You said in your video that the maximum hoop stress is found at 1 foot from the bottom of the course. Can you please provide me with the reference that said that
First reference is the code itself. In order to get the maximum thickness u need to put height where u will get the maximum stress. Code assumed it 1 ft above the bottom. Actually we have welding on top and bottom of course and hydrostatic forces in between that forces the shell outwards
Nice Work. Can you clarify - to convert thickness from m to mm - you need to multiply by 1000 (not divide) - so you get 10^6 on the numerator. Allowable stress S needs to be in Pa in Barlows equation. A typ value for A516M material would be 137 MPa. So when we convert S from MPa to Pa you get 137 x 10^6 on the numerator. So the 10^6 terms in the numerator and denominator cancel each other.
Actually first I brought the right side 1000 to left side. So it goes in divide I.e denominator. Still the thickness is in meters. Then in 2nd step I converted m into mm. Then both 1000 cancels each other
Thank you for the nice explanation. I would like to understand the design stresses bit though. How do we come to arrive at the design stresses during calculation?
If it is steam heated. It is like a multi pass u loop pipe. it can be employed both inside api 650 tank or asme viii vessel. There will be 2 design calc. 1. Heat transfer calc (thermal design) 2. Strength calc as per asme viii div 1 (but done manually for each of coil components like pipe, flange etc.
Let suppose u hv a tank with total height 10m With 5 shell courses each having 2m height 1. For bottom most shell put height H = 10M 2. For 2nd shell from bottom put height H = 8m 3. For 3rd shell from bottom put height H = 6m .........
Why is Hydrotest thickness different. The hydrotest shall be performed on the same tank and not a different tank, right? And the tank must have been manufactured with the design thickness, right?
Let suppose u have to store crude oil SG = 0.8 And hydrotest is usually with water SG = 1 Then u will calculate both tt and td. Finally u have to take the higher value of thickness of tt and td. This means u covered design for both cases.
Dear Sir, If I want to calculate the Thickness of the shell from the second course from the bottom, Will that make any difference in the value of Height ? (Will the height of the bottom or base plate be deducte in this case ?)
If tank height is 10m with 5 courses of 2m each, then H for bottom course will be 10m For 2nd course from bottom it will be 8m 3rd course from bottom will be 6m and so on. In the formula by api 650: 1-ft or 0.3 is subtracted for each course height due to the reason explained in the video. Bottom plate thickness is not included in height calculation
I have a question. Please help me understand this formula in front of me to calculate the shell retirement thickness. 160' tank. H= 38'. G=0.85. Sd= 20,000psi. Each shell course is 8ft wide. what is the retirement thickness for the 3rd course? td= 2.6 X 160' X (38' - 1) X 0.85 / 20,000psi is as far as I can go... where do I add in that each course is 8ft wide and to calculate for the 3rd course retirememt thickness? do i replace the H=38ft for H=24ft to satisfy the 3rd course?
Please help me answer the question.. Tank diameter 200 ft,Total height 56,Thickness hidrotes of shell of shell course 1(inch)=1.239 which material could fulfill the requirement of api 650 for above condition A. astm a-285m grade c B. Astm a-516m grade 58 C. Astm a-573m grade 65 D. Astm a36 Please answer and reason
Great job...Simply Well explained
Nice job sir, good explanation 👏👏👏
very good and clear information
You said in your video that the maximum hoop stress is found at 1 foot from the bottom of the course. Can you please provide me with the reference that said that
First reference is the code itself.
In order to get the maximum thickness u need to put height where u will get the maximum stress. Code assumed it 1 ft above the bottom. Actually we have welding on top and bottom of course and hydrostatic forces in between that forces the shell outwards
As a result the max effect of stress is at 1 ft on the course
Second reference is STORAGE TANK GUIDE by BOB LONG
3rd reference is ABOVE GROUND STORAGE TANK by Philip Myer chapter 7
@@codes.standards Thank you very much but can you provide me with links for these references because i am having trouble to find them
Nice Work. Can you clarify - to convert thickness from m to mm - you need to multiply by 1000 (not divide) - so you get 10^6 on the numerator. Allowable stress S needs to be in Pa in Barlows equation. A typ value for A516M material would be 137 MPa. So when we convert S from MPa to Pa you get 137 x 10^6 on the numerator. So the 10^6 terms in the numerator and denominator cancel each other.
Actually first I brought the right side 1000 to left side.
So it goes in divide I.e denominator. Still the thickness is in meters.
Then in 2nd step I converted m into mm.
