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Industrial automation pe 5421 weeks 2 3 4 10 20 2015

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Industrial Automation PE- 5421 Prof. Charlton S. Inao Defence University 06/21/2022 1 For Manufacturing and Metallurgy Group
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Page 1: Industrial automation pe 5421  weeks 2 3 4  10 20 2015

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Industrial AutomationPE- 5421

Prof. Charlton S. InaoDefence University

For Manufacturing and Metallurgy Group

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Pneumatic System

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Component parts of Pneumatic System

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Circuit Diagram of Pneumatic Elements

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Pneumatic Air Receiver

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Air Supply System

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Pneumatic Service Units(Filter, Regulator,Lubricator)

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DIRECTIONAL CONTROL

VALVES

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Directional Control Valves

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Logic AND function

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Logic AND Application

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Logic OR Valve

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Or Valve Application

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Pneumatic Valves

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Methods of Actuation

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Week 3

Pressure Control ValvesFlow Control ValvesCylinders

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Flow Control Valve

Restrictor Valve

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Flow Control Valve

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Meter In Circuit

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Meter Out circuit

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Cylinders• Single Acting Cylinders•Double Acting Cylinders•Rodless Cylinders•Ram Cylinders

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Single Acting Cylinder

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Double Acting Cylinder

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Rodless Cylinder

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Used in Robotics, hi speed manufacturing, semi conductor assembly and manufacturing and proportional pneumatics

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Ram Cylinder

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Rotary Actuators

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Rotary Actuators

• Pneumatic motors- Torque motors

• Hydraulic motors- winch, cement mixer

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Rotary Actuator

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Torque Motor(Turbine Flow Motors)

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Mono Directional

• Single Direction or Rotation Actuator

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Bi -Directional

• Double or Reversible Rotary Actuators

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Pneumatic Gear Motor

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Pneumatic Sliding Vane Motor

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Rotary Actuators-Vane Type

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Pneumatic Vane MotorsBalanced Unbalanced

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Pneumatic Piston Motor

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Axial piston In Line with Swash Plate

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Axial Piston (Bent Type)

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Process Control Valves

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Diaphragm operated control valve

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Formula for Process Control Valves

• Change in the flow rate=k(change in stem displacement)

• Q/Qmax=S/Smax

• ∆Q/Q=k ∆S

• Q/Qmin=(Qmax/Qmin)(S-Smin)/(Smax-Smin)

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Where• k=constant• Q=is the flow rate at a valve displacement S• Qmax=the maximum flow rate at the

maximum displacement Smax• Q/Qmax=S/Smax• Therefore the percentage change in the

flowrate equals the percentage change in the stem displacement.

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Sample Problem

• An actuator has a stem movement which at full travel is 40 mm. It is mounted on a process control valve with an equal percentage plug and which has a mimimum flow rate of 0.2 m3/s and a maximum flow rate of 4 m3/s. What will be the flow rate when the stem movement is a) 10 mm b) 20 mm.

• Given:• Smax=40mm• Smin=0 mm• Qmin=0.2 m3/s• Qmax=4m3/s

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• Required: What will be the flow rate when the stem movement is a) 10 mm b) 20 mm.

• Solution A: Q/Qmin=(Qmax/Qmin)(S-Smin)/(Smax-Smin)

• Q/0.2=(4/0.2) (10/40)

• Answer a) 0.423 m3/sec• Solution B• Q/0.2=(4/0.2) (20/40)

• Answer b) 0.89 m3/sec

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Flow Coefficients and Valve Sizes

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Control Valve Sizing Formula

• Q=Av√ ∆P/ρ

Where

Av= valve flow coefficient

∆P = the pressure drop across the valve

ρ=density of the fluid

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Sample problems-Control Valve• Determine the valve size that is required to control the flow of

water when the minimum required flow is 0.012 m3/s and the permissible pressure drop across the valve at this flow rate is 300 kPa.

• Using the equation:

Q=Av√ ∆P/ρ

Av=0.012√1000kg/m3/300x103

Av=69.3 x10-5

Then using Table 5.1 Flow Coefficients and Valve Sizes, the valve size is 960 mm.

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Pneumatic Applications

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Package Transfer


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