Pressure Vessel Maximum Diameter Calculation

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SUMMARY

The maximum diameter of a pressure vessel cylinder tank under an internal pressure of 150 psi is calculated to be 3.06 feet (36.67 inches). This is derived from the longitudinal joint strength of 33 kips/ft and the formula D = 2w/p, where w is the load per foot and p is the pressure converted to lb/ft². The pressure is expressed as 21600 lb/ft², calculated from 150 psi multiplied by 144 in²/ft². The discussion emphasizes the importance of understanding the underlying principles of stress calculations in pressure vessels.

PREREQUISITES
  • Understanding of pressure vessel design principles
  • Familiarity with stress equations: σ = Pr/2t and σ = Pr/t
  • Ability to convert pressure units (psi to lb/ft²)
  • Knowledge of longitudinal and circumferential stress concepts
NEXT STEPS
  • Study pressure vessel design codes and standards (e.g., ASME Boiler and Pressure Vessel Code)
  • Learn about material selection for pressure vessels under varying loads
  • Explore advanced stress analysis techniques for cylindrical structures
  • Investigate software tools for pressure vessel design simulations (e.g., ANSYS, SolidWorks)
USEFUL FOR

Engineering students, mechanical engineers, and professionals involved in pressure vessel design and analysis will benefit from this discussion, particularly those preparing for exams or seeking to deepen their understanding of stress calculations in pressure vessels.

tsukuba
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Homework Statement


The strength of longitudinal joint in Fig. 1-17 is 33 kips/ft, whereas for the girth is 16 kips/ft. Calculate the maximum diameter of the cylinder tank if the internal pressure is 150 psi. - See more at:

http://www.mathalino.com/reviewer/m...blem-138-pressure-vessel#sthash.G52XwZOd.dpuf

Homework Equations


σlongitudinal = Pr / 2t

The Attempt at a Solution


I have no way of calculating this since I have 2 unknowns. If you go to the link you can see the steps they've done but it doesn't make sense. Also, where does that 21600 come from?
 
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Look at their solution. The 21600 is in the place p in an expression of the form
pD/(2t). This is the 150 psi expressed in lb/ft^2 to be dimensionally consistent
 
isnt p =125 psi? and why is there a second t under 33000
 
tsukuba said:
isnt p =125 psi? and why is there a second t under 33000

p = 150 psi, the internal pressure, as is stated clearly in the problem statement.

The longitudinal joint can withstand a maximum loading of 33 kip/ft according to the problem statement. The stress on this joint is then the loading divided by the thickness of the pressure vessel, t, so

σ = w / t = 33,000 / t

But also σ = pD/2t, so these equations can be combined thus

33,000 / t = σ = pD / 2t

The t's cancel, leaving pD/2 = 33,000, or D = 2 * 33,000 / (150 * 144) = 3.06 ft. = 36.67 in.
 
Check your units.
 
There are 2 formulas to calculate pressure in a vessel. Pr / 2t and Pr/t
when it doesn't say calculate the circumferential or the longitudinal stress how do I know which one to use?
 
SteamKing said:
p = 150 psi, the internal pressure, as is stated clearly in the problem statement.

The longitudinal joint can withstand a maximum loading of 33 kip/ft according to the problem statement. The stress on this joint is then the loading divided by the thickness of the pressure vessel, t, so

σ = w / t = 33,000 / t

But also σ = pD/2t, so these equations can be combined thus

33,000 / t = σ = pD / 2t

The t's cancel, leaving pD/2 = 33,000, or D = 2 * 33,000 / (150 * 144) = 3.06 ft. = 36.67 in.

p = 150 lbf/in2*144 in2/ft2 = 21600 lbf/ft2

w = 33 kip / ft = 33000 lbf / ft

D = 2 w / p = 2 * (33000 lbf / ft) / 21600 lbf / ft2 = 3.06 ft = 36.67 in

Units checked.
 
SteamKing, the comment to check units was intended for the OP. No one doubts that you know how to work this simple problem. I was under the perhaps mistaken impression that our purpose here was to help the OP learn, to be able to think through the problem, rather than to just display our own brilliance.

I see in this case, that even given a full solution, the OP is not inclined work through it, but rather simply comes back with more simple questions. I think that, in cases like that, it is better not to give full answers, but only to offer hints.
 
Dr. D, this is not my homework. I have a test this week and I look for other sources to help me. I was not given much help at school from the teacher so I have to find my own way of learning and practicing. These could be simple to you but they aren't to me. SteamKing has been nice enough, and the only one that has helped me understand a lot, and I feel that without his help I'd still be stuck.
 
  • #10
  1. tsukuba, you may think that SteamKing has done you a great favor, but I don't think so. It makes no difference whether this is homework, review for a test, or what; it is clearly an academic sort of problem. You started this whole discussion with a fully worked solution available to you, one that any well prepared engineering student should have been able to follow. The whole point of a learning exercise, whether for HW or test review, is for you to learn how to think through the problem. With all the help you had available to you (full solution with the original problem, tips I offered trying to lead you to think through the problem), and you still did not get it until SteamKing spelled out every minute detail for you, I doubt that you have learned very much at all. If this exact problem appears on the test, you may be able to get an answer. If a similar, but slightly different problem appears, I doubt greatly that you will be able to do much with it. That's why I don't think SteamKing has done you any great favor. You have not been pushed to think the problem through as you needed. Good luck with your test.
 

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