Then both 1000 cancels each other
At 9.16 the right side 1000 goes to left side of equation so it goes in divide
At 9.16 the right side 1000 goes to left side of equation so it goes in divide
Thank you for the nice explanation. I would like to understand the design stresses bit though. How do we come to arrive at the design stresses during calculation?
Sorry for delayed reply as I was on annual vacation.
See the design stress or allowable design stress is mention in api 650 std itself.
See api 650 clause 5.6.2
See below link for design or allowable stress explanation.
th-cam.com/video/mSyz7FSVE1Y/w-d-xo.html
API 650
5.6.2.1 The design stress SD, shall 2/3rd of yield or 2/5th of tensile strength which ever is lower.
See table 5.2a/b on API 650 for design stress and hydrostatic test stress.
Thank you for nice explanations.
How do we do coils design in storage tank?
DID ASME Or API 650 talk about coil design in storage tanks?
Steam coil like u loop is used in both tanks and vessels.
It can be designed using asme viii div 1 individually for components.
If it is steam heated. It is like a multi pass u loop pipe.
it can be employed both inside api 650 tank or asme viii vessel.
There will be 2 design calc.
1. Heat transfer calc (thermal design)
2. Strength calc as per asme viii div 1 (but done manually for each of coil components like pipe, flange etc.
This calculated thickness is for the lowest section of the shell, how do you determine the thicknesses for the each upper subsequent shell🤔
Let suppose u hv a tank with total height 10m
With 5 shell courses each having 2m height
1. For bottom most shell put height H = 10M
2. For 2nd shell from bottom put height H = 8m
3. For 3rd shell from bottom put height H = 6m
.........
Also remember there is a 1 ft or 0.3 m height minus in api 650 formula for 1 ft method from height for that shell
@@codes.standards thanks 👍
Why is Hydrotest thickness different. The hydrotest shall be performed on the same tank and not a different tank, right? And the tank must have been manufactured with the design thickness, right?
Let suppose u have to store crude oil SG = 0.8
And hydrotest is usually with water SG = 1
Then u will calculate both tt and td.
Finally u have to take the higher value of thickness of tt and td.
This means u covered design for both cases.
@@codes.standards Got it. Thank you!!
Pls, I need to know the formula for the fabrication of a tank.
Nicely presented
is it for lower plate thickness or for all side
Can u elaborate more.
The thickness formula is same for all
Only height taken for each course will decrease so u need lesser thicknesses at upper courses
Excellent
Dear Sir, If I want to calculate the Thickness of the shell from the second course from the bottom, Will that make any difference in the value of Height ? (Will the height of the bottom or base plate be deducte in this case ?)
For 1 foot method:
Height H for each course is measured from bottom of that shell course to top of the shell or where ur maximum liquid level is
If tank height is 10m with 5 courses of 2m each, then H for bottom course will be 10m
For 2nd course from bottom it will be 8m
3rd course from bottom will be 6m and so on.
In the formula by api 650: 1-ft or 0.3 is subtracted for each course height due to the reason explained in the video.
Bottom plate thickness is not included in height calculation
Kindly share my videos in your links
@@codes.standards thank you!!!!
Very good design tips. What about construction guidlines
Thanks
That's as per code clauses
I have plan for api 650 code presentation later
what is s ?
Dear,
As per api 650 spec. clause 5.6.3
S = allowable stress
Sd = allowable stress for design case
St = allowable stress for design case
Correction St = allowable stress for test case
@@codes.standards will s be given or should we calculate? what is the formula to calculate
It is given in code tables 5.2a and 5.2b
I have a question. Please help me understand this formula in front of me to calculate the shell retirement thickness.
160' tank. H= 38'. G=0.85. Sd= 20,000psi. Each shell course is 8ft wide. what is the retirement thickness for the 3rd course?
td= 2.6 X 160' X (38' - 1) X 0.85 / 20,000psi is as far as I can go... where do I add in that each course is 8ft wide and to calculate for the 3rd course retirememt thickness? do i replace the H=38ft for H=24ft to satisfy the 3rd course?
3rd course from bottom
1. 38-1
2. 30-1
3. 22-1
Kindly share my video in ur links
Sir, section VIII div 1 & 2 ke interview questions and answers pe ekk descriptive video banayiye
Sure.
But currently working on FFS and HTRI topics
Please help me answer the question..
Tank diameter 200 ft,Total height 56,Thickness hidrotes of shell of shell course 1(inch)=1.239 which material could fulfill the requirement of api 650 for above condition
A. astm a-285m grade c
B. Astm a-516m grade 58
C. Astm a-573m grade 65
D. Astm a36
Please answer and reason
What is the design metal temperature?
this video barely explains anything....
Good presentation
Thank you