HYDRAULIC SYSTEMS
EFMII
Adil Loya
Objectives
By the end of this lecture you will:
• Understand how hydraulic power is generated on an
mechanical system (automotive or aircraft systems)
• Know how hydraulic supplies are distributed and
controlled
• Have seen and discussed various hydraulic actuators
Advantages of Hydraulics
• High power/weight ratio
• High transmission efficiency
Pumps and hydraulic motors 85-92%
Linear actuators over 95%
• High reliability
• High controllability
Low compressibility gives accurate response
Low system inertia
• Flexibility of installation
Constraints of Hydraulics
• The system must be leak-free
• The system must have low flammability
• The system must be free of contaminants: air and debris
Basic Hydraulic Principles
1.Liquid is “incompressible”, i.e. if you squeeze a quantity
of fluid, it will flow rather than shrink.
2.Pressure is transmitted throughout a volume of liquid
Pressure = Force / Area
3.You can use a small force applied to a small area to move
a large weight
Basic Hydraulic Principles
Example 1 ;
A car is placed on a hydraulic lift. The car is raised up by
pushing down on a plunger with a force of 10kg. If the
plunger has a cross sectional area of 0.1m^2, and the car is
sitting on a platform with area 10m^2, what weight of car
can be raised by the lift?
Basic Hydraulic Principles
Example 1;
• Pressure exerted by plunger: -
Pressure = Force/ Area = 10 /0.1 = 100 kga/m2
• Force exerted on lift: - Force = Pressure x Area = 100 x 10
= 1000 kga
A Typical Hydraulic System
1.Pumps deliver FLOW not pressure
2.Each circuit needs a pressure line and a return
3.Some spare fluid must be stored, as the amount required
will vary
4.Systems operate at a specified maximum pressure
5.Difficult to store large quantities of energy
A Typical Hydraulic System
Basic Hydraulic Principles
• Pressure = Force / Area Atmospheric pressure
approximately = 1 bar = 105 Pascals Pressure in a bicycle
tyre is around 4.5 barG (bar Gauge)
1.When pressure is measured relative to atmospheric
pressure it is termed “gauge pressure”.
2.When pressure is measured relative to absolute zero (a
perfect vacuum) it is termed “absolute pressure”.
absolute pressure = gauge pressure + atmospheric
pressure
Hydraulic Fluid Requirements
1.Long life
2.Compatible with the materials used in the system
3.Resistant to boiling /sludging /water take-up
4.Wide operating temperature
5.Consistent viscosity
Hydraulic Fluid Standards
1.Mineral Oils
•DTD585 in the UK
•MIL-H-5606 in the USA
•AIR320 in France
•H515 in NATO
2.Synthetic Oils (Fire Resistant)
•Skydrol
•SAE: AS 1241
•MIL-H-83282 and MIL-PRF-46170C
Hydraulic Reservoir
1.Stores fluid
•Expansion
•Variable capacity
• in actuators
•Leaks
2.Must keep air out
3.Aircraft attitudes and manoeuvres
4.Pressurised
Hydraulic Fluid Standards
1.Mineral Oils
•DTD585 in the UK
•MIL-H-5606 in the USA
•AIR320 in France
•H515 in NATO
2.Synthetic Oils (Fire Resistant)
•Skydrol
•SAE: AS 1241
•MIL-H-83282 and MIL-PRF-46170C
Hydraulic Pumps
1.Engine driven pumps
•Fixed displacement
•Variable displacement
2.AC motor driven pumps
3.Power transfer unit (PTU), auxiliary power unit (APU),
ram air turbine (RAT)
Hydraulic Pumps
Three different types of hydraulic pumps exist:
Hydraulic Gear Pumps
• External gear pump
• Internal gear pump
• Gear ring pump
• Screw spindle pump
Hydraulic Vane Pumps
• Single chamber vane pump
• Double chamber vane pump
Hydraulic Piston Pumps
• Axial piston pump
• Radial piston pump
Note: while gear pumps operate with fixed displacement volume, vane and piston pumps may operate with fixed or
variable displacement volumes.
