Lecture 25

HYDRAULIC CIRCUIT DESIGN AND ANALYSIS [CONTINUED]

1.14 Circuit for Fast Approach and Slow Die Closing

A machine intended for high volume production has a high piston velocity. If not controlled, the high-

speed platen approaching the job instead of making a smooth contact will bang on the job. This is not

desirable. In all such cases,“rapid traverse and feed circuits” are employed.

Figure 1.15 Rapid traverse and feed circuit.

In the circuit shown in Fig. 1.15, pump delivery normally passes through FCV(3). During fast

approach,the solenoid-operated DCV (4) is energized. This diverts pump delivery to the cap end of the

cylinder through valve (4). Full flow is thus available for the actuator to advance at the rated speed. A few

millimeters before the platen makes contact with the die, solenoid valve (4) is de-energized forcing the

pump delivery to pass through FCV (3). The platen now approaches the die at a controlled speed because

the flow to cylinder (6) is now regulated. Directional valves (4) and (5), however, must be energized

simultaneously for the approach phase to begin.

Valves (4) to (5) are solenoid-controlled pilot-operated valves intended for handling large flows with

minimum pressure drop. While valve (5) requires a 4.5 bar check valve (6) in the return line to develop

the pilot pressure required to move the main spool, no such facility is required in the case of valve(4)

because the back pressure generated by valve (3) would serve as the pilot pressure for this valve.

1.15Rapid Traverse and Feed, Alternate Circuit

(4)

(6)

(5)

(3)

2

In this circuit (Fig. 1.16), full flow from the pump is allowed to the cap end through the directional valve

(3)for fast approach and the rod end oil freely passes through the normally open deceleration valve back

to the tank. Near about the end of the stroke, a cam depresses a roller attached to the deceleration valve

spool and therefore the valve shifts blocking the flow from the rod end. The flow now has only one

pathway back to the tank and that is through FCV (4). The approach speed is now governed by the setting

of this valve. During piston retraction stroke, full flow is allowed to the rod end through check valve (6).

Figure 1.16 Rapid traverse and feed circuit – alternate circuit.

Example 1.1

A double-acting cylinder is hooked up in a regenerative circuit. The relief-valve setting is 105 bar. The

piston area is 130 cm2 and the rod area is 65 cm

2. If the pump flow is 0.0016 m

3/s, find the cylinder speed

and load-carrying capacity for the

(a) Extending stroke.

(b) Retracting stroke.

Solution:

(a) We have

p

xt

r

4e

0.0016

65 100.246 m/s

Qv

A

load-extension r

5 4105 10    65 1  0  

68250 N

F pA

(b) We have

p

ret

p r

4

0.00160.246 m/s

(13 6 )0 5 10

Qv

A A

(6)

(5)

(4)

(3)

(2)

(1)

3

load-exten

5 4

sion p r )

105 10 (130 65) 10

68250 N

(F p A A

Example 1.2

What is wrong with the circuit diagram given in Fig. 1.17?

Figure 1.17

Solution: A check valve is needed in the hydraulic line just upstream from where the pilot line to the

unloading valve is connected to the hydraulic line. Otherwise the unloading valve would behave like a

pressure-relief valve and thus, valuable energy would be wasted.

Example 1.3

What unique feature does the circuit of Figure 1.18 provide in the operation of the hydraulic cylinder?

Unloading Valve

4

Figure 1.18

Solution:

1. It provides mid-stroke stop and hold of the hydraulic cylinder (during both the extension and

retraction strokes) by deactivation of the four-way, three-position DCV.

2. It provides two speeds of the hydraulic cylinder during the extension stroke:

When the three-way, two-position DCV is unactuated in spring offset mode, extension speed is

normal.

When this DCV is actuated, extension speed increases by the regenerative capability of the circuit.

Example 1.4

For the circuit of Fig. 1.19, give the sequence of operation of cylinders 1 and 2 when the pump is turned

ON. Assume that both cylinders are initially fully retracted.

