ME502: Programmable Logic Controllers Systems Solution Analysis

This document presents a comprehensive solution to an Electrical Engineering assignment focused on Programmable Logic Controllers (PLC) systems. It begins with a detailed explanation of hazards and risks within PLC environments, including the Heinrich model and methods to mitigate major risks. The solution then explores the Manufacturing Message Specification (MMS) protocol and compares communication bus topologies such as Mesh, Star, and Bus. Further, the assignment analyzes different frameworks used in PLC projects, offering merits and shortcomings of each. It proposes a concurrent design framework and compares its advantages and disadvantages. The document also covers the 7 layers of the ISO OSI model, elaborating on the responsibilities of layers 3 and 4 in the TCP/IP protocol stack. Finally, it describes various earthing grid systems, including Series Single Point, Parallel Single Point, and 3D Earthing grids, with illustrative figures. This assignment provides a valuable resource for students studying PLC systems.

Electrical Engg.
Programmable logic controllers systems
Student Name –
Student ID –
Solution 1)
a) Hazard –
A hazard may be defined as the reason that can harm any target which is vulnerable. It can
lead to a damage or may harm the human beings, the environment or the property.
A hazard has no associated risk if no one is exposed to that hazard. A hazard has an incident
associated with it. If this incident leads to any unwanted results and its probability is taken
into account, the risk associated can be found.
Examples of hazards — soil acidification, increasing carbon dioxide levels, terrorism, drug
abuse. A hazard causes damage due to a stored energy present like electrical, mechanical,
thermal, chemical etc. or due to hazardous situations. A hazard is actually some source which
can have a damaging effect on someone or something. In PLC, the hazard is electrical in
nature.

An example of workplace hazard is electricity which causes harm like shocks etc. Risk may
be referred to as the probability that some person will experience the negative effect if he has
an exposure to some hazard.
b) Risk –
A risk may be defined as the probability that when something is exposed to a hazard then it
will have some negative results.
Risk consists of 2 factors — probability of occurrence of harm and severity of the harm. For
example – smoking the cigarette can lead to a risk to develop cancer.
Hence, a hazard is the agent causing harm and risk is the chance that the hazard will harm
someone.
c) Difference between a hazard and a risk –
A hazard has no associated risk if no one is exposed to that hazard. A hazard has an incident
associated with it. If this incident leads to any unwanted results and its probability is taken
into account, the risk associated can be found.
Examples of hazards — soil acidification, increasing carbon dioxide levels, terrorism, drug
abuse etc. A hazard causes damage due to a stored energy present like electrical, mechanical,
thermal, chemical etc. or due to hazardous situations. A hazard is actually some source which
can have a damaging effect on someone or something.

An example of workplace hazard is electricity which causes harm like shocks etc. Risk may
be referred to as the probability that some person will experience the negative effect if he has
an exposure to some hazard.
Risk consists of 2 factors — probability of occurrence of harm and severity of the harm. For
example – smoking the cigarette can lead to a risk to develop cancer.
Hence, a hazard is the agent causing harm and risk is the chance that the hazard will harm
someone.
d) Heinrich model ( probability of occurrence of a major risk ):
The Heinrich’s Law is given as follows:
At a workplace, for every accident which leads to a major injury, there exist 29 accidents
leading to minor injuries and 300 accidents leading to no injuries. Hence, it gives us the
relative values of cases in which major risk is involved. The Law helps to find the ratio as
well as probability of the number of cases with major risk as compared to the total number of
cases.
e) Efforts to decrease the probability of a major risk ( Heinrich model ) :
The Heinrich’s triangle (accident triangle) gives a theory for the prevention of industrial
accidents. It shows that if the count of the minor accidents is decreased, then the count of
serious accidents also reduces. The triangle gives a relationship between the count of
accidents which lead to major, minor and no injury . It can be used in the health and safety
programmes in various industries.

