# Homework Help: Pressure vessels, wall thickness

1. Oct 10, 2012

### silver7767

1. The problem statement, all variables and given/known data

20 m3 of gas at a pressure of 25 bar is to be stored in a cylindrical
pressure vessel 2 m long. Given the following information :
The yield strength of the vessel material is 14,000 psi
If a factor of safety of 5 is to be used, determine:
(a) Whether the vessel should be treated as a thin or thick cylinder.
(b) The wall thickness required for the pressure vessel.

Hi , so far i got,

r=1.784

hoop stress
Qc=19.3*10^6 pa

wall thickness
t=0.231 m

Ri=1.553

thickness ratio
6.72

So this vessel should be treated as a thick wall vessel and minimun wall thickness should be 0.231 m

have i got it right. please let me know.

Last edited: Oct 10, 2012
2. Oct 11, 2012

Anyone ?

3. Nov 9, 2012

### captain666

I've got the same question as an assigmnent and am a bit lost. Just wondered how u got your answers - looking at the info they've given me I can't fit it to any formulas and without knowing whether the vessel is thin or thick walled I don't know which ones to use - that's obviously part of the question but i feel it's like 'chicken and egg' & without knowing nay radius' I'm in the dark.
Any help from yourself as to how u did it would be greatly appreciated.

4. Feb 15, 2013

### JimmyTheBlue

He calculated the radius via the volume and the height. He converted the yield strength to PA then calculated the hoop stress from that using the safety factor. From that he used Hoop stress = pr/t to find the thickness. The rest is self explanatory.

He used the thin cylinder equations as I did to get to that point obviously which as you conclude may be an issue. However they just contain a degree of inaccuracy rather than complete inaccuracy......so having deduced it's a thick cylinder....maybe we're meant to recalc using the relevant related thick equations and find the true thickness?

I also think the inner radius was 1.74 and the outer 1.74 + 0.231. Though that still implies it's thick walled via the ratio consideration.

5. Feb 15, 2013

### captain666

I did the question differently in the end & here's what I did :
Volume of air in vessel = 20m^3.
p1 = air pressure in vessel = 25 bar.
length = 2m.
yield strength = 14,000 PSI.
Factor of safety = 5.
Formula for compressed air in cylinders : P1V1 = P2V2.
P1 is pressure of compressed gas & 25 bar = 362.595 PSI.
V1 is internal volume of cylinder in litres.
P2 is atmospheric pressure (1 ATM = 14.7 PSI).
V2 is volume of gas at pressure p2 : 20m^3 = 20000 litres.

therefore as P1V1 = P2V2,
362.595 x V1 = 14.7 x 20,000
Transposes to V1 = (14.7 x 20,000) divided by 362.95.
Which ends up as V1 = 810.8219915 Litres = 0.811m^3.

And volume of cylinder = L x Pi x R^2
which goes to 0.811 = 2Pi x r^2
which ends up as Inside radius = 0.3593m or 14.15 inches.

Using this radius,if it is treated as a thick walled cylinder, the wall thickness ends up as 1.97m which using the ratio ends up as 7.18 which is less than 10 which indicates a thin walled cylinder.

using the radius and treating it as a thin walled cylinder,the thickness ends up as 1.83 inches which fed unto the ratio equation gives an answer of 7.722 which is again less than 10 therefore both answers indicate that it is a thin walled cylinder.

6. Oct 9, 2013

### LDC1972

So who was correct?

So who was correct on this? I have only briefly looked so haven't much input yet, but the reason I "Googled" was because I took the obvious route (we have length and volume) so using these / ignoring pressure the Dia is 3.56m which is just mad! It'd be a big cumbersome flat tub.........
So I got to thinking surely the gas is compressed and this reduces the physical volume, as the last poster has done.

7. Oct 9, 2013

### Staff: Mentor

The problem statement clearly indicates that the compressed volume of gas within the cylinder is 20 M3.

8. Oct 11, 2013

### LDC1972

Thanks Chester, have clicked your thanks button.

The tutor also confirmed this yesterday...

9. Nov 13, 2013

### oxon88

is the above approach correct?

10. Nov 13, 2013

### LDC1972

Nope!!

It is a big flat tub if memory serves. I got it right anyway. (In the end LOL)

You'll get there, the info's there and if I remember rightly, there are online calculator to check your results. But all this was ages ago!!!

11. Nov 13, 2013

### LDC1972

This is basically what to do

Then re-do with thick walled equations with now known figures and get precise thickness.

Cheers!

12. Nov 13, 2013

### oxon88

ok i will give it a go. thank you for the response.

13. Nov 14, 2013

### oxon88

can you provide any guidance with calculating the hoop stress?

14. Nov 14, 2013

### LDC1972

V = ∏r^2 x length

simplify to r^2 = v / ∏ x length, that's your radius.

σ1 = yield strength / factor of safety

and as σ1 = Pr / t then t = Pr / σ1 and that's your wall thickness using thin wall.

Cool?

Once you get 231mm or so you've cracked it, then move onto thick walled with now known values...

15. Nov 14, 2013

### Staff: Mentor

The equation for the hoop stress is obtained by integrating the stress-equilibrium equation (differential force balance), subject to appropriate boundary conditions.

16. Nov 14, 2013

### oxon88

Many Thanks. just one question.

The thickness ratio for thin walled: r / t ≥ 10 or d / t ≥ 20

i've calculated r to be 1.784m. Is this the inner radius?

I see you have subtracted the calculated wall thickness, then calculated the ratio as:
(1.784 - 0.231) / 0.231 = 6.723.

Is this correct? Or should the ratio be:
1.784 / 0.231 = 7.723

17. Nov 18, 2013

### oxon88

can anyone help with my last question?

18. Dec 4, 2013

### Big Jock

I got the radius to be 1.784 M also Oxon88 but that's as far as I have got so far. Do you have any more light to shed on these two questions?

19. Dec 4, 2013

### oxon88

m still pondering about what the correct ratio should be...

20. Dec 4, 2013

### Big Jock

I cant seem to calculate the hoop stress. Trying to find a formula where I can use the safety factor to find the hoop stress to then work out the wall thickness as I don't fully believe or follow some of the workings above....