Figure 1.19

5

Solution: Cylinder 1 extends, cylinder 2 extends. Cylinder 1 retracts, cylinder 2 retracts.

Example 1.5

What safety features does Fig. 1.20possess in addition to a pressure-relief valve. If the load on cylinder 1

is greater than the load on cylinder 2, how will the cylinder move when DCV is shifted into the extending

or retracting mode? Explain your answer.

Figure 1.20

Solution: Both solenoid-actuated DCVs must be actuated in order to extend or retract the hydraulic

cylinder.

Cylinder 2 will extend through its complete stroke receiving full pump flow while cylinder 1 will not

move. The moment cylinder 2 will extend through its complete stroke, cylinder 1 will receive full pump

flow and extend through its complete stroke. This is because the system pressure builds up until load

resistance is overcome to move cylinder 2 with the smaller load. Then the pressure continues to increase

until the load on cylinder 1 is overcome. This then causes cylinder 1 to extend. In retraction mode, the

cylinders move in the same sequence.

Example 1.6

Assuming that the two double-rod cylinders of Fig. 1.21 are identical, what unique feature does the circuit

in Fig. 1.21 possess.

6

Figure 1.21

Solution: Both cylinder strokes would be synchronized.

Example 1.7

For the hydraulic system is shown in Fig. 1.22

(a) What is the pump pressure for forward stroke if the cylinder loads are 22000 N each and cylinder 1

has the piston area of 65 cm2 and zero back pressure?

(b) What is pump pressure for retraction stroke (loads pull to right), if the piston and rod areas of cylinder

2 equal to 50 cm2 and 15 cm

2, respectively, and zero back pressure?

(c) Solve using a back pressure p3 of 300 kPa instead of zero, the piston area and rod area of cylinder 2

equal 50 and 15 cm2, respectively.

Solution:

(a)Pressure acting during forward stroke is

1 21 4

p1

22000 220006.77 MPa

65 10

F Fp

A

(b)For cylinder 2 we can write

3 p2 r2 2 p2 2( )p A A p A F

For cylinder 1, force balance gives

2 p1 r1 1( )p A A F

But p2 p1 r1A A A . So we can write

2 p2 1 p A F

and rod side pressure of second cylinder is given by

7

1 23 4

p2 r2

22000 2200012570000 Pa 12.57 MPa

35 10

F Fp

A A

Figure 1.22

(c)For cylinder 1, we have

1 p1 2 p1 r1 1( )p A p A A F

Similarly for cylinder 2, we have

2 p2 3 p2 r2 2( )p A p A A F

Adding both equations and noting that p2 p1 r1A A A yield

1 p1 3 p2 r2 1 2( )p A p A A F F

1 2 3 p2 r2

1

p1

( )F F p A Ap

A

2 2 4 2

1 2 4 2

22000 N 22000 N 300000 N / m (50 15) cm 1  0  m

65 cm 1  0  mp

1 6.93 MPap

Ap1

F1 F2

p1 p2 Ap2

Ar1 Ar2

p2 p3

8

Example 1.8

For the double-pump system in Fig. 1.23, what should be pressure setting of the unloading valve and

pressure-relief valve under the following conditions:

(a) Sheet metal punching operation requires a force of 8000 N.

(b) A hydraulic cylinder has a 3.75 cm diameter piston and a 1.25 cm diameter rod.

(c) During the rapid extension of the cylinder, a frictional pressure loss of 675 kPa occurs in the line

from the high-flow pump to the blank end of the cylinder. During the same time, a 350 kPa

pressure loss occurs in the return line from the rod end of the cylinder to the oil tank. Frictional

pressure losses in these lines are negligibly small during the punching operation.

(d) Assume that the unloading valve and relief-valve pressure setting (for their full pump flow

requirements) should be 50% higher than the pressure required to overcome frictional pressure

losses and the cylinder punching load, respectively.

