Shell Design Help
‎Jul 06, 2009
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‎Jul 06, 2009
03:00 AM
Shell Design Help
Hi
I have a unique problem when calculating the tank shell weight in the attached sheet.
The shell weight which is calculated is then required above to determine if the liquid level should be increased which might require the chosen thickness to be increased, which if true would change the tank shell weight.
Hope that makes sense.
The table in the spreadsheet in an inserted excel component.
Any suggestions on how to modify the calculation of the shell weight automatically instead of manually inserting as indicated by the arrows would be welcomed.
Thanks in advance.
Regards
Mike
I have a unique problem when calculating the tank shell weight in the attached sheet.
The shell weight which is calculated is then required above to determine if the liquid level should be increased which might require the chosen thickness to be increased, which if true would change the tank shell weight.
Hope that makes sense.
The table in the spreadsheet in an inserted excel component.
Any suggestions on how to modify the calculation of the shell weight automatically instead of manually inserting as indicated by the arrows would be welcomed.
Thanks in advance.
Regards
Mike
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Mathcad Usage
51 REPLIES 51
‎Jul 06, 2009
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‎Jul 06, 2009
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May be this picture help you
http://twt.mpei.ac.ru/ochkov/Mathcad_12/1_66_Plot_V_V.png
Val
http://twt.mpei.ac.ru/ochkov/v_ochkov.htm
http://twt.mpei.ac.ru/ochkov/Mathcad_12/1_66_Plot_V_V.png
Val
http://twt.mpei.ac.ru/ochkov/v_ochkov.htm
‎Jul 06, 2009
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‎Jul 06, 2009
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>I have a unique problem when calculating the tank shell weight in the attached sheet.<<br> ___________________
There is nothing to calculate first. You purchase the tank for the nominal pressure application, from supplier that met all ASME codes (for instance). Then you calculate the weight or get it from the supplier. Will look at your work sheet later.
jmG
There is nothing to calculate first. You purchase the tank for the nominal pressure application, from supplier that met all ASME codes (for instance). Then you calculate the weight or get it from the supplier. Will look at your work sheet later.
jmG
‎Jul 06, 2009
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‎Jul 06, 2009
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Hello Mike,
I'm not sure I understood your problem. However, if it is about feeding in the results as start values then this attached file may help.
All I did, was to send the result to an Excel file and I read it in at the beginning. It will update every time you recalculate the page. This way, you can use the reult in external applications as well.
Otherwise, an iteration might be easier to achieve than by an external call.
Cheers
Raiko
I'm not sure I understood your problem. However, if it is about feeding in the results as start values then this attached file may help.
All I did, was to send the result to an Excel file and I read it in at the beginning. It will update every time you recalculate the page. This way, you can use the reult in external applications as well.
Otherwise, an iteration might be easier to achieve than by an external call.
Cheers
Raiko
‎Jul 06, 2009
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‎Jul 06, 2009
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Raiko
Thanks for the response.
I can see the logic behind your thinking but I�m a little confused on how to extract the number into an external Excel file and recall it back in.
Regards
Mike
Thanks for the response.
I can see the logic behind your thinking but I�m a little confused on how to extract the number into an external Excel file and recall it back in.
Regards
Mike
‎Jul 06, 2009
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‎Jul 06, 2009
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Hi Mike,
that's elementary my dear Watson 😉
#1 right-click a place in the work sheet (below the variable you wish to export) and choose "insert" - file/output from the context menu
#2 choose the file format: Excel in your case. Click on browse and type in the file name. This works even when the file is not existing. Mathcad creates such a file. Finish that dialogue
#4 an icon appears with a placeholder (i.e. the small balck rectangle on the lower left side) This is the place you can type in the variable you'd like to export. It could be a s well an array.
#5 repaet the steps above the first definition of the variable and do the same with file input.
#6 delete your variable as this is now read in from the Excel file
Hope this helped
Raiko
that's elementary my dear Watson 😉
#1 right-click a place in the work sheet (below the variable you wish to export) and choose "insert" - file/output from the context menu
#2 choose the file format: Excel in your case. Click on browse and type in the file name. This works even when the file is not existing. Mathcad creates such a file. Finish that dialogue
#4 an icon appears with a placeholder (i.e. the small balck rectangle on the lower left side) This is the place you can type in the variable you'd like to export. It could be a s well an array.
