]543210

0.1

0.2

0.3

Fig. 122. Pump-accumulator system (Problem 4.28)

Pump - accumulator system 185

Answer: Required pump delivery �ow, for a system without an accumulator, is:

Qp =Vc

t1=0:6

2= 0:3 Ls¡1

however when we include an accumulator in the system, which will be charged duringtime t2 between subsequent cycles, then required pump �ow Qp can be calculatedfrom equation:

(Q¡Qp)t1 = Qpt2

thus:

Qp = Qt1

t1 + t2= 0:3£

2

2+0:5= 0:24 Ls¡1 answer!

We add approx. 5% for volumetric losses in the pump and valves and select froma catalogue a pump which delivers Qp = 0:25 Ls¡1. The selection of accumulatorsize starts with calculation of �uid volume Va which will �ow into the accumulatorbetween two consecutive working cycles. Thus:

Va = V3 ¡ V2 = Qpt2 = 0:25£ 0:5 = 0:125 L

where:V3 - volume of �uid in the accumulator at minimum pressure p3V2 - volume of �uid in the accumulator at maximum pressure p2

We assume that the accumulator charging and discharging is a polytropic process(compression and expansion of the gas), so we can write:

p1VÂ1 = p2V

Â2 = p3V

Â3 = const:

where:Â - polytropic exponent, Â = 1:5V1 - gas volume in the accumulator at pressure p1

Volume V1 is equal to the capacity of the accumulator, as at pressure p1 (initialcharge pressure) the gas occupies the whole volume of the accumulator. In practice,we accept that p1 = 0:9p3 and the isothermal process (Â = 1) during transition fromp1V1 to p2V2 (i.e. during start-up) thus, the volume of the accumulator is [2]:

V1 =

Vap3

p1

1¡µp3

p2

¶ 1Â

(a)

and if we assume that:


p1 = 0:9p3 = 0:9£ 9 = 8:1 MPa

then the required accumulator volume is:

V1 = 0:125

9

81

1¡µ9

12

¶1Â

= 0:8 L answer!

Problem 4.29 Pump unloading system

A hydraulic system shown in �g. 123 is designed to unload the �xed displace-ment pump when the closed-centre directional control valve is in a neutral position.Calculate the volume V1 of an accumulator, which works with a pump deliveringQp = 0:3 Ls

¡1, necessary to drive a double acting, double rod, hydraulic cylinder.

The cylinder has piston area A = 200 cm2 and load F = 0:3 kN. Assume thatthe cylinder power stroke h = 250mm will be completed in time t = 3 s, and thefull working cycle in time tc = 60 s. The maximum pressure in the system (set bythe relief valve) is p2 = 20 MPa. Plot pressures and �ows against time, and showthat, by incorporating an accumulator into the system, a substantial reduction ofthe pump input power can be achieved.

Answer: Cylinder �ow demand Q during the power stroke is:

Q =Ah

t=200£ 25

3= 1:66 Ls¡1

and cylinder pressure p is:

p =F

A=3£ 105

2£ 10¡2= 15 MPa

thus the input power to the system without an accumulator is equal to:

P1 = Qp = 1:66£ 15£ 106 = 24:9 kW

In the system shown in �g. 123 pump delivery �ow Qp is supplemented by accu-mulator �ow Qa, which is the average �ow from the accumulator during a powerstroke:

Qa = Q¡Qp = 1:66¡ 0:3 = 1:36 Ls¡1

From the above we calculate the accumulator volume to be:

Va = V3 ¡ V2 = Qat = 4:08 L

Pump unloading system 187

Qp

A

h

10 20 30 40 50

60

t [s]

v

vavg

vavg

0

Fig. 123. Pump unloading system an accumulator. Lower diagram shows a cylindercycle (Problem 4.29)

We use the equation for polytropic expansion (Â = 1:5), and for given values ofp2 = 20 MPa, p3 = p = 15 MPa, and also assuming p1 = 0:9p3 = 13:5 MPa (aminimum working pressure), the accumulator minimum volume is calculated usingeq. (a) Problem 4.28:

V1 =

Vap3

p1

1¡µp3

p2

¶1=Â =4:08£

15

13:5

1¡µ15

20

¶2=3 = 26 L answer!

Thus as we require an accumulator which has a minimum volume of 26 L. Therequired input power to the system is then:

P2 = Qpp2 = 0:3£ 20£ 106 = 6kW

and this power must be supplied to the system during charging time, which is:

t1 =Va

Qp=4:08

0:3= 13:6 s


The system will be requiring only minimum power during remaining time tp as thepump �ow is directed to the reservoir at the reservoir pressure:

tp =tc

2¡ t¡ t1 = 30:0¡ 3:0¡ 13:6 = 13:4 s

The variations of �ows and pressures during a cycle are shown in �g. 124. Theabove calculations show that by including an accumulator in the system the size ofthe power unit may be reduced and we also obtain considerable energy savings whichis equal to:

¢P = P2 ¡ P1 = 24:9¡ 6:0 = 18:9 kW : answer!

Q [L/s]

0

5

15

20

10

0.30

1.66

1.00

p [MPa]

t [s]

t [s]

Q

Qp

Qa

pa

pp

t

tc/2

t1

Fig. 124. Time plots of system�s �ows and pressures

Problem 4.30 Load sensing system

A hydraulic system with a variable displacement pump with a load sensing controlsystem, which compensates for the e¤ects of varying load, is shown in �g. 125.Discuss the principle of operation of such a system and show that the inherent lossesdue to the design concept of this circuit are much lower than for systems using other

Load sensing system 189

types of control, e.g. pressure control p = const: Develop design of an analogoussystem which uses a �xed displacement pump and discuss its advantage.

