revision sept 2014 elx305 topics reliability scada pneumatics boolean algebra timers and timing...
TRANSCRIPT
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