Revision Sept 2014

ELX305

Topics• Reliability• Scada• Pneumatics• Boolean Algebra• Timers and timing diagrams• PLC programming

RELIABILITY

Definitions

• RELIABILITY, R(t), is defined as ‘the probability that a system will operate to an agreed level of performance for a specified period, subject to specific environmental conditions.’

(Probability of survival)• UNRELIABILITY, U(t), is ‘the probability that the system will fail in a

specified time period’.(Probability of failure)

• Since there are only two possibilities statistically : R(t) + U(t) = 1

• The definitions for Reliability and Unreliability are generally accepted, but do not take into account the age of a system.

• It is possible to distinguish three periods in the operating life of a system - the “bathtub” curve

Bathtub Curve

Time

“Bathtub”

Burn in Useful life Wear out

Failure rate

Region where l is ~ constant

Constant Failure Rate

• Research into the failure rate of a large range of aerospace industry products and components, showed that many electronic devices could be represented by a l(t) characterised by a short early failure region and an extended ‘constant failure’ region. (no ‘wear out’ region) (Bentley 1999)

• Many electronic components fall into this category• Through quality techniques such as ‘burning in components

and equipment, early failure can be significantly reduced or eliminated

• So given this type of equipment and these quality interventions (burn in), failure rate can be considered to be constant

MTTF and MTBF

• The term Mean Time Before Failure (MTBF) relates to equipment or systems that are repairable. It is measured by testing it for a total period of time (T) and recording the number of faults (N). Each fault is repaired and the equipment put back onto to test. The observed MTBF is given by M=T/N

• In the case of MTTF this is a reliability measure which is suitable for ‘throw away’ items such as capacitors, resistors, transistors i.e. items that cannot be repaired. If a number of items were tested to failure the MTTF is again M= T/N

Reliability formulae R(t) = 1 + U(t) ……………………….(1)

R(t) = e- lt ……………………………(2)

MTBF = ………………….(3)

and therefore l = 1/M ………….……(4)

General formula to be used when l is not constant

……………….(5)

(See Lesson 9 student guide for proofs)

1

M

0

dtRM ss

• If a system consists of two or more units and for 'system success’ both must work, then in reliability terms the units are considered to be in series

i.e. It is assumed that the failure of any unit occurs independently of the failure of the others

Analysis of ‘Series’ systems

Series Systems

Reliability of the series system

From equation (2)

Series SystemsUsing Equations (3) M = 1/l therefore (4) l = 1/M failure rate of the system ls and Mean time between Failure Ms can be calculated

Series Systems

EXAMPLE 1 (Tutorial sheet)

In the following example, assuming a series

reliability, find the system MTBF and Reliability in

1000 hours.

Unit Type 1 2 3

Number of fails (N) 10 4 5

MTBF (hours) 4x105 106 2x106

Series Systems

hours31700

105.31102

5

10

4

104

101 6665

s

s

M

M

97.0

105.311000exp1000 6

sR

Failure rate Number of failures/Time For series system s = 1+2+3+4+n

Series Systems

EXAMPLE 2.

A system consists of 4 elements, each having a probability of survival to 1000 hours R(1000) = 0.95 . What is the combined 1000 hour reliability of the system.

Rs = R1x R2x R3

= .95 x .95 x .95

= 0.857

Analysis of Parallel Systems• If a system consists of two or more units

which ‘normally’ contribute to the systems operation, and if one unit fails, the system continues to function, then the system is considered to be in parallel.

• In series situation, ‘system success’ requires both units to succeed, whereas in parallel ‘system failure’ requires both units to fail, that is, the reliability logic is reversed.