12

3

68

7

4 9

5

∆ps

Fig. 125. Load sensing system - variable displacement pump (Problem 4.30)

Answer: The basic element of the �load sensing� control is compensating valve(6), which takes the pilot pressure from pump delivery line (8) and cylinder pres-sure line (7). Compensating valve (6) provides a control input to pump actuator(2), which adjusts the displacement of pump (1) and maintains constant pressuredi¤erence ¢ps, usually about 1:5 MPa, between the delivery and cylinder ports ofdirectional proportional control valve (4). When the cylinder is stationary, valves(4) and (6) are in positions as shown, and pump delivery pressure p (it has value¢ps) acting on pump actuator (2) decreases the pump displacement close to zero(the pump maintains some �ow to compensate for internal volumetric losses in thesystem).

When proportional valve (4) is actuated, pressure in pilot line (7) causes shiftingof valve (6) to the right which in turn connects the pump actuator to the reservoirallowing the actuator�s spring to increase pump displacement until the pressure dif-ference on directional proportional valve (5) reaches ¢ps. If, due to the action ofvalve (4), pump �ow increases above the value of �ow which is necessary to maintain¢ps then the spool of valve (6) will move to the left and will decrease pump dis-placement. Thus �ow capacity of the pump is automatically controlled in responseto the position of the proportional valve (4).

Thus in this system, as the pressure di¤erence across the proportional valve is main-tained constant, the �ow through this valve and therefore the speed of the cylinder isproportional to the position of control spool of the valve (4). Comparison with othertypes of speed control, either resistive (valve control) or volumetric (pump control)


show that the inherent losses in this system are smallest when the power balanceover the full range of load is considered. Inherent power loss Ploss in this system is:

Ploss = P1 ¡ P2

where:P1 - pump output hydraulic powerP2 - cylinder input hydraulic power

Power P1 is de�ned by equation:

P1 = p1Q1

where:p1 - pump pressureQ1 - pump �ow rate

Power P2 is de�ned by:

P2 = p2Q2

where:p2 - cylinder load pressureQ2 - �ow through the proportional directional control valve (4)

and we note that:

Q1 = Q2 and p1 = p2 +¢ps

thus inherent power loss is:

Ploss = (p2 +¢ps)Q2 ¡ p2Q2 = ¢psQ2

Due to a constant value of ¢ps an inherent power loss Ploss depends only on �ow Q2which is also demand �ow of the cylinder, i.e. on the required speed of the cylinder.For comparison, inherent power loss in a system using p = const: control, is:

Ploss = p0Q2 ¡ p2Q2 = (p0 ¡ p2)Q2

where p0 is maximum system pressure set by the relief valve. Thus in this casethe inherent power loss depends on �ow Q2 and load (i.e. p0 ¡ p2). A graphicalinterpretation of inherent losses in a load sensing and p = const: control systems isshown in �g. 126.

The above analysis shows that power losses in a load sensing system are much lowerthan in a conventional system and only when load sensing system operates at loadpressure p2 which is close to pressure p0 (maximum pressure set on relief valve) arethe power losses similar. A �load sensing� system which uses a �xed displacementpump is shown in �g. 127.

Load sensing system 191

Q

Q1maxQ1=Q2

P1

Ploss

Q

Q1maxQ1=Q2

p0

p2

p

P2

Ploss

p0

p2

p

p1

Fig. 126. Graphical interpretation of inherent power loss in a load sensing systemand constant pressure system.

13

68

7

4 9

5

∆ps

Fig. 127. Load sensing system - �xed displacement pump

The basic hydraulic element is again, as in previous system, compensator valve 6which takes the pilot pressure from pump delivery line 8 and cylinder pressure line(7). When proportional valve 4 is actuated, the pressure in pilot line 7 shifts thevalve 6 to the right. This in turn allows some of the pump �ow to be discharged tothe reservoir until the pressure di¤erence on the proportional valve reaches ¢ps. Thecompensator valve controls the pump delivery pressure by allowing the return of theexcess �uid to the reservoir and thus maintains constant pressure di¤erence ¢psonproportional valve 4 which is independent of the cylinder load (we are ignoring smalllosses in lines 8 and 9). From the point of view of power losses this system does not


o¤er full bene�ts of load sensing over the whole control range as the excess �uid isreturned to the reservoir at pressure dictated by the load.

A graphical interpretation of the losses in this system is shown in �g. 128. Inputpressure to the cylinder is p2, the cylinder demand �ow is Q2, and the pump deliv-ery �ow Q1 = const. The shaded area represents inherent power losses (losses inhydraulic lines, the pump and the cylinder are ignored). A major characteristic ofthis system is very accurate control of cylinder speed, independent of load variations,due to a constant pressure drop across the directional proportional valve (as is wellknown in systems without load compensation the �ow rate through the directionalcontrol valve will vary with load).

Q

Q1=constQ2

p2

p

p1

∆ps

Fig. 128. Interpretation of losses in load sensing system

In practice other types of load sensing systems are also used. Load sensing systemscan also be used in systems employing hydraulic motors as output devices.

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