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 highspeed 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.
(5)
(6)
(3)
(4)
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
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).
(5)
(4)
(6)
(3)
(2)
(1)
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 cm2. If the pump flow is 0.0016 m3/s, find the cylinder speed
and load-carrying capacity for the
(a) Extending stroke.
(b) Retracting stroke.
Solution:
(a) We have
Qp
0.0016
vext 

 0.246 m/s
Ar 65  104
Fload-extension  pAr
 105  105  65 104
 68250 N
(b) We have
Qp
0.0016
vret 

 0.246 m/s
Ap  Ar (130  65)  104
2
Fload-extension  p( Ap  Ar )
 105  105  (130  65)  104
 68250 N
Example 1.2
What is wrong with the circuit diagram given in Fig. 1.17?
Unloading Valve
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?
3
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
4
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.
5
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 cm2, 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
p1 
F1  F2 22000  22000

 6.77 MPa
Ap1
65  104
(b)For cylinder 2 we can write
p3 ( Ap2  Ar2 )  p2 Ap2  F2
For cylinder 1, force balance gives
p2 ( Ap1  Ar1 )  F1
But Ap2  Ap1  Ar1 . So we can write
p2 Ap2  F1
and rod side pressure of second cylinder is given by
6
p3 
F1  F2
22000  22000

 12570000 Pa  12.57 MPa
Ap2  Ar 2
35  104
F2
F1
Ar2
Ar1
p2
p3
Ap2
p1
p2
Ap1
Figure 1.22
(c)For cylinder 1, we have
p1 Ap1  p2 ( Ap1  Ar1 )  F1
Similarly for cylinder 2, we have
p2 Ap2  p3 ( Ap2  Ar2 )  F2
Adding both equations and noting that Ap2  Ap1  Ar1 yield
p1 Ap1  p3 ( Ap2  Ar2 )  F1  F2

 F1  F2  p3 ( Ap2  Ar 2 ) 

 p1  

Ap1




 22000 N  22000 N  300000 N / m 2 (50  15) cm 2  104 m 2 
 p1  

65 cm2  104 m2


 p1  6.93 MPa
7
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.
4/3 DCV (solenoid operated)
Pressure relief
valve
CV1
Highpressure
line
Pressure
unloading
valve
Highpressure
low-flow
pump
Electric motor
Low-pressure
high-flow pump
Low-pressure line
Figure 1.23
Solution:
8
Unloading valve: Back pressure force on the cylinder equals pressure loss in the return line times the
effective area of the cylinder ( Ap  Ar ) :
Fback pressure  350000
N π
 (0.03752  0.01252 ) m2  344 N
m2 4
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:
344 N
pcyl blank end 
 311 kPa
π
(0.03752 )m2
4
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:
8000 N
ppunching 
 7240 kPa
π
(0.03752 )m2
4
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
10000
 160 kgf/cm2  160 bar
2
2
 (10  4.5 )
4
The relief-valve setting pressure is 175 bar.
The cap end area =.7584(10)2 = 75.84 cm2.
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
9
PR
pC
(4)
(4)
10000 kgf
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 reliefvalve 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.
(2)
(1)
(3) Sensor
(1)
(3)
(2)
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
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
V  Vi
a f
t
where final velocity Vf  30 mm/s , initial velocity Vi  100 mm/s , 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:
2
Kinematic viscosity of oil
0.0000930 m /s
3
Specific weight of oil
7840 N/m
Cylinder piston diameter
Cylinder rod diameter
All elbows are 90 with k factor
Pipe lengths and diameters are given:
Pipe Number
Length (m)
Diameter
1
1.83
0.0508
2
9.15
0.0317
3
6.10
0.0317
4
3.05
0.0254
5
3.05
0.0254
0.203 m
0.102 m
0.75
Pipe number
6
7
8
9
(b)Determine the heat generation rate.
(c) Determine the extending and retracting speeds of cylinder.
11
Length(m)
3.05
3.05
12.2
12.2
Diameter
0.0254
0.0254
0.0317
0.0317
Cylinder 1
Cylinder 2
F1
F2
Tee k=1.8
1
6
7
8
5
4
3
Tee k=1.8
DCV, k=5
9
Elbow
2
Check valve
k=4
PRV
1
Elbow
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
v
Q (m3 /s)
A (m2 )
Head loss in the systems is given by
13  f L
 v2
p
HL  
K