Parallel Systems

Therefore for a simple active parallel system the probability Us that the system is in the failed state is: Us = U1 x U2

Parallel Systems

Since U(t) = 1 – R(t)

It follows that:

Reliability of the parallel network

Parallel Systems

Parallel Systems

Missing steps

a

k

a

kee

a

k

ea

kdtKe atat

100

00

Using this information

1

0

))1(()0(

))(()(

0

0

00

e

e

MMM

eMeM

eMdteM

sss

ss

M

t

sM

t

sss

MM

MdteeMTBF MtMt 5.12

220

/2/

Parallel Systems - ExampleTwo identical PLC units are connected in parallel. The reliability requirement is such that both units have to fail before the complete system fails. Each unit has a Mean Time Between Failure (MTBF) =105 hours. Calculate the reliability of the system RS in 104 hours

Solution

21211 RRRRUR SS

MteR / therefore substituting in

5

4

5

4

5

4

5

4

10

10

10

10

10

10

10

10

2121 .eeeeRRRRRS

99.08187.0809.1904.0904.0904.0904.0

SCADA

ELX305

Introduction

• SCADA – Supervisory, Control And Data Acquisition

• Generic term given to a computer based system which provides a user interface to connect to PLC’s and other devices

• SCADA software packages enable data to be shown graphically in ‘real time’ and some cases can be used to supervise/control processes and equipment

SCADA Software

• Many suppliers of SCADA software

e.g. Wonderware, Microscan, Labview• Usually charged according to number of

tags – (inputs/outputs)• Developers or design capability versus a

user license • Some systems protected by a ‘dongle’, a

hardware key

Example of Communications for Electric Power Distribution

Networks• PLC’s or SCADA pc’s can be connected together or

interconnected using a ‘network’• The need to have a common data communications standard

came about due to the rapid growth of PLC control in industrial applications.

• PLC manufacturers initially developed their own individual network communications

• End users faced problems when they had to interface products from different manufacturers

• This led to the development of the Open Systems Interconnection (OSI) framework which identified the main features and function of communication networks that manufacturers have to adhere

Why network PLC’s

• Allows ‘non-critical’ data to be transmitted and shared between controllers and computers.

Typical applications include:– taking quality readings with a PLC and sending the

data to a database computer– distributing recipes or special orders to batch

processing equipment– remote monitoring and control of equipment

SCADA

Types of Network Topologies

Ring and Bus Topologies

• In the Ring and Bus topologies the network control is distributed between all of the PLC’s on the network.

• The wiring only uses a single loop or run of wire.• Only one wire means that the network will slow down

significantly as traffic increases.• It also requires more sophisticated network interfaces

to determine when a PLC is allowed to transmit messages.

• It is also possible for a problem on the network wires to halt the entire network.

Ring and Bus TopologiesBUS• Advantages - Easy to implement/extend, Less cabling,

cheap• Disadvantages - Administration difficult, Limited

length/stations, loss of cable-loss of network.

RING• Advantages - Easy to extend, Network collisions

prevented (token passing)• Disadvantages - Slow, Fault Detection difficult, must

break to extend

Star Topologies

• The Star topology requires more cabling to connect each computer to an intelligent hub.

• The network interfaces in the PLC’s are simpler, and the network is more reliable.

• If one remote device fails then normally the rest of the network continues to function

• Because of this it is often easier to fault find• Said to be deterministic –performance can be predicted. • This can be important in critical applications, especially

in cases where the signal sampling and control output has to be strictly controlled.

Star TopologySTAR• Advantages - Easy to expand, Fault detection easy,

Deterministic, Non-centralised failures handled, • Disadvantages – Relatively expensive, Extra hardware

needed, loss of hub catastrophic

Pros and Cons

• For a factory environment the bus topology is popular. • The large number of wires required for a star

configuration can be expensive and confusing.• The loop of wire required for a ring topology is also

difficult to connect, and it can lead to ground loop problems.

• Smaller bus net works can be connected into a ‘tree’ structure using repeaters. (see study guide lesson 8 pg.4) These boost the signal strength and allow the network to be made larger.

Network hardware

• Various types of hardware are used to connect the devices together

• These include routers, repeaters, hubs, bridges, gateways which transmit and control data Physical connection via cabling are terminated with 10base2 (BNC),10baseF (fibre optic) connectors.