Figure 1.23

Solution:

4/3 DCV (solenoid operated)

CV1

Pressure

unloading

valve

Pressure relief

valve

High-

pressure

low-flow

pump Low-pressure

high-flow pump

High-

pressure

line

Low-pressure line

Electric motor

9

Unloading valve: Back pressure force on the cylinder equals pressure loss in the return line times the

effective area of the cylinder p r( )A A :

2 2 2

back pressure 2

N 350000  (0.0375 0.0125 ) m 344 N

m 4

πF

Pressure at the blank end of the cylinder required to overcome back pressure force equals the back

pressure force divided by the area of the cylinder piston:

cyl blank end2 2

344 N   311 kPa π

(0.0375 )m4

p

Thus,the pressure setting of unloading valve equals

1.50 675 311 kPa 1480 kPa

Pressure relief valve: Pressure required to overcome the punching operation equals the punching load

divided by the area of the cylinder piston:

punching 2 2

8000 N   7240 kPa π

(0.0375 )m4

p

Thus, the pressure setting of pressure-relief valve equals

1.50 ×7240 kPa = 10860 kPa

Example1.9

Design a suitable hydraulic circuit to raise and lower a load of magnitude 10000 kgf at a speed of 100

mm/s. The speed must be equal both during raising and lowering of the load. The load is essentially

overrunning. The load must be lowered gradually onto the platform. Calculate the flow through the

control valves and indicate the pressure gauge readings both at the cap end and at the rod end during

raising and lowering. Explain your reasons for your choice of the hydraulic components. Neglect

mechanical and hydraulic losses. Assume 100 mm bore for the cylinder and a rod diameter = 45 mm.

Solution: In any double-acting cylinder the rod end area is smaller than the cap end area to the extent of

the piston rod cross-sectional area and so the pressure required to raise/lower the load is derived from the

rod end area. Accordingly, the pressure required to raise the load is

2

2 2

10000160 kgf/cm 160 bar

(10 4.5 )

4

The relief-valve setting pressure is 175 bar.

The cap end area =.7584(10)2 = 75.84 cm

2.

If the cylinder has to extend and retract at the rate of 100 mm/s,the flow required at the cap end of the

cylinder is

(75.84 ×10) = 758.4 cm3/s or 47 LPM

Flow required at the rod end would be (62.6 ×10) = 626 cm3/s or 37.5 LPM.

Referring to the circuit in Fig. 1.24, a constant delivery pump with a pressure rating of 175 bar capable of

delivering 50 LPM has been chosen. A variable delivery pump would not help because both velocity and

pressure are constant throughout the cycle. It is required that the piston must travel both during extension

10

and retraction at the same speed. Thus, flow control valve (1) is used on the cap end of the cylinder

because pump supply is constant but cap end and rod end areas differ. Since it is an overrunning load, a

flow control valve became necessary (2) on the rod end (meter-out flow control).

Because it is required that the load must be positioned on the platform gradually, flow control (3) and

solenoid-operated DCV (4) become necessary. Toward the end of the stroke, the load makes contact with

a limit switch. This energizes valve (4) to divert rod end flow through the flow control valve (3) so that

the load is decelerated from 100 mm/s to 30 mm/s. In order to ensure accurate speed control pressure- and

temperature-compensated flow, flow control valves were chosen. Flow through the flow control valve (3)

during deceleration would be = 188 cm3/s or 11 LPM.

While raising the load, the required flow to the rod end of the cylinder is 37.5 LPM. But the pump is

supplying 50 LPM. The excess flow must pass over the relief valve which is set at 175 bar. The relief-

valve setting pressure of 175 bar creates a retracting force on the rod end equaling

(175 × 62.6) = 10955 kgf

Of this, 10000 kgf is required just to balance the load.

Figure 1.24

The remaining 955 kgf acting in the retracting direction has to be balanced by the backpressure due to the

flow control valve (1). Consequently, the pressure gauge Pc at the cap end during retraction would read

955/62.6 = 15 bar. When the load is lowered at a speed of 100 mm/s, the cylinder extends at a velocity of

100 mm/s. The flow entering the cap end of the cylinder is 47 LPM, which is less than the pump delivery,

which is 50 LPM.

The pressure gauge Pc would read 175 bar because the extra flow must be dumped over the relief valve.

