Criteria For Continued Operation

[canadianinspector]

Hello all,
API 653 Section 4.3.2
I'm a little confused on how to apply the T1 and T2 values determined by UT and averaging over the critical length L calculated.
T1 is min Taverage within L
T2 is your least minimum thickness in the corroded area exclusive of pits.


I have a corroded shell of my tank and Tmin shall not be less than 0.100" for any tank course. (calculated tmin is not above the 0.100" so my Tmin is 0.100")

If Criteria for continued operation is as follows:
1. T1 shall be greater than or equal to Tmin.
How could this ever fail since averaging will always raise the value above 0.100"?

2. T2 shall be greater than or equal to 60% tmin (and add CA if any to both).
How is this applied does this mean my T2 can actually be 0.060" (60%ofTmin) for this one isolated area?

Thank you for the help in advance.

 

[HTURKAK]

I have copied and pasted the relevant section;

e. The criteria for continued operation is as follows:
I. The value t1 shall be greater than or equal to tmin (see 4.3.3 or 4.3.4), subject to verification of all other loadings
listed in 4.3.3.5; and

ii. The value t2 shall be greater than or equal to 60 percent of tmin; and

iii. Any corrosion allowance required for service until the time of the next inspection shall be added to tmin and 60 percent of tmin.

If you determine the t1 and t2 as per 4.3.2, and calculated the tmin as per (4.3.3 Minimum Thickness Calculation for Welded Tank Shell)

i) t1 shall be greater or equal to tmin,
ii) t2 shall be greater or equal to 60% tmin.

If you fulfill the both criterias and determine the tmin adding the CA until the next future inspection ( normally 10 yrs later ) the assessment will pass..


How did you calculate the tmin=0.100" ?. If this is a real case, pls share some details (tank size,material , content, t min, t1 ,t2) to get better responds.

 

[canadianinspector]

Hello again,
Ok I think I have enough information to show you.

Tank material is SA-283-C at upper course corrosion
H=Actual height is 24.208 Ft, used H as 23.208 FT for thinned area.
D=43Ft
G=0.9
E=1.0
St upper 26000
CA is 0.0625" not required until next interval
Current corrosion rate of 3.1846 MPY (0.0031846)
My next UT external inspection interval is calculated at 1.88 years, but visual is 0.9 years
T2 - 0.112"
T1 - 0.130"

Tmin - 2.6HDG/SE - using Locally thinned shell course calculated as 0.0898", therefore using 0.100"(code min)


Correct me if I am wrong....
T1 - Despite averaging in "L" as per 4.3.2 which typically will raise your isolated minimum T2 UT reading making it thicker than your TMIN calculation. I understand the intent is really to look at the corrosion rate and see if it will make it to the next proposed inspection date when using the averaged thickness. My T1 is 0.130" My next UT interval of inspection has been calculated at 1.88 years.
So - 1.88x 0.0031846= 0.00598 further loss in that interval.
So - 0.130-0.00598=0.124" I know it passes either way, but is this compared to code min of 0.100" or do I use calculated tmin 0.0898"?

T2- 60% Tmin, so T2 can be 60% of 0.100"(code min) which is 0.060"? or is it 60% of the calculated Tmin which is 60% x 0.0898" = 0.054" ??
I'm confused by this one because no course shall be under 0.100".

Thank you for all that may respond.

 

[HTURKAK]

I looked to your last response ..

I think you are in confusion with 4.3.3,
4.3.3.1 The minimum acceptable shell plate thickness for continued service shall be determined by one or more of the methods noted herein. These methods are limited to tanks with diameters equal to 200 ft or less.
a. When determining the minimum acceptable thickness for an entire shell course, tmin is calculated as follows:
tmin= 2.6 *(H-1) DG/SE for your case, for lowest course tmin =2.6*(23.2-1)*43*0.9/23600 = 0.094 in.

b. When determining the minimum acceptable thickness for any other portions of a shell course (such as a locally thinned area or any other location of interest), tmin is calculated as follows:
tmin = 2.6HDG/SE Three cases are defined for H;

H = height from the bottom of the shell course under consideration to the maximum liquid level when evaluating an entire shell course, in ft;
for this case , tmin =2.6*23.2*43*0.9/23600 =0.0989 in

2nd case, H=height from the bottom of the length L (see 4.3.2.1) from the lowest point of the bottom of L of the locally
thinned area to the maximum liquid level, in ft;
If we assume the corroded area at the bottom course ,and L around bottom, tmin= 2.6*23.2*43*0.9/23600 =0.0989 in.

3rd case ,height from the lowest point within any location of interest to the maximum liquid level, in ft,
for this case, with knowing the location of corroded area , you may calculate tmin.

The assumption of tmin=0.1 in conservatively correct.

i) t1 shall be greater or equal to tmin, T1 = 0.130" > 0.1"

ii) t2 shall be greater or equal to 60% tmin. T2= 0.112" >> 0.6*0.1


Both requirements are fulfilled ..the assesmnet will pass !!

 

[HTURKAK]

I was planning to look preview , but with mistake i pressed submit..

When the second internal inspection will be performed ? and what is the current corrosion rate of for internal surface ?

With assuming exterior corrosion will govern, , 3.1846 MPY (0.0031846)

CA=0.0031846*1.88 CA is 0.006"

Tmin=0.0989 + 0.006 = 0.105 > The use of Tmin= 0.1 in is O.K.

 

[canadianinspector]

Thank you Hturkak
just a couple more clarifications
My tank is only 3 courses high. 23,600 btm stress, 26,000 upper course stress.

