Assignments Electrical Power Safety

Electrical power 1
SAFETY SYSTEMS
Authors Name/s per 1st Affiliation (Author)
Dept. name of the organization
Name of organization, acronyms acceptable
City, Country
mail address
Authors Name/s per 2nd Affiliation (Author)
Dept. name of the organization
Name of organization, acronyms acceptable
City, Country
e-mail address
Question 1

Electrical power 2
What is a proof-test and what is proof test coverage?
Proof test is a form of stress test which is conducted to show the fitness of a load-bearing
structure. Proof testing is very vital part of the lifecycle safety and it is very significant to ensure
the system achieve its specific safety integrity level in the safety lifecycle. While proof test
coverage is simply a measure of the number of undetected dangerous failure which can be
detected through the use of proof test. And when these failures are detected it becomes easier to
be tackled. And a perfect example of a proof test is nondestructive test like ultrasonic.
Question 2.
What is a Process Hazard Analysis (PHA) and who conducts them?
Process Hazard Analysis normally known as PHA is a systematic technique of
identifying as well as evaluating all the risks which can be involved in any particular industrial
process so as to control, reduce or prevent the hazard from happening completely [1]. And most
cases it is conducted by facilitator or team leader who work with workers who understands all
the processes of doing the Process Hazard Analysis [2]. Performing process hazard analysis is
good for the engineering practice and the company which processes very dangerous chemicals.
This will highly help to protect the employees, environment where this company is situated and
the public from causing accident. The PHA helps protecting against the property damage,
process downtime, product quality issues and the adverse publicity accident. For the PHA, a
good example is the use of chemicals to help in prevention of some accidents like fire [2]. There
are some chemicals which are highly inflammable while others can result to fire when reacted
with others.

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Question 3
What is MTBF and can it provide useful data for the calculation of PFDavg (average
probability of failure upon demand) considering perfect inspection, and if a constant
failure rate is assumed for the device?
Mean Time Between Failure is a measure of reliability of a hardware product or component.
And for several component this value is always given in thousands and some cases in tens of
thousands of hours between the failure [1]. This unit of measurement involves only the operation
duration between failures but it cannot include the duration for repair. And it is impossible for
the Mean Time Between Failure to provide useful data for the calculation of PFDavg. This is
because Probability of Failure on Demand average (PFDavg) is actually a probability which a
system will fail dangerously and not capable to conduct its function safety when needed. For
example a hard drive can have an average time between failure to be 270 000 hours Therefore t a
suitable MTFB can be employed as a quantifiable objective while designing another new hard
disk drive.
Question 4
a) An instrument has a MTTF of 18,000 hours and a MTTR of 24 hours. What is the
MTBF?
From the formula of availability;
A = MTTF
MTBF = MTTF
( MTTF+ MTTR ) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
. . . 1

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And since the values of MTTF and MTTR are given we can substitute them in equation 1
above and when that is done we obtain the following;
18,000
MTBF = 18,000
( 1800+24 )
18,000
MTBF = 18,000
(1824 )
MTBF= 18,000× 1824
( 1800 )
MTBF= 1824 hours
b. Availability
Using the above equation 1 we can modify it to have equation 2 as the availability as below;
A = MTTF
MTBF . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2
Given that
MTBF= 1824 hours
MTTF = 1800 hours
Then we can substitute these values in the equation 2 as below;
A = 1800
1824
A= 0.9868
Question 5: