14 replies to this topic

[fallah]

Dears,

Just clarification of reliability calculation by means of an example:

Suppose for a specified capacity we select two options of pumping systems as follows:

1-one running pump (100%)+one stand by pump (100%)
2-two running pumps (2*50%)+one stand by pump (50%)

and also suppose individual reliability of each pump would be 0.9

Now we want to compare the reliability of two above systems:

For system 1: 1-(1-0.9)^2=0.99 means 0.09 value of increasing the reliability with stand by pump

For system 2: Two 50% pumps give reliability of 0.81 (0.9*0.9) since both pump are required.If we add 2*50% pump reliability would increase to 0.964 {1-(1-0.81)^2=0.964} and since we actually add one 50% pump we can gain half of the reliability increasing value means (0.964-0.81)/2=0.077
Thus the reliability of system 2 would be 0.81+0.077=0.887

Above calculations shows reliability decreasing in system 2 by around 0.1 with respect to system 1

I am not sure the above calculations/conclusion is completely right.Any comment/advise would be welcomed.

Regards

[kkala]

Assuming P=0.9 as probability of pump (A) to be operable, probability of same pump to be non operable (A*) is 1-P=0.9.
System 1, pumps A, B, calculated probabilities:
Both operable AxB=0.9x0.9 = 0.810
Only one operable AxB* + A*xB = 2x0.9x0.1=0.180
None operable A*xB* =0.1x0.1= 0.010
Total of above probabilities 1.000
At least one pump operable 0.810+0.0180 = 0.990 (agrees to figure by fallah)
System 2 , pumps A, B, C, calculated probabilities:
All three operable AxBxC = 0.9x0.9x0.9 = 0.729
Only two operable AxBxC*+AxB*xC+A*xBxC = 3x0.9x0.9x0.1 = 0.243
Only one operable AxB*xC*+A*xBxC*+A*xB*xC = 3x0.9x0.1x0.1 = 0.027
None operable A*xB*xC* = 0.1x0.1x0.1 = 0.001
Total of above probabilities 1.000
At least two pumps operable 0.729+0.243=0.972

So System 1 seems to have an advantage over System 2, on the condition that 100% of full flow is necessary. If part of this flow is useful, System 2 can be more advantageous, seeing that in this case relevant probability is 0.972+0.027=0.999 for System 2 (versus 0.990 for System 1).

Note: I remember a similar to System 1 example (for Plant reliability) using normal distribution, with results slightly different to the above procedure .

Edited by Art Montemayor, 04 April 2012 - 01:44 PM.

[fallah]

As i understood from above, you mean probability of operation of number of operating pumps among a pumps system is equal to mentioned pumps system reliability.Do you confirm?

Edited by Art Montemayor, 04 April 2012 - 01:44 PM.

[kkala]

As i understood from above, you mean probability of operation of number of operating pumps among a pumps system is equal to mentioned pumps system reliability.Do you confirm?

Not being familiar with a quantitative expression of reliability, I have assumed it as the probability of the system to be operable (able to operate). I think this complies with your text.
Reliability of pump A = 0.9 is considered as probability = 0.9 of this pump to be operable (thus probability of this pump to be not operable = 0.1). This concept of probability alone would be sufficient for yesterday calculations. And I believe it is same as operational availability of pump A = 0.9.
Calculations are based on this Probability only, e.g. in system 1 (two pumps): the event of one pump operable and one not (only one operable) has a probability of 0.9*0.1 + 0.1*0.9 = 0.18 ; the event of both pumps operable has a probability of 0.9*0.9 = 0.81 ; the event of at least one pump operable has a probability of 0.99 (sum of above); it can be also calculated as 1 - 0.001 (none operable) = 0.99. System 2 has 3 pumps, so more combinations are needed..

Your statement is confirmed (does not say something contrary to the above, hopefully being useful clarifications).

[fallah]

As i understood from above, you mean probability of operation of number of operating pumps among a pumps system is equal to mentioned pumps system reliability.Do you confirm?

Not being familiar with a quantitative expression of reliability, I have assumed it as the probability of the system to be operable (able to operate). I think this complies with your text.
Reliability of pump A = 0.9 is considered as probability = 0.9 of this pump to be operable (thus probability of this pump to be not operable = 0.1). This concept of probability alone would be sufficient for yesterday calculations. And I believe it is same as operational availability of pump A = 0.9.
Calculations are based on this Probability only, e.g. in system 1 (two pumps): the event of one pump operable and one not (only one operable) has a probability of 0.9*0.1 + 0.1*0.9 = 0.18 ; the event of both pumps operable has a probability of 0.9*0.9 = 0.81 ; the event of at least one pump operable has a probability of 0.99 (sum of above); it can be also calculated as 1 - 0.001 (none operable) = 0.99. System 2 has 3 pumps, so more combinations are needed..

Your statement is confirmed (does not say something contrary to the above, hopefully being useful clarifications).

First,small correctin on above highlighted:

1-0.01 (nonoperable)=0.99 that 0.01=0.1*0.1

Second,i presented system reliability calculation in my first post as : System Reliability=1-(1-A)^B

A=Reliability of one member of the system,here pumps system
B=Number of members in the system

Of course,both calculations of mine and yours result in the same value for system 1.What i am not sure to be correct is my reliability calculation for system 2 which i want to know about.

Edited by fallah, 03 April 2010 - 04:09 AM.