Your comments:
a. When determining the minimum acceptable thickness for an entire shell course, tmin is calculated as follows:
tmin= 2.6 *(H-1) DG/SE for your case, for lowest course tmin =2.6*(23.2-1)*43*0.9/23600 = 0.094 in (which will be 0.100")
Yes I agree with this calculation for the bottom course but my issue is at the top of the tank.

My tank actual height as stated above is 24.208' and the corrosion for my case is at the top course at 23.208' (1 foot from the top)
Would I not use the course upper stress of 26,000 as I originally posted?
Therefore for the entire 3rd shell course tmin =2.6*(8.208-1)*43*0.9/26000 = 0.02789 in (again Tmin shall be code of 0.100")

Again using the second case:
And same applies to upper course locally thinned S=26000
H=height from the bottom of the length L (see 4.3.2.1) (which is 23.2 in my case)from the lowest point of the bottom of L of the locally thinned area to the maximum liquid level, in ft; in this case is 1 foot difference from the top of the tank.
tmin = 2.6HDG/SE. The corroded area is at the top course, and bottom of L,
tmin= 2.6*1*43*0.9/26000 =0.00387in. (Tmin shall be code of 0.100")

Got confused at what you wrote. Is this what you meant?
Loss of wall thickness expected in 1.88 years is 0.006 due to corrosion within that interval
i) t1 shall be greater or equal to tmin, T1 = 0.130" - 0.006" = 0.124" therefore is > 0.1" and will make it to the next interval

ii) t2 shall be greater or equal to 60% tmin. T2= 0.112" - 0.006"=0.106" >> 0.6*0.1 and will make it to the next interval

There is no Corrosion Allowance left as it is fully consumed.

No internal inspection at this time.

Thank you for your comments.

 

[HTURKAK]

When determining the minimum acceptable thickness for an entire shell course, tmin is calculated as follows:
tmin= 2.6 *(H-1) DG/SE for your case, for lowest course tmin =2.6*(23.2-1)*43*0.9/23600 = 0.094 in (which will be 0.100")
Yes I agree with this calculation for the bottom course but my issue is at the top of the tank.

Each course ht should be 96 in. For the top course , tmin =2.6*(7.06)*43*0.9/26000 =0.027 in for the entire shell course . Notice that i did not reduce ( H ) one foot again, since the all courses have the same thk.( one foot method is valid for the courses course where the wall thk. increases at lower course ).

For this tank, the top course and middle course SE= 26000 psi.


This calculation is true for the described locally thinned area .


This calculation is based on the corrosion rate that you have provided and the check is performed for the base course..

As far as i understand , the routine inspection included only inspection of the tank's exterior surfaces. If you find evidence of
corrosion tou can repair the exterior surface with paint coatings...what about inner surface ? .

 

[canadianinspector]

Hello Again,
Thank you so much I am more an track than I thought after you clarified.

Yes only external inspection has been performed at this time and the internal corrosion was found by external UT methods.
Currently there is no external corrosion and the surface is easily scanned when access is given.
Internal inspection is not available at this time.

One last question which revolves around the t2 shall be greater or equal to 60% tmin, and in my case T2= 0.112" >> 0.6*0.1

How does the "tmin shall not be less than 0.1" for any tank course" apply for T2?
Am I allowed for this corrosion location to reach 0.060" which is 60%of the minimum course tmin (0.100")

Thank you,

 

[canadianinspector]

Sorry I don't see how you got 7.06"
Each course ht should be 96 in. For the top course , tmin =2.6*(7.06)*43*0.9/26000 =0.027 in for the entire shell course . Notice that i did not reduce ( H ) one foot again, since the all courses have the same thk.( one foot method is valid for the courses course where the wall thk. increases at lower course ).

Yes my 3 courses are 96" (8 feet) and thicker at the bottom and thinner at the top.
I took the tank height as per the dwg right to the top of rafters which is the added 2.5" (0.208 feet)

 

[HTURKAK]

Definition of t2 = least min. thickness in entire area, exclusive of pits.

tmin=(2.6 HDG)/SE

tmin = the minimum acceptable thickness, in in. for each course as calculated from the above formula; however, tmin shall not be less than 0.1 in. for any tank course.
4.3.2.1 (.....e. The criteria for continued operation is as follows:
I. The value t1 shall be greater than or equal to tmin (see 4.3.3 or 4.3.4), subject to verification of all other loadings
listed in 4.3.3.5; and
ii. The value t2 shall be greater than or equal to 60 percent of tmin; and
iii. Any corrosion allowance required for service until the time of the next inspection shall be added to tmin and
60 percent of tmin).


Your Tank Diameter 43 ft < 50 ft Nominal Plate Thickness shall be min. 3/16 in.
H = height from the bottom of the shell course under consideration to the maximum liquid level when evaluating
an entire shell course, in ft;

As per your explanation,( H=Actual height is 24.208 Ft, used H as 23.208 FT for thinned area....and 3 courses are 96" (8 feet) and thicker at the bottom and thinner at the top.) For the top course ,in this case, H =23.08-2*8-1.0=6.08 tmin =2.6*(6.06)*43*0.9/26000 =0.023 in for the entire shell course .However, the min acceptable tmin=0.1 in.

 

[canadianinspector]

Thank you again for taking the time out of your day and being so patient.

 

[HTURKAK]

You are welcome.. next time , provide more info to get clear respond... the full picture was clear for me, after i saw your fourth respond...