An operator has to monitor the pressure in a pressure vessel in a distillery. Th

Question

An operator has to monitor the pressure in a pressure vessel in a distillery. The level and the pressure of the contents of the vessel have to be kept within a narrow range of safety. Pressure is indicated by dials on a control panel. If the pressure increases above the safe limit, a relief valve opens automatically and lowers the pressure. The valve stays open until the pressure is safe. If the relief valve fails to open when it should, an auditory alarm sounds to alert the operator. The automatic system might fail for several reasons:

* The valve is stuck in the closed position: probability 0.0025

* There is a power failure to the valve mechanism: probability 0.0003

It is the operator’s job to monitor the displays and respond to any alarms. The operator might fail for a number of reasons:

* The auditory alarm fails: probability 0.0009

* Operator responds incorrectly: probability 0.01

* Operator absent from the control room at the time: probability 0.005

a) What is the probability that the safety system will fail on those occasions when the pressure exceeds the operating range?

b) If the pressure exceeds the safe limits approximately 3 times per week, what is the annual probability of safety system failure? If the consequences of failure are life threatening, do you find this acceptable?

c)The company decides to improve the safety of its pressure vessel monitoring and control system. It does this by introducing automatic level control system that drains the tank when the pressure exceeds critical (which will happen if the relief valve fails). So, if the relief valve is stuck closed or if there is a power failure to the valve, the tank will be drained. If the probability of failure of the level control system is 0.004, how will this improve the safety of the overall system?

Explanation / Answer

(a)

The probability of automatic system failure will be

P(automatic system failure) = 0.0025+0.0003 = 0.0028

The probability of manual system failure will be

P(manual system failure) = 0.0009+0.01+0.005 = 0.0159

The probability that the safety system will fail on those occasions when the pressure exceeds the operating range is

P(safety system fails) = 0.0028*0.0159 = 0.00004452

(b)

Since pressure failure approximetaly 3 times a week so weekly probability of system failure is 0.00004452*3= 0.00013356

And since there are 52 weeks in a year so annual probability of sustem failure is 0.00694512.

(c)

Now the probability of automatic system failure will be

P(automatic system failure) = 0.0025+0.0003 +0.004 = 0.0068

The probability of manual system failure will be

P(manual system failure) = 0.0009+0.01+0.005 = 0.0159

So

P(safety system fails) = 0.0068*0.0159 = 0.00010812

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