Solution 2 )
a) MMS replaces the TCP / IP protocol in industrial operations on an IEEE 802.3
network base.
The MMS ( Manufacturing Message Specification ) is a standard at the international
level. It deals with message system to transfer the real time process’s data and supervises
the information ( related to control ) in the various devices present on the network and
other computer applications too ( Alphonsus, 2016 ).
MMS stack did not become famous as it is difficult to implement the OSI ( Open Systems
Interconnection ) protocols. However , in the year 1999, a new version was created for
MMS by the use of IP ( Internet Protocol ). With the new structure of stack, MMS
standard has been accepted globally.
b) Communication bus topologies ( 3 ) and their sketches :
A total of 5 types of topology are present. They are: Mesh, Star, Bus, Ring and Hybrid.
Mesh Topology: In the mesh topology, every device is connected with every other device
present on the network using a point- to- point link. A single link is present between 2
devices. The number of links = n ( n – 1 ) / 2 if n is the number of devices present on the
network. It is reliable, has no data traffic issues and is secure. It needs large no of wires and
also there is problem of scalability.
Star Topology: In the star topology, every device is connected with the hub which is the
central device. It does not allow a direct communication between the devices. It has to do
communication using a hub. It is cheap, easy for installation , uses less number of cables , is

robust and the fault can be easily detected. But in case the hub stops working , the system
also stops . Also, the hub requires more maintenance .
Bus topology : In the bus topology , all the devices on the network are connected to a main
cable using drop lines . There is a restriction on number of drop lines and the distance . It is
easy to install and needs lesser number of wires . The fault is difficult to detect and has a
problem in scalability ,
Mesh
Star
Bus

Solution 3)
The 2 Frameworks which can be used by the individuals in the industry to complete the PLC
projects have been described here.
a)
Merits and Shortcomings of
2 given frameworks
Advantages ( Merits ) Disadvantages
( Shortcomings)
Framework A 1. Testing phase is good 1. No layout before
programming phase
Framework B 1. Predictive analysis is
used.
2. The cause and effect
understanding used.
1. Framework is not
structured.
2. The documentation
phase must come at
last. If the
documentation is
done at an earlier
stage, then if changes
are done at further
stage then document
may become wrong if
not updated.
b) Our framework :

A concurrent method can be used for the design framework. The various domains are :
Customer domain, functional domain, Product domain, process domain, resource domain and
control domain. All these domains have 3 types of designing concurrently – Product design
( PD ) , Manufacturing System Design ( MSD ) and Control System Design ( CSD ). This
framework is very efficient as it gives a clear distinction between the various domains. If the
process is made concurrent , then a lot of time is saved. The parallel processing helps to
decrease the time of execution as the various processes are executed parallely. But the design
may be complex due to the presence of parallel systems.
c)
Merits and Shortcomings of
given vs own frameworks
Advantages ( Merits ) Disadvantages
( Shortcomings)
Framework A & B 1. Its easy to design.
2. The steps are well
defined and easy to
understand.
3. Its easy to design
them.
1. It is a sequential
process.
2. It takes a long time in
the execution. Hence,
it is a slow process.
My Framework 1. It can be executed in
a parallel manner.
2. The time consumed is
lesser.
1. Its design is complex.
2. It requires more
hardware elements
and hence it is costly.

Solution 4 )
a) 7 Layers of the ISO OSI model:
The 7 layers of the ISO ( International Organization of Standardization ) OSI ( Open
System Interconnection ) model are :
Application Layer, Presentation Layer, Session Layer, Transport Layer, Network
Layer, Data link Layer and Physical Layer.
The various layers are described here.
1. Physical Layer :
It is positioned at the lowest level in the OSI Model. It starts , maintains and stops
a physical connectivity. The raw data is transmitted and received over a network
using this layer. In this layer, the data rates as well as the voltage levels are
defined. It also carries out the conversion of data in digital or analog form into
electrical or optical format.
2. Data Link Layer :
It helps in the synchronization of the information for transmission by the physical
layer. It ensures that error free data transmission occurs from a node to the next.
The data transmission and reception occurs in the form of frames.
Acknowledgement of receiving and sending the frame is also employed. It
establishes a control on the network by indicating once the frame buffer is full.
3. Network Layer :
It helps in routing a given signal through various channels. It helps to control the
network traffic. It makes a decision regarding the route to be followed by the data.