#5 repaet the steps above the first definition of the variable and do the same with file input.
#6 delete your variable as this is now read in from the Excel file
Hope this helped
Raiko
‎Jul 06, 2009
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‎Jul 06, 2009
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Got it cheers.
Mike
Mike
‎Jul 06, 2009
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‎Jul 06, 2009
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On 7/6/2009 3:31:08 AM, Armo wrote:
>Any suggestions on how to
>modify the calculation of the
>shell weight automatically
>instead of manually inserting
>as indicated by the arrows
>would be welcomed
You can create a user function!
And second
Why do you not use "normal" units - Po:=7000*kg/m^3 but use units as remarks?
Val
http://twt.mpei.ac.ru/ochkov/v_ochkov.htm
>Any suggestions on how to
>modify the calculation of the
>shell weight automatically
>instead of manually inserting
>as indicated by the arrows
>would be welcomed
You can create a user function!
And second
Why do you not use "normal" units - Po:=7000*kg/m^3 but use units as remarks?
Val
http://twt.mpei.ac.ru/ochkov/v_ochkov.htm
‎Jul 06, 2009
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‎Jul 06, 2009
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The reason I have not included units when calculating the tanks is because some of the calculations in API 650 have formulas which are raised to the power of 2,3�.etc.
Which Mathcad doesn�t like.
If you know a work around it would be of great benefit.
Regards
Mike
Which Mathcad doesn�t like.
If you know a work around it would be of great benefit.
Regards
Mike
‎Jul 06, 2009
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‎Jul 06, 2009
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Ther picture shows a ratio of thicknesses raised to a fractional power. That is not a problem when using units. The units cancel out and the ratio is dimensionless.
__________________
� � � � Tom Gutman
__________________
� � � � Tom Gutman
‎Jul 06, 2009
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‎Jul 06, 2009
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Sorry bad example.
The formula attached is used to calculate wind pressure in accordance with API 650.
Where V = kph � wind speed
The answer should come out in kPa, but as the equation is empirical units cannot be used�� I think�.
Regards
Mike
The formula attached is used to calculate wind pressure in accordance with API 650.
Where V = kph � wind speed
The answer should come out in kPa, but as the equation is empirical units cannot be used�� I think�.
Regards
Mike
‎Jul 06, 2009
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On 7/6/2009 6:11:05 AM, Armo wrote:
>Sorry bad example.
>
>The formula attached is used
>to calculate wind pressure in
>accordance with API 650.
>
>Where V = kph � wind speed
>
>The answer should come out in
>kPa, but as the equation is
>empirical units cannot be
>used�� I think�.
You've already done half of what is required: dividing V by kph. To get the result in the correct units just multiply the entire rhs by the expected units: kPa.
Richard
>Sorry bad example.
>
>The formula attached is used
>to calculate wind pressure in
>accordance with API 650.
>
>Where V = kph � wind speed
>
>The answer should come out in
>kPa, but as the equation is
>empirical units cannot be
>used�� I think�.
You've already done half of what is required: dividing V by kph. To get the result in the correct units just multiply the entire rhs by the expected units: kPa.
Richard
‎Jul 06, 2009
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‎Jul 06, 2009
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What version of Mathcad are you using? If MC13 or MC14, my define variables worksheet may be an answer.
__________________
� � � � Tom Gutman
__________________
� � � � Tom Gutman
‎Jul 06, 2009
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‎Jul 06, 2009
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Mathcad 14
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‎Jul 06, 2009
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Not sure if I've fully understood, but the attached might be of some use (I've included weight calculations for max and min thicknesses - these might well be superfluous!).
stv
stv
‎Jul 06, 2009
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‎Jul 06, 2009
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Hi
You have understood perfectly, I have attached the part of your spreadsheet which I needed.
Thanks a lot for the help.
Regards,
Mike
You have understood perfectly, I have attached the part of your spreadsheet which I needed.
Thanks a lot for the help.
Regards,
Mike
‎Jul 06, 2009
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‎Jul 06, 2009
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I misunderstood it was partially filled pipeline and not an horizontal tank. If you need the formula to calculate the volume vs the level, that I have ... unit volume/length also means unit weight/length.
jmG
jmG
‎Jul 06, 2009
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‎Jul 06, 2009
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If it has been posted previously, I already have it cheers.