• Fibre optic networks are highly tolerant to interference but can need specialist to test/commission system. Cables can be prone to damage if not ran correctly, loss of or reduced signal faults can be difficult to trace

Types of transmission

The transmission type determines the communication speed and noise immunity.

• Baseband, simplest, where voltages are switched off and on according to signal bit states . This method is subject to noise, and operates at low speeds e.g. RS-232

• Carrierband transmission uses FSK (Frequency Shift Keying) that will switch a signal between two frequencies to indicate a true or false bit. Provides higher transmission speeds, with reduced noise effects

• Broadband networks transmit data over more than one channel by using multiple carrier frequencies on the same wire.

Bus Network Topology

• Only uses a single transmission wire for all nodes.

• If all of the nodes decide to send messages simultaneously, the messages would be corrupted (a collision occurs).

• There are a variety of methods for dealing with network collisions, and arbitration

Data Transfer

• Data is sent of the network in ‘packets’• Packet sizes can be different depending on the

particular network• Data is organised in a ‘frame’ which is like a long serial

byte.• Each bit of data and its position in the ‘serial byte’ has a

specific purpose and meaning (protocol) • Error detection is normally provided to ensure data

security using either checksum or CRC

Bus Network Collision Techniques

• CSMA/CD (Collision Sense Multiple Access/Collision Detection) – if two nodes start talking and detect a collision then they will stop, wait a random time, and then start again. The time period is usually 1 cycle time (typically 50 microsecs), after which another attempt is made to transmit

• CSMA/BA (Collision Sense Multiple Access/Bitwise Arbitration) – if two nodes start talking at the same time the will stop and use their node addresses to deter mine which one goes first.

Bus Network Collision Techniques

Master-Slave – one device on the network is the master and is the only one that may start communication. Slave devices will only respond to requests from the master.

Token Passing – A token, or permission to talk, is passed sequentially around a net work so that only one station may talk at a time.

Bus Network Collision Techniques

Pros and Cons The token passing method is deterministic, but it

may require that a node with an urgent message to wait to receive the token.

The master-slave method will put a single machine in charge of sending and receiving. This can be restrictive if multiple controllers are to exist on the same network.

The CSMA/CD and CSMA/BA methods will both allow nodes to talk when needed. But, as the number of collisions increase the network perfor mance degrades quickly.

Pneumatics

Control Valve Symbols

2/2 Way Directional Control Valve (Flow

Switch)

• The ports are indicated (on the initial status) :-

Output Port ( Top )Inlet Port ( Bottom )

• Flow is indicated by an arrow ( No flow by lines at right angles )

• For every control valve status a square is drawn

Actuation of Control Valves :-1. Mechanical

• General• Pushbutton• Lever Operated• Foot Pedal• Spring Return• Spring Centered• Roller• Idle Return Roller

Actuation of Control Valves :-2. Electrical / Pneumatic

• Direct Pneumatic• Indirect Pneumatic• Pressure Release• Single Solenoid• Double Solenoid• Electro-Pneumatic

Describe this Control Valve ?

4/2 Way Directional Pushbutton Control Valve

with Spring Return

• The control valve has four ports

• The control valve is operated by a Pushbutton and returned by a Spring

• The control valve has two positions

• The control valve changes the flow direction at the output ports

Linear Actuators

• Single Acting Cylinder (SPRING RETURN)

• Double Acting Cylinder

• Double Acting Cylinder with double ended piston

• Double Acting Cylinders with non-adjustable and adjustable cushioning on one or both ends

3/2 Pneumatic Control Valve

2Actuate Spring Return

Actuate Spring Return2

3/2 Pneumatic Control Valve driving a Linear Actuator

• Single Acting Cylinder (SPRING RETURN)

This is the simplest and cheapest form of cylinder. It only uses 1 port so can be controlled with a 3/2 valve. Particularly useful where many cylinders are needed e.g. automation applications where on/off push type actions are needed. Also used for controlling valves and flaps where on/off states are needed without speed control.