In this operating condition, the extension force (175 ×75.84) = 13744 kgf, together with the load force =

10000 kgf, tries to extend the cylinder.

10

00

0 k

gf

(3) (1)

(2)

PR

(4)

pC

(2)

(4)

(3)

(1)

Sensor

11

According to Newton’s first law of motion the net force must be equal to zero for an object moving at a

constant velocity, neglecting friction. So the balancing force at the rod end should be 23744 kgf.

Therefore, the pressure gauge Pr at the rod would read 23774 / 62.6 = 379 bar. At the end of high-speed

extension solenoid valve (2) is energized to decelerate the load from 100 mm/s to 30 mm/s. The time

duration around which this occurs depends on the valve response time that can be assumed as 20 ms or

0.020 s.

The deceleration

f iV Va

t

where final velocity f 30 mm/sV , initial velocity i 100 mm/sV , δtis the time element = 0.020 s during

which time the change occurs. Substituting the relevant values, we obtain the value of a as 3500

mm/s2.Therefore, the force required to decelerate the cylinder is

F = ma

Now m= 10000/9.81 = 1019 kg . So

F = 1019 ×3.5 = 23566 kgf

Therefore

pR = (23744 + 3566)/62.6 = 436 bar

Example 1.10

For the fluid power system shown in Fig. 1.25,

(a) Determine the external loads F1 and F2 that each hydraulic cylinder can sustain while moving in an

extending direction. Take frictional pressure losses into account. The pump produces a pressure of

increase of 6.90 MPa from the inlet port to the discharge port and a flow rate of 0.00252 m3/s. The

following data are applicable:

Kinematic viscosity of oil 0.0000930 2m /s

Specific weight of oil 7840 3N/m

Cylinder piston diameter 0.203 m

Cylinder rod diameter 0.102 m

All elbows are 90 with k factor 0.75

Pipe lengths and diameters are given:

Pipe Number Length (m) Diameter Pipe number Length(m) Diameter

1 1.83 0.0508 6 3.05 0.0254

2 9.15 0.0317 7 3.05 0.0254

3 6.10 0.0317 8 12.2 0.0317

4 3.05 0.0254 9 12.2 0.0317

5 3.05 0.0254

(b)Determine the heat generation rate.

(c) Determine the extending and retracting speeds of cylinder.

12

Figure 1.25

Solution:

(a) Cylinders 1 and 2 are identical and are connected by identical lines. Therefore, they receive equal

flows and sustain equal loads (F1 = F2).

Velocity is calculated from discharge and area as

3

2

(m /s)

(m )

Qv

A

Head loss in the systems is given by

213

p

L

1 p

 

2

f L vH K

D g

Reynolds number is given by

PRV

Elbow

4

F1

1

5

8

7 6

3

1

2

9

F2

Elbow

Check valve

k=4

Tee k=1.8

Tee k=1.8

Cylinder 1 Cylinder 2

DCV, k=5

13

Re/

VD VD VD

Flow through path 4 (Fig. 1.26) is given by

3

4

0.002520.00126m /s

2Q

Flow through path 6 (Fig 1.26) is given by

2 2

3

6

(0.203 0.102 )0.00126 0.000945 m /s

0.203Q

Similarly for paths 8 and 9 we can write

3

8 9 2  0.000945 0.00189 m /sQ Q

Velocity calculation:

1   22

30.00252 ( )1.24 m/s

(0.0508

s

( )

m

)m

4

/v

2, 3 22

30.00252 ( )3.19  m/s

(0.0317)

s

)

m

(m4

/v

3

4 22

0.00126 ( )2.49 m/s

(0.0254)(m )

4

m /sv

3

6 22

0.000945 ( )1.86 m/s

(0.0254

s

( )

m

)m

4

/v

8, 9 22

30.00189 ( )2.39 m/s

(0.0317)

s

)

m

(m4

/v

Reynolds number calculation:

(1)

1.24 0.0508Re 677

0.000093

2, 3

3.19 0.0317Re 1087

0.000093

(4)