Mike
Mike
‎Jul 06, 2009
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‎Jul 06, 2009
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On 7/6/2009 1:47:54 PM, Armo wrote:
>If it has been posted previously,
>I already have it ... cheers.
>Mike
______________________________
Yes, the CPi is in circulation for at least 9 years in this collab.
jmG
>If it has been posted previously,
>I already have it ... cheers.
>Mike
______________________________
Yes, the CPi is in circulation for at least 9 years in this collab.
jmG
‎Jul 06, 2009
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‎Jul 06, 2009
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Here's a solution using the pushbutton define variables. Note that trying to use the text box form (which automatically does the definitions) results in an endless loop, as the variable being defined appears to be a factor in calculating the value.
__________________
� � � � Tom Gutman
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� � � � Tom Gutman
‎Jul 06, 2009
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Here's a variant using the text box form, with some loop breaking code inserted.
You may have a units problem (I really recommend using Mathcad units consistently, and adjusting any cookbook formulae to use properly united values). In looking to see how Wsh affects the data (it barely does, being used only to determine if Lcri is to be used or not) I noticed that Lcri is described as being in mm, but it is then added to values described as being in m. Inconsistent. And the calculation of Lcri is a bit strange. If one assumes that the factor of 9.8 is really supposed to be g, then the dimensions do not work out. The expression gives an area mass density, not a length.
__________________
� � � � Tom Gutman
You may have a units problem (I really recommend using Mathcad units consistently, and adjusting any cookbook formulae to use properly united values). In looking to see how Wsh affects the data (it barely does, being used only to determine if Lcri is to be used or not) I noticed that Lcri is described as being in mm, but it is then added to values described as being in m. Inconsistent. And the calculation of Lcri is a bit strange. If one assumes that the factor of 9.8 is really supposed to be g, then the dimensions do not work out. The expression gives an area mass density, not a length.
__________________
� � � � Tom Gutman
‎Jul 07, 2009
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‎Jul 07, 2009
03:00 AM
Cheers for the help Tom.
Firstly Lcri should be in mm as it only increases the design Liquid by a small amount.
When designing aboveground storage tanks to API 650 using metric units, there are constant values throughout the code which make the use of units non-consistent.
It really frustrates me that the calculations have to be done in this manner, but if units are inserted into the design data at the start the results change significantly.
The code clearly states a unit for every variable throughout the standard.
Mike
Firstly Lcri should be in mm as it only increases the design Liquid by a small amount.
When designing aboveground storage tanks to API 650 using metric units, there are constant values throughout the code which make the use of units non-consistent.
It really frustrates me that the calculations have to be done in this manner, but if units are inserted into the design data at the start the results change significantly.
The code clearly states a unit for every variable throughout the standard.
Mike
‎Jul 07, 2009
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‎Jul 07, 2009
03:00 AM
It makes no difference whether it increases the design liquid by a large amount or a small amount. You cannot (meaningfully) add a number of meters to a number of millimeters. Both large and small lengths can be expressed in either meters or millimeters. The only difference is the size of the numbers.
Considering that you are working with lengths on the order of meters (about 20, overall) if Lcri is indeed .697mm it is not just small, it is negligible -- way smaller than your tolerances, or probably your measurement errors. If it is actually .697m (the unit is wrong rather than the value) it is still adding just a small amount.
If including units causes the answers to change, then either the inclusion of units was done incorrectly (the empirical equations were not properly adjusted) or the answers calculated without units were wrong. Possibly both.
I quite understand that including units in a calculation that was done without units is not a trivial task. The problem is that many of the constants in the formulae have, in fact, units, units which are unstated but implied by the units for the variables. To get correct equations these missing units must be restored. And restored correctly.
You can do that simply by adding units that cancel the input units and add the output units. But better would be to go back to the derivations of the equations and determining the proper units from there. Often the given constant is a combination of a number of meaningful physical quantities. The calculation of Lcri might include a hidden factor of the density of water, which is, in some common units, just one.
But if you don't do that then you have to manually verify, for every calculation, that not only are the dimensions correct but also that the units are compatible. No adding of lengths in meters to lengths in millimeters. Not a trivial task either.
As a separate side note, there is a question as to the input thicknesses. Are they thicknesses before or after corrosion? Some calculations seem to imply that they are before corrosion, other calculations seem to imply that they are after corrosion.
