Fluid Power Systems and Industrial Automation

hydraulics and pneumatics

1. Fluid Power Systems and
Industrial Automation
Unit 1,2,3
Anand Jayakumar A
Assistant Professor Mechanical Engineering
Suguna College of Engineering
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2. UNIT – I
FLUID POWER SYSTEM GENERATION AND
ACTUATORS
•Need For Automation, Classification of Drives - Hydraulic, Pneumatic
and Electric –Comparison – ISO Symbols for their Elements, Selection
Criteria.
•Generating Elements- Hydraulic Pumps and Motor Gears, Vane,
Piston Pumps – Motors - Selection and Specification - Drive
Characteristics – Utilizing Elements - Linear Actuator – Types,
Mounting Details, Cushioning – Power Packs – Accumulators.
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3. Software
Automation Studio 3.0.5
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4. Automation Studio - Library
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5. Pneumatic Components
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6. Hydraulic Components
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7. Pneumatic Single Acting Cylinder
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8. Pneumatic Double Acting Cylinder
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9. Hydraulic – Single Acting Cylinder
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10. Hydraulic – Double Acting Cylinder
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11. Pneumatic – Pilot Operated
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12. Pneumatic - Reciprocating
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13. Pneumatic – Reciprocating with Stop Switch
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14. Pneumatic Circuit – with flow control with
throttle valve
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15. Pneumatic Circuit – Quick Return Mechanism
using Non Return Throttle Valve
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16. Pneumatic Circuit with AND Valve
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17. Pneumatic Circuit with Shuttle (OR) Valve
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18. Pneumatic Electric Circuit (US)
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19. Pneumatic Reciprocating Circuit – Proximity
Sensor (US)
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20. Pneumatic Reciprocating Circuit – Proximity
Sensor (US)
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21. Pneumatic Electric Circuit (Europe)
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22. Pneumatic Reciprocating Circuit – Proximity
Sensor (Europe)
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23. Pneumatic PLC Circuit
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24. Pneumatic PLC Reciprocating Circuit
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25. On Force, Pressure and Head
1.Find the pressure at a depth of 10m below the free surface of water
in a reservoir.
2.Find the height of water column corresponding to a pressure of
70kPa.
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26. Solution
1.We know that,
•P = ρ*g*h
• = 1000*9.81*10
• = 0.981 x 10^5 N/m2 = 0.981 bar
2.We know that
•P = ρ*g*h
•70 x 10^3 = 1000*9.81*h
•h = 7.135m

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27. On Atmospheric, Gauge and Absolute
Pressure
•If a mercury barometer reads 700mm and Bourdon gauge at a point
in a flow system reads 500 kPa, what is the gauge, atmospheric and
absolute pressures at the point. Take specific gravity of mercury =
13.6
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28. Solution
1.Gauge Pressure
•We know that the gauge pressure is the pressure measured with the
help of pressure measuring instrument – Bourdon gauge.
•Therefore the gauge pressure = 500 kPa or 5 bar.
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29. Solution
2. Atmospheric pressure:
•Barometer reading (atmospheric pressure) = 700mm of mercury
•h = 700 x 13.6 mm of water
• = (700 x 13.6)/1000 = 9.52 m of water
•Atmospheric pressure, P = ρ*g*h
•= 1000*9.81*9.52
•= 93.39 kN/m2
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30. Solution
3. Absolute pressure
• = Gauge pressure + Atmospheric pressure
•= 500 + 93.39 = 593.39 kN/m2
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31. On Atmospheric, Gauge and Absolute
Pressures
•Convert -3.5kPa pressure to an absolute pressure in kPa.
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32. Solution
•Pgauge = -3.5kPa
•Absolute pressure = Gauge pressure + Atmospheric pressure
• = -3.5 + 101.3
• = 97.8kPa
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33. On Work and Power
•A person applied 200N force to move a load 2.5m in 90s.
1.How much work is done?
2.What is the power exerted by the person?
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34. Solution
•Work done = F*d = 200(2.5) = 500 N-m
•Power = (F*d)/t = 200(2.5)/90 = 5.55 watt
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35. On Application of Pascal’s Law
•The diameter of plunger and
ram of a hydraulic press are
50mm and 125mm respectively.
Find the weight lifted by the
hydraulic press when the force
applied at the plunger is 750N.
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36. Solution
• •For a hydraulic press
•F2/F1 = A2/A1
•F2/750 = 0.01227/1.963x10^-3
•F2 = 4.688kN
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37. On Reynolds Number and Fluid Flow
•The kinematic viscosity of a hydraulic oil is 0.9 stroke. If it is flowing in
a 35mm diameter pipe at a velocity of 4m/s, what is the Reynolds
number? Is the flow laminar or turbulent?
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38. Solution
•v = 0.9 stroke = 0.9cm2/s
•= 0.9x10^-4m2/s
•D = 35mm
•V = 4m/s

•Reynolds number Re = VD/v
•Re = 4(35x10^-3)/0.9x10^-4
•= 1555.55
•Since Re<2000, the flow is
laminar.
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39. On Flow Rate and Continuity Equation
•A hydraulic pump delivers fluid at 50 Lpm through a 30 mm diameter
pipe. Determine the fluid velocity.
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40. Solution
•Q = 50 Lpm = 50x10^-3/60m3/s
•D = 30x10^-3m
•Area of pipe A
•= π/4(30x10^-3)^2
•=7.07x10^-4m2
•Discharge Q = AxV
•50x10^-3/60 = 7.07x10^-4V
•V = 1.179m/s
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41. PUMPS
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42. What is a Pump?
•A pump is a device which converts mechanical energy into hydraulic
energy.
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43. Uses of Pumps
1.To transport fluid
2.To develop pressure
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44. Classification of Pumps
Pumps
Non –
Positive
Displacement
Pumps
Centrifugal
flow
Axial flow Mixed Flow
Positive
Displacement
Pumps
Rotary Reciprocating
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45. Classification of Rotary Pumps
Rotary Pumps
Gear
External Gear Internal Gear Lobe Gerotor
Vane
Fixed
Displacement
Balanced
Vane Pump
Unbalanced
Vane Pump
Variable
Displacement
Screw
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46. Classification of Reciprocating Pump
Reciprocating
Pump
Axial Piston
Bent Axis Inline
Variable Plate
Inclinable Swash
Plate
Cam / Crankshaft
driven piston
pump
Radial Piston
Stationary
Cylinder Block
Rotating Cylinder
Block
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47. Centrifugal Flow Pump
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48. Axial Flow Pump
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49. Mixed Flow Pump
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50. External Gear Pump
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51. Internal Gear Pump
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52. Lobe Pump
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53. Gerotor Pump
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54. Fixed Displacement Balanced Vane Pump
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55. Fixed Displacement unbalanced Vane Pump
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56. Variable Displacement Vane Pump
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57. Axial Piston Bent Axis Pump
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58. Inline Variable Plate Piston Pump
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59. Inline Inclinable Swash Plate Piston Pump
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60. Cam / Crankshaft Driven Piston Pump
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61. Radial Piston Stationary Cylinder Block
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62. Radial Piston Rotating Cylinder Block
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63. Thank You
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