3/2 Pneumatic Control Valve driving a Linear Actuator

What Happens if the Pressure Supply is lost ?

Open 22

• Non-Actuated • Actuated

Open2 2

5/2 Pneumatic Control Valve

4 2Actuate Spring Return

4 2Actuate Spring Return

5/2 Pneumatic Control Valve driving a Linear Actuator

What Happens if the Pressure Supply is lost ?

• Actuated

4 2Open

• Non-Actuated

4 2Open

Cascade Valves for higher flows

How does this improve cylinder flow rates ?

4 2

2 2

Flow Control Elements - 1

• Check Valve

• Spring Loaded Check Valve

• Accumulator

• Adjustable Flow Control

• One-Way Flow Control

Actuator Flow Rate Control(One way control valve)

Used to vary air flow in one direction only, typical application would be to vary the extension or contraction of an air cylinder. Imagine this element placed into the flow line between the directional control valve and the input port of the actuator. The response of the actuator would be rapid in one direction and slow in the reverse direction, an action that could not easily be achieved using direct PLC control.

Actuator Flow Rate Control

What is restricted by each flow valve ?

4 2

Sequential OperationFrom the previous Circuit note :• If the sequence is to commence on a start

signal a manual start valve is needed• An Actuator needs an individual

pneumatically driven control valve• Each identified sequence position needs a

switch driven valve• Connections between valves programme the

desired sequence, i.e.A- A+

a-

a+A+ Actuator direct.a+ Valve

Where :-

Two Actuator Operation

For the following sequence

then we require :• a start valve, and• two cylinders driven by pneumatically

operated 5/2 control valves, and• four switch driven 3/2 directional valves

A+

B+a+ b+

A-a-

B-b-Start

Solution :

A+

a+

a-

A-

A-A+

B+B-

B-B+

b-

b+

Start

Boolean Algebra examples

• Simplify

i) ).(. BAAX [4 marks]

ii) CBAY ).( [4 marks]

iii) CACBBAZ ..)..().( [4 marks]

TIMERS

4-60

Timer Bits Enable Bit (EN)

• True when the rung input logic is true.• False when the rung input logic is false.• When true the timer accumulator is

incrementing at the rate set by the timer time base.

Timer BitsTimer Timing Bit (TT)

• True only when the accumulator is

incrementing.

• Remains true until the accumulator reaches

the preset value.

• Returned to a false condition when the

accumulator value is equal to or greater

than the preset value.

4-62

Timer BitsTimer Done Bit (DN)

• Signals the end of the timing process by

changing states from false to true or from

true to false depending on the type of timer

instruction used.

• Operation is dependent upon the type of

timer instruction used.

TON—On Delay Timer

TON Timing diagram example

TON - Timer on - examples

(a)

(b)

4-66

TOF—Off Delay Timer

TOF Timer Example

TOF Timer Example

4-69

RTO—Retentive Timer

RTO Timer Example

EXAMPLE USING ALL THREE TYPES OF TIMER

EXAMPLE – TIMING DIAGRAM

T4:0/TT

T4:1/TT

T4:2/DN

0 5 10 15 20 25 30 35

SECONDS

Input I:1/0

T4:2/TT

5 SEC

5 SEC

10 SEC

10 SEC

10 SEC

10 SEC

RTO TIMED OUT

PLC PROGRAMMING EXAMPLE

The following diagram shows a conveyor and sorting system which is used to feed boxes into three separate chutes. The system loads 6 boxes into each chute. There are two pneumatically operated gates and a fixed gate. These are used to divert the boxes from the conveyor down the chutes. There are three through beam photo electric sensors S1, S2 and S3 which are used to detect the boxes as they pass along the conveyor.