__________________
� � � � Tom Gutman
Considering that you are working with lengths on the order of meters (about 20, overall) if Lcri is indeed .697mm it is not just small, it is negligible -- way smaller than your tolerances, or probably your measurement errors. If it is actually .697m (the unit is wrong rather than the value) it is still adding just a small amount.
If including units causes the answers to change, then either the inclusion of units was done incorrectly (the empirical equations were not properly adjusted) or the answers calculated without units were wrong. Possibly both.
I quite understand that including units in a calculation that was done without units is not a trivial task. The problem is that many of the constants in the formulae have, in fact, units, units which are unstated but implied by the units for the variables. To get correct equations these missing units must be restored. And restored correctly.
You can do that simply by adding units that cancel the input units and add the output units. But better would be to go back to the derivations of the equations and determining the proper units from there. Often the given constant is a combination of a number of meaningful physical quantities. The calculation of Lcri might include a hidden factor of the density of water, which is, in some common units, just one.
But if you don't do that then you have to manually verify, for every calculation, that not only are the dimensions correct but also that the units are compatible. No adding of lengths in meters to lengths in millimeters. Not a trivial task either.
As a separate side note, there is a question as to the input thicknesses. Are they thicknesses before or after corrosion? Some calculations seem to imply that they are before corrosion, other calculations seem to imply that they are after corrosion.
__________________
� � � � Tom Gutman
‎Jul 07, 2009
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‎Jul 07, 2009
03:00 AM
Tom
You are correct Lcri should have the units of m cheers for pointing that out.
I have attached a Mathcad sheet with the unit suggestions you made, you can see that the units and answers are correct until I reach the required shell thickness calculations.
The units are meant to come out in mm.
Any suggestions on how to solve this problem?
Mike
You are correct Lcri should have the units of m cheers for pointing that out.
I have attached a Mathcad sheet with the unit suggestions you made, you can see that the units and answers are correct until I reach the required shell thickness calculations.
The units are meant to come out in mm.
Any suggestions on how to solve this problem?
Mike
‎Jul 07, 2009
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‎Jul 07, 2009
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If the standard states what the units should be for every variable, what does it say the units of ID, Level, SG (presumably dimensionless), Sd and CA should be? If those are empirical formulae then you need to divide out each term by the stated units and then multiply the entire rhs by the desired final units (as I said in my last post). I assume they are empirical, because if they are are derived based on real physical principles then they are dimensionally inconsistent and cannot be correct.
Richard
Richard
‎Jul 07, 2009
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‎Jul 07, 2009
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ID = m
CA = mm
Sd = MPa
Level = m
SG = dimensionless
CA = mm
Sd = MPa
Level = m
SG = dimensionless
‎Jul 07, 2009
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‎Jul 07, 2009
03:00 AM
So like this.
If Level is in m then you can't subtract a value in mm. I have assumed that this should really be 0.3m, since subtracting 0.3mm from values on the order of 10m is a miniscule correction.
Richard
If Level is in m then you can't subtract a value in mm. I have assumed that this should really be 0.3m, since subtracting 0.3mm from values on the order of 10m is a miniscule correction.
Richard
‎Jul 07, 2009
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‎Jul 07, 2009
03:00 AM
Ok cheers for that.
I have done that before but it looks nasty.
Mike
I have done that before but it looks nasty.
Mike
‎Jul 07, 2009
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‎Jul 07, 2009
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On 7/7/2009 10:38:11 AM, Armo wrote:
> but it
>looks nasty.
Yes, but it works nice 🙂
Better to do it and avoid possible wrong answers than not do it and have a nice looking equation.
In this case a little rearrangement of the expression makes it work and look OK.
Richard
> but it
>looks nasty.
Yes, but it works nice 🙂
Better to do it and avoid possible wrong answers than not do it and have a nice looking equation.
In this case a little rearrangement of the expression makes it work and look OK.
Richard
‎Jul 07, 2009
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‎Jul 07, 2009
03:00 AM
STV sorry to request your help directly but you completed the attached spreadsheet the other day and i have had a few problems trying to proceed with it.
Would you be able to have a look at the comments i have made on the calculation.
Cheers Mike
Would you be able to have a look at the comments i have made on the calculation.
Cheers Mike