1. HYDRAULICS & PNEUMATICS
Basic Concepts
FluidSIM-H
Presented by: Dr. Abootorabi

2. Hydrostatic pressure
Hydrostatic pressure is the pressure created above a certain level within a liquid as a result of the weight of the liquid mass. Hydrostatic pressure is not dependent on the shape of the vessel concerned but only on the height and density of the column of liquid.
Hydrostatic pressure can generally be ignored for the purpose of studying hydraulics.

3. Pressure propagation
If a force F acts on an area A of an enclosed liquid, a pressure p is produced which acts throughout the liquid (Pascal's Law).

4. Power transmission
If a force F1 is applied to an area A1 of a liquid, a pressure p results. If, as in this case, the pressure acts on a larger surface A2, then a larger counter-force F2 must be maintained. If A2 is three times as large as A1, then F2 will also be three times as large as F1.
Hydraulic power transmission is comparable to the mechanical law of levers.

5. Displacement transmission
If the input piston of the hydraulic press travels a distance s1, a volume of fluid will be displaced. This same volume displaces the output piston by the distance s2. If the area of this piston is larger than that of the input piston, the distance s2 will be shorter than s1.

6. Displacement transmission

7. Pressure transfer
The fluid pressure p1 exerts a force F1 on the surface A1 which is transferred via the piston rod to the small piston. The force F1 thus acts on the surface A2 and produces the fluid pressure p2 . Since the piston area A2 is smaller than the piston area A1, the pressure p2 must be larger than the pressure p1.
The pressure-transfer (pressure-intensification) effect is put to practical use in pneumatic/hydraulic pressure intensifiers and also in purely hydraulic systems when extremely high pressures are required which a pump cannot deliver.

8. Pressure transfer

9. Types of flow
A distinction is made between laminar flow and turbulent flow. In the case of laminar flow, the hydraulic fluid moves through the pipe in ordered cylindrical layers. If the flow velocity of the hydraulic fluid rises above a critical speed, the fluid particles at the center of the pipe break away to the side, and turbulence results.
Turbulent flow should be avoided in hydraulic circuits by ensuring they are adequate sized.

10. Types of flow
Laminar
Turbulent

11. Cavitation
Motion energy is required for an increase in the flow velocity of the oil at a restriction. This motion energy is derived from the pressure energy. If the vacuum which results is smaller than -0.3 bar, air dissolved in the oil is precipitated out. When the pressure rises again due to a reduction in speed, the oil bursts into the gas bubbles.
Cavitation is a significant factor in hydraulic systems as a cause of wear in devices and connections.

12. Types of flow

13. Cavitation
Local pressure peaks occur during cavitation. This causes the erosion of small particles from the wall of the pipe immediately after the reduced cross-section, leading to material fatigue and often also to fractures. This effect is accompanied by considerable noise.

14. Cavitation

15. Energy
Potential energy:
W=m.g.h
Press with elevated reservoir:

16. Energy
Pressure energy:
W=p.ΔV

17. Energy
Motion energy:
Thermal energy:

18. Power

19. Power

20. Efficiency

21. Calculation of input and output power

22. Throttle points

23. Physical Properties of Hydraulic Fluids
Specific Weight
Also known as unit weight, is the weight per unit volume of a material.
 = m V
kg.m-3
Volume
Density =
Mass
Volume
 =
Specific weight =
Weight W V
N.m-3
Specific weight – Density relationship:
W = mg
 =  g [N. m-3]

24. Specific Gravity (SG) water SGoil = oil = oil water
Specific gravity is a dimensionless unit defined as the specific weight of the fluid divided by the specific weight of water.
Specific weight of the oil [N/m3]
water
SGoil =
oil
9800 N/m3 =
oil
water
Density of the oil [kg/m3]
Density of water [kg/m3]
1000 kg/m3

25. Pressure Head p H =  g Pressure - head relationship: p =  H
It represents the height of a fluid column that produces the static pressure.
 g p
H =
1 ft
0.433 psi
In other words, due to its weight, a 1-ft column of water develops at its base a pressure of psi. The 1-ft height of water is commonly called a pressure head.
Pressure - head relationship: p =  H
Pressure - force relationship: p = F / A

26. The Continuity Equation for Hydraulic Systems
Use of Volume Flow Rate Q
Q1(m3/s) = A1(m2) 1(m/s) = A2 2 = Q2
Flow rates are frequently specified in units of liters per second (Lps) or liters per minute (Lpm). 1m3 = 1000 L

27. The Continuity Equation for Hydraulic Systems
Q is the volume flow rate ( volume of fluid passing a given station per unit time).
Hence, for hydraulic systems, the volume flow rate is also constant in a pipe line. The continuity equation for hydraulic systems can be rewritten as follows:
(/4) D12
(/4) D22 1 2 = A2 A1
Continuity equation for
hydraulic system
Where D1 and D2 are the pipe diameters at stations 1 and 2, respectively.
The final result is: 1 2 = D1 ( D2 ) 2
This equation shows the smaller the pipe size, the greater the velocity.

28. Hydraulic Power Hydraulic Cylinder
Rod
F Load
Barrel p Q
Piston
Hydraulic power is the power delivered by a hydraulic fluid to a load-driving device such as hydraulic cylinder.
Let’s analyze the hydraulic cylinder (above figure) by developing equations that will allow us to answer the following 3 questions:

29. Hydraulic Power Hydraulic Cylinder QUESTIONS
How do we determine how large a piston diameter is required for the cylinder?
What is the pump flow rate required to drive the cylinder through its stroke in a specific time?
How much hydraulic horsepower does the fluid deliver to the cylinder?

30. ANSWER 1 – Piston Size F load A = p
Rod
F Load
Barrel p Q
Piston
ANSWER 1 – Piston Size
Pressure p acts on the area of the piston to produce the force required to overcome the load:
F load
A = p
Load is known from the application
Pressure is established based on the pump design

31. ANSWER 2 – Pump Flow Rate Q [m3/s] = A [m2] ×  [m/s]
Piston velocity
Piston area
Q [m3/s] = A [m2] ×  [m/s]
The larger the piston area and velocity, the greater must be the pump flow rate. A S
Volume displacement VD of the hydraulic cylinder = A X S
Q = VD / t
= (A X S) / t
= A X 
Pump flow rate in a specific time
Piston velocity

32. ANSWER 3 – Hydraulic power
= Force X Distance
Energy
= F X S
Power =
= p A X S
Time
p A S = t =
p A 
or Pa =
p Q
Hydraulic power (W) = p [N/m2] X Q [m3/s]
Hydraulic power (kW) = (p [bar] X Q [lit/min])/600

33. Mechanical Power
The mechanical output power delivered by a hydraulic motor:
Power (W) =
T (N.m) X  (rad/sec)
If RPM is given , must change to rad/sec by X 2/60

34. Introduction to FluidSIM-H Demo

35. Introduction to Simulating and Creating Circuits

36. Introduction to Simulating and Creating Circuits

37. The end.