[kkala]

1. Correction accepted (apparent error).
2. You have not explained "reliability" in words, mathematical formula has now explanations. Calculation done through probabilities seems correct to me, for following reasons.
a. For three pumps (System 2) reliability is 1-(1-0.9)^3=0.999, according to above formula. This means probability=0.999 of "at least one pump operable" (as previously indicated through "probabilities") and would be good if you had two spare pumps installed. But you want at least two out of three pumps operable, which gives probability=0.972 according to "probabilities". Above reliability formula seems not to cover this case.
b. Probability of at least two pumps operable can be also calculated as 1 - probability of (none or only one pump operable) = 1 - (0.001+0.027) = 0.972 (same result).
c. Concerning the original calculation (figures can be operational probability or reliability):
- 0.81 is good, if two 50% pumps are only installed.
- 0.9639 ~ 0.964 would be good, if two "isolated" pairs of 50% pumps were installed ("isolated" = a pump from one pair could not be combined with a pump from the other).
- interpolation between correct results for 2 and 4 installed pumps does not ensure precise value for 3 installed pumps.
-I guess "probability to be operable" or "reliability" or "operational availability" are same as concepts, at least in the way used here. So there may be another reliability definition covering mentioned System 2.
3. It is noted that System 2 has an advantage over System 1, in case that part of 100% flow is still useful (see bottom of probability calculations).

[fallah]

b. Probability of at least two pumps operable can be also calculated as 1 - probability of (none or only one pump operable) = 1 - (0.001+0.027) = 0.972 (same result).

After further investigation seems what i want to obtain regarding the reliability of two out of three identical pumps (3*50% with one as spare) could be resulted in according to your above procedure,i.e. 0.972 considering reliability of 0.9 for individual pump.It can be observed reliability reduced by the amount of 0.018 (0.99-0.972) with regard to 2*100% pumps.

Edited by fallah, 04 April 2010 - 05:08 AM.

[kkala]

After further investigation seems what i want to obtain regarding the reliability of two out of three identical pumps (3*50% with one as spare) could be resulted in according to your above procedure,i.e. 0.972 considering reliability of 0.9 for individual pump.It can be observed reliability reduced by the amount of 0.018 (0.99-0.972) with regard to 2*100% pumps.

Also note that System 2 (3x50% pumps) has an additional probability = 0.027 of only one 50% pump operable (for centrifugal pumps this can result in, say, 60% of flow), while System 1 (2x100% pumps) does not have this chance. It is an advantage of System 2 in case that this reduced flow can be used.
In other words: Probability to have 50% (or a bit more) of flow is 0.990 for System 1 versus 0.999 for System 2.

Edited by kkala, 04 April 2010 - 09:44 AM.

[fallah]

Probability to have 50% (or a bit more) of flow is 0.990 for System 1

Is above statement could be correct for 2*100% pump system?

[kkala]

Probability to have 50% (or a bit more) of flow is 0.990 for System 1

Above statement could be correct for 2*100% pump system?

Yes, according to "probabilities" calculation. The statement is equivalent to "at least one pump is operable" in the specific case under consideration for both Systems 1 and 2. Reliability formula gives same results, too.
In other words: To realize 50% (or a bit more) of flow, one out of two pumps must be operable for System 1, versus one out of three pumps for System 2.
This differs to the requirement of realizing 100% of flow, where one out of two pumps must be operable for System 1, versus two out of three pumps for System 2.

Edited by kkala, 05 April 2010 - 03:47 AM.

[fallah]

In other words: To realize 50% (or a bit more) of flow, one out of two pumps must be operable for System 1, versus one out of three pumps for System 2.
This differs to the requirement of realizing 100% of flow, where one out of two pumps must be operable for System 1, versus two out of three pumps for System 2.

Do you mean probability to have 0%-100% flow by system 1 would be 0.99?

[kkala]

Do you mean probability to have 0%-100% flow by system 1 would be 0.99?

Any calculated probability here concerns the satisfaction of the relevant requirement stated.
Concerning specifically System 1 (two pumps):
-Required flow 100% : probability =0.990
-Required flow 50% (or a bit more): probability =0.990
Both are realized through the event "at least one pump operable", irrespectively of the fact a discharge valve has to partially close to obtain 50% flow. Flow will be zero if "no pump operable".

Specifically for System 1, "probability=0.99 to have flow from minimum to 100%" is true. It is realized through the event "at least one pump operable" (playing with discharge valve).

[Zauberberg]

Hello Fallah

Sorry to come late on this one, I have consulted with Reliability Engineer just to make things 100% clear to me as well. I'm quoting from the "Reliability of Mechanical Systems", published by The Institution of Mechanical Engineers, London, UK, 1994:

- 2x100% case ("100% spare"): The system availability is = 1 - (1-0.9)*(1-0.9) = 99.9%
- 3x50% case ("2 out of 3"): The system availability is = (0.9)^3 + 3*(0.9)^2*(1-0.9) = 97.2%, the formula is A = P^3 + 3P^2F where

P = probability that the item is working
F = probability that the item has failed, F = 1 - P

Availability is accounted only in those cases when 100% performance is available. The same approach we use for plant availability - we are interested in 100% production, anything less than that means that the plant is not available.

Best regards,

[fallah]

Dear Zauberberg,

Thanks for your response,

First let me make a small correction on above highlighted 1 - (1-0.9)*(1-0.9) =99.0%

Second,despite of this fact that the values obtained by above procedure are the same as what we (KKALA and i) conveged on in previous posts,i think availability differs from reliability in concept and following that in value.

Regards

Edited by Art Montemayor, 06 April 2010 - 06:31 AM.

[Zauberberg]

First let me make a small correction on above highlighted 1 - (1-0.9)*(1-0.9) =99.0%

Haha thank you Fallah, it looks like my Math skills have gone for good

Cheers