# Thread: Finding design life from condition

1. ## Finding design life from condition

Hi, I have an pipe which can have a condition rating from 1-5. 1 being perfect and 5 being worst. It is at 8yrs old and has a condition rating of 3.5.
What formula can I use to calculate to calculate where this pipe design life finshes? It is not straight line, it is an ever decreasing curve.

Any help is most appreciated.
Regards Hayden

2. Hello, Hayden!

I have an pipe which can have a condition rating from 1-5.
1 being perfect and 5 being worst.
It is at 8 yrs old and has a condition rating of 3.5.

What formula can I use to calculate to calculate where this pipe design life finshes?
It is not straight line, it is an ever decreasing curve.

With no more information, we must guess the form of the function.

Assuming it is an exponential function, we can solve it.

The condition function would have the form: . $C \:=\:C_o\,e^{at}$
. . where $C_o$ is the initial rating and $a$ is a constant to be determined.

Assuming that a brand-new pipe is perfect, we have: $C_o = 1$
. . So the function is: . $C \:=\:e^{at}$

Your pipe is 8 years old $(t = 8)$ and has a rating of $C = 3.5$
. . The equation becomes: . $3.5 \:=\:e^{8a} \quad\Rightarrow\quad 8a \:=\:\ln(3.5)$
. . Hence: . $a \:=\:\frac{\ln(3.5)}{8} \:\approx\:0.1566$

The condition function is: . $\boxed{C \:=\:e^{0.1566t}}$

I assume that, when the pipe's condition has a rating of 5, it is replaced.
When does this happen? .When is $C = 5$ ?

We have: . $e^{0.1566t} \:=\:5 \quad\Rightarrow\quad 0.1566t \:=\:\ln5 \quad\Rightarrow\quad t \:=\:\frac{\ln5}{0.1566} \:=\:10.27738131$

Hence, we can expect to retire any new pipe in about 10.3 years.

Your pipe has about 2.3 more years of useful life.

3. ## Finding design life from condition

Thanks Soroban!

I'm not sure what one part means though:

1n(3.5) what does the 'n' do?

Regards Hayden

4. Originally Posted by ill_comms
Thanks Soroban!

I'm not sure what one part means though:

1n(3.5) what does the 'n' do?

Regards Hayden
No, it NOT "1n(3.5)". It is "ln(3.5)"- the letter l, not the number 1. And it means "natural logarithm".