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Ksivusya [100]
3 years ago
6

1. Using a typical frequency value for the initiating event and PFD values provided in class lectures, estimate the mishap or co

nsequence frequency for a cooling water failure if the system to be cooled is provided with the following IPLs: (1) Personnel action with a 10 min response time, and (2) A basic process control system (BPCS).
Engineering
1 answer:
algol133 years ago
5 0

Answer:

a basic process control system is necessary as long as the work by the personnel in charge is efficient, this reduces by 90% the errors or failures to calculate the system to operate

Explanation:

Production staff human error

 Another reason why a breakdown can occur is due to an error of the production personnel. This error, in turn, can originate from:

Error of interpretation of an indicator during the normal operation of the equipment, which causes the operator or driver of the installation to make the wrong decision

Incorrect performance in the event of a machine failure. For example, introducing water into a hot boiler in which water has been lost visually; as it is not known how much water is inside, it may be empty and hot, so when water is introduced it will produce instant vaporization, with the consequent increase in pressure that can even cause the boiler to explode.

Physical factors of the operator: this may not be in perfect condition to carry out his work, due to dizziness, sleep, fatigue accumulated by long working hours, illness, etc.

Psychological factors, such as demotivation, problems external to work, etc., greatly influence the proliferation of operational errors

Lack of clear systematic instructions, such as procedures, technical instructions, etc., or poor implementation of these  Lack of training

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Read 2 more answers
Which of the following ranges depicts the 2% tolerance range to the full 9 digits provided?
Lyrx [107]

Answer:

the only one that meets the requirements is option C .

Explanation:

The tolerance of a quantity is the maximum limit of variation allowed for that quantity.

To find it we must have the value of the magnitude, its closest value is the average value, this value can be given or if it is not known it is calculated with the formula

         x_average = ∑ x_{i} / n

The tolerance or error is the current value over the mean value per 100

         Δx₁ = x₁ / x_average

         tolerance = | 100 -Δx₁  100 |

bars indicate absolute value

let's look for these values ​​for each case

a)

    x_average = (2.1700000+ 2.258571429) / 2

    x_average = 2.2142857145

fluctuation for x₁

        Δx₁ = 2.17000 / 2.2142857145

        Tolerance = 100 - 97.999999991

        Tolerance = 2.000000001%

fluctuation x₂

        Δx₂ = 2.258571429 / 2.2142857145

        Δx2 = 1.02

        tolerance = 100 - 102.000000009

        tolerance 2.000000001%

b)

    x_average = (2.2 + 2.29) / 2

    x_average = 2,245

fluctuation x₁

         Δx₁ = 2.2 / 2.245

         Δx₁ = 0.9799554

         tolerance = 100 - 97,999

         Tolerance = 2.00446%

fluctuation x₂

          Δx₂ = 2.29 / 2.245

          Δx₂ = 1.0200445

          Tolerance = 2.00445%

c)

   x_average = (2.211445 +2.3) / 2

   x_average = 2.2557225

       Δx₁ = 2.211445 / 2.2557225 = 0.9803710

       tolerance = 100 - 98.0371

       tolerance = 1.96%

       Δx₂ = 2.3 / 2.2557225 = 1.024624

       tolerance = 100 -101.962896

       tolerance = 1.96%

d)

   x_average = (2.20144927 + 2.29130435) / 2

   x_average = 2.24637681

       Δx₁ = 2.20144927 / 2.24637681 = 0.98000043

       tolerance = 100 - 98.000043

       tolerance = 2.000002%

       Δx₂ = 2.29130435 / 2.24637681 = 1.0200000017

       tolerance = 2.0000002%

e)

   x_average = (2 +2,3) / 2

   x_average = 2.15

   Δx₁ = 2 / 2.15 = 0.93023

   tolerance = 100 -93.023

   tolerance = 6.98%

   Δx₂ = 2.3 / 2.15 = 1.0698

   tolerance = 6.97%

Let's analyze these results, the result E is clearly not in the requested tolerance range, the other values ​​may be within the desired tolerance range depending on the required precision, for the high precision of this exercise the only one that meets the requirements is option C .

4 0
3 years ago
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