Hydraulic Pumps – Engine Driven
• Fixed displacement pumps – constant delivery per
revolution
• Variable displacement pumps - more efficient
Pumps – Fixed Displacement
Gear Pump
Pumps – Fixed Displacement
Vane Pump
Pumps – Fixed Displacement
Bent-Axis Piston Pump
Pumps – Variable Displacement
PISTON or SWASH PLATE Pump
Pumps – Variable Displacement
Pump Flow Characteristics
Actuator Force
Force = Pressure X Area
Actuator Force
Force = Pressure X Area
A hydraulic cylinder has an internal cross section of 5 x
10-3m^2 and a piston which is required to exert a force
of 3kN. What is the system pressure required?
Actuator Force
System pressure = Force / Area = 3 x 103 / 5 x 10-3 =
0.6 x 106 Nm-2 = 0.6 x 106 Pa = 6 bar
Actuator
Actuators (the “muscles” of the system) Three broad
categories of actuator: -
1.Linear actuators (jacks)
2.Semi-rotary actuators
3.Hydraulic motors
Actuator
Actuators (the “muscles” of the system) Three broad
categories of actuator: -
1.Linear actuators (jacks)
2.Semi-rotary actuators
3.Hydraulic motors
Actuator
Advantages: -
•Constant force throughout stroke
•Compact
•Relatively immune to interference
Accumulators
Like pressure intensifier another fluid
system, called accumulator, is very useful to
use energy of the pump effectively during
the period when fluid power is not used in
applications in which large quantity of fluid
is required during working stroke. For this,
the fluid is stored in a vessel during no-
operation period. Figure on right shows two
types of hydraulic accumulator. In simple
accumulator, a sliding ram moves up in a
fixed cylinder when the machine does not
consume fluid.
Accumulators
A load is placed on the ram so that it is equal to the product of the fluid
pressure from pump and area of the ram. Thus accumulator pressure can be
fixed by using appropriate weights on the ram.
When machine needs liquid then it is supplied partly by the pump and the
rest by the accumulator during which the ram moves down. The volume of
the liquid collected in the cylinder is equal to that supplied during no
operation period. If p, a and h are liquid pressure, ram area and
displacement of the ram from minimum to maximum position, then
Energy supplied by fluid = Energy stored by accumulator = p x a x h = p x
cylinder volume
Accumulators (differential)
Weight required in a simple accumulator is large. This reduced in
differential accumulator, which is shown in earlier figure; on slide 28 (b). It
consists of a central plunger around which inverted cylinder is placed that
slides up and down along the plunger axis; a brass sleeve is used between
them to prevent wear and tear of the cylinder surfaces. Load is placed on
the flange of the sliding cylinder. To reduce this load, a fixed arm is
provided, as shown in the figure. During non-operation period, water
enters the accumulator and exerts pressure on the cylinder surface around
fixed ram and pushes it up. The cylinder moves down when liquid flows
from it. We may write following equations considering equilibrium of the
cylinder and relation for work transfer:
Accumulators
W = / 4( D − d ) p h / t
2 2
Where W is the work transfer of cylinder
(W, p, D, d, h, and t are its own weight,
liquid pressure, cylinder diameter, ram
diameter, and vertical displacement of
cylinder in time t, respectively. Load can be
applied by sliding standard weights
around the inverted cylinder.
Accumulators
W = / 4( D − d ) p h / t
2 2
Example 1; The total weight including the self-weight of ram placed on the sliding ram
of a hydraulic accumulator is 40kN. The diameter of the ram is 500 mm. If the frictional
resistance against the movement of the ram is 5% of the total weight, determine the
intensity of pressure of water when (i) the ram is moving up with a uniform speed , and
(ii) the ram is moving down with uniform speed. If the stroke of the ram is 10m and the
ram falls through the full stroke in 4 min steadily, find the work done by the
accumulator per second. If the pump, connected to the inlet of the accumulator, supplies
0.01 m^3/s at the same time, determine the work supplied by the pump per second and
also the power delivered by the accumulator to the hydraulic machine, connected at the
outlet of the hydraulic accumulator, when ram is moving downwards.
Accumulators
Solution 1;
Given: W= 40kN, D = 0.5 m → A=0.1963 m^2, frictional resistance FR = 0.05 x 40000 = 2000 N, t = time taken to travel
=240 s, stroke L = 10 m, Q = 0.01 m^3/s. See Figure –
When ram moves up then the force required, F = weight + FR = 42000N
Therefore; p, intensity of pressure = 42000/0.1963 =214kN/m^2
When ram moves down then the force required, F = weight - FR = 38000N
Therefore; p, intensity of pressure = 38000/0.1963 =193.581kN/m^2
Discharge from the accumulator/s= q = A x L/t = 0.1963 x 10/240 = 0.00818 m^3/s
Work done by accumulator/s = 38000 x 10/240 = 1583.33 W
Pressure head of the pump during working stroke, H = 193,581/w=19.733m; Where w=9810N
Workdone by pump = w Q H = 9810 x 0.01 x 19.733 =1935.81 W
Total work delivered to the machine = 1583.33 +1935.81 =3519W or 3.519 kW
Accumulators
Example 2; It is required to transmit 36.75kW power from an accumulator through a
pipeline of 10 cm diameter and 1500 m long. The ram is loaded with a weight of 1250 k
N and the friction loss in the pipeline equals 2.5% of the total power being transmitted.
Work out the flow rate and velocity of the ram if the friction factor f = 0.01 for the
pipeline.
Solution;
Give: P =36.75 kW, d = 0.1 m, L = 1500m, W = 1250 kN
Power loss = 0.025 x36.75 = 0.9187 kW
Friction head loss 4 fl U 2 4 0.011500 2
Friction head loss h f = = U = 30.58U 2
d 2g 2 9.81 0.1
Power loss in friction = wQh f = 9810 0.7854 0.12 U 30.58U 2 = 2356.2U 3 = 918.7
→ U, flow velocity in pipe = 0.731m/s and Q = 0.7854 x (0.1^2) x U = 5.741x10^-3 m^3/s
Hydraulic cranes and lifts
Hydraulic cranes and lifts mechanism
Hydraulic cranes and lifts
We will not go into details of these simple machines. If there are n pulleys at each end of
the pullet blocks, then the vertical distance moved by the crane will be 2n times the
travel of the ram. Let F be force on the ram, ram movement be y, load be W and the
vertical distance moved by load x and η be the efficiency of the crane then from the basic
machine principle…
W y
F . y. = W .x → = . or , M . A = V .R
F x
Note that x is equal to 2n times the ram movement. Thus more is the number of the
pulleys less is the Velocity Ratio (V.R). In other words for given displacement of the load
the movement of the effort will be less. In hydraulic crane and lift, the load displacement
hydraulic power. In such machines, Mechanical Advantage (M.A) is always less than 1,
since electric motor is being used as power source. Therefore the main aim is to have
large load movement for a small displacement of the effort.
Hydraulic cranes and lifts
Example 3; The efficiency of a hydraulic crane, which is supplied water under a pressure
of 70N/cm^2 for lifting a weight through a height of 10m, is 60%. If the diameter of the
ram is 150mm and velocity ratio 1/6, find (i) the weight lifted, and (ii) the volume of the
water required in litres to lift the weight.
Solution;
Given: = 0.6, p = 7 105 N / m 2 , x = 10 m, D = 0.15 m → A = 0.0176 m 2 and
c
F = p A = 12320 N ,VR = 1/ 6
W dis tance moved by weight W
c = = → W = 1232 N
Fy F VR
Volume of water required for lift = A y = 0.0176 10 / 6 = 0.02933m 3 = 29.33l