2.49 0.0254Re 680

0.000093

(6)

1.86 0.0254            Re 508

0.000093

(8, 9)

2.39 0.0317               Re 815

0.000093

All flows are laminar; hence we can calculate the losses in each branch. The general formula is

213

p

L

1 p

 

2

f L vH K

D g

where

14

64

Ref

Hence the losses are

2

(1)

64 1.83 1.247.5 0.33 m 7840 0.33 Pa 2560 Pa

677 0.0508 2 9.81LH

2

(2)

64 9.15 3.194 10.9 m 85500 Pa

1087 0.0317 2 9.81LH

2

(3)

64 12.2 3.196.8 15.3 m 120000 Pa

1087 0.0317 2 9.81LH

2

(4)

64 3.05 2.491.8 4.14 m 32500 Pa

680 0.0254 2 9.81LH

2

(6)

64 3.05 1.860 2.67 m 20900 Pa

508 0.0254 2 9.81LH

2

(8) (9)

64 6.1 12.2 2.395.75 12.8 m 100500 Pa

815 0.0317 2 9.81L LH H

Total force can now be calculated as

2

1 2

(0.203 )[(6900000) (2560 85500 120000 32500)]

4F F

–

2 2(0.203 0.102 )

(20900 100500)4

1 2 [216000] [2940]F F =213000 N

(b)We have

Heat generation rate (power loss in W) = Pressure × Discharge

={(2560 85500 120000) (0.00252) (2 20900 0.000945) (2 32500 0.00126) (100500 0.00189)}

= 524 + 39.5 + 81.9 + 190 = 835W = 0.835 kW

(c) Cylinder piston diameter = 0.203 m

Area of pistonp (  )A =

22(0.203 )

m4

Cylinder rod diameter = 0.102 m

Area of rod = 2

2(0.102 )m

4

Annulus area annulusA =

2 2(0.203 ) (0.102 )

4 4

Now

15

3

cyl

2

(m /s)

(m )

 

Qv

A

where each cylinder receives one half of pump flow because of the configuration of cylinder. Extension

velocity is given by

3

blank e

p

nd

ext 2

2

(m /s)

(m )

0.001260.0389 m/s

(0.203 )

4

Qv

A

Retracting velocity is given by

3

ret 2

annulus

2 2

(m )

(m )

0.00253

(0.203 ) (0.102 )

4 4

0.0521 m/s

/sQv

A

Example 1.11

Figure 1.26 shows a regenerative circuit in which an 18.65 kW electric motor drives a 90% efficient

pump. The pump discharge pressure is 6897 kPa. Take frictional pressure losses into account.

(a)Determine the external load F that the hydraulic cylinder can sustain in the regenerative mode (spring-

centered position of DCV).

(b) Determine the heat generation rate due to frictional pressures losses in the regenerative mode.

(c) Determine the cylinder speed for each position of the DCV.

The following data are applicable:

Kinematic viscosity of oil 0.0000930 2m /s

Specific weight of oil 7850 3N/m

Cylinder piston diameter 0.203 m

Cylinder rod diameter 0.102 m

All elbows are 90 with k factor 0.75

Pipe lengths and diameters are given

Pipe number Length (m) Diameter

1 0.61 0.0508

2 6.10 0.0445

3 9.15 0.0445 4 9.15 0.0445 5 6.10 0.0445

16

Figure 1.26

Solution: (a) Determination of external load, considering all losses: Let us first calculate the flow rate at different

branches as shown in Fig. 1.27. Before we calculate the losses, we calculate the pump power as

pumpPump power 0.90 18.65 16.79 kWη P

The flow rate is given by

3

pump

16.79 kW0.00243 m

6897 kPa/sQ

We can write the force balance

regen blank end P rod end annulusF p A p A

Now

3

pump 1 2 0.00243 m /sQ Q Q

From the derivation of regenerative circuits, we can write

p

3 pump

r

2

2

3

(0.203) / 40.00243

(0.102)