It converts the message which is outgoing into packet form and also joins the
incoming data in packet form to a message form.
4. Transport Layer :
The decision regarding the single path or parallel path for data transmission is
made here. Several functions are implemented. For example – segmentation,
Splitting etc. It collects messages sent by the Session layer and converts it to
smaller parts and sends to the network layer , so that it can easily handle the data.
It is complex.
5. Session Layer :
It carries out the synchronization and management of the communication between
2 applications. It helps to avoid any data losses.
6. Presentation Layer :
It helps to send data in a format that is understandable by the user. It acts as a
translator for languages. Various functions performed are : Data compression,
Data encryption, data conversion etc.
7. Application Layer :
It is present on top. It contains the application programs. The various functions are
: mail service, network resources etc.
b) Responsibilities of layer 3 and 4 in a TCP/IP protocol stack:
The Layor 3 ( Network layer ) decides the manner in which the data is sent towards the
receiver. Its functions are – routing, addressing, packet forwarding. It is also known as the

host to host layer. It transmits the data as packets. It finds out the best path out of various
options available for delivering the data. It helps in routing a given signal through various
channels. It helps to control the network traffic. It makes a decision regarding the route to be
followed by the data. It converts the message which is outgoing into packet form and also
joins the incoming data in packet form to a message form.
The layer 4 ( Transport layer ) helps in establishing the co-ordination for transferring the data
between the hosts and the systems. It helps in checking the errors and recovering the data. It
helps in segmentation of the data . The segment consists of source and destination number, a
data element and a sequence number. It ensures that segment will reach the right application
and also various segments are delivered in the right order. It can change the transmission rate
of data if the 2 devices do not have same speed. The decision regarding the single path or
parallel path for data transmission is made here. Several functions are implemented. For
example – segmentation, Splitting etc. It collects messages sent by the Session layer and
converts it to smaller parts and sends to the network layer , so that it can easily handle the
data. It is complex.
Solution 5)
a) Series Single Point , Parallel Single Point and 3D Earthing grid systems :
The figure 1 shown below represents a Series Single Point Earthing grid system. The
figure 2 shown below represents a Parallel Single Point Earthing grid system. The
figure 3 shown below represents a 3D Earthing grid system.

Figure 1
Figure 2
Figure 3
b) All the earthing cables in a Parallel Single Point earthing system must be insulated.

The earthing cables in the parallel single point earthing system need to be insulated.
This is done to prevent corrosion and transposition, when the cables have not been
transposed. This is done to prevent any circulating currents or any losses ( if normal
conditions of operation are used ).
c) Ground loops :
Ground loops lead to the production of noise or interference in the computer systems.
To provide protection against the ground loops, some wiring practices are used. It
means that all the signal circuits which are vulnerable must be referenced to one point
only ( as the ground ).
A ground loop is formed due to any interconnection in the electrical equipment used.
This can generate different paths to the ground. Hence , a closed conducting loop may
be formed . Any stray AC magnetic fields can also cause Ground loop to be induced.
d) Hardwired Safety Circuits : Hardwired safety circuits are being replaced by safety PLCs
and safety networks. This is being done to reduce costs and errors as well as to increase
flexibility. The process of troubleshooting the hardwired safety circuits was a time taking
process. Any change in logic was not so easy task.
d) Static and Dynamic tests : Static test examines the code without actually executing it.
It is done at the verification stage. Dynamic test executes and tests the code without
actually examining it. It is done at the validation stage.