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1

Chapter 8 

Hydraulic valves

1

Introduction

There are three types of valves

Directional control valves

Determines the path through which a fluid traverses a given circuit.

Pressure control valves

Protects against overpressure which may occur due to excessive actuator loads or due to the 
closing of a valve 
(pressure relief, pressure reducing, sequence unloading, and counterbalance valves)

Noncompensated flow control valves are used where precise speed control is not required 
 flow rate varies with pressure drop across a flow control valve

Pressure‐compensated flow control valves automatically adjust to changes in pressure drop to 
produce a constant flow rate

2

The most important considerations in any fluid power system is control. The controlling elements 
are called valves

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2

3

Directional Control Valves (DCV)

A valve is a device that receives an external signal (mechanical, fluid pilot signal, electrical or
electronics) to release, stop or redirect the fluid that flows through it.

DCV is to control the direction of fluid flow in any hydraulic system

List the DCV

Directional Control Valves (DCV)

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Pressure control Valves

Pressure Reducing valve

It is used to maintain reduced pressures in specified locations of hydraulic systems

It is actuated by downstream pressure and tends to close as this
pressure reaches the valve setting

It uses a spring loaded spool to control the downstream pressure

If the downstream pressure is below the valve setting, the fluid flows
freely

When the outlet pressure increases to the valve setting, the spool
moves to partially block the outlet port.

If the valve closes completely, leakage past the spool causes downstream pressure to build up
above the valve setting

This is prevented from occuring because a continuous bleed to the tank is permitted via a
seperate drain line to the tank.

Pressure control Valves

Pressure Reducing valve

Two cylinder are connected in parallel

The circuit is designed to operate at a maximum pressure
P1, which is determined by the PRV

It is the maximum pressure at which cylinder 1 operates

By the function of this machine, cylinder 2 is limited to
pressure P2 (P2< P1). This can accomplished by placing a pressure reducing valve

If the pressure in the cylinder 2 circuit rises above P2, the pressure reducing valve closes partially
to create a pressure drop across the valve.

The disadvantage is that the pressure drop represents the lost energy that is being converted into
heat.

Example: the primary part of the circuit is operating at 180 bar. A secondary circuit supplied from
the primary via a pressure reducing valve requires a constant flow of 30 LPM at 100 bar. Find the
power loss over PRV.  180 100 30

4kW
600

 


4 kW cannot be dissipated by natural cooling and heat exchanger may be required.

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Pressure control Valves

UnloadingValve

It is used to dump excess fluid to tank at littlet or no pressure

A common application is in high low pump circuits where two pumps
move an actuator at a high speed and low pressure.

The circuit the shifts to a single pump providing a high pressure to perform work

Pressure control Valves

CounterbalanceValve

It is normally closed valves. Used to maintain a back pressure on a vertical cylinder to prevent it
from failling due to gravity

During the downward movement of the cylinder, the counterbalance velveis set to open at
slightly above the pressure required to hold the piston up

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Hydraulic fuses

It is analogue to an electric fuse because they both are one shot devices

It prevents hydraulic pressure from exceeding an allowable value in order to protect circuit
components from damage

Exceeds a design valuethe thin metaldisk ruptures to relieve the pressure as oil is drained back
to the oil tank

Partial circuit : pressure compensated pump
hydraulic fuse

37

Problem 8-43

inx

inlbKs

inAp

15.0

/2000

²65.0





psi
A

xks
A
F

P
pressurecracking

5.6461
65.0

300





psi
A

Sks
A
F

P
pimp

769
65.0

)15.01.0(2000








Problem 8-47

gpmQ

psip

25

2000




HP
PQ

HP 2.29
1714

252000
1714





38

Page 20

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20

Problem 8-51

SG
p

CAQ 1.38 C= 0.8 (sharp edge)

22 14.32
4

inA gpmQ 3.713
9.0

50
8.01.38

Problem 8-57

9.0

687

2.2

SG

kPap
kPa

LpmC
V

Lpm
SG

p
CQ

V
8.60

9.0
687

2.2

39

Problem 8-58

PsiP

rodpistonAPWAP

1848
12

4

20002
4

750

)(

22

2

2

2

PsiP

rodpistonAPAP

1000
12

4

2
4

750

)(

22

2

2

2

Problem 8-59

singpmQ /23.87
60
231

65.22
9.0

1848
5.0 3

AVQ sin
A
Q

V /2.37
12

4

23.87
22

singpmQ /16.64
60
231

67.16
9.0

1000
5.0 3

sin
A
Q

V /23.27
12

4

16.64
22

40

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