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

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...

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...

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...

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...

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...

Thank you for all the help guys. Much appreciated. Cheers.

Hello there, Basic question for some, but confusing for me on logic and I am wondering if I am calculating wrong. I understand as per API 650 Section 5.5.2 the radial width of the ring shall be from the inside edge of the shell radially as follows: L=2tb Sqrt Fy/2YGH Our tank: Height 56foot...