 2g
1  Dp

Reynolds number is given by
12
Re 
VD 


VD
VD

/ 
Flow through path 4 (Fig. 1.26) is given by
Q4 
0.00252
 0.00126 m3 /s
2
Flow through path 6 (Fig 1.26) is given by
Q6  0.00126 
(0.2032  0.1022 )
 0.000945 m3 /s
0.203
Similarly for paths 8 and 9 we can write
Q8  Q9  2  0.000945  0.00189 m3 /s
Velocity calculation:
0.00252 (m3 /s)
 1.24 m/s
 (0.0508)2
2
(m )
4
0.00252 (m3 /s)
v2, 3 
 3.19 m/s
 (0.0317)2 2
(m )
4
0.00126 (m3 /s)
v4 
 2.49 m/s
 (0.0254)2 2
(m )
4
0.000945 (m3 /s)
v6 
 1.86 m/s
 (0.0254)2 2
(m )
4
0.00189 (m3 /s)
v8, 9 
 2.39 m/s
 (0.0317)2 2
(m )
4
v1 
Reynolds number calculation:
1.24  0.0508
 677
0.000093
3.19  0.0317
Re 2, 3 
 1087
0.000093
2.49  0.0254
Re(4) 
 680
0.000093
1.86  0.0254
Re(6) 
 508
0.000093
2.39  0.0317
Re(8, 9) 
 815
0.000093
Re(1) 
All flows are laminar; hence we can calculate the losses in each branch. The general formula is
13  f L
 v2
p
HL  
K

 2g
1  Dp

where
13
f 
64
Re
Hence the losses are
2
 64 1.83
 1.24
H L (1)  
 7.5 
 0.33 m  7840  0.33 Pa  2560 Pa
 677 0.0508
 2  9.81
2
 64 9.15
 3.19
H L (2)  
 4
 10.9 m  85500 Pa
 1087 0.0317
 2  9.81
2
 64 12.2
 3.19
H L (3)  
 6.8 
 15.3 m  120000 Pa
 1087 0.0317
 2  9.81
2
 64 3.05
 2.49
H L (4)  
 1.8 
 4.14 m  32500 Pa
 680 0.0254
 2  9.81
H L (6)
2
 64 3.05
 1.86

 0
 2.67 m  20900 Pa
 508 0.0254
 2  9.81
2
 64 6.1  12.2
 2.39
H L (8)  H L (9)  
 5.75 
 12.8 m  100500 Pa
 815 0.0317
 2  9.81
Total force can now be calculated as
F1  F2  [(6900000)  (2560  85500  120000  32500)] 
 (0.2032 )
4
–

 (0.2032  0.1022 ) 
(20900  100500) 

4


F1  F2  [216000]  [2940] =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
 (0.2032 ) 2
Area of piston ( Ap ) =
m
4
Cylinder rod diameter = 0.102 m
 (0.1022 ) 2
Area of rod =
m
4
 (0.2032 )  (0.1022 )
Annulus area  Aannulus =

4
4
Now
14
v
Qcyl (m3 /s)
A (m 2 )
where each cylinder receives one half of pump flow because of the configuration of cylinder. Extension
velocity is given by
vext 

Qblank end (m3 /s)
Ap (m 2 )
0.00126
 (0.2032 )
 0.0389 m/s
4
Retracting velocity is given by
vret 

Q (m3 /s)
Aannulus (m 2 )
0.00253
 (0.2032 )  (0.1022 )

4
 0.0521 m/s
4
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 (springcentered 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
Specific weight of oil
Cylinder piston diameter
Cylinder rod diameter
All elbows are 90 with k factor
0.0000930 m2 /s
7850 N/m3
0.203 m
0.102 m
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
15
Cylinder
4
2
Elbow
Elbow
3
1
Elbow
5
K=5
Strainer in the
tank K = 10
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
Pump power  η  Ppump  0.90 18.65  16.79 kW
The flow rate is given by
Qpump 
16.79 kW
 0.00243 m3 /s
6897 kPa
We can write the force balance
Fregen  pblank end AP  prod end Aannulus
Now
Qpump  Q1  Q2  0.00243 m3 /s
From the derivation of regenerative circuits, we can write
Q3 

