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

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.

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).

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.

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.

Displacement transmission

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.

Pressure transfer

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.

Types of flow Laminar Turbulent

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.

Types of flow

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.

Cavitation

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

Energy Pressure energy: W=p.ΔV

Energy Motion energy: Thermal energy:

Power

Power

Efficiency

Calculation of input and output power

Throttle points

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]

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

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 0.433 psi. The 1-ft height of water is commonly called a pressure head. Pressure - head relationship: p =  H Pressure - force relationship: p = F / A

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

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.

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:

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?

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

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

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

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

Introduction to FluidSIM-H Demo

Introduction to Simulating and Creating Circuits

Introduction to Simulating and Creating Circuits

The end.