0.00972 m

/ 4

/s

AQ Q

A

Cylinder

Elb

ow

4

Elbow 3 2

1 5

Elbow

Strainer in the

tank K = 10

K = 5

17

p r

4 pump

r

2 2

2

3

[(0.203) (0.102) / 40.00243

(0.102) / 4

0.00729 m

]

/s

A AQ Q

A

Velocity calculation

3

1    22

0.00243 (m )1.20 m/s

(0.0508 )(m )

4

/sv

3

2   22

0.00243 (m )1.56 m/s

(0.0445 )(m )

4

/sv

3

3 22

0.00972 (m )6.24 m/s

(0.0445 )(

s

m )4

/v

3

4 22

0.00729 (m )4.69 m/s

(0.0445 )(

s

m )4

/v

Reynolds number calculation

(1)

1.20 0.0508Re 655

0.000093

(2)

1.56 0.0445Re 746

0.000093

(3)

6.24 0.0445Re 2990

0.000093

(4)

4.69 0.0445            Re 2240

0.000093

Assume that all flows are laminar; head losses can be calculated as follows:

2

(1)

64 6.10 1.2010 0.74 m 7850 0.33 Pa 5840 Pa

655 0.0508 2 9.81LH

2

(2)

64 6.10 1.565 2.08 m 16300 Pa

746 0.0445 2 9.81LH

2

(3)

64 9.15 6.240.75 10.2 m 80300 Pa

2990 0.0445 2 9.81LH

18

2

(4)

64 9.15 4.690.75 7.43 m 58400 Pa

2240 0.0445 2 9.81LH

The force is given by

2(0.203)

[(6897 kPa) (5.84 16.3 80.3)]4

F

–

2 2[(0.203) (0.102) ][6897kPa (5.84 16.3 58.4)]

4

Solving we get 220 168F =52 kN (b) Determination of heat generation rate

Power loss = ΔQ p

= Pipe 1 loss + pump loss+ pipe 2 loss + pipe 3 loss + pipe 4 loss

= 0.00243 × 5.84 + (18.7−16.8) + 0.00243 × 16.3 +0.00972 × 80.3 + 0.00729 × 58.4

= 0.014+1.9+0.04+0.78+0.43

Power loss = Heat generation rate = 3.16 kW

(c) Cylinder speed for each position of DCV

3

pump 1 2 0.00243 m /sQ Q Q

Upper position of DCV

3

pump

ext 22

p

 (m ) 0.002430.0751  m/s

(0.203 ) (m )

s

4

/Qv

A

Spring-centered position of DCV

3

pump

ext 22

rod

 (m /s) 0.002430.297 m/s

(0.102 ) (m )

4

Qv

A

Lower position of DCV

3

pump

ret 2 22

annulus

(m ) 0.002430.100 m/s

(0.203 ) (0.102 ) (m )

4

s

4

/Qv

A

19

Example 1.12

For the meter-in flow control valve system of Fig. 1.27, the following data are given:

Desired cylinder speed 0.254 m/s

Cylinder piston diameter 0.508 m

Cylinder load 13340 N

Specific gravity of oil 0.9

Pressure-relief valve setting 6895 kPa

Determine the required capacity coefficient of flow control valve.

Figure 1.27

Solution: We have

cyl piston

V

PRV load piston/ 

(

SG

)

V AC

p F A

(1.7)

V C are LPM / kpa .Therefore, we have the following units for the terms in the above equation:

cyl p PRV load piston) / (LPM,  kPa, ) Pressu( ke ar PQ v A p F A

13340 N

20

The flow rate is given by

cyl p

2

3

m 1 L 60 s0.254  0.00203m

s 0.001 m 1  min

30.9 LPM

Q v A

Now it is given that PRV 6895 kPap and

load

2 2

piston

13340 N 1 kPa6570 kPa

0.00203 m 1000 N/m

F

A

Substituting values in Eq. (1.7), we get

cyl piston

V

PRV load piston

 (

SG

30.9

6895 6570

0.9

30.9 LPM1.63 

19.0 kp

/

a

)

v AC

p F A