Ap
Ar
Qpump
 (0.203)2 / 4
 0.00243
 (0.102)2 / 4
 0.00972 m3 /s
16
Q4 

Ap  Ar
Ar
Qpump
 [(0.203) 2  (0.102) 2 ] / 4
 0.00243
 (0.102) 2 / 4
 0.00729 m3 /s
Velocity calculation
0.00243 (m3 /s)
 1.20 m/s
 (0.05082 ) 2
(m )
4
0.00243 (m3 /s)
v2 
 1.56 m/s
 (0.04452 ) 2
(m )
4
0.00972 (m3 /s)
v3 
 6.24 m/s
 (0.04452 ) 2
(m )
4
0.00729 (m3 /s)
v4 
 4.69 m/s
 (0.04452 ) 2
(m )
4
v1 
Reynolds number calculation
Re(1) 
1.20  0.0508
 655
0.000093
1.56  0.0445
 746
0.000093
6.24  0.0445
Re(3) 
 2990
0.000093
4.69  0.0445
Re(4) 
 2240
0.000093
Re(2) 
Assume that all flows are laminar; head losses can be calculated as follows:
2
 64 6.10
 1.20
H L (1)  
 10 
 0.74 m  7850  0.33 Pa  5840 Pa
 655 0.0508
 2  9.81
2
 64 6.10
 1.56
H L (2)  
 5
 2.08 m  16300 Pa
 746 0.0445
 2  9.81
2
 64 9.15
 6.24
H L (3)  
 0.75 
 10.2 m  80300 Pa
 2990 0.0445
 2  9.81
17
2
 64 9.15
 4.69
H L (4)  
 0.75 
 7.43 m  58400 Pa
 2240 0.0445
 2  9.81
The force is given by
F  [(6897 kPa)  (5.84  16.3  80.3)] 
 (0.203)2
–
4

 [(0.203)2  (0.102)2 ] 
[6897
kPa

(5.84

16.3

58.4)]



4


Solving we get
F  220  168 =52 kN
(b) Determination of heat generation rate
Power loss = QΔp
= 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
Qpump  Q1  Q2  0.00243 m3 /s
Upper position of DCV
vext 
Qpump (m3 /s)
2
Ap (m )

0.00243
 0.0751 m/s
 (0.2032 )
4

0.00243
 0.297 m/s
 (0.1022 )
4
Spring-centered position of DCV
vext 
Qpump (m3 /s)
2
Arod (m )
Lower position of DCV
vret 
Qpump (m3 /s)
2
Aannulus (m )

0.00243
 0.100 m/s
 (0.2032 )  (0.1022 )

4
4
18
Example 1.12
For the meter-in flow control valve system of Fig. 1.27, the following data are given:
Desired cylinder speed
Cylinder piston diameter
Cylinder load
Specific gravity of oil
Pressure-relief valve setting
0.254 m/s
0.508 m
13340 N
0.9
6895 kPa
Determine the required capacity coefficient of flow control valve.
13340 N
Figure 1.27
Solution: We have
CV 
Vcyl  Apiston
pPRV  ( Fload / Apiston )
SG
CV are LPM / kpa .Therefore, we have the following units for the terms in the above equation:
Q  vcyl Ap  LPM, pPRV  kPa, (Fload ) / ( Apiston )  Pressure  kPa
19
(1.7)
The flow rate is given by
Q  vcyl Ap
 0.254
m
1L
60 s
 0.00203m 2 
s
0.001 m3 1 min
 30.9 LPM
Now it is given that pPRV  6895 kPa and
Fload
13340 N
1kPa


 6570 kPa
Apiston 0.00203 m2 1000 N/m 2
Substituting values in Eq. (1.7), we get
CV 
vcyl  Apiston
pPRV  ( Fload / Apiston )
SG

30.9
6895  6570
0.9
30.9
LPM

 1.63
19.0
kpa
20