PLC PROGRAMMING EXAMPLE

CH

UT

E 1

CH

UT

E 2

CH

UT

E 3

S3

GATE 2 FLAP

OPERATED BY OUTPUT

O:1/2

GATE 1 FLAP

OPERATED BY OUTPUT

O:1/1

GATE 2 GATE 1

S2 S1

INPUT I:1/3

INPUT I:1/2

INPUT I:1/1

CONVEYOR

PRODUCTION BOXES

FIXED GATE

START(PB1)

STOP(PB2)

INPUT I:1/4

INPUT I:1/5

MOTORCM

O:1/3

RESET (PB3)

INPUT I:1/6

OPERATORSTATION

PLC PROGRAMMING EXAMPLEThe sequence of operation is as follows:

 1. When start button PB1 (I:1/4) on the operator control panel is pressed the conveyor motor CM (O:1/3) starts. (This is a 3 phase 415volt delta connected motor).

2. Gates 1 and 2 are in the open position, boxes are conveyed past sensor S1 (I:1/1) which counts 6 boxes into chute 1

via the fixed gate

3. Then Gate 1 (O:1/1) is closed and Sensor S2 (I:1/2) counts 6 boxes into chute 2.

4. Gate 2 (O:1/2) closes and Sensor S3 (I:1/3) counts a further 6 boxes in to chute3

5. The conveyor is then stopped automatically

PLC PROGRAMMING EXAMPLEAt any time the loading of the conveyor can be stopped via a manual push button PB2 (I:1/5) which is situated on the operator control panel. A reset button PB3 (I:1/6) is also provided to reset counters that are used in the system.

1) Sketch the wiring diagram between the system hardware and the appropriate PLC rack cards.

 2) Design the ladder logic required to implement the above sequence.

PLC PROGRAMMING EXAMPLEWIRING DIAGRAM

1) CREATE AN I/O LIST WITH ALLOCATIONS PROVIDED IN TEXT

2) DRAW THE INPUT AND OUTPUT MODULES SHOWING THE INPUTS AND OUTPUTS CORRECTLY CONNECTEDAS PER I/O MAP

3) REMEMBER THE 3 PHASE MOTOR CANNOT BE DIERECTLY CONNECTED TO THE PLC SO NEEDS TO BE INTERFACED VIA AN OUTPUT RELAY AND CONTACTOR

4) SHOW A BASIC 3 PHASE DOL DIAGRAM

PLC PROGRAMMING EXAMPLEPLC PROGRAM

FROM THE SEQUENCE DESCRIPTION PROVIDED IT SHOULD BE CLEAR THAT THE FOLLOWING ELEMENTS ARE NEEDED

1) A STOP /START CIRCUIT FOR THE MOTOR

2) 3 X ‘UP’ COUNTERS SET AT A COUNT OF ‘6’

3) SOME CONTROL LOGIC TO LIFT AND LOWER CHUTE 2 AND 3 GATES (CHUTE 1 IS FIXED)

 

HAVE A GO AT THIS QUESTION! TEST YOUR SOLUTION WITH LOGIXPRO

PLC PROGRAMMING EXAMPLE1) I/O LIST

I nput/ output list

I NPUT CARD DEVI CE OUTPUT CARD FI ELD I :1/ 1 SENSOR1 O:1/ 1 GATE 1

SOLENOI D I :1/ 2 SENSOR2 O:1/ 2 GATE 2

SOLENOI D I :1/ 3 SENSOR3 0:1/ 3 CONVEYOR

MOTOR I :1/ 4 START PB1 I :1/ 5 STOP PB2 I :1/ 6 RESET PB3

PLC PROGRAMMING EXAMPLESCHEMATIC WIRING DIAGRAM

I1Card 24V DC

Input

0

1

2

3

4

5

6

7

COM

24V DC

Output

01

3

4

5

6

7

COM

24V +

DC -

Start Stop

Reset

O1 Card

+ 24V

- DC

Conv. R 1

110V

A C

R1

M

3Phase415 volt

Motor Control Center

R1

To MCCConveyor

Motor starter

PLC

2

GATE 1

Green

Sensor1 Sensor 2

Sensor 3

SOL

SOL

CONVEYOR MOTOR CM

PLC PROGRAMMING EXAMPLE